Proof of Theorem 3.1

The main issue in this proof is that maps of the form $L_y\otimes t_x^*$ are not, in general, well-defined as automorphisms on $K_A(v)$ . Given $\mathcal F\in K_A(v)$ , $L_y\otimes t_x^*\mathcal F$ has the same Mukai vector as $\mathcal F$ but may not have the same image under the Albanese morphism. For instance, pullback by $t_x^*$ , in general, preserves Néron–Severi classes of line bundles and acts trivially on the structure sheaf, but it does not act trivially on all line bundles.

We therefore seek the $(x,y)\in A\times \hat {A}$ that satisfy the following conditions:

$$ \begin{align*} L&=\det(\mathcal F)=\det(L_y\otimes t_x^*\mathcal F)=L_y^{\otimes r}\otimes t_x^*\det(\mathcal F)= L_y^{\otimes r}\otimes t_x^*L\\ M&=\det(\Phi_P(\mathcal F))=\det(\Phi_P(L_y\otimes t_x^*(\mathcal F)))= \det(t_y^*(L_{-x}\otimes \Phi_P(\mathcal F)))\\ &\phantom{{}=\det(\Phi_P(\mathcal F))}= t_y^*(L_{-x}^{\otimes s}\otimes\det(\Phi_P(\mathcal F))) =t_y^*(L_{-x}^{\otimes s}\otimes M ) =L_{-x}^{\otimes s}\otimes t_y^* M. \end{align*} $$

We may rewrite these conditions as the equations (3.1). Equivalently, these $(x,y)$ are the kernel of the map $\phi $ in (3.2).

Precomposing the map $\phi $ with $\psi \colon A\times \hat {A} \to A\times \hat {A}$ , where $\psi (x,y)= (\phi _m(y)-rx,\phi _l(x)+sy)$ , and applying Lemma 3.2, we have

$$ \begin{align*} \phi\circ\psi(x,y)&= \phi\circ(\phi_m(y)-rx,\phi_l(x)+sy)\\ &=(\phi_m(\phi_l(x)+sy)-s(\phi_m(y)-rx),\phi_l(\phi_m(y)-rx)+r(\phi_l(x)+sy))\\ &=(-\chi\cdot x+rsx,-\chi\cdot y+rsy)=-n(x,y). \end{align*} $$

Thus, $\phi \circ \psi =[-n]$ , so $\phi $ is surjective and is hence an isogeny. Similarly, $\psi \circ \phi =[-n]$ and $G_A(v)\leqslant (A\times \hat {A})[n]$ .

We show $G_A(v)\cong ({\mathbb Z}/n{\mathbb Z})^4$ in Lemma 3.6. This will require an understanding of preimages of elements under $\phi _l$ and $\phi _m$ , which we study in Claims 3.4 and 3.5.

Proof. Since $\ker \phi _l \cong \ker \phi _{-l}$ and $\chi (L) \neq 0$ , we may assume that L is ample. The proof of Riemann–Roch for abelian varieties in [Reference Naruki47, §16] implies that the degree of $\phi _l$ is $\chi ^2$ . The structure of $\ker \phi _l \cap A[p^q]$ is then determined by Lemma 3.2 and the fact that the Weil pairing $e^L$ on the p-torsion is skew-symmetric [Reference Naruki47, §20, Thm. 1]. Since $\phi _m$ is the negative of the dual of $\phi _l$ [Reference Birkenhake and Lange6, §2], the group structure of p-power torsion points in $\ker \phi _m$ is isomorphic to that in $\ker \phi _l$ . The last statement is a consequence of [Reference Naruki47, §23, Thm. 3].

Proof. (a) By Lemma 3.2, the composition $\phi _l\circ \phi _m$ is given by multiplication by $-\chi $ . Thus, $\phi _m\circ \phi _l$ acts on $A[p^d]$ as the zero map, and hence,

$$ \begin{align*}\operatorname{\mathrm{im}} \phi_m|_{\hat{A}[p^d]} \subseteq A[p^d]\cap \ker \phi_l.\end{align*} $$

By Claim 3.4, $A[p^d]\cap \ker \phi _l$ has $p^{2d}$ elements and $\phi _m$ acting on $\hat {A}[p^d]$ is a $p^{2d}$ -to- $1$ map. It follows by counting that $\operatorname {\mathrm {im}} \phi _m|_{\hat {A}[p^d]} = A[p^d]\cap \ker \phi _l$ . By Claim 3.4, the preimage of u is as stated.

Part (b) follows analogously.

(c) By Lemma 3.2, the preimage of $\phi _l(u)$ under $\phi _l\circ \phi _m$ consists of elements of order $p^{c+q}$ . By Lemma 3.4, the result follows.

Part (d) follows analogously.

Proof. Case 1: $\chi =0$ .

Both L and M must have degree $0$ , so $\phi _l$ and $\phi _m$ are both the $0$ -morphism. The equations (3.1) simplify to

$$\begin{align*}0=-ry \quad\text{and}\quad 0=sx. \end{align*}$$

Furthermore, $n=-rs$ . Since $v=(r,l,s)$ is positive and $v^2\geq 4$ , we must have $r>0$ and $s<0$ . The solutions consist of all products of $|s|$ -torsion points on A and r-torsion points on $\hat {A}$ .

The group of solutions is isomorphic to $({\mathbb Z}/r{\mathbb Z})^4 \times ({\mathbb Z}/|s|{\mathbb Z})^4$ , hence $({\mathbb Z}/n{\mathbb Z})^4$ , since in this case, primitivity of the Mukai vector implies r and s are relatively prime.

Now, let p be a prime divisor of n and $p^q$ be the highest power of p dividing n. We treat the remaining cases by analyzing solutions in $(A\times \hat {A})[p^q]$ . We may then conclude by using the Sun Zi Remainder Theorem.

Case 2: $\chi \neq 0$ and at least one of r or s is relatively prime with p.

Suppose r is relatively prime with p. Fix an arbitrary $x\in A[p^q]$ . The equation $\phi _l(x)=-ry$ then has exactly one solution y because multiplication by $-r$ acts bijectively on $\hat {A}[p^q]$ .

Now we check that $(x,y)$ is a solution to (3.1): Applying $\phi _m$ , we have $\phi _m\circ \phi _l(x)=-r\phi _m(y)$ . Using Lemma 3.2, we then have $-rsx=-r\phi _m(y)$ . Since x and y are $p^q$ -torsion, multiplication by $-r$ acts bijectively, implying $sx=\phi _m(y)$ .

Thus, for each $x\in A[p^q]$ , there is one $y\in \hat {A}[p^q]$ so that $(x,y)$ is a solution to (3.1). The projection map $(x,y)\mapsto x$ gives an isomorphism from solutions to (3.1) to $A[p^q]\cong ({\mathbb Z}/p^q{\mathbb Z})^4$ .

If s is relatively prime with p, an analogous argument shows there is exactly one solution $(x,y)$ to (3.1) for each $y\in \hat {A}[p^q]$ and that, again, the group of all solutions is isomorphic to $({\mathbb Z}/p^q{\mathbb Z})^4$ .

Cases 1 and 2 have covered all cases where r and s are not both divisible by p. Going forward, we assume $p\mid r$ and $p\mid s$ . If $\operatorname {\mathrm {char}}(k)\neq 0$ , our assumption in Setting 2.4 that $\operatorname {\mathrm {char}}(k)\nmid n$ implies in the following cases that $\operatorname {\mathrm {char}}(k)\neq p$ , and so we may apply Claim 3.4. By the primitivity of the Mukai vector, $n_1=0$ and $n_2$ is equal to the highest power of p dividing $\chi $ .

Let j be the highest power of p dividing r and k be the highest power of p dividing s. If r or s is $0$ , we choose the convention that j or k is $\infty $ .

In each of Cases 3, 4, 5, we handle in stages the situations where q becomes higher and higher relative to j and k. From now on, we assume $j\geq k$ . If $k>j$ , the argument is analogous.

Case 3: $\chi \neq 0$ , $0<k\leq j$ and $q\leq j$ . We observe that $p^q$ is the highest power of p that divides $\chi $ .

Solutions $(x,y)\in (A\times \hat {A})[p^q]$ to the first equation in (3.1) are precisely those where $\phi _l(x)=0$ . By Claim 3.4, the group of such x is isomorphic to $({\mathbb Z}/p^q{\mathbb Z})^2$ .

Fix such an x. We observe that $sx\in A[p^q]$ and $\phi _l(sx)=0$ . By Claim 3.5(a), the preimage of $sx$ under $\phi _m$ in $\hat {A}[p^q]$ is of the form $b+({\mathbb Z}/p^q{\mathbb Z})^2$ for some $b\in \hat {A}[p^q]$ . Thus, there are $p^{4q}$ total solutions.

The projection $(x,y)\mapsto x$ gives a surjective group homomorphism $G_A(v)\twoheadrightarrow ({\mathbb Z}/p^q{\mathbb Z})^2$ . The kernel of this map consists of all solutions where $x=0$ , which by Claim 3.4 is isomorphic to $({\mathbb Z}/p^q{\mathbb Z})^2$ . Since $G_A(v)\leqslant (A\times \hat {A})[p^q]\cong ({\mathbb Z}/p^q{\mathbb Z})^8$ , this short exact sequence shows it must be isomorphic to $({\mathbb Z}/p^q{\mathbb Z})^4$ .

In Cases 4 and 5, we make a reduction argument. We observe that for any $(x,y)\in G_A(v)$ , $(sx,sy)\in G_A(v)\cap (A\times \hat {A})[p^{q-k}]$ . In each of Cases 4 and 5, we will show that the map

(3.3) $$ \begin{align} G_A(v)\xrightarrow{\cdot s} G_A(v)\cap (A\times\hat{A})[p^{q-k}] \end{align} $$

given by multiplication by s is surjective and $p^{4k}$ -to- $1$ . This argument may be repeated to reduce each case to previous cases.

Case 4: $\chi \neq 0$ , $0<k \leq j< q$ and $q\leq j+k$ .

We note that $p^q\mid \chi $ . Since $q-k\leq j$ , the argument in Case 3 shows that

(3.4) $$ \begin{align} G_A(v)\cap(A\times\hat{A})[p^{q-k}]\cong ({\mathbb Z}/p^{q-k}{\mathbb Z})^4. \end{align} $$

Let $(u,v)\in G_A(v)\cap (A\times \hat {A})[p^{q-k}]$ . We seek $(x,y)\in G_A(v)< (A\times \hat {A})[p^{q}]$ , where $(sx,sy)=(u,v)$ . First, we search for elements y, where $\phi _m(y)=u$ and $sy=v$ . Thus, we look at the preimage of u under $\phi _m$ and analyze which of those elements give v when multiplied by s.

Note that $\phi _l(u)=-rv=0$ . Since u is $p^{q-k}$ -torsion, it is also $p^q$ -torsion. So by Claim 3.5(a), the preimage of u under $\phi _m$ in $\hat {A}[p^q]$ is of the form $b+({\mathbb Z}/p^q{\mathbb Z})^2$ , where $b\in \hat {A}[p^{q}]$ . Multiplying by s gives a $p^{2k}$ -to- $1$ map on the following cosets:

$$ \begin{align*} b+({\mathbb Z}/p^q{\mathbb Z})^2\xrightarrow{\cdot s} sb+({\mathbb Z}/p^{q-k}{\mathbb Z})^2. \end{align*} $$

We will now show that v is in the image of this map: the preimage of $su$ under $\phi _m$ in $\hat {A}[p^q]$ is of the form $v+({\mathbb Z}/p^q{\mathbb Z})^2$ . The preimage of $su$ under $\phi _m$ that is $p^{q-k}$ -torsion is thus of the form $v+({\mathbb Z}/p^{q-k}{\mathbb Z})^2$ and has exactly $p^{2(q-k)}$ elements. Now, the elements of $sb+({\mathbb Z}/p^{q-k}{\mathbb Z})^2$ are $p^{q-k}$ -torsion, there are $p^{2(q-k)}$ of them and their image under $\phi _m$ is $su$ . Thus, these sets are equal, implying $v\in sb+({\mathbb Z}/p^{q-k}{\mathbb Z})^2$ . Thus, there are $p^{2k}$ elements $y\in \hat {A}[p^q]$ with the desired properties.

Now we search for elements x, where $\phi _l(x)=-ry=-\frac {r}{s}v$ and $sx=u$ . Note that since $j\geq k$ , $-\frac {r}{s}=\frac {cp^e}{d}$ for some c, d relatively prime with p. We may define multiplying by $\frac {1}{d}$ on p-power torsion points by taking the preimage under multiplication by d since it is a bijection on such points. We examine the preimage of $-\frac {r}{s}v$ under $\phi _l$ and analyze which of those elements give u when multiplied by s.

Note $\phi _m(-\frac {r}{s}v)=-ru=0$ . So by Claim 3.5(b), the preimage of $-\frac {r}{s}v$ under $\phi _l$ is of the form $a+({\mathbb Z}/p^q{\mathbb Z})^2$ , where $a\in {A}[p^{q}]$ .

Multiplying by s gives a $p^{2k}$ -to- $1$ map on the following cosets:

$$ \begin{align*} a+({\mathbb Z}/p^q{\mathbb Z})^2\xrightarrow{\cdot s} sa+({\mathbb Z}/p^{q-k}{\mathbb Z})^2. \end{align*} $$

We will now show that u is in the image of this map: The preimage of $-rv$ under $\phi _l$ in $A[p^q]$ is of the form $u+({\mathbb Z}/p^q{\mathbb Z})^2$ . The preimage of $-rv$ under $\phi _l$ that is $p^{q-k}$ -torsion is thus of the form $u+({\mathbb Z}/p^{q-k}{\mathbb Z})^2$ . The elements of $sa+({\mathbb Z}/p^{q-k}{\mathbb Z})^2$ are $p^{q-k}$ -torsion and their image under $\phi _l$ is $-rv$ . Thus, these sets are equal, implying $u\in sa+({\mathbb Z}/p^{q-k}{\mathbb Z})^2$ . In summary, there are $p^{2k}$ elements $x\in A[p^q]$ , where $\phi _l(x)=-\frac {r}{s}v=-ry$ and $sx=u$ .

This shows that (3.3) is a surjective $p^{4k}$ -to- $1$ map. Since multiplication by s decreases the order of the p-power torsion of an element by exactly $p^k$ , by (3.4) we may conclude that $G_A(v)\cong ({\mathbb Z}/p^q{\mathbb Z})^4$ .

Case 5: $\chi \neq 0$ , $0<k\leq j< q$ and $j+k<q$ . In this case, $p^{j+k}$ divides $\chi $ and no higher powers of p may divide $\chi $ .

By the argument in Case 4, we have

(3.5) $$ \begin{align} G_A(v)\cap(A\times\hat{A})[p^{j+k}]\cong ({\mathbb Z}/p^{j+k}{\mathbb Z})^4. \end{align} $$

We will first extend our result for solutions of order up to $p^{j+2k}$ . For convenience, define $t:=\min \{q,j+2k\}$ .

Let

$$ \begin{align*}(u,v)\in G_A(v)\cap ((A\times\hat{A})[p^{t-k}]\setminus (A\times\hat{A})[p^{j}]). \end{align*} $$

We seek

$$ \begin{align*}(x,y)\in G_A(v)\cap (A\times\hat{A})[p^{t}] \end{align*} $$

so that $(sx,sy)=(u,v)$ . First, we search for elements y, where $\phi _m(y)=u$ and $sy=v$ ; thus, we look at the preimage of u under $\phi _m$ and analyze which of those elements give v when multiplied by s.

If $\phi _l(u)=0$ , then the argument from Case 4 shows that there are $p^{2k}$ elements $y\in \hat {A}[p^{t}]$ , where $\phi _m(y)=u$ and $sy=v$ .

If $\phi _l(u)\in \hat {A}[p^k]\setminus \{0\}$ , then by Claim 3.5(c), the preimage of u under $\phi _m$ in $\hat {A}[p^{t}]$ is of the form $b+({\mathbb Z}/p^{j+k}{\mathbb Z})^2$ , where $b\in \hat {A}[p^{t}]\setminus \hat {A}[p^{j+k}]$ .

Note that $\phi _l(\phi _m(b))=-rv$ and by Lemma 3.2, $-rsb=-(n+rs)b=-rv$ . Multiplication by s gives a $p^{2k}$ -to- $1$ map on the following cosets:

$$ \begin{align*} b+({\mathbb Z}/p^{j+k}{\mathbb Z})^2\xrightarrow{\cdot s}sb+({\mathbb Z}/p^{j}{\mathbb Z})^2. \end{align*} $$

We will now show that v is in the image of this map.

The preimage of $su$ under $\phi _m$ in $\hat {A}[p^{j+k}]$ is of the form $v+({\mathbb Z}/p^{j+k}{\mathbb Z})^2$ . The part of $v+({\mathbb Z}/p^{j+k}{\mathbb Z})^2$ whose image under multiplication by $-r$ is $-rv$ is of the form $v+({\mathbb Z}/p^{j}{\mathbb Z})^2$ . Since $\phi _m$ maps $sb+({\mathbb Z}/p^{j}{\mathbb Z})^2$ to $su$ and multiplying this coset by $-r$ gives $-rv$ , by counting elements, $sb+({\mathbb Z}/p^{j}{\mathbb Z})^2=v+({\mathbb Z}/p^{j}{\mathbb Z})^2$ . Hence, $v\in sb+({\mathbb Z}/p^{j}{\mathbb Z})^2$ . Thus, there are $p^{2k}$ elements $y\in b+({\mathbb Z}/p^{j+k}{\mathbb Z})^2$ , where $\phi _m(y)=u$ and $sy=v$ .

Now we search for elements x where $\phi _l(x)=-ry=-\frac {r}{s}v$ and $sx=u$ . We examine the preimage of $-\frac {r}{s}v$ under $\phi _l$ and analyze which of those elements give u when multiplied by s.

If $\phi _m(-\frac {r}{s}v)=-ru=0$ , we may conclude using the arguments in Case 4. Otherwise, by Claim 3.5(d), the preimage of $-\frac {r}{s}v$ under $\phi _l$ in $A[p^{t}]$ is of the form $a+({\mathbb Z}/p^{j+k}{\mathbb Z})^2$ where $a\in {A}[p^{t}]\setminus {A}[p^{j+k}]$ . Note that $\phi _m(\phi _l(a))=-ru$ and by Lemma 3.2, $-rsa=-ru$ .

Multiplying by s gives a $p^{2k}$ -to- $1$ map on cosets:

$$ \begin{align*} a+({\mathbb Z}/p^{j+k}{\mathbb Z})^2 \xrightarrow{\cdot s} sa+({\mathbb Z}/p^{j}{\mathbb Z})^2. \end{align*} $$

We will now show that u is in the image of this map. The preimage of $-rv$ under $\phi _l$ is of the form $u+({\mathbb Z}/p^{j+k}{\mathbb Z})^2$ . The part of $u+({\mathbb Z}/p^{j+k}{\mathbb Z})^2$ whose image under multiplication by $-r$ is $-ru$ is of the form $u+({\mathbb Z}/p^{j}{\mathbb Z})^2$ . We have shown that $\phi _l$ maps $sa+({\mathbb Z}/p^{j}{\mathbb Z})^2$ to $-rv$ and multiplying this coset by $-r$ gives $-ru$ . By counting elements, we have the set equality $sa+({\mathbb Z}/p^{j}{\mathbb Z})^2=u+({\mathbb Z}/p^{j}{\mathbb Z})^2$ . Hence, $u\in sa+({\mathbb Z}/p^{j}{\mathbb Z})^2$ . Thus, there are $p^{2k}$ elements $x\in a+({\mathbb Z}/p^{j+k}{\mathbb Z})^2$ , where $\phi _l(x)=-\frac {r}{s}v$ and $sx=u$ . Thus, the following map is surjective and $p^{4k}$ -to- $1$ :

(3.6) $$ \begin{align} G_A(v)\cap (A\times\hat{A})[p^{t}]\xrightarrow{\cdot s} G_A(v)\cap (A\times\hat{A})[p^{t-k}]. \end{align} $$

If $q\leq j+2k$ , we may now conclude, in combination with (3.5), that $G_A(v)\cong ({\mathbb Z}/p^q{\mathbb Z})^4$ .

If $q> j+2k$ , (3.6) shows that $G_A(v)\cap (A\times \hat {A})[p^{j+2k}]\cong ({\mathbb Z}/p^{j+2k}{\mathbb Z})^4$ . The above argument may be repeated for solutions of orders up to $p^{j+3k}$ and then upward inductively to conclude that $G_A(v)\cong ({\mathbb Z}/p^{q}{\mathbb Z})^4$ .

Proof. Let m and $m_0$ be the respective Néron-Severi classes determined by $\Phi _P$ . We see from the definition of $\phi _l$ that $\phi _l=d\cdot \phi _{l_0}$ and likewise, $\phi _m=d\cdot \phi _{m_0}$ : if we choose $(L_0,M_0)\in \operatorname {\mathrm {Pic}}^{l_0}(A)\times \operatorname {\mathrm {Pic}}^{m_0}(\hat {A})$ and $L:=L_0^{\otimes d}$ , $M:=M_0^{\otimes d}$ , then for any $x\in A$ , we have

$$\begin{align*}\phi_l(x):=t_x^*L\otimes L^{-1}=t_x^*L_0^{\otimes d}\otimes (L_0^{\otimes d})^{-1} =(t_x^*L_0\otimes L_0^{-1})^{\otimes d} =d\cdot \phi_{l_0}(x). \end{align*}$$

Since $\phi _{l_0}$ and $\phi _{m_0}$ are group homomorphisms, we have, for any $(x,y)\in A \times \hat {A}$ ,

$$\begin{align*}\phi_l(x)=\phi_{l_0}(dx)\quad\text{and}\quad \phi_m(y)=\phi_{m_0}(dy). \end{align*}$$

Thus, a pair $(x,y)\in A\times \hat {A}$ is a solution to (3.7) if and only if

$$\begin{align*}\phi_{l_0}(dx)=-r_0dy \quad\text{and}\quad \phi_{m_0}(dy)=s_0dx, \end{align*}$$

or equivalently, $(dx,dy)$ solves the equations (3.1) given by $v_0$ . By Theorem 3.1, the set of solutions to the equations (3.1) given by $v_0$ is isomorphic to $({\mathbb Z}/n{\mathbb Z})^4\cong G_A(v_0)$ . We may conclude by observing that the set of solutions to (3.7) is given by exactly the elements of $A\times \hat {A}$ that are in $G_A(v_0)$ after being multiplied by d.

Proof. Recall that for any $(x,y)\in A\times \hat {A}$ , applying $L_y\otimes t_x^*$ to an element $\mathcal F\in K_A(v)$ , we have

$$ \begin{align*}\det(L_y\otimes t_x^*\mathcal F)=L_y^{\otimes r}\otimes t_x^*L, \quad\text{and}\quad \det(\Phi_P(L_y\otimes t_x^*\mathcal F))=L_{-x}^{\otimes s}\otimes t_y^*M. \end{align*} $$

We also recall that the morphism $\phi \colon A\times \hat {A} \to A \times \hat {A}$ from (3.2) is an isogeny defined over k and sends sends $(x,y)$ to

$$ \begin{align*}(t_y^*M\otimes M^{-1}\otimes L_{-x}^{\otimes s}, t_x^*L \otimes L^{-1}\otimes L_y^{\otimes r} ). \end{align*} $$

The element $(a,b)$ desired is precisely a preimage of $(L'\otimes L^{-1},M'\otimes M^{-1})\in (A\times \hat {A})(k)$ under $\phi $ . Finally, $L_b\otimes t_a^*: K_A(v) \to K_A(v)'$ is an isomorphism since it has an inverse $L_{-b}\otimes t_{-a}^*$ .

Construction 3.10. Applying $\iota ^*$ gives an isomorphism from $K_A(v)$ to $\operatorname {\mathrm {alb}}^{-1}(\iota ^*L,\iota ^*M)$ . By Lemma 3.9, there is an $(a,b)\in (A\times \hat {A})(k)$ such that $L_b\otimes t_a^*$ maps isomorphically from $\operatorname {\mathrm {alb}}^{-1}(\iota ^*L,\iota ^*M)$ back to $K_A(v)$ . So we have the following automorphism defined over k:

$$ \begin{align*} \kappa\colon K_A(v)&\to K_A(v)\\ \mathcal F&\mapsto L_b\otimes t_a^*\iota^*{\mathcal F}. \end{align*} $$

Proof. Let $\mathcal F\in K_A(v)$ be in the fixed locus of $\iota ^*$ . Pick $(u,w)\in G_A(v)$ so that $2w=y$ and $2u=x$ , which is possible since n is odd. For instance, when $\frac {v^2}{2}=3$ , $K_A(v)$ is a fourfold and the elements of $G_A(v)$ are all three-torsion, so we may choose $(-x,-y)$ .

Then, $L_{w}\otimes t_{u}^*\mathcal F$ must be fixed by the involution

$$\begin{align*}L_{w}\otimes t_{u}^*\iota^*(L_{-w}\otimes t_{-u}^*)=L_{2w}\otimes t_{2u}^*\iota^*. \end{align*}$$

The other direction of the containment is similar, as is the case with $\iota ^*$ replaced by $\kappa $ .

Proof. By Lemma 3.9, there is an $(x,y)\in A\times \hat {A}$ so that $L_y\otimes t_x^*$ gives an isomorphism from $K_A(v)$ to $K_A(v)'$ . The composition

$$ \begin{align*}(L_{-y}\otimes t_{-x}^*)\circ \iota^*\circ (L_y\otimes t_x^*) \end{align*} $$

may be rearranged to $L_{-2y}\otimes t_{-2x}^*\iota ^*$ , where $L_{-2y}\otimes t_{-2x}^*$ gives an isomorphism from $\operatorname {\mathrm {alb}}^{-1}(\iota ^*L,\iota ^*M)$ to $K_A(v)$ . Thus, by the uniqueness statement of Lemma 3.9, there is an element $(u,w)\in G_A(v)$ for which $L_w\otimes t_u^*\circ L_{-2y}\otimes t_{-2x}^*$ is equal to the map $L_{b}\otimes t_{a}^*$ in the definition of $\kappa $ . Then, $L_{-y}\otimes t_{-x}^* \operatorname {\mathrm {Fix}}(\iota ^*)$ is equal to $\operatorname {\mathrm {Fix}}(\kappa _{(-u,-w)})$ , which is isomorphic to $\operatorname {\mathrm {Fix}}(\kappa )$ by Lemma 3.12.

Proof. Suppose $\mathcal F$ is fixed by $\kappa _{(x,y)}$ . We use the equality $t_x\circ \sigma =\sigma \circ t_{\sigma ^{-1}x}$ and the observation that $\sigma $ commutes with $\iota $ and, moreover, $\kappa $ , since $\kappa $ is defined over the ground field k, to simplify the following equation:

$$\begin{align*}\sigma^*\mathcal F \simeq \sigma^*(L_y\otimes t_x^*\kappa\mathcal F) \simeq \sigma^*L_y\otimes \sigma^*t_x^*\kappa\mathcal F \simeq \sigma^*L_y\otimes t_{\sigma^{-1} x}^*\kappa(\sigma^*\mathcal F). \end{align*}$$

Then, we have $\sigma ^*L_y\simeq L_{\sigma ^{-1}y}$ , which we may verify using $\Phi _P\colon D(A)\to D(\hat {A})$ :

$$ \begin{align*} \Phi_P(L_{-\sigma^{-1}y}\otimes \sigma^*L_y) &\simeq t_{-\sigma^{-1}y}^*\sigma^*k(-y)[-g] \simeq \sigma^*t_{-y}^*k(-y)[-g]\\ &\simeq\sigma^*k(0_{\hat{A}})[-g]\simeq k(0_{\hat{A}})[-g].\\[-38pt] \end{align*} $$

Proof. Elements of $(x,y)\in A\times \hat {A}$ act on $M_A(v)$ via $L_{y}\otimes t_{x}^*$ . Abelian varieties are path-connected, so the action of any element in $A\times \hat {A}$ is homotopic to the identity, which implies the induced action on $H^2(M_A(v),{\mathbb Z})$ is trivial. If $(x,y)\in G_A(v)$ , then Theorem 3.1 shows that the action of $L_{y}\otimes t_{x}^*$ restricts to $K_A(v)$ . By [Reference Yoshioka56, Thm. 0.2(2)], the restriction map $H^2(M_A(v),{\mathbb Z})\to H^2(K_A(v),{\mathbb Z})$ is a surjection. Therefore, $L_{y}\otimes t_{x}^*$ acts trivially on $H^2(K_A(v),{\mathbb Z})$ as well.

By [Reference Yoshioka56, Thm. 0.2(2)], there is an isomorphism

$$ \begin{align*} H^2(K_A(v),{\mathbb Z})\cong v^{\perp}, \end{align*} $$

where $v^{\perp } \subset H^{even}(A,{\mathbb Z})$ is the orthogonal complement to v under the Mukai pairing. Since $\iota ^*$ acts by $-1$ on $H^1(A,{\mathbb Z})$ , it acts trivially on $H^{even}(A,{\mathbb Z})$ .

If we assume L and M are symmetric, $\iota ^*$ is an automorphism of $K_A(v)$ and therefore must act trivially on $H^2(K_A(v),{\mathbb Z})$ . If L and M are not both symmetric, since we are working over an algebraically closed field, we observe that $\kappa $ is a composition of translation to an Albanese fiber over symmetric bundles, application of $\iota ^*$ on that fiber and translation back (cf. proof of Proposition 3.13). Thus, $\kappa $ will act trivially on $H^2(K_A(v),{\mathbb Z})$ as well.

By the discussion above, $\ker \nu $ contains $2n^4$ elements, so by Theorem 3.1, we have identified all of them. The automorphisms in this kernel are clearly symplectic as the symplectic form generates part of $H^2(K_A(v),{\mathbb C})$ .

For any nontrivial choice of $(x,y)\in G_A(v)$ , $L_y\otimes t_x^*$ is not an involution, but by Section 3.2, $\kappa _{(x,y)}$ is an involution on $K_A(v)$ .

Finally, suppose $\dim K_A(v)=4$ . By [Reference Kass31, Thm. 7.5(i)], all of the symplectic involutions of $K_A(v)$ act trivially on $H^2(K_A(v),{\mathbb Z})$ .

Proof. Let $(a,\alpha )\in G_{A_{\bar {k}}}(v)$ . By Remark 3.3, to prove that $(b,\beta ):=f_{\bar {k}}(a,\alpha )\in G_{B_{\bar {k}}}(v')$ , it suffices to produce an element $\mathcal H\in D(B)$ , where $v(\mathcal H)=v'$ , $\det (\mathcal H)=\det (L_{\beta }\otimes t_b^*\mathcal H)$ , and $\det (\Phi _P(\mathcal H))=\det (\Phi _P(L_{\beta }\otimes t_b^*\mathcal H))$ .

Under condition (a), for any $\mathcal F\in M_A(v)$ , we may take $\mathcal H:=F(\mathcal F)$ . In this case, we have $L_{\beta }\otimes t_b^*\mathcal H=F(L_{\alpha }\otimes t_a^*\mathcal F)$ . Since

$$ \begin{align*} \det(\mathcal F) = \det( L_{\alpha}\otimes t_a^*\mathcal F) \quad\text{and}\quad \det(\Phi_P(\mathcal F)) = \det(\Phi_P( L_{\alpha}\otimes t_a^*\mathcal F)), \end{align*} $$

condition (a) allows us to conclude that $\mathcal H$ has the needed property.

Under condition (b), F restricts to an isomorphism $M_A(v)\to M_B(v')$ and by the universal property of the Albanese morphism, there is a commutative diagram as follows:

Thus, F satisfies condition (a).

By [Reference Honigs21, Prop. 4.3] if $K_A(v)$ is a fourfold, the intersection of the fixed loci of $\kappa $ and $(L_{\alpha }\otimes t_a^*)\kappa $ acting on $K_A(r,l,s)$ is nonempty. For instance, in $K_2(A)$ , the intersection of $\operatorname {\mathrm {Fix}}(\kappa )$ and $\operatorname {\mathrm {Fix}}(\kappa _{(\tau ,0)})$ where $\tau \in A[3]$ (cf. Lemma 3.12) contains $(0,\tau ,-\tau )$ .

Let $\mathcal G$ be an element in this intersection. It is thus fixed by $L_{\alpha }\otimes t_a^*$ . Following the diagram above, we see that $\mathcal H:=F(\mathcal G)$ is fixed by $L_{\beta }\otimes t_b^*$ and thus, F satisfies the needed condition.

Proof. Let $w:=(1,0,-n)$ for $n:=\frac {v^2}{2}\geq 1$ , and note that

$$ \begin{align*} w^{\perp} = H^2_{\mathrm {\acute{e}t}}(A_{\bar{k}},{\mathbb Q}_{\ell}(1))\oplus {\mathbb Q}_{\ell}\langle (1,0,n)\rangle. \end{align*} $$

We will show that $v^{\perp } \cong w^{\perp }$ . For any $y \in \tilde {H}(A_{\bar {k}}, {\mathbb Q}_{\ell })$ with $y^2\neq 0$ , let reflection through y be given by

$$\begin{align*}x \mapsto x - \frac{2\langle x,y\rangle}{y^2} y.\end{align*}$$

Observe that $(v-w)^2\neq 0$ or $(v+w)^2 \neq 0$ , and so reflection through $v-w$ or $v+w$ gives an isometry $\tilde {H}(A_{\bar {k}}, {\mathbb Q}_{\ell }) \xrightarrow {\sim } \tilde {H}(A_{\bar {k}}, {\mathbb Q}_{\ell })$ which sends v to $\pm w$ . Thus, the isometry restricts to a Galois equivariant isomorphism $v^{\perp } \xrightarrow {\sim } w^{\perp }$ .

Proof. By [Reference Yoshioka56, Thm. 0.2(2)], along with the comparison theorem for singular and étale cohomology and the smooth base change theorem, we have a Galois equivariant isomorphism $H^2_{\mathrm {\acute{e}t}}(K_A(v)_{\bar {k}},{\mathbb Q}_{\ell }(1))\cong v^{\perp }$ (In fact, this isomorphism exists over ${\mathbb Z}_{\ell }$ , while the isomorphism of Lemma 5.6 may only exist over ${\mathbb Q}_{\ell }$ ). This combined with Lemma 5.6 gives the result.

Proof of Proposition 5.5

Suppose that $K_A(v)$ and $K_B(v')$ are derived equivalent, so they have isomorphic sums of even cohomologies after Tate twists [Reference Honigs, Achter, Casalaina-Martin, Honigs and Vial22, Lem. 3.1]: $\widetilde {H}(K_A(v)_{\bar {k}},{\mathbb Q}_{\ell })\cong \widetilde {H}(K_B(v')_{\bar {k}},{\mathbb Q}_{\ell })$ . We know that the zeroth and top cohomologies of $K_A(v)$ and $K_B(v')$ are trivial Galois representations, and Lemma 5.7 gives that

$$\begin{align*}H^2_{\mathrm {\acute{e}t}}(K_A(v)_{\bar{k}}, {\mathbb Q}_{\ell}(1))\cong H^2_{\mathrm {\acute{e}t}}(A_{\bar{k}}, {\mathbb Q}_{\ell}(1))\oplus {\mathbb Q}_{\ell}. \end{align*}$$

By Theorem 4.4 and Poincáre duality (cf. [Reference Hosono, Lian, Oguiso and Yau23]), it follows that there is an isomorphism of Galois modules

$$\begin{align*}\widetilde{H}(K_A(v),{\mathbb Q}_{\ell})\cong {\mathbb Q}_{\ell}^{\oplus 4} \oplus H^2_{\mathrm {\acute{e}t}}(A_{\bar{k}}, {\mathbb Q}_{\ell}(1))^{\oplus 2} \oplus \operatorname{\mathrm{Sym}}^2H^2_{\mathrm {\acute{e}t}}(K_A(v)_{\bar{k}},{\mathbb Q}_{\ell}(1)) \oplus V_A,\end{align*}$$

where $V_A:=V$ from Theorem 4.4. There is a similar isomorphism for $\widetilde {H}(K_B(v'),{\mathbb Q}_{\ell })$ involving $V_B$ . We will check that these representations are semisimple, so that we can reduce to a comparison of $V_A$ and $V_B$ .

By [Reference Fogarty13, Thm. 3] and its extension to finitely generated fields of characteristic 0 in [Reference Zarhin57, Thm. 4.3], $H^2_{\mathrm {\acute{e}t}}(A_{\bar {k}}, {\mathbb Q}_{\ell })$ is a semisimple representation, and thus so is $\operatorname {\mathrm {Sym}}^2H^2_{\mathrm {\acute{e}t}}(K_A(v)_{\bar {k}},{\mathbb Q}_{\ell }(1))$ . The $\operatorname {\mathrm {Gal}}({\bar {k}}/k)$ -representation ${\mathbb Q}_{\ell }[G_{A_{\bar {k}}}(v)]$ factors through a finite group representation, determined by the finite extension of k over which $G_{A_{\bar {k}}}(v)$ is defined, and so by Maschke’s theorem, it is also semisimple. Thus, the representation $\widetilde {H}(K_A(v),{\mathbb Q}_{\ell })$ is semisimple. The same also holds for $\widetilde {H}(K_B(v'),{\mathbb Q}_{\ell })$ , so applying Schur’s Lemma, this allows us to cancel isomorphic representations in the direct sums for $\widetilde {H}(K_A(v),{\mathbb Q}_{\ell })$ and for $\widetilde {H}(K_B(v'),{\mathbb Q}_{\ell })$ . Since A and B are isogenous, there is an isomorphism $H^2_{\mathrm {\acute{e}t}}(A_{\bar {k}},{\mathbb Q}_{\ell })\cong H^2_{\mathrm {\acute{e}t}}(B_{\bar {k}},{\mathbb Q}_{\ell })$ , so along with the observations above, we are reduced to an isomorphism $V_A\cong V_B$ . This extends to an isomorphism ${\mathbb Q}_{\ell }[G_{A_{\bar {k}}}(v)] \cong {\mathbb Q}_{\ell }[G_{B_{\bar {k}}}(v')]$ , as desired.

Proof. We have $G_{A_{\bar {k}}}(1,0,-3)=A[3]$ and by the discussion in Example 5.4(a), $G_{A_{\bar {k}}}(3,0,-1)=G_{\hat {A}_{\bar {k}}}(1,0,-3)=\hat {A}[3]$ . The result then follows by Proposition 5.5.