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For an additive category $\mathcal C$ there is the notion of a Serre functor on $\mathcal C$, i.e. a an autoequivalence $S$ of $C$ such that there exist isomorphisms $$Hom(A, S(B)) \cong Hom(B, A)^*$$ natural in $A,B \in \mathcal C$.

If $\mathcal D$ is a full subcategory of $\mathcal C$ that is preserved by $S$, when is the restriction of $S$ a Serre functor of $\mathcal D$?

The particular case I am wondering about is the following: In the paper [1], the authors compute the Serre functor S of $D^b(\mathcal O_\lambda)$ of the derived category of an integral block of the BGG category $\mathcal O$. Let $\mathcal O^{\mathfrak p}_\lambda$ be the corresponding block of the parabolic category $\mathcal O$. They authors state that $D^b(\mathcal O^{\mathfrak p}_\lambda)$ is a full triangulated subcategory of $D^b(\mathcal O_\lambda)$ that is preserved by $S$. They then prove that the restriction of $S$ shifted by a non-zero integer $n$ is a Serre functor of $D^b(\mathcal O^{\mathfrak p}_\lambda)$.

Doesn't this imply that for complexes $A, B \in D^b(\mathcal O^{\mathfrak p}_\lambda)$ the above equality of Hom-spaces holds both for $S$ and $S[n]$?

[1] Volodymyr Mazorchuk and Catharina Stroppel. “Projective-injective modules, Serre functors and symmetric algebras”. Journal für die reine und angewandte Mathematik 616 (2008), pp. 131–165.

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It's not true that $D^b(\mathcal{O}^{\mathfrak{p}})$ is a full subcategory of $D^b(\mathcal{O})$. Think about the case $\mathfrak{g}=\mathfrak{p}=\mathfrak{sl}_2$. The category $\mathcal{O}^{\mathfrak{p}}$ is finite-dimensional modules and thus semi-simple, whereas in $D^b(\mathcal{O})$, we have that $\mathrm{Ext}^2(\mathbb{C},\mathbb{C})\cong \mathbb{C}$. That's why the restriction of the Serre functor is not itself a Serre functor; it's a minor miracle that a shift of it is.

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I never looked into [1], so I cannot comment on it, let me write only some generalities about the Serre functors.

I will assume that the categories $C$ and $D$ are triangulated. Then if you know that both $S$ and $S^{-1}$ preserve $D$ then you can conclude that $S_{|D}$ is the Serre functor of $D$. Indeed, $S_{|D}$ is an autoequivalence of $D$ (here you need to know that $D$ is preserved by $S^{-1}$), and of course the defining isomorphism also is true.

A priori, there are examples (if I remember correctly due to Van den Bergh) when a functor which is not an autoequivalence enjoys the defining isomorphism.

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  • $\begingroup$ Thank you. I thought I was overlooking something obvious, but as Ben said the problem was with my understanding of the specific situation. $\endgroup$ – Clemens Koppensteiner Aug 5 '14 at 17:29

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