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](http://arxiv.org/pdf/1312.6358)). But this property seems to be so intuitive that it "should" hold in more general situations.



Rank vanishing in tensor categories
No, I think this often fails. One way to fail would be if your category is linear over a field k of characteristic p. Then the rank of a direct sum of p copies of any fixed object will be zero. But you probably meant to exclude that kind of example. Then again there's another kind of example, the most basic of which would be Z/2-graded vector spaces with the Koszul sign rule. Then the rank behaves more like an Euler characteristic, and it can be certainly be zero without the object being zero.


