3
$\begingroup$

Let $X$ be a compact Riemann surface, and let $E \to X$ be a stable vector bundle of rank $r+1$. Then we know that $P = \mathbb{P}(E)$ comes from an irreducible representation of $\pi_1(X)$ into the projective unitary group $PSU(r+1)$, that is, $E$ admits a projectively flat Hermitian structure. Moreover, there exists a projective unitary flat structure on $P$. It means that there exists a collection of trivializations $\psi_{\alpha}\colon P|_{U_{\alpha}} \rightarrow U_{\alpha} \times \mathbb{P}^r$ such that the corresponding transition functions $$ g_{\alpha\beta} \colon U_{\alpha} \cap U_{\beta} \rightarrow {\rm PSU}(r+1) \subseteq {\rm PSL}(r+1, \mathbb C) $$ are all constants, where $$ \psi_{\alpha} \circ \psi_{\beta}^{-1}(x, v) = \big(x, g_{\alpha\beta}(x) \cdot v \big) \quad \forall x \in U_{\alpha} \cap U_{\beta}, v\in \mathbb{P}^{r}. $$ Denote by $\operatorname{Aut}_X^h(P)$ the holomorphic automorphism group of the projective bundle $P$. A holomorphic automorphism $\sigma \in \operatorname{Aut}_X^h(P)$ is called a unitary flat automorphism if $\sigma$ is defined by a collection of maps $\phi_{\alpha}\colon U_{\alpha} \times \mathbb{P}^r \rightarrow U_{\alpha} \times \mathbb{P}^r$ with respective to the trivializations $\psi_{\alpha}$ such that

  • $\phi_{\alpha}(x, v)=\big( x, g_{\alpha}(x) \cdot v \big)$, where $g_{\alpha}\colon U_{\alpha} \rightarrow {\rm PSU}(r+1)$ is constant.
  • For any $\alpha$ and $\beta$, $\phi_{\alpha} \circ \psi_{\alpha} \circ \psi_{\beta} ^{-1} = \psi_{\alpha} \circ \psi_{\beta} ^{-1}\circ \phi_{\beta}$ on $\psi_{\beta}(P|_{U_{\alpha} \cap U_{\beta}})$.

Denote by $\operatorname{Aut}_X^u(P)$ all unitary flat automorphisms of $P$. There is a natural inclusion $\operatorname{Aut}_{X}^u(P) \subset \operatorname{Aut}_{X}^h(P)$.

My questions are

  1. Does $\operatorname{Aut}_{X}^u(P) = \operatorname{Aut}_{X}^h(P)$ hold?
  2. $\operatorname{Aut}_{X}^u(P) = e$ for all stable $E$?

(Remark: They are very extreme two cases. In fact, I know that for very general stable $E$, $\operatorname{Aut}_{X}^h(P) = e$)

$\endgroup$
0

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.