Non existence of a preferred Horizontal subspace.

Question

If I choose a principal bundle, let us say $G\rightarrow P \rightarrow B$, with $G=U(1)$, $P=T^2$ (2-torus) and $B=S^1$. Can I choose follow the identity element for every $x$ of the base space? Doesn't this imply that there is a preferred Horizontal subspace (the one tangent to every point of the curves I have described), and thus a canonical connection ?

Answer 1

I will try to be precise but informal.

Let $\pi:P\to B$ be the trivial principal $S^1$-bundle over $S^1$ and let $x\in B$ be any point. Then:

1. There is no preferred point of the fiber $\pi^{-1}(x)$.
2. Given a point $y\in \pi^{-1}(x)$, there is no preferred point of nearby fibers; if $U\ni x$ is a neighborhood of $x\in B$, there is no preferred section $s$ of $\pi$ over $U$ such that $s(x)=y$.

Why? As a first approximation to the answer, just imagine a perfect unit circle $C$ in some strange position in space. It is diffeomorphic, even isometric, to our "standard" copy of $S^1 = \{(x,y)\in \mathbb{R}^2 \mid x^2 + y^2 = 1\}$ in many different ways (this "standard" $S^1$ has a "standard" group structure, fixed once and for all, in which the identity is $(1,0)$. We could have chosen the identity differently, but we didn't.) We know what it means to rotate the circle $C$, so there is an $S^1$-action on it. But we don't have a preferred "identity point" on it! If you choose some $p\in C$ and I choose some different $q\in C$, neither of us has a better choice. There is an isometry $C\to S^1$ taking $p$ to $(1,0)$ but there is also an isometry taking $q$ to $(1,0)$.

The above paragraph is all you need to understand point (1) above: the fiber over a point is just something that looks like a circle and has a circle action, so without additional information it has no preferred point. If $\pi:P\to B$ is given in some abstract manner, you can't "choose" a point in the fiber that correponds to your favorite point on some standard copy of $S^1 \subset \mathbb{R}^2$, because there is no preferred correspondence.

Even if $\pi:P\to B$ is given concretely by an atlas of charts you may not be able to choose a distinguished point in each fiber. The atlas may identify the fibers with the circle $C$ from above, for instance.

Point (2) is kind of similar.

That $\pi$ is trivial means the following: let $$p: S^1 \times S^1 \to S^1$$ $$p(a,b)=a.$$ This is a trivial principal $S^1$-bundle over $S^1$. Then since $\pi$ is trivial also, there is an isomorphism of principal bundles. This is a diffeomorphism $f:P\to S^1 \times S^1$ such that $$p\circ f = \pi$$ and which is equivariant with respect to the $S^1$-action, meaning $$f(y.g) = f(y).g$$ for all $y\in P$ and all $g\in S^1$.

However, there are many many different isomorphisms of this kind. Indeed, let $h:S^1 \to S^1$ be any smooth function (I don't care about injectivity or any other condition). Then the function $$P \to S^1 \times S^1$$ $$y \mapsto f(y.h(\pi(y)))$$ is also an isomorphism of principal bundles from $P$ to the trivial bundle. Any two isomorphisms from $P$ to the trivial bundle are related by such an $h$. Who is to say which $h$ gives the best isomorphism?

If $h_1(t) = 1$ and $h_2(t)=t$ are two functions $S^1 \to S^1$, and $f$ is some isomorphism from $P$ to the trivial bundle, we get two different isomorphisms in this way, and we can't deformed one into the other continuously. Of course the isomorphism constructed using $f$ and $h_1$ is $f$ itself, but that doesn't make it better than the other.