So I'm trying to understand in detail the story of gauge theory (or Yang-Mills theory) from a mathematical perspective. Much of the subject works for arbitrary Lie group $G$, but at some point, one usually asks for $G$ to also be compact in order to have a gauge invariant Yang-Mills functional.

However, I'm ultimately interested in gauge theory overlapping with algebraic geometry, so I would like to stay in the world of complex vector bundles, possibly with a metric reducing the structure group. Obviously, a bare complex vector bundle $E$ without any metric has structure group $GL_{n}(\mathbb{C})$ and if $\text{det}(E) \cong \mathcal{O}_{X}$, then the structure group is $SL_{n}(\mathbb{C})$. The existence of a Hermitian metric breaks the structure group down to the compact Lie group $U(n)$ or $SU(n)$ if the determinant is trivial.

So my question is, how many other structure groups can I get this way by reducing the structure group of a complex vector bundle? If there are indeed more, do they all arise by imposing a metric, and is there the corresponding notion of a "$G$-connection"? (i.e. a connection compatible with the metric structure)


It is a fundamental result on compact Lie groups that for a representation of a compact Lie group $G$ on a finite dimensional vector space, there is always an invariant inner product on the space that is invariant under the action of $G$. Applyin this to $\mathbb C^n$, you conclude that $G$ is conjugate to a closed subgroup of $U(n)$.

Thus the compact Lie groups which act effectively on $\mathbb C^n$ are exactly the closed subgroups of $U(n)$. Thus you always have a Hermitian metric preserved by the action of $G$ and any $G$-connection (there is a general notion) also preserves that metric. I don't think that there is a reasonable description of all closed subgroups of $U(n)$.

  • $\begingroup$ Maybe this is my own ignorance here, but I don't think I can name any positive dimensional closed subgroups of $U(n)$, except for $SU(n)$, and I guess lower dimensional unitary groups. I see you mention that there may not be any nice description of all such subgroups, but are you aware of any other examples? $\endgroup$ – Benighted Feb 9 '18 at 1:56
  • $\begingroup$ @StephenPietromonaco: Pick your favorite compact Lie group, say, the compact $G_2$, or $E_8$, etc; it will have some faithful finite-dimensional unitary representation. Or, just consider $SU(2)$: It has an irreducible unitary representation on complex vector space of every dimension $\ge 2$: Just let it act on the space $P_d$ of homogeneous polynomials in two complex variables, of degree $d=n-1$, by precomposition $g\cdot p= p\circ g^{-1}$. $\endgroup$ – Moishe Kohan Feb 9 '18 at 3:40
  • $\begingroup$ @MoisheCohen Interesting, thanks! Okay, so given any compact Lie group $G$ and any complex vector space $V$, we have a representation $\rho : G \to GL(V)$ with image in the unitary group. But I guess my question would be, given a complex vector bundle $E$ with fiber $V$, must there exist a metric on $E$ reducing the structure group to the image of $\rho$? $\endgroup$ – Benighted Feb 9 '18 at 3:59
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    $\begingroup$ The fact that the image of $\rho$ is contained in the unitary group exactly says that the Hermitian inner product is preserved by the action of $G$. $\endgroup$ – Andreas Cap Feb 9 '18 at 5:37

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