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§ ADDITIONAL PROPERTIES OF INTERWOVEN MATRICES

Here,  we present further properties of interwoven matrices which have particular applications in dynamical network theory of mixed kernels.  Specifically,  an interwoven matrix $\bm{P}$ is composed of the sequence of real matrices $\bm{\mathcal{M}} = \{ \bm{M}_{i} \,  | \,   \bm{M}_{i} \in \R^{N \times N} \}$ called constructor matrices such that the rows and columns of each matrix $\bm{M}_{i}$ are uniformly separated by zero elements,  preserving the structure of $\bm{M}_{i}$,  where the order of $\bm{\mathcal{M}}$ defines the sequence of spacing.  Formally we define an interwoven matrix $\bm{P}$ by

\begin{equation}\label{eqn: interwoven_Q}
\bm{P} = \sum_{i = 1}^{r} \bm{M}_{i} \otimes \text{diag} \lp \delta_{i,1},...,\delta_{i,r} \rp,
\end{equation}

where $\otimes$ is the Kronecker product and $\delta_{i,j}$ is the Kronecker delta function (<ref>) as defined in Section <ref>.  However,  here we do not assume that the constructor matrices $\bm{M}_{i}$ are nonnegative.   The following result for the diagonal decomposition of the $\bm{P}$.

Let $\bm{P}$ be the interwoven matrix defined as defined in equation<ref> and let $\bm{Q}$  be the permutation matrix such that

\begin{equation}
\bm{Q} = [  \bm{I}_{N} \otimes \bm{e}_{1} , ...,     \bm{I}_{N} \otimes \bm{e}_{r} ].
\end{equation}

where $\bm{e}_{i} =  \begin{bmatrix} \delta_{i,1}, \,  \cdots,  \,  \delta_{i,r}\end{bmatrix}^{T}$.  Then

\begin{equation}
\bm{Q}^{T} \bm{P} \bm{Q} = \bigoplus_{i= 1}^{r} \bm{M}_{i},
\end{equation}

where $\oplus$ is the direct sum of tensors.

Consider the permutation map $\tau: \{1,...,  rN\} \rightarrow \{1,...,rN \}$ such that

\begin{equation}\label{eqn:tau_map}
\tau \lp x \rp = \lp \lp x-1 \rp \,  \text{mod}\,  r  \rp N + \left\lfloor\dfrac{x-1}{r}\right\rfloor + 1
\end{equation}

which permutes the rows and columns of $\bm{P}$ so any row and columns $\bm{M}_{i}$ of $\bm{P}$ become adjacent for $1 \leq i \leq r $.  In cycle notation,  $\tau \lp x \rp $ defines the mapping

\begin{equation}\label{eqn:cycle_perm}
\begin{pmatrix}
1 & 2& \cdots & r & r+1 & \cdots & rN \\
1 & N + 1 &  \cdots & (r-1)N + 1 & 2 & \cdots & (r-1)N + N \\
\end{pmatrix}
\end{equation}

which represents the column and row permutation of $\bm{P}$.  The cycle (<ref>) defined by equation (<ref>) yields the following matrix representation

\begin{equation}
\bm{Q} = [  \bm{I}_{N} \otimes \bm{e}_{1} , ...,     \bm{I}_{N} \otimes \bm{e}_{r} ],
\end{equation}

namely,  $ \bm{Q} _{i, \tau\lp i  \rp} = 1$ and zero entries else.  Therefore applying the transformation $\bm{Q}$ to $\bm{P}$ produces the block diagonal representation where

\begin{equation}
\bm{Q}^{T} \bm{P} \bm{Q} = \text{diag}\lp \bm{M}_{1},...,\bm{M}_{r} \rp
\end{equation}

which is by definition the direct sum of matrices $\bm{M}_{i}$.

The block diagonal representation of $\bm{P}$ following from <ref> motivates the subsequent properties involving the spectra and inverse of the interwoven matrix $\bm{P}$.

Let $\bm{P}$ be the interwoven matrix as defined in equation (<ref>).  Then $\bm{P}$ has the following properties:

\begin{equation}
\Sp \lp \bm{P} \rp   =  \bigcup_{i = 1}^{r} \Sp \lp \bm{M}_{i} \rp
\end{equation}

including multiplicities;

* if $\bm{M}_{i}$ is invertible for all $1 \leq i \leq k $,  then the inverse of the interwoven matrix $\bm{P}$ is the interweave of the inverse of the construction matrices. That is,

\begin{equation}
\bm{P}^{-1} = \sum_{i = 1}^{r} \bm{M}_{i}^{-1} \otimes \text{diag} \lp \delta_{i,1},...,\delta_{i,r} \rp,
\end{equation}

* the trace of the interwoven matrix is the sum of the traces of the constructor matrices

\begin{equation}
\Tr \lp \bm{P} \rp = \sum_{i=1}^{r} \Tr \lp  \bm{M}_{i }\rp,
\end{equation}

* the determinant of the interwoven matrix is the product of the determinant of the constructor matrices

\begin{equation}
\det \lp \bm{P} \rp = \prod_{i=1}^{r} \det \lp  \bm{M}_{i }\rp.
\end{equation}

Let $\lambda_{k,j} \in \Sp \lp \bm{M}_{k} \rp$ with its associated eigenvector $\bm{v}_{k,j}$.  Then define the interweave extension of $\bm{v}_{k,j}$ by

\begin{equation}
\tilde{\bm{v}}_{k,j} = \bm{v}_{k,j} \otimes \begin{bmatrix}
\delta_{j,1}\\
\vdots\\
\delta_{j,r}
\end{bmatrix}.
\end{equation}

For brevity,  denote the Kronecker diagonal matrix by $\bm{D}_{i} = \text{diag} \lp \delta_{i,1},...,\delta_{i,r} \rp$ and then by direct computation we have

\begin{align}
\bm{P} \tilde{\bm{v}}_{k,j}  =& \lp  \sum_{i = 1}^{r} \bm{M}_{i} \otimes \bm{D}_{i}   \rp  \tilde{\bm{v}}_{k,j}, \nonumber \\
=& \lp  \sum_{i = 1}^{r} \bm{M}_{i} \otimes \bm{D}_{i}  \rp   \lp   \bm{v}_{j}^{k} \otimes \begin{bmatrix} \delta_{j,1}, \,  \cdots,  \,  \delta_{j,r}\end{bmatrix}^{T}   \rp  , \nonumber\\
=&  \lp  \sum_{i = 1}^{r} \bm{M}_{i} \bm{v}_{k,j} \otimes \bm{D}_{i} \begin{bmatrix} \delta_{j,1}, \,  \cdots,  \,  \delta_{j,r}\end{bmatrix}^{T}   \rp , \nonumber \\
=& \,  \bm{M}_{j} \bm{v}_{k,j} \otimes   \begin{bmatrix} \delta_{j,1}, \,  \cdots,  \,  \delta_{j,r}\end{bmatrix}^{T} ,
\end{align}

where the last two equalities follow from the mixed product property of the Kronecker product and that direct multiplication of the Kronecker matrix and vector are non-zero only if $i = j$.  Therefore we have that

\begin{equation}
\bm{P} \tilde{\bm{v}}_{k,j}  = \bm{M}_{j} \bm{v}_{k,j} \otimes   \begin{bmatrix} \delta_{j,1}, \,  \cdots,  \,  \delta_{j,r}\end{bmatrix}^{T}  = \lambda_{k,j}  \bm{v}_{k,j} \otimes  \begin{bmatrix} \delta_{j,1}, \,  \cdots,  \,  \delta_{j,r}\end{bmatrix}^{T} = \lambda_{k,j} \tilde{\bm{v}}_{k,j} ,
\end{equation}

thus $ \lambda_{k,j} $ is an eigenvalue of $\bm{P}$ with associated eigenvector $\tilde{\bm{v}}_{k,j} $.  Next,  there exists $\bm{M}_{i}^{-1}$ for all $1 \leq i \leq r$ from the assumption in (ii).  Note that

\begin{equation}\label{kronecker_mat_prop}
\bm{D}_{i} \bm{D}_{j} = \left\{
\begin{array}{ll}
\bm{D}_{i}   &  i = j, \\
\bm{0}_{r \times r} &   i \neq j,
\end{array}
\right.
\end{equation}

then consider the following matrix $\bm{R}$ defined by the multiplication

\begin{align}
\bm{R} =& \lp  \sum_{i = 1}^{r} \bm{M}_{i} \otimes \bm{D}_{i} \rp \lp  \sum_{i = 1}^{r} \bm{M}_{i}^{-1} \otimes \bm{D}_{i} \rp, \nonumber \\
=& \lp \bm{M}_{1} \otimes \bm{D}_{1} \rp \lp  \sum_{i = 1}^{r} \bm{M}_{i}^{-1} \otimes \bm{D}_{i} \rp + ...  +  \lp \bm{M}_{r} \otimes \bm{D}_{r} \rp \lp  \sum_{i = 1}^{r} \bm{M}_{i}^{-1} \otimes \bm{D}_{i} \rp \label{inverse_comp} .
\end{align}

From the mixed-product property of the Kronecker product and equation<ref>,  we have that<ref> reduces to

\begin{align}
\bm{R} =& \,  \bm{M}_{1}    \bm{M} _{1}^{-1} \otimes \bm{D}_{1} + ... + \bm{M}_{r}    \bm{M} _{r}^{-1} \otimes \bm{D}_{r} , \nonumber \\
=& \sum_{i = 1} ^{r} \bm{I}_{n} \otimes \bm{D}_{i}  = \bm{I}_{rN} ,
\end{align}

hence the inverse of $\bm{P}$ is given by $\bm{P}^{-1} =  \sum_{i = 1}^{r} \bm{M}_{i}^{-1} \otimes \bm{D}_{i}  $ as required for (ii).

The trace of a Kronecker product is the product of the trace [42] such that $\Tr \lp \bm{M}_{i}  \otimes \bm{D}_{i}\rp = \Tr \lp \bm{M}_{i} \rp \Tr \lp \bm{D}_{i} \rp$.  Therefore applying the trace to the definition of $\bm{P}$ (<ref>) yields

\begin{align}
\Tr \lp  \bm{P} \rp &= \Tr \lp \sum_{i= 1}^{r} \bm{M}_{i} \otimes \bm{D}_{i} \rp = \sum_{i= 1}^{r} \Tr \lp \bm{M}_{i} \otimes \bm{D}_{i} \rp = \sum_{i= 1}^{r} \Tr \lp \bm{M}_{i}  \rp \Tr \lp \bm{D}_{i} \rp = \sum_{i= 1}^{r} \Tr \lp \bm{M}_{i}  \rp ,
\end{align}

where the second equality holds by the trace of the sum of matrices [42] and the fourth holds by $\Tr \lp \bm{D}_{i} \rp = 1$ for all $1 \leq i \leq r$.

Property (iv) follows immediately from (i) by expressing the determinant of a matrix as the product of the eigenvalues including multiplicities [42].  From (i) we have that $ \Sp \lp \bm{P} \rp   =  \Sp \lp \bm{M}_{1} \rp \cup ...\cup \Sp \lp \bm{M}_{r} \rp$ including multiplicities and so we know the eigenvalues of $\bm{P}$ are all the eigenvalues of each $\bm{M}_{i}$.  Subsequently,  the determinant of $\bm{P}$ must be the product of all these eigenvalues which leads to the required representation

\begin{equation}
\det{ \lp \bm{P} \rp} = \lp \prod_{j=1}^{N} \lambda_{1,j} \rp... \lp \prod_{j=1}^{N} \lambda_{r,j} \rp = \det \lp \bm{M}_{1} \rp... \det \lp \bm{M}_{r} \rp = \prod_{i= 1}^{r} \det \lp \bm{M}_{i} \rp.
\end{equation}

A direct consequence of <ref> is that if $\bm{M}_{j}$ are nonnegative,  then the spectral radius, $ \rho$,  of interwoven matrix $\bm{P}$ is a real eigenvalue and is defined by

\begin{equation}
\rho \lp \bm{P} \rp = \max_{j} \lp  \rho \lp\bm{M}_{j}  \rp \rp  =  \max \lp  \bigcup_{i = 1}^{r} \Sp \lp \bm{M}_{i} \rp  \rp
\end{equation}

from the Perron-Frobenius theorem for nonnegative matrices [44].

In addition to its spectral properties,  the interwoven matrix (<ref>) also has the following exponent property.

Let $\bm{P}$ be the interwoven matrix defined as defined in equation (<ref>).  Then for all $n \in \N$

\begin{equation} \label{eqn:induction_equation}
\bm{P}^{n} =  \sum_{i = 1}^{r} \bm{M}_{i}^{n} \otimes \bm{D}_{i}.
\end{equation}

The result follows by induction.  Assume for some $k \in \N$ that equation<ref> holds.  Consider the case for $k+1$,

\begin{align}
\lp  \sum_{i = 1}^{r}  \bm{M}_{i} \otimes \bm{D}_{i}   \rp^{k+1} &= \lp \sum_{i = 1}^{r}  \bm{M}_{i} \otimes \bm{D}_{i}  \rp \lp \sum_{i = 1}^{r}  \bm{M}_{i} \otimes \bm{D}_{i} \rp ^{k},  \nonumber \\
&=  \lp \sum_{i = 1}^{r}  \bm{M}_{i} \otimes \bm{D}_{i}  \rp  \lp \sum_{i = 1}^{r}  \bm{M}_{i}^{k} \otimes \bm{D}_{i} \rp,  \label{eqn:hyp_induction}
\end{align}

where the second equality follows from the inductive hypothesis.  Applying the multiplication property of the of Kronecker matrix<ref>,  expansion of equation<ref> and the mixed-product property of the Kronecker product leads to the following cancellations,

\begin{align}
\lp \sum_{i = 1}^{r}  \bm{M}_{i} \otimes \bm{D}_{i}  \rp  \lp \sum_{i = 1}^{r}  \bm{M}_{i}^{k} \otimes \bm{D}_{i} \rp &= \lp \bm{M}_{1} \otimes \bm{D}_{1} \rp\lp \sum_{i = 1}^{r}  \bm{M}_{i}^{k} \otimes \bm{D}_{i} \rp  + \cdots + \lp \bm{M}_{r} \otimes \bm{D}_{r} \rp\lp \sum_{i = 1}^{r}  \bm{M}_{i}^{k} \otimes \bm{D}_{i} \rp,  \nonumber \\
&= \bm{M}_{1} \bm{M}_{1}^{k} \otimes \bm{D}_{1} + \cdots +  \bm{M}_{r} \bm{M}_{r}^{k} \otimes \bm{D}_{r} , \nonumber \\
&=  \sum_{i = 1}^{r} \bm{M}_{i}^{k+1} \otimes \bm{D}_{i} .
\end{align}

That is,  the inductive hypothesis is satisfied and thus equation (<ref>) holds for all $n \in \N$ by the principle of induction.

§ COMPUTATIONAL METHODS

The ODE systems in this study were solved numerically using the  solver in Matlab (R2021a).  Simulations were performed over a total of 1000 time units in addition to an stopping event applied to the ODE solver to check for solution convergence.  Namely,  if all trajectories varied less than $1 \times 10^{-4}$ over four consecutive iterations,  then we assume that the system has converged to a steady state.  We note that all simulations presented in this study satisfied the convergence criteria.  The intracellular kinetics parameter values of the IO system (<ref>-<ref>) used in all simulations are given in Table <ref> below.

Parameter     $a_{1}$     $a_{2}$     $b_{1}$     $b_{2}$     $b_{3}$     $k_{1}$     $k_{2}$     $h_{1}$     $h_{2}$     $h_{3}$
Value         0.01        1           100         100         100         2           2           2           2           1

Parameter values used in the simulations of the IO system (<ref>-<ref>).

Random initial conditions were sampled from a uniform distribution using the  function.  The homogeneous steady state of the system was calculated using the  function that implements the trust-region-dogleg minimisation algorithm [65].  In addition,  heterotypic weighting parameters was set to $w_{2}^{[k]} = 1$ for all simulations.  Both quotient and large-scale ODE systems where solved using the same kinetics functions where respective adjacency matrices were introduced as an argument to these kinetics functions to ensure solution consistency.

To visualise the approximate cell membranes in the large-scale simulation Voronoi diagrams were drawn around graph vertices using the  and  functions within the Computational Geometry toolbox in Matlab (R2021a).  Ghost vertices were introduced to ensure that each graph vertex has a closed boundary.

Eigenvalues of the adjacency matrices were calculated using the  function from the Linear Algebra toolbox in Matlab (R2021a).  The edge structures of the semi-regular non-bipartite graphs used in the numerical spectral investigation are given in Table <ref>.  These graphs were confirmed non-bipartite by violating the spectral symmetry property of bipartite graphs.

$n_{1,\cL_{1}}$    $n_{2,\cL_{1}}$    $n_{1,\cL_{2}}$       $n_{2,\cL_{2}}$
$\cG_{1}$     2     2    2     2

$\cG_{2}$      2   3      2    3

$\cG_{3}$      2     4     2   4

$\cG_{4}$      4     3     4     3

Summary of the bilayer edge connectivity structures for the graphs used in the numerical investigation of non-bipartite spectra in Figure <ref>.

Source code for the simulations presented in this study can be found at <https://github.com/joshwillmoore1/Mixed_Signal_mechanisms>.

§ PROOF OF $\PI_{2}$-DEPENDENT SPECTRAL GAP FOR $\CG_{\TEXT{2D}}$ AND $\CG_{\TEXT{3D}}$ FROM FIGURE <REF>B

Bipartite graphs have many particularly convenient algebraic properties due to the existence of a simple canonical form of the respective adjacency matrix.  Namely,  for any bipartite graph $\cG_{k}$ with adjacency matrix $\bm{W}_{k}$ there exists a permutation matrix $\bm{U}$ that re-indexes the vertices with respect to the independent sets $V_{1}$ and $V_{2}$ such that

\begin{equation}
\bm{U}^{T} \bm{W}_{k} \bm{U}  =\begin{bmatrix}
\bm{0} & \bm{X}_{k} \\
\bm{X}_{k}^{T} & \bm{0}
\end{bmatrix},
\end{equation}

where $\bm{X}_{k}$ is the biadjacency matrix [48].  Subsequently,  the spectra of the bipartite graphs have a distinct structure such that there is a symmetries of eigenvalues respective to the biadjacency matrices,  i.e.,  $\Sp \lp \bm{W}_{k} \rp = \Sp \lp \bm{X}_{k} \rp \cup \Sp \lp - \bm{X}_{k} \rp$.
Leveraging the spectral symmetry of bipartite graphs and the spectral retention of equitable partitions,  we demonstrate that for the bipartite bilayer graphs $\cG_{\text{2D}}$ and $\cG_{\text{3D}}$ in Figure <ref>,  the smallest eigenvalue of $\bm{X}_{k}$ is $-\overline{\lambda}_{k,2}$,  the polarity driven eigenvalue associated with laminar pattern template $\pi_{2}$.

Let $\cG_{k}$ be a regular bipartite bilayer graph with associated row-stochastic weighted adjacency matrix $\bm{W}_{k}$ (<ref>) for 2D or 3D tissues as shown in Figure <ref> ($k = \text{2D},\text{3D}$).  Consider the equitable partition $\pi_{2}$ such that the quotient graph,  $\cG_{k,\pi_{2}}$,  consists of only two representative vertices in each layer of $\cG_{k}$ and has the reduced adjacency matrix $\overline{\bm{W}}_{k}$ (<ref>).  Then biadjacency matrix $\bm{X}_{k}$ associated with $\bm{W}_{k}$ satisfies

\begin{equation}
-\overline{\lambda}_{k,2} = \min \lp \mu_{k,j} : \mu_{k,j} \in  \Sp \lp \bm{X}_{k} \rp \rp,
\end{equation}

where $\overline{\lambda}_{k,2}$ is the smallest eigenvalue of $\overline{\bm{W}}_{k}$ with associated eigenvector $\overline{\bm{v}}_{k,2}$.

The proof for $\cG_{\text{2D}}$ is given as the argument follows identically for $\cG_{\text{3D}}$.  As we make use of the biadjacency form of $\bm{W}_{\text{2D}}$,  we first construct the biadjacency transformation $\bm{U}$.  The bipartite bilayer graph $\cG_{\text{2D}}$ has vertex indices in layer-wise order as defined in Section <ref> with block adjacency matrices given in Example <ref>.  To reorder the vertices of $\cG_{\text{2D}}$ such that vertex groups $V_{1}$ and $V_{2}$ are ordered consecutively,  we define the permutation matrix

\begin{equation}
\bm{U} = \begin{bmatrix}
\bm{I}_{|\cL_{1}|/2} \otimes \bm{D}_{1} & \bm{I}_{|\cL_{1}| /2} \otimes \bm{D}_{2}  \\
\bm{I}_{|\cL_{2}|/2} \otimes \bm{D}_{2} & \bm{I}_{|\cL_{2}|/2} \otimes \bm{D}_{1}
\end{bmatrix},
\end{equation}

where $|\cL_{1}| = |\cL_{2}|$ as each layer has the same number of vertices.  In particular we have the biadjacency form

\begin{align}\label{eqn:biadj_form}
\bm{U}^{T} \bm{W}_{\text{2D}} \bm{U} &=  \begin{bmatrix}
\bm{I}_{|\cL_{1}|/2} \otimes \bm{D}_{1} & \bm{I}_{|\cL_{1}| /2} \otimes \bm{D}_{2}  \\
\bm{I}_{|\cL_{2}|/2} \otimes \bm{D}_{2} & \bm{I}_{|\cL_{2}|/2} \otimes \bm{D}_{1}
\end{bmatrix}
\begin{bmatrix}
\widehat{\bm{W}}_{1,\cL_{1}} & \widehat{\bm{W}}_{2,\cL_{1}} \\
\widehat{\bm{W}}_{2,\cL_{1}}^{T} & \widehat{\bm{W}}_{1,\cL_{2}}
\end{bmatrix}
\begin{bmatrix}
\bm{I}_{|\cL_{1}|/2} \otimes \bm{D}_{1} & \bm{I}_{|\cL_{1}| /2} \otimes \bm{D}_{2}  \\
\bm{I}_{|\cL_{2}|/2} \otimes \bm{D}_{2} & \bm{I}_{|\cL_{2}|/2} \otimes \bm{D}_{1}
\end{bmatrix}, \nonumber \\
\begin{bmatrix}
0 & \bm{X}_{\text{2D}}  \\
\bm{X}_{\text{2D}} & 0
\end{bmatrix},
\end{align}

for $\bm{X}_{\text{2D}}$ in cyclic tridiagonal form

\begin{equation}\label{eqn:X2d}
\bm{X}_{\text{2D}} =  \begin{bmatrix}
\hat{w}_{2}^{[\text{2D}]} & \hat{w}_{1}^{[\text{2D}]} & 0 & \cdots & 0 & \hat{w}_{1}^{[\text{2D}]} \\
\hat{w}_{1}^{[\text{2D}]} &  \hat{w}_{2}^{[\text{2D}]}  & \hat{w}_{1}^{[\text{2D}]} & & &\\
& & \ddots & & & \\
& & & \ddots & & \\
& &  &\hat{w}_{1}^{[\text{2D}]} & \hat{w}_{2}^{[\text{2D}]}  &\hat{w}_{1}^{[\text{2D}]} \\
\hat{w}_{1}^{[\text{2D}]}&0 & \cdots  &0 &\hat{w}_{1}^{[\text{2D}]} &   \hat{w}_{2}^{[\text{2D}]}
\end{bmatrix},
\end{equation}

noting that $\bm{X}_{\text{2D}}^{T} =\bm{X}_{\text{2D}} $ by the regularity of $\cG_{\text{2D}}$ and therefore $\bm{U}^{T} \bm{W}_{\text{2D}} \bm{U}$ is symmetric.

As the laminar pattern template partition $\pi_{2}$ is equitable there exists a lifting matrix $\bm{L} \in \{0,1\}^{N\times2}$ that maps the large-scale adjacency matrix $\bm{W}_{\text{2D}}$ into its reduced form $\overline{\bm{W}}_{\text{2D}}$ such that

\begin{equation}\label{eqn:lifting_relation}
\bm{W}_{\text{2D}} \bm{L} = \bm{L} \overline{\bm{W}}_{\text{2D}},
\end{equation}

as demonstrated in [66].  The lifting transformation is constructed by grouping vertices associated with the partition $\pi_{2}$ on $\bm{W}_{\text{2D}}$ for example $\bm{L}_{ij} = 1$ if $v_{i} \in \cL_{j}$.  Owing to the block structure of $\bm{W}_{\text{2D}}$ (<ref>) which follows from the layer-wise vertex indexing,  we have that

\begin{equation}
\bm{L} = \begin{bmatrix}
\bm{1}_{| \cL_{1} |, 1} & \bm{0}_{| \cL_{1} |,1} \\
\bm{0}_{| \cL_{2} |,1} & \bm{1}_{| \cL_{2} |,1} \\
\end{bmatrix}.
\end{equation}

Critically,  the lifting matrix $\bm{L}$ provides the algebraic link between the quotient and large-scale graphs.

Following from the regular structure of $\cG_{\text{2D}}$ and direct computation,  the eigenvector associated with $\overline{\lambda}_{\text{2D},2}$ has the form $\overline{\bm{v}}_{\text{2D},2} = [1,  -1]^{T}$.  The spectral retention property of the equitable partition, $\pi_{2}$,  guarantees that $\overline{\lambda}_{\text{2D},2} \in \Sp \lp \bm{W}_{\text{2D}} \rp$ where $\bm{L}\overline{\bm{v}}_{\text{2D},2}$ is the corresponding eigenvector for the large-scale graph $\cG_{\text{2D}}$ (by Theorem 9.3.3 in [48]).  Explicitly,  we have that the lifted eigenvector is of the form

\begin{equation}
\bm{L}\overline{\bm{v}}_{\text{2D},2} = \begin{bmatrix}
\bm{1}_{| \cL_{1} |, 1} \\
-\bm{1}_{| \cL_{2} |, 1}
\end{bmatrix},
\end{equation}

with associated eigenvalue $\overline{\lambda}_{\text{2D},2}$.  In the biadjacency matrix form (<ref>),  the corresponding eigenvector has the transformed representation

\begin{equation}
\bm{\nu} \coloneqq \bm{U}^{T} \bm{L}\overline{\bm{v}}_{\text{2D},2}  = \begin{bmatrix}
\smash[b]{\blockTwo{|\cL_{1}|}} &  \smash[b]{\blockTwo{|\cL_{2}|}}
\end{bmatrix}^{T}.
\vspace{0.9em}
\end{equation}

The spectral symmetry of bipartite graphs ensures that if there exists an eigenpair  $ \lp \overline{\lambda}_{\text{2D},2} , \bm{\nu}  = [\bm{x},\bm{y}^{T}]\rp$ then there must also exist the eigenpair $ \lp -\overline{\lambda}_{\text{2D},2}, \tilde{ \bm{\nu} } = [\bm{x},-\bm{y}^{T}]\rp$  [48]. Therefore the eigenvector associated with $- \lambda_{\text{2D},2}$ has biadjacency form

\begin{equation}
\tilde{\bm{\nu}}  = \begin{bmatrix}
\smash[b]{\blockTwo{|\cL_{1}|}} &  \smash[b]{\blockOne{|\cL_{2}|}}
\end{bmatrix}^{T}.
\vspace{1.2em}
\end{equation}

which negates the signs of those entries associated with the latter half of the vertices in $\cG_{\text{2D}}$.  Subsequently,  the first $|\cL_{1}|$ entries of $\tilde{\bm{\nu}}$ are an eigenvector of $\bm{X}_{\text{2D}}$ with eigenvalue $- \overline{\lambda}_{\text{2D},2}$ following from the canonical bidjacency representation of $\bm{W}_{\text{2D}}$ (<ref>).  We denote this reduced eigenvector in normalised form

\begin{equation}
\tilde{\bm{\nu}}_{1}  = \frac{1} { \sqrt{|\cL_{1}|}} \begin{bmatrix}
\smash[b]{\blockTwo{|\cL_{1}|}}
\end{bmatrix}^{T},
\vspace{1.2em}
\end{equation}

and therefore it remains to show that the eigenpair $\lp - \overline{\lambda}_{\text{2D},2},   \tilde{\bm{\nu}}_{1}  \rp$ is minimal in the spectrum of $\bm{X}_{\text{2D}}$.  The Rayleigh quotient for $\bm{X}_{\text{2D}}$ is defined by

\begin{equation}
R_{\bm{X}_{\text{2D}}} \lp \bm{y} \rp  = \frac{\bm{y}^{T} \bm{X}_{\text{2D}} \bm{y}}{\bm{y}^{T} \bm{y}}
\end{equation}

and as $\bm{X}_{\text{2D}}$ is real and symmetric by the Min-Max theorem the Rayleigh quotient is bounded by the maximal and minimal eigenvalues of the matrix,  $R_{\bm{X}_{\text{2D}}} \lp \bm{y} \rp \in [\lambda_{\min}, \lambda_{\max}]$ [48].  In particular,  $R_{\bm{X}_{\text{2D}}} \lp \bm{y} \rp $ generates the eigenvalues of $\bm{X}_{\text{2D}}$ when $\bm{y}$ is the respective eigenvector.  Hence we show that $\tilde{\bm{\nu}}_{1} $ minimises $R_{\bm{X}_{\text{2D}}} \lp \bm{y} \rp$,  namely

\begin{equation}
\argmin _{ \substack {\bm{y} \in  \R^{|\cL_{1}|} \\
|| \bm{y} || = 1}}    \lp R_{\bm{X}_{\text{2D}}} \lp \bm{y} \rp  \rp =\tilde{\bm{\nu}}_{1} .
\end{equation}

where the normality constraint follows from $\bm{X}_{\text{2D}}$ being real and symmetric and so the eigenvectors of $\bm{X}_{\text{2D}}$ are orthonormal with real eigenvalues.

The normalised form of $\tilde{\bm{\nu}}_{1}$ yields $\tilde{\bm{\nu}}_{1}^{T}\tilde{\bm{\nu}}_{1}  = 1$ and therefore the Rayleigh quotient evaluated at $\tilde{\bm{\nu}}_{1}$ simplifies to $R_{\bm{X}_{\text{2D}}} \lp \tilde{\bm{\nu}}_{1} \rp = \tilde{\bm{\nu}}_{1}^{T}  \bm{W}_{\text{2D}}  \tilde{\bm{\nu}}_{1}$.  By direct computation we have that

\begin{align}
\tilde{\bm{\nu}}_{1}^{T}  \bm{X}_{\text{2D}}  \tilde{\bm{\nu}}_{1} &= \sum_{k = 1}^{|\cL_{1}|} \sum_{j=1}^{|\cL_{1}|} \lp  \bm{X}_{\text{2D}}  \rp_{k,j}    \lp  \tilde{\bm{\nu}}_{1} \rp_{k}   \lp \tilde{\bm{\nu}}_{1} \rp_{j} \nonumber \\
&= \sum_{i = 1}^{|\cL_{1}|} \lp   \bm{X}_{\text{2D}} \rp_{ii}  \lp \tilde{\bm{\nu}}_{1} \rp_{i}^{2} +  \sum_{i = 2}^{|\cL_{1}|-1}  \lp \tilde{\bm{\nu}}_{1} \rp_{i}  \lp \lp \bm{X}_{\text{2D}} \rp_{i,i-1}  \lp \tilde{\bm{\nu}}_{1} \rp_{i-1} + \lp \bm{X}_{\text{2D}} \rp_{i,i+1}  \lp \tilde{\bm{\nu}}_{1} \rp_{i+1}  \rp \nonumber \\
&	\quad \quad \quad \quad \quad \quad \quad \quad \quad \,  +   \lp \tilde{\bm{\nu}}_{1} \rp_{1} \lp \lp \bm{X}_{\text{2D}} \rp_{1,2} \lp \tilde{\bm{\nu}}_{1} \rp_{2} + \lp \bm{X}_{\text{2D}} \rp_{1,|\cL_{1}|}  \lp \tilde{\bm{\nu}}_{1} \rp_{|\cL_{1}|} \rp \nonumber \\
&	\quad \quad \quad \quad \quad \quad \quad \quad \quad \,  +  \lp \tilde{\bm{\nu}}_{1} \rp_{|\cL_{1}|} \lp \lp \bm{X}_{\text{2D}} \rp_{|\cL_{1}|,1} \lp \tilde{\bm{\nu}}_{1} \rp_{1} + \lp \bm{X}_{\text{2D}} \rp_{|\cL_{1}|,|\cL_{1}|-1}  \lp \tilde{\bm{\nu}}_{1} \rp_{|\cL_{1}|-1} \rp
\end{align}

by the cyclic tridiagonal form of $\bm{X}_{\text{2D}} $ (<ref>).  Critically as $\lp \bm{X}_{\text{2D}} \rp_{i,j} \geq 0 $ for all $i,j$,  then $\tilde{\bm{\nu}}_{1}^{T}  \bm{X}_{\text{2D}}  \tilde{\bm{\nu}}_{1}$ is minimised when $\text{Sgn}\lp \lp\tilde{\bm{\nu}}_{1}\rp_{k} \rp \neq \text{Sgn}\lp \lp\tilde{\bm{\nu}}_{1}\rp_{k+1} \rp $ for all $k \in \{1,..., |\cL_{1}|-1 \}$ which is satisfied by definition of $\tilde{\bm{\nu}}_{1}$.  Furthermore,  the orthonormal property of the eigenbasis of $\bm{X}_{\text{2D}}$ ensures that no other eigenvector has this alternating sign structure which implies that $-\overline{\lambda}_{\text{2D},2}$ is the smallest eigenvalue of $\bm{X}_{\text{2D}}$.

A consequence of Lemma <ref> is the existence of a spectral gap about the origin for $\cG_{\text{2D}}$ and $\cG_{\text{3D}}$.

Let $\cG_{k}$ and $\cG_{k,\pi_{2}}$ be defined as in Lemma <ref> and let $ \lambda_{k,j} \in \Sp \lp  \bm{W}_{k} \rp$.  If $\overline{\lambda}_{k,2} < 0$ then $\lambda_{k,j}  \not\in  \lp \overline{\lambda}_{k,2} , -\overline{\lambda}_{k,2} \rp $.

From Lemma <ref> we have that $-\overline{\lambda}_{k,2} = \min \lp \mu_{k,j} : \mu_{k,j} \in  \Sp \lp \bm{X}_{k} \rp \rp$ and thus $-\overline{\lambda}_{k,2} > 0$.  From the symmetry of the spectrum of bipartite graphs we have that $\overline{\lambda}_{k,2} < 0$ is the maximum of the negative eigenvalues of $\bm{W}_{k}$ therefore defining a region about the origin bounded by $\overline{\lambda}_{k,2}$ and $-\overline{\lambda}_{k,2}$ that contains no eigenvalues.


T_{\dim_{\mathrm{H}}\Lambda}(t^{*})$. By considering $\theta>\gamma^{-L}$ very
close to $\gamma^{-L}$, we see that $\liminf_{\theta\to
0^{+}}t_{L(\theta)}\dim_{\theta}\Lambda\leqslant t^{*}$. If
$\gamma^{-L}<\theta\leqslant\gamma^{-(L-1)}$ then
$t_{L}(\dim_{\gamma^{-L}}\Lambda)<t_{L}(\dim_{\theta}\Lambda)\leqslant
t_{L}(\dim_{\gamma^{-(L-1)}}\Lambda)$. Therefore

$\displaystyle
t^{*}=\lim_{L\to\infty}t_{L}(\dim_{\theta}\Lambda)\leqslant\liminf_{\theta\to
0^{+}}t_{L(\theta)}(\dim_{\theta}\Lambda)$
$\displaystyle\leqslant\limsup_{\theta\to
0^{+}}t_{L(\theta)}(\dim_{\theta}\Lambda)$
$\displaystyle\leqslant\lim_{L\to\infty}t_{L}(\dim_{\gamma^{-(L-1)}}\Lambda)$
$\displaystyle=T_{\dim_{\mathrm{H}}\Lambda}(t^{*}),$

completing the proof. ∎

###### Proof of part (iii).

For brevity, let us write

$s^{\prime}(\theta)=\frac{\gamma^{L}}{\log
n}\cdot\frac{f_{L}(\theta)}{1+g_{L}(\theta)A_{L}(\theta)}.$ (5.5.5)

Lemma 5.4.5 ensures that
$\underline{t}<t_{1}(\dim_{\mathrm{H}}\Lambda)<t_{\ell}(s(\theta))<T_{\dim_{\mathrm{H}}\Lambda}(t^{\ast})<\overline{t}$
for all $1\leqslant\ell\leqslant L$. Using that $I$ is strictly increasing and
convex, there exist constants $c_{1},c_{1}^{\prime},c_{2},c_{2}^{\prime}>0$
independent of $\theta$ such that for all
$t_{1}(\dim_{\mathrm{H}}\Lambda)\leqslant t\leqslant
T_{\dim_{\mathrm{H}}\Lambda}(t^{\ast})$,

$0<c_{1}<I(t)<c_{1}^{\prime}<\log M-H(\mathbf{P})\;\text{ and
}\;0<c_{2}<I^{\prime}(t)<c_{2}^{\prime}<1.$

Hence, recalling that $\overline{t}\coloneqq\log N-H(\mathbf{P})$ and
$I(\overline{t})=\log M-H(\mathbf{P})$, there exists $c_{3}>0$ such that the
numerator

$f_{L}(\theta)=\overline{t}-t_{L}(s(\theta))-\big{(}I(\overline{t})-I\big{(}t_{L}(s(\theta))\big{)}\big{)}\geqslant
c_{3}>0.$

Furthermore, there also exists $c_{4}>0$ such that $0<c_{4}\leqslant
g_{L}(\theta)\leqslant c_{4}^{-1}<\infty$. Therefore,

$s^{\prime}(\theta)\geqslant\frac{c_{3}}{\log
n}\cdot\frac{\gamma^{L}}{1+c_{4}^{-1}A_{L}(\theta)}\geqslant\frac{c_{3}}{\log
n}\cdot\frac{\gamma^{L}}{1+c_{4}^{-1}\cdot\frac{\gamma^{L}-1}{\gamma-1}}\geqslant\frac{c_{3}}{\log
n\big{(}\gamma^{-1}+\frac{c_{4}^{-1}}{\gamma-1}\big{)}}=:C_{0}>0.$

Next we show that
$\partial_{-}\dim_{\gamma^{-L}}\Lambda<\partial_{+}\dim_{\gamma^{-L}}\Lambda$
for all $L\in\mathbb{N}$ from (5.5.2) and (5.5.3). We can divide both (5.5.2)
and (5.5.3) by $\gamma^{L}/\log n$. We claim that

$\gamma^{-1}\cdot
A_{L+1}(\gamma^{-L})-I^{\prime}\big{(}t_{L}(s(\gamma^{-L}))\big{)}\cdot
A_{L}(\gamma^{-L})=\gamma^{-1},$

implying $\gamma^{-1}\cdot(1+g_{L+1}(\gamma^{-L})\cdot
A_{L+1}(\gamma^{-L}))=1+g_{L}(\gamma^{-L})\cdot A_{L}(\gamma^{-L})$. Indeed,
applying the definition of $A_{L+1}(\gamma^{-L})$ and $A_{L}(\gamma^{-L})$,
observe that it is a telescopic sum with only $\gamma^{-1}$ not cancelling
out. Since we also have

$t_{L+1}(s(\gamma^{-L}))-t_{L}(s(\gamma^{-L}))-\big{(}I\big{(}t_{L+1}(s(\gamma^{-L}))\big{)}-I\big{(}t_{L}(s(\gamma^{-L}))\big{)}\big{)}>0$

for the same reason that $f_{L}(\theta)>0$, it follows that
$\partial_{-}\dim_{\gamma^{-L}}\Lambda<\partial_{+}\dim_{\gamma^{-L}}\Lambda$.
Finally, by Lemma 5.5.2,

$\frac{\partial_{+}\dim_{\gamma^{-L}}\Lambda}{\partial_{-}\dim_{\gamma^{-L}}\Lambda}\xrightarrow[(L\to\infty)]{}\frac{\log
N-t^{*}-\log M+I(t^{*})}{\log N-T_{\dim_{\mathrm{H}}\Lambda}(t^{*})-\log
M+I(T_{\dim_{\mathrm{H}}\Lambda}(t^{*}))}\in(1,\infty).\qed$

###### Proof of part (iv).

The idea is to estimate $L(\theta)$ (which is the number of the
$t_{i}(s(\theta))$) in terms of $s(\theta)\coloneqq\dim_{\theta}\Lambda$. We
do this by using the fact that $t^{\prime}$ is a neutral fixed point of the
function $T_{\dim_{\mathrm{H}}\Lambda}$, and

$t_{1}(s(\theta))\xrightarrow[\theta\to
0^{+}]{}\left(\dim_{\mathrm{H}}\Lambda-\frac{\log M}{\log m}\right)\log
n<t^{\prime}$

and $\liminf_{\theta\to 0^{+}}t_{L(\theta)}(s(\theta))>t^{\prime}$ by Lemmas
5.5.2 and 5.4.2, so most of the $t_{i}(s(\theta))$ lie close to $t^{\prime}$.

By Lemma 5.4.1 and Taylor’s theorem, since $T_{s}^{\prime}(t^{\prime})=1$ and
$T_{s}^{\prime\prime}(t^{\prime})>0$, there exists $c\geqslant 1$ such that
for all $t\in(\underline{t},\overline{t})$,

$\displaystyle\begin{split}T_{s}(t^{\prime})+T_{s}^{\prime}(t^{\prime})(t-t^{\prime})+c^{-1}(t-t^{\prime})^{2}&\leqslant
T_{s}(t)\leqslant
T_{s}(t^{\prime})+T_{s}^{\prime}(t^{\prime})(t-t^{\prime})+c(t-t^{\prime})^{2};\\\
t+c^{-1}(t-t^{\prime})^{2}&\leqslant T_{s}(t)\leqslant
t+c(t-t^{\prime})^{2}.\end{split}$ (5.5.6)

If $L$ is large enough and
$k_{L}\coloneqq\big{\lfloor}\max\left\\{L/10,\frac{2}{\log
2}\log\left(\frac{\overline{t}-\underline{t}}{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}\right)\right\\}\big{\rfloor}$ then

$t^{\prime}-2^{k_{L}}\frac{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}{4}<\underline{t}\;\text{ and
}\;t^{\prime}+2^{k_{L}}\frac{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}{4}>\overline{t}.$

Suppose $k\in\\{1,2,\dotsc,k_{L}\\}$. Then by (5.5.6),

$\displaystyle\\#\Big{\\{}i\in\\{1,2,\dotsc,L(\theta)\\}:t^{\prime}-2^{k}\frac{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}{4}$ $\displaystyle<t_{i}(s(\theta))$
$\displaystyle\phantom{\leqslant}\leqslant
2^{k-1}\frac{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log n}{4}\Big{\\}}$
$\displaystyle\leqslant
1+\frac{16c}{2^{k}L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log n}.$

Summing up, it follows that

$\\#\left\\{i:t_{i}(s(\theta))\leqslant
t^{\prime}-\frac{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}{4}\right\\}\leqslant
k_{L}+\frac{16c}{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log n},$

and we similarly obtain the same bound for the number of $t_{i}(s)$ which are
greater than $t^{\prime}+\frac{L(s(\theta)+\dim_{\mathrm{H}}\Lambda)\log
n}{4}$. But

$\\#\left\\{i:t^{\prime}-\frac{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}{4}<t_{i}(s(\theta))\leqslant
t^{\prime}+\frac{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}{4}\right\\}\leqslant 1+L/2.$

Therefore for $L$ sufficiently large,

$L\leqslant 2k_{L}+\frac{32c}{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}+1+\frac{L}{2}\leqslant
0.9L+\frac{40c}{L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log n}.$

Decreasing $\theta_{0}$ further if required, this tells us that for all
$\theta<\theta_{0}$,

$s(\theta)\leqslant\dim_{\mathrm{H}}\Lambda+\frac{400c}{L^{2}\log
n}\leqslant\dim_{\mathrm{H}}\Lambda+\frac{500c\cdot(\log\gamma)^{2}}{(\log\theta)^{2}\log
n},$

proving the upper bound. This shows in particular that
$L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log n\to 0$ as $\theta\to 0^{+}$.

For the lower bound, we may decrease $\theta_{0}$ further to assume that for
all $\theta<\theta_{0}$,
$t_{1}(\dim_{\mathrm{H}}\Lambda)<t_{1}(s(\theta))<t^{\prime}-3L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n$ and $t_{L}(s(\theta))>t^{\prime}$ by Lemmas 5.5.2 and 5.4.2. Then by
(5.5.6), for large enough $L$,

$\displaystyle L$
$\displaystyle\geqslant\\#\left\\{i:t^{\prime}-3L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n\leqslant t_{i}(s(\theta))\leqslant
t^{\prime}-L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log n\right\\}$
$\displaystyle\geqslant\frac{2L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n}{1.1((s(\theta)-\dim_{\mathrm{H}}\Lambda)\log
n+c(3L(s(\theta)-\dim_{\mathrm{H}}\Lambda)\log n)^{2})}.$

Rearranging, for large enough $L$,

$s(\theta)\geqslant\dim_{\mathrm{H}}\Lambda+\frac{0.9}{L^{2}\cdot 9.9c\log
n}\geqslant\dim_{\mathrm{H}}\Lambda+\frac{(\log\gamma)^{2}}{20c\log
n(\log\theta)^{2}}.\qed$

###### Proof of part (v).

One can differentiate (5.5.5) to obtain

$s^{\prime\prime}(\theta)=\frac{\gamma^{L}\cdot\big{(}f^{\prime}_{L}(\theta)(1+g_{L}(\theta)A_{L}(\theta))-f_{L}(\theta)(g^{\prime}_{L}(\theta)A_{L}(\theta)+g_{L}(\theta)A^{\prime}_{L}(\theta))\big{)}}{(1+g_{L}(\theta)A_{L}(\theta))^{2}\log
n}.$

The sign of $s^{\prime\prime}(\theta)$ is determined by the sign of the term
in parentheses in the numerator. We know that there exists $c>0$ such that
$f_{L}(\theta),g_{L}(\theta),A_{L}(\theta)\geqslant c$. There also exists
$c^{\prime}>0$ such that

$f^{\prime}_{L}(\theta)=\underbrace{\big{(}I^{\prime}\big{(}t_{L}(s(\theta))\big{)}-1\big{)}}_{\leqslant-c^{\prime}<0}\cdot\underbrace{\frac{\mathrm{d}}{\mathrm{d}\theta}t_{L}(s(\theta))}_{\geqslant
c^{\prime}>0\text{ by\leavevmode\nobreak\
\eqref{eq:60}}}\leqslant-(c^{\prime})^{2}<0$

and $A^{\prime}_{L}(\theta)>c^{\prime}$ because all
$I^{\prime}\big{(}t_{\ell}(s(\theta))\big{)}$ and
$I^{\prime\prime}\big{(}t_{\ell}(s(\theta))\big{)}$ are uniformly positive.
Finally,

$g^{\prime}_{L}(\theta)=\gamma^{L}\underbrace{\big{(}1-I^{\prime}\big{(}t_{L}(s(\theta))\big{)}\big{)}}_{\geqslant
c^{\prime}>0}+\gamma\underbrace{(1-\gamma^{L-1}\theta)}_{\geqslant
0}\underbrace{I^{\prime\prime}\big{(}t_{L}(s(\theta))\big{)}}_{>0}\cdot\underbrace{\frac{\mathrm{d}}{\mathrm{d}\theta}t_{L}(s(\theta))}_{\geqslant
c^{\prime}>0}\geqslant\gamma^{L}c^{\prime}>0.$

Hence, $s^{\prime\prime}(\theta)\leqslant C<0$ for some uniform constant
$C>0$, implying that $s(\theta)$ is strictly concave on every interval
$[\gamma^{-L},\gamma^{-(L-1)}]$. ∎

Note that $s^{\prime\prime}(\theta)\leqslant C<0$ holds for a constant $C$
that is independent of both $L$ and $\theta$.

####  5.5.2 Proof of Corollary 5.2.2

We now use the fact that $\dim_{\theta}\Lambda$ is strictly increasing in
$\theta$ to prove Corollary 5.2.2.

###### Proof of Corollary 5.2.2.

Fix $\eta>0$ small enough that
$\dim_{\mathrm{H}}\Lambda+\eta<\dim_{\mathrm{B}}\Lambda$. Since
$\dim_{\theta}\Lambda$ exists and is continuous in $\theta$ including at
$\theta=0$ (see Theorem 5.2.1 and [FFKa, Proposition 4.1]), we can fix
$\theta_{1}<1/\gamma$ small enough that
$\dim_{\theta_{1}}\Lambda+\eta<\dim_{\mathrm{B}}\Lambda$. Let
$s\in(\dim_{\mathrm{H}}\Lambda,\dim_{\theta_{1}}\Lambda+\eta]$, and for each
$\delta\in(0,1)$, let $K=K(\delta)\in\mathbb{N}$ be such that
$n^{-K}\leqslant\delta<n^{-(K-1)}$. For some string $\mathbf{i}$, let
$\mathcal{B}_{\lfloor K/\theta\rfloor}^{K,\mathbf{i}}$ denote the set of
level-$\lfloor K/\theta\rfloor$ approximate squares within the level-$K$
approximate square $B_{K}(\mathbf{i})$. For each $B_{K}(\mathbf{i})$ it is
more cost efficient (in terms of $s$-cost, up to irrelevant multiplicative
constants depending only on $\Lambda$) to subdivide it into level-$\lfloor
K/\theta\rfloor$ approximate squares if and only if

$n^{-Ks}\geqslant\\#\mathcal{B}_{\lfloor K/\theta\rfloor}^{K,\mathbf{i}}\cdot
n^{-sK/\theta}.$ (5.5.7)

To determine $\\#\mathcal{B}_{\lfloor K/\theta\rfloor}^{K,\mathbf{i}}$ for
some $\theta<1/\gamma$, we compare the sequences that define
$B_{K}(\mathbf{i})$ and a level-$\lfloor K/\theta\rfloor$ approximate square
within it:

$\begin{array}[]{c|c|c|ccc}i_{1}\;\dotsb\;i_{K}&\hat{\imath}_{K+1}\;\dotsb\;\hat{\imath}_{\gamma(K)}&&&&\\\
\underbrace{j_{1}\;\dotsb\;j_{K}}_{\text{equal}}&\underbrace{j_{K+1}\;\dotsb\;j_{\gamma(K)}}_{\text{same
column}}&\underbrace{j_{\gamma(K)+1}\;\dotsb\;j_{\lfloor
K/\theta\rfloor}}_{\in[N]\text{ freely}}&\underbrace{\hat{\jmath}_{\lfloor
K/\theta\rfloor+1}\;\dotsb\;\hat{\jmath}_{\gamma(K/\theta)}}_{\in[M]\text{
freely}}.\end{array}$

Thus, $\\#\mathcal{B}_{\lfloor K/\theta\rfloor}^{K,\mathbf{i}}=N^{\lfloor
K/\theta\rfloor-\gamma(K)}\cdot M^{\gamma(K/\theta)-\lfloor
K/\theta\rfloor}\cdot\prod_{\ell=K+1}^{\gamma(K)}N_{\hat{\imath}_{\ell}}$.
Substituting this back into (5.5.7), we get after algebraic manipulations that
it is more cost-efficient to subdivide if and only if

$s\geqslant\frac{\theta}{(1-\theta)}\frac{\gamma-1}{\log
n}\Bigg{(}\frac{1}{\gamma(K)-K}\sum_{\ell=K+1}^{\gamma(K)}\log
N_{\hat{\imath}_{\ell}}\Bigg{)}+\frac{\theta}{1-\theta}(1/\theta-\gamma)\frac{\log
N}{\log n}+\frac{\gamma-1}{1-\theta}\frac{\log M}{\log n}.$

But since $s\leqslant\dim_{\theta_{1}}\Lambda+\eta$, if is more cost-efficient
to subdivide then the following condition for the average must hold:

$\frac{1}{\gamma(K)-K}\sum_{\ell=K+1}^{\gamma(K)}\log
N_{\hat{\imath}_{\ell}}\leqslant\log(N/M)-\Big{(}\frac{1}{\theta}-1\Big{)}\frac{\log
n}{\gamma-1}(\dim_{\mathrm{B}}\Lambda-\dim_{\theta_{1}}\Lambda-\eta).$

As $\theta\to 0$, the right-hand side tends to $-\infty$, so there exists
$\theta_{0}<\theta_{1}/2$ small enough that for all $\theta\leqslant
2\theta_{0}$ it is more cost efficient not to subdivide any of the level-$K$
approximate squares, if using only scales $\delta$ and $\delta^{1/\theta}$.
Now, since $\dim_{\theta}\Lambda$ is strictly increasing in $\theta$ by
Corollary 5.2.3, there exists $\varepsilon<\eta$ small enough that
$\dim_{\theta_{0}}\Lambda+\varepsilon<\dim_{2\theta_{0}}\Lambda$. Then by the
definition of $\theta_{0}$ and since
$\dim_{\theta_{1}}\Lambda+\eta<\dim_{\mathrm{B}}\Lambda$, there exists
$\delta_{1}<1$ such that if $\\{U_{i}\\}$ is a cover of $\Lambda$ using just
two scales $\delta$ and $\delta^{\prime}$ with
$\delta^{\prime}\leqslant\delta^{1/(2\theta_{0})}<\delta\leqslant\delta_{1}$,
then for all $\theta\leqslant 2\theta_{0}$,

$\sum_{i}|U_{i}|^{\dim_{\theta}\Lambda+\varepsilon}\geqslant\sum_{i}|U_{i}|^{\dim_{\theta_{1}}\Lambda+\eta}\geqslant
1.$

Since $\dim_{\theta_{0}}\Lambda+\varepsilon<\dim_{2\theta_{0}}\Lambda$, there
exists $\delta_{0}<\delta_{1}$ such that if $\\{U_{i}\\}$ is a cover using
just two scales $\delta$, $\delta^{\prime}$ with
$\delta^{1/(2\theta_{0})}<\delta^{\prime}\leqslant\delta\leqslant\delta_{0}$,
then for all $\theta\leqslant\theta_{0}$,
$\sum_{i}|U_{i}|^{\dim_{\theta}\Lambda+\varepsilon}\geqslant\sum_{i}|U_{i}|^{\dim_{\theta_{0}}\Lambda+\varepsilon}\geqslant
1$, completing the proof. ∎

####  5.5.3 Proof of Theorems 5.2.4 and 5.2.5

We use primes to denote the parameters of $\Lambda^{\prime}$, and use notation
from Section 5.2.3. In particular, in this section only, $I^{\prime}$ will
denote the rate function associated with $\Lambda^{\prime}$, and not the
derivative of $I$.

######  Lemma 5.5.3.

Let $\Lambda$ and $\Lambda^{\prime}$ be two Bedford–McMullen carpets with non-
uniform fibres defined on grids of size $m\times n$ and $m^{\prime}\times
n^{\prime}$ respectively. If either of (i) or (ii) from Theorem 5.2.5 hold,
then $\frac{\log n}{\log n^{\prime}}=\frac{\log m}{\log
m^{\prime}}\in\mathbb{Q}$.

###### Proof.

First assume that (i) holds. Since the intermediate dimensions of $\Lambda$
and $\Lambda^{\prime}$ have phase transitions at $\gamma^{-1}$ and
$(\gamma^{\prime})^{-1}$ respectively (see Corollary 5.2.3), we must have
$\gamma=\gamma^{\prime}$. To show that $\frac{\log n^{\prime}}{\log
n}\in\mathbb{Q}$, note that Theorem 5.2.1 and the equality of the intermediate
dimensions for $\theta\in(\gamma^{-1},1)$ tells us that for an open interval
of $s$,

$\frac{1}{\log n}I\left(\left(s-\frac{\log M}{\log m}\right)\log
n\right)=\frac{1}{\log n^{\prime}}I^{\prime}\left(\left(s-\frac{\log
M^{\prime}}{\log m^{\prime}}\right)\log n^{\prime}\right).$

Taking Legendre transforms of both sides and using scaling properties of
Legendre transforms, using (5.2.2), for all $\lambda$,

$\frac{1}{\log
n}\log\left(\frac{1}{M}\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}N_{\hat{\imath}}^{\lambda}\right)+\lambda\frac{\log
M}{\log m}=\frac{1}{\log
n^{\prime}}\log\left(\frac{1}{M^{\prime}}\sum_{\hat{\jmath}=1}^{M_{0}^{\prime}}R^{\prime}_{\hat{\jmath}}(N^{\prime}_{\hat{\jmath}})^{\lambda}\right)+\lambda\frac{\log
M^{\prime}}{\log m^{\prime}}.$ (5.5.8)

Fix $K\in\mathbb{N}$ large enough that

$N_{1}^{\frac{\log n^{\prime}}{\log
n}}(N_{2}/N_{1})^{K}<N^{\prime}_{M_{0}^{\prime}}\cdot(n^{\prime})^{\left(\frac{\log
M^{\prime}}{\log m^{\prime}}-\frac{\log M}{\log m}\right)}.$

Then exponentiating (5.5.8),

$\displaystyle\frac{M^{\frac{\log n^{\prime}}{\log
n}}}{M^{\prime}}\sum_{\hat{\jmath}=1}^{M_{0}^{\prime}}R^{\prime}_{\hat{\jmath}}\left(N^{\prime}_{\hat{\jmath}}\cdot(n^{\prime})^{\left(\frac{\log
M^{\prime}}{\log m^{\prime}}-\frac{\log M}{\log
m}\right)}\right)^{\lambda}=\left(\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}N_{\hat{\imath}}^{\lambda}\right)^{\frac{\log
n^{\prime}}{\log n}}$
$\displaystyle\phantom{--}=(R_{1}N_{1}^{\lambda})^{\frac{\log n^{\prime}}{\log
n}}\left(\sum_{k=0}^{K}\frac{\frac{\log n^{\prime}}{\log n}\left(\frac{\log
n^{\prime}}{\log n}-1\right)\dotsb\left(\frac{\log n^{\prime}}{\log
n}-k+1\right)}{k!}\left(\sum_{\hat{\imath}=2}^{M_{0}}\frac{R_{\hat{\imath}}}{R_{1}}\left(\frac{N_{\hat{\imath}}}{N_{1}}\right)^{\lambda}\right)^{k}\right)$
$\displaystyle\phantom{----}+O\left(\left(N_{1}^{\frac{\log n^{\prime}}{\log
n}}\left(\frac{N_{2}}{N_{1}}\right)^{K+1}\right)^{\lambda}\right)$

as $\lambda\to+\infty$, by the generalised binomial theorem. This means that
the coefficient of $N_{1}^{\frac{\log n^{\prime}}{\log
n}}(N_{2}/N_{1})^{K\lambda}$, which is a polynomial equation in $\frac{\log
n^{\prime}}{\log n}$ with rational coefficients, must be 0. So $\frac{\log
n^{\prime}}{\log n}$ is algebraic. But $n^{\frac{\log n^{\prime}}{\log
n}}=n^{\prime}$, so by the Gelfond–Schneider theorem, $\frac{\log
n^{\prime}}{\log n}\in\mathbb{Q}$.

Now we assume only (ii). We first show that $\frac{\log m^{\prime}}{\log
m}\in\mathbb{Q}$. Since the functions $\beta_{\nu}$ and $\beta_{\nu^{\prime}}$
are equal, using (5.2.6) with the change of variable
$\lambda\coloneqq\gamma^{-1}+(1-\gamma^{-1})\xi$,

$\frac{\frac{\lambda-\gamma^{-1}}{1-\gamma^{-1}}\log
N-\log\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}N_{\hat{\imath}}^{\lambda}}{\log
m}=\frac{\frac{\lambda-\gamma^{-1}}{1-\gamma^{-1}}\log
N^{\prime}-\log\sum_{\hat{\jmath}=1}^{M^{\prime}_{0}}{R_{\hat{\jmath}}N_{\hat{\jmath}}^{\prime}}^{(\gamma^{\prime})^{-1}+(1-(\gamma^{\prime}))^{-1}\frac{\lambda-\gamma^{-1}}{1-\gamma^{-1}}}}{\log
m^{\prime}}$

for all $\lambda$. A similar argument to the above using the generalised
binomial theorem and Gelfond–Schneider theorem now gives that $\frac{\log
m^{\prime}}{\log m}\in\mathbb{Q}$. To show moreover that
$\gamma=\gamma^{\prime}$, note that these quantities are bi-Lipschitz
invariants which depend only on the respective carpets, not the choice of
iterated function system (see [FY, Theorem 3.3]). So since $m$ and
$m^{\prime}$ are multiplicatively dependent, we can iterate the IFS to assume
without loss of generality that $\Lambda$ is defined on an $m\times n$ grid,
and $\Lambda^{\prime}$ is defined on an $m\times n^{\prime}$ grid, for the
same $m$. Then by (5.2.6),

$N^{-\xi}\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}N_{\hat{\imath}}^{\gamma^{-1}+(1-\gamma^{-1})\xi}={N^{\prime}}^{-\xi}\sum_{\hat{\jmath}=1}^{M^{\prime}_{0}}R_{\hat{\jmath}}{N_{\hat{\jmath}}^{\prime}}^{(\gamma^{\prime})^{-1}+(1-(\gamma^{\prime})^{-1})\xi}$
(5.5.9)

for all $\xi$. So using a similar argument to the proof of [RYZ, Theorem 1.2],
equating exponential terms and coefficients gives $M_{0}=M^{\prime}_{0}$.
Also, $N_{\hat{\imath}}^{\frac{\log m}{\log
n}}=(N^{\prime}_{\hat{\imath}})^{\frac{\log m}{\log n^{\prime}}}$, so

$N_{\hat{\imath}}^{\prime}=N_{\hat{\imath}}^{\frac{\log n^{\prime}}{\log
n}}\qquad\mbox{for all}\quad\hat{\imath}\in\\{1,\dotsc,M_{0}\\}.$ (5.5.10)

Equating corresponding exponential bases in (5.5.9), applying (5.5.10) and
using that the carpet has non-uniform fibres (so not all $N_{\hat{\imath}}$
are equal) shows that $n=n^{\prime}$. In particular, $\frac{\log
n^{\prime}}{\log n}\in\mathbb{Q}$, as required. ∎

###### Proof of Theorem 5.2.4.

(i) The equality $\log n/\log m=\log n^{\prime}/\log m^{\prime}$ follows from
observing the phase transitions of the Assouad spectrum (see [FY, Theorem
3.3]) or intermediate dimensions. The fact that $\log n/\log
n^{\prime}\in\operatorname{{\mathbb{Q}}}$ follows from Lemma 5.5.3.

(ii) The forward implication follows from (i), and the backward implication
holds by iterating the IFSs (recalling the discussion after the statement of
Theorem 5.2.4). ∎

We now prove Theorem 5.2.5. Since the intermediate dimension and multifractal
spectra obviously do not depend on the grid on which the carpet is defined, in
light of Lemma 5.5.3 and Theorem 5.2.4 it suffices to assume henceforth that
both carpets are defined on the same $m\times n$ grid to begin with. We
already mentioned that the equivalence of (ii) and (vi) was proved in [RYZ,
Theorem 1.2]. To complete the proof, we show that
$\ref{item1}\Rightarrow\ref{itemnow3}\Rightarrow\ref{item4}\Rightarrow\ref{itemnow5}\Rightarrow\ref{itemlq}\Rightarrow\ref{itemnow5}\Rightarrow\ref{item4}\Rightarrow\ref{item1}$.
Of these, the implication $\ref{item1}\Rightarrow\ref{itemnow3}$ is obvious.

###### Proof of $\ref{itemnow3}\Rightarrow\ref{item4}$.

Assume $\dim_{\theta}\Lambda=\dim_{\theta}\Lambda^{\prime}$ on the open
interval $(a,b)\subset[\gamma^{-1},1]$. After rearranging the formula in
Theorem 5.2.1 for $\dim_{\theta}\Lambda$ in the case $L=1$, we obtain that

$\dim_{\theta}\Lambda=\dim_{\mathrm{B}}\Lambda-\Big{(}\frac{1}{\theta}-1\Big{)}\frac{I\big{(}t_{1}(\dim_{\theta}\Lambda)\big{)}}{\log
n}.$

By Corollary 5.2.3 part (i), $\dim_{\theta}\Lambda$ and
$\dim_{\theta}\Lambda^{\prime}$ are real analytic on $(\gamma^{-1},1)$, hence
$\dim_{\theta}\Lambda=\dim_{\theta}\Lambda^{\prime}$ on the whole interval
$[\gamma^{-1},1]$. In particular,
$\dim_{\mathrm{B}}\Lambda=\dim_{\mathrm{B}}\Lambda^{\prime}$, so

$\displaystyle I\big{(}t_{1}(\dim_{\theta}\Lambda)\big{)}$
$\displaystyle=I^{\prime}\big{(}t^{\prime}_{1}(\dim_{\theta}\Lambda^{\prime})\big{)}$
$\displaystyle=I^{\prime}\Big{(}\Big{(}\dim_{\theta}\Lambda^{\prime}-\frac{\log(M^{\prime}M/M)}{\log
m^{\prime}}\Big{)}\log n^{\prime}\Big{)}$
$\displaystyle=I^{\prime}\Big{(}\underbrace{\Big{(}\dim_{\theta}\Lambda-\frac{\log
M}{\log m}\Big{)}\log
n}_{=t_{1}(\dim_{\theta}\Lambda)}-\gamma\log(M^{\prime}/M)\Big{)}.$

Setting $t=t_{1}(\dim_{\theta}\Lambda)$, we see that
$I(t)=I^{\prime}(t-\gamma\log(M^{\prime}/M))$ on the open interval
$(t_{1}(\dim_{a}\Lambda),t_{1}(\dim_{b}\Lambda))$. Since the rate function is
analytic, (v) follows. ∎

###### Proof of $\ref{item4}\Rightarrow\ref{itemnow5}$.

Assume $I(t)=I^{\prime}(t-\gamma\log(M^{\prime}/M))$ on an open interval of
$t$. Without loss of generality we assume that $M^{\prime}\geqslant M$. Using
definition (5.2.2) of the rate function,

$I^{\prime}(t-\gamma\log(M^{\prime}/M))=\sup_{\lambda\in\mathbb{R}}\bigg{\\{}\lambda
t-\log\bigg{(}\frac{(M^{\prime}/M)^{\gamma\lambda}}{M^{\prime}}\sum_{\hat{\imath}=1}^{M^{\prime}_{0}}R^{\prime}_{\hat{\imath}}(N^{\prime}_{\hat{\imath}})^{\lambda}\bigg{)}\bigg{\\}}.$

Since $I$ and $I^{\prime}$ are convex functions, their Legendre transforms
must agree on an open interval, implying that

$\frac{1}{M}\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}N_{\hat{\imath}}^{\lambda}=\frac{1}{M^{\prime}}\sum_{\hat{\imath}=1}^{M^{\prime}_{0}}R^{\prime}_{\hat{\imath}}\big{(}(M^{\prime}/M)^{\gamma}N^{\prime}_{\hat{\imath}}\big{)}^{\lambda}$
(5.5.11)

on an open interval of $\lambda$.

From here the proof follows the idea of the proof of [RYZ, Theorem 1.2].
Taking the $k$-th derivative of both sides of (5.5.11) with respect to
$\lambda$ gives

$\frac{1}{M}\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}N_{\hat{\imath}}^{\lambda}\cdot(\log
N_{\hat{\imath}})^{k}=\frac{1}{M^{\prime}}\sum_{\hat{\imath}=1}^{M^{\prime}_{0}}R^{\prime}_{\hat{\imath}}\big{(}(M^{\prime}/M)^{\gamma}N^{\prime}_{\hat{\imath}}\big{)}^{\lambda}\cdot\big{(}\log((M^{\prime}/M)^{\gamma}N^{\prime}_{\hat{\imath}})\big{)}^{k}.$
(5.5.12)

Recall that the $N_{\hat{\imath}}$ and $N^{\prime}_{\hat{\imath}}$ are ordered
in decreasing order. Since (5.5.12) holds for all $k$, the largest term on
either side must be equal, so $N_{1}/N^{\prime}_{1}=(M^{\prime}/M)^{\gamma}$,
and also its coefficient

$\frac{R_{1}N_{1}^{\lambda}}{M}=\frac{R^{\prime}_{1}\big{(}(M^{\prime}/M)^{\gamma}N^{\prime}_{1}\big{)}^{\lambda}}{M^{\prime}}\;\Longleftrightarrow\;\frac{R^{\prime}_{1}}{R_{1}}=\frac{M^{\prime}}{M}\cdot\left(\frac{M}{M^{\prime}}\right)^{\gamma\lambda}\cdot\left(\frac{N_{1}}{N^{\prime}_{1}}\right)^{\lambda}=\frac{M^{\prime}}{M}.$

After subtracting these terms from both sides of (5.5.12), we repeat the
argument for the next largest term and so on. If $M_{0}\neq M^{\prime}_{0}$
then after $\min\\{M_{0},M^{\prime}_{0}\\}$ steps one side would be 0 and the
other non-zero, a contradiction. Hence, we conclude that
$M_{0}=M^{\prime}_{0}$ and
$N_{\hat{\imath}}/N^{\prime}_{\hat{\imath}}=(R^{\prime}_{\hat{\imath}}/R_{\hat{\imath}})^{\gamma}=(M^{\prime}/M)^{\gamma}$
for all $\hat{\imath}=1,\dotsc,M_{0}$. ∎

###### Proof of $\ref{itemnow5}\Rightarrow\ref{itemlq}$.

If (vi) holds, then substituting
$R_{\hat{\imath}}^{\prime}=R_{\hat{\imath}}M^{\prime}/M$ and
$N_{\hat{\imath}}^{\prime}=N_{\hat{\imath}}(M^{\prime}/M)^{-\gamma}$ gives
$N^{\prime}=N(M^{\prime}/M)^{1-\gamma}$. Substituting into (5.2.9) gives
$T_{\nu}(q)=T_{\nu^{\prime}}(q)$. ∎

###### Proof of $\ref{itemlq}\Rightarrow\ref{itemnow5}$.

Equating the constant term and coefficient of $q$ from (5.2.9) gives that

$\sum_{\hat{\imath}=1}^{M_{0}^{\prime}}R_{\hat{\imath}}^{\prime}(N_{\hat{\imath}}^{\prime})^{q}=\left(\frac{N^{\prime}}{N}\right)^{\frac{\log
n-q\log
n}{\log(n/m)}}\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}(N_{\hat{\imath}})^{q}.$

By the same differentiation argument from [RYZ] that was used in the proof of
$\ref{item4}\Rightarrow\ref{itemnow5}$, $M_{0}=M_{0}^{\prime}$ and
$N_{\hat{\imath}}/N^{\prime}_{\hat{\imath}}=(R^{\prime}_{\hat{\imath}}/R_{\hat{\imath}})^{\gamma}=(N^{\prime}/N)^{\frac{\log
n}{\log(n/m)}}$ for all $\hat{\imath}$. Now observing that

$\frac{M^{\prime}}{M}=\frac{\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}^{\prime}}{\sum_{\hat{\jmath}=1}^{M_{0}}R_{\hat{\jmath}}}=\left(\frac{N^{\prime}}{N}\right)^{\frac{\log
n}{\log(n/m)}}$

shows that (vi) holds. ∎

###### Proof of $\ref{itemnow5}\Rightarrow\ref{item4}$.

Assume that $M_{0}=M^{\prime}_{0}$ and
$N_{\hat{\imath}}/N^{\prime}_{\hat{\imath}}=(R^{\prime}_{\hat{\imath}}/R_{\hat{\imath}})^{\gamma}=(M^{\prime}/M)^{\gamma}$
for all $\hat{\imath}=1,\dotsc,M_{0}$. Then (5.5.11) holds for every
$\lambda\in\mathbb{R}$. Since both sides of (5.5.12) are strictly positive for
$k=2$, both sides of (5.5.11) are convex functions of $\lambda$. Hence, their
Legendre transforms are also equal:

$\sup_{\lambda\in\mathbb{R}}\bigg{\\{}\lambda
t-\log\bigg{(}\frac{1}{M}\sum_{\hat{\imath}=1}^{M_{0}}R_{\hat{\imath}}N_{\hat{\imath}}^{\lambda}\bigg{)}\bigg{\\}}=\sup_{\lambda\in\mathbb{R}}\bigg{\\{}\lambda
t-\log\bigg{(}\frac{(M^{\prime}/M)^{\gamma\lambda}}{M^{\prime}}\sum_{\hat{\imath}=1}^{M^{\prime}_{0}}R^{\prime}_{\hat{\imath}}(N^{\prime}_{\hat{\imath}})^{\lambda}\bigg{)}\bigg{\\}}$

for all $t$, which is precisely $I(t)=I^{\prime}(t-\gamma\log(M^{\prime}/M))$.
∎

###### Proof of $\ref{item4}\Rightarrow\ref{item1}$.

Assume $I(t)=I^{\prime}(t-\gamma\log(M^{\prime}/M))$ for every
$t\in\mathbb{R}$. We claim that for every
$s\in(\dim_{\mathrm{H}}\Lambda,\dim_{\mathrm{B}}\Lambda)$,

$t^{\prime}_{\ell}(s)=t_{\ell}(s)-\gamma\log\left(\frac{M^{\prime}}{M}\right)\quad\text{
for every }\ell\in\mathbb{N}.$ (5.5.13)

The proof goes by induction on $\ell$. For $\ell=1$,
$t_{1}(s)=\big{(}s-\log(M^{\prime}M/M)\log m\big{)}\log
n=t_{1}(s)-\gamma\log(M^{\prime}/M)$. Assuming (5.5.13) for $\ell-1$, we prove
for $\ell$:

$\displaystyle t^{\prime}_{\ell}(s)$
$\displaystyle=T^{\prime}_{s}(t^{\prime}_{\ell-1}(s))$
$\displaystyle=\left(s-\frac{\log M^{\prime}}{\log m}\right)\log n+\gamma
I^{\prime}\left(t_{\ell-1}(s)-\gamma\log\left(\frac{M^{\prime}}{M}\right)\right)$
$\displaystyle\stackrel{{\scriptstyle\ref{item4}}}{{=}}\left(s-\frac{\log(M^{\prime}M/M)}{\log
m}\right)\log n+\gamma I\big{(}t_{\ell-1}(s)\big{)}$
$\displaystyle=T_{s}(t_{\ell-1}(s))-\gamma\log\left(\frac{M^{\prime}}{M}\right),$

which completes the proof of (5.5.13) since
$T_{s}(t_{\ell-1}(s))=t_{\ell}(s)$.

From (5.5.13) and assumption (v) it immediately follows that

$I(t_{\ell}(s))=I^{\prime}(t^{\prime}_{\ell}(s)).$ (5.5.14)

Assumption (v) also implies (vi), thus we know that

$N^{\prime}=\sum_{\hat{\imath}=1}^{M^{\prime}_{0}}R^{\prime}_{\hat{\imath}}N^{\prime}_{\hat{\imath}}\stackrel{{\scriptstyle\ref{itemnow5}}}{{=}}\sum_{\hat{\imath}=1}^{M_{0}}\frac{M^{\prime}R_{\hat{\imath}}}{M}\cdot
N_{\hat{\imath}}\left(\frac{M}{M^{\prime}}\right)^{\gamma}=N\left(\frac{M}{M^{\prime}}\right)^{\gamma-1}.$
(5.5.15)

Writing $s^{\prime}_{\theta}=\dim_{\theta}\Lambda^{\prime}$, using (5.5.13),
(5.5.14), (5.5.15) and algebraic manipulations, we obtain

$\displaystyle 0$ $\displaystyle=-s^{\prime}_{\theta}\log
n+\gamma^{L}\theta\log
N^{\prime}-(\gamma^{L}\theta-1)t^{\prime}_{L}(s^{\prime}_{\theta})+\gamma(1-\gamma^{L-1}\theta)(\log
M^{\prime}-I^{\prime}(t^{\prime}_{L}(s^{\prime}_{\theta})))$
$\displaystyle=-s^{\prime}_{\theta}\log n+\gamma^{L}\theta\log
N-(\gamma^{L}\theta-1)t_{L}(s^{\prime}_{\theta})+\gamma(1-\gamma^{L-1}\theta)(\log
M-I(t_{L}(s^{\prime}_{\theta}))).$

By Theorem 5.2.1, $\dim_{\theta}\Lambda$ is the unique solution to this
equation, so $\dim_{\theta}\Lambda=\dim_{\theta}\Lambda^{\prime}$. This
completes the proof of Theorem 5.2.5. ∎

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