Let $E\longrightarrow B$ be a vector bundle of rank $k$, then its structure group $GL(R^{k})$ acts on the fibre. Now assume that $G$ is a compact Lie group and $E\longrightarrow B$ is a $G$-equivariant vector bundle. From the definition of equivariant bundle we know that $G$ acts on each fibre of $E$ as a linear isomorphism, so that there are two actions of groups on the fibre: $GL(R^{k})$ and $G$, is it true that such two actions commute?

Let $B_{(H)}=\{x\in B:G_{x}\quad is\quad conjugate\quad to\quad H\}$, then every connected component of $B_{(H)}$ is a closed submanifold. Consider the isotropy representation: for any $x$, $y$ contained in the same connected component of $B_{(H)}$ we have representations $$\rho_{x}:G_{x}\longrightarrow GL(R^{k})$$ and $$\rho_{y}:G_{y}\longrightarrow GL(R^{k})$$ is it true that $\rho_{x}=\rho_{y}$?

  • $\begingroup$ Beware of the phrase "the fibre". There is not a canonical action of $GL(R^k)$ on each fibre of a rank $k$ vector bundle. $\endgroup$ Commented Oct 12, 2013 at 12:11
  • $\begingroup$ Also beware that "equivariant bundle" often means a bundle in which $G$ acts on both $E$ and $B$, compatibly. $\endgroup$ Commented Oct 12, 2013 at 12:12
  • $\begingroup$ @yangyang: Yes, they has to commute. You can write out local trivilizations and this become $g(\phi(e))=\phi(g(e))$, which means $g$' action does not depend on the charts. $\endgroup$ Commented Oct 12, 2013 at 21:27

1 Answer 1


Following up on Tom's comment, real G-vector bundles are generally understood to be $O(n)$-bundles and $G$-maps $E\longrightarrow B$ (with local equivariant triviality appropriately defined). Both $G$ and $O(n)$ act on the total space of the associated principal bundle, and it is required that the actions commute. When $G$ acts non-trivially on $B$, it is not true that $G$ acts on fibers. For $b\in B$, the isotropy subgroup $G_b$ of elements that fix $b$ acts on the fiber over $b$.

  • $\begingroup$ Peter, just to be clear, your $E$ is the principal bundle associated to a vector bundle, yes? $\endgroup$ Commented Oct 12, 2013 at 14:13
  • $\begingroup$ Yes, that is what I meant to say. Thanks. Answer edited $\endgroup$
    – Peter May
    Commented Oct 12, 2013 at 17:15
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    $\begingroup$ "Real $G$-vector bundles are generally understood to be $O(n)$-bundles"... I highly doubt that, as this reduction from $GL(n)$ to $O(n)$ is equivalent to choosing a metric. $\endgroup$ Commented Oct 15, 2013 at 17:57

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