## How to see the genus

Let $R$ be a polynomial in two variables; i.e. $R(\lambda,\mu) = \sum_{i,j} a_{ij} \lambda^i\mu^j$ where each $i,j$ is non-negative, and the coefficients $a_{ij}$ are complex numbers which are nonzero for only finitely many pairs $i,j$. For a generic choice of coefficients, the equation $R=0$ determines a smooth complex curve $\Sigma$ in $\mathbb{C}^2$ (i.e. a Riemann surface). How can one see the geometry of the curve directly in the expression for $R$? It turns out that there are several ways to do it, some very old, and some more recent.

The most important geometric invariant of the curve is the genus. To a topologist, this is the number of “handles”; to an algebraic geometer, this is the dimension of the space of holomorphic $1$-forms. One well-known way to calculate the genus is by means of the Newton polygon. In the (real) plane $\mathbb{R}^2$, consider the finite set consisting of the points with integer coordinates $(i,j)$ for which the coefficient $a_{ij}$ of $R$ is nonzero. The convex hull of this finite set is a convex integral polygon, called the Newton polygon of $R$. It turns out that the genus of $\Sigma$ is the number of integer lattice points in the interior of the Newton polygon. In fact, one can find a basis for the space of holomorphic $1$-forms directly from this formulation. Let $R_\mu$ denote the partial derivative of $R$ with respect to $\mu$. Then for each lattice point $(i,j)$ in the interior of the Newton polygon, the $1$-form $(\lambda^i\mu^j/R_\mu) d\lambda$ is a holomorphic $1$-form on $\Sigma$, and the set of all such forms is a basis for the space of all holomorphic $1$-forms.

This is direct but a bit unsatisfying to a topologist, since the connection between the dimension of the space of $1$-forms and the topological idea of handles is somewhat indirect. In some special cases, it is a bit easier to see things. Two important examples are:

1. Hyperelliptic surfaces, i.e equations of the form $\lambda^2 = p(\mu)$ for some polynomial $p(\cdot)$ of degree $n$. The Newton polygon in this case is the triangle with vertices $(0,0), (2,0), (0,n)$ and it has $\lfloor (n-1)/2 \rfloor$ interior lattice points. Geometrically one can “see” the surface by projecting to the $\mu$ plane. For each generic value of $\mu$, the complex number $p(\mu)$ has two distinct square roots, so the map is 2 to 1. However, at the $n$ roots of $p(\cdot)$, there is only 1 preimage. So the map is a double cover, branched over $n$ points, and one can “see” the topology of the surface by cutting open two copies of the complex line along slits joining pairs of points, and gluing.
2. A generic surface of degree $d$. The Newton polygon in this case is the triangle with vertices $(0,0), (d,0), (0,d)$ and it has $(d-1)(d-2)/2$ interior lattice points. One way to “see” the surface in this case is to first imagine $d$ lines in general position (a quite special degree $d$ curve). Each pair of lines intersect in a point, so there are $d(d-1)/2$ points of intersection. After deforming the curve, these points of intersection are resolved into tubes, so one obtains $d$ complex lines joined by $d(d-1)/2$ tubes. The first $d-1$ tubes are needed to tube the lines together into a (multiply)-punctured plane, and the remaining $(d-1)(d-2)/2$ tubes each add one to the genus.

It turns out that there is a nice way to directly see the topology of $\Sigma$ in the Newton polygon, via tropical geometry. I recently learned about this idea from Mohammed Abouzaid in one of his Clay lectures; this point of view was pioneered by Grisha Mikhalkin. The idea is as follows. First consider the restriction of $\Sigma$ to the product $\mathbb{C}^* \times \mathbb{C}^*$; i.e. remove the intersection with the coordinate axes. For generic $R$, this amounts to removing a finite number of points from $\Sigma$, which will not change the genus. Then on this punctured curve $\Sigma$, consider the real valued function $(\lambda,\mu) \to (\log(|\lambda|),\log(|\mu|))$. The image is a subset of $\mathbb{R}^2$, called an amoeba. If one varies the (nonzero) coefficients of $R$ generically, the complex geometry of the curve $\Sigma$ will change, but its topology will not. Hence to see the topology of $\Sigma$ one should deform the coefficients in such a way that the topology of the amoeba can be read off from combinatorial information, encoded in the Newton polygon. The terms in $R$ corresponding to lattice points in a boundary edge of the Newton polygon sum to a polynomial which is homogeneous after a suitable change of coordinates. In the region in which these terms dominate, $\Sigma$ looks more and more like a collection of cylinders, each asymptotic to a cone on some points at infinity. The image in the amoeba is a collection of asymptotically straight rays. If the polynomial were genuinely homogeneous, the preimage of each point in the amoeba would be a circle, parameterized by a choice of argument of (a certain root of) either $\lambda$ or $\mu$. So the amoeba looks like a compact blob with a collection of spikes coming off. As one deforms the coefficients in a suitable way, the compact blob degenerates into a piecewise linear graph which can be read off from purely combinatorial data, and the topology of $\Sigma$ can be recovered by taking the boundary of a thickened tubular neighborhood of this graph.

More explicitly, one chooses a certain triangulation of the Newton polygon into triangles of area $1/2$ and with vertices at integer lattice points (by Pick’s theorem this is equivalent to the condition that each triangle and each edge has no lattice points in the interior). This triangulation must satisfy an additional combinatorial condition, namely that there must exist a convex piecewise linear function on the Newton polygon whose domains of linearity are precisely the triangles. This convex function is used to deform the coefficients of $R$; roughly, if $f$ is the function, choose the coefficient $a_{ij} \sim e^{f(i,j)t}$ and take the limit as $t$ gets very big. The convexity of $f$ guarantees that in the preimage of each triangle of the Newton polygon, the terms of $R$ that contribute the most are those corresponding to the vertices of the triangle. In particular, as $t$ goes to infinity, the amoeba degenerates to the dual spine of the triangle (i.e. a tripod). The preimage of this tripod is a pair of pants; after a change of coordinates, any given triangle can be taken to have vertices $(0,0), (1,0), (0,1)$ corresponding to a linear equation $a\lambda + b\mu = c$ whose solution set in $\mathbb{C}^* \times \mathbb{C}^*$ (for generic $a,b,c$) is a line minus two points — i.e. a pair of pants.

One therefore has a concrete combinatorial description of the degenerate amoeba: pick a triangulation of the Newton polygon satisfying the combinatorial conditions above. Let $\Gamma$ be the graph dual to the triangulation, with edges dual to boundary edges of the triangulation extended indefinitely. The surface $\Sigma$ is obtained by taking the boundary of a thickened neighborhood of $\Gamma$. The genus of $\Sigma$ is equal to the rank of the first homology of the graph $\Gamma$; this is evidently equal to the number of lattice points in the interior of the polygon.

As a really concrete example, consider a polynomial like

$R = 1 + 7z^3 - 23.6w^2 + e^\pi z^3w^2$

(the exact coefficients are irrelevant; the only issue is to choose them generically enough that the resulting curve is smooth (actually I did not check in this case – please pretend that I did!)). The Newton polygon is a rectangle with vertices $(0,0), (3,0), (0,2), (3,2)$. This can be subdivided into twelve triangles of area $1/2$ as in the following figure:

The dual spine is then the following:

which evidently has rank of $H_1$ equal to $2$, equal on the one hand to the number of interior points in the Newton polygon, and on the other hand to the genus of $\Sigma$.

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### 9 Responses to How to see the genus

1. Anon says:

Nice post!

“twelve triangles of unit area” -> “twelve triangles of area one half”

2. Danny Calegari says:

Oops! You’re right (maybe if I say the “units” are “area 1/2” . . .)

3. Francis Davey says:

Is there a similar trick in three variables or where there are several simultaneous polynomials or both? Eg, where you have two polynomials A(p,q) and B(q,r)?

4. abramodj says:

Very nice trick!

5. With havin so much content do you ever run into any issues of plagorism or copyright infringement?
My website has a lot of completely unique content I’ve either written myself or outsourced but it looks like a lot of it is popping it up all over the web without my permission. Do you know any techniques to help reduce content from being ripped off? I’d genuinely appreciate it.

6. Anton Izosimov says:

In fact, it seems that for each lattice point (i,j) in the interior of the Newton polygon, the corresponding holomorphic 1-form should be \lambda^{i-1}\mu^{j-1} d\lambda / R_\mu, not \lambda^{i}\mu^{j} d\lambda / R_\mu. Consider, for instance, the case of a non-singular projective curve of degree n. Then the Newton polygon is a triangle with vertices (0,0), (0,n), (n,0). The “rightmost” lattice points (i,j) in the interior satisfy i + j = n – 1. On other hand, computing the orders of poles at infinite points, we easily see that holomorphic differentials are given by \lambda^{i}\mu^{j} d\lambda / R_\mu with i + j \leq n – 3.

7. Next time I read a blog, I hope that it does not disappoint me just as much as this one. I mean, Yes, it was my choice to read, but I genuinely believed you would probably have something interesting to talk about. All I hear is a bunch of crying about something you could fix if you were not too busy searching for attention.