(Note: I modified the question and instead of looking at simple modules in the question, I look at all indecomposable modules.)

Let $A$ be a finite dimensional algebra over a field $K$ given by an acyclic quiver with relations and assume for simplicity that it has finite global dimension and the field is algebraically closed. It is well known that the projective dimension of the regular module as a bimodule equals the global dimension of the algebra. But what is the injective dimension of the regular module as a bimodule? It seems noone has studied that.

Question:

Do we have that the injective dimension of $A$ as an bimodule is equal to $\max \{ projdim(M)+injdim(M) | M$ is an indecomposable right A-module $ \}$ ?

 

If not, do we at least have that the injective dimension of $A$ is bounded by $\max \{ projdim(M)+injdim(M) | M$ is an indecomposable right A-module $ \}$ ?

It is true for all Nakayama algebras with a linear quiver and at most 6 simple modules, which are altogether 64 algebras (where the value of the injective dimension varies quite alot).

(By the way is the injective dimension of $A$ equal to the projective dimension of $D(A)$ as a bimodule? Im confused about this)

(Note: I modified the question and instead of looking at simple modules in the question, I look at all indecomposable modules.)

Let $A$ be a finite dimensional algebra over a field $K$ given by an acyclic quiver with relations and assume for simplicity that it has finite global dimension and the field is algebraically closed. It is well known that the projective dimension of the regular module as a bimodule equals the global dimension of the algebra. But what is the injective dimension of the regular module as a bimodule? It seems noone has studied that.

Question:

Do we have that the injective dimension of $A$ as an bimodule is equal to $\max \{ projdim(M)+injdim(M) | M$ is an indecomposable right A-module $ \}$ ?

If not, do we at least have that the injective dimension of $A$ is bounded by $\max \{ projdim(M)+injdim(M) | M$ is an indecomposable right A-module $ \}$ ?

It is true for all Nakayama algebras with a linear quiver and at most 6 simple modules, which are altogether 64 algebras (where the value of the injective dimension varies quite alot).

(By the way is the injective dimension of $A$ equal to the projective dimension of $D(A)$ as a bimodule? Im confused about this)

(Note: I modified the question and instead of looking at simple modules in the question, I look at all indecomposable modules.)

Let $A$ be a finite dimensional algebra over a field $K$ given by quiver with relations and assume for simplicity that it has finite global dimension and the field is algebraically closed. It is well known that the projective dimension of the regular module as a bimodule equals the global dimension of the algebra. But what is the injective dimension of the regular module as a bimodule? It seems noone has studied that.

Question:

Do we have that the injective dimension of $A$ as an bimodule is equal to $\max \{ projdim(M)+injdim(M) | M$ is an indecomposable right A-module $ \}$ ?

It is true for all Nakayama algebras with a linear quiver and at most 6 simple modules, which are altogether 64 algebras (where the value of the injective dimension varies quite alot).

(By the way is the injective dimension of $A$ equal to the projective dimension of $D(A)$ as a bimodule? Im confused about this)

(Note: I modified the question and instead of looking at simple modules in the question, I look at all indecomposable modules.)

Let $A$ be a finite dimensional algebra over a field $K$ given by an acyclic quiver with relations and assume for simplicity that it has finite global dimension and the field is algebraically closed. It is well known that the projective dimension of the regular module as a bimodule equals the global dimension of the algebra. But what is the injective dimension of the regular module as a bimodule? It seems noone has studied that.

Question:

Do we have that the injective dimension of $A$ as an bimodule is equal to $\max \{ projdim(M)+injdim(M) | M$ is an indecomposable right A-module $ \}$ ?

If not, do we at least have that the injective dimension of $A$ is bounded by $\max \{ projdim(M)+injdim(M) | M$ is an indecomposable right A-module $ \}$ ?

It is true for all Nakayama algebras with a linear quiver and at most 6 simple modules, which are altogether 64 algebras (where the value of the injective dimension varies quite alot).

(By the way is the injective dimension of $A$ equal to the projective dimension of $D(A)$ as a bimodule? Im confused about this)

Let $A$ be a finite dimensional algebra over a field $K$ given by quiver with relations and assume for simplicity that it has finite global dimension. It is well known that the projective dimension of the regular module as a bimodule equals the global dimension of the algebra. Let $D(A):=Hom_K(A,K)$ the dual of the regular module, which is also a bimodule. It seems that it is not known what the projective dimension of $D(A)$ is in general.

Do we have that the projective dimension of $D(A)$ as an bimodule is larger than or equal to $\max \{ projdim(S)+injdim(S) | S$ is a simple right A-module $ \}$ ?

It is true for all Nakayama algebras with a linear quiver and at most 6 simple modules. Surprisingly, we have equality very often, namely of all 64 Nakayama algebras with a linear quiver with at most 6 simple modules, only the algebra with Kupisch series [ 3, 4, 3, 3, 2, 1 ] had that the projective dimension of $D(A)$ as a bimodule was 4 while $\max \{ projdim(S)+injdim(S) | S$ is a simple right A-module $ \}$=3 and for the other 63 algebras we had equality of the two numbers.

(Note: I modified the question and instead of looking at simple modules in the question, I look at all indecomposable modules.)

Let $A$ be a finite dimensional algebra over a field $K$ given by quiver with relations and assume for simplicity that it has finite global dimension and the field is algebraically closed. It is well known that the projective dimension of the regular module as a bimodule equals the global dimension of the algebra. But what is the injective dimension of the regular module as a bimodule? It seems noone has studied that.

Question:

Do we have that the injective dimension of $A$ as an bimodule is equal to $\max \{ projdim(M)+injdim(M) | M$ is an indecomposable right A-module $ \}$ ?

It is true for all Nakayama algebras with a linear quiver and at most 6 simple modules, which are altogether 64 algebras (where the value of the injective dimension varies quite alot).

(By the way is the injective dimension of $A$ equal to the projective dimension of $D(A)$ as a bimodule? Im confused about this)