# Understanding a statement related to a circle action on a principal bundle found in a paper

I am trying to understand a statement in the paper

http://iopscience.iop.org/0951-7715/3/3/012/pdf/0951-7715_3_3_012.pdf

I give details below so it should not be necessary to look at the paper.

Suppose $Q$ is an $SU(2)$-bundle over $S^4$ which $S^1$-equivariant, and denote by $E$ the rank 2 complex vector bundle associated to $Q$ via the fundamental representation. Over the fixed point of the circle action on $S^4$, an $S^2$, the circle acts fiber-wise.

The statement I have troubles understanding is the following one (on top of page 812 in the paper):

"Assuming that this action [the $S^1$ action] is non-trivial, we see that $E$ splits over $S^2$ as $$\left. E\right|_{S^2}=L\oplus L^*,$$ where $L,L^*\rightarrow S^2$ are line bundles with $S^1$ action. Because the $S^1$ action is by $SU(2)$ transformations, $L$ is the dual of $L^*$."

I cannot see how the fact that the bundle splits is related to the $S^1$ action.

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First I think one can say that a principal bundle automorphism induces an associated bundle isomorphism. If $Q$ is a principal $G$-bundle, $f\in\mathrm{Aut}(Q)$, $E=Q\times_{\rho}F$ is an associated bundle with fibre $F$, define $\hat f:E\rightarrow E$ via $$\hat f([p,u])=[p,\rho(g_f)u],$$ where $[p,u]$ is a point of $E$ and $g_f:Q\rightarrow G$, is the unique $G$-equivariant map such that $f(p)=p\cdot g_f(p)$. By $G$-equivariant I mean that $g_f(p\cdot h)=h^{-1}g_f(p)\,h$ for any $h\in G$, with $\cdot$ the right action of $G$ on $Q$.
Going back to the statement quoted above, the $S^1$ action gives an automorphism on the bundle $\left. Q \right|_{S^2}$ , hence it induces a 2-dimensional representation of $U(1)$ on the $\mathbb{C}^2$ fibres of $\left. E \right|_{S^2}$. This representation must be reducible, hence a priori of the form $$\left(\begin{array}{cc} \exp(ip\alpha)&0\\0&\exp(iq\alpha)\end{array}\right),$$ with $p$,$q$ integers. However, since $G=SU(2)$, this must also be an $SU(2)$ representation, hence $$q=-p.$$