# When is an object in a linear or abelian category simple? Or: How should I define fusion categories?

I'm confused about the notion of simple objects. Now ncatlab says that an object is simple in an abelian category if it only has itself and 0 as subobjects. On another page, it says that the simple objects $X_i$ in $\mathcal{C}$ are those that have $\mathcal{C}(X_i, X_j) \cong \delta_{i,j} k$, where $\mathcal{C}$ is $k$-linear (that does not imply abelian, though)?

Then, some light at the issue: On the first page, it says that being simple in an abelian and $k$-linear category implies that $\mathcal{C}(X_i, X_j) \cong \delta_{i,j} k$.

But then, in a Kuperberg article, he calls an object $X$ "strongly simple" if $\mathcal{C}(X,X) \cong k$, somehow implicitly implying that this property implies simplicity, and not the other way around.

What's the right definition now? In case that abelian categories and linear categories are too different, I care for the case needed to define fusion categories (and only for the fields $\mathbb{R}$ and $\mathbb{C}$).

## 1 Answer

The two definitions are not equivalent, and the first does not imply the second; in a $k$-linear abelian category, any division algebra $D$ over $k$ can occur as the endomorphism $k$-algebra of a simple object $X$ (e.g. in the category of right $D$-modules).

If $k$ is required to be algebraically closed (which I think is a standard simplifying assumption when dealing with fusion categories), then the only finite-dimensional endomorphism algebra that can occur is $k$ itself, and if $X$ is in turn required to be dualizable (which is part of the definition of a fusion category) then I believe this implies that $\text{End}(X)$ is finite-dimensional.

• And for my money, the first definition is the correct one, at least in an abelian category. – Qiaochu Yuan Mar 24 '15 at 17:54
• But you agree that "strongly simple" implies simple? – Turion Mar 25 '15 at 10:59
• @Turion: strongly simple implies simple in a semisimple abelian category; the point is that if $X$ has any nontrivial subobjects then the inclusion $S \to X$ of such a subobject induces a splitting $X \cong S \oplus X/S$ (this requires that the ambient category be semisimple and that we can take cokernels) and hence $\text{End}(X)$ has a nontrivial idempotent in it. – Qiaochu Yuan Mar 25 '15 at 18:54