On birational transformations of Hilbert schemes of points on K3 surfaces

Abstract

We classify the group of birational automorphisms of Hilbert schemes of points on algebraic K3 surfaces of Picard rank one. We study whether these automorphisms are symplectic or non-symplectic and if there exists a hyperkähler birational model on which they become biregular. We also present new geometrical constructions for some of these automorphisms.

Appendix A. Pell’s equations

A generalized Pell’s equation is a quadratic diophantine equation

$$\begin{aligned} P_r(m): X^2 - rY^2 = m \end{aligned}$$

in the unknowns $X, Y \in \mathbb {Z}$, for $r \in \mathbb {N}$ and $m \in \mathbb {Z}{\setminus }\left\{ 0\right\}$. If $m=1$, the equation is called standard. A solution (X, Y) is called positive if $X > 0$, $Y > 0$. The minimal solution of the equation is the positive solution with smallest X. We also use the expression “minimal solution with a property P” to denote the positive solution with smallest X among those which satisfy the property P.

Two solutions (X, Y), $(X',Y')$ of $P_r(m)$ are said to be equivalent if

$$\begin{aligned} \frac{XX' - rYY'}{m} \in \mathbb {Z}, \qquad \frac{XY' - X'Y}{m} \in \mathbb {Z}. \end{aligned}$$

(5)

If the equation is standard, then solutions exist and they are all equivalent.

Inside any equivalence class of solutions, the fundamental solution (X, Y) is the one with smallest non-negative Y, if unique. Otherwise, the smallest non-negative value of Y is realized by two conjugate solutions (X, Y), $(-X,Y)$: in this case, the fundamental solution of the class will be (X, Y) with $X > 0$. If (X, Y) is a fundamental solution of $P_r(m)$, all other solutions $(X',Y')$ in the same equivalence class are of the form

$$\begin{aligned} {\left\{ \begin{array}{ll} X' = aX + rbY \\ Y' = bX + aY \end{array}\right. } \end{aligned}$$

(6)

where (a, b) is a solution of $P_r(1)$.

Let (z, w) be the minimal solution of $P_r(1)$ and $m \in \mathbb {N}$. By [17, Theorem 108], if (X, Y) is a fundamental solution of $P_r(m)$ then:

$$\begin{aligned} 0 < \left| X \right| \le \sqrt{\frac{(z+1)m}{2}}, \qquad 0 \le Y \le w\sqrt{\frac{m}{2(z+1)}}. \end{aligned}$$

(7)

We deduce that the half-open interval $I = \left[ (\sqrt{m}, 0), (z\sqrt{m}, w\sqrt{m})\right)$ on the hyperbola $X^2-rY^2 = m$ contains exactly one solution for each equivalence class of $P_r(m)$. Indeed, if (X, Y) is a fundamental solution with $X > 0$ it is contained in the interval by (7), while if $X < 0$ it is easy to check that the equivalent solution $(X', Y') = (-zX - rwY, -wX - zY)$ lies in I. Moreover, by direct computation using (5), two distinct solutions in I cannot belong to the same equivalence class. Notice that the solutions (X, Y) of $P_r(m)$ contained in I are all the solutions with $X > 0$ and $0 \le \frac{Y}{X} < \frac{w}{z}$.

We often make use of the following lemma.

Lemma A.1

For $t \ge 1$ and $n \ge 2$ such that $t(n-1)$ is not a square, let (z, w) be the minimal solution of $P_{t(n-1)}(1)$ with $z \equiv \pm 1 \pmod {n-1}$. Assume $w \equiv 0 \pmod {2}$.

• (i) $z \equiv 1 \pmod {2(n-1)}$ and $z \equiv 1 \pmod {2t}$ if and only if (z, w) is not the minimal solution of $P_{t(n-1)}(1)$.

• (ii) $z \equiv -1 \pmod {2(n-1)}$ and $z \equiv -1 \pmod {2t}$ if and only if $P_{t(n-1)}(-1)$ is solvable.

• (iii) If $z \equiv 1 \pmod {2(n-1)}$ and $z \equiv -1 \pmod {2t}$ then the equation $(n-1)X^2 - tY^2 = -1$ has integer solutions; if $t \ge 2$ the converse also holds.

• (iv) If $z \equiv -1 \pmod {2(n-1)}$ and $z \equiv 1 \pmod {2t}$ then the equation $(n-1)X^2 - tY^2 = 1$ has integer solutions; if $n \ge 3$ the converse also holds.

Proof

Let $w = 2m$ for $m \in \mathbb {N}$.

• (i)–(ii) Write $z = 2(n-1)p \pm 1 = 2tq \pm 1$ for $p,q \in \mathbb {N}$. Then $p((n-1)p \pm 1) = tm^2$ and $(n-1)p = tq$, hence $r :=\frac{p}{t} \in \mathbb {N}$. We have $r((n-1)rt \pm 1) = m^2$ and it follows that there exist $s,u \in \mathbb {N}$ such that $r = s^2$, $(n-1)tr \pm 1 = u^2$, $m = su$, hence $u^2 - t(n-1)s^2 = \pm 1$. If the sign is $+$, notice that $u < z$, therefore (z, w) is not the minimal solution of $P_{t(n-1)}(1)$. Conversely, assume first that $P_{t(n-1)}(-1)$ is solvable and let (a, b) be the minimal solution. Then by [8, Lemma A.2] the minimal solution of $P_{t(n-1)}(1)$ is $(z,w) = (2t(n-1)b^2 - 1, 2ab)$, which satisfies $z \equiv -1 \pmod {2(n-1)}$ and $z \equiv -1 \pmod {2t}$. Similarly, if the minimal solution (u, s) of $P_{t(n-1)}(1)$ does not satisfy $u \equiv \pm 1 \pmod {n-1}$, then by (6) we have $(z,w) = (2t(n-1)s^2+1,2us)$, hence $z \equiv 1 \pmod {2(n-1)}$ and $z \equiv 1 \pmod {2t}$.

• (iii)–(iv) Write $z = 2(n-1)p \pm 1 = 2tq \mp 1$ for $p,q \in \mathbb {N}$. Then $p((n-1)p \pm 1) = tm^2$ and $(n-1)p \pm 1 = tq$, hence $pq = m^2$. Since $\gcd (p,q) = 1$, there exist $s,u \in \mathbb {N}$ such that $p = s^2$, $q = u^2$ and $m = su$, thus $(n-1)s^2 - tu^2 = \mp 1$. Vice versa, let (a, b) be the integer solution of $(n-1)a^2 - tb^2 = \pm 1$ with smallest $a > 0$. By [8, Lemma A.2], the assumption $t \ge 2$ (if the sign is −) or $n \ge 3$ (if the sign is $+$) implies that the minimal solution of $P_{t(n-1)}(1)$ is $(z,w) = (2(n-1)a^2 \mp 1, 2ab)$, which satisfies $z \equiv \mp 1 \pmod {2(n-1)}$ and $z \equiv \pm 1 \pmod {2t}$.

$\square$

The next two lemmas are used in Sect. 4. They give bounds for the number of equivalence classes of solutions of $P_{8t}(9)$ and $P_{8t}(12)$, depending on $t \in \mathbb {N}$.

Lemma A.2

If $t \equiv 1 \pmod {3}$ or $t \equiv 3,6 \pmod {9}$, then all solutions of $P_{8t}(9)$ are equivalent to $(X,Y) = (3,0)$. If $t \equiv 0 \pmod {9}$, then $P_{8t}(9)$ has either one, two or three classes of solutions. If $t \equiv 2 \pmod {3}$, then $P_{8t}(9)$ has either one or three classes of solutions.

Proof

If $t = 3q+1$, for $q \in \mathbb {N}_0$, then $X^2 \equiv 2Y^2 \pmod {3}$, hence all solutions (X, Y) are of the form $(3X',3Y')$, with $(X')^2 - 8t(Y')^2 = 1$. This is now a standard Pell’s equation, which has only one class of solutions, thus the same holds for $P_{8t}(9)$.

If $t = 3q$ for $q \in \mathbb {N}$ and $(q,3) = 1$, then $X = 3X'$ for some $X' \in \mathbb {Z}$ such that $3(X')^2 - 8qY^2 = 3$. Since $3 \not \mid q$ we need $(X,Y) = (3X', 3Y')$ with $(X')^2 - 24q(Y')^2 = 1$. There exists only one class of solutions $(X',Y')$ for this standard Pell’s equation, thus also the solutions of $P_{8t}(9)$ form a single class.

Assume now that $t = 9q$ for $q \in \mathbb {N}$. Then a solution (X, Y) of $P_{8t}(9)$ is of the form $(X,Y) = (3X',Y)$ with $(X')^2 - 8qY^2 = 1$. Let (z, w) be the minimal solution of $P_{8t}(1): z^2 - 8tw^2 = 1$. Notice that the solutions of $P_{8t}(1)$ are the pairs $(X', \frac{Y}{3})$ for all solutions $(X',Y)$ of $(X')^2 - 8qY^2 = 1$ such that $Y \equiv 0 \pmod {3}$. Let (a, b) be the minimal solution of $(X')^2 - 8qY^2 = 1$. Since this is a standard Pell’s equation, by (6) its next two solutions (for increasing values of $X'$) are $(a^2 + 8qb^2,2ab)$ and $(a^3+24qab^2, 8qb^3+3a^2b)$. We observe that one among these first three positive solutions $(X',Y)$ has $Y \equiv 0 \pmod {3}$. The positive solution $(X',Y)$ with this property and smallest $X'$ is therefore equal to (z, 3w), thus the corresponding solution $(X,Y) = (3X',Y)$ of $P_{8t}(9)$ satisfies $\frac{Y}{X} = \frac{w}{z}$, i.e. it is the first positive solution in the same equivalence class of (3, 0). We conclude that $P_{8t}(9)$ has either one, two or three classes of solutions.

Finally, assume that $t \equiv 2 \pmod {3}$. Let (z, w) be the minimal solution of $P_{8t}(1)$ and let $(u_1,v_1)$, $(u_2,v_2)$ be two positive solutions of $P_{8t}(9)$ such that, for $i=1,2$:

$$\begin{aligned} 0< u_i \le 3\sqrt{\frac{z+1}{2}}, \quad 0 < v_i \le \frac{3w}{\sqrt{2(z+1)}}. \end{aligned}$$

(8)

By (7), this is equivalent to asking that either $(u_i, v_i)$ or $(-u_i, v_i)$ is a fundamental solution of $P_{8t}(9)$, different from (3, 0). Thus $u_1,v_1,u_2,v_2$ are not divisible by 3. From $u_1^2 - 8tv_1^2 = 9$ and $u_2^2 - 8tv_2^2 = 9$ we get

$$\begin{aligned} u_1v_2 \equiv \pm u_2v_1 \pmod {9}, \quad u_1u_2 \equiv \pm 8t v_1v_2 \pmod {9} \end{aligned}$$

(9)

where the signs in the two congruences coincide. If we now multiply $u_1^2 - 8tv_1^2 = 9$ and $u_2^2 - 8tv_2^2 = 9$ member by member, we obtain

$$\begin{aligned} \left( \frac{u_1u_2 \mp 8tv_1v_2}{9} \right) ^2 - 8t\left( \frac{u_1v_2 \mp u_2v_1}{9} \right) ^2 = 1 \end{aligned}$$

(10)

where by (9) the two squares in the LHS term are integers. If we assume that the pairs $(u_1,v_1)$ and $(u_2,v_2)$ are distinct, then $u_1v_2 \mp u_2v_1 \ne 0$. Since (z, w) is the minimal solution of $P_{8t}(1)$, from (10) we have $\left| u_1v_2 \mp u_2v_1 \right| \ge 9w$. However, from (8) we compute $\left| u_1v_2 \mp u_2v_1 \right| < 9w$, which is a contradiction. We conclude $u_1 = u_2$, $v_1 = v_2$. Thus, there are at most three classes of solutions for $P_{8t}(9)$: the class of (3, 0) and possibly the classes of $(u_1,v_1)$ and $(-u_1,v_1)$. Notice that the latter two classes are always distinct: in order for them to coincide we would need

$$\begin{aligned} \frac{u_1^2 + 8tv_1^2}{9} \in \mathbb {Z}, \qquad \frac{2u_1v_1}{9} \in \mathbb {Z}\end{aligned}$$

which does not happen, since $\gcd (u_1,3) = \gcd (v_1,3) = 1$. Hence, the equation has either one or three classes of solutions. $\square$

In an entirely similar way one can prove the following.

Lemma A.3

If $t \equiv 3 \pmod {18}$, then $P_{8t}(12)$ is either not solvable or it has one class of solutions. If $t \equiv 5,11,17 \pmod {18}$, then $P_{8t}(12)$ is either not solvable or it has two classes of solutions. In all other cases, $P_{8t}(12)$ is not solvable.

All cases in Lemma A.2 and Lemma A.3 occur, for suitable values of t.