# Isomorphism for Ext spaces for finite dimensional algebras

Let $$A$$ be an Artin algebra with enveloping algebra $$A^e$$. Then we have $$Hom_{A^e}(X,A^e) \cong Hom_A(D(A) \otimes_A X,A)$$ for a bimodule $$X$$. (see for example in the article "A theorem of Green on the dual of the transpose" by Auslander and Reiten in corollary 4.2.)

Question: When do we have the Ext analogue: $$Ext_{A^e}^i(X,A^e) \cong Ext_A^i(D(A) \otimes_A X,A)$$ for all $$i \geq 1$$?

This holds for example for $$X=A$$. Maybe there is a nice condition and a reference for such isomorphisms.

My guess is that it should be true for $$X=A^{*}$$ (the dual of the bimodule $$A$$) for $$i=1,..,n-2$$ in case $$A$$ is $$n$$-torsionfree for $$n \geq 3$$.

Question 2: Is this true?

This would prove the equivalence of conditions a) and b) for finite dimensional algebras in question 2 in On properties of an algebra as a bimodule

edit:

Here is an example found with the computer that shows that the formula might not hold in general:

Let $$A$$ be the Nakayama algebra with Kupisch series [ 2, 3, 2, 1 ] and $$X=D(A)$$.

Then QPA says that $$Ext_{A^e}^1(D(A),A^e)$$ has dimension 1 while $$Ext_A^1(D(A) \otimes_A D(A) , A)$$ should have dimension 0.

The code: A:=NakayamaAlgebra([2,3,2,1],GF(3));D:=DualOfAlgebraAsModuleOverEnvelopingAlgebra(A);B:=EnvelopingAlgebra(A);RegB:=DirectSumOfQPAModules(IndecProjectiveModules(B)); t:=Size(ExtOverAlgebra(D,RegB)[2]); CoRegA:=DirectSumOfQPAModules(IndecInjectiveModules(A));U:=NakayamaFunctorOfModule(CoRegA);RegA:=DirectSumOfQPAModules(IndecProjectiveModules(A)); tt:=Size(ExtOverAlgebra(U,RegA)[2]);

I used that $$D(A) \otimes_A D(A)$$ is isomorphic to the Nakayama functor applied to $$D(A)$$.

Another example is again $$X=D(A)$$ and A linear oriented of Dynkin type $$A_n$$.

• Can you give an example of when it does not hold please? Commented Apr 26, 2020 at 11:12
• First, I do not understand your first formula. What is $X$ there? You start with $D(A)\otimes_A Tr(X)$ and arrive at something that does not depend on $X$. Second, can you be more specific of how you derive that from the results of the article? Another little remark: in your question, $Ext^i(D(A)\otimes_A X,A)$ is the Ext of right modules (see Corollary 4.2 that you quote); this is not apparent from the way you write it. Commented Apr 26, 2020 at 11:52
• @VladimirDotsenko Sorry, I think I made a mistake. $- \otimes_A Tr(A)$ does lift $\tau^{-1}(-)$ for an Artin algebra (which is the main result of the cited article) but must not take the value 0 when $X$ is injective in it seems (so $\tau^{-1}(X)$=0 when $X$ is injective but $X \otimes_A Tr(A)$ might be nonzero). So I guess I have no explicit counterexample and it seems reasonable that the formula should hold in general with the standard proof. Ill think about that. Sorry for the confusion.
– Mare
Commented Apr 26, 2020 at 12:05
– Mare
Commented Apr 26, 2020 at 12:22
• thanks. I didn't run the calculations myself, but I have no reason to question them. Presumably (since for Hom's everything works), the non-exactness of tensoring with $D(A)$ is the only fathomable issue, right? (And that IS an issue.) Commented Apr 26, 2020 at 14:09

The natural way for proving such an isomorphism would be to take a projective resolution of $$X$$ (over $$A^e$$) and using the Hom-isomorphism to translate to one-sided modules. But then, even if $$\operatorname{Tor}^A_i(D(A),X) = (0)$$ for $$i>0$$, then to get an isomorphism on Ext-groups one would need either that (i) $$D(A)\otimes_A A^e$$ is projective as a left $$A$$-module or (ii) that $$\operatorname{Ext}^1_A(D(A)\otimes_A A^e,A) = 0$$. Then (i) is equivalent to $$A$$ being selfinjective, while (ii) is equivalent to $$\operatorname{Ext}^1_A(D(A), A) = 0$$.
• Thanks, I made my guess 2 more precise, so we do not need the iso for all $i$. Here an example where guess 2 should be true so that we have an iso with $X=A^{*}$ and $i=1$ with $Ext^1(D(A),A)$ nonzero: The Nakayama algebra with Kupisch series [ 2, 2, 3, 2, 2, 2, 1 ] (or any other algebra with dominant dimension at least three and $Ext^1(D(A),A)$ nonzero).
Question 2 has an easy answer in case I made no mistake: Let $$A$$ be $$n$$-torsionfree as an $$A$$-bimodule for $$n \geq 2$$, then also $$X=A^{*}$$ in $$n$$-torsionfree. Choose a minimal projective resolution $$(P_i)$$ of $$X$$, then we get an exact sequence $$0 \rightarrow X^{*} \rightarrow P_0^{*} \rightarrow P_1^{*} \rightarrow ... \rightarrow P_{n-2}^{*}$$. Now we can apply the Hom-isomorphisms to get the result for $$i=1,..,n-2$$.
I was probably just confused because I first thought it should hold for all $$i \geq 1$$.