Euler classes of oriented $2$-dimensional vector bundle, oriented $S^1$-bundle same?

Question

As the question title suggests, are the Euler classes of an oriented $2$-dimensional vector bundle and of an oriented $S^1$-bundle the same?

Answer 1

Given an oriented $2$-dimensional vector bundle $\pi: E \to M$, say we have a Riemannian metric and select a connection $\nabla$ compatible with the metric. If $\omega = (\omega_j^i)$ and $\Omega = (\Omega_j^i)$ are the connection and curvature forms, we have$$\omega = \begin{pmatrix} 0 & \omega_2^1 \\ \omega_1^2 & 0 \end{pmatrix}, \quad \Omega = \begin{pmatrix} 0 & \Omega_2^1 \\ \Omega_1^2 & 0 \end{pmatrix}.$$Moreover, we have$$\omega_1^2 = -\omega_2^1, \quad \Omega_1^2 = -\Omega_2^1, \quad \Omega_2^1 = d\omega_2^1.$$Now, if $P(E)$ denotes the principal $\text{GL}(2, \mathbb{R})$-bundle associated to $E$, then $\nabla$ determines a connection $\tilde{\omega}$ by virtue of the following result, which we will not prove here.

Let $\pi: E \to M$ be an $n$-dimensional vector bundle and $\pi: P \to M$ the associated principal $\text{GL}(n, \mathbb{R})$-bundle. Then there is a natural $1$-to-$1$ correspondence between the set of connections on $E$ and the set of connections on $P$.

On the other hand, if we set$$S(E) = \{u \in E: \|u\| = 1\},$$then the projection $S(E) \to E$ is an oriented $S^1$-bundle. We can also regard $S(E)$ as a submanifold of $P(E)$, i.e. for $u \in S(E)$, let $u'$ be the vector we get by $\pi/2$ rotation of $u$ in the positive direction. We associate the frame $[u, u']$ to $u$. Here, the $(2, 1)$ component of the $1$-form that is the restriction of $\tilde{\omega}$ to $S(E)$, namely, the portion $\omega_1^2$, is a connection form of the principal $S^1$-bundle $S(E)$, since$$\text{exp}\,t\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} = \begin{pmatrix} \cos t & -\sin t \\ \sin t & \cos t\end{pmatrix}.$$Now, by definition, the Euler class of $E$ as an oriented vector bundle is represented by the closed $2$-form ${1\over{2\pi}}\Omega_2^1$. On the other hand, the Euler class of the $S^1$-bundle $S(E)$ is represented by $-{1\over{2\pi}}\Omega_1^2$, and so we are done.