I've been reading about connections between Coxeter groups and preprojective algebras, and I keep running into two operations on the derived categories of preprojective algebras which seem very similar, but I can't tell whether they're the same: spherical twists and tilting.

Let $G$ be a non-Dynkin graph and let $\Pi_G$ be the associated preprojective algebra. This is a particular quotient of the path algebra of the double quiver associated to $G$. As such, it has simple modules $S_i$ associated to the vertices of $G$, obtained by killing every path except the idempotent stationary path $e_i$ at $i$. Let $I_i$ be the kernel of the projection $\Pi_G\to S_i$ (spanned by every path except $e_i$).

On one hand, the $I_i$ are tilting modules, and they satisfy $\operatorname{End}_{\Pi_G}(I_i) \cong \Pi_G$. (Buan-Iyama-Reiten-Scott, "Cluster structures for 2-Calabi-Yau categories and unipotent groups", Proposition II.1.4.) Thus the tilting functors $\operatorname{RHom}_{\Pi_G}(I_i, -)$ and $-\otimes^L_{\Pi_G} I_i$ give inverse autoequivalences of the derived category $D^b(\Pi_G)$. A nice thing about these functors is that the group action they generate on the Grothendieck group of $D^b(\Pi_G)$ corresponds to the reflection representation of the Coxeter group associated to $G$ (e.g., Amiot-Iyama-Reiten-Todorov, "Preprojective algebras and c-sortable words", Lemma 2.6)

On the other hand, the simple modules $S_i$ are 2-spherical objects of $D^b(\Pi_G)$. Thus these define spherical twist functors in the sense of Seidel and Thomas (Definition 2.5). I'm still not fully comfortable with manipulating derived categories, but my rough understanding is that, for an object $F$, the twist $\Phi_{S_i}(F)$ is a cone of a map $$ \operatorname{RHom}_{\Pi_G}(S_i, F) \otimes^L_{\Pi_G} S_i \to F $$ derived from the natural evaluation map. Importantly, these seem to also induce the standard reflection representation of the Coxeter group on the Grothendieck group (e.g., as stated after Corollary 3.6 of Ikeda's "Stability conditions for preprojective algebras and root systems of Kac-Moody Lie algebras".) It's also known that the functors themselves satisfy braid relations.

Hugh Thomas's note "Stability, shards, and preprojective algebras", Lemma 3, seems to conflate these two operations, and in general they seem to fulfill the same vague role in different papers (in particular, through their action on the Grothendieck group). But if they are actually equivalent, I don't see why. It certainly seems relevant that there's an exact sequence $$ 0 \to I_i \to \Pi_G\to S_i\to 0 $$ so I can apply $\operatorname{RHom}_{\Pi_G}(-, F)$ and get an exact triangle $$ \operatorname{RHom}_{\Pi_G}(S_i, F) \to F\to \operatorname{RHom}_{\Pi_G}(I_i, F). $$ This looks very similar to the definition of the spherical twist functor, but not quite the same!

So my questions are:

  • are derived tilting functors and spherical twist functors actually equivalent in this context?
  • if so, why, and in what level of generality does this happen?
  • if not, how are they related?
  • did I make any mistakes in my statements about derived categories above?
  • $\begingroup$ I'm out of my element, but we do know all autoequivalences are spherical twists along a spherical functor ( arXiv:1603.06717, 2.2 ) , however the source category of the construction in that paper is larger than what we want, which is D^b (pt) so that the tilting functor could be seen as coming directly from those spherical objects S_i. However, maybe at the level of DG categories where you actually have a functorial cone ( cf. arXiv:1208.4046 section 3 ), a morphism (RHom(S,F)xF \to F) \to (RHom(I,F)[-1]+F \to F) ( by the triangle you mention ?) can give you an equivalence on the cones? $\endgroup$
    – AT0
    Mar 30, 2021 at 23:11

1 Answer 1


I found it confusing, and hope that this is correct. We work with right modules, and let $K$ be the base field. Representing $F$ by a complex of injective modules, $\operatorname{RHom}_{\Pi_G}(S_i,F)$ becomes a complex of vector spaces; call it $V$. Since $S_i \cong \Pi_G/I_i$ is a $\Pi_G$-$\Pi_G$-bimodule, $\operatorname{RHom}_{\Pi_G}(S_i,F)$ is also a complex of $\Pi_G$-modules; call it $M$. Since the modules in this complex are annihilated by $I_i$, we have $M \cong V \otimes_K S_i$. Since also $\operatorname{Hom}_{\Pi_G}(\Pi_G,F) \cong F$, the triangle $$\operatorname{RHom}_{\Pi_G}(S_i,F)\otimes_K S_i \to F\to \Phi_{S_i}(F)\to$$ giving the spherical twist can be identified with the triangle involving the tilting functor $$\operatorname{RHom}_{\Pi_G}(S_i,F)\to \operatorname{RHom}_{\Pi_G}(\Pi_G,F)\to \operatorname{RHom}_{\Pi_G}(I_i,F)\to.$$

  • $\begingroup$ Yes, this is correct. The key point is that when we define the spherical functor, $ S_i $ is regarded as just a right module, but it is actually a bimodule, and $ Hom_{\Pi_G}(S_i, F) \otimes_K S_i = Hom_{\Pi_G}(S_i, F) $. The $\otimes_K S_i $ doesn't really do anything: it just shows us $\Pi_G $ acts. $\endgroup$ Dec 11, 2021 at 21:09

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