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Critical values of the evaluation map for rational curves in toric surface

Mathematics Asked by Blm on October 3, 2020

I’m looking at rational curves in a toric surface. Such curves have a parametrization of the form
$$tdashrightarrow chi prod_{j=1}^m (t-alpha_j)^{n_j} in (mathbb{C}^*)^2 ,$$
for some scalars $alpha_i$, $chi:mathbb{Z}^2rightarrow mathbb{C}^*$ a morphism, and where $n_j$ are vectors in $mathbb{Z}^2$, and $sum n_j=0$. (it is just the standard parametrization of a rational curve in $( mathbb{C}^*)^2$ without choosing a basis. For a curve of degree $d$ in the projective plane, the vectors would be $(0,1)$, $(1,0)$ and $(-1,-1)$ $d$ times each.) The scalars $alpha_i$ correspond to the points of $mathbb{C}P^1$ which are sent to some toric divisor. One can look at the coordinates of these points with the corresponding toric divisor. The coordinate of $alpha_i$ is
$$mu_i = chi(n_i^T)prod_{jneq i}(alpha_i – alpha_j)^{det (n_i,n_j)},$$
where $n_i^T$ denotes the linear form $det(n_i,-)$. In teh case of the projective plane, one recovers indeed the coordinates of the intersection points with the coordinate axis. The map $f:(chi,alpha_j)mapsto(mu_j)$ is called the evaluation map. I would like to show that this map has no critical value.

In logarithmic coordinates for $mu_i$, the matrix of the differential $df$ has the following form :
$$begin{pmatrix}
n_1^T & sum_{jneq 1}frac{det(n_1,n_j)}{alpha_1-alpha_j} & & & & \
vdots & & ddots & & -frac{det(n_i,n_j)}{alpha_i-alpha_j} & \
vdots & & & sum_{jneq i}frac{det(n_i,n_j)}{alpha_i-alpha_j} & & \
vdots & & -frac{det(n_i,n_j)}{alpha_i-alpha_j} & & ddots & \
n_m^T & & & & & sum_{jneq m}frac{det(n_m,n_j)}{alpha_m-alpha_j} \
end{pmatrix}$$

The first two columns correspond to the derivative regarding the $chi$ coordinates, and the $m$ last columns to the coordinates $alpha_j$. I need to show (although I do not know if it is true) that this matrix is surjective on the hyperplane $sum x_i=0$. We have indeed the obvious relation since $sum n_i=0$ and the right symmetric square matrix has the vector $(1,dots,1)$ in its kernel. This comes from the fact that the product of the $mu_j$ is equal to $pm 1$. Moreover, I already know three vectors in its kernel, which correspond to the reparametrization of the rational curve by $PGL_2(mathbb{C})$ : the first two are $(0,0,1,dots,1)$, corresponding to translations, $(0,0,alpha_1,dots,alpha_m)$, corresponding to dilatation, the last corresponds to inversion. This means that the kernel is already at least $3$-dimensional. It should not be bigger.

My question is, provided that the $alpha_j$ are distincts, is this matrix always surjective to the hyperplane $sum x_i=0$ ? Equivalently, is the kernel always $3$-dimensional ?

  • I already know that for generic values of $alpha_j$, it is surjective, but I would like to have it for all values.
  • I could restrict to the real case ( scalars $alpha_j$ are real or in pairs of complex conjugated points ) and some weaker assumption for the case that interests me, but the question about critical values remains mysterious to me in the complex case.
  • The matrix seems to have a nice form but I do not succeed in using it.

Thanks in advance.

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