Question

I am reading something where this is used extensively, but it is not defined anywhere and no references are given, and I can't find any.

Answer 1

If $\mathcal{C}$ is a $k$-linear category, $X \in \mathcal{C}$, and $V$ is any $k$-vector space (in particular, it could be a hom of two objects in $\mathcal{C}$), then $V \otimes X$ (sometimes written $V \odot Y$ to avoid confusion with a monoidal structure) is the object representing the functor $\mathcal{C} \to \operatorname{Vect}$ sending $Y \in \mathcal{C}$ to $\operatorname{Hom}_{\operatorname{Vect}}(V, \mathcal{C}(X, Y))$, if such an object exists. This is a special case of the notion of copower.

In the case where $\mathcal{C}$ is a tensor category with internal homs, the construction agrees with the tensoring by the internal hom.

Answer 2

Let $\mathcal{A}$ be a k-linear category, $A \in \mathcal{A}$ an object and $V$ a k-vector space. We say that the tensor product of $A$ and $V$ exists if the functor from $\mathcal{A}$ to $\mathbf{Vect}$ given by

$A^{\prime} \mapsto \mathbf{Hom}_{\mathrm{Vect}}(V,\mathcal{A}(A,A^{\prime}))$

is representable. The representing object is sometimes denoted by $V\odot A$ or, more commonly in the k-linear context, by $V\otimes A$. Thus we have a natural isomorphism

$\mathcal{A}(V\odot A,A^{\prime}) \cong \mathrm{Hom}_{\mathrm{Vect}}(V,\mathcal{A}(A,A^{\prime}))$

You can now apply this to the special case where $V$ is the space of homomorphisms between two objects. If all tensor products exist, this is simply saying that $\mathcal{A}(A,-) \colon \mathcal{A} \rightarrow \mathrm{Vect}$ has a left adjoint given by $-\odot A \colon \mathrm{Vect} \rightarrow \mathcal{A}$.

These notions can be generalized to categories enriched in any cosmos $\mathcal{V}$ (a cosmos is a complete and cocomplete symmetric monoidal closed category). These tensor products can then be seen as a special type of weighted colimits.

Answer 3

To add to both the answers of Daniel and Evan, note that, if your category is additive and $V$ is finite dimensional, then $V \odot A$ will always exist. Let $e_1$, $e_2$, ... $e_n$ be be a basis for $V$, then $V \odot A$ is isomorphic to $V^{\oplus n}$.