Return to Question

Given a finite dimensional selfinjective algebra $A$ and an indecomposable module $M$, which is not projective. Let $v=DHom(-,A)$ be the Nakayama functor. In case $Ext^{i}(v(M),M) \neq 0$ for some i, can we say that $Ext^{j}(M,M) \neq 0$ for some $j$ in general?

Special case: A is a Nakayama algebra and M without loss of generality $M=e_0 A/e_0 J^k$, then $Ext^{i}(v(M),M)=Ext^{i+2 \pi(0)}(M,M)$, when $\pi$ is the Nakayama permutation.

This would have some consequences for the Nakayama conjecture, so special cases are welcome too.

Given a finite dimensional selfinjective algebra $A$ and an indecomposable module $M$, which is not projective. Let $v=DHom(-,A)$ be the Nakayama functor. In case $Ext^{i}(v(M),M) \neq 0$ for some i, can we say that $Ext^{j}(M,M) \neq 0$ for some $j$ in general?

This would have some consequences for the Nakayama conjecture, so special cases are welcome too.

Given a finite dimensional selfinjective algebra $A$ and an indecomposable module $M$, which is not projective. Let $v=DHom(-,A)$ be the Nakayama functor. In case $Ext^{i}(v(M),M) \neq 0$, can we say that $Ext^{j}(M,M)$ for some $j$ in general? Special case: A is a Nakayama algebra and M without loss of generality $M=e_0 A/e_0 J^k$, then $Ext^{i}(v(M),M)=Ext^{i+2 \pi(0)}(M,M)$, when $\pi$ is the Nakayama permutation. This would have some consequences for the Nakayama conjecture, so special cases are welcome too.

Given a finite dimensional selfinjective algebra $A$ and an indecomposable module $M$, which is not projective. Let $v=DHom(-,A)$ be the Nakayama functor. In case $Ext^{i}(v(M),M) \neq 0$ for some i, can we say that $Ext^{j}(M,M) \neq 0$ for some $j$ in general?

Special case: A is a Nakayama algebra and M without loss of generality $M=e_0 A/e_0 J^k$, then $Ext^{i}(v(M),M)=Ext^{i+2 \pi(0)}(M,M)$, when $\pi$ is the Nakayama permutation.

This would have some consequences for the Nakayama conjecture, so special cases are welcome too.