# Serre functor on the category $Perf(A)$, $A$ - k-algebra

Consider a finite-dimensional $$k$$-algebra $$A$$ of finite global dimension. Then it is known that the Serre functor on $$D^b(mod-A)$$ exists and is given by the Nakayama functor. The proof goes something like this:

The $$k$$-duality $$(-)^*=R\underline{\text{Hom}}_k(-,k)$$ gives an equivalence $$D^b(mod-A)\to D^b(A-mod).$$ The $$A$$-duality $$(-)^\vee=R\underline{\text{Hom}}_A(-,A)$$ maps the category of perfect complexes $$Perf(A)$$ to $$Perf(A^{op})$$, because $$A^\vee$$ is isomorphic to $$A$$ in $$A-mod$$. Then one proves that for $$M\in Perf(A)$$, $$N\in D^b(mod-A)$$ we have

$$Hom_{D(Mod-A)} (M,N)^*\cong Hom_{D(Mod-A)} (N,M^{\vee*}).$$

Finally, we notice that for finite-dimensional algebras of finite global dimension $$Perf(A)=D^b(A-mod).$$

But what if $$\text{r.gl.dim}(A)=\infty$$? The only reason I see why the above proof might not give us a Serre functor on $$Perf(A)$$ is that $$(-)^*$$ may not map $$Perf(A)$$ to $$Perf(A^{op})$$, since $$A$$ is not necessarily isomorphic to $$A^*$$ as a left $$A$$-module. In this case, is it enough to have $$A\cong A^{op}$$?

The assumption on the finite global dimension is just needed to have $$Perf(A)=D^b(A-mod),$$ which does not hold more generality.
It is indeed true that $$Perf(A)$$ has a Serre functor (given in the same way) when $$A$$ just is a Gorenstein algebra, that is the regular module $$A$$ has finite injective dimension as a left and right $$A$$-modules.
For example any selfinjective non-semisimple algebra (such as $$K[x]/(x^n)$$ for $$n \geq 2$$) is a Gorenstein algebra of infinite global dimension.