I'm looking at rational curves in a toric surface. Such curves have a parametrization of the form $$t\dashrightarrow \chi \prod_{j=1}^m (t-\alpha_j)^{n_j} \in (\mathbb{C}^*)^2 ,$$ for some scalars $\alpha_i$, $\chi:\mathbb{Z}^2\rightarrow \mathbb{C}^*$ 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 $$\mu_i = \chi(n_i^T)\prod_{j\neq i}(\alpha_i - \alpha_j)^{\det (n_i,n_j)},$$ 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 : $$\begin{pmatrix} n_1^T & \sum_{j\neq 1}\frac{\det(n_1,n_j)}{\alpha_1-\alpha_j} & & & & \\ \vdots & & \ddots & & -\frac{\det(n_i,n_j)}{\alpha_i-\alpha_j} & \\ \vdots & & & \sum_{j\neq i}\frac{\det(n_i,n_j)}{\alpha_i-\alpha_j} & & \\ \vdots & & -\frac{\det(n_i,n_j)}{\alpha_i-\alpha_j} & & \ddots & \\ n_m^T & & & & & \sum_{j\neq m}\frac{\det(n_m,n_j)}{\alpha_m-\alpha_j} \\ \end{pmatrix}$$ 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.