Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $C$ be a complex curve. Recall that a Higgs bundle on $C$ is a vector bundle $E$ on $C$ equipped with a morphism $E \to E \otimes K_C$. The space of (stable) Higgs bundles is much studied, and is in particular known to be smooth. Moreover there is a "nonabelian Hodge theorem" giving a diffeomorphism between the moduli of Higgs bundles and a certain character variety of $\pi_1(C)$.

What is known about the moduli space of Higgs bundles with a section, i.e., the space parameterizing triples $(E, s \in \mathrm{H}^{0}(E), \phi: E\to E \otimes K_C)$ ? Is it smooth (after imposing some appropriate stability condition)? Is there an analogue of the "nonabelian Hodge theorem"?

share|improve this question
3  
"..and is in particular known to be smooth": I believe this is only true when the rank and degree of the vector bundle are assumed to be coprime. (In this case, stable and semistable mean the same thing.) –  user5395 Mar 10 '11 at 18:44
    
Do you have a reference for the diffeomorphism in the third sentence? –  Peter Samuelson Mar 14 '11 at 23:44
1  
It is a homeomorphism, and a good reference is Simpson: Moduli of representations of the fundamental group of a smooth projective variety. I. Inst. Hautes Études Sci. Publ. Math. No. 79 (1994), 47–129. and Moduli of representations of the fundamental group of a smooth projective variety. II. Inst. Hautes Études Sci. Publ. Math. No. 80 (1994), 5–79 (1995). –  Richard Wentworth Apr 12 '11 at 2:08
add comment

2 Answers

up vote 3 down vote accepted

There is a fundamental difference between the case of Higgs bundles (where the section lies in a twisted adjoint representation) and the case of a section of the bundle itself (where the section is in the vector representation). In the former, the notion of stability is rigid, whereas in the latter the definition of stability depends on a parameter. This was discovered by Bradlow (J. Differential Geom. 33 (1991), no. 1, 169--213) and Bradlow-Daskalopoulos (Internat. J. Math. 2 (1991), no. 5, 477--513) and exploited by Thaddeus (Invent. Math. 117, no. 2 (1994), 317--353). These papers will point you in the direction of a definition of stability/semistability for the case you're interested in. I guess it will be true that stable points are smooth, though for certain values of the parameter the compactifications will contain strictly semistable (non-smooth) points.

To answer your question, there is apparently no relationship between Bradlow pairs and representations of the fundamental group. Rather, these spaces are more closely related to higher rank generalizations of symmetric products of the Riemann surface (see also J. Amer. Math. Soc. 9 (1996), 529-571).

share|improve this answer
add comment

For stable $E$ of degree $0$ there are no nonzero holomorphic sections. The interessting things must happen at non-stable bundles. As far as I know, there is a desingulaization procedure for the space of semistable holomorphic bundles $V$ similar to your situation: one takes $E=End V$ together with a holomorphic section of $E.$ Again, in the case of stable bundles there are only multiples of the identity, but for nonstable there are more endomorphisms. For details, see Tyurin#s 'red book' on vector bundles over surfaces (Quantization, Classical and Quantum Field Theory and Theta functions) and the references therein. Maybe you can apply these ideas to your situation.

share|improve this answer
add comment

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.