Proof. This can either be proved by a direct computation or by observing that $\text{disc}(\unicode[STIX]{x1D6E4}_{d})$ equals the discriminant of $\unicode[STIX]{x1D6E4}_{d}^{\bot }\subset \widetilde{\unicode[STIX]{x1D6EC}}$ , which is isomorphic to $v^{\bot }\subset A_{2}^{\bot }$ . Similarly, $d=\text{disc}(K_{d})$ equals the discriminant of the lattice $\langle v,h\rangle ^{\bot }$ , which again is just $v^{\bot }\subset A_{2}^{\bot }$ .◻

Proof. Consider a transposition $g:=(12)\in \mathfrak{S}_{3}\subset \text{O}(A_{2})$ . Then $g$ acts trivially on the discriminant $A_{A_{2}}$ (see Remark 2.1) and can, therefore, be extended to $\tilde{g}\in \text{O}(\widetilde{\unicode[STIX]{x1D6EC}})$ acting trivially on $A_{2}^{\bot }$ . Thus, the Hodge structure defined by $x$ admits a Hodge isometry $\tilde{g}$ , which preserves the orientation of the two positive directions given by the $(2,0)$ and $(0,2)$ parts. On the other hand, by construction, it reverses the orientation of the two positive directions in $A_{2}$ .◻

Proof. We prove the second assertion; the first one is even easier. Start with a twisted K3 surface $(S,\unicode[STIX]{x1D6FC})$ and pick a lift $B\in H^{2}(S,\mathbb{Q})$ of $\unicode[STIX]{x1D6FC}$ . Then the algebraic part $\widetilde{H}^{1,1}(S,\unicode[STIX]{x1D6FC},\mathbb{Z})=\exp (B)\cdot \widetilde{H}^{1,1}(S,\mathbb{Q})\cap \widetilde{H}(S,\mathbb{Z})$ contains the lattice $(\mathbb{Z}\cdot (1,B,B^{2}/2)\cap \widetilde{H}(S,\mathbb{Z}))\oplus H^{4}(S,\mathbb{Z})$ , which is isomorphic to $U(n)$ for $n$ minimal with $n(1,B,B^{2}/2)\in \widetilde{H}(S,\mathbb{Z})$ .

Conversely, assume $U(n)\subset \widetilde{\unicode[STIX]{x1D6EC}}$ is of type $(1,1)$ with respect to $x$ . Choosing $n$ minimal, we can assume that the standard isotropic generator $e_{n}=e$ is primitive in $\widetilde{\unicode[STIX]{x1D6EC}}$ . But then $e\in U(n)$ can be completed to a sublattice of $\widetilde{\unicode[STIX]{x1D6EC}}$ isomorphic to the hyperbolic plane $U=\langle e,f\rangle$ , which therefore induces an orthogonal decomposition (usually different from that defining $\widetilde{\unicode[STIX]{x1D6EC}}$ )

With respect to (2.6) the second basis vector $f_{n}\in U(n)$ can be written as $f_{n}=\unicode[STIX]{x1D6FE}+nf+ke$ with $\unicode[STIX]{x1D6FE}\in \unicode[STIX]{x1D6EC}$ . Similarly, a generator of the $(2,0)$ part of the Hodge structure determined by $x$ is orthogonal to $e$ and hence of the form $\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D706}e$ for some $\unicode[STIX]{x1D70E}\in \unicode[STIX]{x1D6EC}\otimes \mathbb{C}$ and $\unicode[STIX]{x1D706}\in \mathbb{C}$ . However, it is also orthogonal to $f_{n}$ and so $(\unicode[STIX]{x1D6FE}.\unicode[STIX]{x1D70E})+n\unicode[STIX]{x1D706}=0$ . Now set $B:=-(1/n)\unicode[STIX]{x1D6FE}$ . Then $\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D706}e=\unicode[STIX]{x1D70E}+B\wedge \unicode[STIX]{x1D70E}$ , where $B\wedge \unicode[STIX]{x1D70E}$ stands for $(B.\unicode[STIX]{x1D70E})e$ .

Eventually, the surjectivity of the period map implies that $\unicode[STIX]{x1D70E}\in \unicode[STIX]{x1D6EC}\otimes \mathbb{C}$ can be realized as the period of some K3 surface $S$ , i.e. there exists an isometry $H^{2}(S,\mathbb{Z})\simeq \unicode[STIX]{x1D6EC}$ identifying $H^{2,0}(S)$ with $\mathbb{C}\cdot \unicode[STIX]{x1D70E}\subset \unicode[STIX]{x1D6EC}\otimes \mathbb{C}$ . Here one uses $(\unicode[STIX]{x1D70E}.\unicode[STIX]{x1D70E})=(\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D706}e.\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D706}e)=0$ and $(\unicode[STIX]{x1D70E}.\bar{\unicode[STIX]{x1D70E}})=(\unicode[STIX]{x1D70E}+\unicode[STIX]{x1D706}e.\bar{\unicode[STIX]{x1D70E}}+\bar{\unicode[STIX]{x1D706}}e)>0$ . Mapping $H^{4}(S,\mathbb{Z})$ to $\mathbb{Z}\cdot e\subset U\subset \unicode[STIX]{x1D6EC}\oplus U$ in (2.6) and defining $\unicode[STIX]{x1D6FC}\in \text{Br}(S)$ as the Brauer class induced by $B$ under  yields a Hodge isometry between $\widetilde{H}(S,\unicode[STIX]{x1D6FC},\mathbb{Z})$ and the Hodge structure defined by $x$ on $\widetilde{\unicode[STIX]{x1D6EC}}$ .◻

Proof. One inclusion follows from the fact that any $U(n)$ contains an isotropic vector. For the converse, assume $x\in e^{\bot }$ for some primitive isotropic $0\neq e\in \widetilde{\unicode[STIX]{x1D6EC}}$ . Then $e\not \in A_{2}^{\bot }$ , as otherwise the positive plane corresponding to $x$ would be contained in the orthogonal complement of $e$ in $A_{2}^{\bot }$ , which has only one positive direction left. Hence, there exists $a\in A_{2}$ with $(a.e)\neq 0$ . Let $f:=(a.e)a-((a)^{2}/2)e$ , which satisfies $(f)^{2}=0$ and $(f.e)=(a.e)^{2}=:n$ . Hence, $e,f$ span a twisted hyperbolic plane $U(n)$ in the $(1,1)$ part of the Hodge structure defined by $x$ .◻

Proof. The first equality is due to Addington and Thomas [Reference Addington and ThomasAT14, Theorem 3.1]. Indeed, they show that $x\in D_{d}$ with $d$ satisfying ( $\ast \ast$ ) if and only if there exists a hyperbolic plane $U\subset \widetilde{\unicode[STIX]{x1D6EC}}$ which is of type $(1,1)$ with respect to $x$ . The latter is in turn equivalent to $x\in D_{\text{K3}}$ , see Lemma 2.6.Footnote ⁴

Maybe surprisingly, the second assertion is easier to prove. We include the elementary argument. Due to Corollary 2.8 we know $D_{\text{K3'}}=D\cap \bigcup e^{\bot }$ with $0\neq e\in \widetilde{\unicode[STIX]{x1D6EC}}$ isotropic. So for one inclusion one has to show that each $D\cap e^{\bot }$ is of the form $D_{d}$ with $d$ satisfying ( $\ast \ast ^{\prime }$ ). Decompose $e=e_{1}+e_{2}\in (A_{2}\oplus A_{2}^{\bot })\otimes \mathbb{Q}$ . Let then $v\in A_{2}^{\bot }$ such that $\mathbb{Q}\cdot e_{2}\cap A_{2}^{\bot }=\mathbb{Z}\cdot v$ and define $K_{d}\subset \text{I}_{2,21}$ as the saturation of the sublattice spanned by $v\in A_{2}^{\bot }\subset \text{I}_{2,21}$ and $h$ . We have to show that the discriminant $d$ of $K_{d}$ satisfies ( $\ast \ast ^{\prime }$ ).

Assume first that $A_{2}\oplus \mathbb{Z}\cdot v\subset \widetilde{\unicode[STIX]{x1D6EC}}$ is primitive. Then $d\equiv 0\,(6)$ and $d=-3(v)^{2}$ , see § 2.2. As $e\in A_{2}\oplus \mathbb{Z}\cdot v$ in this case, the quadratic equation $2(x_{1}^{2}+x_{2}^{2}-x_{1}x_{2})+(v)^{2}x^{2}=0$ admits an integral solution. However, it is a classical result that

for some $w\in A_{2}$ if and only if $n=\prod p_{i}^{n_{i}}$ with $n_{i}\equiv 0\,(2)$ for all $p_{i}\equiv 2\,(3)$ , see [Reference KneserKne02].Footnote ⁵ But clearly this holds for $n=-(v)^{2}/2$ if and only if $d=6n$ satisfies ( $\ast \ast ^{\prime }$ ).

Next assume that $A_{2}\oplus \mathbb{Z}\cdot v\subset \widetilde{\unicode[STIX]{x1D6EC}}$ has index three in its saturation. Hence, $d\equiv 2\,(6)$ and $3d=-(v)^{2}$ . Then argue as before, but now with the isotropic vector $3e\in A_{2}\oplus \mathbb{Z}\cdot v$ and with $n=-(v)^{2}/2=3d/2$ .

Running the argument backwards proves the reverse inclusion. ◻

Proof. Let $\ell :=\text{ord}(\unicode[STIX]{x1D6FC})$ . As proved in [Reference HuybrechtsHuy05], the transcendental lattice $T(S,\unicode[STIX]{x1D6FC})$ is isometric to the kernel of the natural map  defined by $\unicode[STIX]{x1D6FC}$ . Hence,

On the other hand, $\text{disc}(T(S,\unicode[STIX]{x1D6FC}))=\text{disc}(\widetilde{H}^{1,1}(S,\unicode[STIX]{x1D6FC},\mathbb{Z}))=d$ by Lemma 2.2.◻

Proof. Consider a fixed $K_{d}$ and the corresponding generator $v$ of $K_{d}\cap A_{2}^{\bot }$ . As $9|d$ , clearly $d\equiv 0\,(6)$ and so $A_{2}\oplus \mathbb{Z}\cdot v\subset \widetilde{\unicode[STIX]{x1D6EC}}$ is primitive. If there were a $(-2)$ -class $\unicode[STIX]{x1D6FF}\in \widetilde{\unicode[STIX]{x1D6EC}}$ with $D\cap K_{d}^{\bot }=D\cap v^{\bot }\subset D\cap \unicode[STIX]{x1D6FF}^{\bot }$ , then $\unicode[STIX]{x1D6FF}\in A_{2}+\mathbb{Z}\cdot v$ and so $\unicode[STIX]{x1D6FF}=w+kv$ for some $w\in A_{2}$ and $k\in \mathbb{Z}$ . But then $-2=(w)^{2}-k^{2}d/3$ . However, if $9|d$ , then $k^{2}d/3\equiv 0\,(3)$ and hence $(w)^{2}=2m$ with $m\equiv 2\,(3)$ , which contradicts (2.7).◻

Proof. One first observes that, analogously to Lemma 2.6(ii), a period $x\in Q$ is of the type (ii) if and only if the integral $(1,1)$ part of the Hodge structure associated with $x$ contains an isotropic direction. Indeed, if $x$ is of type (ii), i.e. of the form $\unicode[STIX]{x1D70E}+B\wedge \unicode[STIX]{x1D70E}$ , then $H^{4}$ provides an isotropic direction of type $(1,1)$ . For the converse use that any isotropic direction can be completed to a hyperbolic plane $U$ as a direct summand of $\widetilde{\unicode[STIX]{x1D6EC}}$ . Now regard $U$ as $H^{0}\oplus H^{4}$ with $H^{4}$ as the given isotropic direction, which is of type $(1,1)$ . Hence, $x$ is indeed of type (ii).

Now let $x\in D\cap Q$ be of type (ii). It is enough to show that then $x\in D_{\text{K3'}}$ . The integral $(1,1)$ -part of the Hodge structure associated with $x$ contains $A_{2}$ and an isotropic direction, say $\mathbb{Z}\cdot f$ . Then conclude by Proposition 2.8.◻

Proof. By results due to Orlov in the untwisted case and to Canonaco and Stellari [Reference Canonaco and StellariCS07] in the twisted case, any fully faithful functor  is of Fourier–Mukai type, i.e.  $\unicode[STIX]{x1D6F7}\simeq \unicode[STIX]{x1D6F7}_{{\mathcal{E}}}$ for some ${\mathcal{E}}\in \text{D}^{\text{b}}(S\times X,\unicode[STIX]{x1D6FC}^{-1}\boxtimes 1)$ . Therefore, $\unicode[STIX]{x1D6F7}$ induces a homomorphism  see [Reference Hille and van den BerghHvdB07, Remark 3.4].

If $\unicode[STIX]{x1D6F7}$ is induced by an equivalence  then  is an isomorphism and in fact a Hodge isometry  The compatibility with the Hodge structure follows from the twisted Chern character $\text{ch}^{-\unicode[STIX]{x1D6FC}\boxtimes 1}({\mathcal{E}})$ of the Mukai kernel being of Hodge type. See [Reference Huybrechts and StellariHS05, § 1] for the notion of twisted Chern characters. That the quadratic form is respected as well is proved by mimicking the argument for FM-equivalences, see e.g. [Reference HuybrechtsHuy06, § 5.2].

(We are suppressing a number of technical details here. As explained before, the actual realization of the Hodge structure $\widetilde{H}(S,\unicode[STIX]{x1D6FC},\mathbb{Z})$ depends on the choice of a $B\in H^{2}(S,\mathbb{Q})$ lifting $\unicode[STIX]{x1D6FC}$ . Similarly, the Chern character $\text{ch}^{-\unicode[STIX]{x1D6FC}\boxtimes 1}({\mathcal{E}})$ also actually depends on $B$ .)◻

Proof. The argument is an easy modification of the above. ◻

Proof. The proof follows the argument for the analogous statement for K3 surfaces [Reference Bridgeland and MaciociaBM01] closely, but needs a modification at one point that shall be explained.

Due to the proposition, it suffices to prove that up to isomorphism there are only finitely many cubics $X^{\prime }$ such that there exists a Hodge isometry  Any such Hodge isometry induces a Hodge isometry  and an isometry of lattices  We may assume

and similarly for $X^{\prime }$ . Note however that these inclusions need not be respected by $\unicode[STIX]{x1D711}$ . The orthogonal complement of $T({\mathcal{A}}_{X})^{\bot }\subset A_{2}^{\bot }$ is just $N({\mathcal{A}}_{X})\cap A_{2}^{\bot }$ and the two inclusions of $A_{2}^{\bot }$ induce two Hodge structures on $A_{2}^{\bot }$ . Note that if the Hodge isometry $\unicode[STIX]{x1D711}_{T}$ can be extended to a Hodge isometry  which can be interpreted as a Hodge isometry $H^{4}(X,\mathbb{Z})_{\text{prim}}\simeq H^{4}(X^{\prime },\mathbb{Z})_{\text{prim}}$ , then the global Torelli theorem [Reference VoisinVoi86, Reference VoisinVoi08] implies that $X\simeq X^{\prime }$ .

We first show that the set of isomorphism classes of lattices $\unicode[STIX]{x1D6E4}$ occurring as $N({\mathcal{A}}_{X^{\prime }})\cap A_{2}^{\bot }$ is finite. The required lattice theory is slightly more involved than the original in [Reference Bridgeland and MaciociaBM01]. Let us fix two even lattices $\unicode[STIX]{x1D6EC}_{1}$ and $\unicode[STIX]{x1D6EC}$ (in our situation, $\unicode[STIX]{x1D6EC}_{1}=T({\mathcal{A}}_{X})$ and $\unicode[STIX]{x1D6EC}=A_{2}^{\bot }$ ). We show that up to isomorphisms there exist only finitely many lattices $\unicode[STIX]{x1D6EC}_{2}$ occurring as the orthogonal complement of some primitive embedding  For unimodular $\unicode[STIX]{x1D6EC}$ this is standard, but the proof can be tweaked to cover the more general statement. Of course, it suffices to show that only finitely many discriminant forms $(A_{\unicode[STIX]{x1D6EC}_{2}},q_{\unicode[STIX]{x1D6EC}_{2}})$ can occur. Now $G:=\unicode[STIX]{x1D6EC}/(\unicode[STIX]{x1D6EC}_{1}\oplus \unicode[STIX]{x1D6EC}_{2})$ is naturally a finite subgroup of $\unicode[STIX]{x1D6EC}^{\ast }/(\unicode[STIX]{x1D6EC}_{1}\oplus \unicode[STIX]{x1D6EC}_{2})$ of index $d=|\text{disc}(\unicode[STIX]{x1D6EC})|$ . The first projection from $G\subset \unicode[STIX]{x1D6EC}^{\ast }/(\unicode[STIX]{x1D6EC}_{1}\oplus \unicode[STIX]{x1D6EC}_{2})\subset A_{\unicode[STIX]{x1D6EC}_{1}}\oplus A_{\unicode[STIX]{x1D6EC}_{2}}$ defines an isomorphism of $G$ with a finite subgroup of $A_{\unicode[STIX]{x1D6EC}_{1}}$ . This leaves only finitely many possibilities for the finite groups $G$ and $\unicode[STIX]{x1D6EC}^{\ast }/(\unicode[STIX]{x1D6EC}_{1}\oplus \unicode[STIX]{x1D6EC}_{2})$ . Note that $\unicode[STIX]{x1D6EC}/(\unicode[STIX]{x1D6EC}_{1}\oplus \unicode[STIX]{x1D6EC}_{2})\subset A_{\unicode[STIX]{x1D6EC}_{1}}\oplus A_{\unicode[STIX]{x1D6EC}_{2}}$ is isotropic but not necessarily the bigger $\unicode[STIX]{x1D6EC}^{\ast }/(\unicode[STIX]{x1D6EC}_{1}\oplus \unicode[STIX]{x1D6EC}_{2})\subset A_{\unicode[STIX]{x1D6EC}_{1}}\oplus A_{\unicode[STIX]{x1D6EC}_{2}}$ . However, the restriction of the quadratic form to $\unicode[STIX]{x1D6EC}^{\ast }/(\unicode[STIX]{x1D6EC}_{1}\oplus \unicode[STIX]{x1D6EC}_{2})$ takes values only in $(2/d^{2})\mathbb{Z}/2\mathbb{Z}$ . For fixed $G\subset A_{\unicode[STIX]{x1D6EC}_{1}}$ the restriction of $q_{\unicode[STIX]{x1D6EC}_{1}}$ to $G$ can be extended in at most finitely many ways to a quadratic form on $\unicode[STIX]{x1D6EC}^{\ast }/(\unicode[STIX]{x1D6EC}_{1}\oplus \unicode[STIX]{x1D6EC}_{2})$ with values in $(2/d^{2})\mathbb{Z}/2\mathbb{Z}$ . Now use the other projection  to see that there are only finitely many possibilities for the group $A_{\unicode[STIX]{x1D6EC}_{2}}$ and also for the quadratic form $q_{\unicode[STIX]{x1D6EC}_{2}}$ .

To conclude the proof, we can assume that $\unicode[STIX]{x1D6E4}$ is fixed. For two Fourier–Mukai partners realizing the fixed $\unicode[STIX]{x1D6E4}$ , any Hodge isometry $T({\mathcal{A}}_{X_{1}})\simeq T({\mathcal{A}}_{X_{2}})$ can be extended to a Hodge isometry $T({\mathcal{A}}_{X_{1}})\oplus \unicode[STIX]{x1D6E4}\simeq T({\mathcal{A}}_{X_{2}})\oplus \unicode[STIX]{x1D6E4}$ . As the finite index overlattices $T({\mathcal{A}}_{X_{i}})\oplus \unicode[STIX]{x1D6E4}\subset H^{4}(X_{i},\mathbb{Z})_{\text{prim}}$ are all contained in $(T({\mathcal{A}}_{X_{i}})\oplus \unicode[STIX]{x1D6E4})^{\ast }$ , there are only finitely many choices for them, which allows one to reduce to the case that the Hodge isometry extends to a Hodge isometry $H^{4}(X_{1},\mathbb{Z})_{\text{prim}}\simeq H^{4}(X_{2},\mathbb{Z})_{\text{prim}}$ .◻

Proof. The assumption implies that $N({\mathcal{A}}_{X})\simeq A_{2}$ . As any FM-equivalence ${\mathcal{A}}_{X}\simeq {\mathcal{A}}_{X^{\prime }}$ induces a Hodge isometry $\widetilde{H}({\mathcal{A}}_{X},\mathbb{Z})\simeq \widetilde{H}({\mathcal{A}}_{X^{\prime }},\mathbb{Z})$ , also $N({\mathcal{A}}_{X^{\prime }})\simeq A_{2}$ . Moreover, the natural inclusions of the transcendental lattices $T({\mathcal{A}}_{X})\subset A_{2}^{\bot }$ and $T({\mathcal{A}}_{X^{\prime }})\subset A_{2}^{\bot }$ are in fact equalities and the induced Hodge isometry $T({\mathcal{A}}_{X})\simeq T({\mathcal{A}}_{X^{\prime }})$ can therefore be read as a Hodge isometry $H^{4}(X,\mathbb{Z})_{\text{prim}}\simeq H^{4}(X^{\prime },\mathbb{Z})_{\text{prim}}$ , which by the global Torelli theorem [Reference VoisinVoi86] implies that $X\simeq X^{\prime }$ .◻

Proof. As the action on cohomology is independent of the specific cubic $X\subset \mathbb{P}^{5}$ , we can assume that the transcendental lattice $T({\mathcal{A}}_{X})\subset \widetilde{H}({\mathcal{A}}_{X},\mathbb{Z})$ is of odd rank. However, $\pm \text{id}$ are the only Hodge isometries of an irreducible Hodge structure of weight two of K3 type of odd rank, cf. [Reference HuybrechtsHuy16, Corollary 3.3.5], and, as $\unicode[STIX]{x1D6F7}_{0}^{3}\simeq [2]$ acts trivially on $\widetilde{H}({\mathcal{A}}_{X},\mathbb{Z})$ , we must have $\unicode[STIX]{x1D6F7}_{0}^{H}=\text{id}$ on $T({\mathcal{A}}_{X})$ , i.e.  $\unicode[STIX]{x1D6F7}_{0}$ is symplectic.

If $X$ is a cubic with $A_{2}\simeq \widetilde{H}^{1,1}({\mathcal{A}}_{X},\mathbb{Z})$ , then $\unicode[STIX]{x1D6F7}_{0}^{H}$ corresponds to an element in $\text{O}(A_{2})$ . As $\unicode[STIX]{x1D6F7}_{0}$ is symplectic, $\unicode[STIX]{x1D6F7}_{0}^{H}=\text{id}$ on $A_{2}^{\bot }$ and hence $\unicode[STIX]{x1D6F7}_{0}^{H}=\text{id}$ on the discriminant group $A_{A_{2}}$ . Therefore, $\unicode[STIX]{x1D6F7}_{0}^{H}\in \mathfrak{S}_{3}$ , see Remark 2.1. For a cubic $X$ such that ${\mathcal{A}}_{X}\simeq \text{D}^{\text{b}}(S)$ , we know that $\unicode[STIX]{x1D6F7}_{0}^{H}$ must be orientation preserving by [Reference Huybrechts, Macrì and StellariHMS09] and thus $\unicode[STIX]{x1D6F7}_{0}^{H}\in \mathfrak{A}_{3}\simeq \mathbb{Z}/3\mathbb{Z}$ in general.

It remains to show that $\unicode[STIX]{x1D6F7}_{0}^{H}\neq \text{id}$ . One way to see this relies on a direct computation. Another possibility is to use the recent result of Bayer and Bridgeland [Reference Bayer and BridgelandBB13] confirming Bridgeland’s conjecture in [Reference BridgelandBri08] in the case of a K3 surface $S$ of Picard rank one. More precisely, due to [Reference Bayer and BridgelandBB13, Theorem 1.4] for a K3 surface $S$ with $\unicode[STIX]{x1D70C}(S)=1$ the subgroup of $\operatorname{Aut}(\text{D}^{\text{b}}(S))$ of autoequivalences acting trivially on $\widetilde{H}(S,\mathbb{Z})$ is the product of $\mathbb{Z}[2]$ and the free group generated by squares of spherical twists $T_{E}^{2}$ associated with spherical vector bundles $E$ on $S$ . (That this is a reformulation of Bridgeland’s original conjecture for $\unicode[STIX]{x1D70C}(S)=1$ had also been observed by Kawatani [Reference KawataniKaw12].) Hence, if $\unicode[STIX]{x1D6F7}_{0}^{H}=\text{id}$ , then  but then clearly $\unicode[STIX]{x1D6F7}_{0}^{3}$ could not be isomorphic to the double shift $[2]$ .◻

Proof. The argument is a variant of the deformation theory in [Reference Addington and ThomasAT14]. We only indicate the necessary modifications.

As, by assumption, $\unicode[STIX]{x1D6F7}$ is a FM-equivalence, the composition

is a FM-functor with some kernel $P\in \text{D}^{\text{b}}(X^{\prime }\times X)$ contained in ${\mathcal{A}}_{X^{\prime }}(-2)\boxtimes {\mathcal{A}}_{X}$ . It suffices to show that $P$ deforms to $P_{t}\in \text{D}^{\text{b}}({\mathcal{X}}_{t}^{\prime }\times {\mathcal{X}}_{t})$ for $t$ in some open neighbourhood $0\in U\subset T$ , because the conditions for $\unicode[STIX]{x1D6F7}_{P_{t}}$ to factorize via a functor  and for this functor $\unicode[STIX]{x1D6F7}_{t}$ to define an equivalence are both Zariski open. Indeed, $\unicode[STIX]{x1D6F7}_{t}$ takes values in ${\mathcal{A}}_{{\mathcal{X}}_{t}}$ if and only if its composition with the projection  is trivial. The composition, however, is again of FM-type and the vanishing of a FM-kernel is a Zariski open condition. Similarly, whether $\unicode[STIX]{x1D6F7}_{t}$ is an equivalence can be detected by composing it with its adjoints and then checking whether the natural map to the kernel of the identity is an isomorphism, again a Zariski open condition.

The crucial part is to understand the first-order deformations, the higher-order obstructions are dealt with by the $T^{1}$ -lifting property, see [Reference Addington and ThomasAT14, § 7.2] and [Reference Huybrechts, Macrì and StellariHMS09, § 3.2]. First note that by results of Kuznetsov [Reference KuznetsovKuz09] one has

This allows one to compare the first-order deformations

corresponding to some tangent vector $v\in T_{0}T$ of $T$ at $0$ . Due to a result of Toda [Reference TodaTod09] (cf. [Reference Addington and ThomasAT14, Theorem 7.1]) it suffices to show that under  respectively  the classes $\unicode[STIX]{x1D705}_{X^{\prime }}$ and $\unicode[STIX]{x1D705}_{X}$ are mapped to the same class. For this consider the following diagram (cf. [Reference Addington and ThomasAT14, Proposition 6.2]).

By $H^{\ast }(X)\simeq \mathit{HH}_{\ast }(X)$ we denote the HKR isomorphism (see [Reference CăldăraruCăl05]) post-composed with $\sqrt{\text{td}}\wedge (~)$ and so, in particular, $\mathit{HH}_{2}(X)\simeq H^{1}(\unicode[STIX]{x1D6FA}_{X}^{3})$ with chosen generator $\unicode[STIX]{x1D70E}_{X}$ . Similarly for $X^{\prime }$ , where we choose the generator $\unicode[STIX]{x1D70E}_{X^{\prime }}\in H^{1}(\unicode[STIX]{x1D6FA}_{X^{\prime }}^{3})\simeq \mathit{HH}_{2}(X^{\prime })$ such that its image yields $\unicode[STIX]{x1D70E}_{X}$ . Furthermore, $\bar{\unicode[STIX]{x1D705}}_{X}$ denotes the image of $\unicode[STIX]{x1D705}_{X}$ under the projection  see [Reference KuznetsovKuz09], and $\unicode[STIX]{x1D6FC}:=\unicode[STIX]{x1D6F7}^{\mathit{HH}^{\ast }}(\bar{\unicode[STIX]{x1D705}}_{X})$ .

We aim at showing that (1) is commutative. For this note first that (4) is induced by the FM-transform  and hence commutative due to [Reference Macrì and StellariMS09]. The commutativity of (2) is obvious, as Hochschild (co)homology is respected by equivalences, and commutativity of (3) is the analogue of [Reference Addington and ThomasAT14, Proposition 6.1]. (Recall that $\unicode[STIX]{x1D6F7}_{P}$ does not necessarily induce a map $\unicode[STIX]{x1D6F7}_{P}^{\mathit{HH}^{\ast }}$ , as it is not fully faithful.)

The first-order version of the assumption on the Hodge isometry $\unicode[STIX]{x1D711}$ is the statement that the diagram

is commutative. Using the ring-module isomorphism $(\mathit{HH}^{\ast },\mathit{HH}_{\ast })\simeq (H^{\ast }(\bigwedge ^{\ast }T),H^{\ast }(\unicode[STIX]{x1D6FA}^{\ast }))$ for $X^{\prime }$ , this implies that the image in $H^{\ast }(X^{\prime })$ of $\unicode[STIX]{x1D70E}_{X^{\prime }}\in \mathit{HH}_{2}(X^{\prime })$ under contraction with $\unicode[STIX]{x1D705}_{X^{\prime }}$ is mapped to the image of $\unicode[STIX]{x1D70E}_{X}$ under contraction with $\unicode[STIX]{x1D705}_{X}$ . As $\mathit{HH}_{2}(X^{\prime })$ is one-dimensional, this shows that also (1) is commutative.

Therefore, in the diagram

the image of $\unicode[STIX]{x1D705}_{X^{\prime }}\in \mathit{HH}^{2}(X^{\prime })$ under the two compositions  coincide. As the contraction  is injective (as for K3 surfaces), this implies that the image of $\unicode[STIX]{x1D705}_{X^{\prime }}$ under  is indeed $\bar{\unicode[STIX]{x1D705}}_{X}$ as claimed.

As in [Reference Addington and ThomasAT14], the deformation of $P$ to first and then, by $T^{1}$ -lifting property, to higher order is unique, for $\operatorname{Ext}_{X^{\prime }\times X}^{1}(P,P)\simeq \mathit{HH}^{1}({\mathcal{A}}_{X})=0$ by [Reference KuznetsovKuz09].◻

Proof. Let us fix representatives $\unicode[STIX]{x1D6FC}_{t}=\{\unicode[STIX]{x1D6FC}_{t,ijk}\}$ for the Brauer classes on ${\mathcal{S}}_{t}$ and a family $E_{t}$ of locally free $\{\unicode[STIX]{x1D6FC}_{t,ijk}\}$ -twisted sheaves on the fibres ${\mathcal{S}}_{t}$ in a Zariski open neighbourhood of $0\in U\subset T$ .

The proof now consists of copying [Reference Addington and ThomasAT14, §§ 6, 7]. However, the techniques have to be adapted to the twisted case, which sometimes causes additional problems as certain fundamental issues related to Hochschild (co)homology have not been addressed in the twisted setting. For certain parts we choose ad hoc arguments to reduce to the untwisted case, for others we rely on Reinecke [Reference ReineckeRei14].

Section 6 in [Reference Addington and ThomasAT14] deals with Hochschild (co)homology. For a twisted variety $(Z,\unicode[STIX]{x1D6FC})$ one defines $\mathit{HH}^{n}(Z,\unicode[STIX]{x1D6FC}):=\operatorname{Ext}_{(Z,\unicode[STIX]{x1D6FC}^{-1})\times (Z,\unicode[STIX]{x1D6FC})}^{n}({\mathcal{O}}_{\unicode[STIX]{x1D6E5}},{\mathcal{O}}_{\unicode[STIX]{x1D6E5}})$ . Here, $(Z,\unicode[STIX]{x1D6FC}^{-1})\times (Z,\unicode[STIX]{x1D6FC})$ denotes the twisted variety $(Z\times Z,\unicode[STIX]{x1D6FC}^{-1}\boxtimes \unicode[STIX]{x1D6FC})$ . Note that ${\mathcal{O}}_{\unicode[STIX]{x1D6E5}}$ is indeed an $(\unicode[STIX]{x1D6FC}^{-1}\boxtimes \unicode[STIX]{x1D6FC})$ -twisted sheaf. Similarly, one defines $\mathit{HH}_{n}(Z,\unicode[STIX]{x1D6FC}):=\operatorname{Ext}_{(Z,\unicode[STIX]{x1D6FC}^{-1})\times (Z,\unicode[STIX]{x1D6FC})}^{d-n}(\unicode[STIX]{x1D6E5}_{\ast }\unicode[STIX]{x1D714}_{Z}^{-1},{\mathcal{O}}_{\unicode[STIX]{x1D6E5}})$ , where $d=\dim (Z)$ . Composition makes $\mathit{HH}_{\ast }(Z,\unicode[STIX]{x1D6FC})$ a right $\mathit{HH}^{\ast }(Z,\unicode[STIX]{x1D6FC})$ -module. Moreover, there are natural isomorphisms

and

In particular, the HKR isomorphisms post-composed with $\text{td}^{-1/2}\lrcorner (~)$ respectively  $\text{td}^{1/2}\wedge (~)$ yield isomorphisms

Note that these isomorphisms are again compatible with the ring and module structures on both sides, which follows from the fact that the isomorphisms $\mathit{HH}^{\ast }(Z,\unicode[STIX]{x1D6FC})\simeq \mathit{HH}^{\ast }(Z)$ and $\mathit{HH}_{\ast }(Z,\unicode[STIX]{x1D6FC})\simeq \mathit{HH}_{\ast }(Z)$ are. The latter is a consequence of the functoriality properties of $\unicode[STIX]{x1D6E5}_{!}$ , $\unicode[STIX]{x1D6E5}_{\ast }$ and  $\unicode[STIX]{x1D6E5}^{\ast }$ .

For a twisted K3 surface $(S,\unicode[STIX]{x1D6FC})$ one has $\mathit{HH}_{2}(S,\unicode[STIX]{x1D6FC})\simeq H^{0}(\unicode[STIX]{x1D714}_{S})=\mathbb{C}\cdot \unicode[STIX]{x1D70E}_{S}$ and the following diagram commutes.

Let us now consider the fully faithful functor  between the twisted K3 surface $(S,\unicode[STIX]{x1D6FC})$ and the smooth cubic $X$ , where $P\in \text{D}^{\text{b}}((S,\unicode[STIX]{x1D6FC}^{-1})\times X)$ . Then as in [Reference Addington and ThomasAT14, § 6.1] one obtains natural maps

compatible with the module structures, i.e. $\unicode[STIX]{x1D6F7}_{P}^{\mathit{HH}_{\ast }}(a)\circ c=\unicode[STIX]{x1D6F7}_{P}^{\mathit{HH}_{\ast }}(a\circ \unicode[STIX]{x1D6F7}_{P}^{\mathit{HH}^{\ast }}(c))$ for all $a\in \mathit{HH}_{\ast }(S,\unicode[STIX]{x1D6FC})$ and $c\in \mathit{HH}^{\ast }(X)$ . This has been checked by Reinecke in [Reference ReineckeRei14, § 4].

The remaining input in the proof of [Reference Addington and ThomasAT14, Proposition 6.2] is the commutativity of the untwisted version of the following diagram.

Note that defining the induced action on cohomology requires the lift of $\unicode[STIX]{x1D6FC}$ to a class $B\in H^{2}(S,\mathbb{Q})$ , see [Reference HuybrechtsHuy05, Reference Huybrechts and StellariHS05]. Moreover, the usual HKR isomorphism $I$ post-composed with $\text{td}^{1/2}\wedge (~)$ needs to be twisted further to $I^{B}:=\exp (B)\circ I$ .

In principle, one could try to prove the commutativity of (5.1) by rewriting the existing untwisted theory, in particular [Reference CăldăraruCăl05, Reference Macrì and StellariMS09], for the twisted situation. Instead, we follow Yoshioka [Reference YoshiokaYos06] and reduce everything to the untwisted case by pulling back to a Brauer–Severi variety. We briefly review his approach and explain how to apply it to our situation.

Following [Reference YoshiokaYos06] we pick a locally free $\unicode[STIX]{x1D6FC}=\{\unicode[STIX]{x1D6FC}_{ijk}\}$ -twisted sheaf $E=\{E_{i},\unicode[STIX]{x1D711}_{ij}\}$ on a twisted variety $(Z,\unicode[STIX]{x1D6FC})$ and associate to it the projective bundle  which naturally comes with a $\unicode[STIX]{x1D70B}^{\ast }\unicode[STIX]{x1D6FC}^{-1}$ -twisted line bundle $L:={\mathcal{O}}_{\unicode[STIX]{x1D70B}}(1)$ . The pull-back of any $\unicode[STIX]{x1D6FC}$ -twisted sheaf $F=\{F_{i},\unicode[STIX]{x1D713}_{ij}\}$ tensored with $L$ then naturally leads to the untwisted sheaf $\tilde{F}:=\unicode[STIX]{x1D70B}^{\ast }F\otimes L$ . Analogously, any $\unicode[STIX]{x1D6FC}^{-1}$ -twisted sheaf $F$ can be turned into the untwisted sheaf $\unicode[STIX]{x1D70B}^{\ast }F\otimes L^{\ast }$ . The construction yields a functor  which, in fact, defines an equivalence of $\text{D}^{\text{b}}(Z,\unicode[STIX]{x1D6FC})$ with a full subcategory

The construction applied to $E$ itself yields the sheaf $G:=\tilde{E}$ that corresponds to the unique non-trivial extension class in $\operatorname{Ext}_{Y}^{1}({\mathcal{T}}_{\unicode[STIX]{x1D70B}},{\mathcal{O}}_{Y})$ and $\text{D}^{\text{b}}(Y/Z)\subset \text{D}^{\text{b}}(Y)$ can alternatively be described as the full subcategory of all objects $H$ for which the adjunction map  is an isomorphism. Analogously, $\text{D}^{\text{b}}(Z,\unicode[STIX]{x1D6FC}^{-1})$ is equivalent to the full subcategory of objects $H$ for which 

We apply this construction to the twisted K3 surface $(S,\unicode[STIX]{x1D6FC})$ and consider  as above. Assume $\unicode[STIX]{x1D6FC}$ is of order $r$ and choose a lift $B=(1/r)B_{0}$ with $B_{0}\in H^{2}(S,\mathbb{Z})$ of it. The FM-kernel of our given equivalence  which is an object in $\text{D}^{\text{b}}((S,\unicode[STIX]{x1D6FC}^{-1})\times X)$ , is turned into the untwisted sheaf $\tilde{P}:=\unicode[STIX]{x1D70B}^{\ast }P\otimes (L^{\ast }\boxtimes {\mathcal{O}})$ on $Y\times X$ . This leads to the following commutative diagram.

Therefore, the FM-functor  can be written as the composition $\unicode[STIX]{x1D6F7}_{P}=\unicode[STIX]{x1D6F7}_{\tilde{P}}\circ \unicode[STIX]{x1D6F7}_{Q}$ of a twisted FM-functor $\unicode[STIX]{x1D6F7}_{Q}:=\tilde{(~)}$ , with $Q=({\mathcal{O}}_{S}\boxtimes L)|_{\unicode[STIX]{x1D6E4}_{\unicode[STIX]{x1D70B}}}$ , and an untwisted FM-functor $\unicode[STIX]{x1D6F7}_{\tilde{P}}$ :

This allows one to decompose the diagram (5.1) as

The right-hand square is induced by the usual untwisted FM-functor $\unicode[STIX]{x1D6F7}_{\tilde{P}}$ and its commutativity therefore follows from the result of Macrì and Stellari [Reference Macrì and StellariMS09, Theorem 1.2]. Hence, it suffices to prove the commutativity of the left-hand square (which does not involve the cubic $X$ anymore). For greater clarity we split this further by decomposing $\unicode[STIX]{x1D6F7}_{Q}$ as

Let us first consider  and the induced diagram

Note that the usual $\sqrt{\text{td}_{\unicode[STIX]{x1D70B}}}\cdot \unicode[STIX]{x1D70B}^{\ast }$ on the bottom is indeed the map on cohomology induced by the functor  which a priori depends on the choice of the lifts of $\unicode[STIX]{x1D6FC}$ and $\unicode[STIX]{x1D70B}^{\ast }\unicode[STIX]{x1D6FC}$ to classes in $H^{2}(S,\mathbb{Q})$ and $H^{2}(Y,\mathbb{Q})$ , respectively, for which we choose $B$ and $\unicode[STIX]{x1D70B}^{\ast }B$ . The commutativity of the upper and the lower squares is trivial. The commutativity of the middle square is an easy case of [Reference Macrì and StellariMS09, Theorem 1.2]. Next consider  and the following induced diagram (where $\unicode[STIX]{x1D713}$ is defined by the requirement of commutativity).

By definition, $\unicode[STIX]{x1D6F9}^{H,\unicode[STIX]{x1D70B}^{\ast }B}$ is given by multiplication with $\text{ch}^{\unicode[STIX]{x1D70B}^{\ast }(-B)}(L)=\exp (-\unicode[STIX]{x1D70B}^{\ast }B)\,\cdot \,\exp (\text{c}_{1}(L))$ . Here, use that $L^{r}$ is an untwisted line bundle and define $\text{c}_{1}(L):=(1/r)\text{c}_{1}(L^{r})\in H^{1,1}(Y,\mathbb{Q})$ . See [Reference Huybrechts and StellariHS05, § 1] for the conventions concerning twisted Chern classes. In particular, $\unicode[STIX]{x1D6F9}^{H,\unicode[STIX]{x1D70B}^{\ast }B}\circ \exp (\unicode[STIX]{x1D70B}^{\ast }B)=\exp (\text{c}_{1}(L))$ and, therefore, it suffices to prove the commutativity of the diagram

which no longer depends on $B$ and is a special case of Lemma 5.3 below.

This concludes the proof of the commutativity of the diagram (5.1) and hence of [Reference Addington and ThomasAT14, Proposition 6.2] in our twisted setting. More precisely, if a first-order deformation of $X$ in $D_{d}$ given by a class $\unicode[STIX]{x1D705}_{X}\in H^{1}(T_{X})$ corresponds via the interpretation of $D_{d}$ as period domain for $X$ and $S$ to a first-order deformation $\unicode[STIX]{x1D705}_{S}\in H^{1}(T_{S})$ , then  sends $\unicode[STIX]{x1D705}_{X}$ to $\unicode[STIX]{x1D705}_{S}$ .

To conclude the proof one has to prove that the kernel $P\in \text{D}^{\text{b}}((S,\unicode[STIX]{x1D6FC}^{-1})\times X)$ deforms sideways, for which we again apply Yoshioka’s untwisting technique. Instead of attempting to deform the twisted $P$ sideways with $(S,\unicode[STIX]{x1D6FC})\times X$ we deform the untwisted $\tilde{P}$ . As the condition describing the full subcategory $\text{D}^{\text{b}}((S,\unicode[STIX]{x1D6FC}^{-1})\times X)\simeq \text{D}^{\text{b}}((Y\times X)/(S\times X))\subset \text{D}^{\text{b}}(Y\times X)$ is open, any deformation of $\tilde{P}$ will automatically induce a deformation of $P$ .Footnote ⁷ The decomposition (5.2) leads to a diagram

Recall that $\unicode[STIX]{x1D6F7}_{R}^{\mathit{HH}_{\ast }}$ is defined for any FM-functor $\unicode[STIX]{x1D6F7}_{R}$ , whereas in order to define $\unicode[STIX]{x1D6F7}_{R}^{\mathit{HH}^{\ast }}$ one needs $\unicode[STIX]{x1D6F7}_{R}$ to be fully faithful, which is the case for $\unicode[STIX]{x1D6F7}_{P}$ and $\unicode[STIX]{x1D6F7}_{Q}=\tilde{(~)}$ . So, both maps in

are well defined, where as above $\unicode[STIX]{x1D705}_{X}\in H^{1}(T_{X})\subset \mathit{HH}^{2}(X)$ corresponds to $\unicode[STIX]{x1D705}_{S}\in H^{1}(T_{S})\subset \mathit{HH}^{2}(S,\unicode[STIX]{x1D6FC})$ (via their periods or, equivalently, via $\unicode[STIX]{x1D6F7}^{\mathit{HH}^{2}}$ ) and $\unicode[STIX]{x1D705}_{Y}$ is determined by our pre-chosen deformation $E_{t}$ of $E$ .

Now by [Reference Huybrechts and ThomasHT10] the obstruction $o(\tilde{P})$ can be expressed as

(Unfortunately, an analogous formula in the twisted case is not available.) The crucial [Reference Addington and ThomasAT14, Theorem 7.1], which goes back to Toda [Reference TodaTod09], proves that in the untwisted case $o(P)=0$ if $\unicode[STIX]{x1D705}_{X}$ is mapped to $\unicode[STIX]{x1D705}_{S}$ under  However, in the twisted situation one has to face the additional problem that there is no natural map  Nevertheless, the argument in [Reference Addington and ThomasAT14] goes through essentially unchanged as follows. Using the same notation, one writes

The first term is the image of $\unicode[STIX]{x1D70B}_{1}^{\ast }\unicode[STIX]{x1D705}_{Y}\circ \text{At}_{1}({\mathcal{O}}_{\unicode[STIX]{x1D6E5}_{Y}})\in \operatorname{Ext}^{2}({\mathcal{O}}_{\unicode[STIX]{x1D6E5}_{Y}},{\mathcal{O}}_{\unicode[STIX]{x1D6E5}_{Y}})=\mathit{HH}^{2}(Y)$ which is just $\unicode[STIX]{x1D705}_{Y}$ , whereas the second one is the image of $-\unicode[STIX]{x1D70B}_{1}^{\ast }\unicode[STIX]{x1D705}_{X}\circ \text{At}_{2}({\mathcal{O}}_{\unicode[STIX]{x1D6E5}_{X}})\in \operatorname{Ext}^{2}({\mathcal{O}}_{\unicode[STIX]{x1D6E5}_{X}},{\mathcal{O}}_{\unicode[STIX]{x1D6E5}_{X}})=\mathit{HH}^{2}(X)$ which is just $-\unicode[STIX]{x1D705}_{X}$ . Hence, to compare $\unicode[STIX]{x1D705}_{X}$ and $\unicode[STIX]{x1D705}_{Y}$ we do not need a map  (which simply does not exist naturally), as we only need to compare their images in $\operatorname{Ext}^{2}(\tilde{P},\tilde{P})\simeq \mathit{HH}^{2}(S,\unicode[STIX]{x1D6FC})$ . Therefore, it suffices to ensure that under  the class $\unicode[STIX]{x1D705}_{X}$ is mapped to $\unicode[STIX]{x1D705}_{S}$ , which was verified above.

This concludes the argument proving that the FM-kernel $P$ deforms to first order with $(S,\unicode[STIX]{x1D6FC})\times X$ . The arguments in [Reference Addington and ThomasAT14, § 7.2] proving the existence of deformations of $P$ to all orders apply verbatim. Note that at the very end of the argument one needs to apply a result of Lieblich [Reference LieblichLie06] saying that the space of objects with no negative self-Exts in the derived category is an Artin stack of locally finite presentation. Again, the result as such does not seem to be in the literature for the twisted situation, but once again it can be deduced from the untwisted case by Yoshioka’s trick. ◻