3
$\begingroup$

I guess there should be a general category-theory answer to this question, but I'm asking what I'm interested in specifically, since I don't know if it is true:

Let $H$ be a finite-dimensional hereditary algebra. Let $\tau = D \circ Tr: \text{mod}(H) \to \text{mod}(H)$ be the Auslander-Reiten translation and let $\tau^{-} = Tr \circ D: \text{mod}(H) \to \text{mod}(H)$ be the Auslander-Reiten translation in the other direction.

Fact 1: $\tau$ is a left exact additive functor, $\tau^-$ is a right exact additive functor.

Fact 2: Let $\text{mod}(H)_i$ be the category of finite-dimensional $H$-modules without indecomposable injective direct summands and $\text{mod}(H)_p$ be the category of finite-dimensional $H$-modules without indecomposable projective direct summands. Then $\tau: \text{mod}(H)_p \to \text{mod}(H)_i$ is an equivalence of categories with quasi-inverse $\tau^{-}$.

Question: Are $\tau$ and $\tau^-$, restricted as above, exact?

Added after Jeremy's answer: Is there also a proof which only uses the two facts stated by me and not the concrete definition of the Auslander-Reiten translation? So is this somehow just a category theory consequence?

$\endgroup$
2
  • $\begingroup$ Your title mentions (non-)preprojective and preinjective modules, but they don't seem to play a role in the question? $\endgroup$ – Jeremy Rickard Sep 20 '17 at 8:58
  • $\begingroup$ I'm sorry. Originally I thought it would be about non-preprojective (and non-preinjective) modules and while formulating the question I realized it is about non-projective and non-injective modules. I changed the title now. $\endgroup$ – Leon Lang Sep 20 '17 at 18:24
6
$\begingroup$

If $H$ is hereditary, then Auslander-Reiten translation is simply $\tau X=D\text{Ext}^1_H(X,H)$. Also, $\text{Ext}^2_H$ vanishes, and if $X$ has no nonzero projective direct summand then $\text{Hom}_H(X,H)=0$ since every submodule of $H$ is projective. [If $\alpha:X\to H$ is a non-zero map then $\text{im}(\alpha)$ is projective, so $X\to\text{im}(\alpha)$ splits, and so $X$ has a direct summand isomorphic to $\text{im}(\alpha)$.]

So if $0\to L\to M\to N\to 0$ is a short exact sequence of $H$-modules with no projective summands, the long exact $\text{Ext}$ sequence obtained by applying $\text{Hom}_H(-,H)$ reduces to $$0\to\text{Ext}^1_H(N,H)\to\text{Ext}^1_H(M,H)\to\text{Ext}^1_H(L,H)\to0$$ and so, taking vector space duals, we get a short exact sequence $$0\to\tau L\to\tau M\to\tau N\to0.$$

$\endgroup$
1
  • $\begingroup$ Thanks, this answers my question. Do you know if there is also an answer which only uses the two facts I stated? so is the exactness just a categorical corollary from the two facts? I'll edit my question in this regard. $\endgroup$ – Leon Lang Sep 20 '17 at 18:28

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.