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Let algebras be finite dimensional over a field $K$ and let $J$ denote the Jacobson radical (this is the intersection of all maximal right ideals) of an algebra. Being hereditary means that the algebra has global dimension at most 1.

Question:

For the class of algebras $A$ with $Ext_A^1(J,J) \neq 0$, can we explicitly write down a non-split short exact sequence $0 \rightarrow J \rightarrow W \rightarrow J \rightarrow 0$ with a pretty $W$?

A less mathematical question would be:

What is the meaning of $Ext_A^1(J,J)$?

Using findstat, a very beautiful "result" (proof in work) was found: The Nakayama agebras with a linear quiver are in bijection with 321-avoiding permutations and so to every such permutation $\pi$ we can associate a Nakayama algebra $A=A_{\pi}$ with a linear quiver. Now it seems that $Ext_{A_{\pi}}^1(J,J) \cong K^{r_{\pi}}$ when $r_{\pi}$ is the cardinality of the complement of the connectivity set of the permutation as in https://arxiv.org/pdf/math/0509271.pdf.

Let algebras be finite dimensional over a field $K$ and let $J$ denote the Jacobson radical (this is the intersection of all maximal right ideals) of an algebra. Being hereditary means that the algebra has global dimension at most 1.

Question:

For the class of algebras $A$ with $Ext_A^1(J,J) \neq 0$, can we explicitly write down a non-split short exact sequence $0 \rightarrow J \rightarrow W \rightarrow J \rightarrow 0$ with a pretty $W$?

Let algebras be finite dimensional over a field $K$ and let $J$ denote the Jacobson radical (this is the intersection of all maximal right ideals) of an algebra. Being hereditary means that the algebra has global dimension at most 1.

Question:

For the class of algebras $A$ with $Ext_A^1(J,J) \neq 0$, can we explicitly write down a non-split short exact sequence $0 \rightarrow J \rightarrow W \rightarrow J \rightarrow 0$ with a pretty $W$?

Let algebras be finite dimensional over a field $K$ and let $J$ denote the Jacobson radical (this is the intersection of all maximal right ideals) of an algebra. Being hereditary means that the algebra has global dimension at most 1.

Question:

For the class of algebras $A$ with $Ext_A^1(J,J) \neq 0$, can we explicitly write down a non-split short exact sequence $0 \rightarrow J \rightarrow W \rightarrow J \rightarrow 0$ with a pretty $W$?

A less mathematical question would be:

What is the meaning of $Ext_A^1(J,J)$?

Using findstat, a very beautiful "result" (proof in work) was found: The Nakayama agebras with a linear quiver are in bijection with 321-avoiding permutations and so to every such permutation $\pi$ we can associate a Nakayama algebra $A=A_{\pi}$ with a linear quiver. Now it seems that $Ext_{A_{\pi}}^1(J,J) \cong K^{r_{\pi}}$ when $r_{\pi}$ is the cardinality of the complement of the connectivity set of the permutation as in https://arxiv.org/pdf/math/0509271.pdf.

Let algebras be finite dimensional over a field $K$ and let $J$ denote the Jacobson radical (this is the intersection of all maximal right ideals) of an algebra. Being hereditary means that the algebra has global dimension at most 1.

Questions:

1. Do we have Ext_A^1(J,J)=0 iff $A$ is hereditary?

Of course for hereditary $A$, $J$ is projective and thus $Ext_A^1(J,J)=0$ but I wasnt lucky enough to find a counterexample for the convsere but I guess it should be wrong.

2. For the class of algebras $A$ with $Ext_A^1(J,J) \neq 0$, can we explicitly write down a non-split short exact sequence $0 \rightarrow J \rightarrow W \rightarrow J \rightarrow 0$ with a pretty $W$?

Let algebras be finite dimensional over a field $K$ and let $J$ denote the Jacobson radical (this is the intersection of all maximal right ideals) of an algebra. Being hereditary means that the algebra has global dimension at most 1.

Question:

For the class of algebras $A$ with $Ext_A^1(J,J) \neq 0$, can we explicitly write down a non-split short exact sequence $0 \rightarrow J \rightarrow W \rightarrow J \rightarrow 0$ with a pretty $W$?