Is the following property true for every stable holomorphic bundle of rank 2 with trivial determinant on a compact Riemann surface: The space of trace-free Higgs fields, whose determinant have only simple zeros, is open and dense in the space of all trace-free Higgs fields.
I would like to see a proof, a reference or a counter-example.
Motivation: This property is certainly true for a generic stable bundle, as one might see by investigating the Hitchin fibration.
The statement is easily checked to be true for very stable bundles, so when you look for a counter example you will need to look for a wobbly (i.e. stable but not very stable) bundle.
Suppose that $E$ is a rank two bundle with $E$ stable and $\det E = \mathcal{O}$. We are trying to decide if it can happen that for all traceless $\theta : E \to E\otimes K$ it follows that $\det (\theta)$ has a double zero.
This condition looks quadratic but it is in fact linear. Given a point $x \in C$, the traceless condition tells you that $\det (\theta)$ will vanish at $x$ if and only if $\theta(x)$ is a nilpotent $K_{x}$-valued endomorphism of $E_{x}$. Furthermore $\det (\theta)$ will vanish to second order at $x$ if and only if $\theta_{x}$ is the zero endomorphism.
Correction: As the OP points out this statement is incorrect. We can have a multiple zero of the determinant with $\theta_{x}$ still being regular nilpotent. A typical example is a Higgs field which locally looks like $$ \theta = \begin{pmatrix} 0 & 1 \\ z^{2} & 0 \end{pmatrix} dz $$ with $z$ being a local coordinate centered at $x$.
Because of this the approach I suggested previously will not work.
The condition that $\det(\theta)$ has only simple zeroes is equivalent to saying that the spectral curve for $\theta$ is smooth. In other words we want to show that the image of $$ \tag{1} \det : H^{0}(C,\text{End}_{0}(E)\otimes K) \to H^{0}(C,K^{\otimes 2}) $$ is not contained in the discriminant divisor parametrizing singular spectral curves. If $E$ is very stable, then the map (1) is surjective (and in fact is a finite morphism of degree $2^{3g-3}$). So the question is: what happens is when $E$ wobbly? When $E$ is wobbly we have a positive dimensional vector subspace $N \subset H^{0}(C,\text{End}_{0}(E)\otimes K)$ consisting of nilpotent endomorphisms. The map $\det$ induces a rational map $$ f : \mathbb{P}(H^{0}(C,\text{End}_{0}(E)\otimes K) \dashrightarrow \mathbb{P}(H^{0}(C,K^{\otimes 2})) $$ which is given by a $3g-3$ dimensional linear system of quadrics with base locus $\mathbb{P}(N)$. Using the classification of wobbly bundles you can analyze these maps. Most likely blowing up $\mathbb{P}(N)$ will resolve $f$ into a morphism and you may be able to check directly that this morphism is surjective.
In fact, now that I have thought more about this, I think there is a round about way to check the statement by studying the map from a general spectral Prym to the moduli $M$ of semistable vector bundles. If you choose a smooth spectral curve $X$, then the map $\pi : \text{Prym}(X/C) \dashrightarrow M$ is dominant and finite of degree $2^{3g -3}$ over the very stable locus. The locus where $\pi$ is not defined consists of nested Brill-Noether loci parametrizing line bundles with unstable push-forward and you can resolve $\pi$ to a morphism by blowing-up these loci in a sequence starting with the smallest one (which consists of isolated points) and continuing up the nested sequence. At the end you get a morphism $\hat{\pi}$ from the blown-up Prym which is finite and of degree $2^{3g -3}$ onto $M$. The exceptional divisors map onto the wobbly locus and each exceptional divisor is contained in the ramification. You can now check that away from the semistable locus $\hat{\pi}$ is not totally ramified at any of the exceptional divisors which says that the original map $\pi$ is surjective on the stable locus. This seems to prove your statement. However this is too complicated and convoluted and I think that analyzing the Hitchin map as above may be much easier. It will take too long to write the details of the blow up of the Prym here, so if you want to know more about that send me an email and I will be happy to explain.
