what is the circle bundle of $\mathbb{R}^2$ Given dot product on $\mathbb{R}^2$ makes $\mathbb{R}^2$ a Riemannian manifold,
(a) what is the circle bundle $S(\mathbb{R}^2)$ and what is it diffeomorphic to?
(b) And what is the connection on $S(\mathbb{R}^2)$ and its connection $1$-form?
I don't know how to proceed on these questions.
I know that circle bundle is $S(M) = \{(m,v) \in T(M); \langle v,v\rangle =1\}$, where $S(M)$ is the circle bundle.
For $\mathbb{R}^2$, isn't $T(M)$ the tangent bundle just $\mathbb{R}^2$ since at every point in $\mathbb{R}^2$, the tangent plane is same $\mathbb{R}^2$. So would the circle bundle $S(M)$ be same as $T(M)$? Please help; also, I don't know how to proceed on connection $1$-form either.
 A: The circle bundle $S(M)$ cannot be equal to $T(M)$, as the first is a circle bundle and the latter is a vector bundle.
Since $T(\Bbb R^n) = \Bbb R^n\times \Bbb R^n$, you can write $S(\Bbb R^n)$ explicitly as $$S(\Bbb R^n) = \{ (p,v) \in \Bbb R^n \times \Bbb R^n \mid \langle v,v\rangle = 1\} = \Bbb R^n \times \Bbb S^{n-1}.$$It is a fiber bundle with typical fiber $\Bbb S^{n-1}$. In particular, for $n=2$ we have that $S(\Bbb R^n)$ is (diffeomorphic to) a cylinder. The vertical spaces are ${\rm Ver}_{(p,v)}(S(\Bbb R^n)) = \{0\}\times T_v(\Bbb S^{n-1}) = \{0\} \times v^\perp$.
Is there a natural choice of complementary space to ${\rm Ver}_{(p,v)}(S(\Bbb R^n))$? Surely: ${\rm Hor}_{(p,v)}(S(\Bbb R^n)) = T_p\Bbb R^n \times \{0\}$. Thus we have a connection. Everything worked out nicely because $S(\Bbb R^n) \to \Bbb R^n$ was a trivial bundle to begin with (this is not true for arbitrary $(M^n,g)$). A choice of horizontal subspaces is equivalent to choosing a
connection $1$-form when you have a principal bundle, but this is not the case here ($\Bbb S^{n-1}$ is not in general a Lie group), so I'm not sure what they expect of you here.
