Let $\mathcal{F}_{2d}$ be the moduli space of polarized K3 surfaces of degree $2d$ possibly with ADE singularities. Its structure is known as follows. Let $\Lambda_{\rm K3}:=E_{8}(-1)^{\oplus 2}\oplus U^{\oplus 3}$ be the K3 lattice and fix a primitive vector $\lambda_{2d}$ with $(\lambda_{2d},\lambda_{2d})=2d$ and $\Lambda_{2d}:=\lambda_{2d}^{\perp}$. The complex manifold

$$\Omega(\Lambda_{2d}):=\{[w]\in\mathbb{P}(\Lambda_{2d}\otimes\mathbb{C})\mid(w,w)=0,\ (w,\bar{w})>0\}.$$

has two connected components. We choose one component and denote by $\mathcal{D}_{\Lambda_{2d}}$. Let $O(\Lambda_{\rm K3})$ denote the isomorphism group of the lattice $\Lambda_{\rm K3}$ preserving the bilinear form and set

$\displaystyle\tilde{O}(\Lambda_{2d}):=\{g|_{\Lambda_{2d}}:g\in O(\Lambda_{\rm K3}),\,g(\lambda_{2d})=\lambda_{2d}\}.$

The group $\tilde{O}(\Lambda_{2d})$ naturally acts on $\Omega(\Lambda_{2d})$. We define $\tilde{O}^{+}(\Lambda_{2d})$ to be the index two subgroup of $\tilde{O}(\Lambda_{2d})$ consisting of the elements preserving each connected component of $\Omega(\Lambda_{2d})$. Then it is well-known that

$\displaystyle\mathcal{F}_{2d}\simeq\tilde{O}^{+}(\Lambda_{2d})\backslash\mathcal{D}_{\Lambda_{2d}}\simeq\tilde{O}(\Lambda_{2d})\backslash\Omega(\Lambda_{2d}).$

Let $\overline{\mathcal{F}_{2d}}^{{\rm Sat},\tau_{\rm ad}}$ (or simply $\overline{\mathcal{F}_{2d}}^{{\rm Sat}}$ in our papers) be the Satake compactification of $\mathcal{F}_{2d}$ corresponding to the adjoint representation of $O(2,19)$. It decomposes as

$$\overline{\mathcal{F}_{2d}}^{{\rm Sat}}=\mathcal{F}_{2d}\sqcup\bigcup_{l}\mathcal{F}_{2d}(l)\sqcup\bigcup_{p}\mathcal{F}_{2d}(p),$$

where $l$ runs over one-dimensional isotropic subspaces of $\Lambda_{2d}\otimes\mathbb{Q}$, and $p$ runs over two-dimensional isotropic subspaces of $\Lambda_{2d}\otimes\mathbb{Q}$. Also, we simply define the tropical geometric compactification of $\mathcal{F}_{2d}$ as this $\overline{\mathcal{F}_{2d}}^{\rm Sat}$. The boundary component $\mathcal{F}_{2d}(l)$ is given as

$$\mathcal{F}_{2d}(l)=\{v\in(l^{\perp}/l)\otimes\mathbb{R}\mid(v,v)>0\}/\sim.$$

Here $v\sim v^{\prime}$ if $g\cdot v=cv^{\prime}$ for some $g\in\tilde{O}^{+}(\Lambda_{2d})$ and $c\in\mathbb{R}^{\times}$. We have $\mathcal{F}_{2d}(l)=\mathcal{F}_{2d}(l^{\prime})$ if $g\cdot l=l^{\prime}$ for some $g\in\tilde{O}^{+}(\Lambda_{2d})$ and $\mathcal{F}_{2d}(l)\cap\mathcal{F}_{2d}(l^{\prime})=\emptyset$ if otherwise. Since $(l^{\perp}/l)\otimes\mathbb{R}$ has signature $(1,18)$, there is an isomorphism

$$\{v\in(l^{\perp}/l)\otimes\mathbb{R}\mid(v,v)>0\}/\mathbb{R}^{\times}\\ \simeq O(1,18)/O(1)\times O(18)$$

and hence $\mathcal{F}_{2d}(l)$ is an arithmetic quotient of $O(1,18)/O(1)\times O(18)$. The other component $\mathcal{F}_{2d}(p)$ is a point and $\mathcal{F}_{2d}(p)=\mathcal{F}_{2d}(p^{\prime})$ if and only if $g\cdot p=p^{\prime}$ for some $g\in\tilde{O}^{+}(\Lambda_{2d})$. Therefore, if we take representatives of $l$ and $p$ from each equivalence class, we get a finite decomposition:

$$\overline{\mathcal{F}_{2d}}^{{\rm Sat}}=\mathcal{F}_{2d}\sqcup\bigsqcup_{l}\mathcal{F}_{2d}(l)\sqcup\bigsqcup_{p}\mathcal{F}_{2d}(p).$$