# Rank vanishing in tensor categories

Let $\mathcal{C}$ be a cocomplete $\mathbb{Q}$-linear symmetric monoidal category (if convenient, assume that it is locally finitely presentable and/or abelian). Recall that the rank (or dimension) of a dualizable object $V \in \mathcal{C}$ is the composite

$1 \xrightarrow{\mathrm{coev}} V \otimes \check{V} \cong \check{V} \otimes V \xrightarrow{\mathrm{ev}} 1.$

For example, if $\mathcal{C}=\mathsf{Qcoh}(X)$ for some scheme $X$, then the rank of a locally free sheaf $V$ coincides with the usual locally constant function which is called the rank, multiplied with $1$ in $\Gamma(X,\mathcal{O}_X)$. Of course, $V=0$ when the rank of $V$ is zero.

Question. In general, if the rank of a dualizable object $V$ is zero, do we have $V=0$? If not, what is a simple counterexample $(\mathcal{C},V)$?

For example this is true when $\mathcal{C}$ Tannakian (Deligne, Catégories tannakiennes, Lemme 7.3), and more generally when $\mathcal{C}$ is weakly Tannakian (see Schäppi's paper arXiv:1312.6358). But this property seems to be so intuitive that it "should" hold in more general situations.

• Sorry, forgot to mention that everything is over $\mathbb{Q}$ (as in Deligne, section 7). Now it's included. – Martin Brandenburg Mar 30 '14 at 0:08

• I am interested in char. $0$. If $\mathcal{L}$ is an invertible object, then we don't necessarily have $\mathrm{rk}(\mathcal{L})=1$? In fact, the rank equals the signature of $\mathcal{L}$, which is an involution of $1$, and $-1$ is possible e.g. when considering $\mathbb{Z}$-graded objects of $\mathcal{C}$ (twisted symmetry). If $X$ is $1_\mathcal{C}$ concentrated in degree $1$, then every dualizable graded object has the form $M=\sum_n M_n \otimes X^{\otimes n}$ with $M_n$ dualizable (almost all $0$), $\mathrm{rk}(X)=-1$, hence $\mathrm{rk}(M) = \sum_n (-1)^n \mathrm{rk}(M_n)$. Correct? – Martin Brandenburg Mar 30 '14 at 0:25
Even the ranks of irreps can vanish, and even in the semisimple case. For example, consider the representation theory of the cyclic group $C_3$ with three elements over the field $\mathbb F_2$ with two elements. It has two irreps: the trivial, and one of dimension $2$ (in which in a certain basis the generator acts by $\bigl( \begin{smallmatrix} 0 & 1 \\ 1 & 1 \end{smallmatrix}\bigr)$). You might complain that in this example, this irrep splits over the algebraic closure $\overline {\mathbb F_2}$, and in fact over the field $\mathbb F_4 = \mathbb F_2[x]/(x^2=x+1)$ with four elements. Let $q$ be a power of the prime $p$ and $\mathbb F_q$ the field with $q$ elements. The finite group $\mathrm{SL}(2,\mathbb F_q)$ of order $(q^2-1)(q-1)$ acts by permutations of the projective line $\mathbb F_q\mathbb P^1$ of size $q+1$. The linearization of this action over any field $\mathbb F$ of characteristic $p$ contains a trivial submodule, and the quotient of dimension $q$ is absolutely simple, in the sense that it is simple over the algebraic closure $\overline{\mathbb F_p}$. I believe that this is called the Steinberg module, but experts should feel free to correct me.
Here's one well-known positive result in this direction. If V is absolutely simple and $V \otimes V^*$ is semisimple (e.g. if $\mathcal{C}$ is semisimple) then $\mathrm{dim} V \neq 0$.
I learned this from ENO's "On Fusion Categories" but they say it is much older. The proof is as follows. $1$ occurs in $V \otimes V^*$ with multiplicity one (since $1 = \dim \mathrm{Hom}(V,V) = \dim \mathrm{Hom}(1,V \otimes V^*$). Hence any composition of nonzero maps $1 \rightarrow V \otimes V^* \rightarrow 1$ is nonzero.