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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.

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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

3 Answers 3

up vote 8 down vote accepted

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.

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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
Right, the conclusion is already false for chain complexes of rational vector spaces; here the dualizable objects are the bounded chain complexes of finite-dimensional vector spaces, with dimension the Euler characteristic. – Qiaochu Yuan Mar 30 '14 at 1:35

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.

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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.

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