# Does there exist a preferred trivialization of a trivial line bundle?

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 $$C$$?

• No. Why would there be? – Qiaochu Yuan Nov 13 '18 at 19:01
• @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. – A. Haydys Nov 14 '18 at 8:13
• 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. – Qiaochu Yuan Nov 14 '18 at 8:17
• 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! – Qiaochu Yuan Nov 14 '18 at 8:25
• 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. – A. Haydys Nov 14 '18 at 9:19