Patchworking | Dongryul Kim

Let $T$ be a triangle with vertices $(0,0)$ and $(d,0)$ and $(0,d)$.

Definition 1. A convex triangulation of $T$ is a triangulation such that there exists a function $f \colon V \to \mathbb{R}$ (where $V$ is the vertices of the triangulation) such that the convex hull of $(i, j, f(i,j))$ project to $T$. We say that this triangulation is primitive when $V = T \cap \mathbb{Z}^2$.

Now we give each vertex a sign, $\epsilon \colon V \to \lbrace \pm 1 \rbrace$. Then we reflect this triangulation along the $x$-axis and the $y$-axis. For the signs,

when we reflect across the $x$-axis, we switch the sign when $j$ is odd,
when we reflect across the $y$-axis, we switch the sign when $i$ is odd.

Finally, inside each triangle of the triangulation,

if all vertices have the same sign, we don’t to anything,
if two vertices have the same sign, we consider the midpoints of the edges of different signs, and then we connect these two midpoints.

Now we can glue up the both sides of the squares so that this square with vertices $(\pm d, 0)$ and $(0, \pm d)$ becomes $\mathbb{RP}^2$.

Claim 2. This becomes a real algebraic curve of degree $d$ up to homeomorphism. More precisely, it is the curve given by $$ \sum_{i,j} \epsilon(i,j) t^{f(i,j)} x^i y^j $$ when $t \gt 0$ is a very small positive real number.

Tropicalization of equations

Let us normalize this the other way and consider $T = t^{-1}$ very large. The idea is that if we have a polynomial that looks like $$ P(x, y) = \sum_{i,j} T^{a_{i,j}} x^i y^j, $$ and if we write $u = \log_T x$ and $v = \log T_y$ then $$ \log_T P(x,y) = \max_{i,j} (a_{i,j} + ui + vj). $$

Now imagine we have an equation $P(x, y) = 0$ where some terms are positive and some terms are negative. Then we can rearrange the terms so that $$ \sum_{\epsilon=1} T^{a(i,j)} x^i y^j = \sum_{\epsilon=-1} T^{a(i,j)} x^i y^j. $$

Remark 3. Let us consider the tropical equation $$ \sum_{\epsilon=1} T^{a_i} x^i = \sum_{\epsilon=-1} T^{a_i} x^i. $$ Then the roots are going to be the intersection of the two tropical curves $Q^+(u)$ and $Q^-(u)$, which are exactly the corner locus of the sum $Q(u) = \max(Q^+(u), Q^-(u))$ assuming that our curves are generic enough.

So the 1-dimensional recipe is as follows.

Draw the corner locus of $Q(u)$.
Label the edges of $v = Q(u)$ by $+$ if $Q^+(u) \gt Q^-(u)$ and $-$ if $Q^+(u) \lt Q^-(u)$.
The true algebraic variety $P(x) = 0$ is, in the limit, the set of points separating $+$ from $-$.

Let us now move on to the 2-dimensional case. In the limit, we have the equation $P(x,y) = 0$ which corresponds to the corner locus of $$ Q(u,v) = \max_{i,j}(-f(i,j) + ui + vj). $$ Now each region has a corresponding $\pm$ sign. Then we are going to look at the boundary between the $+$ regions and the $-$ regions to obtain the tropical intersection of the two functions.

But this takes care of only the part where $x, y \gt 0$, because we were parametrizing with $u = \log_T x$ and $v = \log_T y$. To take care of all possible $x$ and $y$, we need to reflect $x \mapsto -x$ and $y \mapsto -y$. These three other reflections give rise to the hyperbola.

Some examples

Theorem 4 (Harnack). The maximum number of ovals for a nonsingular real algebraic curve of degree $d$ is $(d-1)(d-2)/2 + 1$.

This curve can be constructed explicitly using patchworking. We consider the triangulation cut out by the equations $$ x \in \mathbb{Z}, \quad y \in \mathbb{Z}, \quad x + y \in \mathbb{Z}. $$ This is indeed convex because $x^2 + y^2 + (x+y)^2$ will have this convex hull. We now put a $+$ sign at $(i, j)$ exactly when both $i, j$ are even.

Here, for every interior point $(i, j)$, there exists a unique quadrant there exists a $(\pm i, \pm j)$ such that the hexagon surrounding this curve is a connected component. There is one extra long curve, and so we do get $$ \frac{(d-1)(d-2)}{2} + 1 $$ components.

Conjecture 5 (Ragsdale, 1906). We say that a connected component is even if it is contained in an even number of connected components and odd otherwise. Let $p$ be the number of even components and $k$ be the number of odd components. Then for a curve of degree $2k$, we have $$ p \le \frac{3k(k-1)}{2} + 1, \quad n \le \frac{3k(k-1)}{2}. $$

But then Itenberg constructed a counterexample using patchworking.

Theorem 6 (Itenberg). There exists a curve with with $$ p = \frac{3k(k-1)}{2} + 1 + \Bigl\lfloor \frac{(k-3)^2+4}{8}\Bigr\rfloor $$ and similarly with $$ n = \frac{3k(k-1)}{2} + \Bigl\lfloor \frac{(k-3)^2+4}{8}\Bigr\rfloor. $$