Title: Strange modules part II
Tags: rt.representation-theory;homological-algebra
Question: Let $A$ be a finite dimensional symmetric algebra over a field (we can also assume that it is connected).
Call a non-projective indecomposable module $M$ strange in case $Ext^i(M,M)=0$ for all but finitely many $i$.
Is it true that strange $M$ have complexity equal to one, which means that the terms in a minimal projective resolution have bounded dimension?
In fact, in all my example $\Omega^{i}(M)$ even had the same dimension as $M$ for all $i \in \mathbb{Z}$.

Here is the accepted answer: Let $M$ be one of the examples you know. Then $M\otimes_kM$ is a strange module for $A\otimes_kA$, but does not have complexity one.
Comment for this answer: Thanks, can you give the argument why it has not complexity equal to one? I guess tensoring the minimal projective resolutions for M gives a projective resolution for the tensor product, but why is it minimal?
Comment for this answer: @Mare Let $X$ be the direct sum of the simple $A$-modules, so $\text{Ext}^i_A(M,X)\neq0$ for all $i$. Then $\text{Ext}^n_{A\otimes A}(M\otimes M,X\otimes X)=\bigoplus_i\text{Ext}^i_A(M,X)\otimes\text{Ext}^{n-i}_A(M,X)$ has unbounded dimension.


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