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wofsy
Oct16-09, 08:09 PM
An oriented surface with a Riemannian metric has a natural action of the unit circle on its tangent bundle. Rotate the tangent vector through the angle theta in the positively direction.

Is there a natural action of SO(n) on the tangent bundle of an oriented Riemannian n-manifold?

Same question for any oriented vector bundle with Riemannian metric over a smooth manifold.

zhentil
Oct17-09, 12:32 AM
In general, the answer is no. I don't see how one could define it independent of basis.

It does act, however, on the frame bundle (kind of like a lift of the tangent bundle to a principal bundle). The 2-dimensional case is quite special, since the frame bundle of an oriented 2-manifold is a circle (i.e. fix an angle - then orientation gives you an orthonormal frame).

wofsy
Oct17-09, 08:53 AM
In general, the answer is no. I don't see how one could define it independent of basis.

It does act, however, on the frame bundle (kind of like a lift of the tangent bundle to a principal bundle). The 2-dimensional case is quite special, since the frame bundle of an oriented 2-manifold is a circle (i.e. fix an angle - then orientation gives you an orthonormal frame).

Right - so what you are saying implies the following. take a unit vector and extend it to an oriented n-frame arbitrarily, keeping the vector as the first element of the frame. Let SO(n) act on this frame. Then the image of v under this action depends on the choice of the orthogonal complement to v.

zhentil
Oct17-09, 09:14 PM
Yes, I believe that's precisely it. Unless the action fixes v (or v itself determines a basis, as in the 2-dimensional oriented case), you have to specify which tangent vector to send it to.