Proof. By definition

\[ \Psi(\gamma^2) = e_{\gamma} \cdot e_{\gamma} (\alpha^n/{n!}) = \frac{\alpha^{(n-2)} \cdot \gamma^{(2)}}{(n-2)!} +q(\gamma,\gamma)\frac{\alpha^{(n-1)} \cdot \beta}{(n-1)!}. \]

The map $\Psi$ is a section of $T$, so $T(\Psi (\gamma ^2)) = \gamma ^2$. Substituting we get

\begin{align*} T(\alpha^{(n-2)} \cdot \gamma^{(2)}) &= (n-2)!\bigg(T(\Psi(\gamma^2)) - q(\gamma,\gamma)\frac{T(\alpha^{(n-1)} \cdot \beta)}{(n-1)!} \bigg)\\ &= (n-2)!(\gamma^2 - q(\gamma,\gamma){\mathsf q}_2), \end{align*}

where we used Lemma 3.7 in the last equality.

Proof. By linearity, we can assume that $q(\lambda,\lambda ) \neq 0$. By definition, we have

\[ \Psi(\lambda^3) = e_{\lambda} \cdot e_{\lambda} \cdot e_{\lambda} \bigg(\frac{\alpha^{(2)}}{2}\bigg) = 3q(\lambda,\lambda)\lambda\beta. \]

Using that $\Psi$ is a section of $T$ we obtain $T(\lambda \beta ) = {\lambda ^3}/{3q(\lambda,\lambda )}$, which is easily seen to satisfy the thesis.

Proof. The symmetric square of $\tilde {v}(E)$ is given by

(6)\begin{equation} \tilde{v}(E)^{(2)} = r^2\alpha^{(2)} + 2r\alpha\lambda + 2rs\alpha\beta + \lambda^{(2)} +2s\lambda\beta + s^2\beta^{(2)} \in \mathrm{Sym}^2 \widetilde{H}(X,{\mathbb{Q}}). \end{equation}

Applying the results above, we obtain

\[ T(\tilde{v}(E)^{(2)}) = 2r^2 + 2r\lambda + 2rs{\mathsf q}_2 + (\lambda^2 - q(\lambda,\lambda){\mathsf q}_2) + 2s\lambda^{\vee} + s^2{\mathsf q}_4. \]

Dividing by $2r$ and rearranging the terms we obtain the formula for the Mukai vector in the statement.

Proof. Taking the $n$th symmetric power of $\tilde {v}(F)$ we get

\begin{align*} \tilde{v}(F)^{(n)} &= r(F)^{n}\alpha^{(n)} + nr(F)^{n-1}\alpha^{(n-1)}c_1(F) + \binom{n}{2}r(F)^{n-2}\alpha^{(n-2)}c_1(F)^{(2)} \\ &\quad + nr(F)^{n-1}s(F)\alpha^{(n-1)}\beta + \cdots. \end{align*}

Using Lemmas 3.7 and 3.8 we get

\begin{align*} T(\tilde{v}(F)^{(n)}) &= n!r(F)^n + n!r(F)^{n-1}c_1(F) + \binom{n}{2}r(F)^{n-2}(n-2)!(c_1(F)^2\\ &\quad - q(c_1(F),c_1(F)){\mathsf q}_2) + n!r(F)^{n-1}s(F){\mathsf q}_2 + \cdots. \end{align*}

The projection onto $\mathrm {SH}(X)$ of the Mukai vector of $F$ is a rational multiple of this class. Since the rank is non-zero, we deduce that $n!r(F)^{n-1}v(F) = T(\tilde {v}(F)^{(n)})$. Dividing by $n!r(F)^{n-1}$ and comparing the terms of degree four, we get

(8)\begin{equation} v_2(F)_{\mathrm{SH}} = \frac{1}{2r(F)}(c_1(F)^2 -q(c_1(F),c_1(F)){\mathsf q}_2) + s(F){\mathsf q}_2. \end{equation}

On the other hand, by definition $v(F) = \operatorname {ch}(F) \cup (\operatorname {td}_X)^{ {1}/{2}}$. By (7) we get

\begin{align*} \operatorname{ch}_2(F)_{\mathrm{SH}} & = v_2(F)_{\mathrm{SH}} - r_Xr(F){\mathsf q}_2 \\ &= \frac{1}{2r(F)}(c_1(F)^2 - q(c_1(F),c_1(F)){\mathsf q}_2) +( s(F) - r_Xr(F) ){\mathsf q}_2. \end{align*}

Substituting $\operatorname {ch}_2(F)_{\mathrm {SH}}$ in the definition of the discriminant we obtain

\begin{align*} \Delta(F)_{\mathrm{SH}} &= (q(c_1(F),c_1(F))+2r_Xr(F)^2-2r(F)s(F)){\mathsf q}_2 \\ & = (\tilde{q}(\tilde{v}(F),\tilde{v}(F)) + 2r_Xr(F)^2){\mathsf q}_2. \end{align*}

Proof. If $F$ is slope semistable for a polarization $H$ on $X$, Bogomolov's inequality gives

\[ \int_X{\Delta(F) \cup H^{n-2}} \geq 0. \]

The thesis follows from the proposition above because $\int _X{H^{n-2}{\mathsf q}_2} = (2n-3)!!q(H)^{n-1} \geq 0$.

Proof. Consider the action of the algebraic group $\mathrm {SO}(\widetilde {H}(X,{\mathbb {C}}))$ on $\mathrm {SH}(X,{\mathbb {C}})$ and $\widetilde {H}(X,{\mathbb {C}})$, obtained integrating the action of $\mathfrak g'_0 \cong \mathfrak {so}(\widetilde {H}(X,{\mathbb {C}}))$ as explained in [Reference TaelmanTae23, Section 5]. It acts by isometries with respect to the bilinear forms $b_{\mathrm {SH}}$ and $\tilde {q}$. Moreover, the projection $T : \mathrm {Sym}^n \widetilde {H}(X,{\mathbb {C}}) \to \mathrm {SH}(X,{\mathbb {C}})$ is $\mathrm {SO}(\widetilde {H}(X,{\mathbb {C}}))$-equivariant, being the projection onto a subrepresentation. Hence, both sides of the equality we want to show are invariant under the action of $\mathrm {SO}(\widetilde {H}(X,{\mathbb {C}}))$.

Write $\tilde {v} = r\alpha + \lambda + s\beta$. Dividing by the rank (we can because both sides are homogeneous of degree $n$) and acting by $\exp (e_{\lambda /r})$ we can assume that $\tilde {v} = \alpha + s\beta$. By definition, we have $\tilde {q}(\tilde {v},\tilde {v}) = -2s$. Moreover, we have

\[ \tilde{v}^{(n)} = \sum{\binom{n}{i}s^i\alpha^{(n-i)}\beta^{(i)}}. \]

Applying 3.7 we obtain

\[ T(\tilde{v}^{(n)}) = \sum{\frac{n!}{i!}s^i{\mathsf q}_{2i}}. \]

By definition of the Mukai pairing $b_{\mathrm {SH}}$, we get

\[ b_{\mathrm{SH}}\big(T(\tilde{v}^{(n)}),T(\tilde{v}^{(n)})\big) = \bigg(\sum{\frac{n!}{i!}\frac{n!}{(n-i)!}\int_X{{\mathsf q}_{2i}{\mathsf q}_{2n-2i}}}\bigg)s^n = C \tilde{q}(\tilde{v},\tilde{v})^n, \]

for some constant $C$ independent of $\tilde {v}$.

Proof. From the Riemann–Roch theorem and the assumption that $v(E)_{\mathrm {SH}} = v(E)$ it follows that $\chi (E,E) = b_{\mathrm {SH}}(v(E),v(E))$. Since $T(\tilde {v}^{(n)}) = n!r^{n-1}v(E)$, from the previous lemma we obtain

\[ b_{\mathrm{SH}}(r^{n-1}v(E),r^{n-1}v(E)) = C\tilde{q}(\tilde{v},\tilde{v})^n \]

for some constant $C$. Dividing both sides by $r^n$ we get

(9)\begin{equation} b_{\mathrm{SH}}\bigg(\frac{v(E)}{r},\frac{v(E)}{r}\bigg) = C\tilde{q}\bigg(\frac{\tilde{v}}{r},\frac{\tilde{v}}{r}\bigg)^n. \end{equation}

To compute the constant $C$, we substitute $\tilde {v}=\alpha + r_X\beta$, the extended Mukai vector of the structure sheaf. Since $r=1$ and $\chi ({\mathcal {O}}_X,{\mathcal {O}}_X) = n+1$ we get $C= (-1)^n ({n+1})/{(2r_X)^n}$. Substituting $C$ into (9) and rearranging we get the result.

Proof. The long exact sequence in cohomology gives the two sequences

\begin{gather*} 0 \to {\mathcal{F}} \to {\mathcal{H}}^0(C) \to {\mathcal{E}} \to 0, \\ 0 \to {\mathcal{H}}^{2}(C) \to {\mathcal{E}} \to 0. \end{gather*}

The rest of the long exact sequence shows that there is no cohomology in degrees different from $0$ and $2$. The first sequence is induced by the evaluation map $\operatorname {Ext}^1({\mathcal {E}},{\mathcal {F}}) \otimes {\mathcal {E}} \to {\mathcal {F}}$. Therefore, it is not split and ${\mathcal {H}}^{0}(C) \cong {\mathcal {G}}$.

Proof. Consider the distinguished triangle

\[ \operatorname{Ext}^*({\mathcal{E}},{\mathcal{F}}) \otimes {\mathcal{E}}[-1] \to C(\operatorname{Ext}^*({\mathcal{E}},{\mathcal{F}}) \otimes {\mathcal{E}} \to {\mathcal{F}}) \to P_{{\mathcal{E}}}({\mathcal{F}}) \]

of Remark 4.1. Applying the long exact sequence of cohomology sheaves and Lemma 4.2 we get the exact sequences

\begin{gather*} 0 \to {\mathcal{G}} \to {\mathcal{H}}^0(P_{{\mathcal{E}}}({\mathcal{F}})) \to 0,\\ 0 \to {\mathcal{H}}^1(P_{{\mathcal{E}}}({\mathcal{F}})) \to {\mathcal{E}} \to {\mathcal{E}} \to {\mathcal{H}}^2(P_{{\mathcal{E}}}({\mathcal{F}})) \to 0, \\ 0 \to {\mathcal{H}}^3(P_{{\mathcal{E}}}({\mathcal{F}})) \to {\mathcal{E}} \to 0. \end{gather*}

If we check that the middle map ${\mathcal {E}} \to {\mathcal {E}}$ in the second sequence is the identity we are done. By definition, it is induced in ${\mathcal {H}}^2$ by the map

\[ \operatorname{Ext}^*({\mathcal{E}},{\mathcal{F}}) \otimes {\mathcal{E}}[-1] \to C, \]

which, in turn, is obtained from the octahedral axiom, composed with the isomorphism in Lemma 4.2. Chasing the definitions and the commutativity in the octahedral axiom one sees that the desired map is induced in ${\mathcal {H}}^2$ by the map

\[ H[-1] : \operatorname{Ext}^*({\mathcal{E}},{\mathcal{F}}) \otimes {\mathcal{E}}[-1] \to \operatorname{Ext}^*({\mathcal{E}},{\mathcal{F}}) \otimes {\mathcal{E}}[1], \]

described explicitly as

in the proof of [Reference Huybrechts and ThomasHT06, Lemma 2.5]. From this description it is clear that the induced map in ${\mathcal {H}}^2$ is the identity.

Proof. From [Reference Huybrechts and ThomasHT06, Lemma $2.5$] we see that $P_{{\mathcal {E}}}({\mathcal {E}}) \cong {\mathcal {E}}[-4]$. Applying the equivalence $P^{-1}_{{\mathcal {E}}}$ to the distinguished triangle

\[ {\mathcal{G}} \to P_{{\mathcal{E}}}({\mathcal{F}}) \to {\mathcal{E}}[-3] \]

of Proposition 4.3 we obtain

\[ P^{-1}_{{\mathcal{E}}}({\mathcal{G}}) \to {\mathcal{F}} \to {\mathcal{E}}[1]. \]

Rotating this triangle gives the thesis.

Proof. Setting ${\mathcal {G}}' := P_{{\mathcal {E}}}^{-1}({\mathcal {G}})$ we get

\begin{align*} \operatorname{Ext}^{k}({\mathcal{E}},{\mathcal{G}}) &= \operatorname{Ext}^k({\mathcal{E}},P_{{\mathcal{E}}}({\mathcal{G}}')) = \operatorname{Ext}^k({\mathcal{E}}[4],{\mathcal{G}}') \\ & =\operatorname{Ext}^{k-4}({\mathcal{E}},{\mathcal{G}}'). \end{align*}

Both objects ${\mathcal {G}}$ and ${\mathcal {G}}'$ are sheaves, so the ext groups above vanish for $k \neq 4$. For $k=4$, the exact sequence

\[ \operatorname{Ext}^4({\mathcal{E}},{\mathcal{E}}) \to \operatorname{Ext}^4({\mathcal{E}},{\mathcal{G}}) \to \operatorname{Ext}^4({\mathcal{E}},{\mathcal{F}}) = 0, \]

shows that it is at most one dimensional. It is non-zero, because of the map ${\mathcal {G}} \to {\mathcal {E}}$, so it is one dimensional.

Proof. Consider the following diagram.

The nine lemma implies that the right vertical map is an isomorphism. From the previous remark we see that

\[ (T_{Z_1}|_W + T_{Z_2}|_W)/T_W \cong (T_W)^{\perp}/T_W, \]

which is a symplectic rank-two bundle, in particular it has trivial determinant. We conclude that

\[ N_{W/Z_1} \otimes N_{W/Z_2} \cong \bigwedge\nolimits^2(N_{W/Z_1} \oplus N_{W/Z_2}) \cong {\mathcal{O}}_X. \]

Proof. The first short exact sequence is given by

\[ 0 \to (T_{Z_1}|_W + T_{Z_2}|_W)/T_{Z_2}|_W \to T_X|_W/(T_{Z_2}|_W) \to T_X|_W/(T_{Z_1}|_W + T_{Z_2}|_W) \to 0, \]

and noting that $(T_{Z_1}|_W + T_{Z_2}|_W)/T_{Z_2}|_W \cong T_{Z_1}|_W / T_W$. The central vertical map in the diagram is induced by the restriction of the isomorphism $\sigma _X : T_X \cong \Omega _X$. The composition

\[ T_{Z_1} + T_{Z_2} \to T_X \cong \Omega_{X} \to \Omega_W \]

vanishes by Remark 4.6. Thus, the central map factors to give the diagram in the statement.

Proof. Consider the long exact sequence obtained applying $\operatorname {Hom}({\mathcal {O}}_{Z_2}(-W),-)$ to

\[ 0 \to {\mathcal{O}}_{Z_2}(-W) \to {\mathcal{O}}_Z \to {\mathcal{O}}_{Z_1} \to 0. \]

Since $Z_2 \subset X$ is a Lagrangian surface, by dimensional reasons the local-to-global spectral sequence vanishes and yields

(11)\begin{equation} \operatorname{Ext}^k({\mathcal{O}}_{Z_2}(-W),{\mathcal{O}}_{Z_2}(-W)) \cong H^k(Z_2,{\mathbb{C}}). \end{equation}

Example 4.10 implies that

(12)\begin{equation} \operatorname{Ext}^k({\mathcal{O}}_{Z_2}(-W),{\mathcal{O}}_{Z_1}) \cong H^{k-1}(W,{\mathbb{C}}). \end{equation}

Therefore, we only need to show that the connecting homomorphisms

\[ \operatorname{Ext}^k({\mathcal{O}}_{Z_2}(-W),{\mathcal{O}}_{Z_1}) \to \operatorname{Ext}^{k+1}({\mathcal{O}}_{Z_2}(-W),{\mathcal{O}}_{Z_2}(-W)) \]

become identified with the pushforwards in cohomology. The Serre dual statement is that the connecting map

\[ \operatorname{Ext}^k({\mathcal{O}}_{Z_2}(-W),{\mathcal{O}}_{Z_2}(-W)) \to \operatorname{Ext}^{k+1}({\mathcal{O}}_{Z_1},{\mathcal{O}}_{Z_2}(-W)) \]

is the restriction $H^k(Z_2,{\mathbb {C}}) \to H^k(W,{\mathbb {C}})$.

The isomorphisms (11) and (12) are induced by the degeneration of the spectral sequences:

\[ H^p\mathop{\mathrm{RHom}}\nolimits({\mathcal{H}}^{-q}(j_2^*j_{2,*}{\mathcal{O}}_{Z_2}(-W)),{\mathcal{O}}_{Z_2}(-W)) \implies \operatorname{Ext}^{p+q}({\mathcal{O}}_{Z_2}(-W),{\mathcal{O}}_{Z_2}(-W)) \]

and

\[ H^p\mathop{\mathrm{RHom}}\nolimits({\mathcal{H}}^{-q}(j_2^*j_{1,*}{\mathcal{O}}_{Z_1}),{\mathcal{O}}_{Z_2}(-W)) \implies \operatorname{Ext}^{p+q+1}({\mathcal{O}}_{Z_1},{\mathcal{O}}_{Z_2}(-W)). \]

The connecting homomorphism is induced by pullback along the map

\[ {\mathcal{O}}_{Z_1} \to {\mathcal{O}}_{Z_2}(-W)[1]. \]

Taking $j_2^*$ and ${\mathcal {H}}^{-q}$ we get the zero map in cohomology for every $q$. This implies that the long exact cohomology sequence induced by $j_2^*{\mathcal {O}}_{Z_1} \to j_2^*cO_{Z_2}(-W)[1]$ is actually a collection of short exact sequences, represented by maps

(13)\begin{equation} {\mathcal{H}}^{-q}(j_2^*j_{1,*}{\mathcal{O}}_{Z_1}) \to {\mathcal{H}}^{-q}(j_2^*j_{2,*}{\mathcal{O}}_{Z_2}(-W))[1]. \end{equation}

Pulling back along those maps gives a map on the $E_2$ page of the spectral sequences, which induces the connecting homomorphism that we wish to understand.

Using [Reference Căldăraru, Katz and SharpeCKS03, Proposition $A.6$] we obtain ${\mathcal {H}}^{-q}(j_2^*j_{1,*}{\mathcal {O}}_{Z_1}) \cong i_{2,*}\bigwedge \nolimits ^q\tilde {N}^{\vee }$, where $i_k : W \to Z_k$ is the inclusion. The map (13) becomes a map

\[ i_{2,*}\bigwedge^q\tilde{N}^{\vee} \to \bigwedge^q N_{Z_2/X}^{\vee} \otimes {\mathcal{O}}_{Z_2}(-W)[1]. \]

Verdier duality gives that $i^!_2 = i_2^* \otimes {\mathcal {O}}_W(W)[-1]$, so the map becomes $\bigwedge ^q\tilde {N}^{\vee } \to \bigwedge ^q N_{Z_2/X}^{\vee }$, which is identified with the restriction map via Lemma 4.8.

The isomorphism

\[ \operatorname{Ext}^k({\mathcal{O}}_{Z_2}(-W),{\mathcal{O}}_Z) \cong H^k(Z_2 - W,{\mathbb{C}}), \]

follows from the five lemma applied to the long exact sequence obtained by Poincaré and Lefschetz duality.

Proof of Proposition 5.2 Since the Poincaré sheaf ${\mathcal {P}} \in \mathop {\mathrm {Coh}}\nolimits (M \times _{{\mathbb {P}}^g} M)$ is flat with respect to both projections by [Reference ArinkinAri13, Theorem A], $\pi _1^{*}({\mathcal {M}}_{L}) \otimes {\mathcal {P}}$ is a sheaf on $M \times _{{\mathbb {P}}^g} M$. Thus, $\Phi (V_L)$ is a complex concentrated in non-negative degrees. To show that it is a locally free sheaf, it suffices to prove

\[ \operatorname{Ext}_M^i(\Phi(V_L),{\mathbb{C}}(x)) = 0 \quad \text{for } i > 0, \]

for every $x \in M$, where ${\mathbb {C}}(x)$ denotes the skyscraper sheaf at $x$. From [Reference ArinkinAri13, Proposition 7.1] it follows that $\Phi ^{-1}({\mathbb {C}}(x)) = {\mathcal {P}}_{M \times \{x^{\vee }\}}[g]$, where $x^{\vee }$ parameterizes the dual sheaf to $x$. Thus, we have

\[ \operatorname{Ext}_M^i(\Phi(V_L),{\mathbb{C}}(x)) = \operatorname{Ext}_M^i(V_L,{\mathcal{L}}_t[g]), \]

where $t:=\pi (x)$, and ${\mathcal {L}}_t := {\mathcal {P}}_{M \times \{x^{\vee }\}}$ is a torsion-free rank-one sheaf supported on $M_t$. Since $V_L$ is a Cohen–Macaulay sheaf of dimension $g$ on $M$, the derived dual $\mathrm {R}\mathcal {H}\!{\it om}_M(V_L,{\mathcal {O}}_M)$ is just $\mathcal {E}xt^g(V_L,{\mathcal {O}}_M)[-g]$. Hence, we have

\begin{align*} \operatorname{Ext}_M^i(V_L,{\mathcal{L}}_t[g]) & \simeq {\mathbb{H}}^i(M, \mathrm{R}\mathcal{H}\!{\it om}_M(V_L,{\mathcal{O}}_M) \otimes^{L} {\mathcal{L}}_t[g]) \\ & \simeq {\mathbb{H}}^i(M, \mathcal{E}xt^g(V_L,{\mathcal{O}}_M) \otimes^{L} {\mathcal{L}}_t). \end{align*}

The sheaf

\[ \mathcal{H}^i(\mathcal{E}xt^g(V_L,{\mathcal{O}}_M) \otimes^{L} {\mathcal{L}}_t) = \mathcal{T}or_k^{{\mathcal{O}}_M}(\mathcal{E}xt^g(V_L,{\mathcal{O}}_M),{\mathcal{L}}_t) \]

vanishes by [Reference SerreSer65, Corollary to Theorem V.4]. Indeed, by [Reference ArinkinAri13, Theorem A(2)] the sheaf ${\mathcal {L}}_t$ is Cohen–Macaulay of dimension $g$ on $M$, and the same holds for $\mathcal {E}xt^g(V_L,{\mathcal {O}}_M)$. Thus, we have

\[ {\mathbb{H}}^i(M, \mathcal{E}xt^g(V_L,{\mathcal{O}}_M) \otimes^{L} {\mathcal{L}}_t)= H^i(M, \mathcal{E}xt^g(V_L,{\mathcal{O}}_M) \otimes {\mathcal{L}}_t) = 0 \quad \text{for } i > 0, \]

because the sheaf $\mathcal {E}xt^g(V_L,{\mathcal {O}}_M) \otimes {\mathcal {L}}_t$ is supported on $M_t \cap L$ which is finite.

Proof. First note that the equivalence $\Phi$ is $({\mathbb {P}}^n)^{\vee }$-linear, because the kernel ${\mathcal {P}}$ is defined on the fiber product. In particular there is an isomorphism of functors

\[ \Phi(-) \otimes {\mathcal{O}}_{M_t} = \Phi(- \otimes {\mathcal{O}}_{M_t}). \]

The projection formula gives the isomorphism of functors $-\otimes {\mathcal {O}}_{M_t} \cong i_{M_t,*}i_{M_t}^*(-)$. To conclude, it remains to prove that

\[ \Phi(i_{M_t,*}(-)) \cong i_{M_t,*}\Phi_{{\mathcal{P}}_t}(-), \]

which follows from the base change theorem as explained in [Reference HuybrechtsHuy06, Lemma $11.30$].

Proof. The statement for $P'$ follows from [Reference MarkmanMar23, Lemma $7.3$]. The statement for the union $P' \cup L$ follows from (16) combined with Theorem 2.5 and Remark 2.6.

Proof of Proposition 5.1 Consider the short exact sequence (14)

\[ 0 \to {\mathcal{L}}(-K^*) \to \overline{{\mathcal{L}}} \to {\mathcal{O}}_{P'} \to 0. \]

After applying the autoequivalence $\Phi$ this becomes

\[ 0 \to \Phi({\mathcal{L}}(-K^*)) \to F \to {\mathcal{O}}_ M \to 0. \]

Here $\Phi ({\mathcal {L}}(-K^*))$ is locally free by Proposition 5.2, and has rank four by Lemma 5.3. It follows that $F$ is locally free of rank five.

Now we compute its extended Mukai vector. Note that since ${\mathcal {L}}$ has degree zero, we have $v({\mathcal {L}}) = v({\mathcal {O}}_L)$, so it suffices to compute the extended Mukai vector of $\Phi ({\mathcal {O}}_Z)$. We start by describing the action of the equivalence $\Phi$ on $\widetilde {H}(X,{\mathbb {Q}})$. We follow the computations done in [Reference BeckmannBec23, Proposition $10.4$] for odd genus.

First, since the skyscraper sheaf of a point goes to a line bundle of degree $0$ on a fiber, we deduce that $\beta \mapsto f$. From the autoduality property of the Poincaré sheaf, described in [Reference ArinkinAri13, Equation $(7.8)$], we see that $f \mapsto \beta$. The choice (15) for the normalization of ${\mathcal {P}}$ implies

\[ \lambda - 3f +3\beta \mapsto -2\big(\alpha + \tfrac{5}{4}\beta\big), \]

where the coefficient $-2$ is determined by imposing the map $\Phi ^{\widetilde {H}}$ to be an isometry. This implies that

\begin{align*} \lambda + f -3\beta &= \lambda - 3f + 3\beta + (4f -6\beta) \mapsto -2 \big(\alpha + \tfrac{5}{4}\beta\big) + (4\beta - 6f)\\ &= -2\alpha -6f +\tfrac{3}{2}\beta. \end{align*}

Since $\tilde {v}({\mathcal {O}}_Z) = \lambda + f -3\beta$ by Lemma 5.4, the formula for the extended Mukai vector follows by [Reference MarkmanMar23, Theorem 1.7(4)]. Since by Lemma 5.3 $F$ has rank five, its normalized extended Mukai vector is

\[ \tilde{v}(F) = 5\alpha +15f - \tfrac{15}{4}\beta. \]

Twisting by ${\mathcal {O}}(-3f)$ we kill the first Chern class, and the extended Mukai vector becomes

\[ \tilde{v}(F_0) = 5\big(\alpha-\tfrac{3}{4}\beta\big). \]

This can be computed, for example, using the fact that tensor product with a line bundle induces an isometry on $\widetilde {H}(X,{\mathbb {Q}})$. Corollary 3.10 then gives

\[ v(F_0) = 5\big(1-\tfrac{3}{4}{\mathsf q}_2 + \tfrac{9}{32}\mathfrak{pt}\big), \]

because if $c_X =1$, then ${\mathsf q}_4 = \mathfrak {pt}$.

Proof. Again, we start with the short exact sequence (14)

\[ 0 \to {\mathcal{L}}(-K^*) \to \overline{{\mathcal{L}}} \to {\mathcal{O}}_{P'} \to 0. \]

The isomorphisms (11) and (12) remain valid for the same reasons. The rest of the proof of Proposition 4.11 is not affected by the twist, so it remains to show that

(17)\begin{equation} \operatorname{Ext}^i(\overline{{\mathcal{L}}},\overline{{\mathcal{L}}}) \cong \operatorname{Ext}^i({\mathcal{L}}(-K^*),\overline{{\mathcal{L}}}) \quad \text{for } i =1,2. \end{equation}

The sheaves ${\mathcal {E}} = {\mathcal {O}}_{P'}$ and ${\mathcal {F}} = {\mathcal {L}}(-K^*)$ satisfy the assumptions of § 4.1 by Theorem 4.9. Thus, (17) is a consequence of the vanishings in Corollary 4.5.

Proof. This relies on the fact that the Lagrangian $L$ is, in fact, the symmetric square of a genus five curve $C$. In fact, we have

\[ H^2(L,{\mathbb{C}}) \cong H^2(C,{\mathbb{C}}) \oplus \bigwedge\nolimits^2 H^1(L,{\mathbb{C}}), \]

where the second summand is embedded via cup product. The fundamental class of $K^* \subset \mathrm {Sym}^2 C$ spans the direct summand $H^2(C,{\mathbb {C}})$. Thus, taking the cokernel as in Lemma 5.5 we get

(18)\begin{equation} \operatorname{Ext}^2(\overline{{\mathcal{L}}},\overline{{\mathcal{L}}}) \cong \bigwedge^2 H^1(L,{\mathbb{C}}). \end{equation}

To conclude, recall that by [Reference MladenovMla19, Theorem 2.1.5], the isomorphism

\[ \operatorname{Ext}^*({\mathcal{L}},{\mathcal{L}}) \cong H^*(L,{\mathbb{C}}) \]

preserves the algebra structure. Hence, the cokernel map

\[ H^2(L,{\mathbb{C}}) \to \operatorname{Ext}^2(\overline{{\mathcal{L}}},\overline{{\mathcal{L}}}) \]

maps the cup product to the Yoneda product. The statement then follows from (18).

Proof. By [Reference O'GradyO'G21, Proposition $3.4$] it is enough to show the statement for a rational ample class in the same chamber as $h$, i.e. we can assume $h = f + \varepsilon \lambda$ for $0 < \varepsilon \ll 1$. We write

\[ c_1(E) = bf + c\lambda, \]

with respect to the decomposition of (4). Let $t \in ({\mathbb {P}}^2)^{\vee }$ be a general point. Lemma 5.3 together with the exact sequence (22), implies that for a general $t$

\[ F''_t = {\mathcal{O}}_{M_t} \oplus L_{t,1} \oplus \cdots \oplus L_{t,4}, \]

where $L_i$ are line bundles of degree zero on $M_t$. Therefore, the restriction $F''_t$ is semistable because it is the sum of line bundles of degree $0$. Thus, we have

\[ 2c\varepsilon = \int_{M_t}{c_1(E_t) \cup h_t} \leq \int_{M_t}{c_1(F''_t) \cup h_t} = 0, \]

which gives $c \leq 0$.

By definition, we have

\[ \mu (E) = \frac{1}{\operatorname{rk}(E)}\int_{M}{(bf+c\lambda)\cup h^3}, \]

where $h^3 = 3\varepsilon (f^2 \cup \lambda ) + 3\varepsilon ^2(f \cup \lambda ^2) + \varepsilon ^3 \lambda ^3$, because $f^3 = 0$. The class $f^2$ is Poincaré dual to a general fiber, so we have

\[ \int_M{f^2 \cup \lambda^2} = \int_{M_t}{\lambda_t^2} >0. \]

In particular, in $\mu (E)$ there is a term in $\varepsilon$, namely $\mu (E) = 3c\varepsilon \int _M{f^2 \cup \lambda ^2} + \varepsilon ^2(\dots )$. On the other hand, since $c_1(F'') = 15f$ and $\operatorname {rk}(F'') = 5$, we have $\mu (F'') = 9\varepsilon ^2(\dots ).$ The assumption that $h$ is destabilizing gives the inequality

\[ \mu(E) \geq \mu(F'') \quad \forall \varepsilon \ll 1. \]

Passing to the limit $\varepsilon \to 0$, we obtain that the term in $\varepsilon$ must be non-negative, i.e. $c \geq 0$. Combining with the previous inequality we get $c = 0$.

Proof. On a general fiber $M_t$ we can write

\[ F''_t = L_{t,0} \oplus L_{t,1} \oplus \dots \oplus L_{t,4}, \]

where $L_{t,0} = {\mathcal {O}}_{M_t}$, and $L_{t,i}$ are non-trivial line bundles of degree zero. The restriction $A_t$ has the same slope as $F''_t$, hence it is a sub-sum of these line bundles,

\[ A_t = L_{t,i_1} \oplus \cdots \oplus L_{t,i_r}. \]

Taking $\Phi ^{-1}$, Lemma 5.3 gives

\[ \Phi^{-1}(A)|_{M_t} = i_{M_t,*}\Phi_{{\mathcal{P}}_t}^{-1}(A_t) ={\mathcal{O}}_{M_t,[L_{t,i_1}]} \oplus \cdots \oplus {\mathcal{O}}_{M_t,[L_{t,i_r}]}. \]

We deduce that over an open $U \subset {\mathbb {P}}^2$, the support $\operatorname {Supp} \Phi ^{-1}(A) \subseteq Z$ and it is finite over the base of degree $r$. Since $r < 5$, by assumption $\operatorname {Supp} \Phi ^{-1}(A)$ it not equal to the whole $Z$. Hence, it must be one of the two components, giving the dichotomy in the statement.

Proof. Recall that $\operatorname {rk}(F'')=5$ and $c_1(F'')=15f$. Let

\[ 0 \to A \to F'' \to B \to 0, \]

be a slope destabilizing short exact sequence. We assume $A$ saturated, so $B$ is torsion-free. By Lemma 6.5 either $\operatorname {rk}(A) = 1$ or $\operatorname {rk}(A) = 4$.

Case 1. Assume that $\operatorname {rk}(A) = 1$. Consider the following commutative diagram with exact rows and columns.

The intersection $A \cap {\mathcal {O}}(2f)$ is defined as the kernel of the map

\[ A \oplus {\mathcal{O}}(2f) \to F'', \quad (a,x) \mapsto a-x. \]

Restricting to the general fiber, both $A$ and ${\mathcal {O}}(2f)$ become trivial, so we see that the intersection is non-trivial, because $F''|_t$ has only one trivial summand. Since $A$ has rank one, the quotient sheaf $A/({\mathcal {O}}_M(2f) \cap A)$ has rank zero. It embeds into $G'$, which is locally free, so it is zero, which gives

\[ A \subset {\mathcal{O}}_M(2f). \]

Using that $B$ is torsion-free, the same argument yields ${\mathcal {O}}_M(2f) \subset A$. We deduce that $A = {\mathcal {O}}_M(2f)$, which is not destabilizing.

Case 2. Assume that $\operatorname {rk}(A) = 4$. The quotient $B$ is a torsion-free rank-one sheaf, so it injects into its double dual $B^{\vee \vee }$, which is a line bundle on $M$ by [Reference HartshorneHar80, Proposition 1.11]. By Proposition 6.4, it suffices to show that $\operatorname {Hom}(F'',{\mathcal {O}}(kf))=0$ vanishes for every $k \leq 3$. By construction, we have

\[ \operatorname{Hom}(F'',{\mathcal{O}}(2f)) = \operatorname{Hom}\big(P^{-1}_{{\mathcal{O}}_{M}(2f)}(F'),{\mathcal{O}}_{M}(2f)\big) = 0. \]

It follows using $h^0(M,{\mathcal {O}}_M(f)) \neq 0$ that $\operatorname {Hom}(F'',{\mathcal {O}}_M(kf))= 0$ for every $k \leq 2$. Hence, it remains to show that $\operatorname {Hom}(F'',{\mathcal {O}}_M(3f)) = 0$.

Let $\varphi : F'' \to {\mathcal {O}}_M(3f)$ be a morphism, and let $D$ be its kernel. Restricting to a general fiber $M_t$ we see that $\operatorname {Hom}({\mathcal {O}}_M(2f),D) = 0$, because $D_t$ splits as a sum of four non-trivial line bundles of degree $0$. This implies that the composition

\[ {\mathcal{O}}_M(2f) \to F'' \to {\mathcal{O}}_M(3f) \]

is not zero, hence it is injective. Applying $\Phi ^{-1}$ we obtain a diagram

where ${\mathcal {O}}_{l}$ is the structure sheaf of a line $l \subset P'$, i.e. of the zero locus of a section of ${\mathcal {O}}_{P'}(1)$. Since $l \not \subseteq L$, the map ${\mathcal {L}}(-2K^*) \to {\mathcal {O}}_{l}(3)$ is zero. Thus, the central map factors through ${\mathcal {O}}_{P'}(2)$, but $\operatorname {Hom}(F'',{\mathcal {O}}(2f)) = 0$ is zero, hence $\varphi$ is zero.

Proof. The proof is simply a refinement of the proof of [Reference MarkmanMar23, Theorem 3.4], where we also keep track of the polarization. Recall that, if $F$ is modular and $\omega \in {\mathcal {K}}(X)$ is a Kähler class such that $F$ is $\omega$-stable, we can deform $F$ along the twistor line ${\mathbb {P}}^1_{\omega }$ spanned by $\omega$ by [Reference VerbitskyVer99, Theorem 3.19]. The assumption $c_1(F) = 0$ implies that even when we deform along a twistor line, the bundle remains untwisted. The deformed bundle is stable with respect to the canonical Kähler class $\omega _t$ on every fiber $X_t$ of the twistor line. Since the walls in the Kähler cone are defined by algebraic classes, we can find a Kähler class $\omega$ with the following properties.

1. (1) The bundle $F$ is slope-stable with respect to $\omega$.

2. (2) The twistor line ${\mathbb {P}}^1_\omega$ is generic, in the sense that the general element has trivial Picard group.

We can do the same for $h'$, obtaining a Kähler class $\omega '$ on $Y$ with the same properties as above. Now, we can deform $F$ along the twistor deformation ${\mathcal {X}}_{\omega } \to {\mathbb {P}}^1_{\omega }$. In particular, if we choose a general element $X_1 \in {\mathbb {P}}^1_{\omega }$, we obtain a modular bundle $F_1$ with the same invariants as $F$ living on $X_1$. Since $X_1$ has trivial Picard $F_1$ is stable with respect to any Kähler class by [Reference MarkmanMar20, Lemma 6.15].

Denote by $X_{\mathrm {last}}$ a general element of ${\mathbb {P}}^1_{\omega '}$. Up to the choice of markings on $X_1$ and $X_{\mathrm {last}}$, we can connect them through a chain of twistor lines whose intersection points have trivial Picard group by [Reference VerbitskyVer96, Theorem 3.2 and 5.2e] (see also [Reference MarkmanMar20, Theorem 6.14]). The bundle $F_1$ deforms on this chain of twistor lines to a bundle $F_{\mathrm {last}}$ on $X_{\mathrm {last}}$, which again is stable with respect to any Kähler class. Deforming $F_{\mathrm {last}}$ back along ${\mathbb {P}}^{1}_{\omega '}$ we get the desired bundle $F'$ on $Y$ which is $\omega '$-stable. Since $\omega '$ and $h'$ live in the same open chamber, $F'$ is also $h'$-stable.

Proof of Theorem 1.1 The vector bundle $F''$ constructed above is atomic and stable with respect to a suitable polarization. By Proposition 5.1 we can twist $F''$ to obtain a stable atomic vector bundle $F_0$ with Mukai vector

\[ v(F_0) = 5\big(1-\tfrac{3}{4}{\mathsf q}_2 + \tfrac{9}{32}\mathfrak{pt}\big). \]

Thus, $F_0$ satisfies the assumptions of the above lemma, and we can deform it to a vector bundle on any Kähler deformation of $X$, which is stable with respect to any $v(F_0)$-generic polarization.

The $\operatorname {Ext}$ algebra remains constant along these deformations by [Reference VerbitskyVer08, Proposition 6.3]. The statement about the Ext groups is Corollary 5.6.

To prove smoothness of the deformation space we argue as follows. The main result in [Reference Meazzini and OnoratiMO23] (or [Reference BeckmannBec22, Theorem 6.1]) gives formality of the algebra $\mathop {\mathrm {RHom}}\nolimits (F_0,F_0)$. Thus, the obstruction to lifting a first-order deformation is the Yoneda square, which vanishes by Corollary 5.6.

Proof. Under the Hochschild–Konstant–Rosenberg (HKR) isomorphism $HH^2(M) \cong HT^2(M)$, the obstruction map $\chi _F(-)$ becomes identified with

\[ HT^2(M) \to \operatorname{Ext}^2(F,F), \quad \eta \mapsto (\mathop{\mathrm{id}}\nolimits_F \otimes \eta) \circ \exp(\mathrm{At}(F)), \]

by [Reference TodaTod09, Proposition 6.1]. We have

\begin{align*} \mathop{\mathrm{Tr}}\nolimits_F(\chi_F(\eta) \circ a \circ b) &= \mathop{\mathrm{Tr}}\nolimits_F((\mathop{\mathrm{id}}\nolimits_F \otimes \eta) \circ \exp(\mathrm{At}(F)) \circ a \circ b)\\ &= \eta \circ \mathop{\mathrm{Tr}}\nolimits_F(\exp(\mathrm{At}(F)) \circ a \circ b), \end{align*}

by linearity of the trace map. Up to changing signs of the graded pieces of $\eta$, this is equal to $\eta \circ \sigma (a \circ b)$. Then, by Poincaré duality there is a class $\omega \in H^*(X,{\mathbb {C}})$, depending only on $\eta$, such that

\[ \int_M{(\eta \circ \sigma(a \circ b)) \cup \sigma^2_ M} = \int_M{\sigma(a \circ b) \cup \omega}. \]

Closedness then follows from [Reference Kuznetsov and MarkushevichKM09, Theorem 2.2].

Proof. The idea is to compare the obstruction map of $\overline {{\mathcal {L}}}$ with the obstruction map of $F_0 = F \otimes {\mathcal {O}}_M(-3f)$. By construction, the equivalence

\[ \Theta' = (- \otimes {\mathcal{O}}(-3f)) \circ \Theta, \]

maps $\overline {{\mathcal {L}}}$ to $F_0$. There is an injective homomorphism

\[ \mu : HH^2(M) \to \widetilde{H}(X,{\mathbb{C}}), \quad \eta \mapsto m_{\eta}(\sigma), \]

where $m_{\eta }$ denotes the action of $\eta \in HH^2(M)$ on $\widetilde {H}(X,{\mathbb {C}})$, see [Reference MarkmanMar23, Section 6]. This action is induced by the action of the LLV algebra via [Reference TaelmanTae23, Theorem A]. If $\Phi$ is any equivalence, the map $\Phi ^{\widetilde {H}}$ is a Hodge isometry compatible with the action of the LLV algebra. Hence, $\mu$ intertwines the actions of $\Phi$, that is there is the following commutative diagram.

A direct computation as in the proof of Proposition 5.1 shows

\begin{gather*} \Theta'^{HH} (\sigma^{\vee}) \in H^1(M,T_M) \oplus H^2(M,{\mathcal{O}}_M),\\ \Theta'^{HH} (f) \in H^2(M,{\mathcal{O}}_M), \end{gather*}

where $\sigma ^{\vee }$ generates $H^0(M,\operatorname {\bigwedge \nolimits }^2T_M)$. The vector bundle $F_0$ deforms along every commutative deformation, so

\[ \chi_{F_0}|_{H^1(M,T_M)} \equiv 0, \]

which implies

\[ \chi_{\overline{{\mathcal{L}}}}({\mathbb{C}} \sigma^{\vee}) = \chi_{F_0}({\mathbb{C}} \overline{\sigma}) = \chi_{\overline{{\mathcal{L}}}}({\mathbb{C}} f), \]

where we are identifying $f \in H^1(M,\Omega ^1_M)$ with its image in $H^1(M,T_M)$ via the isomorphism $\Omega ^1_M \cong T_M$.

Proof. By Lemma 7.6 it suffices to compute the image of the restriction map

\[ H^2(M,{\mathbb{C}}) \to H^2(L,{\mathbb{C}}) \]

as explained in [Reference MarkmanMar23, Remark 3.10], and map it into the quotient $H^2(L,{\mathbb {C}})/H^0(C,{\mathbb {C}})$. The restriction can be computed on the other birational model, that is

\[ H^2(S^{[2]},{\mathbb{C}}) \to H^2(\mathrm{Sym}^2C,{\mathbb{C}}) \cong H^2(C,{\mathbb{C}}) \oplus \operatorname{\bigwedge\nolimits}^2 H^1(C,{\mathbb{C}}). \]

The Künneth formula implies that the first summand in

\[ H^2(S^{[2]},{\mathbb{C}}) \cong H^2(S,{\mathbb{C}}) \oplus {\mathbb{C}}\delta \]

maps to $H^2(C,{\mathbb {C}})$. By definition, the class $\delta$ maps to a multiple of the class of the diagonal $\Delta _C \subset \mathrm {Sym}^2 C$. To conclude, note that, for every $\alpha,\beta \in H^1(C,{\mathbb {C}})$, we have

\[ \int_{\Delta_C}{(\pi_1^*\alpha \wedge \pi^*_2\beta - \pi_1^*\beta \wedge \pi^*_2\alpha)|_{\Delta_C}} = 2\int_C{\alpha \wedge \beta}. \]

Hence, the image of $[\Delta _C]$ in $\bigwedge ^2 H^1(C,{\mathbb {C}})$ represents the Poincaré pairing. Since $\alpha _{\eta }$ is defined by pairing two classes in $\operatorname {Ext}^1(\overline {{\mathcal {L}}},\overline {{\mathcal {L}}})$ with the image of the obstruction map, under the identification with $H^1(C,{\mathbb {C}})$ induced by $\theta$, it corresponds to

\[ \alpha_{\eta}(-,-) = \int_C{ - \wedge -}, \]

which is also the symplectic form on $\widetilde {M}_H(0,2H,-4)$. We conclude that $\theta$ preserves the 2-forms and that $\alpha _{\eta }$ is symplectic on the image.