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Im not sure whether this question is appropriate for MO, but I do not have much experience with bimodules (or I forgot many things).

Let $A$ be a finite dimensional (connected) algebra over a field $k$ and $M$ an indecomposable $A$-bimodule (finite dimensional prefered). For simplicity we can assume that $A$ is even a quiver algebra and maybe that the field is algebraically closed in case this is needed.

Is $M^{\otimes n}$ also indecomposable (or zero) for any $n \geq 1$ as an $A$-bimodule? If not, is it true under some extra conditions? What about when $M=D(A)$ ? (it seems to be this should be true at least for acyclic quiver algebras)

What can be said about the endomorphism ring of $M^{\otimes n}$ as an $A$-bimodule when the endomorphism ring of $M$ is known?

Here the tensor product is always over the algebra $A$. You can also give an answer for general rings $A$ and modules $M$ but I have no feeling whether this question is interesting or trivial in such generality.

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  • $\begingroup$ To be clear, do you mean that $A$ is an associative (unital?) algebra of finite dimension over a field $k$, and all tensor products are taken over $k$ (with the induced $(A,A)$-bimodule structure)? $\endgroup$ Commented Nov 16, 2018 at 16:22
  • $\begingroup$ @R.vanDobbendeBruyn Yes associative and unital and the tensor product is over the algebra $A$ and not the field $k$. Over the field $k$ it should be false since $A \otimes_k A$ is isomorphic to $A^{dim A}$ I guess. $\endgroup$
    – Mare
    Commented Nov 16, 2018 at 17:08
  • $\begingroup$ Yes, of course; tensor product over $B$ naturally takes $_A\operatorname{\underline{Mod}}_B \times\!\ _B\operatorname{\underline{Mod}}_C$ to $_A\operatorname{\underline{Mod}}_C$. $\endgroup$ Commented Nov 16, 2018 at 17:09
  • $\begingroup$ @R.vanDobbendeBruyn What do you mean? Why is Mod underlined? $\endgroup$
    – Mare
    Commented Nov 16, 2018 at 17:15
  • $\begingroup$ Sorry, I just meant the categories $(A,B)$-bimodules, etc. $\endgroup$ Commented Nov 16, 2018 at 17:18

1 Answer 1

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Let $Q=( 2_{\circlearrowleft c}\xleftarrow a 1 \xrightarrow b 3_{\circlearrowleft d})$, that is, 3 vertices with 4 arrows, one loop at vertex 2 and 3 and two arrows from vertex 1 to 2 and 3. Consider the relations $\{ac, c^2, d^2, bd\}$ on $Q$. Define $A = \mathbb{Z}_7Q/\langle ac, c^2, d^2, bd\rangle$.

Let $D(A)$ be the dual of $A$ as a bimodule over $A$. Then given that QPA2 doesn't do any mistakes, $DM^{\otimes 2}$ is decomposable. Here are the computations done in QPA2:

gap> Q := RightQuiver( "Q(3)[a:1->2,b:1->3,c:2->2,d:3->3]" );       
Q(3)[a:1->2,b:1->3,c:2->2,d:3->3]
gap> KQ := PathAlgebra(GF(7), Q);
GF(7) * Q
gap> rels := [ KQ.ac,KQ.cc,KQ.dd,KQ.bd ]; 
[ Z(7)^0*(a*c), Z(7)^0*(c*c), Z(7)^0*(d*d), Z(7)^0*(b*d) ]
gap> A := KQ/rels;                        
(GF(7) * Q) / [ Z(7)^0*(a*c), Z(7)^0*(c*c), Z(7)^0*(d*d), Z(7)^0*(b*d) ]
gap> M := AlgebraAsBimodule(A);           
<1,1,1,0,2,0,0,0,2>
gap> DM := DualOfModule(M);               
<1,1,1,0,2,0,0,0,2>
gap> DM2 := TensorProductOfModules(DM,DM);
<0,1,1,0,1,0,0,0,1>
gap> DM3 := TensorProductOfModules(DM,DM2);
<0,1,1,0,1,0,0,0,1>
gap> IsIndecomposableModule(DM2);
false
gap> DecomposeModule(DM2);
[ <0,0,0,0,0,0,0,0,1>, <0,0,0,0,1,0,0,0,0>, <0,0,1,0,0,0,0,0,0>, <0,1,0,0,0,0,0,0,0> ]
gap> IsomorphicModules(DM2,DM3);
true

Here QPA2 claims that $D(A)^{\otimes 2}$ decomposes in 4 simple bimodules. Furthermore, the bimodules $D(A)^{\otimes n}$ are all isomorphic for $n\geq 2$, and $\operatorname{End}(D(A)^{\otimes 2}) \simeq \mathbb{Z}_7^4$.

I hope that these comments are helpful.

The QPA-team.

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  • $\begingroup$ I dont have QPA 2, so I can not test it. But doing some tests with QPA 1, it feels that it is true. (it seems in QPA 1 it is not possible to calculate such tensor products as modules, but just as vector spaces). $\endgroup$
    – Mare
    Commented Mar 25, 2019 at 11:58
  • 1
    $\begingroup$ In QPA1 it is only possible to compute the vector space of $M\otimes_A N$ for a right module $M$ and a left module $N$. In QPA2 one can compute all possibilities, one-sided with one-side, two-sided with one-sided, one-sided with two-sided and two-sided with two-sided, and get as an result a vector space, one-sided module or a two-sided module. QPA2 is available from the above link, but you also have to install the latest versions of the packages it depends on. Furthermore, can also compute different versions of Hom-spaces in QPA2, which were not available in QPA1. $\endgroup$ Commented Mar 25, 2019 at 14:27

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