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.


1 Answer 1


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.

  • $\begingroup$ Thanks. In my proof I used, that the projective and simples form a basis of $K_0(mod-A)$ and the projectives usually do this only when the global dimension is finite (?) (at least I needed that the Cartan matrix is invertible). So your proof seems to be better. By the way, for selfinjective algebras this invariant is the number of fixed points of the Nakayama permutation. Thus this shows as a special case that the number of fixed points of the Nakayama permutation and especially weakly symmetric algebras are invariant under derived equivalences (which was probably noted before?) $\endgroup$
    – Mare
    Dec 19, 2019 at 16:51
  • 1
    $\begingroup$ @Mare In fact, for self-injective algebras you can recover the entire cycle type of the Nakayama permutation from the eigenvalues of the Nakayama functor acting on $K_0\left(K^b(P_A)\right)$. $\endgroup$ Dec 19, 2019 at 16:56
  • $\begingroup$ One may also ask whether this trace being well defined (meaning that all such Ext vanish for i high enough) and then its value is invariant under derived equivalences. For example the Nakayama algebra with Kupisch series [3,3,4,4] is not Gorenstein but the trace is well defined. All its tilted algebras have the same trace equal to -1. $\endgroup$
    – Mare
    Dec 19, 2019 at 17:22

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