---
created: 2023-05-07T00:02
updated: 2023-05-08T00:05
aliases: ["Notes on Phil's talk"]
---

<https://youtu.be/DmvYmmTBxUY>

# Notes on Phil's talk

Definition: an integral affine structure on a manifold $M$ is a collection of charts to $\RR^n$ whose transition functions are in the integral affine transformation group $\RR^n \semidirect \SL_n(\ZZ)$. An example is taking $\Lambda$ a rank $n$ lattice in $\RR^n$ and considering $\RR^n/\Lambda$. This is a flat torus, the charts come from a fundamental domain $F$ for $\Lambda \actson \RR^n$ composed with the inclusion $F\injects \RR^n$. The transition functions are then just translations. An example where the part in $\SL_n(\ZZ)$ is not the identity comes from the following: take $\RR^2/\ZZ^2$, take a square and glue the sides by translation by $\tv{1, 0}$ and glue the top to bottom by $\matt 1 1 0 1 \in \SL_2(\ZZ)$ and translating by $\tv{0, 1}$:

![](2023-05-06-17.png)

Definition: an $\IAS^2$ is an integral affine structure on the 2-sphere $S^2$ with 24 $\rm{I}_1$ singularities. This isn't exactly true, one can only define the $\IAS$ on $S^2\smts{p_1,\cdots, p_{24}}$, but around each $p_i$ it will take on a standard model.
The model for the $I_1$ singularity is the following: take $\RR^2$, cut out a triangular wedge, and glue by $\matt 1 1 0 1$:
![](2023-05-06-18.png)

One can get a chart about a point on the edge whose transition function is $\matt 1101$: 
![](2023-05-06-19.png)

This gives an $\IAS$ on $\RR^2\smz$, but the $\IAS$ doesn't extend to zero, so this is really an $\IAS$ with singularities:
![](2023-05-06-20.png)


Another model is to realize this as a branch cut with monodromy $\matt 1101$:
![](2023-05-06-21.png)

Note that $M$ acts as the identity along $\tv{1, 0}$, i.e. there is unique (non-generalized) eigenvector with eigenvalue 1, and here it points along the $x\dash$axis. Note that straight lines in the IAS look like the following:
![](2023-05-06-22.png)
Note that the idea of a straight line makes sense because transition functions in the definition of an $\IAM$ preserve what it means to be a straight line in $\RR^n$.

When this is the base of a LFT, this singularity will have a fiber of type $I_1$ in Kodaira's classification of singular fibers of elliptic fibrations. The singularities in the $\IAS$ correspond to singular fibers. This is also called a focus-focus singularity.

We modify the definition of an $\IAS^2$ to allow for limits of $I_1$ singularities when they collide. There is a Gauss-Bonnet type theorem which says that a singular $\IAS$ on a sphere with standard $I_1$ singularities must have $24 = 12 \chi(S^2)$ such singularities.

One can collide singularities to get simpler $\IAS$s. Take a cube, the local charts to $\RR^2$ are the flat charts into the square lattice $\ZZ^2$. Making a sphere out of papers, for every non-corner point one could lay the paper flat.
![](2023-05-06-23.png)

Each of the 8 corners of the cube is a collision of 3 $I_1$ singularities.

Definition: a triangulation of an $\IAS^2$ is a decomposition into lattice triangles of area $1/2$. In the previous example, one such triangulation is the following: 
![](2023-05-06-24.png)
The notion of area is well-defined since $\SL_n(\ZZ)$ preserves area. Note a necessary condition for such a triangulation to exists is that one must have the singularities at integral points, and the transition functions must be in $\ZZ^2\times \SL_2(\ZZ)$. Why this group: these precisely make well-defined the notion of integral points in a manifold, and this group preserves the notion of integral points.

Note that there are 8 singularities, and each corresponds to a toric surface with 3 nontoric blowups. Triangulating in the following way yields this pair at each corner:

![](2023-05-06-33.png)

The singularity here factors into 3 $I_1$ singularities, one along each direction of the cube.

A singularity being at an integral point precisely means that the standard local chart puts the singularity at $0\in \RR^n$ with this standard form:

![](2023-05-06-25.png)

Let $B$ be a triangulated $\IAS^2$, we'll describe how to build from this a singular K3 surface. Recall that a K3 $X$ is such that $K_X \cong \OO_X$, so $X$ is holomorphic symplectic and admits a nonvanishing 2-form, and $h^1(X) = 0$. There will in fact be two ways to build a singular K3.

Let $v_i$ be a vertex of the triangulations and suppose it is not singular. Drawing a local chart, we see a local picture of the triangulation as follows:
![](2023-05-06-26.png)

This is the chart around $v_i$. Taking the fan defined by $\Star(v_i)$ we get a toric surface: 
![](2023-05-06-27.png)
This yields a smooth Looijenga pair/a log CY pair $(V_i, \sum_j D_{ij})$ where $V_i$ is a surface and here $D_{ij}$ is the toric boundary consisting of 8 curves.

If $v_i$ is singular, there is a modified recipe. We'll skip the details, but they key idea is nontoric blowups. This produces a log CY pair / Looijenga pair, but is not necessarily toric.

Note that unfortunately the notion of a cone singularity is not enough to recover the deformation type of pairs, one needs extra combinatorial data, e.g. the factorization into $I_1$ singularities.

We set $\cX_0 \da \union (V_i, \sum D_{ij})$ where $i$ ranges over all vertices of the triangulation of $B$.

For example, consider this local picture of a triangulation around $v_i$ and $v_j$:
![](2023-05-06-28.png)
The intersection complex is dual to the original complex, so we see $\PP^1\times \PP^1$ at $v_j$:
![](2023-05-06-29.png)
 This yields a pair where $D$ is a square which is a sum of two sections and two fibers: 
![](2023-05-06-30.png)

At $v_i$ we see $\FF_2$ with its anticanonical divisor:

![](2023-05-06-31.png)

This  is a local picture of what's happening in the $\IAS^2$:

![](2023-05-06-32.png)

Note that each white triangle corresponds to a triple point. Note $\cX_0$ is singular, since e.g. in the picture above we've glued two smooth surfaces but along a closed curve. It is an SNC surface.

A theorem of Friedman (83) says $\cX_0$ deforms to a K3 surface. There are some ambiguities, e.g. gluing two copies of $\PP^1$ needs to ensure $0\mapstofrom 0$ and $\infty\mapstofrom \infty$, but with the correct choice this smoothing will exist.

Note that triangles of area $1/2$ ensure the surfaces are smooth and thus $\cX_0$ is SNC. One can get around the area condition, but we won't discuss this.

This yields $\cX\to \Delta$ with $\cX_t$ a smooth K3.

Definition: a polarization of a triangular $\IAS^2$ $B$ is a weighted balanced grah supported on the edges of $B$.

For example, take a triangulated square lattice:
![](2023-05-06-34.png)

Here is a graph supported on its edges:

![](2023-05-07.png)

Each edge corresponds to a primitive integral vector, and the balancing condition says their linear combination must be zero. Here is a point where it is balanced:

![](2023-05-07-1.png)

To balance here, one can assign the downward vector weight 2:

![](2023-05-07-3.png)

For singular vertices, the balanced condition is that the sum of all vectors is parallel to the monodromy ray:

![](2023-05-07-4.png)

This notion is chart-invariant: changing the ray by a shear only adds some multiple of the monodromy direction. The triple point formula will be automatic here, ensuring the total space is smooth and $K\dash$trivial. A full triangulation with areas $1/2$ is necessary to get a smooth total space.

Recall from toric geometry that if a fan has rays $\RR_+ v_{ij}$ then there exists a line bundle $L_i \to V$ with $L_i \cdot D_{ij} = n_{ij}$ $\iff$ $\sum n_{ij} v_j = 0$.

Changing the triangulation, we see a $\PP^2$:
![](2023-05-07-6.png)

The line bundle whose intersection with the  3 toric boundariy components with $1$: it's the bundle $\OO_{\PP^2}(1)$.

The weighted balanced graph this gives $L_i$ a line bundle on $V_i$.

These components $(V_i, L_i)$ can be glued to get a surface with bundle $(X_0, L_0)$ which smooths to a polarized K3 and we get a family $(\cX, \cL) \to \Delta$  by Friedman-Scattone. How mirror symmetry appears: the complex geometry is encoded in the triangulated $\IAS^2$, and the symplectic geometry is in the weighted balanced graph. The data of the polarized $\IAS^2$ actually encodes a second, completely separate degeneration.

The discrete Legendre transform: take an $\IAS^2$ with a weighted balanced graph: 
![](2023-05-07-7.png)

Call this $\IAS^2$ $B$ and let $L_{\IAS}$ be the weighted graph.  Note that we previously interpreted it as a fan. The main idea of mirror symmetry: re-interpret fans as polytopes and vice-versa. How we interpret this picture as a polytope: regard each region in in the regions bounded by $\supp(L_{\IAS})$ (edges on which weights are positive) as the base of a Lagrangian torus fibration. Consider one such region $P_i$:
![](2023-05-07-8.png)

There exists some other surface fitting into a Lagrangian torus fibration $\mu_i: (\hat V_i, \sum_j \hat D_{ij}, \omega_i) \to P_i$. The singularities on the $\IAS^2$ correspond to singular fibers in the fibration. 

Note that one gets a Symington polytope in the $I_1$ case, although it is less clear what to do when singularities collide. See the nodal slide operation on singularities.

What one does: replace every vertex of the complete triangulation with the Symington polytope for the line bundle defined by the red edges:

![](2023-05-07-9.png)

E.g. the vertex from before gets replaced with the polytope for $\OO(1)$:
![](2023-05-07-6.png)

At a singularity, one replaces this with a Symington polytope with a singularity inside.

Note that if all singularities are at vertices, all components are toric varieties.

There is a question of uniqueness for these symplectic structures, but ultimately we only extract its class in cohomology and let $\hat L_i \da [\omega_i] \in H^2(V; \ZZ)$, which defines an ample line bundle. One then forms $(\hat X_0, \hat L_0) = \union (\hat V_i, \sum_j \hat D_j, \hat L_j)$ and smooth to get $\hat(\cX, \hat \cL) \to \Delta$ a degeneration of K3s. This is the discrete Legendre transform, and realizes mirror symmetry tropically in the Gross-Siebert program -- the transform can be thought of as re-interpreting the base $B$ as a polytope.

Theorem (EF, AET): The Picard-Lefschetz transform of the degeneration $\cX\to \Delta$, say $T: H^2(\cX_t; \ZZ)\selfmap$ induced by parallel transport of the Gauss-Manin connection, has a logarithm $M = \log T$ and certain class $\lambda, \delta \in H^2(\cX_t; \ZZ)$ such that $Nx = (x \cupprod \delta)\lambda - (x\cupprod \lambda)\delta$,
where these classes can be understood in terms of a certain diffeomorphism  $\phi: \cX\to \hat \cX$ such that $\delta = [\mu\inv(\pt)]$ is the fiber class of a point in the LFT, and $\phi_* \lambda = [\omega_t]$ is the class of the symplectic form.

Punchline: if you want to understand a degeneration of K3 surfaces, you need to understand how to fill in a punctured family no matter what the Picard-Lefschetz transformation is. I.e. if you give me a Picard-Lefshetzs transformation, I need to give a singular K3 surface to fill in, and this gives a compactification of moduli of K3s.

Note: in the LFT,  the tori collapse to circles on edges and points over vertices. In the smoothing, though, the LFT has torii over all edges and points. This undoes the symplectic boundary reduction along the $L_{\IAS}$ when you smooth.

Upshot: to get a singular K3 $\cX_0$ for all possible PL transformations, encoded in pairs $(\delta, \lambda)$, one only needs to write down an appropriate family of LFTs over the sphere.

An example for $F_2$: let $(X, L)$ be a degree 2 K3, so $c_1(L) = 2$. The genneric such surface admits a branched double cover $\pi: X\mapsvia{2:1}\PP^2$ over a sextic curve. There are 19 dimensions of complex moduli, correspond to moduli of sextic curves in $\PP^2$. Given a degeneating family, what $\cX_0$ to fill in? Note $\exists R\subset X$ the ramification locus of $\pi$, so $R\in \abs{3L}$.

Form an 18-gon out of 18 (carefully chosen) fixed vectors $v_i \in \ZZ^2$ with varying coefficients $\ell_i$:
![](2023-05-07-10.png)

Thus closes iff $\sum \ell_i v_i = 0$, which is 2 conditions. Now do 3 surgeries: cut out 3 triangles and add singularities; this introduces 3 new parameters $\ell_{19}, \cdots, \ell_{21}$ for the heights of those triangles.

![](2023-05-07-11.png)

This gives an $\IAS^2$ $B(\ell_1, \cdots, \ell_{21})$, and the two linar conditions yields $\vec \ell \in \RR^{19}$. Call this polytope $P$:
![](2023-05-07-16.png)

Note that $P$ does not have the right charge, so glue $P$ to $P\op$ along the equator to get 18 $I_1$ singularities on the equator  and $6$ $I_1$ singularities in the hemispheres, and thus an $\IAS^2$ of charge 24 defining $B = P\union P\op$.

Drawing this picture, one can describe **any** degeneration of degree 2 K3s. When the $\vec \ell\in \ZZ^n$ instead of $\RR^n$, this describes all possible degenerations. Note that $\vec \ell$ encodes $\lambda$.

Note that we regard $B$ as the base $( \hat{X}, \mu() \to B$ of an LFT on the mirror K3. The classes of $\omega$ range over all possibilities, since in the theorem $\phi_* \lambda = [\omega]$ and $\lambda$ is an invariant of the PL transforamtion. Because we wrote down the $\IAS^2$ for all possible classes of $\omega$, we get a procedure to describe the central fiber of any degeneration in $F_2$:

- Any degeneration has a PL transformation $T$ and $N$ with invariant $\lambda$
- There is a corresponding LFT we built
- Because $\vec \ell$ (corresponding to $\lambda$ is integral), we can triangulate into basis triangles
- The triangulation defines the central fiber,

This is the recipe $\cX^\circ \leadsto \cX_0, \cX$.

Why is this the right $\IAS^2$? Degree 2 K3s have an involution. What is the divisor to take? The involution flipping to two hemispheres, and the ramification divisor $L_{\IAS}$ is the sum of 3 bits along the equator: 

![](2023-05-07-13.png)

Why is there an $I_1$ singularity along the edges on the equator?

![](2023-05-07-14.png)

Note that the gluing doesn't work at the left point; there **must** be a singularity at this point. So choose the vectors $v_{i-1}, v_i, v_{i+1}$ such that the left edges glue by a matrix conjugate to $\matt 1101$:

![](2023-05-07-15.png)

Now one can clearly glue by putting an $I_1$ singularity at the point.


Note there are two conditions necessary: singularities factor into $I_1$ singularities, and one needs a triangulation into basis triangles. With these, you can always building a singular K3 surface -- interpret each vertex as the fan of a toric surface, glue them, doing a modification at the singular vertices.

Note that we are omitting a step: when $\vec \ell\in \ZZ^{19}$, it admits a lattice triangulation. So choose one that is invariant under the involution and contains the equator.

![](2023-05-07-17.png)

Choosing the triangulation builds $\cX_0$, which does depend on the triangulation. How do we get a unique limit? We actually get $(\cX_0, \cL_0)$ and we really need to take the canonical model $\Proj \oplus_k H^0(\cL_0^k)$. Note that the line bundle is trivial in the interior of $P$, so each hemisphere collapses to a single point when we take the canonical model.

Note there is an additional moduli: how the double curves are glued. There was a $\CC^*$ of freedom for each double curve. All of the possible $\CC^*$ of regluing the double curves correspond to moduli at $\bd F_2$. The boundary components are thus parameterized by a torus $(\CC^*)^m$ for some $m$.

Note: why is this different from Seibert's work? (Audience question, maybe Seibert?) This doesn't use scattering diagrams. But note that a log smooth K3 of degree 2 or 3 has unobstructed deformations. This is a modern version of the Friedman-Scattone theorem. The new idea here: compactify by writing down concretely and explicitly exactly which $\IAS$s are relevant. This construction explciitly constructs the central fiber.