# Integral locus of Hitchin morphism

Let $$\Sigma$$ be a Riemann surface of genus $$g$$. To it, we can associated $$M_{Dol}$$ be the Higgs moduli space of rank $$n$$ and degree $$d$$. Fo simplicity let us take $$(n,d)=1$$. This quasiprojective variety admits a morphism $$h:M_{Dol}\rightarrow \bigoplus_{i=0}^nH^0(\Sigma,\Omega^i_{\Sigma})=A_n$$ where $$\Omega^1_{\Sigma}$$ is the usual sheaf of differentials.

Moreover, to each element $$x \in \bigoplus_{i=0}^nH^0(\Sigma,\Omega^i_{\Sigma})$$ we can associated a so called spectral curve $$C_x \rightarrow C$$ and the fiber of $$h$$ are related to torsion free sheaves over such a curve.

I've read in lots of articles that the so called elliptic locus $$A_{n,ell}$$ given by $$x$$ such that $$C_x$$ is integral is open and of big codimension. Does anyone know a simple proof of the openness part? Also, is there any explicit bound for the codimension?

To get the codimension, if the curve corresponding to $$x = (x_i \in H^0 (\Sigma, \Omega_\Sigma^i))_{i=1}^n$$ fails to be integral, then the equation $$T^n - x_1 T^{n-1} + x_2 T^{n-2} + \dots$$ defining the curve splits as a product of two such equations, say $$T^k - y_1 T^{k-1} + \dots + (-1)^k y_k$$ and $$T^{n-k} - z_1 T^{n-k-1} + \dots + (-1)^{n-k} z_{n-k}$$. We have $$y_i, z_i \in H^0 ( \Sigma, \Omega_{\Sigma}^i))$$ so the dimension of the locus where this occurs is
$$\sum_{i=1}^k \dim H^0 ( \Sigma, \Omega_{\Sigma}^i) + \sum_{i=1}^{n-k} \dim H^0 ( \Sigma, \Omega_{\Sigma}^i) = k^2 (g-1) +1 + (n-k)^2 (g-1) + 1$$ and since the total dimension of the Hitchin base is $$n^2 (g-1) +1$$, the codimension is $$(n^2 - k^2 - (n-k)^2)(g-1) -1 = 2 k (n-k) (g-1) -1$$ which is minimized with $$k=1$$ or $$k=n-1$$, giving a total codimension of $$2 (n-1) (g-1)-1$$.
We can also use this factorization to check that the non-integral locus is closed. It is the union over $$k$$ of the image of the map from $$\prod_{i=1}^k H^0 ( \Sigma, \Omega_{\Sigma}^i)) \times \prod_{i=1}^{n-k} H^0 ( \Sigma, \Omega_{\Sigma}^i))$$ that sends $$y_1,\dots, y_k, z_1,\dots, z_{n-k}$$ to $$(T^k - y_1 T^{k-1} + \dots + (-1)^k y_k ) (T^{n-k} - z_1 T^{n-k-1} + \dots + (-1)^{n-k} z_{n-k}).$$ It suffices to check that this map is proper, from which it follows that the image is closed.
In fact, it is finite, because the $$y_1,\dots, y_k, z_1,\dots, z_{n-k}$$ all satisfy monic polynomial equations with coefficients polynomials in $$x_1,\dots, x_n$$. Indeed, we can express the $$y_1, \dots , y_k$$ formally as polynomials in the roots of $$T^n - x_1 T^{n-1} + x_2 T^{n-2} + \dots$$, and the roots certainly satisfy a monic equation, so any polynomial in the roots does as well.
• The comment above was really confuse. I did not really get what you meant when you wrote that $C_x$ is a specialization of $C_y$ and whether this prove that the integral locus is open or not. May 26, 2021 at 9:45