MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(E,\phi)$ be a $G$-Higgs bundle $\phi\in H^{0}(X,ad(E)\otimes D)$ where $D$ is a divisor on X.

I suppose that $(E,\phi)\in \mathcal{M}^{ani}$ the anisotropic locus.

In particuler, this bundle is stable as a Higgs bundle because, it doesn't have any reduction to a parabolic.

Does it imply that the underlying bundle $E$ is itself stable?

More generally, when a stable Higgs bundle has a stable underlying bundle.

share|cite|improve this question
A detailed answer to your question in the case of Riemann surfaces is given in Proposition 3.3 of Hitchin's "Self-duality equations on a Riemann surface". – Sebastian Dec 6 '12 at 7:42
up vote 4 down vote accepted

First, in the standard definition $D = K_X$, so I will give an example in this case. Let $X$ be a curve of genus 2 and $E = O \oplus O(P)$ for a point $P \in X$. Clearly $E$ is unstable with $O(P)$ being the only destabilizing subbundle. Define $\phi$ to be the composition $$ O \oplus O(P) \to O(P) \to O(K_X) \to O(K_X) \oplus O(K_X + P), $$ where the first map is the projection, the second is the embedding given by the point $P' \in |K_X - P|$, and the third is the embedding into the first summand. It is clear that $O(P)$ does not extend to a Higgs subbundle, so $(E,\phi)$ is stable.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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