Physicists often speak of "Connections modulo gauge transformations" as the natural configuration space of a gauge field. In this sense, the fundamental object of study in gauge theory is the space of pairs,

$$\frac{(E,\nabla)}{\text{gauge transformations}}$$

Where $E$ is a complex vector bundle, and $\nabla$ is a suitable connection. In the case of gauge group $U(1)$ and line bundle $E=L$, there is a natural identification of this configuration space with the set of holomorphic structures on $L$, i.e.,

$$\frac{(L,\nabla)}{\text{gauge transformations}}\cong \text{Pic}^0(X)$$

Where the zero is there because there are no magnetic monopoles in nature.

This is immensely useful, because the space on the RHS is amazingly concrete and visualizable, computable, etc...

The question is, does this correspondence continue for bundles of rank higher than one, and general compact gauge group $G$ (i.e. interpretation as holomorphic structures)? This would allow for a similar analysis of general Yang-Mills theory, and other non-abelian gauge theories of the Standard Model.

Also, does anyone have references for these results?

  • 2
    $\begingroup$ Does Donaldson's theorem help? Note also that he has a theorem that the moduli space of irreducible ASD connections on an $SU(2)$ bundle over a compact Kahler surface is the same (as sets) as the isomorphism classes of stable holomorphic rank 2 bundles. I believe this should have extensions to $SU(n)$ but do not currently remember a reference. $\endgroup$ – user98602 Mar 18 '16 at 7:34
  • 1
    $\begingroup$ You might find the discussion in Chapter 5 of this review article helpful. $\endgroup$ – user149792 Apr 7 '16 at 22:39

For the case of $U(n)$, the correspondence continues for higher dimensional complex vector bundles. See page 3 of http://www.homepages.ucl.ac.uk/~ucahjde/YM-lectures/lecture10.pdf for details.

  • 1
    $\begingroup$ I failed to say in my original comment that the extension to all bundles over arbitrary manifolds is known as the Uhlenbeck-Yau theorem. The version in your link is specifically for Riemann surfaces. $\endgroup$ – user98602 Apr 7 '16 at 3:40
  • $\begingroup$ Ahh very sorry about that - thanks for pointing this out! $\endgroup$ – David Roberts Apr 7 '16 at 3:47
  • $\begingroup$ No need to apologize, I just wanted to clarify if it wasn't clear! $\endgroup$ – user98602 Apr 7 '16 at 3:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.