Proof. Let $r$ be an $(n-1)$-dimensional simplex of $\Delta$, and thus a ridge. Then $C_r$ is a curve, and $D_r$ is the union of points corresponding to facets containing $r$, and thus is non-trivial if and only if there are any such facets. If the divisor $D_r$ is non-empty, then it is ample, and so it is big and also $C_r \setminus D_r$ is affine. On the other hand, if $D_r$ is the trivial divisor, then it is clearly not big, and $C_r \setminus D_r$ is a projective curve and so not affine.

Proof of Proposition 1.2 We fix a component $C_v$ of the special fiber $\mathfrak X_0$. By assumption, $C_v$ is projective, so, by Bertini's theorem, we can choose smooth and irreducible elements $H_1, \ldots , H_n$ from the linear system of a very ample divisor on $C_v$, such that the $H_i$ intersect both each other and $D_v$ transversely. We now blow up the points of the intersection $H_1 \cap \cdots \cap H_n$. Then, for each integer $k$ from $1$ to $n-1$, we blow up the strict transforms of the $k$-dimensional varieties in $H_{i_1} \cap \cdots \cap H_{i_{n-k}}$ for all tuples of indices $1 \leq i_1 < \cdots < i_{n-k} \leq n$. For each $k$, these strict transforms are disjoint and thus the order of the blow-ups within a fixed dimension does not matter.

We then repeat the above blow-ups at the strict transform of each component of $\mathfrak X$ to obtain the degeneration $\mathfrak X'$ from the statement. Since $\mathfrak X'$ is the iterated blow-up of a regular degeneration at smooth subvarieties, it is also regular. The centers of the blow-ups intersect the singular locus of the special fiber transversely, so the special fiber $\mathfrak X'_0$ is also reduced and simple normal crossing. Thus, $\mathfrak X'$ is a strict degeneration.

In order to show that the locally closed strata of $\mathfrak X'$ are all affine, we first consider the case of a stratum $C_s'$ of $\mathfrak X_0'$ which maps birationally onto its image in $\mathfrak X$. We let $C_s$ be its image in $\mathfrak X$, and then $C_s'$ is formed by blowing up $C_s$ at the restrictions of intersections of very ample divisors from a component $C_v$ for each vertex $v$ in $s$. Each of these blow-ups produces a new component for $\mathfrak X'$ and thus the difference $C_s' \setminus D_s'$ is an open subset of $C_s$ minus the very ample divisors. This containment may be proper because of the intersection of $C_s$ with other components of $\mathfrak X_0$. However, the complement of a very ample divisor is affine, and the complement of a Cartier divisor in an affine variety is affine, so $C_s' \setminus D_s'$ is affine, as desired.

Let us now consider the components introduced by the blow-ups. The center for such a blow-up is a variety $Y$ within a single component $C_v$. Since we have blown up the intersection of $Y$ with very ample divisors in the previous step, the complement $Y \setminus (Y \cap D_v)$ is affine, as in the previous paragraph. The blow-up of $Y$ is a projective bundle over $Y$ which intersects $C_v$ along a section. Let $C_w$ denote this blow-up and then $C_w \setminus D_w$ is an affine space bundle over an affine variety $Y \setminus (Y \cap D_v)$, and thus it is affine. Further blow-ups only intersect $C_w$ along $D_w$ and therefore do not affect the difference $C_w \setminus D_w$, which remains affine in $\mathfrak X'$. We conclude that the locally closed strata of $\mathfrak X'$ are all affine. As noted above, since the locally closed strata are affine, $\mathfrak X'$ is robust in all dimensions

Proof. We first assume that any two terms of $Z$ intersect in dimension less than $n-1$ and that the coefficients of $Z'$ are positive. Let $p$ be a point of an arbitrary term $T_j$ of $Z$. By definition, the difference $Z - Z'$ is a finite sum of expressions of the form $[P] - [Q_1] - \cdots - [Q_r]$, where $P = Q_1 \cup \cdots \cup Q_r$, and the intersections $Q_i \cap Q_j$ are proper faces of each. We can assume that $p$ is not contained in any facet of the polyhedra appearing in these expressions or facet of the $T_i$. Therefore, if $p$ is in the polyhedron $P$ of one of these expressions, then $p$ is in exactly one of the corresponding $Q_i$. Thus, equivalence between $Z$ and $Z'$ does not change the sum of the coefficients of the polyhedra containing $p$, meaning that

(3)\begin{equation} \sum_{P_i \ni p} a_i =\sum_{P_i' \ni p} a_i'. \end{equation}

Since the coefficients $a_i'$ are positive, the right-hand side of (3) is non-negative. By our assumption on $Z$, the left-hand side consists of just the coefficient $a_j$, which is therefore non-negative.

Second, we assume that the terms of $Z$ and $Z'$ are arbitrary, but the coefficients $a_i$ and $a_i'$ are all positive. Again, let $p$ be a point in one term $T_j$ of $Z$, such that $p$ is not contained in any of the facets of the polyhedra appearing in the equivalence between $Z$ and $Z'$, and we again have equality (3). The left-hand side of (3) is positive since $p \in T_j$, so $p$ must also be contained in some term $T_k'$ of $Z'$. Since we were able to choose $p$ from a dense open subset of ${\bigcup} P_i$, and the union ${\bigcup} P_i'$ is closed, we have an inclusion ${\bigcup} P_i \subset {\bigcup} P_i'$. By symmetry, we have the reverse inclusion, and thus the desired equality.

Proof. Recall that the rank of $D$ is the largest number $r$ such that, for any $r$ points $p_1, \ldots , p_r$ in $\Gamma$, the difference $D - [p_1] - \cdots - [p_r]$ is linearly equivalent to an effective divisor $D'$. Thus, $D' + [p_1] + \cdots + [p_r]$ is an effective divisor linearly equivalent to $D$ and it clearly contains the $p_i$, so $h^0(D) \geq r(D) + 1$. To show the reverse inequality, we assume that there exist points $p_1, \ldots , p_{r(D) + 1}$ such that $D$ is not linearly equivalent to $D' + [p_1] + \cdots + [p_{r(D) + 1}]$ for any effective divisor $D'$. In other words, if we write $\lvert D \rvert$ for the subset of $\Gamma ^d$ which are effective divisors linearly equivalent to $D$, then $\lvert D \rvert$ does not intersect the set

\[ \{p_1\} \times \cdots \times \{p_{r(D) + 1}\} \times \Gamma \times \cdots \times \Gamma \subset \Gamma^d. \]

However, $\lvert D \rvert$ is a closed subset [Reference Mikhalkin and ZharkovMZ08, Theorem 6.2]. Therefore, since $\Gamma$ is not a point, we can perturb the points $p_i$ slightly and the intersection with $\lvert D \rvert$ will still be empty. In particular, we can make the $p_i$ distinct and rational. Thus, $h^0(D) \leq r(D) + 1$, so we have proved the proposition.

Proof. Let $\phi$ be a PL function on $\Delta$ such that $\operatorname {div}(\phi ) + mD$ is effective. Let $C$ be a horizontal circle on the cylinder $\Delta$ parallel to $D$, but not equal to $D$ or the bottom of the cylinder. Let $p$ be a point of $C$ at which the restriction $\phi \vert _C$ achieves its maximum. The key point is that the linear functions on a neighborhood $U$ of $p$ embed $U$ in an open subset of $\mathbb {R}^2$ and $C$ remains a straight line in this embedding. Since the divisor of $\phi$ is effective in $U$, $\phi \vert _{U}$ is a convex function of $\mathbb {R}^2$. Thus, the restriction to $C$, which is a line segment, is also convex, but $\phi \vert _C$ achieves its maximum at $p$ and so $\phi \vert _{C \cap U}$ must be constant. Since $\phi$ is constant in any neighborhood of a point in $C$ where $\phi \vert _C$ achieves its maximum, $\phi$ is constant on $C$.

Thus, $\phi$ is a function solely of the vertical coordinate on $\Delta$, and $\phi$ defines a linear equivalence between $mD$ and $m$ horizontal circles, possibly not distinct. For any $m$ points, these circles can be chosen to contain them, but for $m+1$ points chosen to have $m+1$ distinct heights, no sum of $m$ circles will contain all of the points. Therefore, $h^0(\Delta , mD) = m+1$, as claimed.

Proof Theorem 1.4 Suppose that $\mathfrak X$ is a degeneration over a discrete valuation ring whose weak tropical complex is $\Delta$. Since all the maximal simplices of $\Delta$ are two-dimensional, and it can be checked that $\Delta$ is a tropical complex, $\mathfrak X$ must be robust in dimensions $1$ and $2$ by Propositions 2.3 and 2.5, respectively. The general fiber $X$ is a smooth proper surface and therefore projective. We let $A$ be an ample divisor on $X$. By Proposition 2.14, $\rho (A)$ is a divisor, and from the definition of the specialization (2), we know that $\rho (A)$ is a linear combination of the ridges of $\Delta$.

We now examine the possibilities for $\rho (A)$. The set of all formal sums of the eight ridges of $\Delta$ forms a free Abelian group of rank 8, and those satisfying the balancing condition are a subgroup. In a neighborhood of each vertex $v$ of $\Delta$, the space of linear functions can be computed by considering functions which are linear on the interior of each two-dimensional simplex, and satisfy the condition in Definition 2.10 along each ridge containing $v$. In each case, the space of linear functions has only one parameter for the slopes. For example, let $v$ be the vertex indicated in Figure 1, and $e_1$, $e_2$, $e_3$, and $e_4$ be the edges incident to $v$, beginning with $e_1$ to the left and continuing counterclockwise. Then the linear functions are constant along the diagonal edge $e_2$, and affine linear under the embedding shown. Therefore, the slopes along $e_3$ and $e_4$ are equal, and are the negative of the slope along $e_1$. Therefore, if $c_1$, $c_3$, and $c_4$ are the coefficients of these three edges in a divisor, then $c_1 - c_3 -c_4 = 0$. Similarly, the other three vertices of $\Delta$ also give linear conditions on the coefficients of a divisor, and one can check that these are independent, so the group of formal sums of the ridges which are divisors is a free Abelian group of rank 4.

Next, one can compute the subgroup of divisors which are linearly equivalent to the trivial one by considering PL functions which are linear on the interior of each two-dimensional simplex. Using the properties in Proposition 2.15, one can then check that the group of divisor classes which are linearly equivalent to a formal sum of ridges is isomorphic to $\mathbb {Z}$, with the divisor $D$ from Lemma 3.4 representing twice a generator.

Therefore, $\rho (2A) = 2 \rho (A)$ is linearly equivalent to $lD$ for some integer $l$. Thus, by Theorem 1.1 and Lemma 3.4, we have the inequality

\[ h^0(X, \mathcal{O}(2mA)) \leq h^0(\Delta, mlD) = ml + 1. \]

In particular, $h^0(X, \mathcal {O}(2mA))$ grows at most linearly in $m$, which contradicts the fact that $A$ is an ample divisor on a surface, whose global sections should grow quadratically. Therefore, we conclude that no degeneration $\mathfrak X$ with weak tropical complex $\Delta$ can exist.

Proof. We need to show that the $\alpha (v', r')$ from Construction 4.1 are integral and that they satisfy the required identity (1) for each ridge. We retain the same notation as in Construction 4.1 and we first consider a ridge $r'$ meeting the interior of a facet of $\Delta$. Then, because of the unimodularity, the coordinates of $(w_1 + w_2)/2$ will be half-integers, and so the $c_i$ are also half-integers, and $\alpha (v_i', r') = 2c_i$ is an integer. Moreover, the sum of the $\alpha (v_i', r')$ will equal $2$, which is the degree of $r'$ in $\Delta '$.

Figure 2.

The gray lines show an order $2$ subdivision of two triangles glued along a common edge. In both the original and subdivided complex, $\alpha (v,r) = 1$ for all ridges $r$ which are contained in two facets.

Now consider a ridge $r'$ of the subdivision $\Delta '$ contained in a ridge $r$ of $\Delta$. The fact that the vectors $v_1', \ldots , v_n'$ form the vertices of one simplex of a unimodular triangulation implies that the differences between pairs of the vectors $(y_{i,1}, \ldots , y_{i,n})$ span the integral vectors $\mathbb {Z}^{n}$ whose sum is $0$. Moreover, each vector satisfies the affine linear equation $y_{i,1} + \cdots + y_{i,n} = m$, so the span of these vectors includes all vectors in $\mathbb {Z}^{n}$ whose sum is a multiple of $m$. Thus, to check that the coefficients defined in (7) are integral, it is sufficient to check that the sum of the entries of the vector from the right-hand side of (7) is a multiple of $m$. Indeed, we know that $\alpha (v_1, r) + \cdots + \alpha (v_n, r) = d$ and that $x_{i, 1} + \cdots + x_{i,n} = m-1$, so the total sum is $dm$. Moreover, this implies that $\alpha (v_1', r') + \cdots + \alpha (v_n', r') = d$, which completes the proof that the subdivision is a weak tropical complex.

Proof. The agreement of PL functions is clear, so long as the scaled function $m\phi$ on $\Delta '$ has integral slopes on its domains of linearity. However, each facet $f'$ of $\Delta '$ is identified with a $(1/m)$-scaled copy of the standard unit simplex. If we instead identify $f'$ with the standard unit simplex, then the slopes of $\phi$ will be scaled by $1/m$, and so the slopes of $m\phi$ will all be integers.

For linear functions, we let $r'$ be a ridge of $\Delta '$ and we need to show that linear functions on a neighborhood of $r'$ in $\Delta$, scaled by $m$, agree with linear functions defined by the map $\phi _{r'} \colon N_{r'} \rightarrow L_r \cong \mathbb {R}^{d+n-1}$ from Definition 2.10. First suppose that $r'$ intersects the interior of a facet $f$ of $\Delta$. Then $N_{r'}$ is the union of the two facets containing $r'$, and naturally embeds in $\mathbb {R}^{2+n}$ by sending each vertex to a coordinate vector, as in Definition 2.10. The inclusion of $N_{r'}$ into $f \subset \mathbb {R}^n$ factors through a linear map from $\mathbb {R}^{2+n}$ to $\mathbb {R}^n$, whose kernel contains $(1, 1, -\alpha (v_1', r'), \ldots , -\alpha (v_n', r'))$ by the choice of those structure constants in Construction 4.1. Therefore, the inclusion of $N_{r'}$ into $\mathbb {R}^n$ factors through the map $\phi _{r'} \colon N_{r'} \rightarrow L_{r'}$, defining linear functions on the open subset $N_{r'}^o$, and so the linear functions on $\Delta$ and $\Delta '$ agree.

Now suppose that $r'$ is contained in a ridge $r$ of $\Delta$, and we let $v_i$, $v_i'$, $x_{i,j}$ and $y_{i,j}$ be as in Construction 4.1, and $\phi _r$ as in Definition 2.10. Then, using the fact that $\phi _r$ is affine linear on each facet of $\Delta$, we can evaluate

\begin{align*} \sum_{i=1}^d \phi_r(w_i') &= \sum_{i=1}^d \frac{1}{m} \bigg (\phi_r(w_i) + \sum_{j = 0}^n x_{i,j} \phi_r(v_j)\bigg) \\ &= \frac{1}{m}\sum_{j=0}^n \alpha(v_j, r) \phi_r(v_j) + \frac{1}{m}\sum_{i=1}^d\sum_{j=0}^n x_{i,j}\phi_r(v_j) \\ &= \frac{1}{m} \sum_{j=0}^n \bigg (\alpha(v_j, r) + \sum_{i=1}^d x_{i,j} \bigg)\phi_r(v_j) \\ &= \sum_{j=0}^n \alpha(v_j', r') \phi_r(v_{j}'), \end{align*}

where the last step is by the change of variables in (7), together with the linearity of $\phi _r$ on $r$. Thus, $\phi _r$ satisfies the defining linear relation of $\phi _{r'}$, and so $\phi _{r'}$ is isomorphic to the restriction of $\phi _r$, as in the previous paragraph.

Proof. By Lemma 2.12, passing to equivalent formal sums of polyhedra does not change whether or not the formal sum is balanced, so we can assume that the terms of the formal sum of polyhedra are each contained in the facets of $\Delta '$. Then Lemma 4.4 shows the linear functions on $\Delta$ and $\Delta '$ are the same, up to multiplication by $m$, so we just need to check that the multiplicity $\operatorname {mult}_{Z,Q}(\phi )$ of a linear function along $Q$ of a formal sum of polyhedra $Z$ vanishes in $\Delta$ if and only if it vanishes in $\Delta '$.

Here, the key difference is between two different embeddings of a single facet $f'$ of $\Delta '$ into $\mathbb {R}^n$. First, we can identify $f'$ with a standard unit simplex in $\mathbb {R}^n$. Second, we identify $f'$ with a simplex contained in a facet $f$ of $\Delta$, and $f$ is identified with a standard unit simplex in $\mathbb {R}^n$. Because the subdivision of $\Delta$ is assumed to be unimodular in Construction 4.1, in the latter identification, $f'$ is a standard unit simplex scaled by $1/m$. Thus, if we transform the vectors $\mathbf {w}_i$ and $\mathbf {v}_i$ used to compute the multiplicity in $\Delta$ into the coordinates where $f$ is identified with a standard unit simplex, then the transformed vectors have a common multiple of $m$ in their coordinates. Scaling these vectors by $1/m$ gives suitable vectors for computing the multiplicity in $\Delta '$, and so the multiplicity of a given linear function along $Q$ of $Z$ in $\Delta '$ is $1/m$ times the multiplicity in $\Delta$, which completes the proof.

Proof. To show that linear equivalence on $\Delta$ agrees with that on $\Delta '$, we need to show that, if $\phi$ is a PL function on $\Delta$, then $\operatorname {div}(\phi ) = \operatorname {div}'(m \phi )$, where $\operatorname {div}$ and $\operatorname {div}'$ are the divisor functions from Proposition 2.15 on $\Delta$ and $\Delta '$, respectively. Since Proposition 2.15 uniquely characterizes $\operatorname {div}$ and $\operatorname {div}'$ in terms of four properties, we need to check that those properties coincide for $\Delta$ and $\Delta '$. The first two properties are that the divisor is linear in the PL function and local, which are independent of the subdivision. Moreover, the third property is that linear functions are characterized by having trivial divisor, and by Lemma 4.4, the linear functions on $\Delta$ and $\Delta '$ agree, up to scaling. The fourth property normalizes the divisor function via a PL function supported on a single facet, which also agrees, because when passing to $\Delta '$ we scale both the coordinates on the complex and the PL function by the same factor of $m$.

Proof. In a neighborhood of any point of $C_s$ not in a smaller stratum, $\mathfrak X$ is étale over

\[ \operatorname{Spec} R[t_0, \ldots, t_m, t_{m+1}^{{\pm}}, \ldots, t_{n}^{{\pm}}]/\langle t_0 \cdots t_m - \pi \rangle, \]

with the divisor $C_v$ defined by the element $t_0$. Then, the open stratum of $C_{v'}$ is defined by the fiber product with the spectrum of $R'[t_0^{\pm }, \ldots , t_n^{\pm }]^{v'}$, which is defined to be the $R'$-algebra generated by

\[ \{a t_0^{u_0} \cdots t_n^{u_n} \mid a \in R', \operatorname{val}(a) + x_0 u_0 + \cdots + x_k u_k \geq 0\}, \]

where $\operatorname {val}$ is the discrete valuation on $R'$, by the display equation on [Reference Kempf, Knudsen, Mumford and Saint-DonatKKMS73, p. 76]. The ideal of the component $C_{v'}$ is generated by any $a t_0^{u_0} \cdots t_n^{u_n}$ such that $\operatorname {val}(a) + x_0 u_0 + \cdots + x_k u_k = 1$ and the valuation of $a t_0^{u_0} \cdots t_n^{u_n}$ along $C_{v'}$ is $\operatorname {val}(a) + x_0 u_0 + \cdots + x_k u_k$. In particular, the valuation of $t_0$ is $x_0$ and so the multiplicity of $C_{v'}$ in $\pi ^*(C_v)$ is $x_0$.

Proof. By [Reference Kempf, Knudsen, Mumford and Saint-DonatKKMS73, Chapter II, Theorem 8*], $\mathfrak X' \rightarrow \mathfrak X \times _{\operatorname {Spec} R} \operatorname {Spec} R'$ is proper, so $\mathfrak X'$ is a proper scheme over $\operatorname {Spec} R'$. In addition, $\mathfrak X'$ is regular by [Reference Kempf, Knudsen, Mumford and Saint-DonatKKMS73, Chapter II, Theorem 4*] and because each simplex in the order $m$ subdivision $\Delta '$ is unimodular. Since each vertex is integral, each component of the special fiber of $\mathfrak X'$ is reduced [Reference Kempf, Knudsen, Mumford and Saint-DonatKKMS73, Chapter II, Theorem 1*], and the special fiber is simple normal crossing because $\Delta '$ consists of simplices. Therefore, $\mathfrak X'$ is a regular, semistable degeneration, with dual complex $\Delta '$ and it remains to show that its weak tropical complex is as given in Construction 4.1.

Now let $r'$ be any facet of $\Delta '$ with vertices $v_1', \ldots , v_n'$. First suppose that $r'$ is a ridge of $\Delta '$ meeting the interior of a facet $f$ of $\Delta$. Let $v_0, \ldots , v_n$ be the vertices of $f$. As in Lemma 4.7, use coordinates on $f$ consisting of non-negative integers summing to $m$ such that $v_0$ is $(m,0,\ldots ,0)$, $v_1$ is $(0,m,0,\ldots , 0)$, and so on. Let $(x_{i,0}, \ldots , x_{i,n})$ be the coordinates of $v_i'$ and let $(y_{1,0}, \ldots , y_{1,n})$ and $(y_{2,0}, \ldots , y_{2,n})$ be the coordinates of the vertices $w_1'$ and $w_2'$ in the facets on either side of $r'$. Then we compute the structure constants $\alpha (v_i', r') = C_{v_i'} \cdot C_{r'}$ of $\mathfrak X'$ by computing $\psi ^*(C_{v_i}) \cdot C_{r'}$ in two different ways, where $\psi$ is the map $\psi \colon \mathfrak X' \rightarrow \mathfrak X$. First, by Lemma 4.7, the coefficient of $C_{v_i'}$ in $\psi ^*(C_{v_j})$ is $x_{i, j}$ and the coefficient of $C_{w_i'}$ is $y_{i,j}$. Since these are the only components of the special fiber of $\mathfrak X'$ meeting $C_{r'}$,

(8)\begin{equation} \psi^*(C_{v_j}) \cdot C_{r'} = y_{1,j} + y_{2,j} - \alpha(v_1', r')x_{1,j} - \cdots - \alpha(v_n, r')x_{n,j}. \end{equation}

On the other hand, $\psi (C_{r'})$ is the point $C_{f}$ and so its pushforward is trivial. Therefore, by the projection formula,

(9)\begin{equation} \psi^*(C_{v_j}) \cdot C_{r'} = C_{v_j} \cdot \psi_*(C_r') = 0. \end{equation}

Combining (8) and (9) for all the different indices $j$, we get the vector equation

\[ 0 = w_1 + w_2 - \alpha(v_1', r')v_1' - \cdots - \alpha(v_n', r')v_n'. \]

Rearranging and dividing by $2$,

\[ \frac{w_1 + w_2}{2} = \frac{\alpha(v_1', r')}{2} v_1' + \cdots + \frac{\alpha(v_n', r')}{2} v_n', \]

which shows that the $\alpha (v_i', r')$ are the same in the definition of the structure constants on the subdivision $\Delta '$ in (6).

Second, we consider the case when $r'$ is contained in a ridge $r$ of $\Delta$. We adopt the notation of Construction 4.1, so that $v_1, \ldots , v_n$ are the vertices of $r$. We use coordinates similar to those above and let $(y_{i,1}, \ldots , y_{i,n})$ denote the coordinates of $v_i$ and $(x_{i,1}, \ldots , x_{i,n},1)$ denote the coordinates of the vertices of $\Delta '$ in facets containing $r'$, where the last coordinate is $1$ because the vertex is a distance $1$ from $r'$ by unimodularity. Again, we compute the intersection number $\psi ^{-1}(C_{v}) \cdot C_{f'}$ in two different ways. First, using the Lemma 4.7, we can represent $\psi ^{-1}(C_{v})$ as a linear combination of divisors in $\mathfrak X'$. The ones which intersect $C_{r'}$ correspond to the vertices $v_i$ of $r'$, together with the other vertices of facets of $\Delta '$ containing $r'$. All together, these give

\[ \psi^*(C_v) \cdot C_{r'} = x_{1,i} + \cdots + x_{d,i} - \alpha(v_0', f') y_{0,i} - \cdots - \alpha(v_n', f') y_{n,i}. \]

On the other hand, using the projection formula, $\psi ^{-1}(C_{v_i}) \cdot C_{r'} = C_{v_i} \cdot \psi (C_{r'})$, and we know that $\psi *(C_{r'})$ is $C_r$, so the intersection number is $-\alpha (v_i, r)$. Putting these equations together for all $i$, we get a matrix equation

\[ \begin{pmatrix} y_{0,0} & \cdots & y_{n,0} \\ \vdots & & \vdots \\ y_{0,n} & \cdots & y_{n,n} \end{pmatrix} \begin{pmatrix} -\alpha(v_0', F') \\ \vdots \\ -\alpha(v_n', F') \end{pmatrix} + \begin{pmatrix} x_{1,0} + \cdots + x_{d,0} \\ \vdots \\ x_{1,n} + \cdots + x_{d,n} \end{pmatrix} =\begin{pmatrix} -\alpha(v_0, F) \\ \vdots \\ -\alpha(v_n, F) \end{pmatrix}. \]

We can solve for the $\alpha (v_i', F')$ and we get (7). We conclude that the structure constants of the weak tropical complex of $\mathfrak X'$ are those given in Construction 4.1.

Proof. Our strategy for constructing $\mathfrak X'$ is to first associate a Newton polyhedron to the local equation of $D$ near each stratum of $\mathfrak X$. These Newton polyhedra define a polyhedral subdivision of $\Delta$, which can be further subdivided to a unimodular simplicial subdivision, which then gives a regular semistable degeneration $\mathfrak X'$ by Lemma 4.8. We then check that $\mathfrak X'$ satisfies the desired properties.

In order to define a local Newton polyhedron, we let $s$ be any simplex of $\Delta$, and let $k$ be the dimension of $s$. Let $A_s$ be the local ring of $\mathfrak X$ at the generic point of $C_s$. By assumption, $\mathfrak X$ is regular, so $A_s$ is a regular local ring, and because $\mathfrak X_0$ is a simple normal crossing divisor, the maximal ideal of $A_s$ is generated by $x_0, \ldots , x_k$, each of which is the local equation of one of the components of $\mathfrak X_0$. The completion $\widehat A_s$ of $A_s$ is isomorphic to $S[[x_0, \ldots , x_k]]/\langle x_0 \cdots x_k - \pi \rangle$, where $S$ is a complete, discretely valued extension of $R$ and $\pi$ is a uniformizer in $S$. We let $f \in A_s \subset \widehat A_s$ be the local equation of the closure $\bar D$. Using the isomorphism above, we can write

\[ f = \sum_{a \in I} s_a x_0^{a_0} \cdots x_k^{a_k} \in \widehat R[[x_0, \ldots, x_k]] / \langle x_0 \cdots x_k - \pi \rangle, \]

where $I \subset \mathbb {Z}_{\geq 0}^{k+1}$ is a set of exponent vectors, and we can assume that each coefficient $s_a$ is non-zero. Moreover, by the relation $x_0 \cdots x_k = \pi$, we can rewrite the expression for $f$ so that each coefficient $s_a$ is a unit in $R$, and then this representation is unique. We define the Newton polyhedron of $f$ to be the union of translated orthants,

\[ \operatorname{Newt}_s(f) = \operatorname{conv} \bigg( \bigcup_{a \in I} (a_0, \ldots, a_k) + \mathbb{R}_{{\geq} 0}^{k+1} \bigg)\subset \mathbb{R}_{{\geq} 0}^{k+1}, \]

where $\operatorname {conv}$ denotes the convex hull. Note that the vertices of $\operatorname {Newt}(f)$ are all non-negative integer vectors which are pairwise incomparable, and therefore the set of vectors is finite for the same reason that monomial ideals are finitely generated.

Next, we define the dual subdivision of the $k$-dimensional $s$, which we identify with the convex hull of unit coordinate vectors in $\mathbb {R}^{k+1}$. For any face $Q$ of $\operatorname {Newt}_s(f)$, we define

(10)\begin{equation} Q^* = \Big\{u \in s \subset \mathbb{R}^{k+1}_{{\geq} 0} \mid \mbox{for all } a \in Q,\ a \cdot u = \min_{a' \in \operatorname{Newt}(f)} a' \cdot u \Big\}. \end{equation}

Then $Q^*$ is a rational polytope and, as $Q$ ranges over the faces of $\operatorname {Newt}(f)$, the $Q^*$ form a polyhedral subdivision of $s$, because every point of $s$ defines a linear functional which is minimized on some face of $\operatorname {Newt}(f)$.

In order to build a toroidal modification out of this subdivision, we need to show that the subdivisions of different simplices are compatible. It suffices to show that the subdivision of a simplex $s$ is compatible with that of one of its faces $s'$, which we can take to be the face indexed by $k$. Thus, in the local ring $A_s$, the component $C_{s'}$ is defined by the ideal $\langle x_0, \ldots , x_{k-1}\rangle$, and so $A_{s'}$, the local ring of $C_{s'}$, is the localization of $A_s$ at this ideal. We consider the image of $f$, the local equation of $D$ in the tensor product $\widehat A_s \otimes _{A_s} (A_{s})_{\langle x_0, \ldots , x_{k-1}\rangle }$. We can now regroup the terms of $f$ as

(11)\begin{equation} \sum_{(a_0, \ldots, a_{k-1}) \in I'} \bigg(\sum_{a_k \in I_{(a_0, \ldots, a_{k-1})}} s_ax_k^{a_k} \bigg) x_0^{a_0} \cdots x_{k-1}^{a_{k-1}}, \end{equation}

where $I'$ is the projection of the index set $I$ onto the first $k$ coordinates and $I_{(a_0, \ldots , a_{k-1})}$ is the preimage of $(a_0, \ldots , a_{k-1})$ in this projection. Expression (11) also holds in the completion of the local ring $A_{s'}$ of $C_{s'}$. Since each $s_a$ is not divisible by $\pi$ and $x_k$ is transcendental over $\widehat R$, it follows that $\sum _{a_k} s_{a} x_k^{a^k}$ is not divisible by $\pi$ and so

\[ \operatorname{Newt}_{s'}(f) = \operatorname{conv} \bigg( \bigcup_{(a_0, \ldots, a_{k-1}) \in I'} (a_0, \ldots, a_{k-1}) + \mathbb{R}_{{\geq} 0}^k \bigg), \]

which is the projection of the Newton polyhedron $\operatorname {Newt}_s(f)$ to the first $k$ coordinates. Therefore, a linear functional on $\operatorname {Newt}_{s'}(f)$ is the same as a linear functional on $\operatorname {Newt}_s(f)$ with the last coordinate equal to zero, and so the dual subdivision of $s$ defined by $\operatorname {Newt}_s(f)$ restricted to its face $s'$ agrees with the dual subdivision defined by $\operatorname {Newt}_{s'}(f)$.

Now we have a polyhedral subdivision of $\Delta$. If $m$ is the least common multiple of the denominators of the vertices of the subdivision, then it corresponds to a ramified degree $m$ extension $R'' \supset R$ followed by a toroidal map $\mathfrak X'' \rightarrow \mathfrak X \times _R R''$. Although $\mathfrak X''$ is not semistable, we claim that the closure $\bar D''$ of the pullback of $D$ to $\mathfrak X''$ does not contain any of the strata. Fix a polytope $Q^*$ in the subdivision of $\Delta$, which is dual to a face $Q$ of the Newton polygon $\operatorname {Newt}_s(f)$ of the local equation $f$ of $\bar D$ on a stratum $C_s$ of $\mathfrak X$. Let $a$ be an integer point in $Q$ such that $s_a t_0^{a_0} \ldots t_k^{a_k}$ is a term in $f$. Then an open neighborhood of $\mathfrak X''$ corresponding to $Q^*$ is étale over $R[t_0^\pm , \ldots , t_k^\pm ]^{Q^*}$, which is the $R$-algebra generated by

\[ \{s t_0^{a_0} \cdots t_k^{a_k} \mid \operatorname{val}(s) + u \cdot a \geq 0 \mbox{ for all } u \in Q^*\}. \]

If $a'$ is an integer point in $\operatorname {Newt}_s(f)$, then, by the definition of $Q^*$ in (10), $a' \cdot u \geq a \cdot u$ for all $u \in Q^*$ and so $s_{a'} x^{a' - a}$ is in $R[x_0^\pm , \ldots , x_k^\pm ]^{Q^*}$ for any $s_{a'} \in R$. Thus, $f / x^{a}$ is contained in $R[x_0^\pm , \ldots , x_k^\pm ]^{Q^*}$, and it is the local equation of $\bar D''$. The intersection of $\bar D''$ with the closed stratum is defined by an equation including the term $s_a x^a$ and thus non-zero, so $\bar D''$ does not contain the stratum corresponding to $Q$.

Since $\mathfrak X''$ is not regular, we use [Reference Kempf, Knudsen, Mumford and Saint-DonatKKMS73, Theorem III.4.1], which shows that there exists a further extension and toroidal map $\mathfrak X' \rightarrow \mathfrak X'' \times _{R''} R'$ such that $\mathfrak X'$ is semistable. Since each stratum of $\mathfrak X'$ maps to a stratum of $\mathfrak X''$, the pullback of $\bar D''$ to $\mathfrak X'$ does not contain any strata of the special fiber either. This implies that the pullback of $\bar D''$ is equal to the closure of the pullback of $D$, which is the property claimed in the statement.

Proof of Proposition 1.3 By Lemma 5.2, we can assume that our degeneration $\mathfrak X$ is chosen such that $\bar D$ does not contain any of the closed strata of $\mathfrak X$. Because projection of $X^{\operatorname {an}}$ to the skeleton is compatible with the specialization map, we know that the image of $D^{\operatorname {an}}$ is contained in those simplices $s$ such that $\bar D$ intersects $C_s$. By hypothesis, $\bar D$ does not intersect any zero-dimensional strata $C_f$, and whenever it intersects a curve $C_r$ for some ridge $r$, it does so properly and thus $r$ has positive coefficient in the specialization of $D$. Therefore, the projection of the analytification is contained in the union of the support of the refined specialization of $D$, together with some simplices of dimension $0 \leq n-2$.

It now only remains to show that the projection map is surjective onto the ridges $r$ in the support of the specialization. For this we suppose that $r$ is a ridge such that $\bar D$ intersects $C_r$ and let $x$ be a point in the intersection. In a neighborhood of $x$, we can take local defining equations for the divisors containing $C_r$ and use them to define a map to $\operatorname {Spec} S$ where $S = R[x_1, \ldots , x_{n}] / \langle x_1 \cdots x_{n} - \pi \rangle$. As in [Reference NicaiseNic11, § 3.3, Case 1], the $(n-1)$-dimensional simplex, which we identify with $r$, embeds in the analytification of $\operatorname {Spec} S$ as the skeleton. Since the fiber over $x_1 = \cdots = x_n = 0$ is finite, the map from $\bar D$ to $\operatorname {Spec} S$ is dominant and so by Proposition 3.4.6(7) of [Reference BerkovichBer90], we can lift any norm corresponding to a point in this skeleton to a point in the analytification of $D$. Since the projection onto the skeleton of $X$ is defined in terms of the norms of the $x_i$, we have produced a norm in the analytification of $\bar D$ which projects onto any point of $r$.