2·1. The twistor space of a complex K3 surface

A complex K3 surface X admits a sphere of different complex structures. More explicitly, one can attach a set of different complex structures ${I_\zeta }$ to the underlying differentiable manifold X, so that $\left( {X,{I_\zeta }} \right)$ is again a K3 surface, and these structures are parameterised by a sphere $\zeta \in {S^2} \simeq \mathbb{P}_\mathbb{C}^1$ . These K3 surfaces can be patched together into a 3-dimensional complex manifold $\mathcal{X}$ , called twistor space (see [ Reference Hitchin, Karlhede, Lindström and RočekHKLR87 , section 3·F], [ Reference HitchinHit92 ] or [ Reference JoyceJoy00 , chapter 7] for more on the twistor space and the details of its construction). $\mathcal{X}$ comes with a holomorphic map $\mathcal{X} \to \mathbb{P}_\mathbb{C}^1$ with the property that, for $\zeta \in \mathbb{P}_\mathbb{C}^1 \simeq {S^2}$ , the fibre ${\mathcal{X}_\zeta }$ is the K3 surface $\left( {X,{I_\zeta }} \right)$ . In fact, this construction is non-canonical, and depends on the choice of a Kähler class of X (any K3 surface is Kähler, [ Reference SiuSiu83 ]). The original K3 surface corresponds to one of the poles of $\mathbb{P}_\mathbb{C}^1 \simeq {S^2}$ .

Even if the original K3 surface X is algebraic (or equivalently, projective), $\mathcal{X}$ is no longer algebraic, and has to be thought of as of transcendental nature. The twistor space of a K3 surface plays an important role in several situations. Among all, it provides an example of K3-fibration over a compact base, and it is used in a modern proof of Torelli Theorem (any two K3 surfaces are connected by a finite path of twistor lines, see [ Reference HuybrechtsHuy16 , section 7·3]).

2·2. Hodge structures of K3 type

By polarised Hodge structure of K3 type we mean the data of a vector space T over $\mathbb{Q}$ of dimension $r \geq 2$ , endowed with a symmetric bilinear form $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ of signature $\left( {2,r - 2} \right)$ , and a decomposition

\begin{align*}{T_\mathbb{C}} = {T^{2,0}} \oplus {T^{1,1}} \oplus {T^{0,2}}\end{align*}

such that the $\mathbb{C}$ -linear extension of $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ satisfies:

1. (i) the subspaces ${T^{1,1}}$ and ${T^{2,0}} \oplus {T^{0,2}}$ are orthogonal;

2. (ii) $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ is positive definite on ${P_T} = \left( {{T^{2,0}} \oplus {T^{0,2}}} \right) \cap {T_\mathbb{R}}$ and ${T^{2,0}},{T^{0,2}} \subseteq {T_\mathbb{C}}$ are isotropic;

3. (iii) complex conjugation on ${T_\mathbb{C}}$ preserves ${T^{1,1}}$ and exchanges ${T^{2,0}}$ and ${T^{0,2}}$ ;

4. (iv) ${\rm{di}}{{\rm{m}}_\mathbb{C}}^{2,0} = 1$ .

We will denote by $\sigma $ a $\mathbb{C}$ -generator of ${T^{2,0}}$ . Note that $\bar \sigma $ generates ${T^{0,2}}$ , and the required conditions give $\left( {\sigma .\sigma } \right) = 0$ , $\left( {\sigma .\bar \sigma } \right) \gt 0$ , ${(\Re \!\left( \sigma \right)\!)^2} = {(\Im \!\left( \sigma \right)\!)^2}$ and $\left( {\Re \!\left( \sigma \right).\Im \!\left( \sigma \right)} \right) = 0$ . The plane ${P_T}$ is considered with the orientation given by the basis $\left\{ {\Re \!\left( \sigma \right),\Im \!\left( \sigma \right)} \right\}$ . We will also assume that T is irreducible.

Define ${K_T}$ to be the ring of endomorphism of the Hodge structure T. As T is irreducible, ${K_T}$ is a division algebra. It was pointed out by Zarhin [ Reference ZarhinZar83 ] that ${K_T}$ is indeed a number field, endowed with an embedding ${K_T} \to \mathbb{C}$ . This embedding is defined in the following way: as $\varphi $ preserves the $\left( {2,0} \right)$ -part of T, $\varphi \!\left( \sigma \right) = \varphi \cdot \sigma $ , where we identify $\varphi $ with a scalar in $\mathbb{C}$ . Zarhin proved also that ${K_T}$ is either totally real or complex multiplication (CM). Notice that T is naturally a ${K_T}$ -module and, therefore, it becomes a ${K_T}$ -vector space.

The last condition seems technical, but turns out to be extremely useful in several situations. In fact, one can define the CM type case by requiring the sole condition ${\rm{di}}{{\rm{m}}_{{K_T}}}T = 1$ to be satisfied. Indeed, van Geemen [ Reference van GeemenvGe08 , lemma 3·2] proved that, whenever ${K_T}$ is totally real, ${\rm{di}}{{\rm{m}}_{{K_T}}}T \geq 3$ holds. Of course, the condition ${\rm{di}}{{\rm{m}}_{{K_T}}}T = 1$ is equivalent to $\left[ {{K_T}\;:\;\mathbb{Q}} \right] = r$ .

As the form $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ is non-degenerate, we can define the transpose $\varphi '$ of $\varphi $ by the condition

\begin{align*}\!\left( {\gamma .\varphi \!\left( \delta \right)} \right) = \left( {\varphi '\!\left( \gamma \right).\delta } \right),\end{align*}

for any $\gamma ,\delta \in T$ . $\varphi '$ is in fact an element of ${K_T}$ , and corresponds to the complex conjugate of $\varphi $ via the embedding ${K_T} \hookrightarrow \mathbb{C}$ , that is: $\varphi '\!\left( \sigma \right) = \bar \varphi \cdot \sigma $ (see [ Reference HuybrechtsHuy20 , remark 2·6] or [ Reference HuybrechtsHuy16 , chapter 3]). In particular, an element $\varphi \in {K_T}$ is an isometry for $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ if and only if its image in $\mathbb{C}$ has unitary norm. We will denote by $K_T^0 = {K_T} \cap \mathbb{R}$ the real part of ${K_T}$ , once seen ${K_T}$ as a subfield of $\mathbb{C}$ via the prescribed embedding. Notice that $K_T^0$ can be characterised by the subfield of ${K_T}$ of self-transpose endomorphisms.

The period field ${k_T}$ of T is defined in the following way. Let $\gamma $ vary in T. Among the periods $\left( {\sigma .\gamma } \right)$ , at least one is not zero, say for $\tilde \gamma $ , since $\sigma \ne 0$ and $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ is non-degenerate.Footnote ¹ The period field is defined to be the subfield of $\mathbb{C}$ generated over $\mathbb{Q}$ by the quotients $\left( {\sigma .\gamma } \right)/\left( {\sigma .\tilde \gamma } \right)$ . Clearly, it is enough to consider these quotients only for $\gamma $ varying in a $\mathbb{Q}$ -basis of T. For a proof of the following result, see [ Reference HuybrechtsHuy20 , lemma 2·10].

2·3. Useful facts on CM fields

First of all, we recall a result proven by Blanksby and Loxton [ Reference Blanksby and LoxtonBL78 ].

We now present other elementary properties carried by CM fields.

This Lemma assumes a particular form in the case of a CM field.

Proof. It is enough to show that there exist elements $\alpha \in E$ lying on the circle of arbitrary small non-zero argument (density follows taking powers of these elements). Let $\delta $ be an element of $E\backslash \mathbb{R}$ . For $r \in \mathbb{Q}$ , define

\begin{align*}{\alpha _r} = \frac{{\delta + r}}{{\bar \delta + r}} \in E.\end{align*}

Then $\left| {{\alpha _r}} \right| = 1$ and ${\alpha _r}$ has arbitrary small non-zero argument, since ${\rm{li}}{{\rm{m}}_{r \to \infty }}{\alpha _r} = 1$ .

4·1. Excessive Picard jump values

We have already remarked that the original extended Hodge structure of K3 type on $T \oplus \mathbb{Q}\ell $ is no longer irreducible, being $\ell $ a $\left( {1,1} \right)$ -class. Given $z \in \mathbb{P}_\ell ^1$ , we define the Picard number ${\rho _z}$ of the Hodge structure corresponding to z to be the $\mathbb{Q}$ -dimension of the space of $\left( {1,1} \right)$ -classes of $T \oplus \mathbb{Q}\ell $ , that is

\begin{align*}{\rho _z} = {\rm{di}}{{\rm{m}}_\mathbb{Q}}\!\left( {P_z^ \bot \cap \!\left( {T \oplus \mathbb{Q}\ell } \right)} \right).\end{align*}

For instance, as the original T is irreducible, we deduce that ${\rho _x} = {\rho _{\bar x}} = 1$ .

We are interested in understanding how points of Picard jump distribute on the sphere $\mathbb{P}_\ell ^1 \simeq S_\ell ^2$ , and the possible relative Picard numbers. Huybrechts proved the following result [ Reference HuybrechtsHuy20 , proposition 3·2].

The fact that the Noether–Lefschetz locus is countable is a particular instance of a more general fact (see [ Reference HuybrechtsHuy16 , chapter 6, proposition 2·9]). However, there is a very simple argument that applies in this case. Assume that z is of Picard jump, and let $\ell ' \in T \oplus \mathbb{Q}\ell $ be a $\left( {1,1} \right)$ -class for the Hodge structure induced by z. Then ${\ell ^{{\rm{\prime}} \bot }}$ identifies a line in $\mathbb{P}\!\left( {{T^{2,0}} \oplus {T^{0,2}} \oplus \mathbb{C}\ell } \right)$ , that cuts the conic $\mathbb{P}_\ell ^1$ in two points: z and $\bar z$ . Thus, $\ell '$ is a $\left( {1,1} \right)$ -class only for finitely many $z \in \mathbb{P}_\ell ^1$ , and therefore the Noether–Lefschetz locus is countable, as $T \oplus \mathbb{Q}\ell $ is.

Assume that $z = \left[ {a\sigma + b\bar \sigma + c\ell } \right]$ . The condition defining ${\ell ^{{\rm{\prime}} \bot }}$ can be translated as

(4·1) \begin{align}a\!\left( {\sigma .\ell '} \right) + b\!\left( {\bar \sigma .\ell '} \right) + \left( {\ell .\ell '} \right) = 0.\end{align}

Lemma 4·3. Assume that $\ell ' \in T \oplus \mathbb{Q}\ell $ , $\ell ' \notin \mathbb{Q}\ell $ and choose z to be one of the two points orthogonal to $\ell '$ . Then the point z is contained in the equator $S_\ell ^1$ if and only if $\ell ' \in T$ .

Proof. Write $z = \left[ {\sigma '} \right]$ . Assume that $z \in S_\ell ^1$ , i.e. $\ell \in {P_z} = {\langle \Re \!\left( {\sigma '} \right),\Im \!\left( {\sigma '} \right)\rangle _\mathbb{R}} \subseteq {\langle \sigma ',\overline {\sigma '} \rangle _\mathbb{C}}$ . Then $\ell = a\sigma ' + b\overline {\sigma '} $ for some $a,b \in \mathbb{C}$ ; hence $\left( {\ell .\ell '} \right) = 0$ since $\left( {\sigma '.\ell '} \right) = 0$ , which means $\ell ' \in T$ . Conversely, assume that $\ell ' \in T$ , i.e. $\left( {\ell .\ell '} \right) = 0$ . Write $\sigma ' = a\sigma + b\bar \sigma + c\ell $ . After rescaling $\sigma $ , we may assume that $\left( {\sigma .\ell '} \right) = 1$ , so that $\left( {\bar \sigma .\ell '} \right) = \overline {\!\left( {\sigma .\ell '} \right)} = 1$ as well. The condition $\left( {\sigma '.\ell '} \right) = 0$ implies,

\begin{align*}a\!\left( {\sigma .\ell '} \right) + b\!\left( {\bar \sigma .\ell '} \right) + c\!\left( {\ell .\ell '} \right) = 0,{\rm{\;i}}.{\rm{e}}.{\rm{\;}}b = - a.\end{align*}

Therefore, the condition (3·1)

\begin{align*} - 2{a^2}\!\left( {\sigma .\bar \sigma } \right) + {c^2}d = 0\end{align*}

implies $c \ne 0$ (and therefore we may assume $c = 1$ ) and $a \in \mathbb{R}$ (since $d \gt 0$ and $\left( {\sigma .\bar \sigma } \right) \gt 0$ ). Therefore

\begin{align*}\overline {\sigma '} = a\bar \sigma - a\sigma + \ell \end{align*}

and

\begin{align*}\ell = \frac{{\sigma ' + \overline {\sigma '} }}{2} = \Re \!\left( {\sigma '} \right)\end{align*}

and the proof is concluded, as $\ell \in {P_z}$ .

If the original Hodge structure T has CM, all points of Picard jump on the equator share the same Picard number.

Proof. Firstly, note that $P_z^ \bot \cap \!\left( {T \oplus \mathbb{Q}\ell } \right)$ is all contained in T. This follows from Lemma 4·3 applied to any non-zero element $\ell ' \in P_z^ \bot \cap \!\left( {T \oplus \mathbb{Q}\ell } \right)$ . Hence $P_z^ \bot \cap \!\left( {T \oplus \mathbb{Q}\ell } \right) = P_z^ \bot \cap T$ . The key remark is that $P_z^ \bot \cap T$ admits an action of the real part $K_T^0 = {K_T} \cap \mathbb{R}$ of the endomorphism field ${K_T}$ . To show this, choose $\beta \in K_T^0$ and $\ell ' \in P_z^ \bot \cap T$ . Write $z = \left[ {\sigma ' = a\sigma + b\bar \sigma + \ell } \right]$ . Then, using that $\ell $ is orthogonal to T, we obtain ( $\beta \in K_T^0$ is

self-adjoint for $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ )

\begin{align*}\!\left( {\sigma '.\beta \!\left( {\ell '} \right)} \right) & = \left( {a\sigma + b\bar \sigma + \ell .\beta \!\left( {\ell '} \right)} \right)\\ &= \left( {a\sigma + b\bar \sigma .\beta \!\left( {\ell '} \right)} \right)\\ &= \left( {\beta \!\left( {a\sigma + b\bar \sigma } \right).\ell '} \right)\\ & = a\!\left( {\beta \cdot \sigma .\ell '} \right) + b\!\left( {\beta \cdot \bar \sigma .\ell '} \right)\\ & = \beta \cdot \left( {a\sigma + b\bar \sigma + \ell .\ell '} \right)\\ & = \beta \cdot \left( {\sigma '.\ell '} \right) = 0.\end{align*}

As a consequence, $P_z^ \bot \cap T$ is a $K_T^0$ -vector space. As $\left[ {K_T^0\;:\;\mathbb{Q}} \right] = r/2$ (we are in the CM case by assumption, so that $\left[ {{K_T}\;:\;K_T^0} \right] = 2$ and $\left[ {{K_T}\;:\;\mathbb{Q}} \right] = r$ ), ${\rm{di}}{{\rm{m}}_{K_T^0}}P_z^ \bot \cap T$ can only take three different values: $0,1,2$ . We immediately exclude the case ${\rm{di}}{{\rm{m}}_{K_T^0}}P_z^ \bot \cap T = 0$ since z is a point of Picard jump. Assume by contradiction that ${\rm{di}}{{\rm{m}}_{K_T^0}}P_z^ \bot \cap T = 2$ . In fact, this equality would imply $P_z^ \bot \cap T = T$ . This would force $\sigma '$ to belong to $\mathbb{C}\ell $ , or equivalently to have zero ${T_\mathbb{C}}$ -part (indeed the form $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ in non-degenerate on T, and therefore on ${T_\mathbb{C}}$ ). On the other hand, this is a contradiction, since any non-zero $\sigma ' \in \mathbb{C}\ell $ does not satisfy (3·1). In conclusion, ${\rm{di}}{{\rm{m}}_{K_T^0}}P_z^ \bot \cap T = 1$ , or equivalently ${\rm{di}}{{\rm{m}}_\mathbb{Q}}P_z^ \bot \cap T = \left[ {K_T^0\;:\;\mathbb{Q}} \right] = r/2$ , i.e. ${\rho _z} = r/2$ .

The following result enriches Proposition 4·2. To prove it, it is not necessary to assume that T has CM, or that ${K_T}$ is CM. However, in Section 5·2 we will give a better description of the distribution of points of Picard jump on the equator in the CM case.

Proof. Thanks to Lemma 3·1, a point on the equator $S_\ell ^1 \subseteq \mathbb{P}_\ell ^1$ corresponds via the isomorphism $\mathbb{P}_\ell ^1 \simeq S_\ell ^2$ to the normalisation of the vector

\begin{align*} v\!\left( {a,b} \right) = \left( {\Re \!\left( {b - \bar a} \right),\Im \!\left( {b - \bar a} \right),0} \right).\end{align*}

Lemma 3·1 also gives a condition on z to belong to $S_\ell ^1$ , namely $\left| a \right| = \left| b \right|$ . Write R to denote these absolute values; then $a = R{e^{i\theta }},b = R{e^{i\tau }}$ for some $\theta ,\tau \in \mathbb{R}$ . Moreover, (3·1) gives

\begin{align*}2{R^2}{e^{i\left( {\theta + \tau } \right)}}\!\left( {\sigma .\bar \sigma } \right) = - d,\end{align*}

from which we deduce that $\theta + \tau { \equiv _{2\pi }}\pi $ . Hence, $b = - R{e^{ - i\theta }} = - \bar a$ . Then

\begin{align*} v\!\left( {a,b} \right) = \left( { - 2R{\rm{cos}}\!\left( { - \theta } \right), - 2R{\rm{sin}}\!\left( { - \theta } \right),0} \right).\end{align*}

Therefore, in order to prove the statement, it is enough to show that the set of possible complex arguments assumed by the periods $\left( {\sigma .\gamma } \right)$ , $\gamma \in T$ is dense in the circle $\{ z \in \mathbb{C}\;|\;\left| z \right| = 1\} $ (the point of Picard jump corresponding to such a $\gamma $ would be determined by $a = R{e^{i\theta }}$ such that $\theta $ (or $\pi + \theta $ ) is the opposite of the argument of $\left( {\sigma .\gamma } \right)$ ). However, the $\mathbb{Q}$ -vector space P of the periods $\left( {\sigma .\gamma } \right)$ , $\gamma \in T$ is not contained in an $\mathbb{R}$ -line of $\mathbb{C}$ ; indeed, if this were the case, then the period field ${k_T} = \mathbb{Q}\!\left( {\left( {\sigma .{\gamma _i}} \right)/\left( {\sigma .{\gamma _1}} \right)} \right)$ (for a basis $\left\{ {{\gamma _i}} \right\}$ of T) would be contained in $\mathbb{R}$ , contradiction. As a consequence, P is dense in $\mathbb{C}$ , and then the set of the complex arguments of elements of $P\backslash \left\{ 0 \right\}$ is dense in the circle.

5·1. Outside the equator

Denote by $\mathcal{U}$ the upper-half sphere of $\mathbb{P}_\ell ^1 \simeq S_\ell ^2$ (defined by the condition $\left| a \right| \gt \left| b \right|$ on $z = \left[ {\sigma ' = a\sigma + b\bar \sigma + \ell } \right]$ , see Lemma 3·1), and by $\mathcal{Q}$ the set of points of Picard jump in $\mathcal{U}$ (that is, the Noether–Lefschetz locus of $\mathcal{U}$ ), deprived of the north pole x. Notice that a point $z \in \mathcal{Q}$ satisfies ${\rho _z} = 1$ , by Proposition 4·2, and hence is orthogonal to a unique (up to a rational scalar) non-zero element $\ell ' \in P_z^ \bot \cap \!\left( {T \oplus \mathbb{Q}\ell } \right)$ . Denote by ${\mathcal{Q}^ + }$ the set of $z \in \mathcal{Q}$ for which the corresponding $\ell '$ is such that $\left( {\ell '.\ell '} \right) \gt 0$ .

For a point $z \in \mathbb{P}_\ell ^1$ , we define its altitude as the last coordinate of the vector defined by Lemma 3·1, namely the third coordinate of the corresponding point in $S_\ell ^2$ for the basis $\left\{ {\Re \!\left( \sigma \right),\Im \!\left( \sigma \right),\ell } \right\}$ of ${P_T} \oplus \mathbb{R}\ell $ .

Recall that $d = \left( {\ell .\ell } \right) \in {\mathbb{Z}_{ \gt 0}}$ . Choose an element $\ell ' = {\gamma _1} + \left( {m/d} \right)\ell \in T \oplus \mathbb{Q}\ell $ , $\ell ' \notin T$ , $\ell ' \notin \mathbb{Q}\ell $ , so that the two corresponding orthogonal points are neither the poles nor on the equator (see Lemma 3·5 and Lemma 4·3). Denote by ${z_1} \in \mathcal{Q}$ the only point of Picard jump in the upper-half sphere orthogonal to $\ell '$ . Taking a scalar multiple of $\ell '$ identifies the same ${z_1}$ ; we may assume $m = d$ , so that $\ell ' = {\gamma _1} + \ell $ . For an element $A \in K_T^ \times $ , we define ${z_2} = A\;{\rm{*}}\;{z_1}$ , where ${z_2} \in \mathcal{Q}$ is the only point of Picard jump in the upper-half sphere orthogonal to $\ell '' = {\gamma _2} + \ell $ , where ${\gamma _2} = A\!\left( {{\gamma _1}} \right)$ . This association defines an action of $K_T^ \times $ on $\mathcal{Q}$ .

Remark 5·5. Recall that the homeomorphism $\mathbb{P}_\ell ^1 \simeq S_\ell ^2$ of Lemma 3·1 sends ${z_1} = \left[ {\sigma ' = a\sigma + b\bar \sigma + \ell } \right]$ to the normalisation of the vector

\begin{align*}v\!\left( {a,b} \right) = \left( {\Re \!\left( {b - \bar a} \right),\Im \!\left( {b - \bar a} \right),\!\left( {a\bar a - b\bar b} \right)\frac{{\left( {\sigma .\bar \sigma } \right)}}{{2d}}} \right).\end{align*}

This allows us to deduce that the tangent of the angle between a vector of a point $S_\ell ^2$ and the plane of zero-altitude is given, up to sign, by

\begin{align*}\frac{{\left( {\sigma .\bar \sigma } \right)}}{{2d}}\left| {\bar a + b} \right|.\end{align*}

Indeed, it is suffices to notice that

\begin{align*}a\bar a - b\bar b = \left( {\bar a + b} \right)\!\left( {a - \bar b} \right),\end{align*}

since $ab \in \mathbb{R}$ by (3·1).

Proof of the Proposition. Firstly, assume that $\left| A \right| = 1$ . We are showing that ${z_2} = \left[ {\sigma ''} \right]$ actually belongs to the conic $\mathbb{P}_\ell ^1$ , i.e. $\left( {\sigma ''.\sigma ''} \right) = 0$ . For, it is enough to observe that the (3·1), that here takes the form

\begin{align*}2\!\left( {aA} \right)\!\left( {b\bar A} \right)\!\left( {\sigma .\bar \sigma } \right) + d = 0,\end{align*}

is satisfied since $\left| A \right| = 1$ and ${z_1} \in \mathbb{P}_\ell ^1$ . If $\left| a \right| \gt \left| b \right|$ then $\left| {aA} \right| \gt \left| {b\bar A} \right|$ , so that ${z_2}$ still belongs to the upper-half sphere. To prove the first statement, it suffices to show that $\left( {\sigma ''.\ell ''} \right) = 0$ holds true, where $\ell '' = A\!\left( {{\gamma _1}} \right) + \ell $ . Recall that, since $\left| A \right| = 1$ , A is an isometry for $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ ; therefore we obtain

\begin{align*}\!\left( {\sigma ''.\ell ''} \right) &= \left( {aA\!\left( \sigma \right) + bA\!\left( {\bar \sigma } \right) + \ell .A\!\left( {{\gamma _1}} \right) + \ell } \right)\\ &= a\!\left( {A\!\left( \sigma \right).A\!\left( {{\gamma _1}} \right)} \right) + b\!\left( {A\!\left( {\bar \sigma } \right).A\!\left( {{\gamma _1}} \right)} \right) + d\\ & = a\!\left( {\sigma .{\gamma _1}} \right) + \left( {\bar \sigma .{\gamma _1}} \right) + d\\ & = \left( {\sigma '.\ell '} \right) = 0.\end{align*}

For the statement concerning the action by rotation and the same altitude of ${z_1}$ and ${z_2}$ , it is enough to compare the tangents of the angles between the vectors corresponding to ${z_1}$ and ${z_2}$ , as done in Remark 5·5.

Remark 5·6. One could expect, for a general $A \in K_T^ \times $ not necessarily of norm one, the point ${z_2} = A*{z_1}$ to be given by

\begin{align*}{z_2} = \left[ {\sigma '' = a{{\bar A}^{ - 1}}\!\left( \sigma \right) + b{{\bar A}^{ - 1}}\!\left( {\bar \sigma } \right) + \ell = a{{\bar A}^{ - 1}}\sigma + b{A^{ - 1}}\bar \sigma + \ell } \right],\end{align*}

so that $\left( {\sigma ''.\ell ''} \right) = 0$ by a similar argument. Nonetheless, (3·1) is no longer satisfied and ${z_2} \notin \mathbb{P}_l^1$ , so this guess is not true. It is indeed more complicated to deduce an explicit expression for ${z_2}$ when A does not have norm one.

We continue the proof of Proposition 5·4. Assume now that $\left| A \right| \ne 1$ . Even without computing explicitly ${z_2}$ we can prove that ${z_2}$ has different altitude from ${z_1}$ . Define $B = {\bar A^{ - 1}}$ and consider

\begin{align*}\sigma '' = aB\sigma + b\bar B\bar \sigma + \ell .\end{align*}

It corresponds to a point in $\mathbb{P}\!\left( {{T^{2,0}} \oplus {T^{0,2}} \oplus \mathbb{C}\ell } \right)$ orthogonal to $\ell ''$ . Then $0 = \overline {\!\left( {\sigma ''.\ell ''} \right)} = \left( {\overline {\sigma ''} .\ell ''} \right)$ , where

\begin{align*} \overline {\sigma ''} = \bar bB\sigma + \bar a\bar B\bar \sigma + \ell .\end{align*}

Note that $\sigma '',\overline {\sigma ''} $ are linearly independent over $\mathbb{C}$ since $a \ne \bar b$ (see Remark 3·3). Therefore, the line in $\mathbb{P}\!\left( {{T^{2,0}} \oplus {T^{0,2}} \oplus \mathbb{C}\ell } \right)$ passing through these points is exactly the line of elements orthogonal to $\ell ''$ ; among these points there are ${z_2}$ and $\overline {{z_2}} $ , given by the intersection with the conic $\mathbb{P}_\ell ^1$ . The general point of this line has the form $z = \left[ {\lambda \sigma '' + \mu \overline {\sigma ''} } \right]$ . Since $\lambda + \mu = 0$ does not give an element of $\mathbb{P}_\ell ^1$ , we may assume $\lambda + \mu = 1$ , and $z = \left[ {\lambda \sigma '' + \left( {1 - \lambda } \right)\overline {\sigma ''} } \right]$ .

Assume that $z \in \mathbb{P}_\ell ^1$ . We claim that $\lambda \in \mathbb{R}$ . Firstly, note that $\lambda \ne 1/2$ , otherwise the $\sigma $ -coefficient and the $\bar \sigma $ -coefficient of $\lambda \sigma '' + \left( {1 - \lambda } \right)\overline {\sigma ''} $ would have the same norm, forcing z to be on the equator $S_\ell ^1$ , contradiction. Since $z \in \mathbb{P}_\ell ^1$ , then

\begin{align*}0 = {(\lambda \sigma '' + \left( {1 - \lambda } \right)\overline {\sigma ''} )^2} = {\lambda ^2}\!\left( {\sigma ''.\sigma ''} \right) + 2\lambda \!\left( {1 - \lambda } \right)\!\left( {\sigma ''.\overline {\sigma ''} } \right) + {(1 - \lambda )^2}\!\left( {\overline {\sigma ''} .\overline {\sigma ''} } \right).\end{align*}

Note that $\left( {\sigma ''.\sigma ''} \right) = \left( {\overline {\sigma ''} .\overline {\sigma ''} } \right)$ . Therefore, also the point $z'$ identified by

\begin{align*}z' = \left[ {\!\left( {1 - \lambda } \right)\sigma '' + \lambda \overline {\sigma ''} } \right]\end{align*}

is in the intersection of the conic $\mathbb{P}_\ell ^1$ with the line of elements orthogonal to $\ell ''$ . Since $\lambda \ne 1/2$ , it cannot coincide with z, and thus $z' = \bar z$ , where

\begin{align*}\bar z = \left[ {\overline {\lambda \sigma ''} + \left( {1 - \bar \lambda } \right)\sigma ''} \right].\end{align*}

Hence, $\lambda = \bar \lambda $ .

As noticed in Remark 5·5, to discuss the difference of altitude we are interested in the quantity $\left| {\bar a + b} \right|$ for the new coefficients. We have

\begin{align*}\lambda \sigma '' + \left( {1 - \overline {\sigma ''} } \right) = \left( {\lambda a + \left( {1 - \lambda } \right)\bar b} \right)B\sigma + \left( {\lambda b + \left( {1 - \lambda } \right)\bar a} \right)\bar B\bar \sigma + \ell \end{align*}

and, using $\lambda \in \mathbb{R}$ ,

\begin{align*}\left| {\!\left( {\lambda \bar a + \left( {1 - \lambda } \right)b} \right)\bar B + \left( {\lambda b + \left( {1 - \lambda } \right)\bar a} \right)\bar B} \right| = \left| {\bar a + b} \right|\left| {\bar B} \right| = \left| {\bar a + b} \right|{\left| A \right|^{ - 1}}.\end{align*}

This relation tells us exactly that, for points in the upper-half sphere, acting by an element of $K_T^ \times $ of norm smaller than one increases the altitude, while the action of elements of $K_T^ \times $ of norm greater than one decreases the altitude.

Finally, we prove the third statement. If $A \in \mathbb{R}$ then $B = {\bar A^{ - 1}} = {A^{ - 1}} \in \mathbb{R}$ , too. Arguing as above, we deduce that ${z_2}$ is represented by the element

\begin{align*}\lambda \sigma '' + \left( {1 - \lambda } \right)\overline {\sigma ''} = \left( {\lambda a + \left( {1 - \lambda } \right)\bar b} \right)B\sigma + \left( {\lambda b + \left( {1 - \lambda } \right)\bar a} \right)B\bar \sigma + \ell \end{align*}

for some $\lambda \in \mathbb{R}$ . Set

\begin{align*} \tilde{a}=\big(\lambda a + (1-\lambda)\overline{b}\big) B, \quad \tilde{b}=\big(\lambda b + (1-\lambda) \overline{a}\big) B. \end{align*}

For this element, the quantity $\tilde{b}-\overline{\tilde{a}}$ is equal to

\begin{align*}\!\left( {\left( {2\lambda - 1} \right)b + \left( {1 - 2\lambda } \right)\bar a} \right)B = \left( {b - \bar a} \right)\!\left( {2\lambda - 1} \right)B.\end{align*}

As $\left( {2\lambda - 1} \right)B \in \mathbb{R}$ , the angle determined by $\Re\big(\tilde{b}-\overline{\tilde{a}}\big)$ and $\Im\big(\tilde{b}-\overline{\tilde{a}}\big)$ is the same as the angle determined by $\Re \!\left( {b - \bar a} \right)$ and $\Im \!\left( {b - \bar a} \right)$ , so that ${z_2}$ and ${z_1}$ lie on the same meridian.

Huybrechts proved that, for a point ${z_1} \in {\mathcal{Q}^ + }$ orthogonal to $\ell '$ , the corresponding polarised irreducible Hodge structure of K3 type, namely $T' = {\ell ^{{\rm{\prime}} \bot }}$ , is of CM type, and the real parts $K_T^0$ and $K_{T'}^0$ of ${K_T}$ and ${K_{T'}}$ , respectively, coincide [ Reference HuybrechtsHuy20 , proposition 3·8]. Moreover, he gave an explicit description of ${K_{T'}}$ [ Reference HuybrechtsHuy20 , corollary 3·10]: it is the quadratic extension of $K_T^0 = K_{T'}^0$ described by

\begin{align*} {X^2} + \gamma X + \delta = 0,\end{align*}

whereFootnote ²

\begin{align*} \gamma = m\!\left( {\alpha + {\alpha ^{ - 1}}} \right),{\rm{\;}}\delta = {m^2} - \frac{d}{{2\!\left( {\sigma .\bar \sigma } \right)}}\!\left( {{\alpha ^2} + {\alpha ^{ - 2}} - 2} \right),\end{align*}

${K_T} = \mathbb{Q}\!\left( \alpha \right)$ , $d = \left( {\ell .\ell } \right)$ , $m = \left( {\ell .\ell '} \right)$ and $\sigma $ is such that $\left( {\sigma .\ell '} \right) = 1$ .

Proof. To prove that ${z_2} \in {\mathcal{Q}^ + }$ , it is enough to notice that

\begin{align*}\!\left( {\ell ''.\ell ''} \right) &= \left( {A\!\left( {{\gamma _1}} \right) + \ell .A\!\left( {{\gamma _1}} \right) + \ell } \right)\\ &= \left( {A\!\left( {{\gamma _1}} \right).A\!\left( {{\gamma _1}} \right)} \right) + \left( {\ell .\ell } \right)\\ &= \left( {{\gamma _1}.{\gamma _1}} \right) + \left( {\ell .\ell } \right)\\ &= \left( {\ell '.\ell '} \right) \gt 0,\end{align*}

as A is an isometry for $\left( {{\rm{\;}}.{\rm{\;}}} \right)$ and ${z_1} \in {\mathcal{Q}^ + }$ .

We now prove that the coefficients $\gamma ,\delta $ are the same for the two points ${z_1},{z_2}$ , so that ${K_{T'}} = {K_{T''}}$ . For both points $m = d$ , since we fixed the $\ell $ -part of $\ell ',\ell ''$ ; thus $\gamma $ is the same. Note that $\sigma $ depends on $\ell '$ : it is chosen such that $\left( {\sigma .\ell '} \right) = 1$ . On the other hand, if we choose such a $\sigma $ for $\ell '$ , then $A\!\left( \sigma \right) = A\sigma $ satisfies

\begin{align*}\!\left( {A\!\left( \sigma \right).\ell ''} \right) = \left( {A\!\left( \sigma \right).A\!\left( {{\gamma _1}} \right) + \ell } \right) = \left( {A\!\left( \sigma \right).A\!\left( {{\gamma _1}} \right)} \right) = \left( {\sigma .{\gamma _1}} \right) = \left( {\sigma .\ell '} \right) = 1,\end{align*}

since A is an isometry. Hence, we may choose $A\!\left( \sigma \right) = A\sigma $ for $\ell ''$ . Then

\begin{align*}\!\left( {A\sigma .\overline {A\sigma } } \right) = \left( {A\sigma .\bar A\bar \sigma } \right) = A\bar A\!\left( {\sigma .\bar \sigma } \right) = \left( {\sigma .\bar \sigma } \right)\end{align*}

so that $\delta $ is the same as well.

Endow $K_T^ \times $ with the topology induced as a subspace of ${\mathbb{C}^ \times }$ by the fixed embedding ${K_T} \hookrightarrow \mathbb{C}$ . Notice that the proof of Proposition 5·4 also puts in evidence the continuity of the action of $K_T^ \times $ on $\mathcal{Q}$ . We summarise all the results in the following corollary.

Proof. The only statements left to prove are the ones concerning the density of points of $\mathcal{Q}$ at the same altitude and the density of points of ${\mathcal{Q}^ + }$ . The first assertion follows from the fact that, for a CM field E, the set of points in $E \cap {S^1}$ is dense in ${S^1}$ , according to Corollary 2·7. For the second assertion: let $\gamma \in T$ such that $\left( {\gamma .\gamma } \right) \gt 0$ , so that $\left( {\gamma + \ell .\gamma + \ell } \right) = \left( {\gamma .\gamma } \right) + d \gt 0$ as well. Set $\ell ' = \gamma + \ell $ . Consider an element in ${\mathbb{Q}^ \times } \cdot \left( {{S^1} \cap {K_T}} \right) \subseteq K_T^ \times $ , say $A = \lambda \alpha $ , for $\lambda \in {\mathbb{Q}^ \times }$ and $\alpha \in K_T^ \times $ satisfying $\left| \alpha \right| = 1$ . Then $\ell '' = A\!\left( \gamma \right) + \ell $ satisfies

\begin{align*}\!\left( {\ell ''.\ell ''} \right) = \left( {A\!\left( \gamma \right).A\!\left( \gamma \right)} \right) + d = {\lambda ^2}\!\left( {\gamma .\gamma } \right) + d \gt 0.\end{align*}

The density of ${\mathcal{Q}^ + }$ in $\mathcal{U}$ follows, therefore, from the density of ${\mathbb{Q}^ \times } \cdot \left( {{S^1} \cap {K_T}} \right)$ in ${\mathbb{C}^ \times }$ ( ${S^1} \cap {K_T}$ is dense in ${S^1}$ thanks to Proposition 2·6).

5·2. On the equator

Now we focus on the equator $S_\ell ^1$ . In the case $\mathbb{P}_\ell ^1\backslash S_\ell ^1$ we had at our disposal a clean way to choose one of the two points orthogonal to a certain $\ell ' \in T \oplus \mathbb{Q}\ell $ : picking the one in the upper-half sphere (subject to the condition $\left| a \right| \gt \left| b \right|$ ). Here this choice is not so clear; therefore, we will consider pairs of antipodal points on $S_\ell ^1$ . Define $\mathcal{R}$ to be the set of points of Picard jump on the equator modulo the relation $\left\{ \pm \right\}$ . Pick an element $\ell ' = {\gamma _1} \in T$ , ${\gamma _1} \ne 0$ , and consider ${z_1} = \left[ {\sigma ' = a\sigma + b\bar \sigma + \ell } \right]$ on $S_\ell ^1$ which is orthogonal to ${\gamma _1}$ . By our choice of $\mathcal{R}$ , here we mean at the same time ${z_1}$ and $\overline {{z_1}} $ . For $A \in K_T^ \times $ , we define ${z_2} = A{\rm{*}}{z_1}$ to be the (pair of) point orthogonal to $\ell '' = {\gamma _2} = A\!\left( {{\gamma _1}} \right)$ . This defines an action of $K_T^ \times $ on $\mathcal{R}$ . This action is no longer free: indeed, as explained in the proof of Proposition 4·4, two non-zero elements $\gamma ,\delta $ in T are orthogonal to the same point if there exists a $\beta \in {(K_T^0)^ \times } = K_T^ \times \cap \mathbb{R}$ such that $\beta \!\left( \gamma \right) = \delta $ . Then, ${(K_T^0)^ \times }$ is contained in the kernel of this action, or equivalently in the stabiliser of each point. This induces an action of the quotient group $K_T^ \times /{(K_T^0)^ \times }$ on $\mathcal{R}$ . If T has CM, the action of the quotient is free, i.e. ${(K_T^0)^ \times }$ is exactly the stabilizer of each point: this follows from the fact that ${\rho _z} = r/2$ for points of Picard jump on the equator (Proposition 4·4). However, it will be clear later that the action of $K_T^ \times /{(K_T^0)^ \times }$ is free even if T is not of CM type. Notice that the quotient group is trivial if T is of totally real type. Again, if T has CM then the action is transitive (as ${\rm{di}}{{\rm{m}}_{{K_T}}}T = 1$ ).

We have an explicit way to compute ${z_2} = A\;{\rm{*}}\;{z_1}$ . Pick $\sigma $ satisfying $\left( {\sigma .\ell '} \right) = 1$ . If ${z_1} = \left[ {\sigma ' = a\sigma + b\bar \sigma + \ell } \right]$ , (4·1) gives $a + b = 0$ , i.e. $b = - a$ . On the other hand, we rewrite (3·1) as $ - 2{a^2}\!\left( {\sigma .\bar \sigma } \right) + d = 0$ , which forces $a \in \mathbb{R}$ . Hence,

\begin{align*}\sigma ' = a\sigma - a\bar \sigma + \ell .\end{align*}

Assume that $A = R{e^{i\theta }}$ ; then ${z_2} = \left[ {\sigma ''} \right]$ , where

\begin{align*}\sigma '' = a{e^{i\theta }}\sigma - a{e^{ - i\theta }}\bar \sigma + \ell .\end{align*}

Indeed ${z_2} \in \mathbb{P}_\ell ^1$ holds since (3·1) is satisfied, and

\begin{align*} \!\left( {\sigma ''.A\!\left( {{\gamma _1}} \right)} \right)& = \left( {a{e^{i\theta }}\sigma .A\!\left( {{\gamma _1}} \right)} \right) + \left( { - a{e^{ - i\theta }}\bar \sigma .A\!\left( {{\gamma _1}} \right)} \right)\\ &= a{e^{i\theta }}(\bar A(\sigma).{\gamma _1}) - a{e^{ - i\theta }}\!\left( {\bar A\!\left( {\bar \sigma } \right).{\gamma _1}} \right)\\ & = a{e^{i\theta }}\bar A - a{e^{ - i\theta }}A = 0.\end{align*}

Looking at the isomorphism $\mathbb{P}_\ell ^1 \simeq S_\ell ^2$ defined in Lemma 3·1, we see that $A = R{e^{i\theta }}$ acts on $\mathcal{R}$ by a rotation of angle $ - \theta $ . Note that if $A \in {(K_T^0)^ \times }$ then $\theta = 0$ or $\theta = \pi $ , and ${z_2} = {z_1}$ (in $\mathcal{R}$ ), as already mentioned.

We add a topological flavour to this discussion. Endow $K_T^ \times $ with the topology induced by the topology of ${\mathbb{C}^ \times }$ under the fixed embedding ${K_T} \hookrightarrow \mathbb{C}$ , and ${(K_T^0)^ \times }$ with the subspace topology. Then the topological quotient group $K_T^ \times /{(K_T^0)^ \times }$ is a subspace of the topological quotient group ${\mathbb{P}^1}\!\left( \mathbb{R} \right) = {\mathbb{C}^ \times }/{\mathbb{R}^ \times }$ . Moreover, $\mathcal{R}$ is a subspace of $S_\ell ^1/\left\{ \pm \right\}$ and the action of $K_T^ \times /{(K_T^0)^ \times }$ on it is given by rotation.

In retrospect, we have proven the following result.