# Adjoints of exact functors between semisimple abelian categories

Motivated by the answer to this question, I will ask the following question: Let $$\mathcal{A}$$ and $$\mathcal{B}$$ be small semisimple abelian categories. Let $$U:\mathcal{A} \to \mathcal{B}$$ be a functor that preserves and reflects exact sequences.

Is this enough in general to give an adjoint functor such that the unit of the adjunction is injective? If not, then what are examples of assumptions we can add to make this true?

Edit: In response to the comments below, the definition of semisimple is the one taken from nLab, so yes I am assuming finite direct sums.

An abelian category is called semisimple if every object is a semisimple object, hence a direct sum of finitely many simple objects.

Edit: I am happy to assume that my categories are linear, that is enriced over vector spaces.

– rvk
Oct 31, 2021 at 12:56
• I am asking about for existence of the adjoint as part of the question. Oct 31, 2021 at 13:02
• But I guess the existence should follow from some version of the adjoint functor theorem. Oct 31, 2021 at 13:43
• Could you say what you mean by a “semisimple abelian category”? I know of three (contradictory) definitions. Oct 31, 2021 at 14:21
• I have added some comments above to clear up the quastions addressed here. Oct 31, 2021 at 15:01

This was too big to fit as a comment. Here is a cute, completely trivial, but incredibly useful fact.

Suppose $$(f^*, f_*)$$ is an adjoint pair of functors between additive categories (not necessary abelian or anything of that nature).

If $$X$$ is an object such that $$f^*X \neq 0$$, then the unit map $$\eta\colon X \to f_*f^*X$$ is not zero. This is because this map occurs in the defining relation for unit/counit of an adjunction: the composition

$$f^*X \to f^*f_*f^*X \to f^*X$$

is the identity.

There is a dual result using $$f_*$$ too.

The relevance of this is that if, additionally, $$X$$ is a "simple object" (I leave it to you to pin the meaning of this down), then $$\eta$$ will be automatically injective. If in your category "every object is finite length", then induction will give injectivity on everything.

There are all sorts of fun games you can play with this completely trivial fact and get non-trivial results (for eg., you can deduce auto-equivalences at the level of derived categories using the complex $$id \to f_*f^*$$ from just observing what the functors are doing at the $$K_0$$ level - I am assuming a lot of things about the category in question, abelian, finite length, etc - so don't take this too literally).

I digress though. Coming back to the original question. The interesting bit about the question, in my opinion, is the existence of the adjoint in the first place. As I hope is clear from the above, it won't take much more additional hypothesis to make the unit of adjunction to be injective (or even an isomorphism - assume $$f^*$$ preserves "simples").

In this vein, the "obstruction" to the adjoint existing has a lot to do with if the category "has enough objects". That's very vague, so let me amplify with the example I alluded to in the comments.

Consider the category $$\mathcal{C}$$ of finite dimensional representations for a semisimple Lie algebra over $$\mathbb{C}$$ ($$\mathfrak{sl}_2$$ will do just fine). Every representation is completely reducible, the endomorphism ring of simples is $$\mathbb{C}$$, etc. - everything is about as nice as you can get.

Now consider the forgetful functor from $$\mathcal{C}$$ to vector spaces. This is about as nice as can be: exact, reflects exact sequences, etc. I don't think this has a left adjoint. Morally, such an adjoint would be the free module for the universal enveloping algebra (this is not finite dimensional at all) over a vector space (this is the adjoint in the big category of all modules).

This isn't a proof that such an adjoint can't exist, but should indicate the difficulty in producing one if there aren't "enough objects".

• Thanks a lot. This answer was very helpful. Oct 31, 2021 at 15:21
• Ok, let's try to show that $\mathrm{Forget}: \mathrm{Mod}(\mathfrak{sl}(2)) \to \mathrm{Vec}$ fails to have a left adjoint (where both $\mathrm{Mod}$ and $\mathrm{Vec}$ mean the finite-dimensional versions). Well, since the codomain is generated under direct sums by $\mathbb{C}^1$, and since every additive functor preserves direct sums, the question turns on whether there is or isn't an object $V := \mathrm{Forget}^L(\mathbb{C})$. The defining property of this object $V$, from unpacking the adjunction, is $\hom_{\mathfrak{sl}(2)}(V, W) = \mathrm{Forget}(W)$ for every finite-dimensional $W$. Nov 1, 2021 at 0:58
• Take $W$ simple. Using semisimplicity, you can see that $\hom_{\mathfrak{sl}(2)}(V,W)$ will be of the same dimension as the number of copies of $W$ in a direct sum decomposition of $V$. In other words, if there is an object $V$ with this defining property, then it will need to be $V \overset?= \bigoplus_I \mathrm{dim}(I) I$, where the sum ranges over isomorphism classes of simple objects $I$. Since we know that there are infinitely many non-isomorphic simple $\mathfrak{sl}(2)$, this sum diverges, and there is no such $V$. Nov 1, 2021 at 1:02
• Note that along the way we just proved the Peter–Weyl theorem. Nov 1, 2021 at 1:02