2
$\begingroup$

The space $SL_3(\mathbb{R})/SO_3(\mathbb{R})$ can be though of as some kind of 5-(real)dimensional generalized upper half-space.

Fix any copy of $SO_2(\mathbb{R})$ sitting inside $SO_3(\mathbb{R})$, then the coset space $SL_3(\mathbb{R})/SO_2(\mathbb{R})$ can be though of as a bundle over $SL_3(\mathbb{R})/SO_3(\mathbb{R})$ whose fibers are all $SO_3(\mathbb{R})/SO_2(\mathbb{R}) \cong S^2$, the two-sphere.

My question is whether or not this sphere bundle is trivial. By which I mean, is it the case that $SL_3(\mathbb{R})/SO_2(\mathbb{R})$ is diffeomorphic to $SL_3(\mathbb{R})/SO_3(\mathbb{R})\times S^2$.

$\endgroup$

1 Answer 1

10
$\begingroup$

As you remark, the quotient $SL_n(R)/SO_n(R)$ is contractible, and so any smooth bundle over it is smoothly trivial. So, in particular your bundle is diffeomorphic to a product.

The contractibility of $SL_n(R)/SO_n(R)$ stems from the Gram-Schmidt orthogonalization process; presumably you can use this to exhibit a specific trivialization of your bundle.

$\endgroup$
1
  • $\begingroup$ Great! Thank you. I was pretty sure this was the case, I figured I was overlooking something fundamental. $\endgroup$
    – anon776
    Commented Jan 6, 2014 at 20:13

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .