# Boundary 

Goal for this section: describe how Coxeter-Vinberg diagrams are used to get models at BB cusps.

## The case of abelian varieties

Let us consider the setup first for moduli of principally polarized abelian varieties. 
We have the following:

:::{.theorem}
There is an isomorphism
\[
\eta: \torcptf{\cA_g} \iso (\ksbacpt{\cA_g})^\nu
\] 
for $F$ the second Voronoi fan. \dzg{Is there always a natural morphism from toric compactifications to KSBA? Not in general! We can talk about it when we meet.}

:::

:::{.corollary}
As a result, any punctured 1-parameter family $\cX^\circ \to \Delta^\circ$ has a unique limit $\cX_0$ which can combinatorially be described as a tropically polarized abelian variety $(X_{\mathrm{trop}}, \Theta_{\mathrm{trop}})$ with a tropical $\Theta$ divisor $\Theta_{\mathrm{trop}}$. 

More is true: the fan $F$ is itself a moduli space for such tropical abelian varieties.
:::


:::{.remark title="Motivation from abelian varieties"}
To see how this works, consider a 1-parameter family of abelian varieties. These are tori of the form 
\[
\cX_t \da \coker(\phi_t: \ZZ^g\injects (\CC^*)^g)
,\]
where $\phi_t$ are embeddings that vary in the family. Write this embedding as a matrix $M$; this is a matrix of periods. Then for $t\approx 0$ one exponentiates $M_{ij}$ to get a symmetric positive-definite $g\times g$ matrix $B$.
There is a cone $C \subseteq \ts{B = B^t > 0}$ in $\GL_g(\ZZ)$ and a Coxeter fan $F$ supported on its rational closure $\rc{C}$ which corresponds to affine Dynkin diagram $\tilde A_2$:
\[
\tilde A_2: \quad \dynkin[extended, edge length=1cm]A2
\]
This corresponds to a triangular fundamental domain for a reflection group that acts on $C$. For a cartoon picture, think of a hyperbolic disc $\DD$ and let the fundamental domain be a triangle with ideal vertices:

\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/tesselate_disc.jpg}
\caption{Tessellating a hyperbolic disc by triangles}
\label{fig:tesselateone}
\end{figure}

Note that the straight lines forming the edges should "really" be curved hyperbolic geodesics. One continues reflecting in order to tessellate the hyperbolic disc, then puts this disc in $\RR^3$ at height one and cones it to the origin to get an infinite-type fan $F$:
\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/tesselate_disc2.jpg}
\caption{Coning off the hyperbolic disc to form a fan.}
\label{fig:tesselatetwo}
\end{figure}


Note that the entire fan $F$ admits an $\SL_2(\ZZ)$ action. Now if one rewrites $B$ as a form $B(x,y) = ax^2 + by^2 + c(x+y)^2$ with $a,b,c\in \ZZ_{\geq 0}$, the coordinate vector $\vec v \da (a,b,c)$ defines a point in some chamber of $\rc{C}$. In turn, $\vec v$ defines a 1-parameter degeneration of abelian varieties, and thus a pair $(X_{\mathrm{trop}}, \Theta_{\mathrm{trop}})$. When $B$ is integral it defines an embedding $B:\Lambda\injects \Lambda\dual$\dzg{What is $\Lambda$?} and thus one can construct a torus $T \da \Lambda\dual_\RR/\Lambda \cong \RR^2/\ZZ^2$ which has finitely many integral points defined by $\Lambda$.

Recall that for any lattice $L$ there is an associated Voronoi tessellation by polytopes $P_i$, one such $P_i$ centered around each lattice point $\ell_i$. Let $\mathrm{Vor}_B$ be the Voronoi tessellation of $\Lambda$; this can be identified with a hexagonal honeycomb tessellation of $\RR^2$:

\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/Vor_tess.jpg}
\caption{The Voronoi tesselation associated to $\tilde A_2$, the triangular lattice.}
\label{fig:tesselatethree}
\end{figure}


One then defines $\Theta_{\mathrm{trop}} \da B(\mathrm{Vor}_B)/\Lambda$, a quotient of the image of the hexagonal tessellation. Although the blue vertices in $\mathrm{Vor}_B$ generally have vertices with fractional coordinates, the vertices in the image have integral coordinate vertices with respect to $\Lambda\dual$. The image of a regular hexagon is now a hexagon with side lengths $a,b,c$, and since we've quotiented by $\Lambda$, $\Theta_{\mathrm{trop}}$ is determined by two hexagons with side lengths determined by $\vec v = (a,b,c)$ which are glued together: 

\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/honeycomb_sidelengths.jpg}
\caption{A picture of two relevant polytopes in the image of the Voronoi tessellation, which tessellates the entire dual lattice. After quotienting, these will be the only two relevant polygons.}
\label{fig:honeycomb}
\end{figure}

The claim is that this picture describes an entire degeneration $\cX$ of abelian varieties. To see the central fiber: every vertex $w_i$ in this new tessellation defines an honest fan via $\Star(w_i)$; here there are 2 vertices of valence 3 and 3 edges in the quotient, so the central fiber $\cX_0$ is two copies of $\PP^2$ corresponding to $w_1, w_2$ glued together along three curves corresponding to $a,b,c$. To see the entire family: put this entire picture at height 1, cone to the origin to get a fan, and quotient that fan by a $\ZZ^2$ action to get $\cX$.

Note that in the K3 case, things are harder because the combinatorics only describes $\cX_0$ and not the entire family $\cX$, so one has to appeal to abstract smoothing results to obtain the existence of a family $\cX$ extending $\cX_0$.

Moreover, the original fan $F$ is a moduli of these polyhedral pictures. One can degenerate $\Theta_{\mathrm{trop}}$ by sending some coordinates $a,b,c$ to zero. This degenerates the honeycomb 6-gons into 4-gons if just one side goes to zero. For example, if $a\to 0$, this corresponds to being on a wall in \Cref{fig:tesselatetwo}. If two coordinates degenerate, say $a,b\to 0$, this corresponds to being on a ray. This can be read off by recalling $B(x,y) = ax^2 + by^2 + c(x+y)^2$ and labeling the ideal vertices with monomials $x^2, y^2, (x+y)^2$ as in \Cref{fig:tesselateone}. Thus varying $\vec v = (a,b,c)$ corresponds to varying the side lengths of hexagons and correspondingly moving through $\rc{C}$. Staying in the fundamental chamber doesn't change the overall combinatorial type of \Cref{fig:honeycomb}, but passing through a wall will flip the hexagonal tiling in various ways.



:::

## To the K3 case

The claim is that a similar story more or less goes through for K3s: the Coxeter diagram is much more complicated, and the relevant combinatorial device is an $\IAS^2$ with 24 singularities instead of a tropical variety. We have the following:

:::{.theorem}
There is a morphism
\[
\eta: \torcptf{F_2} \to (\ksbacpt{F_2})^\nu
\]
where $F$ is fan of a Coxeter diagram associated to a cusp of $F_2$, and the Stein factorization of $\eta$ is through a semitoroidal compactification.

:::


:::{.corollary}
Any punctured 1-parameter family $\cX^\circ\to \Delta^\circ$ has a unique limit $\cX_0$ which can be combinatorially described as a singular integral-affine sphere with an integral-affine divisor $(\IAS^2, R_{\IAS})$.
:::

:::{.remark}
This is a much harder theorem than the $\cA_g$ case: periods of K3 surfaces are highly transcendental, and the period map is not well-understood. Also note that the relevant Coxeter diagram for $\cA_g$ was relatively simple, while the diagram for $F_2$ is the following:
\begin{figure}[H]
\centering
\input{tikz/Coxeter_19_1_1}
\caption{The Coxeter diagram for type $(19, 1, 1)$.}
\label{fig:coxeter1911}
\end{figure}
Nodes in \Cref{fig:coxeter1911} correspond to roots spanning a hyperbolic lattice 
\[
N\da U \oplus E_8(-1) \sumpower{2} \oplus A_1(-1),\qquad \signature(N) = (1, 18) 
\] 
which is the Picard lattice of the Dolgachev-Nikulin mirror K3. Decorated nodes $v_i$ record self-intersection numbers $v_i^2$, and edges between $v_i$ and $v_j$ record the intersection numbers $v_i.v_j$. Note that the Coxeter diagram also captures the data of all $(-2)$ curves on the mirror K3 surface and their intersections.

This diagram again describes the fundamental chamber of a reflection group, and the cone in this case $C = \ts{v^2 > 0}$. Toroidal compactifications of $F_2$ correspond to fans whose support is $\rc{C}$ (i.e. the interior, plus rational rays on the boundary). There is a natural fundamental chamber defined by $\ts{v \mid v.r_i \geq 0}$ where $\ts{r_i}$ are roots, the difference is that now some vertices of the fundamental chamber may be ideal vertices:

\begin{figure}[H]
\centering
\includegraphics[width=0.5\textwidth]{figures/F2HyperbolicReflection.jpg}
\caption{A fundamental chamber $F$ for a reflection group. Reflecting over walls of $F$ successively generates a tiling of the hyperbolic disc by copies of $F$. Note that one vertex is an ideal vertex, i.e. it is in $\partial{\overline{\HH^n}}$.}
\label{fig:hyperbreflection}
\end{figure}

Proceeding similarly to take the cone over this picture and allow rational boundary points yields the cone $\rc{C}$ and a corresponding infinite-type fan $\cF$ -- this is a fan since the faces are rationally generated, $F$ is a fundamental chamber for the reflection group $W(N)$, the fan is $W\dash$invariant by construction and moreover invariant under $\Orth(N)$. Since there is a short exact sequence \[0\to W(N) \to \Orth(N)\to S_3\to 0\] the index of $W(N)$ is finite and thus $F$ is finite volume.

Points in this fan can naturally be interpreted as period points, so a choice of a point in the fan yields a degenerating family of K3 surfaces by the Torelli theorem. Let $v\in F$ be a point in the fundamental chamber, we will next consider how this corresponds to a combinatorial object, the same way $\vec v = (a,b,c)$ did in the case of $\cA_g$.

First consider a fan with 18 rays, corresponding to a toric surface $\Sigma$ with 18 curves. Note that the rays alternative between long and short vectors:

\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/toric18.jpg}
\caption{The starting point: a toric surface with 18 rays.}
\label{fig:toric18rrays}
\end{figure}

This corresponds to a polytope $P_\Sigma$ which is an 18-gon (not necessarily regular) which is the moment polytope for $X_\Sigma$ where $\Sigma$ is the fan in \Cref{fig:toric18rrays} and has edge lengths $\ell_0, \cdots, \ell_{17}\in \RR$, which determines a polarization $L$ for $X_\Sigma$. Although not shown in the picture here, we can call each edge "long" if it was dual to a long vector, and similarly "short" if dual to a short vector. 

Note also that each edge can be written as $\ell_i v_i$ for $v_i$ some unit vectors, and it is a nontrivial condition on $\vec \ell$ that this polygon closes. In particular, one needs $\sum_{i=0}^{17} \ell_i v_i = 0$.

Now cut triangles out of sides $0, 6, 12$ and call the resulting polygon non-convex polygon $P$. Each triangle cut corresponds to a non-toric blowup of $X_\Sigma$, i.e. a blowup at a point $p$ which is not $T\dash$invariant. This introduces three new length parameters $\ell_{18}, \ell_{20}, \ell_{20}$ corresponding to the heights of these three triangles. Each will introduce an $\rm{I}_1$ singularity to the moment polytope.

\begin{figure}[H]
\centering
\includegraphics[width=0.4\textwidth]{figures/toric18polytope.jpg}
\includegraphics[width=0.4\textwidth]{figures/toric18polytope2.jpg}
\caption{The Symington polytope: an 18-gon, before and after a nontoric blowup corresponding to cutting out triangles.}
\label{fig:symblowup}
\end{figure}

Regarding such polytopes as the Symington polytopes, which are bases of Lagrangian torus fibrations, these are in particular elliptic fibrations and these singularities precisely correspond to introducing singular type $\rm{I}_1$ fibers in Kodaira's classification.

Take two copies of $P$, say $P$ and $P^{\mathrm op}$, and glue them together along the outer edges and call the result $B$. This is topolologically the gluing of two discs, and thus $B$ is homeomorphic to $S^2$. Each gluing along the outer edges introduces a new $\rm{I}_1$ singularity, yielding $3+3 = 6$ singularities in the hemispheres and $18$ singularities along the equator for a total of $24$ singularities of type $\rm{I}_1$ and thus an $\IAS^2$ with charge $24$. 

Note that there are now 24 length parameters: $\ell_0,\cdots, \ell_{17}$ along the equator, $\ell_{18}, \ell_{19}, \ell_{20}$ in the northern hemisphere, and $\ell_{21}, \ell_{22}, \ell_{23}$ in the southern hemisphere. The tuple $\vec \ell = (\ell_1, \cdots, \ell_{23})$ turns out to correspond to 24 vectors in a 19 dimensional space, and there are enough conditions to ensure the polygons actually close.

This produces the tropical sphere $\IAS^2$, so one also needs to describe its tropical divisor $R_{\IAS}$. The above construction works for any K3 with a nonsymplectic involution, e.g. an elliptic K3, and the $\IAS^2$ is naturally equipped with an involution $\iota$ that swaps $P$ and $P^\opp$. The ramification divisor of $\iota$ is the equator, highlighted in blue in the following cartoon picture of $B$, and one takes $R_{\IAS}$ to be the sum of the blue edges with coefficient 2 for even (short?) sides and coefficient 1 for odd (long?) sides:

\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/IASPuPop.jpg}
\caption{Caption}
\label{fig:iaspupop}
\end{figure}

We now describe how one obtains a degeneration $\cX$ of K3 surfaces from this combinatorial picture. One must first extend this $\IAS^2$ to a complete triangulation by basis triangles. 
This triangulation should be done on $P$ first, before the doubling construction, so that the vertices and edges in the northern hemisphere are perfectly matched with those in the southern. Here is a cartoon of what this might look like on one copy of $P$, before gluing:

\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/IAScompletetriang.jpg}
\caption{The $\IAS$ on $P$ extended to a complete triangulation by basis triangles.}
\label{fig:IAS_complete_triangulation}
\end{figure}

This is again a cartoon picture, meant to show how vertices and triangles in the hemispheres should match in pairs exchanged by the involution $\iota$. Here e.g. the blue triangles are meant to match, as well as $\Star(\tilde w_1)$ and $\Star(\tilde w_1^\opp)$:

\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/IASdoubletriang.jpg}
\caption{A completely triangulated $\IAS^2$ defined by $B \da P\union P^\opp$.}
\label{fig:iasdoubletriang}
\end{figure}

This final picture describes the central fiber $\cX_0$ of a Kulikov degeneration of K3 surfaces in the following way: there are many non-singular vertices $p_i$, and exactly 24 singular vertices $w_i$ and $\tilde w_i, \tilde w_i^\opp$. For the $p_i$, there is a fan defined by $\Star(p_i)$ which defines a toric surface $V_i$. For the 24 singular vertices, there is a modified recipe to cook up a semi-toric surface -- since the singularity is type $\rm{I}_1$, this will be a charge 1 surface, and thus realizable as a toric surface with a single non-toric blowup, a \textit{semitoric} surface. How to make this blowup is uniquely determined by an additional omitted decoration called the \textit{monodromy ray} at the singular vertex. Roughly, this is a preferred ray cooked up from the Picard-Lefschetz monodromy operator around the singular vertex. One can think of this as a "singular fan".

\begin{figure}[H]
\centering
\includegraphics[width=0.9\textwidth]{figures/IASmonodromy.jpg}
\caption{$\Star(\tilde w_2)$ in \Cref{fig:IAS_complete_triangulation} with the extra data of a monodromy vector.}
\label{fig:iascompletetriangulation}
\end{figure}

So \[ \cX_0 = \Union_i V_i \union \Union_{j=1}^{24} W_j\] where the $V_i$ are all toric surfaces and the $W_j$ are all semitoric surfaces of charge 1, and the triangulation determines how they are all glued together. To see how this gluing is done, consider the following local picture in the triangulation:

\begin{figure}[H]
\centering
\includegraphics[width=0.5\textwidth]{figures/IASgluing.jpg}
\caption{Local gluing in the $\IAS^2$ of two toric surfaces $\Sigma_7$ and $\PP^2$}
\label{fig:iasgluing}
\end{figure}

At the orange vertex, taking the star we see three rays and thus a copy of $\PP^2$. At the green vertex, we see 7 rays, and thus some toric surface $\Sigma_7$ which is probably something like a Hirzebruch surface $\FF_n$ with 3 toric blowups. Since the orange and green vertices are adjacent by exactly one edge, this means we glue $\PP^2$ to $\Sigma_7$ along the the curves determined by rays pointing along that edge. Moreover, whenever there is a triangle, this corresponds to three surfaces glued together along a triple point.

The general case is that the $\IAS^2$ has 24 copies of $\rm{I}_1$ singularities; these singularities can collide to produce semitoric surfaces with multiple nontoric blowups.

:::

:::{.remark}
Note that this \textit{only} describes $\cX_0$ and not an entire family $\cX$. Friedman solved this problem: there is a technical condition called \textbf{$d\dash$semistability}, and if this is satisfied then $\cX_0$ is smoothable. Moreover the smoothing will have the correct period and/or monodromy vector $\lambda$.\dzg{Haven't discussed $\lambda$ here yet!}

To obtain all degenerations, one considers all of the ways this combinatorial object can degenerate. Sending some $\ell_i\to 0$ causes the 18-gon to collapse into a small polygon, or causes some hemispherical singularities to descend into the equator. This corresponds to moving an interior point of original fundamental chamber $F$ onto a wall, and wall-crossing mutates the $\IAS^2$ in some other ways.
:::

:::{.remark}
Some miscellaneous remarks:

- The Kulikov models are highly non-unique, differing by flops. Adding the divisor $R_{\IAS}$ fixes this and pins down $\cX$ uniquely.
- It seems one can read off the stable model from the $\IAS^2$. For the honeycombs in the $\cA_g$ case, everything was contracted down to two $\PP^2$s glued along their 3 boundary curves in a $\Theta\dash$graph. In the $F_2$ case, one contracts everything in the $\IAS^2$ except for the equator, i.e. the interiors of the hemispheres are contracted. The most general degeneration is 18 copies of $\PP^2$ glued in a cycle; one can then send some $\ell_i\to 0$ to collide the vertices and get fewer than 18 surfaces.
- It seems one can also read off Type II degenerations from the $\IAS^2$. Here there are 4 Type II cusps, 3 correspond to collapsing the 18-gon in the $\IAS^2$ in the equatorial plane to an interval. The 4th involves collapsing the 18-gon to a point with bits sticking out. Type I degenerations correspond to collapsing everything to a point.
- Why everything works simply here: there is only one relevant cusp in $F_2$, and the involution propagates to everything including the $\IAS^2$. The Coxeter diagram is also highly symmetric, hinting at how to make the right toric and $\IAS$ construction.


:::


## Notes from Phil's talk

:::{.remark}
For $\Sigma_g$ a compact complex curve of genus $g$, choose a symplectic basis $\ts{\alpha_i, \beta_i}_{i\leq g}$ of $H_1(\Sigma_G; \ZZ)$, then there is a unique basis $(\omega_1,\cdots, \omega_g)$ of $H^0(\Omega_{\Sigma_g})$ such that $\int_{\alpha_i} \omega_j = \delta_{ij}$.
In this basis, form the \textbf{period matrix} $\tau = (\int_{\beta_i} \omega_j )_{i,j=1}^g$. This satisfies $\tau^t = \tau$ and $\Im(\tau) > 0$ is positive-definite, and is thus an element in The \textbf{Siegel upper half-space}
\[
\cH_g \da \ts{\tau \in \Sym_{g\times g}(\CC) \mid \Im(\tau) > 0}
.\]
The \textbf{Jacobian} of $\Sigma$ is defined as $\mathrm{Jac}(\Sigma) \da \CC^g / (\ZZ^g\oplus \tau \ZZ^g)$.
Note that we made a choice of "marking" by choosing the symplectic basis $\ts{\alpha_i, \beta_i}$, and any two such choices are related by $\matt ABCD \in \Sp_{2g}(\ZZ)$, the isometry group of $\ZZ^{2g}$ with the standard symplectic form, where the action is $\matt ABCD \cvec {I_g} \tau = \cvec{A+B\tau}{C+D\tau}$, the analogue of a linear fractional transformation. To renormalize the 1-forms, we change basis to get a similar matrix $\cvec{I_g}{(A+B\tau)\inv (C+D\tau)}$.
Thus to get an invariant, we consider
\[
[\tau]\in \dcosetl{\Sp_{2g}(\ZZ)}{\cH_g} \da \cA_g
,\]
the moduli space of PPAVs. Here one can realize the polarization on $A$ as a symplectic form on $H_1(A;\ZZ)$ which is represented by a holomorphic line bundle $L\in \Pic(A)$, i.e. identifying a symplectic form on $H_1(A;\ZZ)$ as an element of $H^2(A;\ZZ)$ which we want to be a $(1, 1)$ form.
:::
:::{.remark}
Now $\cA_g$ is not compact, so we consider degenerations over $\cX^\circ\to \Delta^\circ$ and let $\Delta^\circ\to \cA_g$ be the associated period mapping -- how does the period map degenerate as $t\to 0$? The answer is that a certain isotropic subspace $I \leq (\ZZ^{2g}, \omega_\std)$ becomes distinguished by the fact that periods against $I^\perp$ remain finite.
:::

:::{.example}
Let $y^2 = x^3 + x^2 + t$ be a family of elliptic curves over $\AA^1\setminus{0}$. At $t=0$ this degenerates to a nodal cubic. There is a vanishing cycle $\alpha$, and the distinguished isotropic subspace is precisely $I\da \ZZ\alpha$. One shows $I^\perp = \ZZ\alpha$ as well. In this case, to be in $I^\perp$ means to be a curve that does not pass through the thinning neck of the torus that degenerates; any curve that does pass through should intuitively have a period that blows up. We normalize by picking a $c_t$ such that $\int_\alpha c_t {dx\over y} = 1$, then $\int_\beta \omega_t = \tau_t \in \CC$. However, this isn't well-defined: one can parallel-transport $\beta$ around $t=0$ and the monodromy action will be a Dehn twist, so integrals against $\beta$ are only well-defined up to $\ZZ p$ where $p$ are periods against $\alpha$, here $p$ is normalized to 1. So $\int_\beta \omega_t\in \tau_h + \ZZ \in \CC/\ZZ$.
As $t\to 0$, one was $\tau_t \to +i\infty$ if $\alpha, \beta$ are oriented properly. We can fix this ambiguity by exponentiation, getting a well-defined invariant $\exp(2\pi i \int_\beta \omega_t) \in \CC^*$.
:::

:::{.remark}
How this works for $g\geq 1$: assume $I$ is Lagrangian, so $I^\perp = I$, corresponding to a maximally unipotent degeneration. If this were a genus $g$ curve, we could pinch $\leq g$ disjoint cycles simultaneously, and a maximal degeneration will pinch exactly $g$. Since $\Sp_{2g}(\ZZ)$ acts transitively on Lagrangian subspaces in  $\mathrm{LGr}(V)$, we can assume $I = \bigoplus_i \ZZ \alpha_i$ is generated by the $\alpha$ curves. Generalizing the $\CC^*$ embedding in the previous case, we obtain a torus embedding
\begin{align*}
E: \cH_g &\injects (\CC^*)^{g\choose 2} \\
\tau &\mapsto 
\left[
\begin{matrix}
    \exp(2\pi i \tau_{11}) & \cdots & \\
    \vdots & \ddots & \vdots \\
        & \cdots & \exp(2\pi i \tau_{gg} )
\end{matrix}
\right]
\end{align*}
Since $\tau$ is symmetric, the image $E(\tau)$ is again symmetric. Note that the $\beta_i$ cycles are well-defined up to translation in $I$, but because the 1-form was normalized so that integrals of $\alpha_j$ along $\beta_i$ were 1 or 0, the entries in this matrix are well-defined up to integers.
Thus we can exponentiate every entry in the period matrix to get a well-defined symmetric matrix. The unipotent orbit theorem of Schmid gives an asymptotic estimate
\[
E(\tau_t) \sim_{t\to 0}
\left[
\begin{matrix}
    c_{11} t^{n_{11}} & c_{12} t^{n_{12}} & \cdots & \\ 
    \vdots & \ddots & & \vdots \\
        & \cdots & &c_{gg} t^{n_{gg}}
\end{matrix}
\right] \in \Mat_{n\times n}(\CC^*)
\]
which is a cocharacter of $(\CC^*)^{g\choose 2}$, i.e. an inclusion $\CC^* \injects (\CC^*)^{g\choose 2}$ which is a composition of a group morphism and a translation. Here the $c_{ij}$ are the translation parts, and if $c_{ij} = 1$ for all $i,j$ this yields an honest group morphism. Such a cocharacter is called a unipotent orbit. This asymptotic estimate is quantified, so there is a precise speed at which the period matrix approaches the cocharacter.

Setting $N \da (n_{ij})$, we have $N\in \Sym_{g\times g}(\ZZ)$ and $N > 0$. These entries capture the relative speeds at which the various cycles are collapsing.
Since the $c_{ij}$ are ultimately just translations, we'll omit them from here onward.
:::

:::{.remark}
Define a cone
\[
P_g \da \ts{N\in \Sym_{g\times g}(\ZZ) \mid N > 0} \subseteq \ZZ^{g\choose 2}
\]
and consider the family
\[
(\CC^*)^g \over \bracket{ 
(t^{n_{11}}, t^{n_{12}}, \cdots, t^{n_{1g}}), 
(t^{n_{21}}, t^{n_{22}}, \cdots, t^{n_{2g}}),
\cdots ,
    (t^{n_{g1}}, t^{n_{g2}}, \cdots, t^{n_{gg}})
}
\]
over $\Delta^\circ$. How does one extend this family over $t=0$? If $N$ has full rank, this entire expression is isomorphic to $(\CC^g/\ZZ^g)/\tau\ZZ^g$.
There are two answers, one by fans and one by polytopes.
:::

:::{.remark}
The following is the fan construction due to Mumford, which most easily generalizes to K3 surfaces.

Consider the example
\[
N = \matt 2113 \leadsto 
{(\CC^*)^2 \over 
\bracket{
(t^2, t),
(t, t^3)
}}
\]

Note $N > 0$, since $\det N > 0, \mathrm{Trace} N > 0$, and $N(\ZZ^g)$ is generated by the vector $(2,1)$ and $(1, 3)$. First quotient $\RR^2$ by this lattice to get a flat real 2-torus, then take a polyhedral tiling whose vertices are integer points. Here we take a tiling of the fundamental domain and translate it everywhere. This gives a tiling $\cF_0$ on the universal cover $\RR^g$. Now put this picture at height 1 in $\RR^{g+1}$ to get a tiling $\tilde \cF_0$ of $\RR^g\times\ts{1}\subseteq \RR^{g+1}$, and let $\tilde \cF \da \Cone(\tilde \cF_0) \subset \RR^{g+1}$ be its cone. Taking the toric variety $X(\tilde \cF)$, and define $X(\cF) \da X(\tilde \cF)/N(\ZZ^g)$, where the quotient makes sense precisely because $N(\ZZ^g)$ acts on $\RR^g\times \ts{1}$ by translation, and this extends to a linear action on $\RR^g$, which moreover preserves $\tilde \cF$ and thus acts on the toric variety. There is a morphism $\phi: X(\cF)\to \AA^1$ induced by the morphism of fans given by the height function: projection in $\RR^{g+1}$ onto the last coordinate, whose image is in $\RR_{\geq 0}$. This map descends to the quotient since the linear action preserves the height function. 

This produces a degenerating fan of abelian varieties.
A fiber $\phi\inv(t)$ of $X(\tilde F)$ for $t\neq 0$ yields $(\CC^*)^g$, and the action $N(\ZZ^g)$ acts by translations of the form $(t^{n_{i1}}, t^{n_{i2}}, \cdots, t^{n_{ig}})$ in the original family. Thus we recover the original family as an infinite quotient of a toric variety. But the toric variety has a toric boundary, encoded in the tiling.
The fiber $\phi\inv(0)$ has dual complex $\Gamma(X_0(\cF)) = \cF_0$ equal to the original tiling, and $X_0(\cF)$ is a union of toric varieties.

In the original lattice, in the quotient there are precisely 3 0-cells, and we interpret the star of each 0-cell as the fan of a toric surface. They are glued according to the tiling.

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:::{.remark}
The polytope construction, which builds the projective coordinate ring instead. One defines $Q$ to be the hull of certain points, constructs a theta function, and takes Proj of a certain graded algebra generated by such functions with an explicit multiplication rule and structure constants. These define a certain PL function with a "bending locus" which gives a polyhedral decomposition of $\RR^g/\ZZ^g$. For any $N\in P_g$ one can define the Delaunay decomposition $\mathrm{Del}(N)$, and the central fiber $\cX_0$ of the family will have \textbf{intersection} complex $\mathrm{Del}(N)$ -- the loci where the PL function is linear will be polytopes which are the cells of the Delaunay decomposition. The second Voronoi fan $F^\mathrm{Vor}$ is a decomposition of $P_g$ into loci where $\mathrm{Del}(N)$ is constant. One then takes $\dcosetl{ \Sym_{g\times g}(\ZZ) }{\cH_g} \injects (\CC^*)^{g\choose 2} \to X(F^\mathrm{Vor})$. One the quotients by conjugation in $\GL_g(\ZZ)$ to get $X(F^\mathrm{Vor}) /\GL_g(\ZZ) \injects \overline{\cA_g}^{\mathrm{Vor}}$. Correspondingly, for any $\cX^\circ$ in $\cA_g$, tracing through this construction gives a proper family $\cX$ in $\overline{\cA_g}^{\mathrm{Vor}}$ -- note that we've only described what toric compactification to take for the maximally unipotent degenerations, but one can carry out similar constructions for the other cusps of $\bbcpt{\cA_g}$.
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:::{.remark}
One should ask if $\overline{\cA_g}^{\mathrm{Vor}}$ actually solves a moduli problem, and the answer is yes (up to normalization) by a theorem of Alexeev. The moduli problem is the moduli of semi-abelic pairs. Define $\overline{\cA_g}^\Theta$ to be the closure of pairs $(X, \eps R)$ where $R$ is their theta divisors, then Alexeev shows
\[
\overline{\cA_g}^{\mathrm{Vor}} = (\overline{\cA_g}^\Theta)^\nu
\]
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:::{.remark}
How do we do something similar for K3 surfaces? Fix $v\in \lkt$ primitive with $v^2 = 2d$ and define
\begin{align*}
\Omega_{L} &\da \ts{\CC x \in \Gr_1(\lkt_\CC) \mid (x,x) = 0, (x, \bar x) > 0 } \\
\Omega_{2d} &\da v^{\perp \Omega_{\lkt} } = \ts{x\in \lkt_\CC \mid (x, v) = 0} \\
\Gamma_{2d} &\da \Stab_{\Orth(\lkt)}(v) = \ts{\gamma\in \Orth(\lkt) \mid \gamma(v) = v} \\
F_{2d} &\da \dcosetl{ \Gamma_{2d}}{\Omega_{2d} }
\end{align*}

Here $\Omega_{2d}$ plays the role of $\cH^g$ in the abelian variety case, and is a Hermitian symmetric domain of type IV or $\SO_{2, n}$, and $F_{2d}$ is an arithmetic quotient. Fixing a marking $\phi: H^2(X;\ZZ) \to \lkt$, the period map for a family $\cX^\circ \to \Delta^\circ$ is given by taking $H^{2, 0}(X) = \CC \omega$ and looking at $[\omega] \da \phi(\omega) \in F_{2d}$, since $[\omega]\in \Omega_{2d}$ but is ambiguous up to change of marking (elements of $\Gamma$). This is a map $\Delta^\circ\to F_{2d}$.

Given a degenerating family, there is a distinguished isotropic lattice $I \leq v^\perp$ where $\signature v^\perp = (2, 19)$. Note $I$ can only have rank 1 or 2. The rank 1 case (Type III degenerations) is a maximally unipotent degeneration; the central fiber is as singular as possible, and $\cX_0$ will always have 0-strata. In contrast, in the rank 2 case (Type II degenerations) there are models of the degeneration with no 0-strata.

In the rank 1/Type III case, there is a vanishing cycle $\delta$ associated to a 0-stratum in $\cX_0$ which is topologically a 2-torus. It turns out that $\delta$ is an isotropic vector that spans the isotropic lattice, so we can write $I = \ZZ\delta\subseteq v^\perp$. In the degeneration, the 2-torus collapses to a point.

In the rank 2/Type II case, there are two linearly independent isotropic vectors $\delta$ and $\lambda$ in $v^\perp$ corresponding to 2-tori collapsing simultaneously not to isolated points as in the previous case, but rather to circles in $\cX_0$. They are in the singular locus of $\cX_0$, which is an elliptic curve.

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:::{.remark}
We henceforth assume $\rank_\ZZ I = 1$ and write $I = \ZZ\delta$ for $\delta$ the isotropic vanishing cycle. Normalize $\omega_t$ so that $\int_\delta \omega_t = 1$ for $t\neq 0$. Let $\ts{\gamma_i}_{i=1}^{19}$ be a basis of $\delta^\perp/\delta$. Since $\delta \in v^\perp$ was isotropic of signature $(2, 19)$, we have $\signature(\delta^\perp/\delta) = (1, 18)$ and this gives us a hyperbolic lattice of rank 19. Consider the integral $\int_{\gamma_i} \omega_t \in \CC$. For this to make sense, one needs to lift the $\delta_i$ from $\delta^\perp/\delta$ to $\delta^\perp$, and the choice of lift is ambiguous up to a multiple of $\delta$. By the normalization of the integral, we get a well-defined period
\[
\int_{\gamma_i} \omega_t\in \CC/\ZZ
\]
As in the PPAV case, we use the exponential to get rid of the quotient by $\ZZ$. Letting $U_\delta\leq \Gamma_{2d}$ be the unipotent subgroup stabilizing $\delta$, we get the following torus embedding
\begin{align*}
\dcosetl{U_\delta}{\Omega_{2d}} \mapsvia{\psi} (\CC^*)^{19} \\
\CC [\omega_t] &\mapsto \qty{ \exp\qty{2\pi i \int_{\gamma_1}\omega_t}, \cdots, \exp\qty{2\pi i \int_{\gamma_{19}} \omega_t }
}
\end{align*}
and a nilpotent orbit theory yielding an asymptotic estimate 
\[
\psi_t \sim (c_1 t^{\lambda _1}, \cdots, c_{19} t^{\lambda_{19}} )
\]
with $c_i\in \CC^*$, so the periods are approximated by a cocharacter where the $\lambda_i$ measure how fast the periods degenerate.
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:::{.remark}
Degenerations: a theorem of KPP shows that after a finite base change and birational modifications, any degeneration of K3s has a model where

- $X$ is smooth
- $X_0$ is RNC
- $K_X = \OO_X$

The most famous degeneration of K3s is the Fermat degeneration is a non-example, since smoothness fails:
\[
V(x_0 x_1x_2 x_3 = t(x_0 ^4 + x_1^4 + x_2^4 + x_3^4))
\]
This threefold has precisely 24 conifold singular points.
The central fiber at $t=0$ is a tetrahedron, 4 planes $\PP^2$ in $\PP^3$, and the singular points come from intersecting each edge of the tetrahedron with the residual quartic. One can get a smooth threefold by taking a small resolution of the singular points. There are choices for the resolutions, differing by flops, so here is a heuristic of a symmetric choice where along each edge there are two resolutions extending into each component:

image

The result has four components $V_i$ which are isomorphic to $\Bl_6\PP^2$, 2 points on each of 3 lines in $\PP^2$.

An observation originally due to GHK: there is an IAS on $\Gamma(X_0)$, i.e. there are charts to $\RR^2$ up to post-composition with $\SL_2(\ZZ) \semidirect \RR^2$. 

Here is an example of $\cX_0 = \union V_i$ for a Kulikov degeneration (written as a decomposition into irreducible components). Each unlabeled edge has an implicit label of $-1$:

image

This forms a tiling of the sphere. Each tile corresponds to an irreducible component $V_i$ of $\cX_0$. The edges correspond to components $V_i, V_j$ glued along an anticanonical cycle of rational curves $V_{ij}$. The edge numbers record the self-intersection numbers of the cycles $V_{ij}$ regarded as a cycle in $V_i$ and $V_{ji} = V_{ij}$ regarded as a cycle in $j$.

A general fact about degenerations of CYs: $\cX_0$ is generally a union of log CY varieties, i.e. there are meromorphic 2-forms on components and they are glued along their poles so that the residues agree. The red lines in the image denotes the pole locus of these forms. Each triple point is where 3 surfaces are glued. Since the overall variety is a SNC surface, there are only double curves and triple points.
Note that this picture is the \textbf{intersection complex} of $\cX_0$, and not the dual complex $\Gamma(\cX_0)$. To obtain the dual complex, take the dual tiling, regard each integral 0-cell in the result as a fan, and glue the fans.

Here are the fans:

image


Here is how this interacts with the original diagram:

image

This works fine at most vertices, but at most 24 components are non-toric. Note that from toric geometry, if $(V, D)$ is a toric pair then $-D_i^2 v_i = v_{i-1} + v_{i+1}$ and so one can enforce this formula on such components. For example, the following pair has all $-1$ curves since $v_2 = v_1 + v_3$:

image

Enforcing this formula locally, non-toric points force some $\SL_2(\ZZ)$ monodromy in the IAS. 

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:::{.remark}
This is the analogue of the Mumford fan construction.
Note that in the PPAV case, the lattice didn't specify a Kulikov degeneration since it was not a complete triangulation. But completing this to a complete triangulation of the corresponding real 2-torus does yield a Kulikov model. For K3s, instead of a complete triangulation on $T^2$, we're taking a complete triangulation of an $\IAS^2$. Note that unlike the PPAV case, a triangulated $\IAS^2$ only gives $\cX_0$ (glued from ACPs) and not the entire family $\cX$. An abstract theorem of Friedman says it smooth to a K3, but one does not get an explicit construction of the smoothing $\cX_t$. 

There is also no polytope construction here whatsoever, only the fan construction for the central fiber. GHK and Siebert have been working on the polytope side. It's hard: it's not clear what the multiplication rule for theta functions should be.
We represent an $\IAS^2$ with the following data:

\dzg{Missing, see video.}

This recovers $\cX_0$ by taking fans at vertices.
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:::{.remark}
Joint work with Valery: a polarizing divisor is a divisor $R$ in the generic K3 surface in $F_{2d}(\CC)$. This corresponds to a choice of ample divisor on a Zariski open subset of $F_{2d}(\CC)$. For such a choice, we define $\overline{F_{2d}}^R$ to be the closure of K3 pairs $(X, \eps R)$ in the space of KSBA stable pairs. A generic K3 has Picard rank 1, and it's in the ample class, so any divisor on the generic K3 is automatically ample. Thus $K_X + \eps R > 0$ since $K_X = 0$. The pair also has slc singularities. Note that we've allowed all K3s to have ADE singularities, these are examples of slc, and taking $\eps$ small enough resolves any problems. One needs $R$ not to pass through log canonical centers, and there are no log canonical centers on an ADE K3.
Their theorem gives an explicit description of such a moduli space.
:::

:::{.remark}
We say $R$ is recognizable if it extends to a unique divisor $R_0$ on any Kulikov surface.
Idea: for $\cX_0$ there are many different smoothing families $\cX_i$ and choices of divisors $R_i$. For any 1-parameter family, taking the Zariski closure of $R_i$ yields a flat limit $R_{i, 0}$ on $\cX_0$. If $R$ is recognizable, these flat limits do not vary, so the choice of divisor can be made on \textit{any} K3, even a smooth K3. If $R$ is a recognizable polarizing divisor, there is a unique semifan $F_R$ such that
\[
\overline{F_{2d}}^{F_R} = (\overline{F_{2d}}^R)^\nu
\]
This relates a Hodge-theoretic compactification on the left with a geometric compactification on the right.
:::

:::{.remark}
A semitoroidal compactification simultaneously generalizes toroidal and BB compactifications. 

Recall that assocaited to a degeneration of K3s we had $\vec \lambda \da (\lambda_1, \cdots, \lambda_{19}) \in \delta^\perp/\delta$, a signature $(1, 18)$ lattice. Friedman-Scattone show that $\vec \lambda^2$ is the number of triple points in $\cX_0$.
The semifan $F_R$ is a locally polyhedral $\Gamma_\delta \da \Stab_\Gamma(\delta)$ invariant decomposition of the positive cone $C^+ \subset \delta^\perp/\delta$. This is the future light cone in the corresponding hyperbolic space. Roughly $\overline{F_{2d}}^{F_R}$ is $X(F_R)/\Gamma_\delta$. 

Why this? We had a torus embedding of the first partial quotient $\dcosetl{U_\delta}{\DD} \to (\CC^*)^{19}$ and the latter is canonically identified with $\delta^\perp/\delta\tensor \CC^*$. The monodromy invariant $\vec \lambda$ was approximated by the cocharacter $\lambda \tensor \CC^*$. We extend that torus by a toric variety whose fan has support in $\delta^\perp/\delta$. 

Note that here semitoroidal corresponds to \textit{locally} polyhedral. A globally polyhedral tiling condition would just yield a usual fan. For instance, the cones here might have infinitely many rational polyhedral walls. On the other hand, the BB compactification corresponds to the trivial compactification of $C^+$ which is just the entirety of $C^+$. 
:::

:::{.remark}
AE prove that recognizable divisors $\rc{R} \da \sum_{C\in \abs L, C^\nu \cong \PP^1} C$ exist. The rational curve divisor is always recognizable for any degree $2d$, so this exhibits some semitoroidal compactifications with geometric meaning.

AET give some explicit examples for $F_2$. Degree 2 K3s are generically 2-to-1 covers $\pi:X\to \PP^1$ branched over a sextic, take $L \da \pi^* \OO_{\PP^1}(1)$. One takes the $R$ to be the ramification divisor $R\in \abs{3L}$; it is a recognizable divisor. They construct a semifan $F_R$ which is a coarsening of the Coxeter fan for the root system in $\delta^\perp/\delta$; one takes a subset of the root mirrors.

The construction of the singular K3 surface: start with the heart $\IAS^2$, triangulate completely, double this construction, replace each vertex with the surface defined by the star. Note the cuts introducing shears along the boundary. The cuts introduce 3 singularities in each hemisphere, and angular defects of the polygonal gluing introduce 18 singularities along the equator. This $\IAS^2$ has an involution, and this $\cX_0$ naturally has an involution. The ramification divisor of the $\IAS^2$ is in blue, it's a tropical ramification divisor. It is the dual complex of the limit of ramification divisors.

Why is this recognizable? $\cX_0$ admits an involution $\iota_0$. From this we can determine the limit of $\Fix(\iota)$. Note that $\cX_0$ alone determines $R_0$, the limit of $R_t$, and $R_0 = \Fix(\iota_0)$. This implies recognizability since the choice of divisor $R$ can be made on any Kulikov surface.
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:::{.remark}
On joint work with ABE for elliptic K3s. Take $X\to \PP^1$ an elliptic fibration with fiber $f$ and section $s$. This is not of the form $F_{2d}$, since here one takes $H = \ZZ s \oplus \ZZ f$ for the polarization. Generically the fibration has 24 singular fibers. They show $R \da s + m\sum f_i$ for $f_i$ the singular fibers is recognizable for any multiple $m$. The lattice ${\rm II}_{1, 17}$ is reflective, and $F_R$ here refines the Coxeter chamber into 9 subchambers. This is a fan which is strictly not a semifan. There is a corresponding tropical elliptic K3 given by the following $\IAS^2$.

image

Here one glues the top too the bottom, identifying the segments by a vertical shear.
Note it has an $S^1$ fibration which tropicalizes the elliptic fibration. The blue vertical lines are limits of singular fibers, the blue horizontal is the limit of the section. 


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