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?


1 Answer 1


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.

  • $\begingroup$ Sorry I've not understood what you said about the geometrical properties of the integral locus $\endgroup$ May 25, 2021 at 15:33
  • $\begingroup$ 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. $\endgroup$ May 26, 2021 at 9:45
  • $\begingroup$ @TommasoScognamiglio I now give an alternate, more algebraic argument. $\endgroup$
    – Will Sawin
    May 26, 2021 at 21:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.