# A linear category is a Vect-module

I would like to know how to show that any linear category is a $\mathrm{Vec}$-module. Here $\mathrm{Vect}$ denotes a category of finite dimensional vector spaces.

More general statement can be found in the paper Dualizable tensor categories, Example 3.2.9

## 1 Answer

Let $\mathsf{C}$ be a linear category. To give $\mathsf{C}$ the structure of a $\mathsf{Vect}$ (left) module essentially amounts to giving a bifunctor $$\otimes : \mathsf{Vect} \times \mathsf{C} \to \mathsf{C}$$ plus associativity and unitality constraints satisfying a bunch of conditions. I will only give an idea of the definition of $\otimes$, constructing the constraints and checking the conditions is tedious but doable.

On the level of objects, given a vector space $V$ and an object $X \in \mathsf{C}$, we must give something called $V \otimes X$. Since $V$ is finite-dimensional, $V \cong k^n$ for some integer $n$. Intuitively, $k \otimes X$ must be $X$ by the unitality constraint, and $(V \oplus W) \otimes X$ must be $(V \otimes X) \oplus (W \otimes X)$, where now $\oplus$ is the direct sum (biproduct) in the category $\mathsf{C}$. So we define $k^n \otimes X$ to be $X \oplus \dots \oplus X = X^{\oplus n}$.

On the level of morphism, given $f : k^n \to k^m$ a linear map and $g : X \to Y$ a morphism in $\mathsf{C}$, we must give $f \otimes g : k^n \otimes X \to k^m \otimes Y$. The linear map $f$ is represented by a matrix $A = (a_{ij})$, and we have $k^n \otimes X = X^{\oplus n}$, $k^m \otimes Y = Y^{\oplus m}$.

By definition of the biproduct, giving a morphism $X^{\oplus n} \to Y^{\oplus m}$ is the same thing as giving $nm$ morphisms $X \to Y$, in the form of an $m \times n$ matrix. Since $\hom_{\mathsf{C}}(X,Y)$ is a $k$-vector space, we can consider $a_{ij} \cdot g : X \to Y$; these will exactly the entries of the matrix defining the morphism $X^{\oplus n} \to Y^{\oplus m}$.

(I'm not very precise here – I'm using the fact that $\mathsf{Vect}$ is equivalent to its full subcategory spanned by the objects $k^n$.)