A uniform rank and action-multiplicity bound for module categories

On the real vector space with orthonormal basis $b_i=[X_i]$, rigidity gives Frobenius reciprocity $N_{ij}^k=N_{i^{\ast }k}^j=N_{k{j}^{\ast }}^i$, and $N_{ij}^0={\delta }_{j,i^{\ast }}$. Indeed the adjunctions $\mathrm{Hom}(X_i\otimes X_j,X_k)\cong \mathrm{Hom}(X_j,X_i^{\ast }\otimes X_k)$ and $\mathrm{Hom}(X_i\otimes X_j,X_k)\cong \mathrm{Hom}(X_i,X_k\otimes X_j^{\ast })$ give the first equalities by dimensions; simplicity of the unit gives the second. Left and right multiplication matrices $L_i,R_i$ therefore have transposes $L_{i^{\ast }},R_{i^{\ast }}$, respectively, and $L_i{R}_j=R_j{L}_i$ by associativity. The matrix $S={\sum }_i{L}_i$ is real symmetric and strictly positive entrywise: for any $a,b$, the nonzero object $X_b\otimes X_a^{\ast }$ has a simple summand $X_i$, and reciprocity then gives $N_{ia}^b>0$. The object is nonzero because tensoring by a nonzero rigid object is faithful: the unit is a summand of its dual times itself, so a zero tensor product would force the other factor to be zero.

Here is the required positive-matrix argument without an unproved Perron theorem. A real symmetric strictly positive matrix $T$ has a largest eigenvalue $\lambda$ by the finite-dimensional spectral theorem, equivalently the maximum of $v^{t}Tv$ on the real unit sphere. Replacing a maximizing vector by its coordinatewise absolute value cannot decrease this expression; take a nonnegative maximizer $v$. The equation $Tv=\lambda v$ implies $\lambda >0$ and every coordinate of $v$ is positive. Every maximizing vector has coordinates of one sign: if both signs occur, taking absolute values strictly increases its quadratic form because every cross entry is positive. Hence the top eigenspace is one-dimensional, for an independent top eigenvector could be chosen orthogonal to $v$ and would necessarily have both signs. Apply this to $S$ and normalize its positive top eigenvector $v$ by $v_0=1$.

Since $R_i$ commutes with $S$, $R_{i}v=d_{i}v$ for a scalar $d_i$; it is positive because $R_i$ is nonzero and nonnegative and $v$ is strictly positive. Transposition and the scalar product give $d_i⟨v,v⟩=⟨v,R_{i}v⟩=⟨R_{i^{\ast }}v,v⟩=d_{i^{\ast }}⟨v,v⟩$. The coefficient of $b_0$ in $v{b}_i$ is $v_{i^{\ast }}$, so $d_i=v_{i^{\ast }}=v_i$. Right multiplication by a product now gives $d_i{d}_j={\sum }_k{N}_{ij}^k{d}_k$, and $R_0=1$ gives $d_0=1$. These numbers are fixed by the fusion data, since $S$ and the normalization of its unique positive top eigenvector are fixed. Writing $\mathrm{\Omega }={\sum }_i{d}_i{b}_i=v$, we have $\mathrm{\Omega }{b}_i=d_i\mathrm{\Omega }$ and therefore ${\mathrm{\Omega }}^2=D\mathrm{\Omega }$ for $D={\sum }_i{d}_i^2$.

For a module category let $A_i$ have entry $(b,a)$ equal to $n_{ai}^b$. Adjunction for the action, obtained by applying its associator to the evaluation and coevaluation of $X_i$, gives $\mathrm{Hom}(m_a\odot X_i,m_b)\cong \mathrm{Hom}(m_a,m_b\odot X_i^{\ast })$, hence $A_i^t=A_{i^{\ast }}$. The matrices represent the fusion ring in the opposite order, which does not affect the equation for the square of a single element. Thus $B={\sum }_i{d}_i{A}_i$ is symmetric, has nonnegative entries, satisfies $B^2=DB$, and has diagonal entries at least one since $A_0=1$. If the category is indecomposable, $B$ is strictly positive: the undirected graph joining simples whenever some $A_i$ has a positive connecting entry is connected, since its connected components are exactly action-stable summands. A path from $a$ to $b$ gives a positive entry in the product of its action matrices. Associativity expands that product as a nonnegative integer sum of the $A_i$, so some $A_i$ already connects $a$ to $b$.

Every eigenvalue of symmetric $B$ is either $0$ or $D$, by $B^2=DB$. In the indecomposable case strict positivity and step 2.1 make its top eigenspace one-dimensional; since $B\mathrm{\ne }0$, its nonzero eigenvalue is $D$ and it has rank one. Thus $r\le \mathrm{tr}B=D$, as each of its $r$ diagonal entries is at least one. For the multiplicity bound in any module category, no indecomposability is needed: the same polynomial identity and symmetry show that the operator norm of $B$ is $D$, so $0\le d_i(A_i{)}_{ba}\le B_{ba}\le D$ by applying the norm inequality to the unit coordinate vectors. Each entry is therefore a nonnegative integer bounded by $D\mathrm{/}{d}_i$. At rank $r$ there are only $\mathrm{\mid }I\mathrm{\mid }{r}^2$ entries and finitely many such arrays. For indecomposable arrays $1\le r\le \lfloor D\rfloor$, so their union is finite. The empty category, if allowed, adds the single rank-zero array and is not indecomposable.