Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $X$ be a compact Riemann surface with genus $2$. Can you give me examples of stable principal $SL(2)$-bundles on $X$?

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

Take any irreducible flat $SU(2)$-bundle over your surface.

Edit 1: Explicitly, take two noncommuting rotations $a, b$ of order $>2$ in $SO(3)$ and lift them (arbitrarily) to elements $A, B$ in $SU(2)$, which is a 2-fold cover of $SO(3)$. Now, let $\pi$ be the fundamental group of the genus 2 surface $S$, $$ \pi=<a_1, b_1, a_2, b_2| [a_1,b_1][a_2,b_2]=1>. $$ Let $F_2$ be the free group on two generators $x, y$. Define epimorphism $\pi\to F_2$ by sending $a_1$ to $x$, $b_1$ to $1$, $a_2$ ot $y$ and $b_2$ to $1$. Lastly, consider the homomorphism $F_2\to SU(2)$ sending $x$ to $A$ and $y$ to $B$. Let $h: \pi\to SU(2)$ be the composition of the above homomorphisms. Using the associated bundle construction, see my answer in

Compatibility of a connection and metric

we obtain a flat principal $SU(2)$-bundle over the surface $S$ with holonomy $h$. Embedding $SU(2)$ in $G=SL(2,C)$, we obtain a holomorphic principal $G$-bundle over $S$. In order to see that this bundle is stable note that the flat $SU(2)$-bundle is irreducible (due to our choice of $A, B$); we can now apply the theorem of Narasimhan-Seshadri that the corresponding rank 2 holomorphic bundle is stable. Therefore, the corresponding principal $G$-bundle is stable as well.

Edit 2: You can find the statement and references to proofs of Narasimhan-Seshadri theorem in http://en.wikipedia.org/wiki/Narasimhan-Seshadri_theorem (google is your friend!). I am using above the "easy" direction of the theorem, namely that vector bundles associated with unitary representations of $\pi_1$ are semistable; they are stable provided that the unitary representation is irreducible, i.e., admits no proper invariant subspace. In other words, the associated flat bundle is irreducible as a flat bundle.

share|improve this answer
    
Could you describe explicitly it? –  ArthurStuart Oct 8 '13 at 13:07
    
I'm sorry, I have some questions: are $a,b$ in $SU(2)$? What is surface group? Is $F_2 \simeq \mathbb{Z}_2$? What type of associate bundle have I to consider? Thank's. –  ArthurStuart Oct 8 '13 at 13:48
    
I added details with answers to your questions. –  studiosus Oct 8 '13 at 17:52
    
I'm sorry, but why we have to take the order of $a,b$ $>2$? –  ArthurStuart Oct 8 '13 at 20:14
    
I'm sorry, but why we have to take the order of $a,b$ $>2$? And in you construiction of associated bundle you obtain a flat vector bundle not a principal $G$-bundle... –  ArthurStuart Oct 8 '13 at 22:46
show 11 more comments

Your Answer

 
discard

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.