i\sin(\alpha-\beta) & \cos(\alpha-\beta)\\
\end{bmatrix}
\]

using <Ref>. Since $\alpha-\beta\notin\{0,\pm\pi\}$, we see that $C$ fails to be self-adjoint, and consequently by <Ref> the proof is completed.

The Pimsner-Popa constant $\lambda(R_u,R_v)$ for the pair of subfactors $R_u\mbox{ and }R_v$ of the hyperfinite $II_1$ factor $R$ is $1/2$.

Let $k\geq 2$. We claim the following,
λ(u_2k+1(M_2⊗M_2^(k))u_2k+1^ * , v_2k+1(M_2⊗M_2^(k))v_2k+1^*)=1/2 .

Recall from Theorem <ref> that

\[
u_{2k+1}=\mbox{bl-diag}\left\{u_{2k}\big(I_2\otimes I_2^{(k)}\big),u_{2k}\big(\sigma_3\otimes I_2^{(k)}\big)\right\}
\]

as an element in $\Delta_2\otimes M_2\otimes M_2^{(k)}$. Therefore,

\[
u_{2k+1}^{\,*}v_{2k+1}=\mbox{bl-diag}\left\{u_{2k}^{\,*}v_{2k}\,,\big(\sigma_3\otimes I_2^{(k)}\big)u_{2k}^{\,*}v_{2k}\big(\sigma_3\otimes I_2^{(k)}\big)\right\}\,.
\]

λ(u_2k+1(M_2⊗M_2^(k))u_2k+1^ * , v_2k+1(M_2⊗M_2^(k))v_2k+1^*)

=    λ(M_2^(k+1) , u_2k+1^ *v_2k+1M_2^(k+1)v_2k+1^*u_2k+1)

with $u_{2k+1}^{\,*}v_{2k+1}\in\Delta_2\otimes M_2\otimes M_2^{(k)}$. We now wish to apply <Ref>, and for that first consider the following matrix
v_2k^*u_2k(σ_3⊗I_2^(k))u_2k^ *v_2k(σ_3⊗I_2^(k)) .

By <Ref>, <ref> we get that the matrix $A(u_4,v_4)$ is not diagonal. Hence, by <Ref> we get that the matrix in <Ref> is not diagonal. As a consequence of <Ref>, our claim has been established. By <Ref>, $\,\lambda(R_u,R_v)$ is the limit of a decreasing sequence involving $\lambda$ at each step of the tower of basic construction starting with the complex Hadamard matrices $u,v$. Since we have produced a constant subsequence in <Ref>, the proof is concluded.

§.§ Factoriality of $R_u\cap R_v$ and the quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$

An astute reader must have noticed that  the intersection $R_u\cap R_v$ need not be a factor a priori. Another difficulty is to determine whether $[R:R_u\cap R_v]$ is finite or not. Our goal in this section is to address both these issues. We prove that $R_u\cap R_v$ is a $II_1$ subfactor of $R$ with $[R:R_u\cap R_v]$=4. We also characterize when the quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$ of $II_1$ factors is a commuting (and consequently, co-commuting) square. For the sake of brevity, we introduce some notations.

* For $k\in\bbn\cup\{0\}$, let

\begin{align*}
A_{2k} &= M_2\otimes M_2^{(k)}\,,\cr
B^u_{2k} &= u_{2k}(\Delta_2\otimes M_2^{(k)})u_{2k}^*\,,\cr
B^v_{2k} &= v_{2k}(\Delta_2\otimes M_2^{(k)})v_{2k}^*\,,\cr
C_{2k} &= B^u_{2k}\cap {B}^v_{2k}\,.
\end{align*}

These are the even steps in the tower of basic constructions in <Ref>. Note that

\[
\overline{\cup_k\,A_{2k}}^{\mbox{ sot}}=R,\,\,\overline{\cup_k\,B^u_{2k}}^{\mbox{ sot}}=R_u,\,\,\overline{\cup_k\,B^v_{2k}}^{\mbox{ sot}}=R_v,\,\mbox{ and }\,\overline{\cup_k\,C_{2k}}^{\mbox{ sot}}=R_u\cap R_v.
\]

* For $k\in\bbn$, let $W_{2k}:=E_{11}\otimes I_2^{(k)}+E_{22}\otimes\sigma_1^{(k)}$ in $\Delta_2\otimes M_2^{(k)}$. Each $W_{2k}$ is unitary in $M_2^{(k+1)}$. Recall that $A^{(k)}\in M_2^{(k)}$ denotes the matrix $A^{\otimes\,k}=A\otimes A\otimes\cdots\otimes A$ in $M_2^{(k)}$.

We begin by computing the Sano-Watatani angle between $B^u_0\mbox{ and }B^v_0$ (see Notation <ref>), and show that the following quadruple

B^u_0    ⊂    A_0

∪        ∪

C_0    ⊂    B^v_0

may not be a commuting square. This is an example which shows that the floor (and consequently the roof due to <Ref>) in <Ref> need not be a commuting square. The difficulty starts from here.

If $u= \frac{1}{\sqrt{2}}\begin{bmatrix}
e^{i\alpha_1} & e^{i (\alpha_1+\alpha_3)}\\
e^{i\alpha_2} & -e^{i(\alpha_2+\alpha_3)}
\end{bmatrix}$ and $v= \frac{1}{\sqrt{2}}\begin{bmatrix}
e^{i\beta_1} & e^{i (\beta_1+\beta_3)}\\
e^{i\beta_2} & -e^{i(\beta_2+\beta_3)}
\end{bmatrix}$, then the cosine of the angle between $u\Delta_2u^*\mbox{ and }v\Delta_2v^*$ is the set $\big\{0,\big|\cos((\alpha_1-\alpha_2)-(\beta_1-\beta_2))\big|\big\}$, and the quadruple in <Ref> is a commuting square if and only if $\alpha_1-\alpha_2=\beta_1-\beta_2\pm\frac{\pi}{2}$.

It is clear that $u\Delta_2u^*\cap\,v\Delta_2v^*=\bbc$, as $\mbox{dim}(u\Delta_2u^*)=\mbox{dim}(v\Delta_2v^*)=2$ and $u\Delta_2u^*=v\Delta_2v^*$ implies that $u\sim v$ by <Ref>. Consider the non-negative operator $E_{u\Delta_2u^*}E_{v\Delta_2v^*}E_{u\Delta_2u^*}$, and observe that as an element of $\mathcal{B}(M_2,tr)=M_4$, this is same as the operator $\mbox{Ad}_uE_{\Delta_2}\mbox{Ad}_{u^*}\mbox{Ad}_vE_{\Delta_2}\mbox{Ad}_{v^*}\mbox{Ad}_uE_{\Delta_2}\mbox{Ad}_{u^*}$. Put $w=(w_{ij})=u^*v$, a unitary in $M_2$. Then, we have the operator $\mbox{Ad}_uE_{\Delta_2} E_{w\Delta_2w^*}E_{\Delta_2}\mbox{Ad}_{u^*}$ in $\mathcal{B}(M_2)$. First consider $E_{\Delta_2}E_{w\Delta_2w^*}E_{\Delta_2}\in\mathcal{B}(M_2)$. Since $w$ is a $2\times 2$ unitary matrix, we have $|w_{12}|^2=|w_{21}|^2$. We put $c= \sum_{i=1}^2{|w_{1i}|}^4$ and $s= {2}\prod_{i=1}^2 {|w_{i1}|}^2$. It is easy to observe that $\sum_{i=1}^2 {|w_{2i}|}^4=c$ and $2\prod_{i=1}^2 {|w_{i2}|}^2=s$. Therefore, using the fact that $|w_{11}|^2+|w_{12}|^2=1$ (as $w^*$ is a unitary) we get the following,

\[
c+s=|w_{11}|^4+|w_{12}|^4+ 2|w_{11}|^2|w_{21}|^2=|w_{11}|^4+|w_{12}|^4+ 2|w_{11}|^2|w_{12}|^2=1.
\]

Now, it is easy to verify that

\[
E_{\Delta} E_{w\Delta w^*} E_{\Delta}(A)=\mathrm{diag}\{c a_{11}+ (1-c) a_{22}\,,\,(1-c) a_{11}+ c a_{22}\}
\]

for any matrix $A=(a_{ij})$ in $M_2$. Therefore, with respect to the ordered orthonormal basis $\mathfrak{B}:=\big\{\sqrt{2}E_{11},\sqrt{2}E_{22},\sqrt{2}E_{12},\sqrt{2}E_{21}\big\}$ of $M_2$, the operator $E_{\Delta_2}E_{w\Delta_2 w^*}E_{\Delta_2}:M_2\longrightarrow M_2$, when viewed as a linear map from $\bbc^{4}$ to $\bbc^{4}$, is of the form $\mbox{bl-diag}\{C,0\}$, where
${C}= \begin{bmatrix}
c & 1-c \\
1-c &~ c
\end{bmatrix}$. The conditional expectation from $M_2$ onto $u\Delta_2u^*\cap v\Delta_2v^*$ is simply the trace $tr$. Therefore, with respect to the ordered orthonormal basis $\mathfrak{B}$, the operator $tr:M_2\to\bbc\subset M_2$ given by $A\mapsto tr(A)I_2$ is of the form $\mbox{bl-diag}\{J_2,0\}$, where $J_2$ is the projection $\frac{1}{2}\sum_{i,j=1}^2E_{ij}$ in $M_2$. Hence, $E_{u\Delta_2u^*}E_{v\Delta_2v^*}E_{u\Delta_2u^*}-tr$ has the form $\mbox{bl-diag}\{C-J_2,0\}$. The quadruple in <Ref> is a commuting square if and only if $E_{u\Delta_2u^*}E_{v\Delta_2v^*}E_{u\Delta_2u^*}-tr$, that is, $C=J_2$. Since $J_2=\frac{1}{2}\sum_{i,j=1}^2 E_{ij}$, we see that $C=J_2$ holds if and only if $c=s=\frac{1}{2}$. A straightforward verification leads us that $c=\frac{1}{2}(1+\cos^2(\beta_1-\alpha_1-\beta_2+\alpha_2))$. Therefore, the quadruple in <Ref> is a commuting square if and only if $\alpha_1-\alpha_2=\beta_1-\beta_2\pm\frac{\pi}{2}$.

Now, suppose that this condition is not satisfied. The spectrum of $\mbox{Ad}_u(E_{\Delta_2}E_{w\Delta_2 w^*}E_{\Delta_2}-tr)\mbox{Ad}_{u^*}$ is simply the spectrum of the matrix $C-J_2\in M_2$. This is because $\mbox{Ad}_u\in\mathcal{B}(M_2)$ is a unitary. Therefore, the spectrum of the operator $E_{u\Delta_2u^*}E_{v\Delta_2v^*}E_{u\Delta_2u^*}-E_{u\Delta_2u^*\,\cap\,v\Delta_2v^*}$ is the spectrum of the matrix $C-J_2$, which is $\{0,2c-1\}$. Observe that $2c-1=\big({|w_{11}|}^2-{|w_{12}|}^2\big)^2$. A straightforward verification shows that $|w_{11}|^2-|w_{12}|^2=\cos\big((\alpha_1-\alpha_2)-(\beta_1-\beta_2)\big)$. Therefore, we obtain that cosine of the angle between $u\Delta_2u^*\mbox{ and }v\Delta_2v^*$ is the set $\big\{0,\big|\cos(\alpha_1-\alpha_2)-(\beta_1-\beta_2))\big|\big\}$.

Let $\,u\mbox{ and }v$ be of the form as described in <Ref>. Then, the cosine of the angle between $\mathrm{Ad}_u(\Delta_2)\mbox{ and }\mathrm{Ad}_v(\Delta_2)$ is the set $\{0,|\cos(\alpha-\beta)|\}$.

By <Ref>, the quadruple $(C_0\subset B^v_0\,,\,B^u_0\subset A_0)$ in <Ref>, and consequently $(R_u\cap R_v\subset R_u,R_v\subset R)$ due to <Ref>, is not a commuting square (see <Ref> in this regard) unless $\alpha-\beta=\pm\frac{\pi}{2}$. With the notations as in <Ref>, we now identify the finite-dimensional grid $C_{2k}$ for $R_u\cap R_v$.

We have the following

\[
{u}^*_{2k}{v}_{2k} \big(\Delta_2\otimes M^{(k)}_2\big){v}^*_{2k}{u}_{2k} =W_{2k}\big(u^*v\Delta_2v^*u\otimes M^{(k)}_2\big) W^*_{2k}
\]

for any $k\in\bbn$, where $W_{2k}$ is as defined in <Ref>.

We first prove it for $k=1$. We have
u^*v    =    1/2(1+e^i(β-α))I_2+1/2(1-e^i(β-α))σ_1 .

Applying <Ref> we see that
u^*v+σ_3u^*vσ_3     u^*v-σ_3u^*vσ_3

u^*v+σ_3u^*vσ_3      u^*v-σ_3u^*vσ_3

Note that

$u^*v+\sigma_3u^*v\sigma_3=\big(1+e^{i(\beta-\alpha)}\big)I_2\quad\mbox{ and }\quad u^*v-\sigma_3u^*v\sigma_3=\big(1-e^{i(\beta-\alpha)}\big)\sigma_1\,.$

Therefore, we get that
u^*_2v_2=1/2(1+e^i(β-α))I_2⊗I_2+1/2(1-e^i(β-α))σ_1⊗σ_1 .

Combining <Ref> and <Ref>, we get the following

\[
u^*_2v_2=\mbox{bl-diag}\{I_2,\sigma_1\}(u^*v\otimes I_2)\mbox{bl-diag}\{I_2,\sigma_1\},
\]

and hence we have

\[
u^*_2v_2(\Delta_2\otimes M_2)v^*_2u_2=\mbox{bl-diag}\{I_2,\sigma_1\}\big(u^*v\Delta_2v^*u\otimes M_2\big) \mbox{bl-diag}\{I_2,\sigma_1\}.
\]

To complete the induction note that for $k\geq 2$, thanks to <Ref>, we see that

\[
\]

In other words, by <Ref> we obtain the following equality

\[
u^*_{2k}v_{2k}=\mbox{bl-diag}\{I^{(k)}_2,\sigma_1^{(k)}\}\big(u^*v\otimes I^{(k)}_2\big)\mbox{bl-diag}\{I^{(k)}_2,\sigma_1^{(k)}\}\,.
\]

Since $W_{2k}=\mbox{bl-diag}\{I^{(k)}_2,\sigma_1^{(k)}\}\in\Delta_2\otimes M^{(k)}_2$, the proof is completed.

For any $k\in\bbn$, we have the following
=    Ad_u_2kW_2k((Δ_2∩(u^*vΔ_2v^*u))⊗M_2^(k))

=    Ad_u_2kW_2k(⊗M_2^(k)) .

We have the following,

=    _u_2k(Δ_2⊗M_2^(k)⋂_u_2k^*v_2k(Δ_2⊗M_2^(k)))

=    _u_2k(Δ_2⊗M_2^(k)⋂_W_2k(u^*vΔ_2v^*u⊗M_2^(k)))    ()

=    _u_2kW_2k(Δ_2⊗M_2^(k)⋂  (u^*vΔ_2v^*u)⊗M_2^(k))

=    _u_2kW_2k((Δ_2∩ (u^*vΔ_2v^*u))⊗M_2^(k)) .

This proves the first equality, and for the second, note that since $\beta\neq\alpha,\pi\pm\alpha$ in our case, we have $\Delta_2\cap \left(u^*v\Delta_2v^*u\right)=\bbc\,$.

We have the following tower of finite von Neumann algebras

\[
\begin{matrix}
A_0 &\subset & A_2 &\subset & A_4 &\subset  &\ldots &\subset & R \cr
\cup &  &\cup &  &\cup & & \ldots &  &\cup \cr
B_0^u &\subset & B_2^u &\subset & B_4^u &\subset  &\ldots &\subset & R_u \cr
\cup &  &\cup &  &\cup &  &\ldots &  &\cup \cr
C_0 &\subset & C_2 &\subset & C_4 &\subset  &\ldots &\subset & R_u\cap R_v
\end{matrix}
\]

In the following diagram

\[
\begin{matrix}
A_0 &\subset & A_2 &\subset  & A_4 &\subset &\ldots &\subset & R \cr
\cup &  &\cup &  &\cup &  & \ldots &  & \cup \cr
C_0 &\subset & C_2 &\subset & C_4 &\subset &\ldots &\subset & R_u\cap R_v
\end{matrix}
\]

each individual quadruple is a commuting square.

Choose any $k\in\bbn$ and consider the following quadruple

\[
\begin{matrix}
A_{2k-2} &\subset & A_{2k} \cr
\cup &  &\cup \cr
C_{2k-2} &\subset & C_{2k}\,.
\end{matrix}
\]

Now, draw the cube $\mathscr{C}_k$ as described in <Ref>.

7 cm!

Commuting cube $\mathscr{C}_k$

Here, the adjacent faces $(B^u_{2k-2}\subset A_{2k-2}\,,\,B^u_{2k}\subset A_{2k})$ and $(B^v_{2k-2}\subset A_{2k-2}\,,\,B^v_{2k}\subset A_{2k})$ are commuting squares by the construction of spin model subfactors $R_u\subset R$ and $R_v\subset R$ respectively. Since the required quadruple is the slice in <Ref>, by <Ref> it is a commuting square.

The von Neumann subalgebra $R_u\cap R_v$ of $R$ is a $II_1$ subfactor with $[R:R_u\cap R_v]=4$ and $[R_u:R_u\cap R_v]=[R_v:R_u\cap R_v]=2$.

By <Ref>, we have $C_0=\bbc$ and $C_2=\mbox{Ad}_{u_2W_2}(\bbc\otimes M_2)$. In <Ref>, putting $k=1$ we have the following commuting square denoted by the symbol $\mathscr{S}$:

⊗M_2    ⊂    M_2⊗M_2 ∪        ∪   ⊂    _u_2W_2(⊗M_2)

which is the slice in the commuting cube $\mathscr{C}_1$ as described in <Ref>. Observe that the norm of the inclusion matrices of both the vertical embeddings in $\mathscr{S}$ are same and equal to $2$. Therefore, by Lemma 3.10 in [5], we see that $\mathscr{S}$ is a non-degenerate commuting square. In other words, the slice in the commuting cube $\mathscr{C}_1$ is non-degenerate, and hence the commuting cube $\mathscr{C}_1$ is a non-degenerate commuting cube (see <Ref>). The tower of the basic construction for the the horizontal inclusion $\mathbb{C}\otimes M_2\subset M_2\otimes M_2$ in $\mathscr{S}$ is given by
ℂ⊗M_2 ⊂M_2⊗M_2 ⊂^f_1 M^(2)_2 ⊗M_2⊂^f_2  M^(3)_2⊗M_2⊂⋯⊂^f_k M^(k)_2⊗M_2⊂⋯
where $f_k=\frac{1}{2}\sum_{i,j=1}^{2}E_{ij}\otimes E_{ij}\otimes I^{(k)}_2\in M^{(k+1)}_2\otimes M_2=M^{(k+2)}_2$ is the Jones projection. Indeed, by our construction in <Ref>, we have the following tower

\[
\bbc\subset\Delta_2\subset^{\,e_1} M_2\subset^{\,e_2}\Delta_2\otimes M_2\subset^{\,e_1\otimes I_2\,} M_2\otimes M_2\subset^{\,e_2\otimes I_2\,}\Delta_2\otimes M_2\otimes M_2\subset\cdots\cdots\subset R
\]

for the hyperfinite factor $R$, where $e_1=J_2,\,e_2=\mbox{diag}\{E_{11},E_{22}\}$. Applying Theorem $3.5$ in [1], for instance, we see the above form of $f_k$. We put $L_{-1}=\bbc$ and $L_0= \mbox{Ad}_{u_2W_2}(\bbc\otimes M_2)$ and for $k \geq 1$, suppose that $L_k=\{L_{k-1},f_k\}^{\dprime}$. As $\mathscr{S}$ is a non-degenerate commuting square, it follows that the following tower of algebras is the Jones' basic construction tower:

\[
\mathbb{C}\subset L_0\subset L_1\subset^{f_2} L_2 \subset \cdots \subset^{f_k} L_k\subset\cdots
\]

Define $R_{u,v}=\overline{\bigcup_k L_k}^{\tiny\, \mbox{SOT}}$. Note that $R_{u,v}$ is a $II_1$ subfactor of $R$ with $[R:R_{u,v}]=4$ (see Corollary $5.7.4$ in [26], for example). That the tower in <Ref> indeed gives us $R$ follows as an application of Theorem $3.5$ in [1]. Thus, the quadruple $(R_{u,v}\subset R_u,R_v\subset R)$ is obtained as iterated basic construction of the non-degenerate commuting cube (see <Ref>) $\mathscr{C}_1$. Notice that $R_{u,v}\subset R_u\cap R_v$ because $f_k\in R_u\cap R_v$ (by <Ref>) and $\mbox{Ad}_{u_2W_2}(\bbc\otimes M_2)\subset R_u\cap R_v\,$. Since $[R:R_{u,v}]=4$, thanks to the multiplicativity of Jones index, we must have $[R_u:R_{u,v}]=2$. Hence, $R_{u,v}\subset R_u$ is irreducible. Since $R_{u,v}\subset R_u\cap R_v\subset R_u$, we conclude that $R_u\cap R_v$ is also a type $II_1$ factor, and $R_{u,v}=R_u\cap R_v$ as $[R_u\cap R_v:R_{u,v}]=1$.

The quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$ of $II_1$ factors is obtained as an iterated basic construction of the non-degenerate commuting cube $\mathscr{C}_1$ described in <Ref>. The iterated basic construction of the non-degenerate commuting cube $\mathscr{C}_1$ at the $k$-th step coincides with the commuting cube $\mathscr{C}_k$ for all $k\in\bbn$.

In <Ref>, the quadruple $(R_{u,v}\subset R_u,R_v\subset R)$ of $II_1$ factors is obtained as an iterated basic construction of the non-degenerate commuting cube $\mathscr{C}_1$, and since $R_{u,v}=R_u\cap R_v$, the first part follows. For the second part, we need to show that the following grid of (finite-dimensional) non-degenerate commuting squares

\[
\begin{matrix}
M_2 &\subset & M_2\otimes M_2 &\subset^{\,f_1} & \cdots &\subset & M_2\otimes M_2^{(k)} & \subset^{\,f_k} & \cdots & \subset R\\
\cup & &\cup & & & &\cup &  & & \cup\\
\bbc &\subset & \mathrm{Ad}_{u_2W_2}(\bbc\otimes M_2) &\subset & \cdots &\subset & \mathrm{Ad}_{u_{2k}W_{2k}}(\bbc\otimes M_2^{(k)}) & \subset & \cdots & \subset R_u\cap R_v
\end{matrix}
\]

is the tower of basic constructions for the subfactor $R_u\cap R_v\subset R$, where $f_k=\frac{1}{2}\sum_{i,j=1}^{2}E_{ij}\otimes E_{ij}\otimes I^{(k)}_2\in M^{(k+1)}_2\otimes M_2=M^{(k+2)}_2$ is the Jones projection. Recall from the proof of <Ref> that the tower $L_{-1}\subset L_0\subset L_1\subset\cdots\subset R_{u,v}$ of finite-dimensional algebras is the tower of basic constructions for the subfactor $R_{u,v}$ and $R_{u,v}=R_u\cap R_v$. We show that $L_k=C_{2k+2}$ for all $k\geq 0$. In the following grid of quadruples

\[
\begin{matrix}
A_0 &\subset & A_2 &\subset^{\,f_1} & A_4 &\subset^{\,f_2} & A_6 &\subset^{\,f_3}  \cdots &\subset & R\\
\cup & & \cup & & \cup & & \cup &  & & \cup\\
C_0 &\subset & C_2 &\subset & C_4 &\subset & C_6 & \cdots &\subset & R_u\cap R_v\\
|| & & || & & \rotatebox{90}{$\subseteq$} & & \rotatebox{90}{$\subseteq$} & & & ||\\
L_{-1} &\subset & L_0 &\subset & L_1 &\subset & L_2 & \subset  \cdots &\subset & R_{u,v}\\
\end{matrix}
\]

where $f_k=\frac{1}{2}\sum_{i,j=1}^{2}E_{ij}\otimes E_{ij}\otimes I^{(k)}_2\in A_{2k+2}$, we have the entire quadruple $(L_k\subset A_{2k+2},L_{k+1}\subset A_{2k+4})$ and the upper quadruple $(C_{2k}\subset A_{2k},C_{2k+2}\subset A_{2k+2})$ are commuting squares for each $k\geq 0$. Therefore, the lower quadruple $(L_k\subset C_{2k+2},L_{k+1}\subset C_{2k+4})$ is also a commuting square. Indeed, we see that $E^{C_{2k+4}}_{C_{2k+2}}\big(L_{k+1}\big)=E^{A_{2k+4}}_{A_{2k+2}}\big(L_{k+1}\big)\subseteq L_k$. By [37], we have $\lambda(R_u\cap R_v,R_{u,v})$ is the limit of the decreasing sequence $\{\lambda(C_{2k+2},L_k)\}_{k\geq 0}$. Since $R_u\cap R_v=R_{u,v}$, we have $\lambda(R_u\cap R_v,R_{u,v})=1$, and hence $\lambda(C_{2k+2},L_k)=1$ for all $k\geq 0$. Thus, $C_{2k+2}=L_k$ for all $k\geq 0$ by <Ref>, which concludes the proof.

The interior and the exterior angle in the quadruple of $II_1$ factors $(R_u\cap R_v\subset R_u,R_v\subset R)$ are equal and given by the following,

\[
\cos\big(\alpha^{R_u\cap R_v}_R(R_u,R_v)\big)=\cos\big(\beta^{R_u\cap R_v}_R(R_u,R_v)\big)=\cos^2(\alpha-\beta)\,.
\]

The quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$ is a commuting (and consequently, co-commuting) square if and only if $\,\alpha-\beta=\pm\frac{\pi}{2}$. Here, $\alpha\mbox{ and }\beta$ are associated with the complex Hadamard matrices $u\mbox{ and }v$ as described in <Ref>.

Recall from <Ref> that the quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$ is obtained as an iterated basic construction of the non-degenerate commuting cube $\mathscr{C}_1$ described in <Ref>, and further the basic construction of $\mathscr{C}_1$ at the $k$-th step coincides with $\mathscr{C}_k$ for all $k\in\bbn$. Stacking all the commuting cubes $\mathscr{C}_k$, we have a commuting cube $\mathscr{C}_\infty$ described in <Ref>.

7 cm!

Commuting cube $\mathscr{C}_\infty$

The adjacent faces in $\mathscr{C}_\infty$ are the commuting squares $(\mbox{Ad}_u(\Delta_2)\subset M_2\,,\,R_u\subset R)$ and $(\mbox{Ad}_v(\Delta_2)\subset M_2\,,\,R_v\subset R)$. Applying <Ref>, we see that the quadruples $(\bbc\subset\mbox{Ad}_u(\Delta_2),R_u\cap R_v\subset R_u)$ (which is one of the remaining faces) and $(\bbc\subset M_2,R_u\cap R_v\subset R)$ (which is the slice) are also commuting squares. It can now be easily checked that $\{\lambda_1:=\mbox{Ad}_u(\sqrt{2}E_{11}),\lambda_2:=\mbox{Ad}_u(\sqrt{2}E_{22})\}$ is a basis for $R_u/R_u\cap R_v$, and similarly $\{\mu_1:=\mbox{Ad}_v(\sqrt{2}E_{11}),\mu_2:=\mbox{Ad}_v(\sqrt{2}E_{22})\}$ is a basis for $R_v/R_u\cap R_v$. Since $\lambda_i,\mu_i\in M_2$ for $i=1,2$, we have $E^R_{R_u\cap R_v}(\lambda_i^*\mu_j)=E^{M_2}_\bbc(\lambda_i^*\mu_j)$ for $i,j\in\{1,2\}$. This is because the quadruple $(\bbc\subset M_2,R_u\cap R_v\subset R)$ is a commuting square. Now, by the formula of the interior angle in <Ref>, we get the following,
cos(α^R_u∩R_v_R(R_u,R_v))    =    ∑_i,j=1^2 tr(E^R_R_u∩R_v(λ_i^*μ_j)μ_j^*λ_i)-1

=    ∑_i,j=1^2 tr(E^M_2_(λ_i^*μ_j)μ_j^*λ_i)-1

=    ∑_i,j=1^2 tr(λ_i^*μ_j) tr(μ_j^*λ_i)-1

=    cos^2(α-β) ,

where the last equality is a simple verification using the form of $u\mbox{ and }v$ described in <Ref>. Finally, recall from [2] that if $(N\subset P,Q\subset M)$ is a quadruple of $II_1$ factors such that $N\subset M$ is extremal with $[M:N]<\infty$ and $[P:N]=[M:Q]$, then $\alpha^N_M(P,Q)=\beta^N_M(P,Q)$, that is, the interior and the exterior angle coincide. In our case, $R_u\cap R_v=R_{u,v}$ (<Ref>) is extremal and all the other hypotheses are also satisfied along with it. Hence, we get the first part of the theorem.

Finally, the quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$ is a commuting (and consequently, co-commuting) square if and only if $\cos\big(\alpha^{R_u\cap R_v}_R(R_u,R_v)\big)=0$, which happens if and only if $\alpha-\beta=\pm\frac{\pi}{2}$.

The quadruple of $II_1$ factors

\[
\begin{matrix}
R_u &\subset & R\cr
\rotatebox{90}{$\subset$} &\ &\rotatebox{90}{$\subset$}\cr
R_{u}\cap R_v &\subset & R_v
\end{matrix}
\]

is always non-degenerate, that is $\overline{R_uR_v}=\overline{R_vR_u}=R$.

For any $k\in\bbn\cup\{0\}$, we claim that the following holds
u_2k(Δ_2⊗M_2^(k))u_2k^*v_2k(Δ_2⊗M_2^(k))v_2k^*=M_2⊗M_2^(k) .

First consider $k=0$. We have

\[
\]

where $\beta\neq\alpha,\pi\pm\alpha$ (<Ref> and discussion after it). Hence, none of the entries of the unitary matrix $u^*v$ are zero. Observe that if any unitary matrix $U$ in $M_2$ is such that none of the entries of it are zero, then $\Delta_2U\Delta_2=M_2$. Therefore, we have $\Delta_2u^*v\Delta_2=M_2$, and consequently $u\Delta_2u^*v\Delta_2v^*=M_2$, which settles Eqn. <ref> for $k=0$. For $k\in\bbn$, by <Ref> and <Ref> we have the following,

\[
{u}^*_{2k}{v}_{2k}\big(\Delta_2\otimes M^{(k)}_2\big){v}^*_{2k}{u}_{2k} =\mbox{bl-diag}\big\{I^{(k)}_2,\sigma_1^{(k)}\big\}\big(u^*v\Delta_2v^*u\otimes M^{(k)}_2\big)\mbox{bl-diag}\big\{I^{(k)}_2,\sigma_1^{(k)}\big\}.
\]

=    u_2k((Δ_2 ⊗M^(k)_2){I^(k)_2,σ_1^(k)})(u^*vΔ_2v^*u⊗M^(k)_2){I^(k)_2,σ_1^(k)}u_2k^*

=    _u_2k((Δ_2 ⊗M^(k)_2)(u^*vΔ_2 v^*u⊗M^(k)_2){I^(k)_2,σ_1^(k)})

=    _u_2k((Δ_2u^*vΔ_2 v^*u⊗M^(k)_2){I^(k)_2,σ_1^(k)})

=    _u_2k((M_2⊗M^(k)_2){I^(k)_2,σ_1^(k)})    ()

=    _u_2k(M_2⊗M^(k)_2).

Here in the above, the second last equality follows from the $k=0$-step, that is, $u\Delta_2u^*v\Delta_2v^*=M_2$. Since ${u}_{2k}\in M_2\otimes M^{(k)}_2$, we finally have the following,

\[
u_{2k}\big(\Delta_2\otimes M_2^{(k)}\big)u_{2k}^*v_{2k}\big(\Delta_2\otimes M_2^{(k)}\big)v_{2k}^*=M_2\otimes M^{(k)}_2
\]

for all $k\in\bbn\cup\{0\}$. Hence, we conclude  that $\overline{R_uR_v}^{\tiny\,\mbox{SOT}}=R$. Indeed, since $R_u=\overline{\bigcup_i\,B_{2i}^u}^{\tiny\,\mbox{SOT}}$ and $R_v=\overline{\bigcup_j\,B_{2j}^v}^{\tiny\,\mbox{SOT}},$ we have

\[
R=\overline{\cup_i\,B_{2i}^uB_{2i}^v}\subseteq\overline{\cup_{i,j}\,B_{2i}^uB_{2j}^v}\subseteq\overline{(\cup_i\,B_{2i}^u)(\cup_j\,B_{2j}^v)}\subseteq\overline{\overline{(\cup_i\,B_{2i}^u)}\,\overline{(\cup_j\,B_{2j}^v)}}=\overline{R_uR_v}\subseteq R\,,
\]

where the closure being taken in the SOT topology. Therefore, the given quadruple is always non-degenerate.

If the condition of commuting square is given for a quadruple of $II_1$ factors $(N\subset P,Q\subset M)$ with $[M:N]<\infty$, then the quadruple is non-degenerate if and only if it is a co-commuting square (see [43]). In the absence of commuting square, if irreducibility of $N\subset M$ is given, then also this holds (see [16]). In <Ref>, we have a family of quadruples of subfactors indexed by $2\times 2$ complex Hadamard matrices $u\mbox{ and }v$ such that $u\nsim v$, where `non-degeneracy' and `co-commuting' are not the same. To the best of our knowledge, this is the first concrete example in the literature (see Remark $2.3$ in this regard).

§.§ Characterization of the quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$

Due to <Ref>, <Ref>, and <Ref>, it is clear that the subfactor $R_u\cap R_v\subset R$ is not irreducible, at least when $\alpha-\beta\neq\pm\frac{\pi}{2}$. We obtain its commutant in $R$, and characterize the subfactor $R_u\cap R_v\subset R$. As a curious fact, we obtain that although we have started with two spin model subfactors $R_u,\,R_v\subset R$, their intersection is a vertex model subfactor. Furthermore, we characterize the quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$ and show that it is generated (in the sense of iterated basic construction) by a single bi-unitary matrix in $M_2\otimes M_2$.

For any $2\times 2$ complex Hadamard matrices $u\mbox{ and }v$ such that $u\nsim v$, the subfactor $R_u\cap R_v\subset R$ is a vertex model subfactor of Jones index $4$.

We have the symmtric commuting square (<Ref> and <Ref>) $\mathscr{S}$:

\[
\begin{matrix}
\mbox{Ad}_{u_2W_2}(\bbc\otimes M_2) &\subset & M_2\otimes M_2 \cr
\cup &  &\cup \cr
\bbc &\subset & \bbc\otimes M_2
\end{matrix}
\]

which is same as the following commuting square

\[
\begin{matrix}
\mbox{Ad}_{u_2W_2V_2}(M_2\otimes\bbc) &\subset & M_2\otimes M_2 \cr
\cup &  &\cup \cr
\bbc &\subset & \bbc\otimes M_2
\end{matrix}
\]

where $\mbox{Ad}_{V_2}:M_2\otimes\bbc\longmapsto\bbc\otimes M_2$ is the flip given by the following permutation matrix

\[
1 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}=\begin{bmatrix}
E_{11} & E_{21}\\
E_{12} & E_{22}
\end{bmatrix}
\]

in $M_4$. It is now clear that $u_2W_2V_2$ is a bi-unitary in $M_4=M_2\otimes M_2$ (can be checked directly also) and $R_u\cap R_v\subset R$ is a vertex model subfactor of index $4$. A simple verification will show the following,
u     uσ_1

u     -uσ_1

where $\sigma_1$ is Pauli spin matrix.

The relative commutant ${(R_u\cap R_v)}^\prime\cap R$ is $\bbc\oplus\bbc$.

Recall from <Ref> that the following grid of (finite-dimensional) non-degenerate commuting squares

\[
\begin{matrix}
M_2 &\subset & M_2\otimes M_2 &\subset^{\,f_1} & \cdots &\subset & M_2\otimes M_2^{(k)} & \subset^{\,f_k} & \cdots & \subset R\\
\cup & &\cup & & & &\cup &  & & \cup\\
\bbc &\subset & \mathrm{Ad}_{u_2W_2}(\bbc\otimes M_2) &\subset & \cdots &\subset & \mathrm{Ad}_{u_{2k}W_{2k}}(\bbc\otimes M_2^{(k)}) & \subset & \cdots & \subset R_u\cap R_v
\end{matrix}
\]

is the tower of basic constructions for the subfactor $R_u\cap R_v\subset R$. By the Ocneanu compactness theorem (see <Ref>), we have the following
(R_u∩R_v)^'∩R=(_u_2W_2(⊗M_2))^'∩(⊗M_2) .

Then, <Ref> implies the following
(R_u∩R_v)^'∩R=_u_2W_2((⊗M_2)^'∩_W_2^*u_2^*(⊗M_2)) ,

where $W_2$ is as defined in <Ref>. Now, for any $A\in M_2$ we have the following using <Ref>,
=    1/2[
(_u^*+_σ_3u^*)(A)     ((_u^*-_σ_3u^*)(A))σ_1

σ_1((_u^*-_σ_3u^*)(A))     _σ_1((_u^*+_σ_3u^*)(A))

] .

Since $\left(\bbc\otimes M_2\right)^\prime\cap (M_2\otimes M_2)=M_2\otimes\bbc$, we immediately see that if any $B\otimes I_2$ in $M_2\otimes\bbc\subseteq M_2\otimes M_2$ has to lie in $\mbox{Ad}_{W_2^*u_2^*}(\bbc\otimes M_2)$, then it is necessary that $B=\left[{\begin{smallmatrix}
z & w\\
w & z\\
\end{smallmatrix}}\right]$ for some $z,w\in\bbc$. This proves the following,
z     w

w     z

]⊗I_2 : z,w∈} .

Conversely, for any $B=\left[{\begin{smallmatrix}
z & w\\
w & z\\
\end{smallmatrix}}\right]\otimes I_2$ consider $C=\mbox{Ad}_u\left(\left[{\begin{smallmatrix}
z & w\\
w & z\\
\end{smallmatrix}}\right]\right)\in M_2$ and observe that $\mbox{Ad}_{W_2^*u_2^*}(I_2\otimes C)=B$. Hence, the inclusion in <Ref> is in fact equality. By <Ref>, using <Ref> we get that
(R_u∩R_v)^'∩R    =    {_u_2W_2([
z     w

w     z

]⊗I_2) : z,w∈}

=    {_u_2({I_2,σ_1} ([
z     w

w     z

]⊗I_2){I_2,σ_1}) : z,w∈}

=    {I_2⊗_u([
z     w

w     z

]) : z,w∈}

=    {I_2⊗_uF_2([

]) : z,w∈} ,
where $F_2$ is the $2\times 2$ Fourier matrix. Therefore, ${(R_u\cap R_v)}^\prime\cap R=\bbc\oplus\bbc$, where the embedding $\bbc\oplus\bbc\xhookrightarrow{} M_2$ is given by $(\gamma,\delta)\mapsto\mbox{Ad}_{uF_2}(\mbox{diag}\{\gamma,\delta\})$.

* The pair of subfactors $R_u$ and $R_v$ of the hyperfinite type $II_1$ factor $R$ are conjugate to each other via a unitary in ${(R_u\cap R_v)}^\prime\cap R$.

* The bi-unitary matrix $\,u_2W_2V_2$ in $M_4$ generates (in the sense of iterated basic construction of finite-dimensional grids) the composition of subfactors $R_u\cap R_v\subset R_u\subset R$, and consequently the quadruple $(R_u\cap R_v\subset R_u,R_v\subset R)$.

Since $u=uv^*v$, we have $B^u_0=\mbox{Ad}_u(\Delta_2)=\mbox{Ad}_{uv^*}\mbox{Ad}_v(\Delta_2)=\mbox{Ad}_{uv^*}(B^v_0)$. We claim that $u_{2k}=(I_2^{(k)}\otimes uv^*)v_{2k}$ for all $k\geq 0$. This follows by induction on $k$, since by <Ref>, we have the following,
u_2k+2    =    (I_2⊗u_2k) {I_2^(k),η_k} (ξ_k⊗I_2^(k))

=    (I_2⊗I_2^(k)⊗uv^*)(I_2⊗v_2k) {I_2^(k),η_k} (ξ_k⊗I_2^(k))

=    (I_2^(k+1)⊗uv^*)v_2k+2 .

Therefore, we have $B^u_{2k}=\mbox{Ad}_{uv^*}(B^v_{2k})$, where the unitary $uv^*\in M_2$ is identified with $I_2^{(k)}\otimes uv^*\in M_2\otimes M_2^{(k)}=A_{2k}$, and consequently $\cup_{k\geq 0}B^u_{2k}=\mbox{Ad}_{uv^*}\big(\cup_{k\geq 0}B^v_{2k}\big)$. Since $R_u=\overline{\cup_k\,B^u_{2k}}^{\mbox{ sot}}$ and $R_v=\overline{\cup_k\,B^v_{2k}}^{\mbox{ sot}}$,  we have $R_u=\mbox{Ad}_{uv^*}(R_v)$, where the unitary $uv^*$ is the diagonal matrix $\mbox{diag}\{1,e^{i(\alpha-\beta)}\}$ (see <Ref>) in $M_2$. Now, $uv^*=\mbox{Ad}_u(v^*u)$ and the matrix $v^*u$ is the following

\[
1+e^{i(\alpha-\beta)} & 1-e^{i(\alpha-\beta)}\\
1-e^{i(\alpha-\beta)} & 1+e^{i(\alpha-\beta)}
\end{bmatrix}\,.
\]

Hence by <Ref> (in particularly, <Ref>), we see that $uv^*\in{(R_u\cap R_v)}^\prime\cap R$. This completes part $(i)$.

For part $(ii)$, using the fact that $W_2=E_{11}\otimes I_2+E_{22}\otimes\sigma_1$ (see <Ref>) lies in $\Delta_2\otimes M_2$ we have $B^u_2=\mbox{Ad}_{u_2}(\Delta_2\otimes M_2)=\mbox{Ad}_{u_2W_2}(\Delta_2\otimes M_2)$. Observe that $\mbox{Ad}_{V_2}:M_2\otimes\Delta_2\mapsto\Delta_2\otimes M_2$, where $V_2$ is the flip as in <Ref>, and hence $B^u_2=\mbox{Ad}_{u_2W_2V_2}(M_2\otimes \Delta_2)$. By induction, it is easy to verify that for all $k\in\bbn$, if we consider the following unitary matrix

\[
V_{2k}:=\prod_{j=0}^{k-1}\big(I_2^{(j)}\otimes V_2\otimes I_2^{(k-j-1)}\big)
\]

in $M_2^{(k+1)}=M_2\otimes M_2^{(k)}$, then $\mbox{Ad}_{V_{2k}}:M_2^{(k)}\otimes\Delta_2\longmapsto\Delta_2\otimes M_2^{(k)}$. Since $W_{2k}=E_{11}\otimes I_2^{(k)}+E_{22}\otimes\sigma_1^{(k)}\in\Delta_2\otimes M_2^{(k)}$, we have

\[
B^u_{2k}=\mbox{Ad}_{u_{2k}}(\Delta_2\otimes M_2^{(k)})=\mbox{Ad}_{u_{2k}W_{2k}}(\Delta_2\otimes M_2^{(k)})=\mbox{Ad}_{u_{2k}W_{2k}V_{2k}}(M_2^{(k)}\otimes\Delta_2).
\]

for all $k\in\bbn$. Therefore, we have the following grid of non-degenerate commuting squares

\[
\begin{matrix}
M_2 &\subset & M_2\otimes M_2 &\subset^{\,f_1} & \cdots &\subset & M_2\otimes M_2^{(k)} & \subset^{\,f_k} & \cdots & \subset R\\
\cup & &\cup & & & &\cup &  & & \cup\\
\mbox{Ad}_u(\Delta_2) &\subset & \mathrm{Ad}_{u_2W_2V_2}(M_2\otimes\Delta_2) &\subset & \cdots &\subset & \mathrm{Ad}_{u_{2k}W_{2k}V_{2k}}(M_2^{(k)}\otimes\Delta_2) & \subset & \cdots & \subset R_u\\
\cup & &\cup & & & &\cup &  & & \cup\\
\bbc &\subset & \mathrm{Ad}_{u_2W_2V_2}(M_2\otimes\bbc) &\subset & \cdots &\subset & \mathrm{Ad}_{u_{2k}W_{2k}V_{2k}}(M_2^{(k)}\otimes\bbc) & \subset & \cdots & \subset R_u\cap R_v\\
\end{matrix}
\]

where $u_2W_2V_2\in M_4$ (see <Ref>) is the bi-unitary obtained in <Ref>, and $f_k=\frac{1}{2}\sum_{i,j=1}^{2}E_{ij}\otimes E_{ij}\otimes I^{(k)}_2\in M^{(k+2)}_2,\,k\in\bbn,$ are the Jones' projections. Finally, since $R_u\mbox{ and }R_v$ are unitary conjugate, and the conjugation is implemented by a unitary belonging at the very first stage (i,e. $M_2$), we see that the single bi-unitary matrix $u_2W_2V_2$ in $M_4$ generates (in the sense of iterated basic construction) the quadruple $(R_u\cap R_v\subset R_u\,,\,R_v\subset R)$ of $II_1$ factors, and we have an explicit grid of finite-dimensional algebras for the entire quadruple as described above.

Before moving further, we pause for an important remark.

It is known that (see [32]) any bi-unitary matrix in $M_4$ is equivalent to $W(\omega)=\mbox{diag}\{1,1,1,\omega\}$ for some $\omega\in\mathbb{S}^1$. Moreover, $W(\omega)$ is equivalent to $W(\omega^\prime)$ if and only if $\Re(\omega)=\Re(\omega^\prime)$. In our situation, using <Ref> one can see that $u_2W_2V_2=(F_2\otimes u)\mbox{diag}\{I_2,\sigma_1\}$. Since $F_2\sigma_1F_2=\mbox{diag}\{1,-1\}=\sigma_3$, we have $u_2W_2V_2=(F_2\otimes u)(I_2\otimes F_2)\mbox{diag}\{I_2,\sigma_3\}(I_2\otimes F_2)=(F_2\otimes uF_2)W(-1)(I_2\otimes F_2)$, that is, the bi-unitary $u_2W_2V_2$ is equivalent to $W(-1)$.

By <Ref> (see also <Ref>) and Section $4$ in [32], we conclude that $R_u\cap R_v\subset R$ is of depth 2, and by [40] we have the following characterization.

The subfactor $R_u\cap R_v\subset R$ is isomorphic to the diagonal subfactor

\[
\left\{\begin{bmatrix}
x & 0\\
0 & \theta(x)
\end{bmatrix}:x\in R\,,\,\theta\in\mathrm{Out}(R)~\mathrm{such~ that}~{\theta}^2=id \right\}\subset M_2(\mathbb{C})\otimes R\,.
\]

Under this isomorphism  $R_u$ becomes $\big(\mathrm{diag}\{1,e^{i(\alpha-\beta)}\}\big)R_v\big(\mathrm{diag}\{1,e^{i(\beta-\alpha)}\}\big)$, where $R_v$ is given by

\[
\Bigg\{\begin{bmatrix}
x & y\\
\theta(y) & \theta(x)
\end{bmatrix} : x,y \in R\Bigg\}\,.
\]

Here, $\alpha\mbox{ and }\beta$ are associated with the complex Hadamard matrices $u\mbox{ and }v$ as described in <Ref>.

§.§ Relative entropy and Sano-Watatani angle between $R_u\mbox{ and }R_v$

We investigate relative entropy and the Sano-Watatani angle between the pair of spin model subfactors $R_u\mbox{ and }R_v$.

For the pair of spin model subfactors $R_u\mbox{ and }R_v$, we have the following.

* $H(R|R_u\cap R_v)=2\log 2\mbox{ and }H(R_u|R_u\cap R_v)=H(R_v|R_u\cap R_v)=\log 2$.

* $h(R_u|R_v)=\eta\big(\cos^2\big(\frac{\alpha-\beta}{2}\big)\big)+\eta\big(\sin^2\big(\frac{\alpha-\beta}{2}\big)\big)\,.$

* $\eta\big(\cos^2\big(\frac{\alpha-\beta}{2}\big)\big)+\eta\big(\sin^2\big(\frac{\alpha-\beta}{2}\big)\big)\leq H(R_u|R_v)\leq\log 2\,,$
and if $\,\alpha-\beta=\pm\frac{\pi}{2}$, then $H(R_u|R_v)=h(R_u|R_v)=\log 2$. Here, $\alpha\mbox{ and }\beta$ are associated with the complex Hadamard matrices $u\mbox{ and }v$ as in <Ref>

Corollary $4.6$ in [37] gives the value of $H(R|R_u\cap R_v),\,H(R_u|R_u\cap R_v)$ and $H(R_v|R_u\cap R_v)$. Since $H(R_u|R_v)\leq H(R_u|R)+H(R|R_v)=H(R|R_v)$, the upper bound for $H(R_u|R_v)$ is clear. Now, recall from [9] that $h(u\Delta_2u^*|v\Delta_2v^*)=h(\Delta_2|u^*v\Delta_2v^*u)=\frac{1}{2}\sum_{i,j=1}^2\eta |(u^*v)_{ij}|^2$. Each $|(u^*v)_{ij}|^2$ can easily be obtained from <Ref>, and since

\[
h(u\Delta_2u^*|v\Delta_2v^*)\leq H(u\Delta_2u^*|v\Delta_2v^*)\leq H(R_u|R_v)
\]

due to <Ref>, the lower bound for $H(R_u|R_v)$ is obtained. We also have

\[
\]

as $h(R_u|R_v)\geq h(u\Delta_2u^*|v\Delta_2v^*)$. To obtain the reverse inequality, recall that by <Ref> we have $R_u=\mbox{Ad}_{uv^*}(R_v)$, where $uv^*=\mathrm{diag}\{1, e^{i(\alpha-\beta)}\}$. Therefore, thanks to Theorem $3.4$ in [10], it is now a straightforward verification that

\[
\]

Finally, for $\alpha-\beta=\pm\frac{\pi}{2}$ the lower bound for $H(R_u|R_v)$ also becomes $\log 2$, which completes the proof.

The Sano-Watatani angle between the subfactors $R_u\mbox{ and }R_v$ is the singleton set $\{\arccos|\cos(\alpha-\beta)|\}$, where $\alpha\mbox{ and }\beta$ are associated with the complex Hadamard matrices $u\mbox{ and }v$ as described in <Ref>.

For $\alpha-\beta=\pm\frac{\pi}{2}$, the result holds as $(R_u\cap R_v\subset R_u,R_v\subset R)$ is a commuting square by <Ref>. Consider now $\alpha-\beta\neq\pm\frac{\pi}{2}$. Recall the commuting cube $\mathscr{C}_k$ introduced in <Ref>, in particularly, see <Ref>. Consider the commuting cube $\mathscr{C}_1$ (for $k=1$). This is a non-degenerate commuting cube as shown in <Ref>. The quadruple of $II_1$ factors $(R_u\cap R_v\subset R_u,R_v\subset R)$ is obtained as an iterated basic construction of this non-degenerate commuting cube (<Ref>), as illustrated in <Ref> (recall <Ref> in this regard).

7 cm!

Iterated basic construction of non-degenerate commuting cube

With the notation as in <Ref>, using <Ref>, we get that the non-negative matrix $E_{B^u_0}^{A_0}E_{B^v_0}^{A_0}E_{B^u_0}^{A_0}-E_{C_0}^{A_0}$ satisfies the following,

\[
\big(E_{B^u_0}E_{B^v_0}E_{B^u_0}-E_{C_0}\big)^2=\cos^2(\alpha-\beta)\big(E_{B^u_0}E_{B^v_0}E_{B^u_0}-E_{C_0}\big)\,.
\]

Since we have the non-degenerate commuting cube $\mathscr{C}_\infty$ in <Ref> (which is the final resulting commuting cube in <Ref>), as an application of <Ref> we get that the spectrum of the cosine of the angle operator $\sqrt{e^R_{R_u}e^R_{R_v}e^R_{R_u}-e^R_{R_u\cap R_v}}$ for the pair of subfactors $R_u,R_v\subset R$ is the singleton set $\{|\cos(\alpha-\beta)|\}$.

Subfactors arising from   complex Hadamard matrices

Recall that there exists a continuous, one-parameter family of inequivalent $4\times 4$ complex Hadamard matrices, and this family is given by the following unitary

\[
u(z)= \frac{1}{2}\,\left[{\begin{matrix}
1 & 1 & 1 & 1\\
1 & iz & -1 & -iz\\
1 & -1 & 1 & -1\\
1 & -iz & -1 & iz\\
\end{matrix}}\right]
\]

where $\,z=e^{i\phi}\in\mathbb{S}^1$ with $\phi\in[0,\pi)$. Consider two distinct elements of this family

\[
1 & 1 & 1 & 1\\
1 & ia & -1 & -ia\\
1 & -1 & 1 & -1\\
1 & -ia & -1 & ia\\
\end{matrix}}\right] \quad\mbox{and}\quad v=\frac{1}{2}\,\left[{\begin{matrix}
1 & 1 & 1 & 1\\
1 & ib & -1 & -ib\\
1 & -1 & 1 & -1\\
1 & -ib & -1 & ib\\
\end{matrix}}\right]
\]

where $a,b\in\mathbb{S}^1$ (semicircle to be precise) and $a\neq b$. Throughout the section, remember that $b\neq\pm a$, since $\phi\in[0,\pi)$. We obtain the pair of spin model subfactors $R_u\subset R$ and $R_v\subset R$ as described in <Ref>. Since $u\mbox{ and }v$ are Hadamard inequivalent, we have $R_u\neq R_v$ due to <Ref>.  That is, we have a pair of irreducible subfactors each with index 4 :

\[\begin{matrix}
R_u &\subset & R\\
& & \cup\\
& & R_v\\
\end{matrix}
\]

The following

\[
\mathbb{C}\subset \Delta_4\subset M_4\subset \Delta_4 \otimes M_4\subset M_4\otimes M_4\subset \Delta_4\otimes M_4\otimes M_4\subset M_4\otimes M_4\otimes M_4\subset\cdots
\]

is a tower of Jones' basic construction, and $R$ is the closure in the SOT topology of the union of these subalgebras. Following <Ref>, the ladder of basic constructions of the following `spin model commuting square'.

\[
\begin{matrix}
u\Delta_4 u^* &\subset & M_4 \cr \cup &\ & \cup\cr \mathbb{C} &\subset &  \Delta_4
\end{matrix}\]

is depicted in <Ref> (note that in our convention $u_0=u$), where the unitary matrices $u_j$ are given in <Ref>.

$R$ $R$
d1$\Delta_4\otimes M_4\otimes M_4$
d2$M_4\otimes M_4$
d3$\Delta_4\otimes M_4$
f_1$u_3(M_4\otimes M_4)u_3^*$
f_2$u_2(\Delta_4 \otimes M_4)u_2^*$
e_1$v_3(M_4\otimes M_4)v_3^*$
e_2$v_2(\Delta_4 \otimes M_4)v_2^*$

A pair of $4\times 4$ Hadamard matrices and basic constructions

Although the basic construction (in the general $n\times n$ case) is well-known (see [26, 34, 29], for instance),  our basic construction is slightly different than the one in the existing literature to make computations easier. We remark that $R_u$ is the closure in the SOT topology of the union of the left vertical algebras in <Ref>, and similarly $R_v$ is that of the right vertical algebras.

We fix the following one-parameter family of unitary matrices throughout this section. For $\,\alpha\in\mathbb{S}^1$,
1     1     1     1

-i     α    i     -α

1     -1     1     -1

i     α    -i     -α

]       ,       W_1^(α)=[
i     0     0     0

0     α     0     0

0     0     -i     0

0     0     0     -α

] ,

1     0     0     0

0     -1     0     0

0     0     1     0

0     0     0     -1

]       ,       W_3^(α)=[
-i     0     0     0

0     α     0     0

0     0     i     0

0     0     0     -α

] .

These unitary matrices satisfy $\big(W_1^{(\alpha)}\big)^*=W_3^{(\overline{\alpha})},\,W_2^*=W_2,\,\big(W_3^{(\alpha)}\big)^*=W_1^{(\overline{\alpha})}$. Let $I_4^{(k)}$ denote the unit element $I_4\otimes\ldots\otimes I_4$ of $M_4^{(k)}:=M_4^{\,\otimes\,k}$. We have the following tower of basic construction (with the convention $u_0=u$).

The tower of the basic construction for $\mathbb{C}\subset u\Delta_4u^*$ is given by

\[
\mathbb{C}\subset u_0\Delta_4u^*_0\subset u_1M_4u^*_1\subset u_2(\Delta_4\otimes M_4)u^*_2\subset u_3(M_4\otimes M_4)u^*_3\subset\cdots
\]

where each unitary matrix $u_i$ is given by the following prescription :

* for $k\in\bbn\cup\{0\}$, we have $\,u_{2k+1}=E_{11}\otimes u_{2k}+\sum_{j=1}^3E_{j+1,j+1}\otimes u_{2k}\big(W_j^{(a)}\otimes I_4^{(k)}\big)$;

* for $k\in\bbn$, we have $\,u_{2k}=u_{2k-1}\big(\xi_a\otimes I_4^{(k)}\big)$.

For the proof of this theorem we request the interested reader to visit the Appendix. We use the following notations throughout this section which are at par with that in part $(1)$ of Notation <ref>. For $k\in\bbn\cup\{0\}$, let

\begin{align*}
A_{2k+1} &= \Delta_4\otimes M_4^{(k+1)}\,,\cr
B^u_{2k+1} &= u_{2k+1}(\bbc\otimes M_4^{(k+1)})u_{2k+1}^*\,,\cr
B^v_{2k+1} &= v_{2k+1}(\bbc\otimes M_4^{(k+1)})v_{2k+1}^*\,,\cr
C_{2k+1} &= B^u_{2k+1}\cap {B}^v_{2k+1}\,.
\end{align*}

These are the odd steps in the tower of basic constructions in <Ref>. Note that

\[
\overline{\cup A_{2k+1}}^{\mbox{ sot}}=R,\,\overline{\cup B^u_{2k+1}}^{\mbox{ sot}}=R_u,\,\overline{\cup B^v_{2k+1}}^{\mbox{ sot}}=R_v\mbox{ and } \overline{\cup\, C_{2k+1}}^{\mbox{ sot}}=R_u\cap R_v.
\]

We shall see latter in <Ref> and <Ref> the merit for considering odd steps in the tower described in <Ref>, as against the even steps considered in <Ref>. The following lemma is useful for the subsequent subsections.

In the following diagram

\[
\begin{matrix}
A_1 &\subset & A_3 &\subset  & A_5 &\subset &\ldots &\subset & R \cr
\cup &  &\cup &  &\cup &  &  &  & \cup \cr
C_1 &\subset & C_3 &\subset & C_5 &\subset &\ldots &\subset & R_u\cap R_v
\end{matrix}
\]

each individual quadruple is a commuting square.

Fix a $k\in\bbn$ and consider the cube in <Ref>.

7 cm!

A commuting cube in $4\times 4$

The adjacent faces are clearly commuting squares by the construction of $R_u,R_v\subset R$. The following quadruple

\[
\begin{matrix}
C_{2k+1} &\subset & A_{2k+1}\\
\cup &  & \cup\\
C_{2k-1} &\subset & A_{2k-1}\\
\end{matrix}
\]

is the slice in the cube. The required proof now follows from <Ref>.

For the sake of reader's benefit, we draw a flowchart in <Ref> to display our plan of actions in this section.

rounded rectangle,
minimum width=2.5cm,
minimum height=1cm] (block1)  Start with distinct $4\times 4$ inequivalent

complex Hadamard matrices $u\mbox{ and }v$;

trapezium left angle = 65,
trapezium right angle = 115,
trapezium stretches,
below=of block1,
minimum width=3.5cm,
minimum height=1cm] (block2)  Obtain the pair of spin model subfactors $R_u\mbox{ and }R_v\,$;

below=of block2,
minimum width=3.5cm,
minimum height=1cm] (block3)  Computation of $\lambda(R_u,R_v)\,$;

below=of block3,
minimum width=3.5cm,
minimum height=1cm] (block40)  Spectral decomposition for the family of unitary

matrices in the tower of basic constructions ;

below=of block40,
minimum width=3.5cm,
minimum height=1cm] (block4)  Identifying two situations, namely

$\,u\sim v\mbox{ and }u\nsim v$ depending on certain

rational or irrational rotations  ;

below=of block4,
minimum width=3.5cm,
minimum height=1cm] (block5)  When $u\sim v$, construction of a family

of finite index subfactors $R^{(m)}_{u,v}$ of $R$;

below=of block5,
minimum width=3.5cm,
minimum height=1cm] (block60)  The Ocneanu compactness and

irreducibility of $R^{(m)}_{u,v}$ in $R$ ;

below=of block60,
minimum width=3.5cm,
minimum height=1cm] (block61)  Factoriality of $R_u\cap R_v\,$

and rigidity of the

interior angle $\alpha_R(R_u,R_v)$  ;

below=of block61,
minimum width=3.5cm,
minimum height=1cm] (block62)  Intermediate subfactor containing $R_u,R_v\,$;

rounded rectangle,
right=of block62,
minimum width=3.5cm,
minimum height=1cm] (block63)  Angle between $R_u,R_v$ and

relative entropy $H(R_u|R_v)\,$;

below right=of block4,
minimum width=3.5cm,
minimum height=1cm] (block6)  When $u\nsim v,\,R_u\cap R_v$ has infinite

Pimsner-Popa index in $R$;

below=of block6,
minimum width=3.5cm,
minimum height=1cm] (block7)  Bounds for $H(R_u|R_v)\,$;

rounded rectangle,
below=of block7,
minimum width=3.5cm,
minimum height=1cm] (block8)  Characteriztion of the commuting

square $\big(R_u\cap R_v\subset R_u,R_v\subset R\big)$

for inequivalent pair of $u\mbox{ and }v$;

[-latex] (block1) edge (block2)
(block2) edge (block3)
(block3) edge (block40)
(block40) edge (block4)
(block4) edge (block5)
(block5) edge (block60)
(block60) edge (block61)
(block61) edge (block62)
(block62) edge (block63)
(block6) edge (block7)
(block61) edge (block8);

[-latex] (block4) -| (block6)
node[pos=0.5,fill=white,inner sep=0];

Flowchart for our plan of actions in <Ref>

§.§ Computation of the Pimsner-Popa constant

In this subsection, we compute the Pimsner-Popa probabilistic constant $\lambda(R_u,R_v)$. We introduce the following matrices
1     0     1     0

0     1     0     1

1     0     1     0

0     1     0     1

]  ,  q=1/2[
1     0     -1     0

0     ab     0     -ab

-1     0     1     0

0     -ab     0     ab

which will play pivotal role throughout the section. Observe that $p$ is a projection (but $q$ is not) and $pq=qp=0$. Also note that $u^*v=p+q$.

One has the following,

* $\big(W_j^{(a)}\big)^*p\,W_j^{(b)}=q^*\mbox{ and } \big(W_j^{(a)}\big)^*q\,W_j^{(b)}=p~\mbox{ for }j=1,3\,;$

* $W_2\,p\,W_2=p$ and $W_2\,q\,W_2=q$;

* $\xi_a^*(E_{11}+E_{33})\xi_b=p\,\mbox{ and } \,\xi_a^*(E_{22}+E_{44})\xi_b=q\,.$

Straightforward verification using <Ref>, and hence we leave it to the reader.

For any $k\in\bbn$, we have the following identity

$u_{2k}^*v_{2k}=p\otimes u_{2k-2}^*v_{2k-2}+q\otimes v_{2k-2}^*u_{2k-2}$

in $M_4\otimes M_4^{(k)}$.

We use induction on $k$. For $k=1$, it is a straightforward verification using Theorem <ref>. Assume that the result holds up to $k-1$. Then, using part (i) of Lemma <ref> we get the following for $j=1,3$,

=    (W_j^(a))^*pW_j^(b)⊗u_2k-4^*v_2k-4+(W_j^(a))^*qW_j^(b)⊗v_2k-4^*u_2k-4

=    q^*⊗u_2k-4^*v_2k-4+p⊗v_2k-4^*u_2k-4

=    v_2k-2^*u_2k-2

Moreover, by part $(ii)$ of <Ref>, we see that
(W_2⊗I_4^(k-1))^*u_2k-2^*v_2k-2(W_2⊗I_4^(k-1))=u_2k-2^*v_2k-2 .

Now, using part $(iii)$ of Lemma <ref>, along with Theorem <ref>, we finally get the following,

=    (ξ_a^*⊗I_4^(k)) {u_2k-2^*v_2k-2,v_2k-2^*u_2k-2,u_2k-2^*v_2k-2,v_2k-2^*u_2k-2} (ξ_b⊗I_4^(k))

=    (ξ_a^*⊗I_4^(k)) ({p⊗u_2k-4^*v_2k-4,q^*⊗u_2k-4^*v_2k-4,p⊗u_2k-4^*v_2k-4,q^*⊗u_2k-4^*v_2k-4}

=    (ξ_a^*⊗I_4⊗I_4^(k-1))((E_11+E_33)⊗p⊗u_2k-4^*v_2k-4+(E_22+E_44)⊗q^*⊗u_2k-4^*v_2k-4

=    p⊗p⊗u_2k-4^*v_2k-4+q⊗q^*⊗u_2k-4^*v_2k-4+p⊗q⊗v_2k-4^*u_2k-4+q⊗p⊗v_2k-4^*u_2k-4

=    p⊗u_2k-2^*v_2k-2+q⊗v_2k-2^*u_2k-2 ,

where the last equality follows from the induction hypothesis.

For any $k\geq 0$, the unitary $u_{2k+1}^*v_{2k+1}$ is of the following form

\[
\mbox{bl-diag}\left\{u_{2k}^*v_{2k}\,,\,v_{2k}^*u_{2k}\,,\,u_{2k}^*v_{2k}\,,\,v_{2k}^*u_{2k}\right\}\,.
\]

We have the following identities

$\big(W_1^{(a)}\otimes I_4^{(k)}\big)^*u_{2k}^*v_{2k}\big(W_1^{(b)}\otimes I_4^{(k)}\big)=\big(W_3^{(a)}\otimes I_4^{(k)}\big)^*u_{2k}^*v_{2k}\big(W_3^{(b)}\otimes I_4^{(k)}\big)=v_{2k}^*u_{2k}\,,$

$\big(W_2\otimes I_4^{(k)}\big)^*u_{2k}^*v_{2k}\big(W_2\otimes I_4^{(k)}\big)=u_{2k}^*v_{2k}\,,$

due to <Ref>. By Theorem <ref>, we have $u_{2k+1}=\mbox{bl-diag}\{u_{2k},u_{2k}\big(W_1^{(a)}\otimes I_4^{(k)}\big),u_{2k}\big(W_2\otimes I_4^{(k)}\big),u_{2k}\big(W_3^{(a)}\otimes I_4^{(k)}\big)\}$, and hence the result follows.

Let $\,U_1,U_2\in M_n$ be two unitary matrices and consider the unital inclusions $M_n\subseteq \Delta_4\otimes M_n$ and $UM_nU^*\subseteq\Delta_4\otimes M_n$, where $U\in\Delta_4\otimes M_n$ is the block diagonal unitary matrix $\mbox{\em{bl-diag}}\{U_1,U_2,U_1,U_2\}$. Then, the Pimsner-Popa constant satisfies the following,

\[
\lambda^{\Delta_4\otimes M_n}\left(M_n,UM_nU^*\right)=\lambda^{\Delta_2\otimes M_n}\left(M_n,VM_nV^*\right)\,,
\]

where $V=\mbox{\em{bl-diag}}\{U_1,U_2\}$ is a block diagonal unitary matrix in $\Delta_2\otimes M_n$.

For all $x\in\left(M_n\right)_+\subseteq\left(\Delta_4\otimes M_n\right)_+$, observe that $E^{\Delta_4\otimes M_n}_{UM_nU^*}(x)\geq tx$ for $t\in[0,1]$ if and only if both the following inequalities

\[
\frac{1}{2}\left(U_1^*xU_1+U_2^*xU_2\right)\geq tU_1^*xU_1
\]

\[
\frac{1}{2}\left(U_1^*xU_1+U_2^*xU_2\right)\geq tU_2^*xU_2
\]

hold simultaneously. Now, observe that for all $y\in\left(M_n\right)_+\subseteq\left(\Delta_2\otimes M_n\right)_+$, the inequality $E^{\Delta_2\otimes M_n}_{VM_nV^*}(y)\geq ty$ for $t\in[0,1]$ holds if and only if the following inequality

\[
\frac{1}{2}\left(U_1^*yU_1+U_2^*yU_2\right)\geq t\,\mbox{bl-diag}\{U_1^*yU_1,U_2^*yU_2\}
\]

holds, which completes the proof.

We are going to invoke Theorem <ref> to finally determine the value of the Pimsner-Popa constant for the $k$-th step of the tower of basic construction. For that we need to determine whether the matrix $\big(v_{2k}^*u_{2k}\big)^2$ in $M_4\otimes M_4^{(k)}$ is not diagonal or scalar. Recall that $u\mbox{ and }v$ are Hadamard inequivalent matrices. That is, if we write $a=e^{i\alpha}$ and $b=e^{i\beta}$, then both $\alpha,\beta\in[0,\pi)$. Since $\alpha-\beta\in(-\pi,\pi)$, we have $b\neq -a$.

For any $k\geq 0$, the matrix $\big(v_{2k}^*u_{2k}\big)^2$ is not diagonal in $M_4\otimes M_4^{(k)}$.

First take $k=0$. We have $v^*u=p+q^*$, and hence $\left(v^*u\right)^2=p+(q^*)^2$ which is not a diagonal matrix as $b\neq -a$. Now for $k\geq 1$, using Proposition <ref> we have the following,
(v_2k^*u_2k)^2    =    (p⊗v_2k-2^*u_2k-2+q^*⊗u_2k-2^*v_2k-2)^2

=    p⊗(v_2k-2^*u_2k-2)^2+(q^*)^2⊗(u_2k-2^*v_2k-2)^2 .

From here, we see that if $\big(v_{2k-2}^*u_{2k-2}\big)^2$ is not self-adjoint then we are done, and even if $\big(v_{2k-2}^*u_{2k-2}\big)^2$ is self-adjoint, then the fact that $(a\overline{b})^2\neq 1$ (otherwise we get $b=-a$) proves the claim.

The Pimsner-Popa constant for the pair of subfactors $R_u\mbox{ and }R_v$ of the hyperfinite $II_1$ factor $R$ is given by $\lambda(R_u,R_v)=1/2$.

Combining <Ref>, <ref>, <ref>, together with <Ref>, we get the following
λ(u_2k+1M_4^(k+1)u_2k+1^* , v_2k+1M_4^(k+1)v_2k+1^*)=1/2

for any $k\geq 0$. This immediately gives us that $\lambda(R_u,R_v)=\frac{1}{2}$, because by <Ref>, we have $\lambda(R_u,R_v)$ is the limit of a decreasing sequence of $\lambda$ at each step of the tower of basic construction, and we have produced a constant subsequence in <Ref>.

§.§ Construction of a family of subfactors of hyperfinite $II_1$ factor $R$

This subsection and its intimately connected sequel contains one of the most important result of the paper. We will construct a family of potentially new integer index subfactors of $R$ in this subsection, and latter in the sequel we show irreducibility of these family of subfactors. For the sake of reader's convenience, we draw a technical roadmap for this subsection and its immediate sequel in <Ref>.

rounded rectangle,
minimum width=2.5cm,
minimum height=1cm] (block0)  Start with $R_u\neq R_v\subset R$ along with

their f.d. grids $B^u_{2k+1},B^v_{2k+1},A_{2k+1}$ ;

below=of block0,
minimum width=2.5cm,
minimum height=1cm] (block1)  Obtain the spectral

decomposition of

$(u_{2k}^*v_{2k})^2$ for all $k\in\bbn\,$;

trapezium left angle = 65,
trapezium right angle = 115,
trapezium stretches,
below left=of block1,
minimum width=3.5cm,
minimum height=1cm] (block2) $\,\theta^{\,2}$ is a root of unity for

$\theta:=\overline{a}b$ and let $\theta^{\,m}=1$ ;

below right=of block1,
minimum width=3.5cm,
minimum height=1cm] (block3) $\,\theta^{\,2}$ is not a root of

unity for $\theta:=\overline{a}b$ ;

below=of block3,
minimum width=3.5cm,
minimum height=1cm] (block4) $\,\texttt{\#}\sigma\big((u_{2k}^*v_{2k})^2\big)=k+2,\,$

increases with $k\,$;

below=of block4,
minimum width=3.5cm,
minimum height=1cm] (block50)  Identify the f.d. grid

$C_{2r+1}\,$ for $R_u\cap R_v\,$;

rounded rectangle,
below=of block50,
minimum width=3.5cm,
minimum height=1cm] (block5)  $R_u\cap R_v$ has infinite

Pimsner-Popa index in $R\,$;

below=of block2,
minimum width=3.5cm,
minimum height=1cm] (block20) $\,\sigma\big((u_{2k}^*v_{2k})^2\big)$ stabilizes beyond

$k=(m-2)/2$, and

for all $j\geq 0$ ;

below=of block20,
minimum width=3.5cm,
minimum height=1cm] (block60)  Identify the f.d. grid

$C_{2r+1}$ for $R_u\cap R_v\,$;

below=of block60,
minimum width=3.5cm,
minimum height=1cm] (block30)  The commuting square

$\big(C_{2k-1}\subset A_{2k-1},C_{2k+1}\subset A_{2k+1}\big)\,$

becomes non-degenerate;

below=of block30,
minimum width=3.5cm,
minimum height=1cm] (block7)  Construction of a family of

subfactors $R^{(m)}_{u,v}\subset R$ with

index $2m$, if $\theta^{\,m}=1$ ;

right=of block7,
minimum width=3.5cm,
minimum height=1cm] (block8)  Irreducibility of $R^{(m)}_{u,v}\subset R\,$;

rounded rectangle,
below=of block8,
minimum width=3.5cm,
minimum height=1cm] (block9)  Factoriality and finite

index for $R_u\cap R_v\subset R\,$;

[-latex] (block0) edge (block1)
(block1) edge (block2)
(block1) edge (block3)
(block3) edge (block4)
(block4) edge (block50)
(block50) edge (block5)
(block2) edge (block20)
(block20) edge (block60)
(block60) edge (block30)
(block30) edge (block7)
(block7) edge (block8)
(block8) edge (block9);

Technical roadmap for subsections $7.2$ and $7.3$

To fulfil our goal, we need to focus on the von Neumann subalgebra $R_u\cap R_v$ of the hyperfinite type $II_1$ factor $R$. To understand the structure of $R_u\cap R_v$, since we already have the finite-dimensional grids for the spin model subfactors $R_u\mbox{ and }R_v$, our target is to obtain the same for $R_u\cap R_v$. It turns out that in doing so a surprising phenomenon occurs. A family of finite index potentially new subfactors of $R$ arises. Moreover, with the help of it, we can also conclude factoriality of $R_u\cap R_v$ along with its index in $R$ in certain situations. This also helps to conclude infinite Pimsner-Popa index of $R_u\cap R_v$ in certain other situations. We begin by fixing some notations to be used throughout this subsection and latter.

Notation:   (i) For complex Hadamard matrices $u\mbox{ and }v$ parametrized by $a\mbox{ and }b$, let $\theta=\overline{a}b\in\mathbb{S}^1$. We replace the matrix $q$ introduced in <Ref> by the notation $q(\theta)$ to stress its dependence on $\theta$. This is required in what follows next.

(ii) Consider the following matrices
1     0     1     0

0     1     0     1

1     0     -1     0

0     1     0     -1

]        C(θ)=[
1     0

0     θ

] .

Note that $W$ is a self-adjoint unitary matrix in $M_4$.

(iii) For a matrix $T$, $\sigma(T)$ will denote its spectrum.

For any $k\in\bbn$, we have the following,

* $\mathrm{Ad}_u(\Delta_4)\cap\mathrm{Ad}_v(\Delta_4)=\mathrm{Ad}_u(\bbc^3)$, where the emebedding $\bbc^3\xhookrightarrow{}\Delta_4\subseteq M_4$ is given by $(\lambda_1,\lambda_2,\lambda_3)\mapsto(\lambda_1,\lambda_2,\lambda_3,\lambda_2)\,,$

* $u_{2k}^*v_{2k}\big(\Delta_4\otimes M_4^{(k)}\big)v_{2k}^*u_{2k}\subseteq\left(u^*v\Delta_4v^*u\right)\otimes M_4^{(k)}\,$,

* $\mathrm{Ad}_{u_{2k}}\big(\Delta_4\otimes M_4^{(k)}\big)\cap\mathrm{Ad}_{v_{2k}}\big(\Delta_4\otimes M_4^{(k)}\big)\subseteq\mathrm{Ad}_{u_{2k}}\big(\bbc^3\otimes M_4^{(k)}\big)\subseteq M_4\otimes M_4^{(k)}$.

We have $\mathrm{Ad}_u(\Delta_4)\cap\mathrm{Ad}_v(\Delta_4)=\mathrm{Ad}_u(\Delta_4\cap\mathrm{Ad}_{u^*v}(\Delta_4))$. Recall that $u^*v=p+q(\theta)$, where $p\mbox{ and }q(\theta)$ are as in <Ref>. Hence, for any $\mbox{bl-diag}\{\mu_1,\mu_2,\mu_3,\mu_4\}\in\Delta_4$, we have the following
μ_1     0     0     0

0     μ_2(2+θ+θ)/4+ μ_4(2-θ-θ)/4     0      μ_2(θ-θ)/4+ μ_4(θ-θ)/4

0     0     μ_3     0

0     μ_2(θ-θ)/4+ μ_4(θ-θ)/4     0     μ_2(2-θ-θ)/4+ μ_4(2+θ+θ)/4

Observe that $\theta\neq\overline{\theta}$. This is because $\theta=\overline{\theta}$ implies that $b=\pm a$, since $\theta=\overline{a}b$. However, in our situation this never happens as already discussed at the beginning of <Ref>. Therefore, the intersection $\Delta_4\cap\mathrm{Ad}_{u^*v}(\Delta_4)$ consists of the diagonal matrices $\mbox{bl-diag}\{\mu_1,\mu_2,\mu_3,\mu_4\}$ such that $\mu_1,\mu_3$ are arbitrary complex numbers, and $\mu_2,\mu_4$ are equal. In other words,

\[
\Delta_4\cap\mathrm{Ad}_{u^*v}(\Delta_4)=\big\{\mbox{bl-diag}\{\mu_1,\mu_2,\mu_3,\mu_2\}\,:\,\mu_1,\mu_2,\mu_3\in\bbc\big\}\,,
\]

which completes part $(i)$. For part $(ii)$, by <Ref> we get the following,

=    (p⊗u_2k-2^*v_2k-2+q(θ)⊗v_2k-2^*u_2k-2)(Δ_4⊗M_4^(k))(p⊗v_2k-2^*u_2k-2+q(θ)^*⊗u_2k-2^*v_2k-2)

⊆    (pΔ_4 p+pΔ_4 q(θ)^*+q(θ)Δ_4 p+qΔ_4 q(θ)^*)⊗M_4^(k)

=    ((p+q(θ))Δ_4(p+q(θ)^*))⊗M_4^(k) .

Since $p+q(\theta)=u^*v\,$, and consequently $p+q(\theta)^*=v^*u\,$, part $(ii)$ is done. Part $(iii)$ follows from part $(i)$ and $(ii)$ together.

For any $k\geq 0$, the intersection

\[
C_{2k+1}=\mathrm{Ad}_{u_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)\cap\mathrm{Ad}_{v_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)
\]

in $A_{2k+1}=\Delta_4\otimes M_4^{(k+1)}$ is given by the following,

\[
\left\{\mathrm{Ad}_{u_{2k+1}}\big(I_4\otimes A\big):A\in\big((u_{2k}^*v_{2k})^2\big)^\prime\cap M_4^{(k+1)}\right\}\,.
\]

First observe the following,

=    Ad_u_2k+1(⊗M_4^(k+1)∩Ad_u_2k+1^*v_2k+1(⊗M_4^(k+1)) .

Recall from <Ref> that $u_{2k+1}^*v_{2k+1}=\mbox{bl-diag}\left\{u_{2k}^*v_{2k},v_{2k}^*u_{2k},u_{2k}^*v_{2k},v_{2k}^*u_{2k}\right\}$. An element $\xi$ is in $\bbc\otimes M_4^{(k+1)}\cap \mbox{Ad}_{u_{2k+1}^*v_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)$ if and only if

\[
\xi=I_4\otimes A=\mbox{Ad}_{u_{2k+1}^*v_{2k+1}}(I_4\otimes B)
\]

for some $A,B\in M_4^{(k+1)}$. This implies that $A=\mbox{Ad}_{u_{2k}^*v_{2k}}(B)=\mbox{Ad}_{v_{2k}^*u_{2k}}(B)$, which means $A$ remains invariant under the conjugation action by the unitary $\big(u_{2k}^*v_{2k}\big)^2\in M_4^{(k+1)}$, and this completes the proof.

Both the inclusions in part $(ii)$ and $(iii)$ of <Ref> is strict for $k>0$. To obtain $\mathrm{Ad}_{u_{2k}}\big(\Delta_4\otimes M_4^{(k)}\big)\cap\mathrm{Ad}_{v_{2k}}\big(\Delta_4\otimes M_4^{(k)}\big)$, one has to further identify the intersection $C_{2k+1}\cap\big(\bbc\otimes M_4^{(k+1)}\big)$ concretely, as $(C_{2k}\subset A_{2k}\,,\,C_{2k+1}\subset A_{2k+1})$ forms a commuting square, where $C_{2k}$ and $A_{2k}$ are in the even steps of the tower of basic constructions in <Ref>. Thus, the intersections in the even steps of the tower are little harder to identify as compared to that in the odd steps obtained in <Ref>. This is where we differ from <Ref>.

A scheme to identify the finite-dimensional grid for $R_u\cap R_v$:  In order to concretely identify the finite-dimensional grid $\{C_{2k+1}:k\geq 0\}$ of the von Neumann subalgebra $R_u\cap R_v$ of $R$, it is now clear from <Ref> that we need to find the Wedderburn-Artin decomposition of the algebra $\big((u_{2k}^*v_{2k})^2\big)^\prime\cap M_4^{(k+1)}$ for each $k$. The following lemma is crucial in this regard, and although it is an elementary fact from Linear algebra, for the sake of completeness we provide a short proof.

Suppose that we have a block-diagonal matrix $A=[A_{i,j}]=A_{1,1}\oplus\cdots\oplus A_{n,n}$ such that each block $A_{i,i},\,1\leq i\leq n,$ has exactly one eigenvalue $\lambda_i$ and $\lambda_i\neq\lambda_j$ for $i\neq j$. Then, the only matrices which commute with $A$ are block-diagonal with the same block sizes as $A$.

Suppose that $A$ is a $k\times k$ matrix and consider a matrix $B$ in $M_k$. Putting $B$ in block form $(B_{i,j})$, let $A_iB_{i,j}=B_{i,j}A_j$ for all $i,j$. We claim that $i\neq j$ implies $B_{i,j}=0$. Note that $A_iB_{i,j}=B_{i,j}A_j$ implies $\mathcal{P}(A_i)B_{i,j}=B_{i,j}\mathcal{P}(A_j)$ for any polynomial $\mathcal{P}$. There exists $r_j\in\bbn$ such that $(A_j-\lambda_j)^{r_j}=0$, and hence $(A_i-\lambda_j)^{r_j}B_{i,j}=0$. Since $A_i-\lambda_j$ is invertible for $i\neq j$ by hypothesis, we get that $B_{i,j}=0$ for $i\neq j$.

We describe the scheme that we are now going to follow. Suppose that we have a normal matrix $x\in M_n$. Let $\{\lambda_1,\ldots,\lambda_r\},\,1\leq r\leq n,$ be the distinct eigenvalues of $x$ with geometric multiplicities $m(\lambda_i),\,1\leq i\leq r$. By the spectral theorem, $n=\sum_{i=1}^rm(\lambda_i)$. That is, there exists a unitary matrix $U\in M_n$ such that $x=U\mbox{diag}\{\lambda_11_{m(\lambda_1)},\ldots,\lambda_r1_{m(\lambda_r)}\}U^*$. By <Ref>, we have the Wedderburn-Artin decomposition of the algebra $\{x\}^\prime\cap M_n$ as $\oplus_{j=1}^rM_{m(\lambda_j)}$ with $n=\sum_{j=1}^rm(\lambda_j)$.

In view of the above, it is now clear that our first job is to find the spectral decomposition of the unitary matrices $(u_{2k}^*v_{2k})^2\in M_4^{(k+1)}$ for each $k\in\bbn\cup\{0\}$, and the next subsection deals with precisely this.

§.§.§ Spectral decomposition of $\big(u_{2k}^*v_{2k}\big)^2$ for $k\in\bbn\cup\{0\}$

We obtain the spectral decomposition for the unitary matrices $\big(u_{2k}^*v_{2k}\big)^2\in M_4^{(k+1)}$ for all $k\in\bbn\cup\{0\}$. Due to the analysis carried out here, we will be able to identify the intersections $C_{2k+1}$ concretely for all $k$ as discussed above. In this regard, recall the tower of basic construction obtained in <Ref>, the matrices $W$ and $C(\theta)$ from <Ref>, and finally the matrices $p,q(\theta)$ from <Ref>.

* For any $n\in\bbn,\,C(\theta)^n=C(\theta^n)$ and $C(\theta)^*=C(\overline{\theta})$.

* We have $WpW=\mathrm{diag}\{I_2,0\}$ and $Wq(\theta)W=\mathrm{diag}\{0,C(\theta)\}$.

* For any $n\in\bbn,\,q(\theta)^n=q(\theta^{\,n})$.

These are straightforward verification, and hence left to the reader.

For any $k\in\bbn$, we have the following,

\[
\sigma\big(\big(u_{2k}^*v_{2k}\big)^2\big)=\sigma\big(\big(u_{2k-2}^*v_{2k-2}\big)^2\big)\cup\sigma\big(\big(v_{2k-2}^*u_{2k-2}\big)^2\big)\cup\sigma\big(\big(\theta\,v_{2k-2}^*u_{2k-2}\big)^2\big).
\]

By <Ref>, we have the identity $u_{2k}^*v_{2k}=p\otimes u_{2k-2}^*v_{2k-2}+q(\theta)\otimes v_{2k-2}^*u_{2k-2}$ in $M_4\otimes M_4^{(k)}$. Since $pq(\theta)=q(\theta)p=0$ and $p$ is a projection, we get that

\[
\big(u_{2k}^*v_{2k}\big)^2=p\otimes \big(u_{2k-2}^*v_{2k-2}\big)^2+q(\theta^2)\otimes\big(v_{2k-2}^*u_{2k-2}\big)^2
\]

in $M_4\otimes M_4^{(k)}$ by part $(iii)$ of <Ref>. Using part $(i),(ii)$ of <Ref>, we have the following identity in $M_4\otimes M_4^{(k)}$,

=    E_11⊗I_2⊗_W⊗W^(k-1)((u_2k-2^*v_2k-2)^2)+E_22⊗C(θ^ 2)⊗_W⊗W^(k-1)((v_2k-2^*u_2k-2)^2).
This immediately says the following,

=    σ(_W^(k)((u_2k-2^*v_2k-2)^2))∪σ(C(θ^ 2)⊗_W^(k)((v_2k-2^*u_2k-2)^2)) ,

which concludes the proof since $W$ is a unitary matrix in $M_4$ and $C(\theta^{\,2})$ is a diagonal matrix in $M_2$.

For any $k\in\bbn\cup\{0\}$, the set of all eigenvalues (including repeated ones possibly) of the unitary matrix $(u_{2k}^*v_{2k})^2$ is $\big\{1,\theta^{\,2},\overline{\theta}^{\,2},\theta^{\,4},\overline{\theta}^{\,4},\ldots,\theta^{\,k},\overline{\theta}^{\,k},\theta^{\,k+2}\big\}$ if $k$ is even, and $\big\{1,\theta^{\,2},\overline{\theta}^{\,2},\theta^{\,4},\overline{\theta}^{\,4},\ldots,\theta^{\,k-1},\overline{\theta}^{\,k-1},\theta^{\,k+1},\overline{\theta}^{\,k+1}\big\}$ if $k$ is odd.

We use induction on $k$. First suppose that $k$ is even. The basis step of $k=0$ follows from the following fact (using <Ref>)

\[
\mbox{Ad}_W\big((u^*v)^2\big)=\mbox{Ad}_W\big((p+q(\theta))^2\big)=\mbox{Ad}_W\big(p+q(\theta^{\,2})\big)=\mbox{bl-diag}\{I_2,C(\theta^{\,2})\}\,,
\]

and hence $\sigma\big((u^*v)^2\big)=\{1,\theta^{\,2}\}$. Assume that the statement holds for $k-2$. That is, we have $\sigma\big(\big(u_{2k-4}^*v_{2k-4}\big)^2\big)=\{1,\theta^{\,2},\overline{\theta}^{\,2},\theta^{\,4},\overline{\theta}^{\,4},\ldots,\theta^{\,k-2},\overline{\theta}^{\,k-2},\theta^{\,k}\}$. Hence, $\sigma\big(\big(v_{2k-4}^*u_{2k-4}\big)^2\big)=\{1,\theta^{\,2},\overline{\theta}^{\,2},\theta^{\,4},\overline{\theta}^{\,4},\ldots,\theta^{\,k-2},\overline{\theta}^{\,k-2},\overline{\theta}^{\,k}\}$. Now, observe that when each element of this set is multiplied by $\theta^{\,2}$, we get back the set $\sigma\big(\big(u_{2k-4}^*v_{2k-4}\big)^2\big)$. That is, we have $\,\sigma\big(\big(\theta\,v_{2k-4}^*u_{2k-4}\big)^2\big)=\sigma\big(\big(u_{2k-4}^*v_{2k-4}\big)^2\big)$. By <Ref> we now have the following,
σ((u_2k-2^*v_2k-2)^2)    =    σ((u_2k-4^*v_2k-4)^2)∪σ((v_2k-4^*u_2k-4)^2)∪σ((θ v_2k-4^*u_2k-4)^2)

=    {1,θ^ 2,θ^ 2,θ^ 4,θ^ 4,…,θ^ k-2,θ^ k-2,θ^ k}∪{1,θ^ 2,θ^ 2,θ^ 4,θ^ 4,…,θ^ k-2,θ^ k-2,θ^ k}

=    {1,θ^ 2,θ^ 2,θ^ 4,θ^ 4,…,θ^ k,θ^ k},

which proves the statement for $k-1$, which is odd. Similar application of <Ref> shows that

\[
\sigma\big(\big(u_{2k}^*v_{2k}\big)^2\big)=\{1,\theta^{\,2},\overline{\theta}^{\,2},\theta^{\,4},\overline{\theta}^{\,4},\ldots,\theta^{\,k},\overline{\theta}^{\,k},\theta^{\,k+2}\},
\]

which proves the statement for $k$, and finishes the induction. Similarly, starting with $k$ odd and using the following fact that (use <Ref> and <Ref>)

\[
\mbox{Ad}_{W\otimes W}\big((u_2^*v_2)^2\big)=WpW\otimes\mbox{Ad}_W(u^*v)^2+Wq(\theta^{\,2})W\otimes\mbox{Ad}_W(v^*u)^2\,,
\]

we can use induction on $k$ to conclude the proof.

Note that we are not claiming in the above lemma that $\sigma\big((u_{2k}^*v_{2k})^2\big)$ consists of distinct eigenvalues. In fact, we are going to investigate this now.

Terminology : For a complex number $z$ of modulus $1$, we will say that $z$ is an even root of unity if $z^n=1$ for some even integer $n$. Note that this is same as saying $z^2$ is a root of unity. Throughout the article, by even root of unity we always mean primitive even root of unity. That is, if $z^n=1$ for some $n\in 2\bbn$, then $z^m\neq 1$ for any $m\in 2\bbn$ with $m<n$.

We now investigate whether $\sigma\big((u_{2k}^*v_{2k})^2\big)$ obtained in <Ref> consists of distinct eigenvalues only, and it turns out that it depends on $\,\theta\in\mathbb{S}^1$. If $\,\theta$ is not an even root of unity, then it is easy to observe that all the elements of the following sets

\[
\big\{1,\theta^{\,2},\overline{\theta}^{\,2},\theta^{\,4},\overline{\theta}^{\,4},\ldots,\theta^{\,k},\overline{\theta}^{\,k},\theta^{\,k+2}\big\}\,,\,\mbox{when $k$ is even}
\]

\[
\big\{1,\theta^{\,2},\overline{\theta}^{\,2},\theta^{\,4},\overline{\theta}^{\,4},\ldots,\theta^{\,k-1},\overline{\theta}^{\,k-1},\theta^{\,k+1},\overline{\theta}^{\,k+1}\big\}\,,\,\mbox{when $k$ is odd}
\]

are distinct and we have cardinality of $\sigma\big((u_{2k}^*v_{2k}\big)^2\big)$ is $k+2$. Therefore, when $k$ increases, number of elements in the spectrum of the unitary matrices $\big((u_{2k}^*v_{2k})^2\big)$ also increase, and $\sigma\big((u_{2k}^*v_{2k}\big)^2\big)\subsetneq\sigma\big((u_{2k+2}^*v_{2k+2}\big)^2\big)$. However, as soon as $\,\theta$ is an even root of unity, that is $\,\theta^{\,m}=1$ for some $m\in2\bbn+2$ (and $\theta^{\,r}\neq 1$ for any $r<m$), the situation changes drastically. We notice that for $k=(m-2)/2$, the last eigenvalue in the enumeration described in <Ref> coincides with the second last eigenvalue, and we get that $\sigma\big((u_{2k}^*v_{2k}\big)^2\big)=\sigma\big((u_{2k-2}^*v_{2k-2}\big)^2\big)$ irrespective of whether $k$ is even or odd. Note that we are not considering the case $\theta^{\,2}=1$ (i,e., $m=2$) because otherwise we would get $b=\pm a$, which is not possible in our situation as already discussed at the beginning of <Ref>. These discussions culminate into the following results.

For any $k\in\bbn\cup\{0\}$, if $\,\theta$ is not an even root of unity, then we have
σ((u_2k^*v_2k)^2)    =
{1,θ^ 2,θ^ 2,θ^ 4,θ^ 4,…,θ^ k,θ^ k,θ^ k+2}       k , {1,θ^ 2,θ^ 2,θ^ 4,θ^ 4,…,θ^ k-1,θ^ k-1,θ^ k+1,θ^ k+1}       k ,

and $\texttt{\#}\,\sigma\big(\big(u_{2k}^*v_{2k}\big)^2\big)= k+2$ for all $k\in\bbn\cup\{0\}$.

Let $\theta^{\,m}=1$ for some $m\in 2\bbn+2$ and choose $k=(m-2)/2$. Then, we have

\[
\sigma\big((u_{2k-2}^*v_{2k-2})^2\big)=\sigma\big((u_{2k+2j}^*v_{2k+2j})^2\big)=\begin{cases}
\{1,\theta^{\,2},\overline{\theta}^{\,2},\ldots,\theta^{\,k-1},\overline{\theta}^{\,k-1},\theta^{\,k+1}\}& \mbox{ if }\,\,k\mbox{ is odd}\,,\cr
\{1,\theta^{\,2},\overline{\theta}^{\,2},\ldots,\theta^{\,k-2},\overline{\theta}^{\,k-2},\theta^{\,k},\overline{\theta}^{\,k}\}& \mbox{ if }\,\,k\mbox{ is even},
\end{cases}
\]

for all $j\in\bbn\cup\{0\}$, and $\texttt{\#}\,\sigma\big((u_{2k+2j}^*v_{2k+2j})^2\big)=k+1$.

If $\theta^{\,m}=1$ for some $m\in 2\bbn+2$, then choose $k=(m-2)/2$. If $k$ is odd, then by <Ref> we have $\theta^{\,k+1}=\overline{\theta}^{\,k+1}$ and $\texttt{\#}\sigma\big((u_{2k}^*v_{2k})^2\big)=k+1$. If $k$ is even, then again by <Ref> we have $\theta^{\,k+2}=\overline{\theta}^{\,k}$ and $\texttt{\#}\sigma\big((u_{2k}^*v_{2k})^2\big)=k+1$. It now follows by induction on $j$ that $\sigma\big((u_{2k-2}^*v_{2k-2})^2\big)=\sigma\big((u_{2k}^*v_{2k})^2\big)=\sigma\big((u_{2(k+j)}^*v_{2(k+j)})^2\big)$ for all $j\in\bbn$ with cardinality of the spectrum being equal to $k+1$.

Our first job of finding the distinct eigenvalues for each element in the family of unitary matrices $\{(u_{2k}^*v_{2k})^2\,:\,k\in\bbn\}$ is completed by <Ref> and <ref>. Our next job is to find the multiplicities of the eigenvalues. Throughout the rest of the article, we fix the above enumeration of the eigenvalues of $\big(u_{2k}^*v_{2k}\big)^2$, that is, the eigenvalue $1$ is followed by $\theta^{\,2}$, then $\overline{\theta}^{\,2}$, and so on. In other words, we always arrange the eigenvalues according to increasing power of $\theta$ immediately followed by $\overline{\theta}$. The reason for this will be clear in the sequel.

We denote the multiplicity of the eigenvalue $1$ by $m_k(1)$; and that of $\theta^{\,2r}\mbox{ and }\overline{\theta}^{\,2r}$ by $m_k(\theta^{\,2r})\mbox{ and }m_k(\overline{\theta}^{\,2r})$ respectively for $r=1,2,\ldots, \frac{\lfloor k+1\rfloor}{2}$. First we shall deal with the case of $\,\theta$ not being an even root of unity.

Let $\,\theta$ be not an even root of unity. The multiplicities of the eigenvalues $1\mbox{ and }\theta^{\,2}$ of $(u^*v)^2$ are respectively given by $m_0(1)=3$ and $m_0(\theta^{\,2})=1$, and that of $(u_{2k}^*v_{2k})^2$ for $k\in\bbn$ are given by the following chain of recurrence relations :

* For $k\in\bbn$ odd,
m_1(1)=10, m_1(θ^ 2)=5, m_1(θ^ 2)=1,

m_k(1)=3m_k-1(1)+m_k-1(θ^ 2),

m_k(θ^ 2r)=2m_k-1(θ^ 2r)+(1-δ_r,k+1/2)m_k-1(θ^ 2r)+m_k-1(θ^ 2r-2)   1≤r≤k+1/2,

m_k(θ^ 2r)=2(1-δ_r,k+1/2)m_k-1(θ^ 2r)+m_k-1(θ^ 2r)+(1-δ_r,k+1/2)m_k-1(θ^ 2r+2)  1≤r≤k+1/2.

* For $k\in\bbn$ even,
m_2(1)=35, m_2(θ^ 2)=21, m_2(θ^ 2)=7, m_2(θ^ 4)=1,

m_k(1)=3m_k-1(1)+m_k-1(θ^ 2),

m_k(θ^ 2r)=2m_k-1(θ^ 2r)+m_k-1(θ^ 2r)+m_k-1(θ^ 2r-2)   1≤r≤k/2,

m_k(θ^ 2r)=2m_k-1(θ^ 2r)+m_k-1(θ^ 2r)+(1-δ_r,k/2)m_k-1(θ^ 2r+2)   1≤r≤k/2,

m_k(θ^ k+2)=m_k-1(θ^ k) .

Here, $\delta_{i,j}$ denotes the Kronecker delta.

The case of $k=0$ follows immediately from <Ref>, and we see that $m_0(1)=3$ and $m_0(\theta^{\,2})=1$ because $(u^*v)^2=p+q(\theta^{\,2})$ (recall $p\mbox{ and }q(\theta)$ from <Ref>). Now, recall from <Ref> the following (also, recall the fact that $C(\theta^{\,2})=\mbox{diag}\{1,\theta^{\,2}\}\in M_2$),

=    [ 1      ;   1    ;     0  ;       0;   ]⊗_W⊗W^(k-1)((u_2k-2^*v_2k-2)^2)+[    0               ;         0          ;              1     ;                θ^ 2;      ]⊗_W⊗W^(k-1)((v_2k-2^*u_2k-2)^2).

The case of $k\in\bbn$ now easily follows from the above equation by induction on $k$ in the same spirit of <Ref>.

Thus, once we know the multiplicities of the eigenvalues $1,\theta^{\,2},\overline{\theta}^{\,2}$ when $k=1$, we know that for all $k\in\bbn$. The reason for enumerating the eigenvalues of $\big(u_{2k}^*v_{2k}\big)^2$ according to increasing power of $\theta$ immediately followed by $\overline{\theta}$ shall be clear now from the following result.

Let $\,\theta$ be not an even root of unity. The multiplicities $m_k(.)$ of the eigenvalues of $\big(u_{2k}^*v_{2k}\big)^2$ satisfy the ordering $m_k(1)>m_k(\theta^{\,2})>m_k(\overline{\theta}^{\,2})>\ldots$ for any $k\in\bbn\cup\{0\}$.

We prove by induction on $k$. The basis step of the induction is clear from <Ref>. Assume that the statement holds for the $k$-th step and $k$ is an odd integer. By <Ref>, we have

\[
\sigma\big((u_{2k}^*v_{2k})^2\big)=\big\{1,\theta^{\,2},\overline{\theta}^{\,2},\ldots,\theta^{\,k-1},\overline{\theta}^{\,k-1},\theta^{\,k+1},\overline{\theta}^{\,k+1}\big\}\,,
\]

and by the induction hypothesis we have

\[
\]

Now to show for the $(k+1)$-th step, first observe that

\[
\]

using <Ref>. Similarly, it follows that for each $1\leq r\leq\frac{k-1}{2}$, we have $m_{k+1}(\theta^{\,2r})>m_{k+1}(\overline{\theta}^{\,2r})>m_{k+1}(\theta^{\,2r+2})$. It remains only to show that $m_{k+1}(\theta^{\,k+1})>m_{k+1}(\overline{\theta}^{\,k+1})>m_{k+1}(\theta^{\,k+3})$. Again by <Ref>, observe that
m_k+1(θ^ k+1)    =    2m_k(θ^ k+1)+m_k(θ^ k+1)+m_k(θ^ k-1)

>    2m_k(θ^ k+1)+m_k(θ^ k+1)

=    m_k+1(θ^ k+1)

and $m_{k+1}(\overline{\theta}^{\,k+1})=2m_k(\overline{\theta}^{\,k+1})+m_k(\theta^{\,k+1})>m_k(\overline{\theta}^{\,k+1})=m_{k+1}(\theta^{\,k+3})$. This completes the induction and we have the following,

\[
\]

Since $k+1$ is an even integer, by <Ref> we have

\[
\sigma\big((u_{2k+2}^*v_{2k+2})^2\big)=\big\{1,\theta^{\,2},\overline{\theta}^{\,2},\ldots,\theta^{\,k+1},\overline{\theta}^{\,k+1},\theta^{\,k+3}\big\},
\]

and hence we are done.

Exact similar argument works if we start with $k$ even in the induction hypothesis, instead of $k$ being odd.

Let $\,\theta$ be not an even root of unity. For $k\in\bbn$ odd, the multiplicity $m_k(\overline{\theta}^{\,k+1})$ of the last eigenvalue $\overline{\theta}^{\,k+1}$ obtained in $\Cref{oo}$ is $1$, and for $k\in\bbn$ even, the multiplicity $m_k(\theta^{\,k+2})$ of the last eigenvalue $\theta^{\,k+2}$ is also $1$.

When $k\in\bbn$ is odd, by induction on $k$ it follows from <Ref> that
m_k(θ^ k+1)    =    2m_k-1(θ^ k+1)+m_k-1(θ^ k+1)

=    m_k-1(θ^ k+1)

=    m_k-2(θ^ k-1)

=    1 .    ()

When $k\in\bbn$ is even, by <Ref> we have $m_k(\theta^{\,k+2})=m_{k-1}(\overline{\theta}^{\,k})$, and the result follows from the first part.

Combining <Ref> and <ref>, we obtain the following spectral decomposition of each element in the family of unitary matrices $\{\big(u_{2k}^*v_{2k}\big)^2\in M_4^{(k+1)}\,:\,k\in\bbn\}$, when $\,\theta$ is not an even root of unity.

Let $\,\theta$ be not an even root of unity. The spectral decomposition for the unitary $\big(u_{2k}^*v_{2k}\big)^2$ in $M_4^{(k+1)}$ is $\sum_{j=1}^{k+2}\lambda_jE_j$, where $\lambda_1=1,\lambda_2=\theta^{\,2},\lambda_3=\overline{\theta}^{\,2}$ and so on, as described in $\Cref{oo}$. The dimension of the eigenspaces $E_j$, that is, the geometric multiplicities of the eigenvalues $\lambda_j$, are determined by the chain of recurrence relations obtained in $\Cref{recurrence}$, and we have $\sum_{j=1}^{k+2}\mbox{dim}(E_j)=4^{k+1}$.

Now, we shall deal with the case of $\,\theta$ being an even root of unity. Recall the eigenvalues from <Ref>. To distinguish the multiplicities of the eigenvalues in this case with that obtained in <Ref>, we denote them by $\widetilde{m}_k$ in this case.

Let $\theta^{\,m}=1$ for some $m\in 2\bbn+2$ and choose $k=(m-2)/2$. Then, up to $\big(u_{2k-2}^*v_{2k-2}\big)^2$, the multiplicities $\widetilde{m}_{k-1}(\lambda_r)$ of the eigenvalues $\lambda_r$ coincide with $m_{k-1}(\lambda_r)$ as defined in <Ref>, and from $\big(u_{2k}^*v_{2k}\big)^2$ onward, we have the following new chain of recurrence relations for the multiplicities $\widetilde{m}_{k+j}(.)$ for $0\leq j\leq k-1\,$:

* If $k\in\bbn$ is odd, then for any $0\leq j\leq k-1$ we have

* if $j\geq 0$ even, then
m_k+j(1)=m_k+j(1), m_k+j(θ^ 2)=m_k+j(θ^ 2),…, m_k+j(θ^ k-(j+1))=m_k+j(θ^ k-(j+1))

m_k+j(θ^ k+1)=m_k+j(θ^ k+1)+m_k+j(θ^ k+1),

m_k+j(θ^ k-r)=m_k+j(θ^ k-r)+m_k+j(θ^ k+r+2)     1≤r≤j-1 ,

m_k+j(θ^ k-r)=m_k+j(θ^ k-r)+m_k+j(θ^ k+r+2)    1≤r≤j-1 .

* if $j\geq 1$ odd, then
m_k+j(1)=m_k+j(1), m_k+j(θ^ 2)=m_k+j(θ^ 2),…, m_k+j(θ^ k-j)=m_k+j(θ^ k-j)

m_k+j(θ^ k+1)=m_k+j(θ^ k+1)+m_k+j(θ^ k+1),

m_k+j(θ^ k-r)=m_k+j(θ^ k-r)+m_k+j(θ^ k+r+2)     1≤r≤j ,

m_k+j(θ^ k-r)=m_k+j(θ^ k-r)+m_k+j(θ^ k+r+2)    1≤r≤j-2 .

* If $k\in\bbn$ is even, then for any $0\leq j\leq k-1$ we have

* if $j\geq 0$ even, then
m_k+j(1)=m_k+j(1), m_k+j(θ^ 2)=m_k+j(θ^ 2),…, m_k+j(θ^ k-j)=m_k+j(θ^ k-j)

m_k+j(θ^ k-r)=m_k+j(θ^ k-r)+m_k+j(θ^ k+r+2)     0≤r≤j ,

m_k+j(θ^ k-r)=m_k+j(θ^ k-r)+m_k+j(θ^ k+r+2)    0≤r≤j-2 .

* if $j\geq 1$ odd, then
m_k+j(1)=m_k+j(1), m_k+j(θ^ 2)=m_k+j(θ^ 2),…, m_k+j(θ^ k-j-1)=m_k+j(θ^ k-j-1)

m_k+j(θ^ k-r)=m_k+j(θ^ k-r)+m_k+j(θ^ k+r+2)     0≤r≤j-1 ,

m_k+j(θ^ k-r)=m_k+j(θ^ k-r)+m_k+j(θ^ k+r+2)    0≤r≤j-1 .

Here, $\,m_{k+j}(.)$ is as defined in $\Cref{recurrence}$.

Recall <Ref> in this regard. First assume that $k$ is an even integer. If $j$ increases as an even integer starting from $0$, then the eigenvalue $\theta^{\,k+j+2}$ coincides with $\overline{\theta}^{\,k-j}$ for all $j$ and $\overline{\theta}^{\,k+j}$ coincides with $\theta^{\,k-j+2}$ for all $j$. Moreover, multiplicities of the eigenvalues from $1$ up to $\theta^{\,k-j}$ remain unchanged. This is because $\theta^{\,m}=\theta^{\,2k+2}=1$. Similarly, if $j$ increases as an odd integer starting from $1$, then the eigenvalues $\theta^{\,k+j+1}$ and $\overline{\theta}^{\,k+j+1}$ coincide with $\overline{\theta}^{\,k-j+1}$ and $\theta^{\,k-j+1}$ respectively for all $j$ because of similar reason, and multiplicities of the eigenvalues from $1$ up to $\overline{\theta}^{\,k-j-1}$ remain unchanged. When $k$ is an odd integer, exact similar analysis holds. Now, the result follows in the same spirit of <Ref> using the following identity (<Ref>)

=    [ 1      ;   1    ;     0  ;       0;   ]⊗_W⊗W^(k-1)((u_2k-2^*v_2k-2)^2)+[    0               ;         0          ;              1     ;                θ^ 2;      ]⊗_W⊗W^(k-1)((v_2k-2^*u_2k-2)^2).

as the basic tool.

If we consider the unitary $\big(u_{2k}^*v_{2k}\big)^2$ and put the condition $\theta^{\,2k+2}=1$ for some $k\in\bbn$, then the multiplicity of only the last eigenvalue changes, irrespective of $k$ being even or odd (compare the situations in <Ref> and putting $j=0$ in <Ref>). The multiplicities of all the remaining eigenvalues do not change. This fact will be crucially used in subsequent sections. This is another reason (apart from <Ref>) of choosing the enumeration of the eigenvalues according to increasing power of $\theta$ immediately followed by $\overline{\theta}$, to keep track of what are the changes when $k$ increases.

Note that in the above we have obtained the multiplicities $\widetilde{m}_{k+j}(.)$ of the eigenvalues only for $0\leq j\leq k-1$. In this article, we do not need $j$ to go beyond $k-1$. This will be clear when we apply this in the subsection $7.2.3$. We remark that for $j\geq k$, with a little more effort, it is possible to obtain all the multiplicities $\widetilde{m}_{k+j}(.)$ in terms of recurrence relations, and write down the full spectral decomposition for this case of $\theta^{\,m}=1$ for $m\in 2\bbn+2$. However, this will be completely unnecessary for the purpose of this paper, and to stay focused to our goal, we leave it to the interested readers.

Combining <Ref> and <ref>, we obtain the following spectral decomposition when $\theta^{\,m}=1,\,m\in2\bbn+2$ (i,e., when $\theta$ is an even root of unity).

Let $\theta^{\,m}=1$ for some $m\in 2\bbn+2$ and choose $k=(m-2)/2$. The spectrum of $(u_{2k+2j}^*v_{2k+2j})^2$ remains the same with cardinality $k+1$ for all $j\in\bbn\cup\{-1,0\}$ as described in <Ref>. The spectral decomposition for the unitary $\big(u_{2k+2j}^*v_{2k+2j}\big)^2,\,0\leq j\leq k-1,$ is $\sum_{i=1}^{k+1}\lambda_iE_i^{(j)}$, where the dimension of the eigenspaces $E_i^{(j)}$, that is, the geometric multiplicities of the eigenvalues $\lambda_i$, are determined by the chain of recurrence relations described in $\Cref{recurrence1}$, and we have $\sum_{i=1}^{k+1}\mbox{dim}(E_i^{(j)})=4^{k+1+j}$ for each $0\leq j\leq k-1$.

§.§.§ A family of von Neumann subalgebras of $R$ with the infinite

Pimsner-Popa index

The spectral analysis carried out in the previous subsection says that if we consider the following subgroup of $\mathbb{S}^1$
Γ:={ω∈𝕊^1:ω^m=1, m∈2+2}∪{1} ,

that is, $\Gamma$ consists of all even roots of unity, then the spectrum of $\big(u_{2k}^*v_{2k}\big)^2$ depends on whether $\theta\in\Gamma$ or $\theta\in\mathbb{S}^1\setminus\Gamma$, where $\theta=\overline{a}b$ (note that $\theta\neq 1$ as $b\neq a$). It turns out that depending on whether $\theta$ is a primitive even root of unity or not, that is, whether $\theta^{\,2}$ is a rational rotation or an irrational rotation, the intersection $R_u\cap R_v$, which is a priori only a von Neumann subalgebra of the hyperfinite $II_1$ factor $R$, has finite or infinite Pimsner-Popa index in $R$. The set $\Gamma$ can also be interpreted in terms of an equivalence relation among the Hadamard matrices $u$ and $v$, which we discuss now.

An equivalence relation : Let $u(a)\mbox{ and }u(b)$ be two distinct $4\times 4$ Hadamard inequivalent complex Hadamard matrices parametrized by circle parameters $a\mbox{ and }b$ respectively (recall that $b\neq\pm a$ in our situation). We define a relation $u(a)\sim u(b)$ if and only if $\theta(a,b):=\overline{a}b\in\Gamma$, that is, $\theta(a,b)^{\,m}=1\mbox{ for some }m\in 2\bbn+2$. Note that $\theta(a,b)\neq 1$. It is easy to check that `$\sim$' is an equivalence relation. Therefore, for $u=u(a)$ and $v=u(b)$, the analysis carried out in previous subsection indicates that the spectrum of $\big(u_{2k}^*v_{2k}\big)^2$ for $k\in\bbn$ depends on whether $u\sim v$ or $u\nsim v$.

Caution: This equivalence relation has no relation with the one defined in <Ref>.

We shall see that this equivalence relation plays a major role in our analysis here. In this subsection, we will deal with the case of $u\nsim v$ (i,e. $\theta\notin\Gamma$) and request the reader to wait until the next subsection, where $u\sim v$ (i,e. $\theta\in\Gamma$) will be dealt.

Recall once again the unitary matrix $W\in M_4$ from <Ref>. Also, note that by $W^{(k)},\,k\in\bbn,$ we mean the matrix $W\otimes W\otimes\ldots\otimes W\in M_4^{(k)}:=M_4^{\,\otimes\,k}$.

We have $\big((u^*v)^2\big)^\prime\cap M_4=\mathrm{Ad}_W(M_3\oplus\bbc)$.

Since $\mbox{Ad}_W(u^*v)^2=\mbox{bl-diag}\{I_2,C(\theta^{\,2})\}=\mbox{diag}\{I_3,\theta^{\,2}\}$ in $M_4$ due to <Ref>, an application of <Ref> and <ref> proves the claim.

If $u\nsim v$, then there exist certain unitary matrices $\,U_k\in M_4^{(k+1)}$ for each $k\in\bbn$ such that the following holds,

Ad_W^(k+1)U_k(M_m_k(1)⊕⊕_r=1^k+1/2(M_m_k(θ^ 2r)⊕M_m_k(θ^ 2r)))       ;

Ad_W^(k+1)U_k(M_m_k(1)⊕⊕_r=1^k/2(M_m_k(θ^ 2r)⊕M_m_k(θ^ 2r))⊕)  .

We use <Ref> throughout. For $k=1$, by <Ref> and <Ref> we have
Ad_W⊗W((u_2^*v_2)^2)    =    E_11⊗I_2⊗Ad_W((u^*v)^2)+E_22⊗C(θ^ 2)⊗Ad_W((v^*u)^2)

=    E_11⊗I_2⊗({1,1,1,θ^ 2})+E_22⊗C(θ^ 2)⊗({1,1,1,θ^ 2}) .

There exists a unitary matrix (a permutation matrix to be precise) $U_1\in M_4^{(2)}$ which leaves $\Delta_4^{(2)}$ invariant and
Ad_W⊗W((u_2^*v_2)^2)    =    Ad_U_1(I_10⊕θ^ 2I_5⊕θ^ 2) .

For example, take $U_1=\big(E_{4,11}+E_{11,4}+\sum_{i\neq 4,11}E_{ii}\big)\big(E_{8,12}+E_{12,8}+\sum_{i\neq 8,12}E_{ii}\big)\big(E_{12,16}+E_{16,12}+\sum_{i\neq 12,16}E_{ii}\big)$. Now, applying <Ref>, <Ref>, and <Ref>, we have the following,
Ad_U_1^*(W⊗W)(((u_2^*v_2)^2)^'∩M_4^(2))    =    (Ad_U_1^*(W⊗W)(u_2^*v_2)^2)^'∩M_4^(2)

=    (I_10⊕θ^ 2I_5⊕θ^ 2)^'∩M_16

=    M_10⊕M_5⊕    ( <Ref>,θ∉Γ)

=    M_m_1(1)⊕M_m_1(θ^ 2)⊕M_m_1(θ^ 2),

which finishes the $k=1$ case. Now, suppose that $k$ is odd and we have a unitary matrix $U_k$ in $M_4^{(k+1)}$ such that the following holds
Ad_W^(k+1)((u_2k^*v_2k)^2)=Ad_U_k(I_m_k(1)⊕⊕_r=1^k+1/2(θ^ 2rI_m_k(θ^ 2r)⊕θ^ 2rI_m_k(θ^ 2r))) .

Using <Ref>, by <Ref> and <Ref> we have the following,

=    E_11⊗I_2⊗Ad_W^(k+1)(u_2k^*v_2k)^2+E_22⊗C(θ^ 2)⊗Ad_W^(k+1)(v_2k^*u_2k)^2

=    E_11⊗I_2⊗Ad_U_k(I_m_k(1)⊕ ⊕_r=1^k+1/2(θ^ 2r I_m_k(θ^ 2r)⊕θ^ 2r I_m_k(θ^ 2r)))

+E_22⊗C(θ^ 2)⊗Ad_U_k(I_m_k(1)⊕ ⊕_r=1^k+1/2(θ^ 2r I_m_k(θ^ 2r)⊕θ^ 2r I_m_k(θ^ 2r)))^*

=    Ad_(I_4⊗U_k)V_k(I_m_k+1(1)⊕⊕_r=1^k+1/2(θ^ 2r I_m_k+1(θ^ 2r)⊕θ^ 2r I_m_k+1(θ^ 2r))⊕θ^ k+3 I_m_k+1(θ^ k+3)) ,

where the last line follows from <Ref>. Here, $V_k$ is a permutation matrix which interchange the diagonal entries suitably to put all the same eigenvalue together at one block which are initially situated in various places at the diagonal. Let $U_{k+1}=(I_4\otimes U_k)V_k$, which is a unitary matrix in $M_4^{(k+2)}$, and we obtain the following by <Ref>,

=    (Ad_U_k+1^*W^(k+2)(u_2k+2^*v_2k+2)^2)^'∩M_4^(k+2)

=    (I_m_k+1(1)⊕⊕_r=1^k+1/2(θ^ 2r I_m_k+1(θ^ 2r)⊕θ^ 2r I_m_k+1(θ^ 2r))⊕θ^ k+3 I_m_k+1(θ^ k+3))^'∩M_4^(k+2)

=    M_m_k+1(1)⊕⊕_r=1^k+1/2(M_m_k+1(θ^ 2r)⊕M_m_k+1(θ^ 2r))⊕M_m_k+1(θ^ k+3)  ( <Ref>,θ∉Γ).

Now, $m_{k+1}(\theta^{\,k+3})=1$ by <Ref>, which completes the induction argument. Exactly similar argument works when we start the induction with $k$ even instead of $k$ being odd.

If $\,u\nsim v$, then we have $\,C_1=\mathrm{Ad}_{u_1(I_4\otimes W)}\big(\bbc\otimes(M_3\oplus\bbc)\big)$, and for any $k\in\bbn$

Ad_u_2k+1(I_4⊗W^(k+1)U_k)(⊗(M_m_k(1)⊕⊕_r=1^k+1/2(M_m_k(θ^ 2r)⊕M_m_k(θ^ 2r))))       ;

Ad_u_2k+1(I_4⊗W^(k+1)U_k)(⊗(M_m_k(1)⊕⊕_r=1^k/2(M_m_k(θ^ 2r)⊕M_m_k(θ^ 2r))⊕))  ;

where the unitary matrices $U_k$'s are as in <Ref>.

By <Ref>, we know that $C_{2k+1}$ for $k\geq 0$ is completely determined by $\big((u_{2k}^*v_{2k})^2\big)^\prime\cap M_4^{(k+1)}$ up to the conjugation by the unitary $u_{2k+1}$. Now, <Ref> proves the case of $C_1$, and <Ref> proves the claim for any $k\in\bbn$.

For the sake of better understanding, we mention below first few commutants $\big((u_{2k}^*v_{2k})^2\big)^\prime\cap M_4^{(k+1)},\,k\in\bbn\cup\{0\},$ concretely up to unitary matrices :

\[
\begin{matrix}
\underline{\mbox{for }k=0}\\
\cup\\
\end{matrix}\qquad\begin{matrix}
\underline{\mbox{for }k=1}\\
M_4\otimes M_4\\
\cup\\
M_{10}\oplus M_5\oplus\bbc\\
\end{matrix}\qquad\begin{matrix}
\underline{\mbox{for }k=2}\\
M_4\otimes M_4\otimes M_4\\
\cup\\
M_{35}\oplus M_{21}\oplus M_7\oplus\bbc\\
\end{matrix}\qquad\begin{matrix}
\underline{\mbox{for }k=3}\\
M_4\otimes M_4\otimes M_4\otimes M_4\\
\cup\\
M_{126}\oplus M_{84}\oplus M_{36}\oplus M_{9}\oplus\bbc\\
\end{matrix}
\]

The first vertical inclusion for $k=0$ is determined by the unitary $W$ in $M_4$, and all the others are determined by the unitary $W^{(k+1)}U_k$ in $M_4^{(k+1)}$ for $k\in\bbn$, where $U_k$'s are the unitary matrices in $M_4^{(k+1)}$ obtained in <Ref>. Therefore, in view of <Ref>, first few stages of the following quadruple for $k\in\bbn$

\[
\begin{matrix}
A_{2k-1} &\subset & A_{2k+1}\\
|| &  & ||\\
\bbc\otimes\Delta_4\otimes M_4^{(k)} & & \Delta_4\otimes M_4^{(k+1)}\\
\cup &  & \cup\\
\bbc\otimes\mbox{Ad}_{u_{2k-1}}\big(\bbc\otimes(M_4^{(k)})\big)\cap\mbox{Ad}_{v_{2k-1}}\big(\bbc\otimes(M_4^{(k)})\big) & & \mbox{Ad}_{u_{2k+1}}\big(\bbc\otimes(M_4^{(k+1)})\big)\cap\mbox{Ad}_{v_{2k+1}}\big(\bbc\otimes(M_4^{(k+1)})\big)\\
|| &  & ||\\
C_{2k-1} &\subset & C_{2k+1}\\
\end{matrix}
\]

are the following :

For $k=1:$

\[
\begin{matrix}
\bbc\otimes\Delta_4\otimes M_4 &\subset & \Delta_4\otimes M_4\otimes M_4\\
\cup &  & \cup\\
\bbc\otimes\mbox{Ad}_{u_1(I_4\otimes W)}(\bbc\otimes(M_3\oplus\bbc)) &\subset & \mbox{Ad}_{u_3\big(I_4\otimes W^{(2)}U_1\big)}(\bbc\otimes(M_{10}\oplus M_5\oplus\bbc))\\
\end{matrix}
\]

For $k=2:$

\[
\begin{matrix}
\bbc\otimes\Delta_4\otimes M_4\otimes M_4 &\subset & \Delta_4\otimes M_4\otimes M_4\otimes M_4\\
\cup &  & \cup\\
\bbc\otimes\mbox{Ad}_{u_3\big(I_4\otimes W^{(2)}U_1\big)}\big(\bbc\otimes(M_{10}\oplus M_5\oplus\bbc)\big) &\subset & \mbox{Ad}_{u_5\big(I_4\otimes W^{(3)}U_2\big)}(\bbc\otimes(M_{35}\oplus M_{21}\oplus M_7\oplus\bbc))\\
\end{matrix}
\]

For $k=3:$

\[
\begin{matrix}
\mbox{Ad}_{u_7\big(I_4\otimes W^{(4)}U_3\big)}(\bbc\otimes(M_{126}\oplus M_{84}\oplus M_{36}\oplus M_9\oplus\bbc)) &\subset & \Delta_4\otimes M_4\otimes M_4\otimes M_4\otimes M_4\\
\cup &  & \cup\\
\bbc\otimes\mbox{Ad}_{u_5\big(I_4\otimes W^{(3)}U_2\big)}(\bbc\otimes(M_{35}\oplus M_{21}\oplus M_7\oplus\bbc)) &\subset & \bbc\otimes\Delta_4\otimes M_4\otimes M_4\otimes M_4\\
\end{matrix}
\]

For $u\nsim v$ and $k\in\bbn\cup\{0\}$, the inclusion matrix for the inclusion $\big((u_{2k}^*v_{2k})^2\big)^\prime\cap M_4^{(k+1)}\subset M_4\otimes M_4^{(k)}$ is the column matrix $(1\,\,1\,\ldots\,1)^{T}$ of order $(k+2)\times 1$.

By <Ref>, we have the spectral decomposition of the unitary matrix $(u_{2k}^*v_{2k})^2$ for all $k\in\bbn\cup\{0\}$. Suppose that $k$ is odd. Then, $\{1,\theta^{\,2},\overline{\theta}^{\,2},\ldots,\theta^{\,k+1},\overline{\theta}^{\,k+1}\}$ is the set of distinct eigenvalues of $(u_{2k}^*v_{2k})^2$, and $m_k(1)+\sum_rm_k(\theta^{\,2r})+m_k(\overline{\theta}^{\,2r})=4^{k+1}$, where $m_k(.)$ are the multiplicities of the eigenvalues. By <Ref> it is now clear that the required inclusion matrix is the column matrix $(1\,\,1\,\ldots\,1)^{T}$ of order $(k+2)\times 1$, as there are $(k+2)$-many eigenvalues of $(u_{2k}^*v_{2k})^2$ by <Ref>. Exactly similar analysis holds for $k$ even.

For $u\nsim v$ and $k\in\bbn\cup\{0\}$, the inclusion matrix $\Lambda_k$ for the inclusion

\[
C_{2k+1}:=\mathrm{Ad}_{u_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)\cap\mathrm{Ad}_{v_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)\subset \Delta_4\otimes M_4\otimes M_4^{(k)}=:A_{2k+1}
\]

is the $(k+2)\times 4$ matrix each of whose entry is $1$. Thus, $||\Lambda_k||^2=4(k+2)$.

Since $\Delta_4\otimes M_4\otimes M_4^{(k)}$ is the direct sum of four copies of $M_4\otimes M_4^{(k)}$, and the unitary matrix $u_{2k+1}(I_4\otimes W^{(k+1)}U_k)$ in <Ref> lies in $\Delta_4\otimes M_4\otimes M_4^{(k)}$, each entry of the inclusion matrix $\Lambda_k$ equals to $1$ is clear by <Ref>. To determine $||\Lambda_k||^2$, observe that $\Lambda_k\Lambda_k^T$ is the matrix $4(k+2)J_{k+2}$, where $J_{k+2}$ is the projection $\sum_{j=1}^{k+2}(k+2)^{-1}E_{jj}$ in $M_{k+2}$.

In the following tower of commuting squares

\[
\begin{matrix}
A_1 &\subset & A_3 &\subset  & A_5 &\subset &\ldots\cr
\cup &  &\cup &  &\cup &  &  & \cr
C_1 &\subset & C_3 &\subset & C_5 &\subset &\ldots
\end{matrix}
\]

the first quadruple $(C_1\subset A_1,C_3\subset A_3)$ is not non-degenerate. This is because the inclusion matrix $\Lambda_{C_1}^{A_1}=\Lambda_0$ is a $2\times 4$ matrix with $||\Lambda_{C_1}^{A_1}||^2=8$ by <Ref>, whereas the inclusion matrix $\Lambda_{C_3}^{A_3}=\Lambda_1$ is a $3\times 4$ matrix with $||\Lambda_{C_3}^{A_3}||^2=12$. That is, $||\Lambda_{C_1}^{A_1}||^2\neq ||\Lambda_{C_3}^{A_3}||^2$. In fact, none of the quadruples are non-degenerate, which follows from <Ref>. This also justifies part $(ii)$ in <Ref>. Moreover, we can also conclude that $C_1\subset C_3\subset C_5\subset\ldots$ is not a basic construction, as otherwise the quadruple $(C_1\subset A_1,C_3\subset A_3)$ would have been non-degenerate by [5].

If $u\nsim v$, then $\lambda(R,R_u\cap R_v)=\lambda(R_u,R_u\cap R_v)=\lambda(R_v,R_u\cap R_v)=0$. That is, $R_u\cap R_v$ is a von Neumann subalgebra of the hyperfinite $II_1$ factor $R$ with infinite Pimsner-Popa index.

First observe that it is enough to only conclude that $\lambda(R,R_u\cap R_v)=0$. This is because by the submultiplicativity property of $\lambda$ for $R_u\cap R_v\subset R_u\subset R$, we have $\lambda(R,R_u\cap R_v)\geq\lambda(R_u,R_u\cap R_v)\lambda(R,R_u)=\frac{1}{4}\lambda(R_u,R_u\cap R_v)$ (similarly for $v$ in place of $u$) because $\lambda(R,R_u)=[R:R_u]^{-1}=\frac{1}{4}$ by [37]. Hence if $\lambda(R,R_u\cap R_v)=0$, then automatically we get that $\lambda(R_u,R_u\cap R_v)=0$.

Now by <Ref>, we have the following

\[
\lambda\left(\Delta_4\otimes M_4^{(k+1)}\,,\,\mathrm{Ad}_{u_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)\cap\mathrm{Ad}_{v_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)\right)^{-1}=\max_\ell\Big(\sum_{r=1}^{k+2}\frac{b_{r\ell}s_r}{t_\ell}\Big)\,,
\]

where $b_{r\ell}=\min\{a_{r\ell},n_r\}$ and $s_r$ denotes the trace of minimal projections in each summand $N_r$ of $\big((u_{2k}^*v_{2k})^2\big)^\prime\cap M_4^{(k+1)}$ described in <Ref>, and $t_\ell$ denotes trace of minimal projections in $\bbc\otimes M_4^{(k+1)}\subset\Delta_4\otimes M_4^{(k+1)}$ for $\ell=1,\ldots,4$. Hence, $t_\ell=\frac{1}{4^{(k+2)}}$ for each $\ell$ (recall the unique trace on $R$ obtained through the basic construction as in <Ref>), and since the inclusion matrix $\Lambda_k$ is the $(k+2)\times 4$ matrix each of whose entry is $1$, we get that $b_{r\ell}=1$ for each $r,\ell$. This says that $s_r=\frac{1}{4^{(k+1)}}$ for each $1\leq r\leq k+2$ because $\Lambda_k(t_1\,\,t_2\,\,t_3\,\,t_4)^T=(s_1\,\,\cdots\,\,s_{k+2})^T$. Thus,

\[
\lambda^{-1}=4^{(k+2)}\sum_{r=1}^{k+2}s_r=4^{(k+2)}\sum_{r=1}^{k+2}\frac{1}{4^{(k+1)}}=4(k+2)\,.
\]

Recall that $R_u\cap R_v$ is the SOT-limit of $C_{2k+1}=\mathrm{Ad}_{u_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)\cap\mathrm{Ad}_{v_{2k+1}}\big(\bbc\otimes M_4^{(k+1)}\big)$. Now, using <Ref> we get that $\lambda(R,R_u\cap R_v)=\lim_{k\to\infty}\frac{1}{4(k+2)}=0$.

Therefore, we obtain the following family of von Neumann subalgebras of $R$

\[
\left\{R_{u(a)}\cap R_{u(b)}\subset R:\,u(a)\nsim u(b);\,a,b\in\mathbb{S}^1\right\}\,,
\]

such that any member of this family has infinite Pimsner-Popa index in $R$.

Question: Is $R_u\cap R_v$ for $u\nsim v$ a factor?

§.§.§ Construction of a family of finite index subfactors of $R$

We now deal with the case of $u\sim v$. Before we jump into the general situation, first let us understand few initial cases to realize where it differs from the corresponding situation in $u\nsim v$. Recall <Ref> in this regard. Consider again the following tower of commuting squares

\[
\begin{matrix}
A_1 &\subset & A_3 &\subset  & A_5 &\subset &\ldots\cr
\cup &  &\cup &  &\cup &  &  & \cr
C_1 &\subset & C_3 &\subset & C_5 &\subset &\ldots
\end{matrix}
\]

In the case of $u\nsim v$, for $k=0$, we have seen that the inclusion matrix $\Lambda_0$ for the inclusion $C_1=\mbox{Ad}_{u_1(I_4\otimes W)}(\bbc\otimes(M_3\oplus\bbc))\subset\Delta_4\otimes M_4=A_1$ is the $2\times 4$ matrix each of whose entry is $1$ (<Ref>), and the square of its norm is $||\Lambda_0||^2=8$. Now let $k=1$. The inclusion matrix $\Lambda_1$ for the inclusion $C_3=\mbox{Ad}_{u_3(I_4\otimes W^{(2)}U_1)}(\bbc\otimes(M_{10}\oplus M_5\oplus\bbc))\subset\Delta_4\otimes M_4\otimes M_4=A_3$ is the $3\times 4$ matrix each of whose entry is $1$, and the square of its norm is $||\Lambda_1||^2=12$. Thus, the following quadruple

\[
\begin{matrix}
A_1:=\bbc\otimes\Delta_4\otimes M_4 &\subset & \Delta_4\otimes M_4\otimes M_4=:A_3\\
\cup &  & \cup\\
C_1:=\bbc\otimes\mbox{Ad}_{u_1(I_4\otimes W)}(\bbc\otimes(M_3\oplus\bbc)) &\subset & \mbox{Ad}_{u_3\big(I_4\otimes W^{(2)}U_1\big)}(\bbc\otimes(M_{10}\oplus M_5\oplus\bbc))=:C_3\\
\end{matrix}
\]

is a commuting square which is not non-degenerate if $u\nsim v$. However, if we assume that $u\sim v$ and $\theta=\overline{a}b$ is such that $\theta^{\,4}=1$, then the situation changes drastically. By <Ref>, we see that $\sigma\big((u_2^*v_2)^2\big)=\{1,\theta^{\,2}\}$ as $\theta^{\,4}=1$ (putting $k=2$ there). By <Ref>, the multiplicity of the eigenvalue $1$ remains the same, i,e. $10$, but that of $\theta^{\,2}$ is now $6$ (as $\theta^{\,2}=\overline{\theta}^{\,2}$) instead of $5$. In view of <Ref>, we now have the following modified quadruple,

\[
\begin{matrix}
A_1:=\bbc\otimes\Delta_4\otimes M_4 &\subset & \Delta_4\otimes M_4\otimes M_4=:A_3\\
\cup &  & \cup\\
C_1:=\bbc\otimes\mbox{Ad}_{u_1(I_4\otimes W)}(\bbc\otimes(M_3\oplus\bbc)) &\subset & \mbox{Ad}_{u_3\big(I_4\otimes W^{(2)}U_1\big)}(\bbc\otimes(M_{10}\oplus M_6))=:C_3\\
\end{matrix}
\]

The inclusion matrix $\tilde{\Lambda}_1$ for the inclusion $\mbox{Ad}_{u_3(I_4\otimes W^{(2)}U_1)}(\bbc\otimes(M_{10}\oplus M_6))\subset\Delta_4\otimes M_4\otimes M_4$ is the $2\times 4$ matrix each of whose entry is $1$. That is, $\tilde{\Lambda}_1=\Lambda_0$ and $||\tilde{\Lambda}_1||^2=||\Lambda_0||^2=8$. Therefore, in this case the quadruple $(C_1\subset A_1,C_3\subset A_3)$, which is already a commuting square, becomes non-degenerate also by Lemma $3.10$ in [5]. Iterating Jones' basic construction on the quadruple $(C_1\subset A_1,C_3\subset A_3)$, we obtain a subfactor $R^{(4)}$ of $R$ such that $R^{(4)}\subseteq R_u\cap R_v$ (by <Ref>) and $[R:R^{(4)}]=||\Lambda_0||^2=8$. Since $[R_u:R^{(4)}]=2$, by the irreducibility of $R^{(4)}$ in $R_u$, it follows that $R^{(4)}=R_u\cap R_v$. That is, $R_u\cap R_v$ is a finite index subfactor of $R$ with $[R:R_u\cap R_v]=8$. This is in sharp contrast with <Ref>.

Similarly, if we assume that $u\sim v$ and $\theta=\overline{a}b$ is such that $\theta^{\,6}=1$, we see that $\sigma\big((u_2^*v_2)^2\big)=\sigma\big((u_4^*v_4)^2\big)=\{1,\theta^{\,2},\overline{\theta}^{\,2}\}$ by <Ref>. In this case, the quadruple $(C_1\subset A_1,C_3\subset A_3)$ remains not non-degenerate as before, however, a similar analysis as above tells us that $\tilde{\Lambda}_2=\Lambda_1$ with the square of their norm being equal to $12$. In other words, the quadruple $(C_3\subset A_3,C_5\subset A_5)$ becomes a non-degenerate commuting square. Therefore, iterating Jones' basic construction on the quadruple $(C_3\subset A_3,C_5\subset A_5)$, we obtain a subfactor $R^{(6)}$ of $R$ such that $R^{(6)}\subseteq R_u\cap R_v$ (by <Ref>) and $[R:R^{(6)}]=||\Lambda_1||^2=12$. Since $[R_u:R^{(6)}]=3$, by the irreducibility of $R^{(6)}$ in $R_u$, it follows that $R^{(6)}=R_u\cap R_v$. That is, $R_u\cap R_v$ is a finite index subfactor of $R$ with $[R:R_u\cap R_v]=12$, which is again in sharp contrast with <Ref>.

It turns out that this phenomena happens in general, that is, for $\theta^{\,m}=1$ where $m\in 2\bbn+2$. We need the following technical lemma first. Recall <Ref> and <Ref> in this regard, along with the notations.

For $u=u(a),v=u(b)$ let $\theta=\overline{a}b$, and suppose that $\theta$ is a primitive even root of unity. That is, $\,\theta^{\,m}=1$ for some $m\in 2\bbn+2$. Then, for $k=(m-2)/2$ we have the following,

Ad_u_2k+1(I_4⊗W^(k+1)U_k)(⊗(M_m_k(1)⊕⊕_r=1^k-1/2(M_m_k(θ^ 2r)⊕M_m_k(θ^ 2r))

⊕M_m_k(θ^ k+1)))     ,

Ad_u_2k+1(I_4⊗W^(k+1)U_k)(⊗(M_m_k(1)⊕⊕_r=1^k/2(M_m_k(θ^ 2r)⊕M_m_k(θ^ 2r))))  .

Here, $U_k$ remains the same unitary obtained in <Ref>.

Consider the unitary $(u_{2k}^*v_{2k})^2$ in $M_4^{(k+1)}$. By <Ref>, the set of distinct eigenvalues are either $\{1,\theta^{\,2},\overline{\theta}^{\,2},\ldots,\theta^{\,k-1},\overline{\theta}^{\,k-1},\theta^{\,k+1}\}$ or $\{1,\theta^{\,2},\overline{\theta}^{\,2},\ldots,\theta^{\,k},\overline{\theta}^{\,k}\}$ depending on $k$ being odd or even respectively, and $\sigma\big((u_{2k}^*v_{2k})^2\big)=\sigma\big((u_{2k-2}^*v_{2k-2})^2\big)$ with cardinality $k+1$. The multiplicities $\widetilde{m}_k(.)$ of the eigenvalues are given by <Ref>. The rest of the proof follows along the similar line of <Ref>.

The above lemma can be thought of as a counterpart to <Ref> for the case of $u\sim v$. We have crucially used here the enumeration of eigenvalues according to increasing power of $\theta$ followed immediately by $\overline{\theta}$ (see also <Ref> and <Ref>) to keep the unitary $U_k$ same with that in <Ref>. Now we are ready to prove our main theorem.

Let $u=u(a)\mbox{ and }v=u(b)$ be $4\times 4$ Hadamard inequivalent complex Hadamard matrices such that $u\sim v$. If $\theta=\overline{a}b$ satisfies $\theta^{\,m}=1$ for $m\in 2\bbn+2$, then there is a finite index subfactor $R^{(m)}_{u,v}$ of the hyperfinite type $II_1$ factor $R$ such that $[R:R^{(m)}_{u,v}]=2m$ and $R^{(m)}_{u,v}\subseteq R_u\cap R_v$.

The cases of $m=4,6$ are already discussed in detail at the beginning of this subsection. For $\theta^{\,m}=1$ with $m\geq 8$, choose $k_m\in\bbn$ such that $m=2k_m+2$. Such a positive integer $k_m$ always exists and $k_m\geq 3$. Consider the following inclusion

\[
C_{2k_m-1}:=\mbox{Ad}_{u_{2k_m-1}}\big(\bbc\otimes M_4^{(k_m)}\big)\cap\mbox{Ad}_{v_{2k_m-1}}\big(\bbc\otimes M_4^{(k_m)}\big)\subset\Delta_4\otimes M_4^{(k_m)}=:A_{2k_m-1},
\]

which is same as $\mbox{Ad}_{u_{2k_m-1}}\big(\bbc\otimes\big((u_{2k_m-2}^*v_{2k_m-2})^2\big)^\prime\cap M_4^{(k_m)}\big)\subset\Delta_4\otimes M_4^{(k_m)}$ due to <Ref>. The set of eigenvalues of $(u_{2k_m-2}^*v_{2k_m-2})^2$ is given by <Ref>, and their multiplicities are given by <Ref>. Observe that up to this stage, the eigenvalues, along with their multiplicities, remain the same as in <Ref> and <Ref> for the case of $u\nsim v$. This is because we always take $\theta$ to be a primitive even root of unity. Therefore, <Ref> remains valid for $C_{2k_m-1}$, and the inclusion matrix $\Lambda_{k_m-1}$ for the inclusion $C_{2k_m-1}\subset A_{2k_m-1}$ is a $(k_m+1)\times 4$ matrix each of whose entry is $1$, and the square of the norm of it is $||\Lambda_{k_m-1}||^2=4(k_m+1)$ (see <Ref>). Now, consider the next inclusion

\[
C_{2k_m+1}:=\mbox{Ad}_{u_{2k_m+1}}\big(\bbc\otimes M_4^{(k_m+1)}\big)\cap\mbox{Ad}_{v_{2k_m+1}}\big(\bbc\otimes M_4^{(k_m+1)}\big)\subset\Delta_4\otimes M_4^{(k_m+1)}=:A_{2k_m+1},
\]

which is same as $\mbox{Ad}_{u_{2k_m+1}}\big(\bbc\otimes\big((u_{2k_m}^*v_{2k_m})^2\big)^\prime\cap M_4^{(k_m+1)}\big)\subset\Delta_4\otimes M_4^{(k_m+1)}$ due to <Ref>. That is, we are considering the following quadruple

\[
\begin{matrix}
A_{2k_m-1} &\subset &A_{2k_m+1}\\
|| &   &  ||\\
\bbc\otimes\Delta_4\otimes M_4^{(k_m)} & & \Delta_4\otimes M_4^{(k_m+1)}\\
\cup &  & \cup\\
\bbc\otimes\mbox{Ad}_{u_{2k_m-1}}\big(\bbc\otimes\big((u_{2k_m-2}^*v_{2k_m-2})^2\big)^\prime\cap M_4^{(k_m)}\big) & & \mbox{Ad}_{u_{2k_m+1}}\big(\bbc\otimes\big((u_{2k_m}^*v_{2k_m})^2\big)^\prime\cap M_4^{(k_m+1)}\big)\\
|| &   &  ||\\
C_{2k_m-1} &\subset &C_{2k_m+1}\\
\end{matrix}
\]

Recall that if $u$ were not related to $v$, then this quadruple is not non-degenerate as explained in <Ref>. This is because in the absence of any such condition on $\theta$, the inclusion matrix for $C_{2k_m+1}\subset A_{2k_m+1}$ is a $(k_m+2)\times 4$ matrix each of whose entry is $1$ by <Ref>. However, we show that putting the condition $\theta^{\,2k_m+2}=1$ forces a row-drop in the inclusion matrix. More precisely, we claim that the inclusion matrix $\Lambda_{k_m}$ for the right vertical inclusion $C_{2k_m+1}\subset A_{2k_m+1}$ is also a $(k_m+1)\times 4$ matrix each of whose entry is $1$, and hence $||\Lambda_{k_m-1}||^2=||\Lambda_{k_m}||^2=4(k_m+1)$. Consider the unitary $(u_{2k_m}^*v_{2k_m})^2$ in $M_4^{(k_m+1)}$. By <Ref>, we see that the number of direct summands in the Wedderburn–Artin decomposition of the algebra $\big((u_{2k_m}^*v_{2k_m})^2\big)^\prime\cap M_4^{(k_m+1)}$ is exactly $k_m+1$. Since $A_{2k_m+1}=\Delta_4\otimes M_4^{(k_m+1)}=\oplus M_4^{(k_m+1)}$, and the unitary $u_{2k+1}(I_4\otimes W^{(k+1)}U_k)$ in <Ref> belongs to $A_{2k_m+1}$, the inclusion matrix $\Lambda_{k_m}$ for the inclusion $C_{2k_m+1}\subset A_{2k_m+1}$ becomes a $(k_m+1)\times 4$ matrix each of whose entry is $1$. This finishes the claim.

Therefore, the following quadruple

A_2k_m-1     ⊂    A_2k_m+1

∪        ∪

C_2k_m-1     ⊂    C_2k_m+1

which is already a commuting square, becomes non-degenerate also by Lemma $3.10$ in [5]. Iterating Jones' basic construction starting with this non-degenerate commuting square, we obtain a subfactor $R^{(m)}_{u,v}$ of the hyperfinite type $II_1$ factor $R$ such that $R^{(m)}_{u,v}\subseteq R_u\cap R_v$ (by <Ref>) and $[R:R^{(m)}_{u,v}]=||\Lambda_{k_m-1}||^2=4(k_m+1)=2m$. That the tower $A_1\subset A_3\subset A_5\subset\cdots$ indeed gives us $R$ follows as an application of Theorem $3.5$ in [1], for instance.

Therefore, we obtain the following family of finite index subfactors

\[
\left\{R^{(m)}_{u(a),u(b)}\subset R:\,u(a)\sim u(b)\mbox{ with }(\overline{a}b)^m=1,\,m\in 2\bbn+2\right\}
\]

such that $\big[R:R^{(m)}_{u(a),u(b)}\big]=2m$, where $u(a),u(b)$ are complex Hadamard matrices of order $4\times 4$ parametrized by the circle parameters $a\mbox{ and }b$ respectively such that $u(a)\sim u(b)$ with $(\overline{a}b)^m=1$ for $m\in 2\bbn+2$. The converse of the above result is also true.

Given any $m\in 2\bbn+2$, there exists a pair $(u,v)$ of Hadamard inequivalent $4\times 4$ complex Hadamard matrices and a subfactor $R_0\subset R$ such that $R_0\subset R_u\cap R_v$ and $[R:R_0]=2m$.

Given any integer $m\geq 4$, choose a complex Hadamard matrix $u(a)$ of the required form, where $a$ is any arbitrary complex number on the semicircle. Consider $b=\omega a$, where $\omega\neq\pm 1$ is any primitive $m$-th root of unity. If $b$ lies in the lower semicircle, consider $-b$ instead of $b$, which lies in the upper semicircle. Obtain the complex Hadamard matrix $u(b)$. For the pair $(u,v):=(u(a),u(b))$, we have $(\overline{a}b)^m=1$. By <Ref>, we have the required subfactor $R_0:=R^{(m)}_{u,v}\subset R$.

We shall see in the next subsection that all these subfactors $R^{(m)}_{u(a),u(b)}\subset R$ are irreducible. We end this subsection with the following question.

Question: Can we characterize the following family of finite index subfactors of $R$
\[\left\{R^{(m)}_{u(a),u(b)}\subset R:\,u(a)\sim u(b)\mbox{ with }(\overline{a}b)^m=1,\,m\in 2\bbn+2\right\}?\]

§.§ Relative commutant and factoriality of $R_u\cap R_v$ when $u\sim v$

This subsection is intimately connected with its immediate prequel. Goal of this subsection is to prove that all the subfactors $R^{(m)}_{u(a),u(b)}\subset R$ constructed there are irreducible in $R$. As a consequence, we will get that $R_{u(a)}\cap R_{u(b)}$ is factor when $u(a)\sim u(b)$. Major tools are the Ocneanu compactness and the spectral decomspoition obtained in subsection $7.2.1$. We begin by recalling the following well-known fact from Linear algebra.

Let $B$ be a block-diagonal matrix, that is $B\equiv B_1\oplus B_2$, and $A$ be a matrix such that $[A,B]=0$. Suppose that $B_1\mbox{ and }B_2$ do not share any eigenvalue over $\bbc$. Then, $A$ has the same block-diagonal structure as $B$, that is, it is block-diagonal with blocks placed at the same positions as those of $B$.

Let $A=\begin{pmatrix}
X & Y\\
Z & W
\end{pmatrix}$. Since $[A,B]=0$, we obtain two equations $B_1Y-YB_2=0$ and and $B_2Z-ZB_1=0$. Since $B_1\mbox{ and }B_2$ do not share any eigenvalue, by the Sylvester criterion, we get that $Y=0$ and $Z=0$.

The commutant of any finite-dimensional algebra $N=\oplus_{j=1}^r N_{n_j}$, where $N_{n_j}$ is the type $I_{n_j}$ factor, in the type $I_m$ factor $M_m$ such that $\sum_jn_j=m$ is $\bbc^{\,r}$.

By the previous <Ref>, if $A\in M_m$ commutes with every $B=\oplus_{j=1}^rB_j$ where $B_j\in N_{n_j}$, then $A$ must be of the form $\oplus_{j=1}^rA_j$ where $A_j\in N_{n_j}$. This is because it is always possible to choose such a block-diagonal matrix $B$ satisfying the hypothesis of <Ref>. Since each $N_{n_j}$ is the type $I_{n_j}$ factor, each $A_j$ must be a scalar matrix of order $n_j$.

Now we prove our main theorem. The crucial ingredients are <Ref> and <ref>, and more precisely <Ref>. We request the readers to recall the corresponding notations also. Recall that the unitary $U_k$ in <Ref> is defined inductively by $U_k=(I_4\otimes U_{k-1})V_{k-1}$, where $V_{k-1}$ is certain permutation matrix in $M_4^{(k+1)}$ that only interchange the diagonal entries (see the proof of <Ref>). This permutation matrix has crucial role in what follows next.

Let $u=u(a)\mbox{ and }v=u(b)$ be $4\times 4$ Hadamard inequivalent complex Hadamard matrices such that $u\sim v$ with $(\overline{a}b)^m=1$. Then, the subfactor $R^{(m)}_{u,v}\subset R$ is irreducible.

Since $u\sim v$, we have $\theta=\overline{a}b\in\Gamma$. If $\theta^{\,m}=1$ ($\theta$ is a primitive even root of unity), take $k=(m-2)/2$. The proof is a bit technical, and hence we break it into several steps for clarity. We separate two situations, namely $k$ being odd or even. First suppose that $k$ is odd. By <Ref>, we have the multiplicities $\widetilde{m}_k(.)$ of the eigenvalues of $(u_{2k}^*v_{2k})^2$ (and similarly $\widetilde{m}_{k-1}(.)$ for $(u_{2k-2}^*v_{2k-2})^2$). To avoid the cumbersome notation, we denote $\widetilde{m}_k(.)$ simply by $m_k(.)$ specifically for this proof. We hope that this does not create any confusion because by <Ref>, we see that $\widetilde{m}_{k-1}(.)$ is anyway equal to $m_{k-1}(.)$, and so is $\widetilde{m}_k(.)$ equal to $m_k(.)$ except for the last eigenvalue $\theta^{\,k+1}$ (that is, only $\widetilde{m}_k(\theta^{\,k+1})\neq m_k(\theta^{\,k+1})$).

Step 1:  The action of the permutation matrix $V_{k-1}$ (<Ref>, <Ref>).

Consider the unitary matrix $(u_{2k}^*v_{2k})^2$. Recall from <Ref> that

\[
(u_{2k}^*v_{2k})^2=p\otimes (u_{2k-2}^*v_{2k-2})^2+q(\theta^{\,2})\otimes(v_{2k-2}^*u_{2k-2})^2
\]

for $k\in\bbn$. Recall from <Ref> the following,

=    E_11⊗I_2⊗Ad_W^(k-1)(u_2k-2^*v_2k-2)^2+E_22⊗C(θ^ 2)⊗Ad_W^(k-1)(v_2k-2^*u_2k-2)^2

=    E_11⊗I_2⊗Ad_U_k-1(I_m_k-1(1)⊕ ⊕_r=1^k-1/2(θ^ 2r I_m_k-1(θ^ 2r)⊕θ^ 2r I_m_k-1(θ^ 2r))⊕θ^ k+1 I_m_k-1(θ^ k+1))

+E_22⊗C(θ^ 2)⊗Ad_U_k-1(I_m_k-1(1)⊕ ⊕_r=1^k-1/2(θ^ 2r I_m_k-1(θ^ 2r)⊕θ^ 2r I_m_k-1(θ^ 2r))⊕θ^ k+1 I_m_k-1(θ^ k+1))

=    Ad_(I_4⊗U_k-1)V_k-1(I_m_k(1)⊕⊕_r=1^k-1/2(θ^ 2r I_m_k(θ^ 2r)⊕θ^ 2r I_m_k(θ^ 2r))⊕θ^ k+1 I_m_k(θ^ k+1)) .

Note that here $\theta^{\,k+1}=\overline{\theta}^{\,k+1}$ as $\theta^{\,m}=1$. Since $C(\theta^{\,2})$ is the diagonal matrix $\mbox{diag}\{1,\theta^{\,2}\}$, we obtain that under the conjugation action by the permutation matrix $V_{k-1}$, the element

\[
\alpha I_{m_k(1)}\bigoplus\bigoplus_{r=1}^{\frac{k-1}{2}}\Big(\beta_{2r}I_{m_k(\theta^{\,2r})}\oplus\gamma_{2r}I_{m_k(\overline{\theta}^{\,2r})}\Big)\bigoplus\delta I_{m_k(\theta^{\,k+1})}
\]

where $\alpha,\beta_{2r},\gamma_{2r},\delta\in\bbc$ for $r=1,\ldots,\frac{k-1}{2}$, becomes the following
(E_11+E_22)⊗{αI_m_k-1(1)⊕⊕_r=1^k-1/2(β_2rI_m_k-1(θ^ 2r)⊕γ_2rI_m_k-1(θ^ 2r))⊕δI_m_k-1(θ^ k+1)}

+E_33⊗{αI_m_k-1(1)⊕⊕_r=1^k-1/2(γ_2rI_m_k-1(θ^ 2r)⊕β_2rI_m_k-1(θ^ 2r))⊕δI_m_k-1(θ^ k+1)}

+E_44⊗{(β_2 I_m_k-1(1)⊕αI_m_k-1(θ^ 2))⊕⊕_r=1^k-3/2(β_2r+2I_m_k-1(θ^ 2r)⊕γ_2rI_m_k-1(θ^ 2r+2))

⊕(δI_m_k-1(θ^ k-1)⊕γ_k-1I_m_k-1(θ^ k+1))} .

This finishes our first step.

Step 2:  The structure of $\,C_{2k+1}^{\,\prime}\cap A_{2k-1}$.

We find the general structure of this commutant. With the convention $W_0=I_4$, we have the following,
C_2k+1^ '∩A_2k-1

=    {_u_2k+1(⊗((u_2k^*v_2k)^2)^'∩M_4^(k+1))}^'∩(⊗Δ_4⊗M_4^(k))

=    {_u_2k+1(⊗((u_2k^*v_2k)^2)^'∩M_4^(k+1))}^'∩_I_4⊗u_2k-1(⊗Δ_4⊗M_4^(k))

=    _u_2k+1{(⊗((u_2k^*v_2k)^2)^'∩M_4^(k+1))^'∩ _u_2k+1^*(I_4⊗u_2k-1)(⊗Δ_4⊗M_4^(k))}

=    _u_2k+1{(⊗((u_2k^*v_2k)^2)^'∩M_4^(k+1))^'∩ _(∑_j=0^3E_j+1,j+1⊗W_j^*ξ^*⊗I_4^(k))(⊗Δ_4⊗M_4^(k))}

=    _u_2k+1{(⊗Ad_W^(k+1)(Ad_U_k(M_m_k(1)⊕M_m_k(θ^ 2)⊕M_m_k(θ^ 2)⊕⋯⊕M_m_k(θ^ k+1))))^'

⋂ _(∑_j=0^3E_j+1,j+1⊗W_j^*ξ^*⊗I_4^(k))(⊗Δ_4⊗M_4^(k))}      (<Ref>)

=    _u_2k+1(I_4⊗W^(k+1)){(⊗Ad_U_k(M_m_k(1)⊕M_m_k(θ^ 2)⊕M_m_k(θ^ 2)⊕⋯⊕M_m_k(θ^ k+1)))^'

⋂ _(∑_j=0^3E_j+1,j+1⊗WW_j^*ξ^*⊗I_4^(k))(⊗Δ_4⊗M_4^(k))}

=    _u_2k+1(I_4⊗W⊗W^(k)U_k-1){(⊗Ad_V_k-1(M_m_k(1)⊕M_m_k(θ^ 2)⊕M_m_k(θ^ 2)⊕⋯⊕M_m_k(θ^ k+1)))^'

⋂ _(∑_j=0^3E_j+1,j+1⊗WW_j^*ξ^*⊗I_4^(k))(⊗Δ_4⊗M_4^(k))} ,
where the last line follows from the fact that $U_s=(I_4\otimes U_{s-1})V_{s-1}$ with $U_{s-1}\in M_4^{(s)}$. Now, we have

=    {_Wξ^*⊗I_4^(k)(x) , _WW_1^*ξ^*⊗I_4^(k)(x) , _WW_2ξ^*⊗I_4^(k)(x) , _WW_3^*ξ^*⊗I_4^(k)(x)}
where $x\in\Delta_4\otimes M_4^{(k)}$. We write $x=\mbox{bl-diag}\{x_1,x_2,x_3,x_4\}$ with $x_j\in M_4^{(k)}$ for $j=1,2,3,4$. Then, using <Ref> it is easy to verify the following four identities,
_Wξ^*⊗I_4^(k)(x)    =    1/2((E_11+E_22)⊗(x_1+x_3)+(E_33+E_44)⊗(x_2+x_4)

_WW_1^*ξ^*⊗I_4^(k)(x)    =    1/2((E_11+E_22)⊗(x_2+x_4)+(E_33+E_44)⊗(x_1+x_3)

_WW_2ξ^*⊗I_4^(k)(x)    =    1/2((E_11+E_22)⊗(x_1+x_3)+(E_33+E_44)⊗(x_2+x_4)

_WW_3^*ξ^*⊗I_4^(k)(x)    =    1/2((E_11+E_22)⊗(x_2+x_4)+(E_33+E_44)⊗(x_1+x_3)

+(E_12+E_21)⊗(x_4-x_2)+(iaE_34-iaE_43)⊗(x_1-x_3)) .

<Ref> describes arbitrary elements of the set in <Ref>.

Step 3:  Proof of $\,C_{2k+1}^{\,\prime}\cap A_{2k-1}=\bbc$.

This is the major step in the proof. In view of <Ref>, it is enough to prove the following,
=    (⊗Ad_V_k-1(M_m_k(1)⊕M_m_k(θ^ 2)⊕M_m_k(θ^ 2)⊕⋯⊕M_m_k(θ^ k+1)))^'

⋂  _(∑_j=0^3E_j+1,j+1⊗WW_j^*ξ^*⊗I_4^(k))(⊗Δ_4⊗M_4^(k)) .

First observe that since the permutation matrix $V_{k-1}$ only interchanges the diagonal elements, we get the following,

\[
\Big(\bbc\otimes\mathrm{Ad}_{V_{k-1}}\Big(M_{m_k(1)}\oplus M_{m_k(\theta^{\,2})}\oplus M_{m_k(\overline{\theta}^{\,2})}\oplus\cdots\oplus M_{m_k(\theta^{\,k+1})}\Big)\Big)^\prime\cap A_{2k+1}\subseteq\Delta_4\otimes\Delta_4^{(k+1)}\,.
\]

To see this, apply <Ref> and observe the following,
(Ad_V_k-1(M_m_k(1)⊕M_m_k(θ^ 2)⊕M_m_k(θ^ 2)⊕⋯⊕M_m_k(θ^ k+1)))^'∩M_4⊗M_4^(k)

=    Ad_V_k-1((M_m_k(1)⊕M_m_k(θ^ 2)⊕M_m_k(θ^ 2)⊕⋯⊕M_m_k(θ^ k+1))^'∩M_4^(k+1))

=    Ad_V_k-1(I_m_k(1)⊕I_m_k(θ^ 2)⊕I_m_k(θ^ 2)⊕…⊕I_m_k(θ^ k+1)) ,

where the last line follows from <Ref>. Therefore, the intersection in the right hand side of <Ref> must consists of purely diagonal matrices because

\[
\mbox{Ad}_{\big(\sum_{j=0}^3E_{j+1,j+1}\otimes WW_j^*\xi^*\otimes I_4^{(k)}\big)}\big(\bbc\otimes\Delta_4\otimes M_4^{(k)}\big)\subseteq A_{2k+1}:=\Delta_4\otimes M_4\otimes M_4^{(k)}\,.
\]

=    {_Wξ^*⊗I_4^(k)(x) , _WW_1^*ξ^*⊗I_4^(k)(x) , _WW_2ξ^*⊗I_4^(k)(x) , _WW_3^*ξ^*⊗I_4^(k)(x)}

for $x=\mbox{bl-diag}\{x_1,x_2,x_3,x_4\}\in\Delta_4\otimes M_4^{(k)}$, we see that any element of the intersection in <Ref> must satisfy $x_1=x_3$ and $x_2=x_4$, with $x_1,x_2\in\Delta_4^{(k)}$ because of the identities established in <Ref> in Step $2$. Thus, to prove that the <Ref> holds, it is enough to prove that the following,
{Δ_4⊗Ad_V_k-1(αI_m_k(1)⊕⊕_r=1^k-1/2(β_2r I_m_k(θ^ 2)⊕γ_2r I_m_k(θ^ 2))⊕δI_m_k(θ^ k+1))) : α,β_2r,γ_2r,δ∈}

⋂ {I_2⊗{x_1,x_1,x_2,x_2,x_2,x_2,x_1,x_1} : x_1,x_2∈Δ_4^(k)}= .

Because of the fact that $\Delta_4=\Delta_2\otimes\Delta_2$, to establish <Ref>, it is enough to prove the following,
{Δ_2⊗Ad_V_k-1(αI_m_k(1)⊕⊕_r=1^k-1/2(β_2r I_m_k(θ^ 2)⊕γ_2r I_m_k(θ^ 2))⊕δI_m_k(θ^ k+1))) : α,β_2r,γ_2r,δ∈}

⋂ {{x_1,x_1,x_2,x_2,x_2,x_2,x_1,x_1} : x_1,x_2∈Δ_4^(k)}= .

Now, observe that the intersection in <Ref> is determined by the following two intersections,
{Ad_V_k-1(αI_m_k(1)⊕⊕_r=1^k-1/2(β_2r I_m_k(θ^ 2)⊕γ_2r I_m_k(θ^ 2))⊕δI_m_k(θ^ k+1))) : α,β_2r,γ_2r,δ∈}

⋂  {x_1,x_1,x_2,x_2} ,

{Ad_V_k-1(α^'I_m_k(1)⊕⊕_r=1^k-1/2(β^'_2r I_m_k(θ^ 2)⊕γ^'_2r I_m_k(θ^ 2))⊕δ^'I_m_k(θ^ k+1))) : α^',β^'_2r,γ^'_2r,δ^'∈}

⋂ {x_2,x_2,x_1,x_1} .

Consider arbitrary element $\xi(\alpha,\beta,\gamma,\delta)$ in <Ref>

\[
\xi(\alpha,\beta,\gamma,\delta):=\mathrm{Ad}_{V_{k-1}}\Big(\alpha I_{m_k(1)}\bigoplus\bigoplus_{r=1}^{\frac{k-1}{2}}\Big(\beta_{2r}I_{m_k(\theta^{\,2r})}\oplus\gamma_{2r}I_{m_k(\overline{\theta}^{\,2r})}\Big)\bigoplus\delta I_{m_k(\theta^{\,k+1})}\Big)\,,
\]

and similarly $\xi(\alpha^\prime,\beta^\prime,\gamma^\prime,\delta^\prime)$ in <Ref>. Then $\xi(\alpha,\beta,\gamma,\delta)$ (similarly $\xi(\alpha^\prime,\beta^\prime,\gamma^\prime,\delta^\prime)$) is of the form described in <Ref> in Step $1$. Therefore, we get the following six equations,
x_1    =    αI_m_k-1(1)⊕⊕_r=1^k-1/2(β_2r I_m_k-1(θ^ 2)⊕γ_2r I_m_k-1(θ^ 2))⊕δI_m_k-1(θ^ k+1)

x_1    =    α^'I_m_k-1(1)⊕⊕_r=1^k-1/2(γ^'_2r I_m_k-1(θ^ 2)⊕β^'_2r I_m_k-1(θ^ 2))⊕δ^'I_m_k-1(θ^ k+1)

x_1    =    (β^'_2 I_m_k-1(1)⊕α^'I_m_k-1(θ^ 2))⊕⊕_r=1^k-3/2(β^'_2r+2I_m_k-1(θ^ 2r)⊕γ^'_2rI_m_k-1(θ^ 2r+2))

⊕(δ^'I_m_k-1(θ^ k+1)⊕γ^'_k-1I_m_k-1(θ^ k+1)) ,

x_2    =    αI_m_k-1(1)⊕⊕_r=1^k-1/2(γ_2r I_m_k-1(θ^ 2)⊕β_2r I_m_k-1(θ^ 2))⊕δI_m_k-1(θ^ k+1)

x_2    =    (β_2 I_m_k-1(1)⊕αI_m_k-1(θ^ 2))⊕⊕_r=1^k-3/2(β_2r+2I_m_k-1(θ^ 2r)⊕γ_2rI_m_k-1(θ^ 2r+2))

⊕(δI_m_k-1(θ^ k+1)⊕γ_k-1I_m_k-1(θ^ k+1))

x_2    =    α^'I_m_k-1(1)⊕⊕_r=1^k-1/2(β^'_2r I_m_k-1(θ^ 2)⊕γ^'_2r I_m_k-1(θ^ 2))⊕δ^'I_m_k-1(θ^ k+1) .

If the associated `matrix order' with each complex number is understood and creates no confusion, writing in terms of tuple, these six equations are the following set of equalities,
x_1    =    (α,β_2,γ_2,β_4,γ_4,…,β_k-1,γ_k-1,δ)

=    (α^',γ^'_2,β^'_2,γ^'_4,β^'_4,…,γ^'_k-1,β^'_k-1,δ^')

=    (β_2^',α^',β^'_4,γ^'_2,β^'_6,γ^'_4,…,β^'_k-1,γ^'_k-3,δ^',γ^'_k-1) ,

x_2    =    (α,γ_2,β_2,γ_4,β_4,…,γ_k-1,β_k-1,δ)

=    (β_2,α,β_4,γ_2,β_6,γ_4,…,β_k-1,γ_k-3,δ,γ_k-1)

=    (α^',β^'_2,γ^'_2,β^'_4,γ^'_4,…,β^'_k-1,γ^'_k-1,δ^') .

Immediately we see that $\alpha=\alpha^\prime$ and $\delta=\delta^\prime$. From the second and third equations associated with $x_1$, we get the following,

\[
\alpha^\prime=\beta^\prime_{2r}=\beta^\prime_{2r+2}=\delta^\prime\,\,\mbox{ and }\,\,\alpha^\prime=\gamma_{2r}^\prime=\gamma^\prime_{2r+2}=\delta^\prime
\]

for $1\leq r\leq\frac{k-3}{2}$. These two equations together with $\alpha=\alpha^\prime$ and $\delta=\delta^\prime$ shows that $x_1=\alpha(1,\ldots,1)$. Moreover, by the third equation associated with $x_2$, we see that $x_2=\alpha(1,\ldots,1)$. Hence, we finally get that $x_1=x_2=\alpha(1,\ldots,1)$, that is, $\xi(\alpha,\beta,\gamma,\delta)=\xi(\alpha^\prime,\beta^\prime,\gamma^\prime,\delta^\prime)=\alpha 1$. This finally establishes the claim in <Ref>, and finishes this step.

Step 4:  The Ocneanu compactness.

This is the final step of the proof. First, briefly recall the construction of $R^{(m)}_{u,v}$ from <Ref>. The following quadruple

\[
\begin{matrix}
A_{2k-1} &\subset & A_{2k+1}\\
|| &  & ||\\
\bbc\otimes\Delta_4\otimes M_4^{(k)} & & \Delta_4\otimes M_4\otimes M_4^{(k)}\\
\cup &  & \cup\\
\bbc\otimes\mbox{Ad}_{u_{2k-1}}\big(\bbc\otimes\big((u_{2k-2}^*v_{2k-2})^2\big)^\prime\cap M_4^{(k)}\big) & & \mbox{Ad}_{u_{2k+1}}\big(\bbc\otimes\big((u_{2k}^*v_{2k})^2\big)^\prime\cap M_4^{(k+1)}\big)\\
|| &  & ||\\
C_{2k-1} &\subset & C_{2k+1}
\end{matrix}
\]

becomes a non-degenerate commuting square, and we have started iterating the basic construction from this stage. If $D_{2k+j}$, for $j\geq 3$ odd integer, denote the finite-dimensional subalgebras of $C_{2k+j}$ obtained through the basic construction, then $R^{(m)}_{u,v}$ is constructed as the SOT-limit of $D_{2k+j}$, whereas $R$ is the SOT-limit of $A_{2k+j}$. Note that we have used the convention that $D_{2k+i}=C_{2k+i}$ for $i=\pm 1$. Since for $j\geq -1$ odd integer, we have $(D_{2k+j}\subset A_{2k+j},D_{2k+j+2}\subset A_{2k+j+2})$ is a tower of symmetric commuting squares with limit $R^{(m)}_{u,v}\subset R$, by the Ocneanu compactness we have the following,

\[
\big(R^{(m)}_{u,v}\big)^{\,\prime}\cap R=D_{2k+1}^{\,\prime}\cap A_{2k-1}=C_{2k+1}^{\,\prime}\cap A_{2k-1}\,.
\]

Now, by the Step $3$ we get the irreducibility of $R^{(m)}_{u,v}\subset R$ when $k$ is odd.

Finally, the exact similar analysis holds when $k$ is even in Step $1$, which completes the proof.

For the pair of spin model subfactors $R_{u(a)}, R_{u(b)}\subset R$ such that $u(a)\sim u(b)$, we have $R_{u(a)}\cap R_{u(b)}$ is a factor and $R_{u(a)}\cap R_{u(b)}=R^{(m)}_{u(a),u(b)}$ if $\,(\overline{a}b)^{\,m}=1$.

Let $u=u(a)\mbox{ and }v=u(b)$ be $4\times 4$ Hadamard inequivalent complex Hadamard matrices such that $u\sim v$. If $\theta=\overline{a}b$ satisfies $\,\theta^{\,m}=1$, we take $k=(m-2)/2$. By <Ref>, we immediately see that $R_u\cap R_v$ is a factor since $R^{(m)}_{u,v}\subset R_u\cap R_v$ and $R^{(m)}_{u,v}\subset R$ is irreducible. For the identification of $R_u\cap R_v$ with $R^{(m)}_{u,v}$, it is enough to prove that $[R_u:R_u\cap R_v]=[R_u:R^{(m)}_{u,v}]$. Note that $[R_u:R^{(m)}_{u,v}]=\frac{m}{2}$.

We claim that $[R_u:R_u\cap R_v]=\lambda^{-1}(R_u,R_u\cap R_v)=\frac{m}{2}$. The first equality is obvious by [37], since $R_u\cap R_v$ is a factor. To prove the second equality, recall that $R_u$ is the SOT-limit of $B^u_{2j+1}$ and $R_u\cap R_v$ is that of $C_{2j+1}$. Given $\theta^{\,m}=1$, if we take $k=(m-2)/2$, then $\sigma\big((u_{2k-2}^*v_{2k-2})^2\big)=\sigma\big((u_{2k+2j}^*v_{2k+2j})^2\big)$ for all $j\in\bbn\cup\{0\}$, and cardinality of the spectrum is $k+1$ (see <Ref>). Since $(u_{2k+2j}^*v_{2k+2j})^2$ is diagonalizable for all $j\in\bbn\cup\{-1\}$, the sum of the multiplicities of the eigenvalues of $(u_{2k+2j}^*v_{2k+2j})^2$ will add up to the order of $M_4^{(k+j+1)}$ for each $j\in\bbn\cup\{-1\}$, which is equal to $4^{k+j+1}$. Note that the eigenvalues remain the same, but the multiplicities are changing with $j$. Since $C_{2k+2j+1}$ is given by the algebra $\big((u_{2k+2j}^*v_{2k+2j})^2\big)^\prime\cap M_4^{(k+j+1)}$ (<Ref>), we see that at each stage of the inclusion $C_{2k+2j+1}\subset B^u_{2k+2j+1}$, we have $(k+1)$-many Wedderburn-Artin direct summands in $M_{4^{k+j+1}}$, and order of the summands add up to $4^{k+j+1}$.

Now, each quadruple $(C_{2r+1}\subset B^u_{2r+1},C_{2r+3}\subset B^u_{2r+3})$ is a commuting square for $r\in\bbn\cup\{0\}$, which follows as an application of <Ref>. By <Ref> and <Ref>, we see that $\lambda(R_u,R_u\cap R_v)$ is the limit of the eventually constant sequence $\{(k+1)^{-1}\}$. Therefore, $\lambda(R_u,R_u\cap R_v)^{-1}=k+1=\frac{m}{2}$, which proves the claim. Thus, we get that $\lambda(R_u,R_u\cap R_v)^{-1}=[R_u:R_u\cap R_v]=\frac{m}{2}=[R_u:R^{(m)}_{u,v}]$. By the multiplicativity of the Jones' index, we immediately get that $R^{(m)}_{u,v}=R_u\cap R_v$.

We remark that using the above theorem, it takes only a little more effort to show that in the following tower of finite-dimensional algebras

\[
C_1\subset C_3\subset C_5\subset\ldots\subset C_{2k-1}\subset C_{2k+1}\subset\ldots\subset R_u\cap R_v
\]

if $\theta^{\,m}=1$, then $C_{2k-1}\subset C_{2k+1}\subset\ldots\subset R_u\cap R_v$ is a tower of basic construction, where $k=(m-2)/2$. To see this, fix a $m$ and let $k=(m-2)/2$. Recall that the subfactor $R^{(m)}_{u,v}\subset R$ is the limit of the tower of symmetric commuting squares $(D_{2k+j}\subset A_{2k+j},D_{2k+j+2}\subset A_{2k+j+2})$, for $j\geq -1$ odd integer, where $D_{2k+j}=C_{2k+j}$ for $j=\pm 1$ and $D_{2k+j}\subset C_{2k+j}$ for $j\geq 3$. Since for $j\geq 3,\,(D_{2k+j}\subset A_{2k+j},D_{2k+j+2}\subset A_{2k+j+2})$ and $(C_{2k+j}\subset A_{2k+j},C_{2k+j+2}\subset A_{2k+j+2})$ both are commuting squares in the following diagram

\[
\begin{matrix}
\subset & A_{2k+3} &\subset & A_{2k+5} &\subset & A_{2k+7} &\subset &\cdots &\subset & R\\
& \cup & & \cup & & \cup & & & & \cup\\
\subset & C_{2k+3} &\subset & C_{2k+5} &\subset & C_{2k+7} &\subset &\cdots &\subset & R_u\cap R_v\\
& || & & \cup & & \cup & & & & \cup\\
& D_{2k+3} &\subset  & D_{2k+5} &\subset & D_{2k+7} &\subset &\cdots &\subset & R^{(m)}_{u,v}\\
\end{matrix}
\]

we have that $(D_{2k+j}\subset C_{2k+j},D_{2k+j+2}\subset C_{2k+j+2})$ is also a commuting square for any $j\geq 3$ odd integer. This follows from the following,

\[
E_{D_{2k+j+2}}^{C_{2k+j+2}}(C_{2k+j})=E_{D_{2k+j+2}}^{B^u_{2k+j+2}}(C_{2k+j})\subset E_{D_{2k+j+2}}^{B^u_{2k+j+2}}(B^u_{2k+j})\subset D_{2k+j}\,.
\]

Now, by Proposition $2.6$ in [37] and <Ref>, we get that $1=\lambda(R_u\cap R_v,R^{(m)}_{u,v})=\lim_{j\to\infty}\lambda(C_{2k+j},D_{2k+j})$ decreasingly. Since $\lambda$ is always less than or equal to $1$, we get that $\lambda(C_{2k+j},D_{2k+j})=1$ for all $j\geq 3$ odd integer. This says that $C_{2k+j}\subseteq D_{2k+j}$ for all $j$ by <Ref>, and since the reverse inclusion is already there, we finally get that $C_{2k+j}=D_{2k+j}$ for all $j\geq 3$ odd integer. Since the tower $D_{2k+3}\subset D_{2k+7}\subset\cdots$ is the iterated basic construction for the inclusion $C_{2k-1}\subset C_{2k+1}$, our claim follows.

For $a,b,c,d\in\mathbb{S}^1$, if $\big(\overline{a}b\big)^m=\big(\overline{c}d\big)^n=1$ and $m\neq n$, then the corresponding subfactors $R_{u(a)}\cap R_{u(b)}\subset R$ and $R_{u(c)}\cap R_{u(d)}\subset R$ are non-isomorphic.

For Hadamard inequivalent complex Hadamard matrices $u\mbox{ and }v$ such that $u\sim v$, the quadruple of $II_1$ factors $(R_u\cap R_v\subset R_u,R_v\subset R)$ is obtained as an iterated basic construction of a non-degenerate commuting cube.

Follows from <Ref> and <ref>, and the discussion in <Ref>.

§.§.§ The interior, exterior angles and its rigidity

We show that the (interior) angle between $R_u$ and $R_v$ in the sense of [2] (see <Ref>) has certain rigidity when $u\sim v$. Recall from <Ref> that if $u\sim v$ and $\theta^{\,m}=1$ for some $m\in 2\mathbb{N}+2$, where $u=u(a)$ and $v=u(b)$ with $\theta=\overline{a}b$, then we have $R^{(m)}_{u,v}=R_u\cap R_v$. Therefore, we get  a family of quadruples of $II_1$ factors

\[
\begin{matrix}
R_u & \subset  & R \cr
\cup &  &\cup  \cr
R_u\cap R_v& \subset & R_v&
\end{matrix}
\]

such that $[R: R_u\cap R_v]=2m$ and $[R_u:R_u\cap R_v]=[R_v:R_u\cap R_v]=m/2$. Recall that given any quadruple of finite index subfactors

\[ \begin{matrix}
P & \subset  & M \cr
\cup &  &\cup  \cr
N & \subset & Q&
\end{matrix}
\]

the interior angle $\alpha^N_M(P,Q)$ and the exterior angle $\beta^N_M(P,Q)$ have the following formulae (see [2]):

\begin{align*}
\alpha^N_M(P,Q) &=\displaystyle \frac{[M:N]\,tr(e_Pe_Q)-1}{\sqrt{[P:N]-1}\sqrt{[Q:N]-1}}\,,\\
\beta^N_M(P,Q) &=\displaystyle \frac{tr(e_Pe_Q)-tr(e_{P}) tr(e_{Q})}{\sqrt{tr(e_P)-{tr(e_P)}^2} \sqrt{tr(e_Q)-{tr(e_Q)}^2}}\,.
\end{align*}

We also know that $\lambda(P,Q)=\displaystyle\frac{tr(e_Pe_Q)}{tr(e_P)}$ if $N\subset M$ is irreducible (Theorem $3.3$ in [4]). Since in our case, $R^{(m)}_{u,v}\subset R$ is irreducible by <Ref>, and $\lambda(R_u,R_v)=1/2$ by <Ref>, we have $tr(e_{R_u}e_{R_v})=1/8$. Thus, by the above formulae of angles we see that $\cos \beta (R_u,R_v)= 1/3$ and $\cos \alpha(R_u,R_v)=\frac{m-4}{2(m-2)}<1/2$. Therefore, we conclude that $\alpha(R_u,R_v)>\pi/3.$ We summarize these findings as the following theorem.

The interior angle $\alpha \big(R_u,R_v\big)$  is strictly  greater than $\pi/3$ and the exterior angle $\beta \big(R_u,R_v\big)=\arccos 1/3$. Furthermore, $\alpha \big(R_u,R_v\big)$ converges to $\pi/3$ decreasingly as $m$ tends to infinity.

As an immediate corollary, using <Ref> we obtain the following.

For any $4\times 4$ Hadamard inequivalent complex Hadamard matrices $u\mbox{ and }v$ such that $u\sim v$, the following quadruple of subfactors

\[
\begin{matrix}
R_u & \subset  & R \cr
\cup &  &\cup  \cr
R_u\cap R_v& \subset & R_v&
\end{matrix}
\]

is a commuting square if and only if $m=4$, that is, $b=\pm ia$. Moreover, the above quadruple is never a co-commuting square.

§.§ The Sano-Watatani angle and Connes-Størmer relative entropy

For inequivalent complex Hadamard matrices $u\mbox{ and }v$, we compute the exact value of relative entropy $H(R_u|R_v)$ (along with its variant $h(R_u|R_v)$) and the Sano-Watatani angle $\mathrm{Ang}_R(R_u,R_v)$ for the case of $u\sim v$. When $u\nsim v$, we give a legitimate upper and lower bounds for $H(R_u|R_v)$. As a crucial tool, we have shown the existence of a proper subfactor of $R$ containing both $R_u$ and $R_v$.

§.§.§ The subfactor $R_u\vee R_v\subset R$

This subsection is devoted to  prove the existence of an index $2$ subfactor of $R$ containing both $R_u\mbox{ and }R_v$. Recall that we have pair of irreducible subfactors $R_u$ and $R_v$ of $R$ each with index $4$. Consider the von Neumann subalgebra $R_u\vee R_v\subset R$. By the irreducibility of $R_u\subset R$ we get that $R_u\vee R_v\subset R$ is a factor. We show that $R_u\vee R_v$ is a proper subfactor $R$. We begin with the following notations which are at par with that in the beginning of <Ref>. For $k\in\bbn\cup\{0\}$, let

\begin{align*}
A_{2k} &= M_4\otimes M_4^{(k)}\,,\cr
B^u_{2k} &= u_{2k}(\Delta_4\otimes M_4^{(k)})u_{2k}^*\,,\cr
B^v_{2k} &= v_{2k}(\Delta_4\otimes M_4^{(k)})v_{2k}^*\,,\cr
\mathcal{I}_{2k} &= B^u_{2k}\vee B^v_{2k}\,.
\end{align*}

Note that these are the even steps in the tower of basic constructions described in <Ref>.

For all $k\in\bbn\cup\{0\},\,\mathcal{I}_{2k}$ is a proper subalgebra of $A_{2k}$.

Since each $A_{2k}$ is factor, it is enough to show that $\bbc\oplus\bbc\subseteq\mathcal{I}_{2k}^{\,\prime}\cap A_{2k}$ for any $k\in\bbn\cup\{0\}$. First observe that
(B^u_2k)^'∩A_2k    =    _u_2k((Δ_4⊗M_4^(k))^'∩_u_2k^*(A_2k))

=    _u_2k((Δ_4⊗M_4^(k))^'∩(M_4⊗M_4^(k)))    (u_2k∈A_2k=M_4^(k+1))

=    _u_2k(Δ_4⊗^(k)) ,

and similarly, $(B^v_{2k})^\prime\cap A_{2k}=\mbox{Ad}_{v_{2k}}(\Delta_4\otimes\bbc^{(k)})$. Now, $\mathcal{I}_{2k}^{\,\prime}\cap A_{2k}=\{B^u_{2k}\cup B^v_{2k}\}^\prime\cap A_{2k}=(B^u_{2k})^\prime\cap(B^v_{2k})^\prime\cap A_{2k}$. Hence, by <Ref> we need to show the following,

or in other words,