Recall that we saw how to degenerate a real algebraic curve into a tropical curve. Patchworking is about trying to go back.
Here is the input for patchworking.
- $S$ is some collection of types of isolated hypersurface singularities,
- a convex $n$-dimensional polytope $\Delta \subseteq \mathbb{R}^n$ with a subdivision $\Delta = \Delta_1 \cup \dotsb \cup \Delta_N$,
- a piecewise linear function $\nu \colon \Delta \to \mathbb{R}$ such that $\nu \vert_{\Delta_k}$ are linear, and satisfying $\nu(\Delta \cap \mathbb{Z}^n) \subseteq \mathbb{Z}$,
- a collection of numbers $a_i$ for $i \in \Delta \cap \mathbb{Z}^n$, where $a_i \neq 0$ for $i$ a vertex of one of the $\Delta_k$, such that $$ f_k(z) = \sum_{i \in \Delta_k \cap \mathbb{Z}^n} a_i z^i $$ is peripherally nondegenerate.
We also assume that $$ f_k^\sigma(z) = \sum_{i \in \Delta_k \cap \mathbb{Z}^n \cap \sigma} a_i z^i $$ are nonsingular for all proper faces $\sigma \subsetneq \Delta$. This translate to the fact that $\operatorname{Sing}(f_k)$ lie in the torus.
Goal 1. Find a polynomial $f$ such that $\lbrace f = 0 \rbrace$ has singularities corresponding to the union $\coprod_k \operatorname{Sing}(f_k)$ with the same types.
The main theorem
Let $G$ be an adjacency graph on $\Delta_1, \dotsc, \Delta_N$, where these are the vertices and two $\Delta_i$ are connected when they share an edge.
Theorem 2. Assume $f_1, \dotsc, f_N$ are all peripherally nondegenerate. Assume also that there exists an orientation $\Gamma$ of the edges of $G$ so that there is no cycle, and satisfying the property $(\Delta_k, \partial \Delta_{k,+}(\Gamma), f_k)$ is $S$-transversal. Then there exists a polynomial $$ f \in \mathcal{P}(\Delta) = \biggl\lbrace \sum_{i \in \Delta \cap \mathbb{Z}^n} b_i z^i : b_i \in \mathbb{C} \biggr\rbrace $$ such that the singularities types of $f$ are the disjoint union of the singularity types of $f_i$.
What is this transversality? Fix a facet $F \in \mathcal{P}(\Delta)$ some union $\partial \Delta_+ \subseteq \partial \Delta$ of some facets of $\Delta$. Then we have a subspace $$ \mathcal{P}(\Delta, \partial \Delta_+, F) = \lbrace G \in \mathcal{P}(\Delta) : G^\sigma = F^\sigma \text{ for all } \sigma \in \Delta_+ \rbrace. $$
Definition 3. We say that $(\Delta, \partial \Delta_+, F)$ is $S$-transversal if the following hold for $d \gg 0$:
- for each $W \in \operatorname{Sing}(F)$, the space $$ M_d(W) = \left\lbrace \varphi \in \mathbb{C}_ d[x] : \begin{matrix} \varphi \text{ has a unique singular point in a} \br \text{neighborhood of } W \text{ of the same type} \end{matrix} \right\rbrace $$ is a smooth analytic subset,
- the intersection $M_d(F) = \bigcap_{W \in \operatorname{Sing}(F)} M_d(W)$ is transversal,
- the intersection $M_d(F) \cap \mathcal{P}(\Delta, \partial \Delta_+, F)$ is transversal.
The subset $\partial \Delta_{k+}(\Gamma)$ is the subset of faces corresponding to the edges that start from $\Delta_k$ in the orientation $\Delta$.
Remark 4. If all the singularities are nodal, then this condition on $\Gamma$ is always true.
Sketch of proof
Basically we want to do a one-parameter deformation $$ f_t = \sum_{i \in \Delta \cap \mathbb{Z}^n} A_i(t) t^{\nu(i)} z^i \in \mathcal{P}(\Delta) \subseteq \mathbb{C}[z] $$ for $t \neq 0$. Here, we will choose these $A_i$ so that $$ A_i = a_i + O(t) $$ as $t \to 0$. Recall that $\nu$ was piecewise linear, so we have $$ \nu \vert_{\Delta_k} = \lambda_k \vert_{\Delta_k} = \langle \alpha_{\bullet,k}, - \rangle + \alpha_{0,k} $$ for numbers $\alpha_{0,k}, \dotsc, \alpha_{n,k}$. Writing $\nu_k = \nu - \lambda_k$, and making a normalization of coordinates $$ T_k^{(t)}(z_1, \dotsc, z_n) = (z_1 t^{\alpha_{1,k}}, \dotsc, z_n t^{\alpha_{n,k}}), $$ we can now write $$ f_{(t),k}(z) = \sum_i A_i(t) t^{\nu_k(i)} z^i = \sum_i a_i z^i + O(t) = f_k(z) + O(t). $$ Here, the exact relation to the previous polynomial is $$ f_{(t)}(z) = f_{(t),k}(T_k^{(t)}(z)) t^{\alpha_{\bullet,k}}. $$
We now find $Q \subseteq (\mathbb{C}^\times)^n$ compact such that the interior of $Q$ contains the union $\bigcup_{k=1}^N \operatorname{Sing}(f_k)$. We can find $t \ll 1$ so that for any $0 \lt \lvert t \rvert \lt t_0$ such that $(T_i^{(t)})^{-1}(Q)$ is disjoint in $(\mathbb{C}^\times)^n$ and $f_{k,(t)}$ is a $S$-equisingular deformation of $f_k$ on $Q$.
Remark 5. Here this peripherally nondegenerate condition guarantees that $f_{(t)}(z)$ has no new singularities in $(\mathbb{C}^\times)^n - Q$.
How do we choose this $A_i(t)$? Here we need to use this transversality condition. We first start with the $\Delta_k$ where there is no edge going out. The transversality condition then tells us that there exists a deformation $A_i(t)$ for the $i$ with the prescribed singularities, up to first order. We don’t have to worry about the higher order terms, because of transversality. Then at this point, we can start fixing the deformations on other $\Delta_j$ up to first order, and inductively doing this.