0
$\begingroup$

I have read several times about $k$-linear categories. I understand in general the construction of enriched categories.

But in the special case of vector spaces I am confused about the vector space addition. If the Hom spaces are vector spaces, where does the addition come from? Composing isn't commutative, so that doesn't work...

In particular, I have a semisimple category $\mathcal{C}$ which is a module category over $\mathrm{vect}_k$ - meaning there exist an action on both objects $V.C$ ($V \in \mathrm{vect}$ and $C \in \mathcal{C}$) and morphisms $f.\phi$ with $f$ vector morphism and $\phi \in \mathcal{C}$.

I want to show that this is $k$-linear i.e. that $\mathrm{Hom}(C, C')$ is a vector space $\forall C, C' \in \mathcal{C}$.

I got the scalar multiplication with help of the action on morphisms, but how does one get the addition?

Thanks in advance

$\endgroup$
  • 1
    $\begingroup$ Isn't your semisimple category already additive? $\endgroup$ – Kevin Carlson Mar 26 at 16:19
1
$\begingroup$

In general, enrichment is extra structure, and you need to specify it. It's worth understanding in detail the one-object case: for $V$ a monoidal category (such as vector spaces with tensor product), a one-object $V$-enriched category is a monoid in $V$. So, for example, a one-object $k$-linear category is a $k$-algebra. It has an underlying category which is given by forgetting the vector space structure and remembering only the multiplication; you need to specify the vector space structure, and in particular the addition, as extra data on top of this.

However, enrichment over commutative monoids has a very peculiar property, which is that under mild hypotheses (the existence of either finite products or finite coproducts) it is actually determined by the bare category structure. See this blog post for a lengthy discussion of how this works.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.