TITLE: Derived invariant for Gorenstein algebras?
QUESTION [6 upvotes]: Let $A$ be a finite dimensional algebra with simple modules $S_i$ and projective indecomposable modules $P_l$ and global dimension $g< \infty$ (and $n$ is the number of simple modules).
I noted that the number $\sum\limits_{l=1}^{n}{\sum\limits_{i=0}^{g}{(-1)^i Ext_A^i(S_l,P_l)}}$ is a derived invariant of the algebra. 
In case I made no mistake, the proof goes by showing that this is the negative trace of the transpose of the inverse of the Coxeter transformation of the algbra, which is a derived invariant.
One can also express the other coefficients of the Coxeter polynomial like this. Did this appear somewhere before?

Question: Is this still true when $A$ is just assumed to be Gorenstein? (that is the injective dimension of $A$ is finite on both sides).

Restricted to selfinjective algebras this would be implies by a positive answer to the question whether the Nakayama permutation is a derived invariant, which I forgot whether it is true.

REPLY [5 votes]: Doesn't basically the same proof work?
Let $A$ be a Gorenstein finite dimensional algebra over an algebraically closed field.
Let $K^b(P_A)$ be the subcategory of the bounded derived category $D^b(\text{mod-}A)$ consisting of perfect objects (i.e., bounded complexes of finitely generated projectives). Since $A$ is Gorenstein, this is also the subcategory consisting of objects isomorphic to bounded complexes of injectives. 
The inverse derived Nakayama functor $\mathbf{R}\text{Hom}_A(DA,-)$ induces a self-equivalence of $K^b(P_A)$, and hence induces an endomorphism of the Grothendieck group $K_0\left(K^b(P_A)\right)$, which has a basis given by the classes of the indecomposable injectives, and its matrix with respect to this basis has entries $\sum_i(-1)^i\dim\text{Ext}^i(S_m,P_l)$, the coefficient of $[I_m]$ when $[P_l]$ is written in terms of the basis of indecomposable injectives. So the trace of this endomorphism is precisely the quantity that you are considering.
This is derived invariant, since $K^b(P_A)$ and the inverse derived Nakayama functor are derived invariant.