MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The real Stiefel manifold $V_{n,k}$ of orthogonal $k$-frames in $\mathbb{R}^n$ can be viewed as the reductive homogeneous space $G/H=O(n)/O(n-k)$. If ${\frak{so}}(n)$ is the Lie algebra of $O(n)$, then we have the reductive decomposition $$ {\frak{so}}(n)={\frak{m}}+{\frak{h}} $$ where $$ {\frak{m}}=\left \{ \begin{pmatrix} A & B\\ -B^T & O \end{pmatrix}: A\>\> \text{is a}\>\> k\times k \>\>\text{skew-symmetric matrix} \right \} $$ and $$ {\frak{h}}=\left \{ \begin{pmatrix} O & O\\ O & C \end{pmatrix}: C\>\> \text{is a}\>\> (n-k)\times (n-k) \>\>\text{skew-symmetric matrix} \right \} $$ with ${\frak{so}}(n)$ and ${\frak{h}}$ a reductive pair.

According to Helgason, there is a local diffeomorphism $$ (\text{exp}X,h)\mapsto (\text{exp}X)h,\quad \text{where}\>\> X\in {\frak{m}}, h\in H. $$ However, there doesn't seem to be anything in the literature where this decomposition is calculated explicitly. I did find the following ``polar decomposition": $$ V_{n,k}\times P_k\rightarrow M_{n,k},\quad (v,r)\mapsto vr^{1/2} $$ where $P_k$ is the set of positive semi-definite symmetric matrices and $M_{n,k}$ is the set of all $n\times k$ real matrices.

Does this last decomposition have any connection with the polar decompositions of Lie groups? If not, is there an explicit description of the polar decomposition involving Stiefel manifolds somewhere?

share|cite|improve this question
For a Cartan decomposition, you need $[\mathfrak{m},\mathfrak{m}]\subset\mathfrak{h}$ and that is certainly not the case here. – Fran Burstall Aug 17 '13 at 13:06
@Fran: I'm not sure of the correct terminology here. I've changed Cartan to polar. – Oliver Jones Aug 17 '13 at 21:22

It is easier for me to examine this geometrically, rather than from the point of view of Lie groups and algebras.

First, the identity matrix $I \in O(n)$ represents the standard orthonormal basis $e_1, \dots, e_n$ of $\mathbb{R}^n$, and its coset $I\cdot O(n-k)$ represents the $k$-plane spanned by the first $k$ basis vectors $e_1, \dots, e_k$.

The local splitting you're looking for is equivalent to extending the standard orthonormal basis to a local map from $V_{n,i}$ to $O(n)$, viewed as the space of orthonormal frames in $\mathbb{R}^n$, such that the span of the first $k$ vectors in each frame is equal to the corresponding $k$-plane in $V_{n,k}$. Helgason is basically pointing out that one way to do this is to use the connection naturally induced by the bi-invariant Riemannian metric on $O(n)$ to parallel transport the initial frame along geodesic rays starting at the identity matrix.

This local section can be described as a solution to a linear second first order system of ODE's along each geodesic (essentially the Jacobi equations along each geodesic), and in principle these ODE's could be integrated to give some kind of formula for this section. Unfortunately, I've never computed this and don't know what it looks like. And it's not clear to me how to transfer this description to the setting of Lie groups and algebras. My wild guess is that you can write the system of ODE's in terms of the Lie algebra, but you won't be able to integrate them explicitly.

One possibly easier way to do this is to using the dual description using invariant differential forms as presented in:

Griffiths, P. On Cartan's method of Lie groups and moving frames as applied to uniqueness and existence questions in differential geometry. Duke Math. J. 41 (1974), 775–814.

share|cite|improve this answer
Thanks for the suggestions; I'll look into it. – Oliver Jones May 21 '14 at 7:26

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.