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Let $A$ be a finite dimensional algebra with enveloping algebra $A^e$.

Auslander and Reiten proved in "On a theorem of E. Green on the dual of the transpose" that $Hom_A(Tr_{A^e}(A),M) \cong \tau(M)$ for any non-projective indecomposable $A$-module $M$.

Question 1:We have $Hom_A(Tr_{A^e}(A),M) \cong D( Tr_{A^e}(A) \otimes_A D(M)) $, is this isomorphic to $M \otimes_A \tau_{A^e}(A)$?

Question 2: In case question 1 is true, when we have an almost split sequence of $A$-bimodules $0 \rightarrow \tau_{A^e}(A) \rightarrow X \rightarrow A \rightarrow 0$ and we tensor it with an $A$-module $M$ over $A$, we get an exact sequence $0 \rightarrow \tau(M) \rightarrow M \otimes_A X \rightarrow M \rightarrow 0$. Is this exact sequence again almost split? If not, does this at least work when $A$ is a symmetric algebra where question 1 has indeed a positive answer?

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1 Answer 1

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Let $A=k[x]/(x^2)$ and $S$ the simple $A$-module for a field $k$. If $\eta\colon 0\to \tau_{A^e}(A) \to X \to A\to 0$ is the almost split sequence ending in $A$ over $A^e$, then $S\otimes_A X$ is a semisimple module. It follows that $S\otimes_A \eta$ is a split exact sequence. Hence Question 2 is not always true.

There is an analogue of this for group algebras $kG$ and Hopf algebras with involutive antipode, with $k$ an algebraically closed field of characteristic $p$. See https://www.sciencedirect.com/science/article/pii/0021869386901730 for the group algebra case. Then, if $0\to \tau(k) \to E\to k\to 0$ is the almost split sequence ending in the trivial module $k$, then tensoring with any module $M$, we obtain an exact sequence $0\to \tau(k)\otimes_k M \to E\otimes_k M \to k\otimes_k M\to 0$, where $k\otimes_k M \simeq M$. If $p \not\mid\dim_k M$, then this sequence is almost split. Hence the tensor product is not always almost split. I would guess the result would be something similar for bimodules. Henning Krause had a PhD-student who address this question at some point. I only remember that it is not true in general, but I don't remember for what reason.

Here is how one can analyze this in QPA2:

gap> Q := Quiver( RIGHT, "Q(1)[a:1->1]" );
Q(1)[a:1->1]
gap> KQ := PathAlgebra( GF( 2 ), Q );
GF(2) * Q
gap> rels := [ One( KQ ) * Q.a * Q.a ];
[ Z(2)^0*(a*a) ]
gap> A := KQ/rels;
(GF(2) * Q) / [ Z(2)^0*(a*a) ]
gap> M := AlgebraAsBimodule( A );
<2>
gap> R := UnderlyingRepresentation( M );
<2>
gap> U := AsModule( LEFT, R );
<2>
gap> TrU := TransposeOfModule( U );
<2>
gap> DTrU := DualOfModule( TrU );
<2>
gap> p := ProjectiveCover( U );
<(4)->(2)>
gap> q := KernelEmbedding( p );
<(2)->(4)>
gap> V := Source( q );
<2>
gap> homVDTrU := Hom( V, DTrU );
Hom(2, 2)
gap> f := BasisVectors( Basis( homVDTrU ) )[ 1 ];
<(2)->(2)>
gap> T := Pushout( f, q );
<4>
gap> RR := UnderlyingRepresentation( T );
<4>
gap> TT := AsBimodule( RR );
<4>
gap> S := SimpleModules( RIGHT, A );
[ <1> ]
gap> MM := TensorProductOfModules( S[ 1 ], TT );
<2>
gap> IsSemisimpleRepresentation( UnderlyingRepresentation( MM ) );
true

The module $T$ is the middle term in the almost split sequence ending in $A$ as a bimodule.

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  • $\begingroup$ Thanks. Is it also possible to test question 1 with QPA 2? The "result" looks like it should naturally hold, but I can not see how an isomorphism might look in general. So maybe it is just wrong for some random algebra like a non-selfinjective Nakayama algebra. $\endgroup$
    – Mare
    Commented May 15, 2020 at 22:21
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    $\begingroup$ I tried a bigger example, but QPA2 has gotten to be very slow compared to earlier, so it didn't seem to finish. For a Nakayama algebra with two vertices it looked like it was true (which one probably can check quite easily by hand). $\endgroup$ Commented May 16, 2020 at 18:01

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