Stiefel manifold

In mathematics, the Stiefel manifold $V_k(\R^n)$ is the set of all orthonormal ''k''-frames in $\R^n.$ That is, it is the set of ordered orthonormal ''k''-tuples of vectors in $\R^n.$ It is named after Swiss mathematician Eduard Stiefel. Likewise one can define the complex Stiefel manifold $V_k(\Complex^n)$ of orthonormal ''k''-frames in $\Complex^n$ and the quaternionic Stiefel manifold $V_k(\mathbb^n)$ of orthonormal ''k''-frames in $\mathbb^n$. More generally, the construction applies to any real, complex, or quaternionic inner product space.

In some contexts, a non- compact Stiefel manifold is defined as the set of all linearly independent ''k''-frames in $\R^n, \Complex^n,$ or $\mathbb^n;$ this is homotopy equivalent, as the compact Stiefel manifold is a deformation retract of the non-compact one, by Gram–Schmidt. Statements about the non-compact form correspond to those for the compact form, replacing the orthogonal group (or unitary or symplectic group) with the general linear group.

Topology

Let $\mathbb$ stand for $\R,\Complex,$ or $\mathbb.$ The Stiefel manifold $V_k(\mathbb F^n)$ can be thought of as a set of ''n'' × ''k'' matrices by writing a ''k''-frame as a matrix of ''k'' column vectors in $\mathbb F^n.$ The orthonormality condition is expressed by ''A''*''A'' = $I_k$ where ''A''* denotes the conjugate transpose of ''A'' and $I_k$ denotes the ''k'' × ''k'' identity matrix. We then have

:$V_k(\mathbb F^n) = \left\.$

The topology on $V_k(\mathbb F^n)$ is the subspace topology inherited from $\mathbb^.$ With this topology $V_k(\mathbb F^n)$ is a compact manifold whose dimension is given by

:$\begin \dim V_k(\R^n) &= nk - \frack(k+1) \\ \dim V_k(\Complex^n) &= 2nk - k^2 \\ \dim V_k(\mathbb^n) &= 4nk - k(2k-1) \end$

As a homogeneous space

Each of the Stiefel manifolds $V_k(\mathbb F^n)$ can be viewed as a homogeneous space for the action of a classical group in a natural manner.

Every orthogonal transformation of a ''k''-frame in $\R^n$ results in another ''k''-frame, and any two ''k''-frames are related by some orthogonal transformation. In other words, the orthogonal group O(''n'') acts transitively on $V_k(\R^n).$ The stabilizer subgroup of a given frame is the subgroup isomorphic to O(''n''−''k'') which acts nontrivially on the orthogonal complement of the space spanned by that frame.

Likewise the unitary group U(''n'') acts transitively on $V_k(\Complex^n)$ with stabilizer subgroup U(''n''−''k'') and the symplectic group Sp(''n'') acts transitively on $V_k(\mathbb^n)$ with stabilizer subgroup Sp(''n''−''k'').

In each case $V_k(\mathbb F^n)$ can be viewed as a homogeneous space:

:$\begin V_k(\R^n) &\cong \mbox(n)/\mbox(n-k)\\ V_k(\Complex^n) &\cong \mbox(n)/\mbox(n-k)\\ V_k(\mathbb^n) &\cong \mbox(n)/\mbox(n-k) \end$

When ''k'' = ''n'', the corresponding action is free so that the Stiefel manifold $V_n(\mathbb F^n)$ is a principal homogeneous space for the corresponding classical group.

When ''k'' is strictly less than ''n'' then the special orthogonal group SO(''n'') also acts transitively on $V_k(\R^n)$ with stabilizer subgroup isomorphic to SO(''n''−''k'') so that

:$V_k(\R^n) \cong \mbox(n)/\mbox(n-k)\qquad\mbox k < n.$

The same holds for the action of the special unitary group on $V_k(\Complex^n)$

:$V_k(\Complex^n) \cong \mbox(n)/\mbox(n-k)\qquad\mbox k < n.$

Thus for ''k'' = ''n'' − 1, the Stiefel manifold is a principal homogeneous space for the corresponding ''special'' classical group.

Uniform measure

The Stiefel manifold can be equipped with a uniform measure, i.e. a Borel measure that is invariant under the action of the groups noted above. For example, $V_1(\R^2)$ which is isomorphic to the unit circle in the Euclidean plane, has as its uniform measure the obvious uniform measure ( arc length) on the circle. It is straightforward to sample this measure on $V_k(\mathbb F^n)$ using Gaussian random matrices: if $A\in\mathbb^$ is a random matrix with independent entries identically distributed according to the standard normal distribution on $\mathbb$ and ''A'' = ''QR'' is the QR factorization of ''A'', then the matrices, $Q\in\mathbb^, R\in\mathbb^$ are independent random variables and ''Q'' is distributed according to the uniform measure on $V_k(\mathbb F^n).$ This result is a consequence of the Bartlett decomposition theorem.

Special cases

A 1-frame in $\mathbb^n$ is nothing but a unit vector, so the Stiefel manifold $V_1(\mathbb F^n)$ is just the unit sphere in $\mathbb^n.$ Therefore:

:$\begin V_1(\R^n) &= S^\\ V_1(\Complex^n) &= S^\\ V_1(\mathbb^n) &= S^ \end$

Given a 2-frame in $\R^n,$ let the first vector define a point in ''S''^''n''−1 and the second a unit tangent vector to the sphere at that point. In this way, the Stiefel manifold $V_2(\R^n)$ may be identified with the unit tangent bundle

When ''k'' = ''n'' or ''n''−1 we saw in the previous section that $V_k(\mathbb^n)$ is a principal homogeneous space, and therefore diffeomorphic to the corresponding classical group:

:$\begin V_(\R^n) &\cong \mathrm(n)\\ V_(\Complex^n) &\cong \mathrm(n) \end$

:$\begin V_(\R^n) &\cong \mathrm O(n)\\ V_(\Complex^n) &\cong \mathrm U(n)\\ V_(\mathbb^n) &\cong \mathrm(n) \end$

Functoriality

Given an orthogonal inclusion between vector spaces $X \hookrightarrow Y,$ the image of a set of ''k'' orthonormal vectors is orthonormal, so there is an induced closed inclusion of Stiefel manifolds, $V_k(X) \hookrightarrow V_k(Y),$ and this is functorial. More subtly, given an ''n''-dimensional vector space ''X'', the dual basis construction gives a bijection between bases for ''X'' and bases for the dual space $X^*,$ which is continuous, and thus yields a homeomorphism of top Stiefel manifolds $V_n(X) \stackrel V_n(X^*).$ This is also functorial for isomorphisms of vector spaces.

As a principal bundle

There is a natural projection

:$p : V_k(\mathbb F^n) \to G_k(\mathbb F^n)$

from the Stiefel manifold $V_k(\mathbb F^n)$ to the Grassmannian of ''k''-planes in $\mathbb^n$ which sends a ''k''-frame to the subspace spanned by that frame. The fiber over a given point ''P'' in $G_k(\mathbb F^n)$ is the set of all orthonormal ''k''-frames contained in the space ''P''.

This projection has the structure of a principal ''G''-bundle where ''G'' is the associated classical group of degree ''k''. Take the real case for concreteness. There is a natural right action of O(''k'') on $V_k(\R^n)$ which rotates a ''k''-frame in the space it spans. This action is free but not transitive. The orbits of this action are precisely the orthonormal ''k''-frames spanning a given ''k''-dimensional subspace; that is, they are the fibers of the map ''p''. Similar arguments hold in the complex and quaternionic cases.

We then have a sequence of principal bundles:

:$\begin \mathrm O(k) &\to V_k(\R^n) \to G_k(\R^n)\\ \mathrm U(k) &\to V_k(\Complex^n) \to G_k(\Complex^n)\\ \mathrm(k) &\to V_k(\mathbb^n) \to G_k(\mathbb^n) \end$

The vector bundles associated to these principal bundles via the natural action of ''G'' on $\mathbb^k$ are just the tautological bundles over the Grassmannians. In other words, the Stiefel manifold $V_k(\mathbb F^n)$ is the orthogonal, unitary, or symplectic frame bundle associated to the tautological bundle on a Grassmannian.

When one passes to the $n\to \infty$ limit, these bundles become the universal bundles for the classical groups.

Homotopy

The Stiefel manifolds fit into a family of fibrations:

:$V_(\R^) \to V_k(\R^n) \to S^,$

thus the first non-trivial homotopy group of the space $V_k(\R^n)$ is in dimension ''n'' − ''k''. Moreover,

:$\pi_ V_k(\R^n) \simeq \begin \Z & n-k \text k=1 \\ \Z_2 & n-k \text k>1 \end$

This result is used in the obstruction-theoretic definition of Stiefel–Whitney classes.

See also

* Flag manifold
*Matrix Langevin distribution

References

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* {{Springer, id=Stiefel_manifold, title=Stiefel manifold

Differential geometry
Homogeneous spaces
Fiber bundles
Manifolds