Let $L\to M$ be a topologically trivial complex Hermitian line bundle (over a manifold of dimension three, if this is of any importance). I assume that $L$ admits a trivialization, however, I do not assume that any particular trivialization is chosen.

The gauge group $\mathcal G:=C^0(M; U(1))$ acts on the space of trivializations in a natural way and this action is transitive. It is easy to see that $\mathcal G$ is not connected in general. In fact, we have a natural homomorphism $\mathcal G\to H^1(M;\mathbb Z), f\mapsto f^*a$, where $a$ is a generator of $H^1(U(1);\mathbb Z)$. This in turn yields an action of $H^1(M;\mathbb Z)$ on the space $C$ of connected components of the space of all trivializations of $L$, i.e., $C$ is an $H^1(M;\mathbb Z)$-torsor. So, the question is whether there is a natural choice of the origin in $C$?

  • 4
    $\begingroup$ No. Why would there be? $\endgroup$ – Qiaochu Yuan Nov 13 '18 at 19:01
  • $\begingroup$ @Qiaochu Yuan Is this "no, I don't see how to do this" or "no, there is an obstruction". In the latter case I would be interested to know what is the obstruction. $\endgroup$ – A. Haydys Nov 14 '18 at 8:13
  • $\begingroup$ You say you want a natural choice of the origin; natural in what? It certainly cannot be natural wrt automorphisms of $L$, since this acts transitively as you say. It doesn't even make sense to ask for it to be natural wrt automorphisms of $M$ since $L$ has not been equipped with any kind of equivariant structure, so automorphisms of $M$ don't even act. $\endgroup$ – Qiaochu Yuan Nov 14 '18 at 8:17
  • $\begingroup$ In general, you're giving us two objects $a, b$ in a category and telling us they're isomorphic (here the category of line bundles, $a$ is the given line bundle, and $b$ is the trivial one). Then $\text{Iso}(a, b)$ is a torsor over either $\text{Aut}(a)$ or $\text{Aut}(b)$ and without more information nothing more can be said. In order to specify a point in $\text{Iso}(a, b)$ you'd have to, y'know, specify an isomorphism between $a$ and $b$, which in your first paragraph you explicitly tell us you aren't doing! $\endgroup$ – Qiaochu Yuan Nov 14 '18 at 8:25
  • $\begingroup$ As far as I see $Aut(M)$ acts on the space of pairs $(L, s)$ consisting of a trivial line bundle and a nowhere vanishing section. Namely, if $f\in Aut (M)$, then $f\cdot (L, s) = (f^*L, f^*s)$. In particular, naturality perfectly makes sense. $\endgroup$ – A. Haydys Nov 14 '18 at 9:19

Your Answer

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

Browse other questions tagged or ask your own question.