Higgs bundle and spectral curve

Question

I have trouble understanding the following part in E. Wittens paper "More On Gauge Theory And Geometric Langlands":

The context is the following: $(E, \varphi)$ is a Higgs bundle with structure group $SU(2)$ on the Riemann surface $C$. We look at the eigenvalue problem $\varphi \phi =y \phi$ where $\phi \in E$ and $y$ is a holomorphic 1-form.

For $N=2$ this is equivalent to $y^2-1/2 Trace(\varphi^2)=0$. For this case, if the eigenvalues are not distinct, it means that $y=0$ and so $Tr(\varphi^2)=0$.

1)My problem starts when it says "As we require $Tr(\varphi^2)$ to have simple zeros...". Why can we require that?

2)Also, if we assume this, why does $\varphi $ have to be of the form ? We know the following:

• $Tr(\varphi)=0$ since $G=SU(2)$.

• $Tr(\varphi^2)=0$ for $z=0$.

• $Tr(\varphi^2)$ has simple zeros

Now let

$\varphi$= $\begin{pmatrix} a & b\\ c & -a \end{pmatrix}$

Then $\varphi^2= \begin{pmatrix} a^2+bc & 0\\ 0 & a^2+bc \end{pmatrix}$ and we conclude that $(2a^2+2bc)(0)=0$ and that it is a simple zero. Can we already follow from that that $\varphi$ has to be of the form in the paper?

3)What does it mean that we regard $y$ as a local parameter? I mean, $y$ is arbitrary, why can we just replace $z$ by $y^2$? This seems so random to me.

Thanks in advance for any answer, I understand that this is a question which is maybe hard to answer because you first need to understand the context here so I really appreciate any help!

Answer 1

I realize this is nearly a year-old and you've maybe/probably found an answer by now. Nevertheless, I will have a go at answering your questions.

1. According to BNR's paper, the Jacobian criterion implies that the spectral curve defined by $y^n + a_1y^{n-1} + ... + a_n = 0$ with $a_i \in H^0(X, K_X^i)$ is smooth iff at every multiple zero of the section $a_n$, the section $a_{n-1}$ is non-zero.

Since $\phi$ is traceless, $\det(y - \phi) = y^2 + \det(\phi)$. Since $a_1 = tr(\phi)$, $\det(\phi)$ must have simple zeroes since the spectral curve is smooth. Since $\det(\phi) = - \frac{1}{2}tr(\phi^2)$, it follows that $tr(\phi^2)$ must have simple zeroes.

2. The ramification points of the spectral curve are given by the divisor $tr(\phi^2)$. Now, assume that $tr(\phi^2) = 0$ at $z = 0$, which makes $z = 0$ a ramification point. Then locally, the spectral 2-cover looks like $y \mapsto y^2 = z$. Sections of line bundles, including $K_X$, are of the form $s(z) = f_0(z) + yf_1(z)$.

By the spectral correspondence, the Higgs field is conjugate to the "multiplication-by-$y$" endomorphism. The latter sends $s(z)$ to $ys(z) = yf_0(z) + y^2 f_1(z) = yf_0(z) + zf_1(z).$

Taking $f_0(z)$, $f_1(z)$ to be local bases, we have that $\phi$ is conjugate to $\begin{pmatrix} 0 & z \\ 1 & 0 \end{pmatrix}.$ One thing that is lacking about this answer is that I am not sure how $tr(\phi^2)$ having simple zeroes help.

3. Perhaps the above answers this question. We replace $z$ with $y^2$ because locally at $z = 0$, $tr(\phi^2) = 0$ and the spectral curve looks like the double cover $y \mapsto y^2 = z$.