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Can there exist two non-equivalent equivariant actions of a group $G$ on vector bundle over a $G$ space?

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3 Answers 3

up vote 5 down vote accepted

Maybe I am miss understanding the question, but it seems the answer is yes.

Take your favorite G-space, mine is $S^1$ with the $\mathbb{Z}/2$-action "flip". Then consider the trivial vector bundles $S^1 \times V$, where $V$ is a $G$-representation. In my favorite example $V = \mathbb{R}$ can be either the trivial representation or the sign representation. Taking the diagonal $G$-action gives an equivariant action of the group on the vector bundle. They are distinct for distinct representations (at least in the $S^1$-example) yet the underlying vector bundles are the same if the representations have the same dimension (they are trivial bundles after all).

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4  
Or more simply, replace $S^1$ with a point? –  Steven Sam Feb 10 '10 at 20:12
    
Of course! yes use a point. It's much easier then the circle. I should have used that from the beginning. –  Chris Schommer-Pries Feb 11 '10 at 1:41

For an example from Algebraic Geometry, $\mathbb{P}^1_\mathbb{C}$ is a homogeneous space for both $GL(2)$ and $SL(2)$. The trivial line bundle and the determinant line bundle are isomorphic as $SL(2)$-equivariant bundles, and therefore as vector bundles, but are non-isomorphic as $GL(2)$-equivariant bundles.

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More generally, one can ask whether a group action on a manifold $M$ (sorry, I work primarily in that category) can be lifted to a fibre bundle over $M$. There are obstructions to the lifting and even if a lifting exists it may not be unique. The earliest results of this nature I am aware of are in the paper Lifting compact group actions in fiber bundles by Hattori and Yoshida. You may also want to look at this more recent article Lifts of smooth group actions to line bundles by Ignasi Mundet, which may contain more recent references.

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