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Given a symmetric finite dimensional algebra $A$ over a finite field with enveloping algebra $A$.

Assume we know that $\Omega_{A^e}^i(A) \cong A_{f}$, where $f$ is some automorphism of the algebra $A$. (so $A_{f}$ is the bimodule $A$ twisted by this automorphism)

Question 1: Is there a way to obtain $f$ using QPA? Note that we can calculate $\Omega_{A^e}^i(A)$ using QPA.

 

Question 2: Is there a good theoretical way how to obtain $f$ in a quick way? Does $f$ have special properties?

 

Quesiton 3: We know that $A_{f}$ is a cyclic $A^e$-module, how can we find an element $x$ in the enveloping algebra $A^e$ with QPA so that we have $A_f = x A^e$?

 

Question 4: Assume we know that $A$ as a bimodule admits a projective resolution (not necessarily minimal) of the form: $... \rightarrow A^e \rightarrow ... A^e \rightarrow A^e \rightarrow A \rightarrow 0$,so that every term can be choosen to be the regular module of the enveloping algebra. This means that $\Omega_{A^e}^i(A)=x_i A^e$ for some elements $x_i \in A^e$. Is there a canonical choice of the $x_i$ or some nice behavior? Can one obtain the $x_i$ in a nice form via QPA?

More generally, can we obtain a minimal free resolution of a module in QPA instead of a minimal projective resolution?

This is possible by filtering through all elements but in practise this takes too long.

Given a symmetric finite dimensional algebra $A$ over a finite field with enveloping algebra $A$.

Assume we know that $\Omega_{A^e}^i(A) \cong A_{f}$, where $f$ is some automorphism of the algebra $A$. (so $A_{f}$ is the bimodule $A$ twisted by this automorphism)

Question 1: Is there a way to obtain $f$ using QPA? Note that we can calculate $\Omega_{A^e}^i(A)$ using QPA.

Question 2: Is there a good theoretical way how to obtain $f$ in a quick way? Does $f$ have special properties?

Quesiton 3: We know that $A_{f}$ is a cyclic $A^e$-module, how can we find an element $x$ in the enveloping algebra $A^e$ with QPA so that we have $A_f = x A^e$?

Question 4: Assume we know that $A$ as a bimodule admits a projective resolution (not necessarily minimal) of the form: $... \rightarrow A^e \rightarrow ... A^e \rightarrow A^e \rightarrow A \rightarrow 0$,so that every term can be choosen to be the regular module of the enveloping algebra. This means that $\Omega_{A^e}^i(A)=x_i A^e$ for some elements $x_i \in A^e$. Is there a canonical choice of the $x_i$ or some nice behavior? Can one obtain the $x_i$ in a nice form via QPA?

More generally, can we obtain a minimal free resolution of a module in QPA instead of a minimal projective resolution?

This is possible by filtering through all elements but in practise this takes too long.

Given a symmetric finite dimensional algebra $A$ over a finite field with enveloping algebra $A$.

Assume we know that $\Omega_{A^e}^i(A) \cong A_{f}$, where $f$ is some automorphism of the algebra $A$. (so $A_{f}$ is the bimodule $A$ twisted by this automorphism)

Question 1: Is there a way to obtain $f$ using QPA? Note that we can calculate $\Omega_{A^e}^i(A)$ using QPA.

Question 2: Is there a good theoretical way how to obtain $f$ in a quick way? Does $f$ have special properties?

Quesiton 3: We know that $A_{f}$ is a cyclic $A^e$-module, how can we find an element $x$ in the enveloping algebra $A^e$ with QPA so that we have $A_f = x A^e$?

Question 4: Assume we know that $A$ as a bimodule admits a projective resolution (not necessarily minimal) of the form: $... \rightarrow A^e \rightarrow ... A^e \rightarrow A^e \rightarrow A \rightarrow 0$,so that every term can be choosen to be the regular module of the enveloping algebra. This means that $\Omega_{A^e}^i(A)=x_i A^e$ for some elements $x_i \in A^e$. Is there a canonical choice of the $x_i$ or some nice behavior? Can one obtain the $x_i$ in a nice form via QPA?

This is possible by filtering through all elements but in practise this takes too long.

Given a symmetric finite dimensional algebra $A$ over a finite field with enveloping algebra $A$.

Assume we know that $\Omega_{A^e}^i(A) \cong A_{f}$, where $f$ is some automorphism of the algebra $A$. (so $A_{f}$ is the bimodule $A$ twisted by this automorphism)

Question 1: Is there a way to obtain $f$ using QPA? Note that we can calculate $\Omega_{A^e}^i(A)$ using QPA.

Question 2: Is there a good theoretical way how to obtain $f$ in a quick way? Does $f$ have special properties?

Quesiton 3: We know that $A_{f}$ is a cyclic $A^e$-module, how can we find an element $x$ in the enveloping algebra $A^e$ with QPA so that we have $A_f = x A^e$?

Question 4: Assume we know that $A$ as a bimodule admits a projective resolution (not necessarily minimal) of the form: $... \rightarrow A^e \rightarrow ... A^e \rightarrow A^e \rightarrow A \rightarrow 0$,so that every term can be choosen to be the regular module of the enveloping algebra. This means that $\Omega_{A^e}^i(A)=x_i A^e$ for some elements $x_i \in A^e$. Is there a canonical choice of the $x_i$ or some nice behavior? Can one obtain the $x_i$ in a nice form via QPA?

More generally, can we obtain a minimal free resolution of a module in QPA instead of a minimal projective resolution?

This is possible by filtering through all elements but in practise this takes too long.

Given a symmetric finite dimensional algebra $A$ over a finite field with enveloping algebra $A$.

Assume we know that $\Omega_{A^e}^i(A) \cong A_{f}$, where $f$ is some automorphism of the algebra $A$. (so $A_{f}$ is the bimodule $A$ twisted by this automorphism)

Question 1: Is there a way to obtain $f$ using QPA? Note that we can calculate $\Omega_{A^e}^i(A)$ using QPA.

Question 2: Is there a good theoretical way how to obtain $f$ in a quick way? Does $f$ have special properties?

Quesiton 3: We know that $A_{f}$ is a cyclic $A^e$-module, how can we find an element $x$ in the enveloping algebra $A^e$ with QPA so that we have $A_f = x A^e$?

This is possible by filtering through all elements but in practise this takes too long.

Given a symmetric finite dimensional algebra $A$ over a finite field with enveloping algebra $A$.

Assume we know that $\Omega_{A^e}^i(A) \cong A_{f}$, where $f$ is some automorphism of the algebra $A$. (so $A_{f}$ is the bimodule $A$ twisted by this automorphism)

Question 1: Is there a way to obtain $f$ using QPA? Note that we can calculate $\Omega_{A^e}^i(A)$ using QPA.

Question 2: Is there a good theoretical way how to obtain $f$ in a quick way? Does $f$ have special properties?

Quesiton 3: We know that $A_{f}$ is a cyclic $A^e$-module, how can we find an element $x$ in the enveloping algebra $A^e$ with QPA so that we have $A_f = x A^e$?

Question 4: Assume we know that $A$ as a bimodule admits a projective resolution (not necessarily minimal) of the form: $... \rightarrow A^e \rightarrow ... A^e \rightarrow A^e \rightarrow A \rightarrow 0$,so that every term can be choosen to be the regular module of the enveloping algebra. This means that $\Omega_{A^e}^i(A)=x_i A^e$ for some elements $x_i \in A^e$. Is there a canonical choice of the $x_i$ or some nice behavior? Can one obtain the $x_i$ in a nice form via QPA?

This is possible by filtering through all elements but in practise this takes too long.