Lecture 3: Lower bounds: from periods to curves
Edgar Costa (MIT)
ICERM: Arithmetic, Geometry and Computations on K3 surfaces
September 16, 2026
Supported by the Simons Foundation
Slides available at edgarcosta.org
Today: produce the classes. A lower bound is a curve you can write down, and enough curves with their intersections is the lattice itself.
For some approaches see [Bouyer-Costa-Festi-Nicholls-West] and, for degree 2 K3 surfaces, [Festi].
Lefschetz (1,1) theorem
A homology class $\gamma \in H_2(X, \mathbf{Z})$ is in $\operatorname{Pic} \overline{X}$ if and only if $\int_\gamma \omega_X = 0$, where $\omega_X$ is the nonzero holomorphic 2-form $\omega_X$ on $X$, unique up to scaling.
$$H_2(X, \mathbf{Z}) \longrightarrow \mathbf{C}, \qquad \gamma \mapsto \int_\gamma \omega_X.$$
Hence, if $\Pi = [\int_\gamma \omega_X]_{\gamma \in H_2(X, \mathbf{Z})} \in \mathbf{C}^{22}$ represents this map, then we are reduced to finding a (saturated) lattice $\Lambda \subset H_2(X, \mathbf{Z})$ of solutions
$$\Pi R = 0, \qquad R \in H_2(X, \mathbf{Z}) \simeq \mathbf{Z}^{22}.$$
$$\operatorname{Pic}(\overline{X})_{\mid B} := \mathbf{Z} \langle \gamma \in \operatorname{Pic} \overline{X} \mid -\gamma^2_{\mathrm{prim}} < B \rangle \subseteq \Lambda \qquad \text{(Lairez-Sert\"{o}z).}$$
Let $C$ be a nice (smooth, projective, geometrically integral) curve over $k$ of genus $g$ given by equations. Let $J$ be the Jacobian of $C$.
Goal
Given the equations of $C$, compute the endomorphism ring $\operatorname{End} \overline{J}$.
$J = \mathbf{C}^g / \Lambda_J$, the period lattice $\Lambda_J$ computed numerically, to high precision, from a basis of $H^0(C, \Omega_C)$.
By picking a $k$-basis for $H^0(C, \Omega_C)$, we have
$$\operatorname{End}(J) = \left\{ T \in M_g(k) \mid T \Lambda_J \subset \Lambda_J \right\}$$
Hence, if $\Pi$ is a period matrix for $C$, i.e., $\Lambda_J = \Pi \mathbf{Z}^{2g}$, then we are reduced to finding a $\mathbf{Z}$-basis of the solutions $(T, R)$ to
$$T \Pi = \Pi R, \qquad T \in M_g(\overline{k}), \quad R \in M_{2g}(\mathbf{Z}).$$
The Galois module structure of $\operatorname{End}(\overline{J})$ is given via its action on $T \in M_g(\overline{k})$.
Heuristically, via lattice reduction algorithms, we can find such a $\mathbf{Z}$-basis.
There is no obvious way to prove that our guesses are actually correct.
$$\begin{aligned} \alpha_C : \ & C \xrightarrow{\ \ AJ\ \ } J \xrightarrow{\ \ \alpha\ \ } J \dashrightarrow \operatorname{Sym}^g(C) \\ & P \mapsto \{Q_1, \ldots, Q_g\} \Longleftrightarrow \alpha([P - P_0]) = \left[ \sum_{i=1}^g Q_i - P_0 \right] \end{aligned}$$
This traces out a divisor on $C \times C$, which determines $\alpha$.
This divisor is a certificate of containment for $\alpha \in \operatorname{End} \overline{J}$.
Theorem (C-Mascot-Sijsling-Voight)
We give an algorithm for nondegenerate $\alpha \in \mathrm{M}_g(\overline{k})$
$$\alpha \mapsto \begin{cases} \texttt{true} & \text{if } \alpha \in \operatorname{End} \overline{J}, \text{ and a certificate} \\ \texttt{false} & \text{if } \alpha \notin \operatorname{End} \overline{J} \end{cases}$$
By interpolation via $\alpha_C$ or by locally solving a differential equation on $C \times C$.
$$\begin{aligned} 0 &= -8x^2 + 8xy + 17y^2 - 34xz - 2yz - 28z^2 - 10xw - 9yw - 18zw + 2w^2, \\ 0 &= 4x^3 - 6x^2 y - 6x y^2 + 12x^2 z + 6xyz + 24y^2 z - 12x z^2 - 24z^3 + 2x^2 w + 7xyw \\ &\qquad + 4y^2 w + 4xzw - 13yzw - 8z^2 w - 20x w^2 - 3z w^2 - 12w^3 \end{aligned}$$
has real multiplication by the maximal order of $\mathbf{Q}(x)/(x^4 - x^3 - 3x^2 + x + 1)$.The first step to show that, under Langlands, it corresponds to a specific Hilbert modular form $f$, i.e., $J_{\mathbf{Q}(\sqrt{3})} \sim A_f$. We used this in a recent project, where we show that the 2-isogeny field of $A_f$ solves the inverse Galois problem for $\operatorname{PSL}_2(\mathbf{F}_{16}) \rtimes C_2 \simeq \texttt{17T7}$.
From the equations of $X$, compute $\operatorname{Pic} \overline{X} \subset H_2(X, \mathbf{Z})$ as a $\operatorname{Gal}(\bar k/k)$-module.
Today we will try to use the period map.
$$H_2(X, \mathbf{Z}) \longrightarrow \mathbf{C}, \qquad \gamma \mapsto \int_\gamma \omega_X.$$
$$\Pi = [\int_\gamma \omega_X]_{\gamma \in H_2(X, \mathbf{Z})} \in \mathbf{C}^{22}$$
$$\operatorname{Pic}(X^{al}) = \ker\!\left(H_2(X, \mathbf{Z}) \xrightarrow{\gamma \mapsto \int_\gamma \omega_X} \mathbf{C}\right)$$
$$X : x^{4} + xyzw + y^{3}z + yw^{3} + z^{3}w = 0 \subset \mathbf{P}^3$$
This example started at a workshop at ICERM about thinking about K3 surfaces on the LMFDB.
$$\Pi \Lambda \approx 0 \qquad \operatorname{Pic}(X^{al})|_B \subseteq \Lambda \overset{?}{\subseteq} \operatorname{Pic}X^{al}.$$
133056
smooth rational quartics spanning $\Lambda$.Turns out one can compute a bit more for hypersurfaces
$$\varphi \colon H_2(X, \mathbf{Z}) \times H^2_{\mathrm{dR}}(X/k) \to \mathbf{C} \qquad (\gamma, \omega) \longmapsto \int_\gamma \omega$$
Note, if $\gamma \in \operatorname{Pic} X^{al}$, then $\frac{1}{2\pi i}\int_\gamma \omega \in k^{al}$ for $\omega \in F^1 H^2_{\mathrm{dR}}(X/k)$.
If $\gamma = [C] \in H_2(X, \mathbf{Z})$ for a curve $C \subset X$ then from $\frac{1}{2\pi i}\left(\int_\gamma \omega\right)_{\omega \in F^1}$ one can construct an ideal $I_\gamma$ such that $I(C) \subsetneq I_\gamma$.
In favorable circumstances we expect low order equations in $I_\gamma$ to span $I(C)$.
For example, smooth rational curves of degree up to 4 in K3s.
No hope to recover rational curves of degree higher than 4. For $d \geq 5$ one needs $a \geq 3$ before $I(C)_a$ is non-zero, and there the Jacobian ideal contributes superfluous equations to $I_{\gamma,a}$, so: $$I(C)_a \subsetneq I_{\gamma,a}.$$
We need "isolated" classes, so elliptic curves are also hard.
An elliptic curve $C$ has $[C]^2 = 0$ and moves in a pencil, so its periods do not determine a single curve.
For a smooth rational quartic curve $C \subset X$ we have that the equation of the quadric surface containing $C$ generates $I_{[C],2}$, i.e., $I(C)_2 = I_{[C],2}$.
$$X : x^{4} + xyzw + y^{3}z + yw^{3} + z^{3}w = 0 \subset \mathbf{P}^3$$
$$\operatorname{Pic}(\overline{X})|_B \subseteq \Lambda \overset{?}{\subseteq} \operatorname{Pic}\overline{X}$$
Reconstruct the quadric surfaces containing some of the 133056 rational quartics in $X$ using the curve classes.
$$I_{[C],2} = \langle a_0 x^2 + \cdots + a_9 w^2 \rangle_{\mathbf{C}}$$
that defines a quadric surface $Q$, such that $Q \cap X = C \cup \overline{C}$. Hence, we expect an orbit of 168 quadrics each containing a pair of quartics.Reconstruct the ten coefficients $a_i$ of these quadrics $a_0 x^2 + \cdots + a_9 w^2$ in a Galois orbit of size 168.
$x^{168} - 10014013832542203812872613924739x^{161} + 171047690745503707515328576627906817785436888130925209472262244x^{154} - 1268317331496745879603035032448157273146519836562713924560050631153969519297207668270922371313x^{147} + 23237703563539410755436556575134206593366430461423708193774287327245213403024087108979694756912313 \cdots$
The abstract isomorphism problem feels hopeless.
Construct $\mathbf{Q}(a_k) \hookrightarrow L$, where $L = \mathbf{Q}(a_0, \dots, a_9) = \mathbf{Q}(a_0)$.
In our case, we have all the compatible embeddings
$$\sigma_i : \mathbf{Q}(a_k) \hookrightarrow L \hookrightarrow \mathbf{C}$$
Thus the isomorphism is given by the solution of the following linear system
$$\{\sigma_i(a_k)^j\}_{i, j} \cdot v = \{\sigma_i(a_0)\}_i, \qquad v \in \mathbf{Q}^{168}$$
Distinct nodes make $\{\sigma_i(a_k)^j\}$ invertible, so the solution $v \in \mathbf{Q}^{168}$ is unique; the denominators of $v$ are bounded a priori, so enough precision pins $v$ down exactly and the isomorphism is then verified exactly.
In practice it is faster to refine the complex embeddings iteratively: their height is 4k digits, not 120k.
$$Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168$$
Show that $Q \cap X$ decomposes into two quartic curves.
$$\begin{aligned} &Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168\\ &\Lambda_Q := \langle [C] : C \subset \sigma(Q) \cap X, \, \sigma : L \hookrightarrow \mathbf{C} \rangle \subseteq \operatorname{Pic}(\overline{X})|_B \subseteq \Lambda \overset{?}{\subseteq} \operatorname{Pic} \overline{X} \end{aligned}$$
The inclusion $\Lambda_Q \subseteq \Lambda$ is not explicit!
Nonetheless, $\operatorname{Pic} \overline{X}$ and $\Lambda$ are saturated in $H_2(X, \mathbf{Z})$.
Hence, it is sufficient to show that $\operatorname{rank} \Lambda_Q = \operatorname{rank} \Lambda = 19$.
We can do this in two ways:
$$\operatorname{Pic} \overline{X} = \Lambda$$
$$Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168$$
$Q \cap X$ decomposes into a pair of quartics over $K$ a quadratic extension of $L$.
Compute $K$ and $\operatorname{Gal}(K/\mathbf{Q})$ acting on $\Lambda_Q$.
Via the identification with the original classes we have $\frac{1}{2 \pi i} \left( \int_C \omega \right)_{\omega \in F^1} \in K^{21}$.
These can be reconstructed in the same fashion as we reconstructed $a_i$.
Unclear how to certify this step! What are the denominators of $\frac{1}{2 \pi i} \int_C \omega$?
Can one compute $K$ using geometry without Gröbner basis?
For $Q$ smooth, $K = L(\sqrt{\operatorname{disc} Q})$ [Costa-Sertöz].
$$Q : a_0 x^2 + a_1 x y + \cdots + a_9 w^2 = 0 \subset \mathbf{P}^3, \quad [L := \mathbf{Q}(\{a_i\}_i):\mathbf{Q}] = 168$$
$Q \cap X$ decomposes into a pair of quartics over $K$ a quadratic extension of $L$.
Compute $K$ and $\operatorname{Gal}(K/\mathbf{Q})$ acting on $\Lambda_Q$.
The direct computation of $\operatorname{Gal}(K/\mathbf{Q})$ looks hopeless.
We guess that $K = F(\sqrt[14]{u})$ where $[F : \mathbf{Q}] = 24$ and $\operatorname{Gal}(F/\mathbf{Q}) = C_3 \times \operatorname{PGL}(2,7)$.
Note, $\#\operatorname{Gal}(F/\mathbf{Q})$ is 14 times smaller than $\#\operatorname{Aut} \operatorname{Pic} \overline{X}$.
There is a new paper about computing Galois groups of this kind of polynomial [Elsenhans-Steel].
The quartic surface $X : x^4 + xyzw + y^3 z + yw^3 + z^3 w = 0 \subset \mathbf{P}^3$ has $\operatorname{Pic} \overline{X} = \Lambda$, generated by quartics over a quadratic extension of $L := \mathbf{Q}(\{a_i\}_i)$.
We are still developing the method and figure out its applications/limitations.
Wanna be a Theorem (C-Sertöz)
There is a practical algorithm to compute the saturation of the lattice generated by rational curves of degree up to 4.
We are trying to cover all the K3 surfaces given by five nomials.
Periods computed for all 161 examples; most work with very little precision.
Mukai ($X_{153}$): 133056 curves, smallest orbit 336, periods to about 1k digits.
the quadric coefficients generate a degree-168 field.
$X_{110}$: 28224 curves, smallest orbit 3456, rank 17, periods to about 11k digits.
reconstructing quadric coefficients in a degree-288 field.