$\mathrm{supp}{\mathrm{T}_{3,-}^{(\mathrm{nc})}}\subseteq\mathrm{supp}{\left(\mathrm{id}-\mathcal{W}_{[(12)]}\right)}$.
In the proof of Lemma 21, we establish that
$\mathcal{W}_{\mathrm{CS}}\in\subset\mathrm{supp}{\mathcal{P}_{\mathrm{S}}}$.
Also
$\mathcal{W}_{\mathrm{CS}}\subset\mathrm{supp}{\left(\mathrm{id}+\mathcal{W}_{{(23)}}\right)}$
using the form of $|w_{\mathrm{tr}}\rangle$ in equation (235). Thus, in fact
$\mathcal{P}_{\mathrm{CS}}\
\mbox{$|w_{\mathrm{tr}}\rangle$}=\mbox{$|w_{\mathrm{tr}}\rangle$}$, using the
form of $\mathcal{P}_{\mathrm{CS}}$ in equation (129). From equation (129), we
also see that $\mathrm{supp}{\left(\mathrm{id}-\mathcal{W}_{\tau}\right)}$ is
orthogonal to $\mathrm{supp}{\mathcal{P}_{\mathrm{CS}}}$ for all
transpositions $\tau\in\mathrm{S}_{3}$. ∎

######  Lemma 19.

The largest singular value of $\mathrm{T}_{1,-}^{(\mathrm{nc})}$ is
$\dfrac{1}{2N}$.

###### Proof.

For $\left\langle a,b\right\rangle=1$, define

$\mbox{$|\bar{a};b\rangle$}\ :=\ \dfrac{\sqrt{2}}{N}\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=1\end{subarray}}\
\left(-1\right)^{\left\langle b,h\right\rangle}\ \mbox{$|a,h,a+h\rangle$}.$
(219)

Firstly, note that
$\mbox{$|\bar{a};b\rangle$}\in\mathrm{supp}{\mathrm{T}_{1,-}^{(\mathrm{nc})}}$,
since $\mbox{$\langle a,h,a+h|\bar{a};{b}\rangle$}=\mbox{$\langle
a,a+h,h|\bar{a};{b}\rangle$}$ for all $h\in{\mathbb{F}_{2}}_{*}^{2m}$. Next,
note that for any $a,b$ such that $\left\langle a,b\right\rangle=1$,
$\mbox{$|\bar{a};b\rangle$}=-\mbox{$|\bar{a};a+b\rangle$}$. Hence at most
$\frac{\left(N^{2}-1\right)\left(N^{2}\right)}{2}$ of the $|\bar{a};b\rangle$
can be linearly dependent. Next, for a given $a,b$ and $c,d$ such that
$\left\langle a,b\right\rangle=\left\langle c,d\right\rangle=1$, we see that

$\mbox{$\langle\bar{a};b|\bar{c};d\rangle$}\ =\ \delta_{a,c}\ \left(\
\delta_{b,c}-\delta_{b+c,a}\right),$ (220)

such that one can obtain an orthonormal basis of
$\frac{\left(N^{2}-1\right)\left(N^{2}\right)}{2}$ vectors from the
$|\bar{a};b\rangle$, which then spans
$\mathrm{supp}{\left(\mathrm{id}-\mathcal{W}_{[(23)]}\right)}$. We show how to
construct such a basis: note that the subset $\left\\{0,a\right\\}$ is an
abelian subgroup of $\mathbb{F}_{2}^{2m}$, and has
$\frac{\left(N^{2}-1\right)\left(N^{2}\right)}{4}$ cosets in
$\mathbb{F}_{2}^{2m}$. Label the coset from $1$ to
$\frac{\left(N^{2}-1\right)\left(N^{2}\right)}{4}$, and choose coset
representatives $b_{j}$. Then the subset
$\left\\{\mbox{$|\bar{a};b_{j}\rangle$}\right\\}$ is an orthonormal basis for
$\mathrm{supp}{\left(\mathrm{id}-\mathcal{W}_{[(23)]}\right)}$. Next, one sees
that

$\mathrm{T}_{1,-}^{(\mathrm{nc})}\ \dfrac{1}{\sqrt{2}}\ \left(\
\mbox{$|a,b_{j},a+b_{j}\rangle$}\ -\mbox{$|a,a+b_{j},b_{j}\rangle$}\right)\ =\
-\dfrac{1}{2N}\ \mbox{$|\bar{a};b_{j}\rangle$}.$ (221)

Thus $\mathrm{T}_{1,-}^{(\mathrm{nc})}$ takes us from one orthonormal basis
for $\mathrm{supp}{\left(\mathrm{id}-\mathcal{W}_{[(23)]}\right)}$ to another
orthonormal basis for
$\mathrm{supp}{\left(\mathrm{id}-\mathcal{W}_{[(23)]}\right)}$, and equation
(221) is essentially the singular-value decomposition of
$\mathrm{T}_{1,-}^{(\mathrm{nc})}$, with the singular values $\frac{1}{2N}$. ∎

######  Lemma 20.

The spectral decomposition of $\mathrm{T}_{1,+}^{(\mathrm{nc})}$ is

$\mathrm{T}_{1,+}^{(\mathrm{nc})}\ =\ \dfrac{1}{4}\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\
\mbox{$|\hat{a}\rangle\\!\langle\hat{a}|$},$ (222)

where for any $a,b\in{\mathbb{F}_{2}}_{*}^{2m}$,

$\displaystyle\mbox{$|\hat{a}\rangle$}\ :=\ \dfrac{\sqrt{2}}{N}\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=1\end{subarray}}\
\ \mbox{$|a,h,a+h\rangle$},$ (223) $\displaystyle\
\mbox{$\langle\hat{a}|\hat{b}\rangle$}\ =\ \delta_{a,b}.$ (224)

###### Proof.

From equation (184), we note that
$\mathrm{supp}{\mathrm{T}_{1,+}^{(\mathrm{nc})}}\subset\mathrm{supp}{\left(\mathrm{id}+\mathcal{W}_{[(23)]}\right)}$.
Define the following vectors in
$\mathrm{supp}{\left(\mathrm{id}+\mathcal{W}_{[(23)]}\right)}$ for any
$a\in{\mathbb{F}_{2}}_{*}^{2m}$ and $b\in\mathbb{F}_{2}^{2m}$, and such that
$\left\langle a,b\right\rangle=0$.

$\displaystyle\mbox{$|\hat{a};{b}\rangle$}\ :=\ \dfrac{\sqrt{2}}{N}\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=1\end{subarray}}\
(-1)^{\left\langle b,h\right\rangle}\ \mbox{$|a,h,a+h\rangle$}.$ (225)

Note that $|\hat{a}\rangle$ really is just $|\hat{a};{0}\rangle$, if we
allowed the second register to take the value $0$ (which we temporarily allow
throughout the body of this proof). This differentiation between both types of
vectors will prove to be convenient. That
$\mbox{$|\hat{a};{b}\rangle$}\in\mathrm{supp}{\left(\mathrm{id}+\mathcal{W}_{[(23)]}\right)}$
is proved next:

$\displaystyle\mathcal{W}_{[(23)]}\ \mbox{$|\hat{a};{b}\rangle$}\ \propto$
$\displaystyle\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\end{subarray}}\ (-1)^{\left\langle b,h\right\rangle}\
\mbox{$|a,a+h,h\rangle$}.$ $\displaystyle=$ $\displaystyle\
\sum_{\begin{subarray}{c}h:\\\ \left\langle
a,a+h\right\rangle=1\end{subarray}}\ (-1)^{\left\langle b,a+h\right\rangle}\
\mbox{$|a,a+h,h\rangle$}$ $\displaystyle=$ $\displaystyle\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=1\end{subarray}}\
(-1)^{\left\langle b,h\right\rangle}\ \mbox{$|a,h,a+h\rangle$}.$ (226)

The second line uses $\left\langle a,b\right\rangle=0$, and to go from the
second line to the third, define $h^{\prime}=a+h$, and note that $\left\langle
a,h\right\rangle=1\Leftrightarrow\left\langle a,h^{\prime}\right\rangle=1$.
Since $h^{\prime}$ is a dummy variable, the prime symbol is then removed.
Invariance is established by noting that the RHS of equations (B.1.4) and
(225) are proportional to one another. Next, for
$a,c\in{\mathbb{F}_{2}}_{*}^{2m}$ and $b,d\in\mathbb{F}_{2}^{2m}$ such that
$\left\langle a,b\right\rangle=\left\langle c,d\right\rangle=0$, we get

$\langle\hat{a};b|\hat{c};{d}\rangle$ $\displaystyle=$ $\displaystyle\
\dfrac{2}{N^{2}}\ \delta_{a,c}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\end{subarray}}\ \left(-1\right)^{\left\langle
h,b+d\right\rangle}$ $\displaystyle=$ $\displaystyle\ \delta_{a,c}\
\left(\left(\delta_{b,d}-\delta_{b+d,a}\right)+\frac{2}{N^{2}}\
\left(1-\delta_{b,d}+\delta_{b+d,a}\right)\left(\ \mathrm{S}_{1}-\
\mathrm{S}_{2}\ \right)\right)$ $\displaystyle=$ $\displaystyle\ \delta_{a,c}\
\left(\delta_{b,d}\ -\delta_{b+d,a}\right),$ (227)

where

$\mathrm{S}_{1}\ :=\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle b+d,h\right\rangle=0\end{subarray}}\ 1,\ \
\ \ \mathrm{S}_{2}\ :=\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle b+d,h\right\rangle=1\end{subarray}}\ 1.$
(228)

One can use Lemma 2 to show that when $a$ and $b+d$ are linearly independent
$\mathrm{S}_{1}\ -\mathrm{S}_{2}=0$.

Equation (B.1.4) informs us that

$\mbox{$|\hat{a};{b}\rangle$}\ =\ \mbox{$|\hat{a};{a+b}\rangle$},$ (229)

which is also seen by directly plugging in $a+b$ instead of $b$ in equation
(225). Thus the vectors $|\hat{a};{b}\rangle$, upto the identification between
$|\hat{a};{b}\rangle$ and $|\hat{a};{a+b}\rangle$, form an orthonormal basis
for $\mathrm{supp}{\left(\mathrm{id}+\mathcal{W}_{[(23)]}\right)}$. Next, we
show that they’re also the eigenvectors of $\mathrm{T}_{1,+}^{(\mathrm{nc})}$
Note that $\mathrm{T}_{1,+}^{(\mathrm{nc})}\
\mbox{$|\hat{a};{b}\rangle$}=\mathrm{T}_{1}^{(\mathrm{nc})}\
\mbox{$|\hat{a};{b}\rangle$}$.

$\displaystyle\ \frac{N}{\sqrt{2}}\times\ \mathrm{T}_{1,+}^{(\mathrm{nc})}\
\mbox{$|\hat{a};{b}\rangle$}$ $\displaystyle=$ $\displaystyle\ \left(\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=1\end{subarray}}\
(-1)^{\left\langle b,h\right\rangle}\
\mathrm{T}_{1,+}^{(\mathrm{nc})}\mbox{$|a,h,a+h\rangle$}\ \right).$
$\displaystyle=$ $\displaystyle\ \left(\ \sum_{\begin{subarray}{c}h:\\\
\left\langle a,h\right\rangle=1\end{subarray}}\ (-1)^{\left\langle
b,h\right\rangle}\ \left(\ \dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}k:\\\
\left\langle a,k\right\rangle=1\\\ \left\langle
h,k\right\rangle=1\end{subarray}}\ \mbox{$|a,k,a+k\rangle$}\ \right)\
\right).$ $\displaystyle=$ $\displaystyle\ \dfrac{1}{N^{2}}\ \
\sum_{\begin{subarray}{c}k:\\\ \left\langle
a,k\right\rangle=1\end{subarray}}\left(\mathrm{S}_{1}\ -\mathrm{S}_{2}\right)\
\mbox{$|a,k,a+k\rangle$},$ (230)

where we define

$\ \mathrm{S}_{1}\ :=\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
h,a\right\rangle=1\\\ \left\langle h,k\right\rangle=1\\\ \left\langle
h,b\right\rangle=0\end{subarray}}\ 1,\;\;\;\;\mathrm{S}_{2}\ :=\
\sum_{\begin{subarray}{c}h:\\\ \left\langle h,a\right\rangle=1\\\ \left\langle
h,k\right\rangle=1\\\ \left\langle h,b\right\rangle=1\end{subarray}}\ 1.$
(231)

First, let’s note that $a,k$ are always linearly independent since
$\left\langle a,k\right\rangle=1$. When $b=0$, then $\mathrm{S}_{2}=0$, and
Lemma 2 tell us that $S_{1}=\frac{N^{2}}{4}$.When $b=a$, $\mathrm{S}_{1}=0$
and $S_{2}=\frac{N^{2}}{4}$. For all other values of $b$, Lemma 2 tells us
that both sums equal and cancel out. When $b=0$, Lemma 2 tell us that the
first sum if $\frac{N^{2}}{4}$, whereas the second sum is $0$, while for $b=a$
the reverse is true. Hence we get.

$\displaystyle\ \mathrm{T}_{1,+}^{(\mathrm{nc})}\
\mbox{$|\hat{a};{b}\rangle$}\ =\ \dfrac{1}{4}\ \left(\delta_{b,0}\
+\delta_{b,a}\right)\mbox{$|\hat{a};{b}\rangle$}.$ (232)

The equation proves that $|\hat{a}\rangle$ are eigenvectors of
$\mathrm{T}_{1,+}^{(\mathrm{nc})}$ with eigenvalue $\frac{1}{4}$, and that
it’s orthogonal complement (which is spanned by $|\hat{a};{b}\rangle$’s for
$b\neq 0,a$) is the eigenspace corresponding to the $0$ eigenvalue. ∎

######  Lemma 21.

$\mathcal{P}_{\mathrm{S}}\ \mathrm{T}_{1}^{(\mathrm{nc})}\
\mathcal{P}_{\mathrm{S}}$ has three eigenvalues, which are as follows:
$\frac{3}{4}$, $\frac{1}{4}\left(1\pm\frac{2}{N}\right)$. All its eigenspaces
are orthogonal. The eigenspace corresponding to $\frac{3}{4}$ is
$\mathcal{W}_{\mathrm{CS}}$.

###### Proof.

We obtain the complete spectral decomposition of
$\mathrm{T}_{1}^{(\mathrm{nc})}$ in Lemma 20. Note that the kernel of
$\mathrm{T}_{1}^{(\mathrm{nc})}$ is contained in the kernel of
$\mathcal{P}_{\mathrm{S}}\mathrm{T}_{1}^{(\mathrm{nc})}\mathcal{P}_{\mathrm{S}}$.
Hence the support of
$\mathcal{P}_{\mathrm{S}}\mathrm{T}_{1}^{(\mathrm{nc})}\mathcal{P}_{\mathrm{S}}$
is spanned by $\mathcal{P}_{\mathrm{S}}\mbox{$|\hat{a}\rangle$}$, for all
$a\in{\mathbb{F}_{2}}_{*}^{2m}$ ($|\hat{a}\rangle$’s are defined in equation
(223)). Hence, using the spectral decomposition of
$\mathrm{T}_{1}^{(\mathrm{nc})}$, which is obtained in Lemma 20, we get

$\ \mathcal{P}_{\mathrm{S}}\ \mathrm{T}_{1,+}^{(\mathrm{nc})}\
\mathcal{P}_{\mathrm{S}}\ =\ \dfrac{1}{4}\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\
\mathcal{P}_{\mathrm{S}}\mbox{$|\hat{a}\rangle\\!\langle\hat{a}|$}\mathcal{P}_{\mathrm{S}}.$
(233)

From equation (233) one clearly sees that
$\mathcal{P}_{\mathrm{S}}\mathrm{T}_{1}^{(\mathrm{nc})}\mathcal{P}_{\mathrm{S}}$
is self-adjoint. Hence all its eigenspaces are mutually orthogonal. The inner
product among the vectors $\mathcal{P}_{\mathrm{S}}\mbox{$|\hat{a}\rangle$}$
is detailed in Section E in the Appendix.

$\displaystyle\
\mbox{$\langle\hat{a}|$}\mathcal{P}_{\mathrm{S}}\mbox{$|\hat{b}\rangle$}\ =\
\frac{1}{3}\ \left(\delta_{a,b}\ +\ \dfrac{4}{N^{2}}\ \delta_{\left\langle
a,b\right\rangle,1}\right).$ (234)

Hence $\mathcal{P}_{\mathrm{S}}\mbox{$|\hat{a}\rangle$}$ aren’t orthonormal.

We decompose $\mathrm{supp}{\left(\mathrm{id}+\mathcal{W}_{[(23)]}\right)}$ as
follows.

$\displaystyle\ \mathcal{W}_{\mathrm{CS}}\ =\ \mathrm{span}\ \left\\{\
\mbox{$|w_{\mathrm{tr}}\rangle$}\
=\frac{N}{\sqrt{2}}\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\
\mbox{$|\hat{a}\rangle$}\right\\},$ (235) $\displaystyle\ \mathcal{W}_{+}\ :=\
\mathrm{span}\ \left\\{\ \mbox{$|w_{+}\rangle$}\ =\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{a}\ \mbox{$|\hat{a}\rangle$},\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{a}\ =0,\
\sum_{\begin{subarray}{c}h:\\\ \left\langle h,a\right\rangle\end{subarray}=1}\
\lambda_{h}\ =\ \frac{N}{2}\ \lambda_{a}\right\\},$ (236) $\displaystyle\
\mathcal{W}_{-}\ :=\ \mathrm{span}\ \left\\{\ \mbox{$|w_{-}\rangle$}\ =\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{a}\ \mbox{$|\hat{a}\rangle$},\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{a}\ =0,\
\sum_{\begin{subarray}{c}h:\\\ \left\langle h,a\right\rangle\end{subarray}=1}\
\lambda_{h}\ =\ -\frac{N}{2}\ \lambda_{a}\right\\}.$ (237)

We note that $|w_{\mathrm{tr}}\rangle$ is the same as was defined in equation
(170). Note that the equations in the definition of $\mathcal{W}_{+}$ and
$\mathcal{W}_{-}$ are consistent and have solutions. We know this because the
same equations are used to decompose the irreducible representations in the
diagonal sector in Subsubsections B.1.2 and B.1.3, and they were also
introduced in the diagonal sector in Helsen et al Helsen2016 . We also know
that $\dim\mathcal{W}_{+}=\frac{N(N-1)}{2}-1$ and
$\dim\mathcal{W}_{-}=\frac{N(N+1)}{2}-1$. And that the three spaces
$\mathcal{W}_{\mathrm{CS}}$, $\mathcal{W}_{+}$ and $\mathcal{W}_{-}$ are
orthogonal to each other, collectively spanning the entire $N^{2}-1$
dimensional support of $\mathrm{id}+\mathcal{W}_{[(23)]}$. Using
$\mbox{$|w\rangle$}_{\mathrm{tr}}$ defined in equation (170), one can directly
verify that
$\mathcal{P}_{\mathrm{S}}\mbox{$|w_{\mathrm{tr}}\rangle$}=\mbox{$|w_{\mathrm{tr}}\rangle$}$666$\mathcal{W}_{\mathrm{CS}}$
is the sum of all basis vectors $|a,b,a+b\rangle$ with $\left\langle
a,b\right\rangle=1$, and hence has to invariant under
$\mathcal{P}_{\mathrm{S}}$.. Also we have that

$\mathcal{P}_{\mathrm{S}}\ \mathrm{T}_{1,+}^{(\mathrm{nc})}\
\mathcal{P}_{\mathrm{S}}\mbox{$|w_{\mathrm{tr}}\rangle$}\ =\ {\small\text{
$\frac{N}{\sqrt{2}}\times\ $}}\mathcal{P}_{\mathrm{S}}\
\left(\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \mathrm{T}_{1,+}^{(\mathrm{nc})}\
\mbox{$|\hat{a}\rangle$}\right)\
=\frac{1}{4}\mbox{$|w_{\mathrm{tr}}\rangle$},$ (238)

where we used Lemma (20). Next, for $\mathcal{P}_{\mathrm{S}}\
\mbox{$|w_{\pm}\rangle$}\in\mathcal{W}_{\pm}$

$\displaystyle\ \mathcal{P}_{\mathrm{S}}\
\mathrm{T}_{1,+}^{(\mathrm{nc})}\mathcal{P}_{\mathrm{S}}\ \
\mathcal{P}_{\mathrm{S}}\mbox{$|w_{\pm}\rangle$}\ $ $\displaystyle=$
$\displaystyle\ \mathcal{P}_{\mathrm{S}}\ \times\ \frac{1}{4}\
\sum_{a,b\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{b}\
\mbox{$\langle\hat{a}|$}\mathcal{P}_{\mathrm{S}}\mbox{$|\hat{b}\rangle$}\ \
\mbox{$|\hat{a}\rangle$}$ $\displaystyle=$ $\displaystyle\ \frac{1}{4}\
\sum_{a,b\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{b}\
\left(\frac{1}{3}\left(\delta_{a,b}+\ \dfrac{4}{N^{2}}\ \delta_{\left\langle
a,b\right\rangle,1}\right)\right)\
\mbox{$|\hat{a}\rangle$},\,\,\left(\mathrm{using}\ \mathrm{equation}\
\mathrm{\eqref{eq:inner_a_ps_b}}\right)$ $\displaystyle=$ $\displaystyle\
\frac{1}{12}\left(1\pm\dfrac{2}{N}\right)\ \mathcal{P}_{\mathrm{S}}\
\mbox{$|w_{\pm}\rangle$},$ (239)

Thus the remaining eigenvalues of
$\mathcal{P}_{\mathrm{S}}\mathrm{T}_{1}^{(\mathrm{nc})}\mathcal{P}_{\mathrm{S}}$
are $\frac{1}{12}\left(1\pm\frac{2}{N}\right)$, and the corresponding
eigenspaces are given by $\mathcal{P}_{\mathrm{S}}\mathcal{W}_{\pm}$. That
$|w_{tr}\rangle$ is orthogonal to
$\mathcal{P}_{\mathrm{S}}\mbox{$|w_{\pm}\rangle$}$ seen from the fact that
$\mbox{$\langle
w_{\mathrm{tr}}|$}\mathcal{P}_{\mathrm{S}}\mbox{$|w_{\pm}\rangle$}=\mbox{$\langle
w_{\mathrm{tr}}|w_{\pm}\rangle$}=0$. ∎

######  Remark 9.

The eigenspace $\mathcal{W}_{\mathrm{CS}}$ corresponds to the uniform sum of
all basis vector $|a,b,a+b\rangle$ (where $\left\langle a,b\right\rangle=1$).
The same eigenspace is also identified in Tan20  (Section 5.3).

######  Lemma 22.

The eigenvalues of
$\mathcal{P}_{\omega}\mathrm{T}_{1}^{(\mathrm{nc})}\mathcal{P}_{\omega}$ and
$\mathcal{P}_{\omega^{*}}\mathrm{T}_{1}^{(\mathrm{nc})}\mathcal{P}_{\omega^{*}}$
are $\frac{1}{12}\left(1\pm\frac{1}{N}\right)$ and $0$.

###### Proof.

We prove this following the same steps as in Lemma 21. We explicitly prove the
lemma for ${\hat{\mathcal{G}_{\mathcal{T}}}}_{+,1}$, and the proof for
${\hat{\mathcal{G}_{\mathcal{T}}}}_{+,2}$ follows with identical steps. In
fact, by taking the complex conjugate of all the steps in the proof body
below, one obtains equations, which together constitute the proof of
${\hat{\mathcal{G}_{\mathcal{T}}}}_{+,2}$.
$\mathrm{supp}{\left(\mathrm{id}+\mathcal{W}_{[(23)]}\right)}$ is decomposed
into an orthogonal direct sum of $\mathcal{W}_{\mathrm{CS}}$,
$\mathcal{W}_{+}$ and $\mathcal{W}_{-}$ as in equation (235), (236) and (237).
In the body of the proof of Lemma 21, we noted that
$\mathcal{W}_{\mathrm{CS}}\subset\mathrm{supp}{\mathcal{P}_{\mathrm{S}}}$, and
since $\mathcal{P}_{\mathrm{S}}$ and $\mathcal{P}_{\omega}$ are orthogonal
projectors,
$\mathcal{P}_{\omega}\mathrm{T}_{1}^{(\mathrm{nc})}\mathcal{P}_{\omega}\
\mbox{$|w_{\mathrm{tr}}\rangle$}=0$. Next, let
$\mbox{$|w_{\pm}\rangle$}\in\mathcal{P}_{\omega}\ \mathcal{W}_{\pm}$, then we
have

$\displaystyle\ \mathcal{P}_{\omega}\
\mathrm{T}_{1,+}^{(\mathrm{nc})}\mathcal{P}_{\omega}\ \ \mathcal{P}_{\omega}\
\mbox{$|w_{\pm}\rangle$}\ $ $\displaystyle=$ $\displaystyle\
\mathcal{P}_{\omega}\times\frac{1}{4}\ \sum_{a,b\in{\mathbb{F}_{2}}_{*}^{2m}}\
\lambda_{b}\
\mbox{$\langle\hat{a}|$}\mathcal{P}_{\omega}\mbox{$|\hat{b}\rangle$}\
\mbox{$|\hat{a}\rangle$}$ $\displaystyle=$ $\displaystyle\
\mathcal{P}_{\omega}\times\ \frac{1}{4}\
\sum_{a,b\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{b}\
\left(\frac{1}{3}\left(\delta_{a,b}-\ \delta_{\left\langle
a,b\right\rangle,1}\right)\right)\
\mbox{$|\hat{a}\rangle$},\,\,\left(\mathrm{using}\ \mathrm{equation}\
\mathrm{\eqref{eq:inner_a_pw_or_pws_b}}\right)$ $\displaystyle=$
$\displaystyle\ \frac{1}{12}\left(1\mp\dfrac{1}{N}\right)\
\mathcal{P}_{\omega}\ \mbox{$|w_{\pm}\rangle$},$ (240)

where $\mbox{$\langle\hat{a}|$}\mathcal{P}_{\omega}\mbox{$|\hat{b}\rangle$}$
is evaluated in Section E. Thus the remaining eigenvalues are
$\frac{1}{12}\left(1\pm\frac{1}{N}\right)$. ∎

####  B.1.5 Action of $\mathcal{G}_{\mathcal{T}}$ on
$\mathcal{V}^{(\mathrm{c})}$

######  Lemma 23.

On $\mathcal{V}^{(\mathrm{c})}$, $\mathcal{G}_{\mathcal{T}}$’s highest
eigenvalue is $+1$ with a one-dimensional eigenspace. It is given by

$\ \mathcal{W}_{\mathrm{CS}}\ :=\ \mathrm{span}\ \left\\{\
\mbox{$|w_{\mathrm{tr}}\rangle$}\ :=\sum_{\begin{subarray}{c}a,b\\\
\left\langle a,b\right\rangle=0\\\ a,b,a+b\neq 0\end{subarray}}\
\mbox{$|a,b,a+b\rangle$}\right\\}.$ (241)

The second highest eigenvalue of $\mathcal{G}_{\mathcal{T}}$ on
$\mathcal{V}^{(\mathrm{c})}$ is upper bounded by
$\frac{1}{2}\left(1+\frac{4}{N}+\frac{2}{N(N-2)}\right)$.

###### Proof.

Structure of proof: we first prove that $\mathcal{G}_{\mathcal{T}}$ can be
written as a sum of the following three linear operators.

$\mathcal{G}_{\mathcal{T}}\ =\ \frac{1}{4}\ \mathrm{id}+\
\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}+\hat{\mathcal{G}_{\mathcal{T}}}_{\left[\mathrm{S}\right]},$
(242)

with the properties

  1. 1.

$\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$ is a self-adjoint linear
operator. It’s highest eigenvalue is $\frac{3}{4}$, and
$\mathcal{W}_{\mathrm{CS}}$ is the corresponding eigenspace. The remaining
non-zero eigenvalues of $\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$ are
$\frac{1}{4}\left(1+\frac{2}{N-2}\right)$,
$\frac{1}{4}\left(1+\frac{2}{N+2}\right)$,
$\frac{1}{4}\left(1-\frac{2}{N-2}\right)$ and
$\frac{1}{4}\left(1-\frac{2}{N+2}\right)$.

  2. 2.

$\hat{\mathcal{G}_{\mathcal{T}}}_{\left[\mathrm{S}\right]}$ is also self-
adjoint. It’s largest eigenvalue is lower or equal to $\frac{3}{2N}$, i.e.,

$\left|\left|\ \hat{\mathcal{G}_{\mathcal{T}}}_{\left[\mathrm{S}\right]}\
\right|\right|_{\mathrm{op}}\ \leq\dfrac{3}{2N}.$ (243)

$\mathrm{supp}{\hat{\mathcal{G}_{\mathcal{T}}}_{\left[\mathrm{S}\right]}}$
lies in the orthogonal complement of $\mathcal{W}_{\mathrm{CS}}$.

We construct the target decomposition of equation (242) in Lemma 24. We obtain
the eigenvalues of $\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$ in Lemma 27
along with their eigenspaces. In Lemma 24 we prove the inequality (243).

∎

######  Lemma 24.

In this lemma, we construct the decomposition given in equation (242).

###### Proof.

Using equation (B.1), one sees that the action of $\mathcal{G}_{\mathcal{T}}$
on $|a,b,a+b\rangle$ when $a,b$ are such that $\left\langle
a,b\right\rangle=0$ is

$\displaystyle\mathcal{G}_{\mathcal{T}}\ \mbox{$|a,b,a+b\rangle$}\ =\ $
$\displaystyle\ \ \ \ \ \ \ \dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\
\left\langle a,h\right\rangle=0\\\ \left\langle
b,h\right\rangle=0\end{subarray}}\ \mbox{$|a,b,a+b\rangle$}$ (244)
$\displaystyle\ +\ \dfrac{1}{N^{2}}\ \ \sum_{\begin{subarray}{c}h:\\\
\left\langle a,h\right\rangle=0\\\ \left\langle
b,h\right\rangle=1\end{subarray}}\
\left|a\;\;\;\;\;b+h\;\;\;\;\;a+b+h\right\rangle$ (245) $\displaystyle\ +\
\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle b,h\right\rangle=0\end{subarray}}\
\left|a+h\;\;\;\;\;b\;\;\;\;\;a+b+h\right\rangle$ (246) $\displaystyle\ +\
\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle b,h\right\rangle=1\end{subarray}}\
\left|a+h\;\;\;\;\;b+h\;\;\;\;\;a+b\right\rangle.$ (247)

In Section F, we prove from equation (244) (and the expressions (245), (246)
and (247)) that the action of $\mathcal{G}_{\mathcal{T}}$, when restricted to
$\mathcal{V}^{(\mathrm{c})}$ can be re-written in the following form

$\mathcal{G}_{\mathcal{T}}\ =\ \dfrac{1}{4}\mathrm{id}\ +\
\mathrm{T}_{1}^{(\mathrm{c})}\ +\ \mathrm{T}_{2}^{(\mathrm{c})}\ +\
\mathrm{T}_{3}^{(\mathrm{c})},$ (248)

where by $\mathrm{id}$ we mean the identity restricted to
$\mathcal{V}^{(\mathrm{c})}$, and we define the linear operators
$\mathrm{T}_{1}^{(\mathrm{c})}$, $\mathrm{T}_{2}^{(\mathrm{c})}$ and
$\mathrm{T}_{3}^{(\mathrm{c})}$ on $\mathcal{V}^{(\mathrm{c})}$ by providing
their action on the basis vectors $|a,b,a+b\rangle$, for $\left\langle
a,b\right\rangle=0$, and $a,b,a+b\neq 0$.

$\displaystyle\mathrm{T}_{1}^{(\mathrm{c})}\ \mbox{$|a,b,a+b\rangle$}\ :=\
\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle b,h\right\rangle=1\end{subarray}}\
\mbox{$|a,h,a+h\rangle$},$ (249) $\displaystyle\mathrm{T}_{2}^{(\mathrm{c})}\
\mbox{$|b,a,a+b\rangle$}\ :=\ \dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\
\left\langle a,h\right\rangle=0\\\ \left\langle
b,h\right\rangle=1\end{subarray}}\ \mbox{$|h,a,a+h\rangle$},$ (250)
$\displaystyle\mathrm{T}_{3}^{(\mathrm{c})}\ \mbox{$|a+b,b,a\rangle$}\ :=\
\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle b,h\right\rangle=1\end{subarray}}\
\mbox{$|a+h,h,a\rangle$}.$ (251)

We define $\hat{\mathcal{G}_{\mathcal{T}}}$ as

$\hat{\mathcal{G}_{\mathcal{T}}}\ :=\ \mathcal{G}_{\mathcal{T}}-\dfrac{1}{4}\
\mathrm{id}.$ (252)

Then from equation (248) we see that

$\hat{\mathcal{G}_{\mathcal{T}}}\ =\ \mathrm{T}_{1}^{(\mathrm{c})}\ +\
\mathrm{T}_{2}^{(\mathrm{c})}\ +\ \mathrm{T}_{3}^{(\mathrm{c})}.$ (253)

Below we give some important properties of $\mathrm{T}_{1}^{(\mathrm{c})}$,
$\mathrm{T}_{2}^{(\mathrm{c})}$ and $\mathrm{T}_{3}^{(\mathrm{c})}$, and also
some relations among them.

  1. 1.

From equation (249) it is seen that

$\mbox{$\langle a,b,a+b|$}\ T\ \mbox{$|a,c,a+c\rangle$}\ =\ \frac{2}{N^{2}}\
\delta_{\left\langle b,c\right\rangle,1}\ =\ \mbox{$\langle a,c,a+c|$}\ T\
\mbox{$|a,b,a+b\rangle$}.$ (254)

Thus, $\mathrm{T}_{1}^{(\mathrm{c})}$ is self-adjoint. Similarly,
$\mathrm{T}_{2}^{(\mathrm{c})}$ and $\mathrm{T}_{3}^{(\mathrm{c})}$ are also
self-adjoint, and all of them admit a spectral decomposition into orthogonal
eigenspaces.

  2. 2.

In Lemma 25, we prove that $\mathrm{T}_{1}^{(\mathrm{c})}$,
$\mathrm{T}_{2}^{(\mathrm{c})}$ and $\mathrm{T}_{3}^{(\mathrm{c})}$ commute
with $\mathcal{W}_{[(23)]}$, $\mathcal{W}_{[(13)]}$ and
$\mathcal{W}_{[(12)]}$, respectively. Thus $\mathrm{T}_{1}^{(\mathrm{c})}$ can
be diagonalized along the $+1$ and $-1$ eigenspaces of $\mathcal{W}_{[(23)]}$,
and similarly for $\mathrm{T}_{2}^{(\mathrm{c})}$ and
$\mathrm{T}_{3}^{(\mathrm{c})}$.

  3. 3.

Lemma 25 also proves that
$\mathrm{T}_{2}^{(\mathrm{c})}=\mathcal{W}_{[(123)]}\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(132)]}=\mathcal{W}_{[(12)]}\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(12)]}$,
and
$\mathrm{T}_{3}^{(\mathrm{c})}=\mathcal{W}_{[(132)]}\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(123)]}=\mathcal{W}_{[(13)]}\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(13)]}$.
Hence, obtaining the spectral decomposition of $\mathrm{T}_{1}^{(\mathrm{c})}$
also gives us the spectral decomposition of $\mathrm{T}_{2}^{(\mathrm{c})}$
and $\mathrm{T}_{3}^{(\mathrm{c})}$.

  4. 4.

In Lemma 26 we obtain the spectral decomposition for
$\mathrm{T}_{1}^{(\mathrm{c})}$. We see the following.

    1. (a)

$\displaystyle\frac{1}{2}\left(\mathrm{id}-\mathcal{W}_{[(23)}\right)\
\mathrm{T}_{1}^{(\mathrm{c})}\ =0.$ (255)
$\displaystyle\frac{1}{2}\left(\mathrm{id}+\mathcal{W}_{[(23)}\right)\
\mathrm{T}_{1}^{(\mathrm{c})}\ =\mathrm{T}_{1}^{(\mathrm{c})}.$ (256)

    2. (b)

Define

$\mathcal{V}_{1}^{(\mathrm{(c)}}\ :=\
\mathrm{supp}{\left(\mathrm{id}+\mathcal{W}_{[(23)]}\right)}.$ (257)

Then, in other words, the
$\mathrm{supp}(\mathrm{T}_{1}^{(\mathrm{c})})=\mathcal{V}_{1}^{(\mathrm{(c)}}$.
Note that $\mathcal{V}_{1}^{(\mathrm{(c)}}$ is
$\frac{\left(N^{2}-1\right)\left(N^{2}-4\right)}{4}$ dimensional.

    3. (c)

$\mathcal{V}_{1}^{(\mathrm{(c)}}$ has an orthogonal decomposition along two
subspaces: ${\mathcal{V}_{1}^{(\mathrm{c})}}$ which is $N^{2}-1$ dimensional
and its orthogonal complement, ${\mathcal{V}_{1}^{(\mathrm{c})}}^{\perp}$,
which is $\frac{\left(N^{2}-1\right)\left(N^{2}-8\right)}{4}$ dimensional,
with the following properties: let ${{\mathcal{P}_{1}^{(\mathrm{c})}}}$ and
${{\mathcal{P}_{1}^{(\mathrm{c})}}}^{\perp}$ be the projectors on
${\mathcal{V}_{1}^{(\mathrm{c})}}$ and
${\mathcal{V}_{1}^{(\mathrm{c})}}^{\perp}$ respectively. Then the following is
true. Then,

$\displaystyle{{\mathcal{P}_{1}^{(\mathrm{c})}}}\
\mathrm{T}_{1}^{(\mathrm{c})}\ =\ $ $\displaystyle\frac{1}{4}\
{{\mathcal{P}_{1}^{(\mathrm{c})}}},$ (258)
$\displaystyle{{\mathcal{P}_{1}^{(\mathrm{c})}}}^{\perp}\
\left(\mathrm{T}_{1}^{(\mathrm{c})}\right)^{2}\ =\ $
$\displaystyle\frac{1}{4N^{2}}\ {{\mathcal{P}_{1}^{(\mathrm{c})}}}^{\perp},$
(259)

Hence the eigenvalues of $\mathrm{T}_{1}^{(\mathrm{c})}$ are $\frac{1}{4}$ and
$\pm\frac{1}{2N}$. We will use the following orthogonal decomposition of
$\mathrm{T}_{1}^{(\mathrm{c})}$.

$\mathrm{T}_{1}^{(\mathrm{c})}\ =\ \mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\
+\mathrm{T}_{1,-}^{(\mathrm{c})},$ (260)

where

$\displaystyle\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\ :=\
{{\mathcal{P}_{1}^{(\mathrm{c})}}}\ \mathrm{T}_{1}^{(\mathrm{c})}$ (261)
$\displaystyle\mathrm{T}_{1,-}^{(\mathrm{c})}\ :=\
{{\mathcal{P}_{1}^{(\mathrm{c})}}}^{\perp}\ \mathrm{T}_{1}^{(\mathrm{c})}.$
(262)

  5. 5.

Since $\mathrm{T}_{2}^{(\mathrm{c})}$ and $\mathrm{T}_{3}^{(\mathrm{c})}$ are
unitary transformations of $\mathrm{T}_{1}^{(\mathrm{c})}$, their
eigendecompositions are identical to $\mathrm{T}_{1}^{(\mathrm{c})}$’s. Hence
we have the following.

$\displaystyle\mathrm{T}_{2}^{(\mathrm{c})}\ =\
\mathrm{T}_{2,\mathrm{D}}^{(\mathrm{c})}\ +\ \mathrm{T}_{2,-}^{(\mathrm{c})},$
(263) $\displaystyle\mathrm{T}_{3}^{(\mathrm{c})}\ =\
\mathrm{T}_{3,\mathrm{D}}^{(\mathrm{c})}\ +\ \mathrm{T}_{3,-}^{(\mathrm{c})},$
(264)

where, for
${{\mathcal{P}_{2}^{(\mathrm{c})}}}=\mathcal{W}_{[(123)]}{{\mathcal{P}_{1}^{(\mathrm{c})}}}\mathcal{W}_{[(132)]}=\mathcal{W}_{[(12)]}{{\mathcal{P}_{1}^{(\mathrm{c})}}}\mathcal{W}_{[(12)]}$
, and
${{\mathcal{P}_{3}^{(\mathrm{c})}}}=\mathcal{W}_{[(132)]}{{\mathcal{P}_{1}^{(\mathrm{c})}}}\mathcal{W}_{[(123)]}=\mathcal{W}_{[(13)]}{{\mathcal{P}_{1}^{(\mathrm{c})}}}\mathcal{W}_{[(13)]}$,
we get

$\displaystyle\mathrm{T}_{2,\mathrm{D}}^{(\mathrm{c})}\ :=\
{{\mathcal{P}_{2}^{(\mathrm{c})}}}\ \mathrm{T}_{2}^{(\mathrm{c})}\ =\
\mathcal{W}_{[(123)]}\ \mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\
\mathcal{W}_{[(132)]}\ =\
\mathcal{W}_{[(12)]}\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\mathcal{W}_{[(12)]}$
(265) $\displaystyle\mathrm{T}_{2,-}^{(\mathrm{c})}\ :=\
{{\mathcal{P}_{2}^{(\mathrm{c})}}}^{\perp}\ \mathrm{T}_{2}^{(\mathrm{c})}\ =\
\mathcal{W}_{[(123)]}\ \mathrm{T}_{1,-}^{(\mathrm{c})}\ \mathcal{W}_{[(132)]}\
=\ \mathcal{W}_{[(12)]}\mathrm{T}_{1,-}^{(\mathrm{c})}\mathcal{W}_{[(12)]}$
(266) $\displaystyle\mathrm{T}_{3,\mathrm{D}}^{(\mathrm{c})}\ :=\
{{\mathcal{P}_{3}^{(\mathrm{c})}}}\ \mathrm{T}_{3}^{(\mathrm{c})}\ =\
\mathcal{W}_{[(132)]}\ \mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\
\mathcal{W}_{[(123)]}\ =\
\mathcal{W}_{[(13)]}\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\mathcal{W}_{[(13)]},$
(267) $\displaystyle\mathrm{T}_{3,-}^{(\mathrm{c})}\ :=\
{{\mathcal{P}_{3}^{(\mathrm{c})}}}^{\perp}\ \mathrm{T}_{3}^{(\mathrm{c})}\ =\
\mathcal{W}_{[(132)]}\ \mathrm{T}_{1,-}^{(\mathrm{c})}\ \mathcal{W}_{[(123)]}\
=\ \mathcal{W}_{[(13)]}\mathrm{T}_{1,-}^{(\mathrm{c})}\mathcal{W}_{[(13)]}.$
(268)

Define

$\displaystyle\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ $
$\displaystyle\>=\ \mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\ +\
\mathrm{T}_{2,\mathrm{D}}^{(\mathrm{c})}\ +\
\mathrm{T}_{3,\mathrm{D}}^{(\mathrm{c})},$ (269)
$\displaystyle\hat{\mathcal{G}_{\mathcal{T}}}_{\left[\mathrm{S}\right]}\ $
$\displaystyle\ :=\ \mathrm{T}_{1,-}^{(\mathrm{c})}\ +\
\mathrm{T}_{2,-}^{(\mathrm{c})}\ +\ \mathrm{T}_{3,-}^{(\mathrm{c})},$ (270)

from which we see that

$\displaystyle\hat{\mathcal{G}_{\mathcal{T}}}\ =\ \
\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ +\
\hat{\mathcal{G}_{\mathcal{T}}}_{\left[\mathrm{S}\right]}.$ (271)

Since $\left|\left|\ \mathrm{T}_{1,-}^{(\mathrm{c})}\
\right|\right|_{\mathrm{op}}$, $\left|\left|\ \mathrm{T}_{2,-}^{(\mathrm{c})}\
\right|\right|_{\mathrm{op}}$ and $\left|\left|\
\mathrm{T}_{3,-}^{(\mathrm{c})}\ \right|\right|_{\mathrm{op}}$ $\leq$
$\frac{1}{2N}$, from equation (270) we have that

$\left|\left|\ \hat{\mathcal{G}_{\mathcal{T}}}_{\left[\mathrm{S}\right]}\
\right|\right|_{\mathrm{op}}\ \leq\ \dfrac{3}{2N}.$ (272)

Finally, Lemma 27 gives the full eigendecomposition of
$\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$, which is as was specified in
Lemma 24. Hence the construction of the decomposition is as was promised in
the proof of Lemma 24. ∎

######  Lemma 25.

$\mathrm{T}_{1}^{(\mathrm{c})}$, $\mathrm{T}_{2}^{(\mathrm{c})}$ and
$\mathrm{T}_{3}^{(\mathrm{c})}$ have the following properties.

  * •

$\mathrm{T}_{1}^{(\mathrm{c})}$, $\mathrm{T}_{2}^{(\mathrm{c})}$ and
$\mathrm{T}_{3}^{(\mathrm{c})}$ commute with $\mathcal{W}_{[(23)]}$,
$\mathcal{W}_{[(13)]}$, and $\mathcal{W}_{[(12)]}$ respectively.

$\displaystyle\mathcal{W}_{[(23)]}\ \mathrm{T}_{1}^{(\mathrm{c})}\
\mathcal{W}_{[(23)]}\ =\ \mathrm{T}_{1}^{(\mathrm{c})},$ (273)
$\displaystyle\mathcal{W}_{[(13)]}\ \mathrm{T}_{2}^{(\mathrm{c})}\
\mathcal{W}_{[(13)]}\ =\ \mathrm{T}_{2}^{(\mathrm{c})},$ (274)
$\displaystyle\mathcal{W}_{[(12)]}\ \mathrm{T}_{3}^{(\mathrm{c})}\
\mathcal{W}_{[(12)]}\ =\ \mathrm{T}_{3}^{(\mathrm{c})}.$ (275)

  * •

$\mathrm{T}_{1}^{(\mathrm{c})}$, $\mathrm{T}_{2}^{(\mathrm{c})}$ and
$\mathrm{T}_{3}^{(\mathrm{c})}$ are permutations of each other:

$\displaystyle\mathrm{T}_{3}^{(\mathrm{c})}\ =$ $\displaystyle\
\mathcal{W}_{[(13)]}\ \mathrm{T}_{1}^{(\mathrm{c})}\ \mathcal{W}_{[(13)]}$
(276) $\displaystyle=$ $\displaystyle\ \ \mathcal{W}_{[(132)]}\
\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(123)]}$ (277) $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(123)]}\
\mathrm{T}_{2}^{(\mathrm{c})}\mathcal{W}_{[(132)]}$ $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(23)]}\ \mathrm{T}_{2}^{(\mathrm{c})}\
\mathcal{W}_{[(23)]},$ $\displaystyle\ \mathrm{T}_{2}^{(\mathrm{c})}\ =\ $
$\displaystyle\ \mathcal{W}_{[(123)]}\
\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(132)]}$ (278) $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(12)]}\ \mathrm{T}_{1}^{(\mathrm{c})}\
\mathcal{W}_{[(12)]}$ $\displaystyle=$ $\displaystyle\ \mathcal{W}_{[(132)]}\
\mathrm{T}_{3}^{(\mathrm{c})}\mathcal{W}_{[(123)]}$ $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(23)]}\ \mathrm{T}_{3}^{(\mathrm{c})}\
\mathcal{W}_{[(23)]}$ $\displaystyle\mathrm{T}_{1}^{(\mathrm{c})}\ =$
$\displaystyle\ \mathcal{W}_{[(123)]}\
\mathrm{T}_{3}^{(\mathrm{c})}\mathcal{W}_{[(132)]}$ (279) $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(13)]}\ \mathrm{T}_{3}^{(\mathrm{c})}\
\mathcal{W}_{[(13)]}$ $\displaystyle=$ $\displaystyle\ \mathcal{W}_{[(132)]}\
\mathrm{T}_{2}^{(\mathrm{c})}\mathcal{W}_{[(123)]}$ $\displaystyle=\ $
$\displaystyle\mathcal{W}_{[(12)]}\ \mathrm{T}_{2}^{(\mathrm{c})}\
\mathcal{W}_{[(12)]},$

###### Proof.

Structure of proof:first we prove equation (273). Then we prove the first
(top-most) relation in equation (276). Equation (277) is proved by conjugating
both sides of equation (273) by $\mathcal{W}_{[(13)]}$, and noting that
$\mathcal{W}_{[(123)}=\mathcal{W}_{[(13)]}\mathcal{W}_{[(23)]}$. The rest of
the relations can be proved using identical steps. Proof of equation (273): in
equation (249) we see that

$\displaystyle\ \mathrm{T}_{1}^{(\mathrm{c})}\ \mbox{$|a,a+b,b\rangle$}$
$\displaystyle\ =\ \dfrac{1}{N^{2}}\ \,\,\sum_{\begin{subarray}{c}k:\\\
\left\langle a,k\right\rangle=0\\\ \left\langle
a+b,k\right\rangle=1\end{subarray}}\ \mbox{$|a,k,a+k\rangle$}$ $\displaystyle\
=\ \dfrac{1}{N^{2}}\ \,\,\sum_{\begin{subarray}{c}k:\\\ \left\langle
a,k\right\rangle=0\\\ \left\langle b,k\right\rangle=1\end{subarray}}\
\mbox{$|a,k,a+k\rangle$}$ $\displaystyle\ =\ \dfrac{1}{N^{2}}\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=0\\\ \left\langle
b,h\right\rangle=1\end{subarray}}\ \mbox{$|a,a+h,h\rangle$},$ (280)

where we go from the first line to the second using

$\left\\{k\in{\mathbb{F}_{2}}_{*}^{2m}\ \big{|}\ \left\langle
a,k\right\rangle=0,\ \left\langle a+b,k\right\rangle=1\right\\}\ =\
\left\\{k\in{\mathbb{F}_{2}}_{*}^{2m}\ \big{|}\ \left\langle
a,k\right\rangle=0,\ \left\langle b,k\right\rangle=1\right\\}.$ (281)

To go from the second line to the third, we change variables from $k$ to
$h=a+k$, by noting that

$\left\\{\ h\ \big{|}\ \left\langle a,h\right\rangle=0,\ \left\langle
b,h\right\rangle=1\right\\}\ =\ a\ +\ \left\\{k\in{\mathbb{F}_{2}}_{*}^{2m}\
\big{|}\ \left\langle a,k\right\rangle=0,\ \left\langle
b,k\right\rangle=1\right\\},\,\,\,\left(\mathrm{using}\ \left\langle
a,b\right\rangle=0\right).$ (282)

Noting that the LHS of equation (280) can be re-written as
$\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(23)]}\ \mbox{$|a,b,a+b\rangle$}$,
and the RHS in (280) can be re-written as the RHS of (283), we get the
following.

$\displaystyle\ \mathrm{T}_{1}^{(\mathrm{c})}\ \ \Bigg{(}\
\mathcal{W}_{[(23)]}\ \mbox{$|a,b,a+b\rangle$}\Bigg{)}$ $\displaystyle\ =\ \
\mathcal{W}_{[(23)]}\ \left(\ \dfrac{1}{N^{2}}\
\,\,\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=0\\\
\left\langle b,h\right\rangle=1\end{subarray}}\
\mbox{$|a,h,a+h\rangle$}\right).$ (283)

Comparing equation (283) with equation (249) we see that the action of
$\mathrm{T}_{1}^{(\mathrm{c})}$ on $|a,b,a+b\rangle$ commutes with the action
of $\mathcal{W}_{[(23)]}$ on $|a,b,a+b\rangle$. Since this is true for all
basis vectors $|a,b,a+b\rangle$ in $\mathcal{V}^{(\mathrm{c})}$, we obtain
equation (273).

Proof of equation (276): apply $\mathcal{W}_{[(13)]}$ to both sides of
equation (249)

$\displaystyle\ \mathcal{W}_{[(13)]}\ \mathrm{T}_{1}^{(\mathrm{c})}\
\mathcal{W}_{[(13)]}\ \mbox{$|a+b,b,a\rangle$}\ =$ $\displaystyle\
\dfrac{1}{N^{2}}\ \,\,\sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle b,h\right\rangle=1\end{subarray}}\
\mbox{$|a+h,h,a\rangle$}$ $\displaystyle=$ $\displaystyle\
\mathrm{T}_{3}^{(\mathrm{c})}\ \mbox{$|a+b,b,a\rangle$},$ (284)

and equation (276) is proved by noting that equation (B.1.5) holds for all
basis vectors $|a+b,b,a\rangle$. ∎

######  Lemma 26.

The spectral decomposition of $\mathrm{T}_{1}^{(\mathrm{c})}$ (defined in
equation (249)) is

$\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\ =\ \dfrac{1}{4}\
{{\mathcal{P}_{1}^{(\mathrm{c})}}}\ +\
\dfrac{1}{2N}{{\mathcal{P}_{1,+}^{(\mathrm{c})}}}\
-\dfrac{1}{2N}{{\mathcal{P}_{1,-}^{(\mathrm{c})}}},$ (285)

where ${{\mathcal{P}_{1}^{(\mathrm{c})}}}$,
${{\mathcal{P}_{1,+}^{(\mathrm{c})}}}$ and
${{\mathcal{P}_{1,-}^{(\mathrm{c})}}}$ are orthogonal projectors defined on
the following orthogonal spaces.

$\displaystyle{\mathcal{V}_{1}^{(\mathrm{c})}}\ :=\ \mathrm{span}\ \left\\{\
\mbox{$|\hat{a}\rangle$}\ \forall\ a\in{\mathbb{F}_{2}}_{*}^{2m}\ \big{|}\
\mbox{$|\hat{a}\rangle$}\ :=\ \dfrac{\sqrt{2}}{\sqrt{N^{2}-4}}\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=0\\\ h\neq
a,0\end{subarray}}\ \ \mbox{$|a,h,a+h\rangle$}\right\\},$ (286)
$\displaystyle{\mathcal{V}_{1,+}^{(\mathrm{c})}}\ :=\ \mathrm{span}\ \left\\{\
\mbox{$|a_{+}\rangle$}\ =\sum_{\begin{subarray}{c}b:\\\ \left\langle
a,b\right\rangle=0\\\ b\neq 0,a\end{subarray}}\lambda_{b}\
\mbox{$|a,b,a+b\rangle$},\ \forall\ a\in{\mathbb{F}_{2}}_{*}^{2m}\ \big{|}\
\lambda_{b}\text{ satisfy equations in \eqref{vc_eq:set_plus}.}\right\\},$
(287) $\displaystyle{\mathcal{V}_{1,-}^{(\mathrm{c})}}\ :=\ \mathrm{span}\
\left\\{\ \mbox{$|a_{-}\rangle$}\ =\sum_{\begin{subarray}{c}b:\\\ \left\langle
a,b\right\rangle=0\\\ b\neq 0,a\end{subarray}}\lambda_{b}\
\mbox{$|a,b,a+b\rangle$},\ \forall\ a\in{\mathbb{F}_{2}}_{*}^{2m}\ \big{|}\
\lambda_{b}\text{ satisfy equations in \eqref{vc_eq:set_minus}.}\right\\},$
(288)

where

$\displaystyle\sum_{\begin{subarray}{c}b:\\\ \left\langle
a,b\right\rangle=0\\\ b\neq 0,a\end{subarray}}\lambda_{b}=0,\ \ \ \ \
\sum_{\begin{subarray}{c}h:\\\ \left\langle h,a\right\rangle=0\\\ \left\langle
h,b\right\rangle=1\end{subarray}}\lambda_{h}\ =\ \dfrac{1}{2N}\lambda_{b},$
(289)

and

$\displaystyle\sum_{\begin{subarray}{c}b:\\\ \left\langle
a,b\right\rangle=0\\\ b\neq 0,a\end{subarray}}\lambda_{b}=0,\ \ \ \ \
\sum_{\begin{subarray}{c}h:\\\ \left\langle h,a\right\rangle=0\\\ \left\langle
h,b\right\rangle=1\end{subarray}}\lambda_{h}\ =\ -\dfrac{1}{2N}\lambda_{b}.$
(290)

$\dim{\mathcal{V}_{1}^{(\mathrm{c})}}=N^{2}-1$,
$\dim{\mathcal{V}_{1,+}^{(\mathrm{c})}}=\frac{\left(N^{2}-1\right)\left(N^{2}-2N-8\right)}{8}$,
and
$\dim{\mathcal{V}_{1,-}^{(\mathrm{c})}}=\frac{\left(N^{2}-1\right)\left(N^{2}+2N-8\right)}{8}$

###### Proof.

From equation (254) we know that $\mathrm{T}_{1}^{(\mathrm{c})}$ is self-
adjoint and has a spectral decomposition with orthogonal eigenspaces.

From the definition of $\mathrm{T}_{1}^{(\mathrm{c})}$ in equation (249), we
see that

$\mathrm{T}_{1}^{(\mathrm{c})}\ \mbox{$|a,b,a+b\rangle$}\
=\mathrm{T}_{1}^{(\mathrm{c})}\ \mbox{$|a,a+b,b\rangle$},$ (291)

since if some $h$ is contained in the set $\left\\{k\in\mathbb{F}_{2}^{2m}\
\big{|}\ \left\langle a,k\right\rangle=0,\left\langle
b,k\right\rangle=1\right\\}$, then so is $a+h$. This further implies

$\displaystyle\mathrm{T}_{1}^{(\mathrm{c})}\ \mbox{$|a,b,a+b\rangle$}\ =\
\mathrm{T}_{1}^{(\mathrm{c})}\ \mathcal{W}_{[(23)]}\ \mbox{$|a,b,a+b\rangle$}$
$\displaystyle\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;=\
\mathcal{W}_{[(23)]}\ \mathrm{T}_{1}^{(\mathrm{c})}\ \mbox{$|a,b,a+b\rangle$}$
$\displaystyle\Longrightarrow$ $\displaystyle\
\left(\mathrm{id}-\mathcal{W}_{[(23)]}\right)\ \mathrm{T}_{1}^{(\mathrm{c})}\
\mbox{$|a,b,a+b\rangle$}\ =\ 0.$ (292)

Since equation (B.1.5) is true for all $|a,b,a+b\rangle$, we get

$\left(\mathrm{id}-\mathcal{W}_{[(23)]}\right)\ \mathrm{T}_{1}^{(\mathrm{c})}\
=\ 0,$ (293)

which informs us that
$\mathrm{supp}{\mathrm{T}_{1}^{(\mathrm{c})}}\subseteq\mathcal{V}_{1}^{(\mathrm{(c)}}$.

Next, we prove that for $|\hat{a}\rangle$, as defined in equation (286), is an
eigenvector of $\mathrm{T}_{1}^{(\mathrm{c})}$, with eigenvalue $\frac{1}{4}$.

$\displaystyle\mathrm{T}_{1}^{(\mathrm{c})}\ \mbox{$|a\rangle$}\ =\ $
$\displaystyle\ \mathrm{T}_{1}^{(\mathrm{c})}\
\frac{\sqrt{2}}{\sqrt{N^{2}-4}}\ \sum_{\begin{subarray}{c}b:\\\ \left\langle
a,b\right\rangle=0\\\ b\neq 0,a\end{subarray}}\ \mbox{$|a,b,a+b\rangle$}$
$\displaystyle=\ $ $\displaystyle\frac{\sqrt{2}}{N^{2}\ \sqrt{N^{2}-4}}\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=0\end{subarray}}\
\mathrm{S}_{h}\ \ \mbox{$|a,h,a+h\rangle$}$ $\displaystyle=\ $ $\displaystyle\
\dfrac{1}{4}\ \mbox{$|\hat{a}\rangle$},$ (294)

where in the second line $\mathrm{S}_{h}=\left|\
\left\\{b\in\mathbb{F}_{2}^{2m}\ \big{|}\ \left\langle
a,b\right\rangle=0,\left\langle h,b\right\rangle=1,\ b\neq 0,a\right\\}\
\right|$. Note that the condition $\left\langle h,b\right\rangle=1$ ensures
that $b\neq 0,a$. $\mathrm{S}_{h}=\left|\ \left\\{b\in\mathbb{F}_{2}^{2m}\
\big{|}\ \left\langle a,b\right\rangle=0,\left\langle
h,b\right\rangle=1\right\\}\ \right|$. Also, had $h=0,a$, then $\left\langle
b,h\right\rangle\neq 1$, so $h\neq 0,a$. Since $a,h$ are line arly
independent, Lemma 2 tells us that this size is $\dfrac{N^{2}}{4}$. This
proves equation (286).

Let $b,c\in{\mathbb{F}_{2}}_{*}^{2m}\setminus\\{a\\}$ such that $\left\langle
a,b\right\rangle=\left\langle a,c\right\rangle=0$ and $c,a+c\neq b$.

Then

$\displaystyle\ {\mathrm{T}_{1}^{(\mathrm{c})}}^{2}\
\left(\left(\mbox{$|a,b,a+b\rangle$}+\mbox{$|a,a+b,b\rangle$}\right)-\left(\mbox{$|a,c,a+c\rangle$}+\mbox{$|a,a+c,c\rangle$}\right)\right)$
$\displaystyle=$ $\displaystyle\
\left(\dfrac{2}{N^{2}}\sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle b,h\right\rangle=1\end{subarray}}\ T\
\mbox{$|a,h,a+h\rangle$}\right)-\left(\dfrac{2}{N^{2}}\sum_{\begin{subarray}{c}h:\\\
\left\langle a,h\right\rangle=0\\\ \left\langle
c,h\right\rangle=1\end{subarray}}\ T\ \mbox{$|a,h,a+h\rangle$}\right)$
$\displaystyle=$ $\displaystyle\ \dfrac{2}{N^{4}}\
\sum_{\begin{subarray}{c}k:\\\ \left\langle a,k\right\rangle=0\end{subarray}}\
\left(\mathrm{S}_{b,k}\ -\mathrm{S}_{c,k}\ \right)\ \mbox{$|a,k,a+k\rangle$},$
(295)

where

$\displaystyle\mathrm{S}_{b,k}=\left|\left\\{h\in\mathbb{F}_{2}^{2m}\ \big{|}\
\left\langle a,h\right\rangle=0,\ \left\langle b,h\right\rangle=\left\langle
k,h\right\rangle=1\ \right\\}\right|$ (296)
$\displaystyle\mathrm{S}_{c,k}=\left|\left\\{h\in\mathbb{F}_{2}^{2m}\ \big{|}\
\left\langle a,h\right\rangle=0,\ \left\langle c,h\right\rangle=\left\langle
k,h\right\rangle=1\ \right\\}\right|$ (297)

Note that if $k=0,a$ then $\mathrm{S}_{b,k},\mathrm{S}_{c,k}=0$ since
$\left\langle h,k\right\rangle\neq 1$ in that case. Using Lemma 2, we see that

$\displaystyle\mathrm{S}_{b,k}\ =\ \dfrac{N^{2}}{4}\
\left(\delta_{k,b}+\delta_{k,a+b}\right)-\dfrac{N^{2}}{8}\
\left(1-\left(\delta_{k,b}+\delta_{k,a+b}\right)\right)$ (298)
$\displaystyle\mathrm{S}_{c,k}\ =\ \dfrac{N^{2}}{4}\
\left(\delta_{k,c}+\delta_{k,a+c}\right)-\dfrac{N^{2}}{8}\
\left(1-\left(\delta_{k,c}+\delta_{k,a+c}\right)\right).$ (299)

Using equations (298) and (299) in equation (B.1.5), we finally get

$\displaystyle{\mathrm{T}_{1}^{(\mathrm{c})}}^{2}\
\left(\left(\mbox{$|a,b,a+b\rangle$}+\mbox{$|a,a+b,b\rangle$}\right)-\left(\mbox{$|a,c,a+c\rangle$}+\mbox{$|a,a+c,c\rangle$}\right)\right)$
$\displaystyle=$ $\displaystyle\ \dfrac{1}{4N^{2}}\
\left(\left(\mbox{$|a,b,a+b\rangle$}+\mbox{$|a,a+b,b\rangle$}\right)-\left(\mbox{$|a,c,a+c\rangle$}+\mbox{$|a,a+c,c\rangle$}\right)\right).$
(300)

Note that ${\mathcal{V}_{1}^{(\mathrm{c})}}^{\perp}$ (orthogonal complement of
${\mathcal{V}_{1}^{(\mathrm{c})}}$ in $\mathcal{V}_{1}^{(\mathrm{(c)}}$) is
spanned by vectors of the form
$\mbox{$|a,b,a+b\rangle$}+\mbox{$|a,a+b,b\rangle$}-\left(\mbox{$|a,c,a+c\rangle$}+\mbox{$|a,a+c,c\rangle$}\right)$.
Hence the eigenvalues of $\mathrm{T}_{1}^{(\mathrm{c})}$ on
${\mathcal{V}_{1}^{(\mathrm{c})}}^{\perp}$ (which is the orthogonal complement
of ${\mathcal{V}_{1}^{(\mathrm{c})}}$ in $\mathcal{V}_{1}^{(\mathrm{(c)}}$)
can be $\frac{1}{2N}$ or $-\frac{1}{2N}$. If $|a_{\pm}\rangle$ is an
eigenvector with eigenvalue $\pm\frac{1}{2N}$, and it has the basis expansion

$\mbox{$|a_{\pm}\rangle$}\ =\sum_{\begin{subarray}{c}b:\\\ \left\langle
a,b\right\rangle=0\\\ b\neq 0,a\end{subarray}}\ \lambda_{b}\
\mbox{$|a,b,a+b\rangle$},$ (301)

then $\lambda_{b}$ should satisfy the following two conditions.

  1. 1.

So that $|a_{\pm}\rangle$ are orthogonal to $|\hat{a}\rangle$, the
$\lambda_{b}$’s should satisfy

$\sum_{\begin{subarray}{c}b:\\\ \left\langle a,b\right\rangle=0\\\ b\neq
0,a\end{subarray}}\ \lambda_{b}\ =\ 0.$ (302)

  2. 2.

So that it has eigenvalue $\pm\dfrac{1}{2N}$, the $\lambda_{b}$’s must satisfy

$\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=0\\\
\left\langle b,h\right\rangle=1\end{subarray}}\ \lambda_{h}\ =\
\pm\dfrac{2}{N}\ \lambda_{b}.$ (303)

Hence we have obtained the eigenspaces in equations (287) and (288).

Using equation (254), we see that
$\operatorname{tr}\mathrm{T}_{1}^{(\mathrm{c})}=0$. If $n_{+}$ is the
degeneracy of $\frac{1}{2N}$, then the degeneracy of $-\frac{1}{2N}$ is
$\mathrm{rank}\
\mathrm{T}_{1}^{(\mathrm{c})}-\dim{\mathcal{V}_{1}^{(\mathrm{c})}}-n_{+}$. We
get the $\operatorname{tr}\mathrm{T}_{1}^{(\mathrm{c})}=0$ equation in terms
of the eigenvalues and degeneracies.

$\dfrac{1}{2N}\ n_{+}-\dfrac{1}{2N}\left(\mathrm{rank}\
\mathrm{T}_{1}^{(\mathrm{c})}-\dim{\mathcal{V}_{1}^{(\mathrm{c})}}-n_{+}\right)+\dfrac{1}{4}\dim{\mathcal{V}_{1}^{(\mathrm{c})}},$
(304)

which can be solved using the fact that
$\mathrm{rank}\mathrm{T}_{1}^{(\mathrm{c})}=\dim\mathcal{V}_{1}^{(\mathrm{(c)}}=\frac{\left(N^{2}-1\right)\left(N^{2}-4\right)}{4}$,
and that $\dim{\mathcal{V}_{1}^{(\mathrm{c})}}=N^{2}-1$. Using this we see
that the multiplicity of the $\frac{1}{2N}$ eigenvalue is
$=\frac{\left(N^{2}-1\right)\left(N^{2}-2N-8\right)}{8}$, whereas the
multiplicity of the $-\frac{1}{2N}$ eigenvalue is
$=\frac{\left(N^{2}-1\right)\left(N^{2}+2N-8\right)}{8}$. ∎

######  Lemma 27.

The largest eigenvalue of $\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$
(defined in equation (269)) is $\frac{3}{4}$, and $\mathcal{W}_{\mathrm{CS}}$
(defined in (241)) is the corresponding eigenspace. All remaining eigenvalues
are less than or equal to $\frac{1}{4}\left(1+\frac{2}{N-2}\right)$, and their
corresponding eigenspaces lie in the orthogonal complement of
$\mathcal{W}_{\mathrm{CS}}$.

###### Proof.

It is convenient to adopt the notation defined in equation (135), with the
identification that $\mathcal{P}_{0}=\mathcal{P}_{\mathrm{S}}$,
$\mathcal{P}_{1}=\mathcal{P}_{\omega}$ and
$\mathcal{P}_{2}=\mathcal{P}_{\omega^{*}}$.

First we obtain an orthogonal decomposition of
$\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$ using the orthogonal projectors
$\mathcal{P}_{k}$: note that due to the permutation relations among
$\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}$,
$\mathrm{T}_{2,\mathrm{D}}^{(\mathrm{c})}$ and
$\mathrm{T}_{3,\mathrm{D}}^{(\mathrm{c})}$ (see Lemma 25 and equations (265)
and (267)), $\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$ commutes with all
$\mathcal{W}_{\pi}$’s, and hence consequently commutes with
$\mathcal{P}_{k}$’s. Noting the completeness relation in equation (132), we
multiply on the right (or left) by
$\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$.

$\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ =\ \sum_{k=0}^{2}\
\mathcal{P}_{k}\ \hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ =\
\sum_{k=0}^{2}\ \mathcal{P}_{k}\ \hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\
\mathcal{P}_{k},$ (305)

where we used the fact that $\mathcal{P}_{k}^{2}=\mathcal{P}_{k}$ and and that
$\mathcal{P}_{k}$ commutes with
$\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$. Define

$\displaystyle\ {\hat{\mathcal{G}_{\mathcal{T}}}}_{\mathrm{D},k}\ :=\
\mathcal{P}_{k}\ \hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\
\mathcal{P}_{k}.$ (306)

Then we get the following orthogonal decomposition of
$\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$

$\displaystyle\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ =$ $\displaystyle\
{\hat{\mathcal{G}_{\mathcal{T}}}}_{+,0}\
+{\hat{\mathcal{G}_{\mathcal{T}}}}_{+,1}\ +\
{\hat{\mathcal{G}_{\mathcal{T}}}}_{+,2}.$ (307)

Hence the eigendecomposition of $\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$
is given by the eigendecomposition of
${\hat{\mathcal{G}_{\mathcal{T}}}}_{+,0}$,
${\hat{\mathcal{G}_{\mathcal{T}}}}_{+,1}$ and
${\hat{\mathcal{G}_{\mathcal{T}}}}_{+,2}$. To obtain these
eigendecompositions, note the following.

$\displaystyle{\hat{\mathcal{G}_{\mathcal{T}}}}_{\mathrm{D},k}\ =\ 3\
\mathcal{P}_{k}\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\mathcal{P}_{k},$ (308)

which we prove using two things: the permutation relations: that
$\mathrm{T}_{2}^{(\mathrm{c})}=\mathcal{W}_{[(123)]}\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(132)]}$
and
$\mathrm{T}_{3}^{(\mathrm{c})}=\mathcal{W}_{[(132)]}\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{W}_{[(123)]}$
(obtained in Lemma 25), and also that

$\mathcal{P}_{k}\mathcal{W}_{\sigma}\
\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\mathcal{W}_{\sigma^{-}1}\
\mathcal{P}_{k}\ =\
\mathcal{P}_{k}\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\mathcal{P}_{k}.$ (309)

Thus one needs to obtain the eigendecomposition of
$\mathcal{P}_{k}\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\mathcal{P}_{k}$. From
Lemma 26, we see that $\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}$ has the
following eigendecomposition

$\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\ =\ \dfrac{1}{4}\
\mbox{$|\hat{a}\rangle\\!\langle\hat{a}|$},$ (310)

where $|\hat{a}\rangle$ was defined in equation (286). We decompose
${\mathcal{V}_{1}^{(\mathrm{c})}}$ (also defined in equation 286) into three
orthogonal subspaces.

$\displaystyle\ \mathcal{W}_{\mathrm{CS}}\ =\ \mathrm{span}\ \left\\{\
\mbox{$|w_{\mathrm{tr}}\rangle$}\
=\sqrt{\frac{N^{2}-4}{2}}\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\
\mbox{$|\hat{a}\rangle$}\right\\},$ (311) $\displaystyle\ \mathcal{W}_{+}\ :=\
\mathrm{span}\ \left\\{\ \mbox{$|w_{+}\rangle$}\ =\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{a}\ \mbox{$|\hat{a}\rangle$},\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{a}\ =0,\
\sum_{\begin{subarray}{c}h:\\\ \left\langle h,a\right\rangle\end{subarray}=1}\
\lambda_{h}\ =\ \frac{N}{2}\ \lambda_{a}\right\\},$ (312) $\displaystyle\
\mathcal{W}_{-}\ :=\ \mathrm{span}\ \left\\{\ \mbox{$|w_{-}\rangle$}\ =\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{a}\ \mbox{$|\hat{a}\rangle$},\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \lambda_{a}\ =0,\
\sum_{\begin{subarray}{c}h:\\\ \left\langle h,a\right\rangle\end{subarray}=1}\
\lambda_{h}\ =\ -\frac{N}{2}\ \lambda_{a}\right\\},$ (313)

We note that $|w_{\mathrm{tr}}\rangle$ is the same as was defined in equation
(241), which can be seen as follows:

$\displaystyle\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\ \mbox{$|\hat{a}\rangle$}\
\propto\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\left(\sum_{\begin{subarray}{c}b\\\
\left\langle a,b\right\rangle=0\\\ b\neq 0,a\end{subarray}}\
\mbox{$|a,b,a+b\rangle$}\right)\ =\ \sum_{\begin{subarray}{c}a,b\\\
\left\langle a,b\right\rangle=0\\\ a,b,a+b\neq
0\end{subarray}}\mbox{$|a,b,a+b\rangle$}\ =\mbox{$|w_{\mathrm{tr}}\rangle$}.$
(314)

Note that the equations in the definition of $\mathcal{W}_{+}$ and
$\mathcal{W}_{-}$ are consistent and have solutions. We know this because
these are the same equations are used to decompose the irreducible
representations in the diagonal sector in Subsubsections B.1.2 and B.1.3, and
they were also introduced in the diagonal sector in Helsen et al Helsen2016
(see Lemma 4 therein). We also know that
$\dim\mathcal{W}_{+}=\frac{N(N-1)}{2}-1$ and
$\dim\mathcal{W}_{-}=\frac{N(N+1)}{2}-1$. And that the three spaces
$\mathcal{W}_{\mathrm{CS}}$, $\mathcal{W}_{+}$ and $\mathcal{W}_{-}$ are
orthogonal to each other, collectively spanning the entire $N^{2}-1$
dimensional support of $\mathrm{id}+\mathcal{W}_{[(23)]}$.

We want to further decompose $\mathcal{W}_{\mathrm{CS}}$, $\mathcal{W}_{+}$
and $\mathcal{W}_{-}$ using $\mathcal{P}_{k}$. We start with
$\mathcal{W}_{\mathrm{CS}}$.

Note that
$\mathcal{W}_{\sigma}\mbox{$|w_{\mathrm{tr}}\rangle$}=\mbox{$|w_{\mathrm{tr}}\rangle$}$
for777That
$\mathcal{W}_{\pi}\mbox{$|w_{\mathrm{tr}}\rangle$}=\mbox{$|w_{\mathrm{tr}}\rangle$}$
is actually true for all $\pi\in\mathrm{S}_{3}$.
$\sigma=\mathrm{id},[(123)],[(132)]$. We demonstrate this for
$\sigma=[(123)]$.

$\displaystyle\mathcal{W}_{[(123)]}\ \mbox{$|w_{\mathrm{tr}}\rangle$}\ $
$\displaystyle=\ \sum_{\begin{subarray}{c}a,b\\\ \left\langle
a,b\right\rangle=0\\\ a,b,a+b\neq 0\end{subarray}}\ \mathcal{W}_{[(123)]}\
\mbox{$|a,b,a+b\rangle$}$ $\displaystyle=\ \sum_{\begin{subarray}{c}a,b\\\
\left\langle a,b\right\rangle=0\\\ a,b,a+b\neq 0\end{subarray}}\
\mbox{$|a+b,a,b\rangle$}$ $\displaystyle=\
\sum_{\begin{subarray}{c}a^{\prime},b^{\prime}\\\ \left\langle
a^{\prime},b^{\prime}\right\rangle=0\\\
a^{\prime},b^{\prime},a^{\prime}+b^{\prime}\neq 0\end{subarray}}\
\mbox{$|a^{\prime},b^{\prime},a^{\prime}+b^{\prime}\rangle$},$
$\displaystyle=\ \mbox{$|w_{\mathrm{tr}}\rangle$},$ (315)

where the dummy variables were changed in the last line as follows:
$b^{\prime}=a$, and $a^{\prime}=a+b$ (so that $a^{\prime}$ varies with $b$ for
fixed $b^{\prime}=a$). This tells us the following.

$\mathcal{P}_{0}\mbox{$|w_{\mathrm{tr}}\rangle$}\ =\
\mbox{$|w_{\mathrm{tr}}\rangle$},\;\;\mathcal{P}_{1}\mbox{$|w_{\mathrm{tr}}\rangle$}\
=\ \mathcal{P}_{2}\ \mbox{$|w_{\mathrm{tr}}\rangle$}\ =\ 0.$ (316)

Finally, we have the following orthogonal decomposition of
${\mathcal{V}_{1}^{(\mathrm{c})}}$.

$\displaystyle{\mathcal{V}_{1}^{(\mathrm{c})}}\ =\
\underbrace{\mathcal{W}_{\mathrm{CS}}\oplus\mathcal{V}_{0,+}\oplus\mathcal{V}_{0,-}}_{\mathrm{supp}{{\hat{\mathcal{G}_{\mathcal{T}}}}_{+,0}}}\oplus\underbrace{\mathcal{V}_{1,+}\oplus\mathcal{V}_{1,-}}_{\mathrm{supp}{{\hat{\mathcal{G}_{\mathcal{T}}}}_{+,1}}}\oplus\underbrace{\mathcal{V}_{2,+}\oplus\mathcal{V}_{2,-}}_{\mathrm{supp}{{\hat{\mathcal{G}_{\mathcal{T}}}}_{+,2}}},$
(317)

where

$\displaystyle\mathcal{V}_{k,\pm}\ :=\
\mathrm{supp}{\mathcal{P}_{k}}\cap\mathcal{W}_{\pm}.$ (318)

Since $\mathcal{W}_{\pm}$ and $\mathrm{supp}{\mathcal{P}_{k}}$ are both
invariant under the action of $\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}$,
$\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}$ can be decomposed as follows.

$\displaystyle\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ =\ \
\mathcal{P}_{\mathrm{tr}}\ \ \hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$
$\displaystyle\ \ +\ \ \mathcal{P}_{0,+}\ \
\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ \ +\ \ \mathcal{P}_{0,-}\ \
\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$ $\displaystyle\ \ +\ \
\mathcal{P}_{1,+}\ \ \hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ \ +\ \ \
\mathcal{P}_{1,-}\ \ \hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ \ $
$\displaystyle\ \ +\ \ \mathcal{P}_{2,+}\ \
\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ \ +\ \ \mathcal{P}_{2,-}\ \
\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}},$ (319)

where

$\mathcal{P}_{j,\pm}\ :=\ \mathcal{V}_{k,\pm}.$ (320)

On $\mathcal{W}_{\mathrm{CS}}$ we see

$\displaystyle\ \mathcal{P}_{\mathrm{S}}\
\mathrm{T}_{1,\mathrm{D}}^{(\mathrm{c})}\mathcal{P}_{\mathrm{S}}\
\left(\mathcal{P}_{\mathrm{S}}\mbox{$|w_{\mathrm{tr}}\rangle$}\right)\ =\
\dfrac{1}{4}\ \mathcal{P}_{\mathrm{S}}\mbox{$|w_{\mathrm{tr}}\rangle$}\ \ \
\Longrightarrow\ \ \ {\hat{\mathcal{G}_{\mathcal{T}}}}_{+,0}\
\left(\mathcal{P}_{\mathrm{S}}\mbox{$|w_{\mathrm{tr}}\rangle$}\right)\ =\
\dfrac{3}{4}\ \mathcal{P}_{\mathrm{S}}\mbox{$|w_{\mathrm{tr}}\rangle$},$ (321)

which proves the first part of Lemma 27.

To obtain the remaining eigenvalues, note that
$\mathcal{P}_{k}\mbox{$|w_{\pm}\rangle$}$ span $\mathcal{V}_{k,\pm}$, where
$\mbox{$|w_{\pm}\rangle$}\in\mathcal{W}_{\pm}$. Using the expansions of
$|w_{\pm}\rangle$ in equations (312) and (313), we obtain the action of
$\mathcal{P}_{k}\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{P}_{k}$ on
$\mathcal{P}_{k}\mbox{$|w_{\pm}\rangle$}$.

$\displaystyle\ \mathcal{P}_{k}\mathrm{T}_{1}^{(\mathrm{c})}\mathcal{P}_{k}\
\left(\mathcal{P}_{k}\ \mbox{$|w_{\pm}\rangle$}\right)$ $\displaystyle=$
$\displaystyle\ \dfrac{1}{4}\ \sum_{a,b\in{\mathbb{F}_{2}}_{*}^{2m}}\
\lambda_{b}\ \left(\frac{1}{3}\
\mbox{$\langle\hat{a}|$}\mathcal{P}_{k}\mbox{$|\hat{b}\rangle$}\right)\
\mbox{$|\hat{a}\rangle$},$ $\displaystyle=$ $\displaystyle\ \begin{cases}\ \ \
\ \dfrac{1}{12}\left(1-\dfrac{2}{N+2}\right)\
\mathcal{P}_{0}\mbox{$|w_{+}\rangle$}\\\ \ \ \ \
\dfrac{1}{12}\left(1+\dfrac{2}{N-2}\right)\
\mathcal{P}_{0}\mbox{$|w_{-}\rangle$}\\\ \ \ \ \
\dfrac{1}{12}\left(1+\dfrac{1}{N+2}\right)\
\mathcal{P}_{1}\mbox{$|w_{+}\rangle$}\\\ \ \ \ \
\dfrac{1}{12}\left(1-\dfrac{1}{N-2}\right)\
\mathcal{P}_{1}\mbox{$|w_{-}\rangle$}\\\ \ \ \ \
\dfrac{1}{12}\left(1+\dfrac{1}{N+2}\right)\
\mathcal{P}_{2}\mbox{$|w_{+}\rangle$}\\\ \ \ \ \
\dfrac{1}{12}\left(1-\dfrac{1}{N-2}\right)\
\mathcal{P}_{2}\mbox{$|w_{-}\rangle$}\end{cases},$ (322)

where we used the inner product between
$\mathcal{P}_{k}\mbox{$|\hat{a}\rangle$}$ and
$\mathcal{P}_{k}\mbox{$|\hat{b}\rangle$}$, which is

$\displaystyle\ \mbox{$\langle\hat{a}|$}\ \mathcal{P}_{k}\
\mbox{$|\hat{b}\rangle$}\ =\ \begin{cases}\frac{1}{3}\ \left(\delta_{a,b}\ +\
\frac{4}{N^{2}-4}\ \left(1-\delta_{a,b}\right)\delta_{\left\langle
a,b\right\rangle,0}\right),\;k=0,\\\ \frac{1}{3}\ \left(\delta_{a,b}\ -\
\frac{2}{N^{2}-4}\ \left(1-\delta_{a,b}\right)\delta_{\left\langle
a,b\right\rangle,0}\right),\;k=1,2,\end{cases}$ (323)

and also used the relation

$\sum_{\begin{subarray}{c}b:\\\ \left\langle a,b\right\rangle=0\\\ b\neq
a\end{subarray}}\ \lambda_{b}\ =-\sum_{\begin{subarray}{c}b:\\\ \left\langle
a,b\right\rangle=1\end{subarray}}\ \lambda_{b}\ \ -\lambda_{a}\ =\
-\left(\dfrac{\pm N\ +\ 2}{2}\right)\lambda,$ (324)

which we obtain from equations (312) and (313).

Hence the full eigen decomposition of
$\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}$ is given by

$\displaystyle\hat{\mathcal{G}_{\mathcal{T}}}_{\mathrm{D}}\ =$ $\displaystyle\
\ \frac{3}{4}\mathcal{P}_{\mathrm{tr}}\ +\
\frac{1}{4}\left(1+\frac{2}{N-2}\right)\ \mathcal{P}_{0,-}\ +\
\frac{1}{4}\left(1+\frac{1}{N+2}\right)\
\left(\mathcal{P}_{1,+}+\mathcal{P}_{2,+}\right)\ $ $\displaystyle+\
\frac{1}{4}\left(1-\frac{1}{N-2}\right)\
\left(\mathcal{P}_{1,-}+\mathcal{P}_{2,-}\right)\ +\
\frac{1}{4}\left(1-\frac{1}{N+2}\right)\ \mathcal{P}_{0,+}.$ (325)

∎

##  Appendix C Proofs of equation (177)

On the RHS of equation (173), we see that the first term is given by

$\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle b,h\right\rangle=0\end{subarray}}\
\mbox{$|a,b,a+b\rangle$}\ =\ \dfrac{1}{4}\ \mbox{$|a,b,a+b\rangle$},$ (326)

where we used Lemma 2 to arrive at the result. We simplify the term in the
expression (174) by observing the following: for each $h$ which satisfies
$\left\langle a,h\right\rangle=0$, $\left\langle b,h\right\rangle=1$, $b+h$
will satisfy $\left\langle a,b+h\right\rangle=\left\langle
b,b+h\right\rangle=1$ (using $\left\langle a,b\right\rangle=1$). Similarly,
for each $h^{\prime}$ that satisfies $\left\langle
a,h^{\prime}\right\rangle=\left\langle b,h^{\prime}\right\rangle=1$,
$a+h^{\prime}$ will satisfy $\left\langle a,b+h^{\prime}\right\rangle=0$, and
$\left\langle b,b+h^{\prime}\right\rangle=1$. We replace $b+h$ in the term
$\left|a\;\;\;\;\;b+h\;\;\;\;\;a+b+h\right\rangle$ with $h^{\prime}=h+b$, thus
obtaining $|a,h^{\prime},b+h^{\prime}\rangle$ and sum over $h^{\prime}$ which
satisfy $\left\langle a,h^{\prime}\right\rangle=\left\langle
b,h^{\prime}\right\rangle=1$. Removing the primed symbol from $h^{\prime}$
(since it is a dummy variable), we see that the expression in equation (174)
is

$\ \dfrac{1}{N^{2}}\ \ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle b,h\right\rangle=1\end{subarray}}\
\left|a\;\;\;\;\;b+h\;\;\;\;\;a+b+h\right\rangle\ =\ \dfrac{1}{N^{2}}\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=1\\\ \left\langle
b,h\right\rangle=1\end{subarray}}\ \mbox{$|a,h,a+h\rangle$}.$ (327)

The expression in equation (175) is simplified in a similar way to give us

$\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle b,h\right\rangle=0\end{subarray}}\
\left|a+h\;\;\;\;\;b\;\;\;\;\;a+b+h\right\rangle\ =\ \dfrac{1}{N^{2}}\ \
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=1\\\ \left\langle
b,h\right\rangle=1\end{subarray}}\ \mbox{$|h,b,b+h\rangle$}.$ (328)

Finally, the expression in equation (176) simplifies to give us

$\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle b,h\right\rangle=1\end{subarray}}\
\left|a+h\;\;\;\;\;b+h\;\;\;\;\;a+b\right\rangle\ =\ \dfrac{1}{N^{2}}\ \
\sum_{\begin{subarray}{c}h:\\\ \left\langle b,h\right\rangle=1\\\ \left\langle
a+b,h\right\rangle=1\end{subarray}}\ \mbox{$|a+b+h,h,a+b\rangle$}.$ (329)

##  Appendix D Permutation relations among
$\mathrm{T}_{1,\pm}^{(\mathrm{nc})}$, $\mathrm{T}_{2,\pm}^{(\mathrm{nc})}$ and
$\mathrm{T}_{3,\pm}^{(\mathrm{nc})}$

We prove the following

  * •

$\mathrm{T}_{1,\pm}^{(\mathrm{nc})}$, $\mathrm{T}_{2,\pm}^{(\mathrm{nc})}$ and
$\mathrm{T}_{3,\pm}^{(\mathrm{nc})}$ commute with $\mathcal{W}_{[(23)]}$,
$\mathcal{W}_{[(13)]}$, and $\mathcal{W}_{[(12)]}$ respectively.

$\displaystyle\mathcal{W}_{[(23)]}\ \mathrm{T}_{1,\pm}^{(\mathrm{nc})}\
\mathcal{W}_{[(23)]}\ =\ \mathrm{T}_{1,\pm}^{(\mathrm{nc})},$ (330)
$\displaystyle\mathcal{W}_{[(13)]}\ \mathrm{T}_{2,\pm}^{(\mathrm{nc})}\
\mathcal{W}_{[(13)]}\ =\ \mathrm{T}_{2,\pm}^{(\mathrm{nc})},$ (331)
$\displaystyle\mathcal{W}_{[(12)]}\ \mathrm{T}_{3,\pm}^{(\mathrm{nc})}\
\mathcal{W}_{[(12)]}\ =\ \mathrm{T}_{3,\pm}^{(\mathrm{nc})}.$ (332)

  * •

$\mathrm{T}_{1,\pm}^{(\mathrm{nc})}$, $\mathrm{T}_{2,\pm}^{(\mathrm{nc})}$ and
$\mathrm{T}_{3,\pm}^{(\mathrm{nc})}$ are permutations of each other:

$\displaystyle\mathrm{T}_{3,\pm}^{(\mathrm{nc})}\ =$ $\displaystyle\
\mathcal{W}_{[(13)]}\ \mathrm{T}_{1,\pm}^{(\mathrm{nc})}\
\mathcal{W}_{[(13)]}$ (333) $\displaystyle=$ $\displaystyle\ \
\mathcal{W}_{[(132)]}\
\mathrm{T}_{1,\pm}^{(\mathrm{nc})}\mathcal{W}_{[(123)]}$ (334)
$\displaystyle=$ $\displaystyle\ \mathcal{W}_{[(123)]}\
\mathrm{T}_{2,\pm}^{(\mathrm{nc})}\mathcal{W}_{[(132)]}$ $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(23)]}\ \mathrm{T}_{2,\pm}^{(\mathrm{nc})}\
\mathcal{W}_{[(23)]},$ $\displaystyle\ \mathrm{T}_{2,\pm}^{(\mathrm{nc})}\ =\
$ $\displaystyle\ \mathcal{W}_{[(123)]}\
\mathrm{T}_{1,\pm}^{(\mathrm{nc})}\mathcal{W}_{[(132)]}$ (335)
$\displaystyle=$ $\displaystyle\ \mathcal{W}_{[(12)]}\
\mathrm{T}_{1,\pm}^{(\mathrm{nc})}\ \mathcal{W}_{[(12)]}$ $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(132)]}\
\mathrm{T}_{3,\pm}^{(\mathrm{nc})}\mathcal{W}_{[(123)]}$ $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(23)]}\ \mathrm{T}_{3,\pm}^{(\mathrm{nc})}\
\mathcal{W}_{[(23)]}$ $\displaystyle\mathrm{T}_{1,\pm}^{(\mathrm{nc})}\ =$
$\displaystyle\ \mathcal{W}_{[(123)]}\
\mathrm{T}_{3,\pm}^{(\mathrm{nc})}\mathcal{W}_{[(132)]}$ (336)
$\displaystyle=$ $\displaystyle\ \mathcal{W}_{[(13)]}\
\mathrm{T}_{3,\pm}^{(\mathrm{nc})}\ \mathcal{W}_{[(13)]}$ $\displaystyle=$
$\displaystyle\ \mathcal{W}_{[(132)]}\
\mathrm{T}_{2,\pm}^{(\mathrm{nc})}\mathcal{W}_{[(123)]}$ $\displaystyle=\ $
$\displaystyle\mathcal{W}_{[(12)]}\ \mathrm{T}_{2,\pm}^{(\mathrm{nc})}\
\mathcal{W}_{[(12)]},$

  * •

$\hat{\mathcal{G}_{\mathcal{T}}}_{\pm}$ commutes with $\mathcal{W}_{\pi}$ for
all $\pi\in\mathrm{S}_{3}$.

###### Proof.

Equation (330) is proved by conjugating both sides of equations (183) and
(184) by $\mathcal{W}_{[(23)]}$, and noting that the RHS of both equations
remain invariant. Equations (331) and (332) can be proved similarly.

Equation (333) is proved by conjugating equations (183) and (184) by
$\mathcal{W}_{[(13)]}$, and by noting that $[(13)][(23)][(13)]=[(12)]$, and
using equation (198), one then notes that the RHS are the same as equations
(187) and (188) respectively. To prove relation (334), conjugate equation
(330) by $\mathcal{W}_{[(13)]}$, and use equation (333) and the relation:
$[(132)]=[(13)][(23)]$. All other equations can be proved using identical
steps.

That $\hat{\mathcal{G}_{\mathcal{T}}}_{\pm}$ commutes with $\mathcal{W}_{\pi}$
for all $\pi\in\mathrm{S}_{3}$ now follows since
$\hat{\mathcal{G}_{\mathcal{T}}}_{\pm}=\mathrm{T}_{1,\pm}^{(\mathrm{nc})}+\mathrm{T}_{2,\pm}^{(\mathrm{nc})}+\mathrm{T}_{3,\pm}^{(\mathrm{nc})}$.
∎

##  Appendix E Inner elements of the orthogonal complement of r products
between vectors in $\mathcal{V}_{1,+}^{(\mathrm{(nc)}}$

We prove

$\displaystyle\
\mbox{$\langle\hat{a}|$}\mathcal{P}_{\mathrm{S}}\mbox{$|\hat{b}\rangle$}\ =\
\frac{1}{3}\ \left(\delta_{a,b}\ +\ \dfrac{4}{N^{2}}\ \delta_{\left\langle
a,b\right\rangle,1}\right).$ (337) $\displaystyle\ \mbox{$\langle\hat{a}|$}\
\mathcal{P}_{\omega}\ \mbox{$|\hat{b}\rangle$}\ =\ \mbox{$\langle\hat{a}|$}\
\mathcal{P}_{\omega^{*}}\ \mbox{$|\hat{b}\rangle$}\ =\ \frac{1}{3}\
\left(\delta_{a,b}\ -\ \dfrac{2}{N^{2}}\ \delta_{\left\langle
a,b\right\rangle,1}\right),$ (338)

by proving it for general $\mathcal{P}_{(}k)$ defined in equation (135).

$\displaystyle\ \mbox{$\langle\hat{a}|$}\ \mathcal{P}_{k}\
\mbox{$|\hat{b}\rangle$}$ $\displaystyle=$ $\displaystyle\
\left(\frac{\sqrt{2}}{N}\sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\end{subarray}}\mbox{$\langle a,h,a+h|$}\ \right)\
\frac{1}{3}\left(\mathrm{id}+\omega^{k}\ \mathcal{W}_{[(123)]}\ +\omega^{2k}\
\mathcal{W}_{[(132)]}\right)\
\left(\frac{\sqrt{2}}{N}\sum_{\begin{subarray}{c}k:\\\ \left\langle
b,k\right\rangle=1\end{subarray}}\mbox{$|b,k,a+k\rangle$}\right)$
$\displaystyle=$ $\displaystyle\
\frac{2}{N^{2}}\sum_{\begin{subarray}{c}h,k:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle
b,k\right\rangle=1\end{subarray}}\frac{1}{3}\left(\delta_{a,b}\
\delta_{h,k}+\omega^{k}\ \delta_{h,b}\ \delta_{k,a+b}\ +\omega^{2k}\
\delta_{k,a}\ \delta_{h,a+b}\right)$ $\displaystyle=$ $\displaystyle\
\frac{1}{3}\ \left(\ \delta_{a,b}\ +\ \left(\omega^{k}+\omega^{2k}\right)\
\frac{2}{N^{2}}\ \delta_{\left\langle a,b\right\rangle,1}\ \right).$
$\displaystyle=$ $\displaystyle\ \begin{cases}\frac{1}{3}\ \left(\delta_{a,b}\
+\ \frac{4}{N^{2}}\ \delta_{\left\langle a,b\right\rangle,1}\right),\;k=0,\\\
\frac{1}{3}\ \left(\delta_{a,b}\ -\ \frac{2}{N^{2}}\ \delta_{\left\langle
a,b\right\rangle,1}\right),\;k=1,2,\end{cases}$ (339)

which proves equations (234) and (338).

##  Appendix F Proofs of equation (248)

On the RHS of equation (244), we see that the first term is given by

$\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle b,h\right\rangle=0\end{subarray}}\
\mbox{$|a,b,a+b\rangle$}\ =\ \dfrac{1}{4}\ \mbox{$|a,b,a+b\rangle$},$ (340)

where we used Lemma 2 to arrive at the result. We simplify the term in the
expression (245) by observing the following: for each $h$ which satisfies
$\left\langle a,h\right\rangle=0$, $\left\langle b,h\right\rangle=1$, $b+h$
will satisfy $\left\langle a,b+h\right\rangle=0$, and $\left\langle
b,b+h\right\rangle=1$ (using $\left\langle a,b\right\rangle=0$). Similarly,
for each $h^{\prime}$ that satisfies $\left\langle
a,h^{\prime}\right\rangle=0$, and $\left\langle b,h^{\prime}\right\rangle=1$,
$a+h^{\prime}$ will satisfy $\left\langle a,b+h^{\prime}\right\rangle=0$, and
$\left\langle b,b+h^{\prime}\right\rangle=1$. We replace $b+h$ in the term
$\left|a\;\;\;\;\;b+h\;\;\;\;\;a+b+h\right\rangle$ with $h^{\prime}=h+b$, thus
obtaining $|a,h^{\prime},b+h^{\prime}\rangle$ and sum over $h^{\prime}$ which
satisfy $\left\langle a,h^{\prime}\right\rangle=0$, $\left\langle
b,h^{\prime}\right\rangle=1$. Removing the primed symbol from $h^{\prime}$
(since it is a dummy variable), we see that the expression in equation (245)
is

$\ \dfrac{1}{N^{2}}\ \ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle b,h\right\rangle=1\end{subarray}}\
\left|a\;\;\;\;\;b+h\;\;\;\;\;a+b+h\right\rangle\ =\ \dfrac{1}{N^{2}}\
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=0\\\ \left\langle
b,h\right\rangle=1\end{subarray}}\ \mbox{$|a,h,a+h\rangle$}.$ (341)

The expression in equation (246) is simplified in a similar way to give us

$\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle b,h\right\rangle=0\end{subarray}}\
\left|a+h\;\;\;\;\;b\;\;\;\;\;a+b+h\right\rangle\ =\ \dfrac{1}{N^{2}}\ \
\sum_{\begin{subarray}{c}h:\\\ \left\langle a,h\right\rangle=1\\\ \left\langle
b,h\right\rangle=0\end{subarray}}\ \mbox{$|h,b,b+h\rangle$}.$ (342)

Finally, the expression in equation (247) simplifies to give us

$\dfrac{1}{N^{2}}\ \sum_{\begin{subarray}{c}h:\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle b,h\right\rangle=1\end{subarray}}\
\left|a+h\;\;\;\;\;b+h\;\;\;\;\;a+b\right\rangle\ =\ \dfrac{1}{N^{2}}\ \
\sum_{\begin{subarray}{c}h:\\\ \left\langle b,h\right\rangle=1\\\ \left\langle
a+b,h\right\rangle=0\end{subarray}}\ \mbox{$|a+b+h,h,a+b\rangle$}.$ (343)

##  Appendix G Proof of Theorem 4

We begin by proving equation (39). For that, first note that
$\dim\mathrm{span}\left\\{\mbox{$|a,a\rangle$}\ |\
a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}=N^{2}-1$. Also, the vectors
$|a,a\rangle$ form an ONB for $\mathrm{span}\left\\{\mbox{$|a,a\rangle$}\ |\
a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}$. From equation (G) in Lemma 28, see
that $\dim\mathrm{span}\left\\{\mbox{$|\hat{a}\rangle$}\ |\
a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}=N^{2}-1$. Also, the vectors
$|\hat{a}\rangle$ are an ONB for
$\mathrm{span}\left\\{\mbox{$|\hat{a}\rangle$}\ |\
a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}$. Hence the dimensions of both spaces
are the same. Next note that

$\mathcal{U}_{h}\ \mbox{$|a,a\rangle$}\ =\mbox{$|a+\left\langle
a,h\right\rangle h,a+\left\langle a,h\right\rangle h\rangle$},\ \forall\
h\in\mathbb{F}_{2}^{2m},$ (344)

which can be directly verified by substituting from equation (64) into
equation (70) for $t=2$. Next from Lemma 30, note that by putting $b=0$ in
equation (354), we get

$\mathcal{U}_{h}\ \mbox{$|\hat{a}\rangle$}\ =\ \mbox{$|\hat{a}+\left\langle
a,h\right\rangle\hat{h}\rangle$}.$ (345)

Thus it is seen that $L^{(d)\rightarrow(\mathrm{nc})}$ satisfies the equation
$\mathcal{U}_{h}\ L^{(d)\rightarrow(\mathrm{nc})}\ \mbox{$|a,a\rangle$}\ =\
L^{(d)\rightarrow(\mathrm{nc})}\ \mathcal{U}_{h}\ \mbox{$|a,a\rangle$}$ for
all $a\in{\mathbb{F}_{2}}_{*}^{2m}$. Note that $\mathcal{C}_{m}$ is generated
by all Clifford transvection and Paulis. The action of the diagonal sector
subspace (Lemma 4 in Helsen2016 ) is invariant under the adjoint action of
Paulis, just like all vectors of the form $|a,b,a+b\rangle$ are (see Lemma 5).
This immediately tells us that $L^{(d)\rightarrow(\mathrm{nc})}$ is an
intertwiner between both representations of the Clifford group. This proves
equation (38).

Next, we prove that the subrepresentations acting on the subspaces
$V_{1}\oplus V_{2}$ in Lemma 4 of Helsen2016  are equivalent to the
subrepresentations acting on $V^{(\mathrm{nc})}_{1}\oplus
V^{(\mathrm{nc})}_{2}$ in equation (40). Lemma 4 in Helsen2016  and Lemma 29
tell us that $\dim\left(V_{1}\oplus
V_{2}\right)=\dim\left(V^{(\mathrm{nc})}_{1}\oplus
V^{(\mathrm{nc})}_{2}\right)=N^{2}-2$. Lemma 30 informs us that for any
transvection Clifford $\mathcal{U}_{h}$, and non-zero vector
$\mbox{$|x\rangle$}\in V^{(\mathrm{nc})}_{1}\oplus V^{(\mathrm{nc})}_{2}$,
with the expansion
$\mbox{$|x\rangle$}=\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}\mbox{$|A(a)\rangle$}$,
we must have that $\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}=0$ (Lemma
29), we get that
$\mathcal{U}_{h}\mbox{$|x\rangle$}=\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}\mbox{$|A(a+\left\langle
h,a\right\rangle h)\rangle$}$ (using equation (354)). Similarly for the same
choice of $\lambda_{a}$, the vector $\mbox{$|v\rangle$}\in V_{1}\oplus V_{2}$
in the diagonal sector subspace, $\mbox{$|v\rangle$}=\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}\mbox{$|a,a\rangle$}$, the
action of the same transvection Clifford is
$\mathcal{U}_{h}\mbox{$|v\rangle$}\ =\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}\mbox{$|a+\left\langle
a,h\right\rangle h,a+\left\langle a,h\right\rangle h\rangle$}$. Define
$K:V^{(\mathrm{nc})}\oplus V^{(\mathrm{nc})}\rightarrow V_{1}\oplus V_{2}$ as

$K\left(\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}\mbox{$|A(a)\rangle$}\right)\
=\ \sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}\mbox{$|a,a\rangle$},$ (346)

with $\lambda_{a}$ such that
$\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}=0$. It is easily verified
that $K$ is linear. Thus $K$ is an intertwiner between both subrepresentations
of the Clifford group.
First note that $\mathrm{P}_{S}$ and $\mathcal{P}_{1}$ are both projectors,
such that $\mathcal{P}_{s}\mathcal{P}_{1}=0$, i.e., they have orthogonal
supports. Next, $\mathcal{G}_{\mathcal{T}}$ commutes with $W_{\sigma}$ for all
$\sigma\in\mathrm{S}_{3}$. Thus, the action of $\mathcal{G}_{\mathcal{T}}$ on
$V^{(\mathrm{nc})}_{\mathrm{null,1}}$ and on
$V^{(\mathrm{nc})}_{\mathrm{null,S}}$ is completely determined by the action
of $\mathcal{G}_{\mathcal{T}}$ on the subspace spanned by
$\mbox{$|a;b\rangle$}+\mbox{$|b;a+b\rangle$}+\mbox{$|a+b;a\rangle$}$ for
$a,b\in{\mathbb{F}_{2}}_{*}^{2m}$ such that $\left\langle a,b\right\rangle=0$
and $a\neq b$. Lemma 31 tells us that $\left(\mathcal{G}_{\mathcal{T}}-\
\frac{1}{4}\mathrm{id}\right)\left(\mbox{$|\hat{a};b\rangle$}+\mbox{$|\hat{b};a+b\rangle$}+\mbox{$|\hat{a+b};a\rangle$}\right)=0$.
Thus vectors of the form
$\mbox{$|\hat{a};b\rangle$}+\mbox{$|\hat{b};a+b\rangle$}+\mbox{$|\hat{a+b};a\rangle$}$
lie in the $1/4$ eigenspace of $\mathcal{G}_{\mathcal{T}}$. Since
$\mathcal{G}_{\mathcal{T}}$ commutes with $\mathcal{U}_{h}$ for each
$h\in\mathbb{F}_{2}^{2m}$, and since $\mathcal{U}_{h}$ generate the action of
the Clifford group on $\mathrm{span}\left\\{\mbox{$|a,b,a+b\rangle$}\ |\
a,b\in{\mathbb{F}_{2}}_{*}^{2m},\ a\neq b\right\\}$, the result follows from a
corollary of Schur’s lemma.

######  Lemma 28.

For any $a,b\in{\mathbb{F}_{2}}_{*}^{2m}$, we have that

$\mbox{$\langle A(a)|A(b)\rangle$}\ =\
\left(\dfrac{N^{2}}{2}-1\right)\delta_{a,b}\ -\ \delta_{\left\langle
a,b\right\rangle,0}.$ (347)

###### Proof.

From equation (37), we get that for any $h,k\in{\mathbb{F}_{2}}_{*}^{2m}$ and
$a,b\in\mathbb{F}_{2}^{2m}$,

$\langle\hat{h};a|\hat{k};b\rangle$ $\displaystyle\ =\ $
$\displaystyle\frac{2}{N^{2}}\left(\sum_{s:\left\langle s,h\right\rangle=1}\
\left(-1\right)^{\left\langle s,a\right\rangle}\ \mbox{$\langle
h,s,h+s|$}\right)\ \left(\sum_{t:\left\langle t,k\right\rangle=1}\
\left(-1\right)^{\left\langle t,b\right\rangle}\
\mbox{$|k,t,k+t\rangle$}\right)$ $\displaystyle\ =\ $
$\displaystyle\frac{2}{N^{2}}\ \sum_{s:\left\langle s,h\right\rangle=1}\
\sum_{t:\left\langle t,k\right\rangle=1}\ \ \left(-1\right)^{\left\langle
s,a\right\rangle+\left\langle t,b\right\rangle}\ \mbox{$\langle
h,s,h+s|k,t,k+t\rangle$}$ $\displaystyle\ =\ $ $\displaystyle\frac{2}{N^{2}}\
\delta_{h,k}\ \sum_{s:\left\langle s,h\right\rangle=1}\
\left(-1\right)^{\left\langle s,a+b\right\rangle},$ $\displaystyle\ =\ $
$\displaystyle\delta_{h,k}\left(\delta_{a,b}\ -\ \delta_{h,a+b}\right).$ (348)

The last line can be explained as follows. If $a=b$, then from Lemma 2 there
are $N^{2}/2$ $s\in\mathbb{F}_{2}$ such that $\left\langle
s,h\right\rangle=1$, and we get that
$\mbox{$\langle\hat{h};a|\hat{k};b\rangle$}=\delta_{h,k}\delta_{a,b}$. If
$h=a+b$, then, using the same arguments as above we see that
$\mbox{$\langle\hat{h};a|\hat{k};b\rangle$}=-\delta_{h,k}\delta_{h,a+b}$. When
$h$ and $a+b$ are linearly independent, then Lemma 2 informs us that there are
$N^{2}/4$ solutions for $s$ for the simultaneous equations $\left\langle
s,h\right\rangle=1$, $\left\langle s,a+b\right\rangle=0$, and again $N^{2}/4$
solutions for the simultaneous equations $\left\langle s,h\right\rangle=1$,
$\left\langle s,a+b\right\rangle=1$. Thus, the summation gives $0$.
Next, from equation (41),

$\langle A(a)|A(b)\rangle$ $\displaystyle\ =\ $ $\displaystyle\left(\
\sum_{h:\begin{subarray}{c}\left\langle h,a\right\rangle=0\\\ h\neq
a,0\end{subarray}}\ \mbox{$\langle\hat{h};a|$}\right)\left(\
\sum_{k:\begin{subarray}{c}\left\langle k,b\right\rangle=0\\\ k\neq
b,0\end{subarray}}\ \mbox{$|\hat{k};b\rangle$}\right)$ $\displaystyle\ =\ $
$\displaystyle\ \sum_{h:\begin{subarray}{c}\left\langle h,a\right\rangle=0\\\
h\neq a,0\end{subarray}}\ \sum_{k:\begin{subarray}{c}\left\langle
k,b\right\rangle=0\\\ k\neq b,0\end{subarray}}\
\delta_{h,k}\left(\delta_{a,b}\ -\ \delta_{h,a+b}\right)$ $\displaystyle\ =\ $
$\displaystyle\ \delta_{a,b}\left(\sum_{h:\begin{subarray}{c}\left\langle
h,a\right\rangle=0\\\ h\neq a,0\end{subarray}}\ 1\right)\
-\left(\delta_{\left\langle a,b\right\rangle,0}-\delta_{a,b}\right)$
$\displaystyle\ =\ $ $\displaystyle\ \delta_{a,b}\
\left(\frac{N^{2}}{2}-1\right)\ -\ \delta_{\left\langle a,b\right\rangle,0}.$
(349)

The second last line can be explained as follows: if $a=b$, we only need to
count the number of $h\in\mathbb{F}_{2}$ such that $\left\langle
h,a\right\rangle=0$, which is $N^{2}/2$ from Lemma 2. Excluding the cases
$h=0,a$, we get that this sum of $N^{2}/2-2$. Note that when $a=b$,
$\left\langle a,b\right\rangle=0$. This verifies the final line. If, instead,
$h=a+b$, we need to count over all $h$, which are non-zero, such that
$\left\langle h,a\right\rangle=\left\langle h,b\right\rangle=0$, and $h=a+b$.
If $\left\langle a,b\right\rangle=1$, then these conditions won’t be
satisfied. If, $\left\langle a,b\right\rangle=0$, this condition is satisfies
only for one choice of $h$, i.e., $h=a+b$. Note that $h\neq 0$. This implies
that $a\neq b$. Thus the sum, in this case simply becomes
$-\delta_{\left\langle a,b\right\rangle,0}$. Thus we see that the final line
is correct in this case. Hence equation (41) is proved. ∎

######  Lemma 29.

The dimension of the subspace spanned by vector $|A(a)\rangle$, which are
defined in equation (41) is

$\dim\mathrm{span}\left\\{\mbox{$|A(a)\rangle$}\ |\
a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}=N^{2}-2.$ (350)

We also have that
$\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\mbox{$|A(a)\rangle$}=0$, and any non-
zero vector $\mbox{$|x\rangle$}\in\mathrm{span}\left\\{\mbox{$|A(a)\rangle$}\
|\ a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}$ of the form $\mbox{$|x\rangle$}\ =\
\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}\mbox{$|A(a)\rangle$}$, has the
property $\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}=0$, and for any
choice of $\lambda_{a}$ such that
$\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}|\lambda_{a}|^{2}>0$, and
$\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}=0$, $|x\rangle$ is non-zero.

###### Proof.

Consider an $\left(N^{2}-1\right)\times\left(N^{2}-1\right)$ matrix $H$, whose
matrix elements are indexed by vectors in ${\mathbb{F}_{2}}_{*}^{2m}$, and

$H_{a,b}\ =\ \delta_{\left\langle a,b\right\rangle,0},\ \forall\
a,b\in{\mathbb{F}_{2}}_{*}^{2m}.$ (351)

Using Lemma 2, we see that $\sum_{b\in{\mathbb{F}_{2}}_{*}^{2m}}H_{a,b}\ =\
\frac{N^{2}}{2}-1$, for all $a\in{\mathbb{F}_{2}}_{*}^{2m}$. Thus
$\frac{N^{2}}{2}-1$ is an eigenvalue of $H$. We compute the matrix elements of
$H^{2}$ as follows.

$\displaystyle\left(H^{2}\right)_{a,b}\ =\ \left|\left\\{\
c\in{\mathbb{F}_{2}}_{*}^{2m}\ |\ \left\langle a,c\right\rangle=\left\langle
b,c\right\rangle=0\right\\}\right|\ =\ \dfrac{N^{2}}{4}\delta_{a,b}\ +\
\dfrac{N^{2}}{4}-1,$ (352)

which we obtain as follows: when $a=b$, Lemma 2 tell us that $\left|\left\\{\
c\in{\mathbb{F}_{2}}_{*}^{2m}\ |\ \left\langle
a,c\right\rangle=0\right\\}\right|=\frac{N^{2}}{2}-1$, whereas when $a\neq b$,
Lemma 2 tell us that $\left|\left\\{\ c\in{\mathbb{F}_{2}}_{*}^{2m}\ |\
\left\langle a,c\right\rangle=\left\langle
b,c\right\rangle=0\right\\}\right|=\frac{N^{2}}{4}-1$. This agrees with the
final line in equation (352). Let $J$ be an
$\left(N^{2}-1\right)\times\left(N^{2}-1\right)$ matrix, whose all matrix
elements are $1$. Then $H$ satisfies the equation:

$H^{2}\ =\ \frac{N^{2}}{4}\mathbbmss{1}_{N^{2}-1}\ +\
\left(\frac{N^{2}}{4}-1\right)\ J.$ (353)

$\mathrm{Rank}J=1$, and the corresponding eigenspace of $H^{2}$ is given by
$\sum_{b\in{\mathbb{F}_{2}}_{*}^{2m}}\left(H^{2}\right)_{a,b}\ =\
\frac{N^{2}-4}{4}$, as expected, since it should be the square of the
eigenvalue of $H$, for the same eigenvector of $H$. Since the eigenspaces of
the other eigenvalues of $H^{2}$ lie in $\mathrm{ker}J$, the other eigenvalues
have to be $\pm\frac{N}{2}$. The multiplicity of the eigenvalue $N^{2}/2-1$ is
simply $1$.
Let $G$ be the $\left(N^{2}-1\right)\times\left(N^{2}-1\right)$ gram matrix of
the vectors $|A(a)\rangle$ defined in equation (41). Then $G_{a,b}\ =\
\mbox{$\langle A(a)|A(b)\rangle$}$. From Lemma 28, we get that
$G=\left(\frac{N^{2}}{2}-1\right)\mathbbmss{1}_{N^{2}-1}\ -\ H$, thus it has
eigenvalues $0$, $\frac{N^{2}\pm N-2}{2}$, where the multiplicity of the
eigenvalue $0$ is $1$. This proves that
$\mathrm{rank}G=\dim\mathrm{span}\left\\{\mbox{$|A(a)\rangle$}\ |\
a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}=N^{2}-1$. This also proves that
$\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\mbox{$|A(a)\rangle$}=0$, since
$||\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\mbox{$|A(a)\rangle$}||^{2}=0$, which
is inferred by observing that $GJ=0$. Since
$\dim\mathrm{span}\left\\{\mbox{$|A(a)\rangle$}\ |\
a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}=N^{2}-2$, this tells us that any vector
$\mbox{$|x\rangle$}\in\mathrm{span}\left\\{\mbox{$|A(a)\rangle$}\ |\
a\in{\mathbb{F}_{2}}_{*}^{2m}\right\\}$ of the form
$\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}\mbox{$|A(a)\rangle$}$, where
$\sum_{a\in{\mathbb{F}_{2}}_{*}^{2m}}\lambda_{a}=0$, is non-zero (because its
norm, which may be computed from $G$ will be non-zero). ∎

######  Lemma 30.

For any $a,h\in{\mathbb{F}_{2}}_{*}^{2m}$, and $b\in\mathbb{F}_{2}^{2m}$, we
have that

$\displaystyle\mathcal{U}_{h}\ \mbox{$|\hat{a};b\rangle$}\ =\
\mbox{$|\hat{a}+\left\langle a,h\right\rangle\hat{h};b+\left\langle
b,h\right\rangle h\rangle$},$ (354) $\displaystyle\mathcal{U}_{h}\
\mbox{$|A(a)\rangle$}\ =\ \mbox{$|A\left(a+\left\langle a,h\right\rangle
h\right)\rangle$}.$ (355)

###### Proof.

We verify equation (354) first. From equation (37), and equation (148) we get
that

$\displaystyle\mathcal{U}_{h}\ \mbox{$|a;b\rangle$}\ \ \propto\ $
$\displaystyle\sum_{k:\left\langle k,a\right\rangle=1}\
\left(-1\right)^{\left\langle k,b\right\rangle}\ \mbox{$|a+\left\langle
h,a\right\rangle h,k+\left\langle k,h\right\rangle k,a+k+\left\langle
h,a+k\right\rangle h\rangle$}\ $ $\displaystyle\ =\ $
$\displaystyle\sum_{k^{\prime}:\left\langle k,a\right\rangle=1}\
\left(-1\right)^{\left\langle k^{\prime},b+\left\langle
h,b\right\rangle\right\rangle}\ \mbox{$|a+\left\langle h,a\right\rangle
h,k^{\prime},a+k^{\prime}+\left\langle h,a\right\rangle h\rangle$},$ (356)

where the last line is observed by noting that if for any $k$, we define
$k^{\prime}=k+\left\langle k,h\right\rangle h$, then $\left\langle
k^{\prime},a+\left\langle a,h\right\rangle h\right\rangle=\left\langle
k,a\right\rangle$. Replacing $a$ with $b$ we get, similarly, that
$\left\langle k^{\prime},b+\left\langle h,b\right\rangle
h\right\rangle=\left\langle k,b\right\rangle$. Summing over the new dummy
variables $k^{\prime}$ instead of the old variable $k$, we get the last line.
Next, from equation (41) and equation (354), we get

$\displaystyle\mathcal{U}_{h}\ \mbox{$|A(a)\rangle$}=\ $
$\displaystyle\sum_{k:\begin{subarray}{c}\left\langle k,a\right\rangle=0\\\
k\neq a,0\end{subarray}}\ \mathcal{U}_{h}\ \mbox{$|\hat{k};a\rangle$}$
$\displaystyle=\ $ $\displaystyle\sum_{k:\begin{subarray}{c}\left\langle
k,a\right\rangle=0\\\ h\neq a,0\end{subarray}}\ \mathcal{U}_{h}\
\mbox{$|\hat{k+\left\langle h,k\right\rangle h};a+\left\langle
a,h\right\rangle h\rangle$}$ $\displaystyle=\ $
$\displaystyle\sum_{k^{\prime}:\begin{subarray}{c}\left\langle
k^{\prime},a+\left\langle h,a\right\rangle h\right\rangle=0\\\ k^{\prime}\neq
0,a+\left\langle h,a\right\rangle h\end{subarray}}\
\mbox{$|\hat{k^{\prime}};a+\left\langle h,a\right\rangle h\rangle$},$ (357)

where we explain the last line as follows. For any $k$, denote
$k^{\prime}\coloneqq k+\left\langle h,k\right\rangle h$. Then we see that
$\left\langle k^{\prime},a\right\rangle=\left\langle k,a\right\rangle$. Also,
$k=\left\langle h,k\right\rangle h$, then $k=\left\langle h,\left\langle
h,k\right\rangle h\right\rangle h=0$. We rule out $k=0$ in the summation over
$k$. Similarly, if $k+a=\left\langle k+a,h\right\rangle h$, then $k+a=0$,
which we rule out in the summation. Thus $k^{\prime}\neq 0,a+\left\langle
h,a\right\rangle h$. Replacing the dummy variable $k$ with the newer dummy
variable $k^{\prime}$ thus gives us the last line. ∎

######  Lemma 31.

We have that

$\left(\mathcal{G}_{\mathcal{T}}-\
\frac{1}{4}\mathrm{id}\right)\left(\mbox{$|\hat{a};b\rangle$}+\mbox{$|\hat{b};a+b\rangle$}+\mbox{$|\hat{a+b};a\rangle$}\right)=0.$
(358)

###### Proof.

Note that

$\displaystyle\mathcal{G}_{\mathcal{T}}\left(\mbox{$|a;b\rangle$}+\mbox{$|b;a+b\rangle$}+\mbox{$|a+b;a\rangle$}\right)$
$\displaystyle\ \propto\ $ $\displaystyle\sum_{h\in\mathbb{F}_{2}^{2m}}\
\left(\begin{aligned} \ \mbox{$|a+\left\langle a,h\right\rangle
h;b+\left\langle b,h\right\rangle h\rangle$}\\\ +\mbox{$|b+\left\langle
b,h\right\rangle h;a+b+\left\langle a+b,h\right\rangle h\rangle$}\\\
+\mbox{$|a+b+\left\langle a+b,h\right\rangle h;a+\left\langle a,h\right\rangle
h\rangle$}\end{aligned}\right)$ $\displaystyle\ =\ $
$\displaystyle\sum_{\begin{subarray}{c}h\in\mathbb{F}_{2}^{2m}\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle
b,h\right\rangle=0\end{subarray}}\left(\mbox{$|a;b\rangle$}+\mbox{$|b;a+b\rangle$}+\mbox{$|a+b;a\rangle$}\right)$
$\displaystyle+$
$\displaystyle\sum_{\begin{subarray}{c}h\in\mathbb{F}_{2}^{2m}\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle
b,h\right\rangle=0\end{subarray}}\left(\mbox{$|a+h;b\rangle$}+\mbox{$|b;a+b+h\rangle$}+\mbox{$|a+b+h;a+h\rangle$}\right)$
$\displaystyle+$
$\displaystyle\sum_{\begin{subarray}{c}h\in\mathbb{F}_{2}^{2m}\\\ \left\langle
a,h\right\rangle=0\\\ \left\langle
b,h\right\rangle=1\end{subarray}}\left(\mbox{$|a;b+h\rangle$}+\mbox{$|b+h;a+b+h\rangle$}+\mbox{$|a+b+h;a\rangle$}\right)$
$\displaystyle+$
$\displaystyle\sum_{\begin{subarray}{c}h\in\mathbb{F}_{2}^{2m}\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle
b,h\right\rangle=1\end{subarray}}\left(\mbox{$|a+h;b+h\rangle$}+\mbox{$|b+h;a+b\rangle$}+\mbox{$|a+b;a+h\rangle$}\right).$
(359)

In the second line we re-express the summation

$\sum_{\begin{subarray}{c}h\in\mathbb{F}_{2}^{2m}\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle
b,h\right\rangle=0\end{subarray}}\mbox{$|a+h;b\rangle$}=\sum_{\begin{subarray}{c}h^{\prime}\in\mathbb{F}_{2}^{2m}\\\
\left\langle a,h^{\prime}\right\rangle=1\\\ \left\langle
b,h^{\prime}\right\rangle=0\end{subarray}}\ \mbox{$|h^{\prime};b\rangle$}$
(360)

by defining a new dummy variables $h^{\prime}\coloneqq a+h$ using the old
dummy variable $h$ and noting that it satisfies the conditions: $\left\langle
a,h^{\prime}\right\rangle=\left\langle a,h\right\rangle$ and $\left\langle
b,h^{\prime}\right\rangle=\left\langle b,h\right\rangle$. Similarly one may
express

$\sum_{\begin{subarray}{c}h\in\mathbb{F}_{2}^{2m}\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle
b,h\right\rangle=0\end{subarray}}\mbox{$|a+b+h;a+h\rangle$}=\sum_{\begin{subarray}{c}h^{\prime}\in\mathbb{F}_{2}^{2m}\\\
\left\langle a,h^{\prime}\right\rangle=1\\\ \left\langle
b,h^{\prime}\right\rangle=0\end{subarray}}\
\mbox{$|h^{\prime};h^{\prime}+b\rangle$},$ (361)

by defining a new dummy variable $h^{\prime}\coloneqq a+b+h$ using the old
dummy variable and noting that $\left\langle
h^{\prime},a\right\rangle=\left\langle h,a\right\rangle$ and $\left\langle
h^{\prime},b\right\rangle=\left\langle h,b\right\rangle$. Using equation (37),
we see that the sum of both term is $0$. The term

$\sum_{\begin{subarray}{c}h\in\mathbb{F}_{2}^{2m}\\\ \left\langle
a,h\right\rangle=1\\\ \left\langle
b,h\right\rangle=0\end{subarray}}\mbox{$|b;a+b+h\rangle$}=0$ (362)

because for any $h\in\left\\{k\in\mathbb{F}_{2}^{2m}\ |\ \left\langle
a,k\right\rangle=1,\ \left\langle b,k\right\rangle=0\right\\}$, $b+h$ will
also belong to the same set. Thus the vectors $|b;a+b+h\rangle$ and
$|b;a+h\rangle$ both occur in the summation and from equation (37) we see that
their sum is $0$. All terms will cancel out this way.
Above we explained in detail while the sum in the second line in equation (G)
is $0$. The third and fourth lines are also $0$, and this can be proved using
the same nature of arguments. This proves the lemma. ∎

##  Appendix H Proof of Theorem 5

Note that all the eigenspaces listed in Theorem 4 span
$\mathrm{supp}\dfrac{1}{2}\left(\mathrm{id}+W_{[(23)]}\right)$. Also note that
$W$ anti-commutes with $\dfrac{1}{2}\left(\mathrm{id}+W_{[(23)]}\right)$,
while commuting with the action of the Clifford group (due to Schur-Weyl
duality). This proves that if an irreducible representation within
$V^{(\mathrm{nc})}_{x}$ isn’t annihilated by $W$, then $W$’s action on takes
it to an orthogonal subspace, which carries the same irreducible
representation. Hence proved.

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