A note on spherical functors

We provide a short and simple proof of the fact that a spherical functor between triangulated categories induces an autoequivalence.


Introduction
Throughout this article, we find it intuitive and convenient to define functors using certain natural triangles of other functors. However, since the operation of taking cones in a triangulated category is not functorial, we first need to clarify what we mean by a triangle of functors. To this end, we assume that all our functors are induced by Fourier-Mukai kernels and we write a triangle of functors to mean the triangle of underlying kernels. At the level of kernels, the cone construction exists and makes sense, and so we can define a functor appearing in a triangle to be the Fourier-Mukai functor induced by the cone on the morphism between the kernels; see the section on conventions below for more details on this.
With this in mind, an exact functor F : A → B between triangulated categories with left adjoint L and right adjoint R is said to be spherical if the cotwist C, defined by the triangle is an autoequivalence of A which identifies the adjoints, that is, we have R ≃ CL[1].
Such functors are ubiquitous in algebraic geometry; particularly in the study of autoequivalences of the derived category since the twist T , defined by the triangle is an autoequivalence of B. The observation that spherical functors F : A → B give rise to interesting autoequivalences of the target category is well documented in the current literature; see [ST01,Rou06,Ann07,Add16,AL16,Kuz15,Seg16].
The aim of this brief note is to provide a short and simple proof of the fact that the twist associated to a spherical functor is an autoequivalence. A fundamental result of Johnstone [Joh02] allows us to plug the gap in Anno's original approach [Ann07] without leaving the framework of triangulated categories. Conventions: To ensure that all our cone constructions make sense, we work in the context of Cȃldȃraru and Willerton's 2-category V ar of Fourier-Mukai kernels [CW10] where the base category of spaces is fixed to be smooth complex projective varieties. That is, we implicitly assume that all our functors are Fourier-Mukai functors, all our maps between functors are induced by maps between their kernels and all our triangulated categories are given by the bounded derived category D(X) of coherent sheaves on some smooth complex projective variety X.
Explicitly, if F = FM P : D(X) → D(Y ) is a Fourier-Mukai functor for some kernel P ∈ D(X×Y ), where X and Y are both smooth and projective, then the right adjoint is given by R = FM Q where Q = P ∨ ⊗ π * X ω X [dim X] and π X : X × Y → X is the projection. Furthermore, the composition is given by RF = FM R where R = π 13 * (π * 12 P ⊗ π * 23 Q) and π ij is the projection from X × Y × X to the product of the i th and j th factor. Now, since [CW10, Section 3.4] shows that the unit η : id X → RF of adjunction is induced by a morphism ϕ : O ∆ X → R of kernels in D(X ×X), we can set S := cone(ϕ)[−1] and define the cotwist of F to be C := FM S . In particular, when we write the triangle C → id X → RF of functors, we are really referring to the triangle S → O ∆ X → R of kernels. Moreover, Cȃldȃraru and Willerton use string diagrams to give an elegant proof that the operation of taking a kernel to its adjoint can be extended to a 2-functor on V ar, which allows us to take adjoints of the triangle C → id X → RF in a precise and rigorous way.
In general, the existence of the twist and cotwist, defined via triangles as above, requires the technical machinery of dg-enhancements. However, the main point of this paper is that the dg-machinery is not required to say whether they are equivalences or not. Our assumptions and choice of conventions are made in order to simplify as much as possible away from existence issues, and put a sharper focus on the huge simplification of the proof of the equivalence.
One can of course replace all our smooth projective varieties and Fourier-Mukai kernels with suitable enhancements and the results will hold in much more generality but we choose to give a simplified approach so that the main ideas of the proof, which are really quite simple, are not obscured by gross technicalities.
Acknowledgements: I thank Michael Wemyss for numerous useful discussions which initiated this project and Alexander Kuznetsov for clarifying the details of [Kuz15]. I also thank Andreas Krug, Andreas Hochenegger and David Ploog for their helpful comments on a previous version.

Preliminaries
Definition 1.1. If F : A → B is a functor with left adjoint L and right adjoint R then we distinguish the units and counits with subscripts as follows: If an argument only deals with one adjoint pair then we shall drop the subscripts.
Definition 1.2. Let F : A → B be an exact functor between triangulated categories with left adjoint L and right adjoint R. With the running conventions, we define the twist T and cotwist C of F by the following exact triangles: Similarly, define the dual twist T ′ and dual cotwist C ′ of F by Proof. First take left adjoints of the triangle C → id A η R −→ RF defining the cotwist C to get an exact triangle LF → id A → C ′ and then observe that the adjunction isomorphism Hom(id A , RF ) ≃ Hom(LF, id A ) maps the unit η R to the counit ε L .
This shows that C ′ is left adjoint to C. Similarly, we can take left adjoints of → F L and use the other 1 adjunction isomorphism to see that Hom(F R, id B ) ≃ Hom(id B , F L) ; ε R → η L . Therefore, T ′ ⊣ T as well.
Lemma 1.4. Let F : A → B be a functor with right adjoint R. Then we have the following natural isomorphisms: Similarly, if F : A → B is a functor with left adjoint L then we have the following natural isomorphisms: Proof. These identities are standard, and all stem from the triangular identities associated to the adjoint pairs. For example, [Mac71, Theorem IV.1.1(ii)] tells us that the composition R splitting of RF R and allows us to complete the following diagram: R using the octahedral axiom to get a functorial isomorphism The following result is the key technical lemma which will allow us to easily deduce that the twist associated to a spherical functor is an autoequivalence. Proof. This statement is the dual of [Joh02, Lemma 1.1.1] and we include the proof for completeness.
Recall that a monad for a category C consists of an endofunctor T : C → C together with natural transformations σ : id C → T and µ : T 2 → T which render the following diagrams commutative: q q q q q q q q q q q q q q q q q q q q q q q q q q q q Every adjunction F ⊣ R defines a monad (RF, η, RεF ) in A, where η and ε are the unit and counit respectively. That is, the monad structure on RF is captured by the following commutative diagrams: Using the given isomorphism RF ≃ id A , we can transfer the monad structure on RF arising from the adjunction to a monad structure (σ, µ) on id A . In particular, we have a commutative diagram of the form: and we see that µ is a one-sided inverse for σ. But the set of natural transformations of the identity functor is a commutative monoid. Indeed, for natural transformations ζ, ξ : id C → id C , we have ζ x ξ x = ξ x ζ x by the naturality of either ζ or ξ. Therefore, a one-sided inverse must be a two-sided inverse. That is, we must have σ • µ ≃ id A as well and we can deduce that σ and µ are inverse isomorphisms.
Transferring back again, we see that the unit η : id A ∼ − → RF is an isomorphism, which is equivalent to F being fully faithful.

Spherical Functors
Definition 2.1. We say that a functor F : A → B with left adjoint L and right adjoint R is spherical if the cotwist C is an autoequivalence of A and the canonical is an isomorphism; compare this with [KS15, Remark 3.20] and the fairly mild conditions therein.
Addington [Add16, §1.5] also works with any isomorphism R ≃ CL[1], but his proof of the fact that the twist is an equivalence is different from ours.
Lemma 2.3. Let F be a spherical functor. Then Proof. (i) First, let us observe that we have a commutative diagram:

RF R
Indeed, the left hand side is just the triangular identity composed with R on both the left and the right, and the square on the right commutes since the arrows act on separate variables. In particular, we have and the canonical isomorphism f := γ R L • Rη L : R → RF L → CL[1] provides us with a commutative diagram: The left square commutes since the arrows act on separate variables and the commutativity of the right square follows from the identity (2.1) observed at the beginning.
In particular, we have an isomorphism is an isomorphism since C is an equivalence by assumption.
(ii) Using the octahedral axiom on the compositions γ R L•Rη L and gives the following commutative diagrams of triangles: Recall, either by [Nee01, Proposition 1.4.6] or [Hub08, Axiom D'], that the top left square in the left diagram gives rise to a triangle of the form: Now, the map in the middle is an isomorphism if and only if Q ≃ 0 which is equiv- Similarly, we can use the isomorphism Remark 2.4. Notice that Q ≃ 0 and Q ′ ≃ 0 is also equivalent to the canonical maps being isomorphisms, respectively.
We are now ready to prove the main result of this article: Proof. (i) As in [Ann07], we use the triangles F R → id B → T and T ′ → id B → F L to construct a commutative diagram: If we consider the top right square of the previous diagram together with the commutative square: consisting of four copies of F L and all maps being the identity, then we can use the natural map F η R L : F L → F RF L to form a commutative cube: Indeed, the left face is just the triangular identity composed with L on the right; the top and bottom faces are clearly commutative and the commutativity of the right face follows from where the first isomorphism uses the commutativity of the back face and the second isomorphism uses the triangular identity composed with L on the right again.
Applying the octahedral axiom to the top and right faces of this commutative cube produces the following commutative diagram of triangles:  (1) T is an autoequivalence of B; (2) C is an autoequivalence of A; ( instead (see Figure 3) then it is easy to see that condition (4) also implies the unit and so on.
Using these, we can take right adjoints of F RF in two different ways to get: Since adjoints are unique up to unique isomorphism, we can infer that RF L ≃ LF R.
We can do the same with F LF and put all this information together as follows:

Now we can take right and left adjoints of the isomorphisms
from Lemma 1.4, to get isomorphisms Notice that these isomorphisms split LF R and RF L into respectively, where the inclusion and projection maps are illustrated in the diagram below. Thus, it remains to show that the cross-maps are zero. More precisely, if we consider the following diagram of triangles: then we have to prove that the natural maps

CL[1]
are all zero. It is actually enough to show that (2.8) and (2.9) are zero since, together, these imply that the canonical map γ R L • Rη L : R → RF L → CL[1] in condition (4) of the theorem is an isomorphism and so by Lemma 2.3(i) we see that the canonical (3) is also an isomorphism.
The strategy is to show that the compositions in (2.8) and (2.9) are adjoint to the zero map and hence zero themselves.
To this end, we compose the diagram of triangles in Figure 2 In particular, it follows that F LF R and F RF L both split as F R ⊕ F L, and hence the isomorphism F LF R ≃ F RF L, which is unique up to unique isomorphism, must be equal to the following composition: (2.10) Furthermore, if we now compose our isomorphism in (2.5) with F , and denote the resulting composition by F (2.5), then we see that This, in turn, implies that the composition F (2.8) of the map in (2.8) with F can be written as which must be zero since it involves the composition of two consecutive maps in an exact triangle.
However, this composition is equal to ε R T ′ [1]•F (2.8) which is zero by our argument above. That is, the composition in (2.8) is adjoint to the zero map and hence must be zero itself.
Turning our attention to (2.9), we first observe that the composition Indeed, if we use the canonical isomorphisms F C ′ R[−1] ≃ F L and F LF R ≃ F RF L established in (2.3) and (2.10) respectively, then we can rewrite (2.11) as Since the composition in (2.10) is an isomorphism, the identity (2.12) implies that This allows us to rewrite the composition F (2.9) as which must be zero since it involves the composition of two consecutive maps in an exact triangle.
Now, just as we did for (2.8), we can observe that Hom(L, R) ≃ Hom(F L, id B ) by adjunction and this isomorphism sends the composition in (2.9) to However, this composition is equal to ε R • F (2.9) which is zero by our argument above. That is, the composition in (2.9) is adjoint to the zero map and hence must be zero itself. Using the splitting (2.6) of LF R and RF L whilst inspecting Figure 2, we see that the compositions (2.8) and (2.9) being zero implies that we have isomorphisms the first of which is condition (4) of the theorem. From here, one can employ a similar argument to the ones above to show that (2.7) is also the zero map and thus gives rise to isomorphisms the first of which is condition (3), or one can use Lemma 2.3(i) to combine the assumed condition (2) with the proven condition (4) to get condition (3).

Appendix A. Identifying adjoints by an autoequivalence
Let us suppose we are working with the same definitions that were introduced in Section 1. Then we can use the triangles C → id A → RF and LF → id A → C ′ to construct a commutative diagram: If we consider the top right square of the previous diagram together with the commutative square: consisting of four copies of RF and all maps being the identity, then we can use the natural map Rη L F : RF → RF LF to form a commutative cube: Indeed, the left face is just the triangular identity composed with R on the left; the top and bottom faces are clearly commutative and the commutativity of the right face follows from where the first isomorphism uses the commutativity of the back face and the second isomorphism uses the triangular identity composed with R on the left again.
Applying the octahedral axiom to the top and right faces of this commutative cube produces the following commutative diagram of triangles: Proof. We first claim that the following diagram is commutative: To see this, recall that Hom and hence taking left adjoints of the triangle C It follows that Hom(C, id) ≃ Hom(id, C ′ ) ; δ R → α L .
Explicitly, the adjunction is given by If C is an equivalence then it follows that C ′ is both left and right adjoint to C.
In particular, taking left adjoints of the diagram on the right of (A.3) gives That is, ε C • Cα L ≃ δ R which is equivalent to Cα L ≃ η C • δ R since C is an equivalence, i.e. η −1 C ≃ ε C . Therefore, the square (A.2) commutes and from this, together with (A.1), we can observe that we have a commutative diagram of triangles: Since the second and third vertical arrows are isomorphisms, it follows that the first vertical arrow is also an isomorphism. Proof. First, let us observe that we have a commutative diagram:

RF L
Indeed, the left hand side is just the triangular identity composed with R on the left and L on the right, and the square on the right commutes since the arrows act on separate variables. That is, (A.4) Next, recall that Hom(id A , RF ) ≃ Hom(F, F ) ; η R → id F . Indeed, we have We have another commutative diagram:

CLF L[1]
Cε Comparing this diagram with our initial observation in (A.4) implies that we must have g ≃ γ R L which completes the proof.