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Complex Stiefel and Grassmann tangent projections

Statement

For X∈St⁡C(n,r), as a real manifold, TXSt⁡C(n,r)={Z:X∗Z+Z∗X=0} and ΠXSt(G)=G−XHerm⁡(X∗G). The complex Grassmann tangent has the unique horizontal representative X∗Z=0, and its quotient-metric projection is ΠXGr(G)=(I−XX∗)G.

Facts & Assumptions

Given: 1≤r≤n, X∗X=Ir, and the real Frobenius metrics of Complex Stiefel and Grassmann optimization conventions.

[F1]

The Stiefel frame set is the level set of X↦X∗X−Ir, and the Grassmann space is its quotient by the right unitary action (Complex Stiefel and Grassmann optimization conventions).

[F2]

The tangent of a regular level set is the kernel of the level-map differential (The tangent space of a regular level set is the kernel).

[F3]

The complex Grassmann quotient has graph charts, and its frame bundle has right unitary fibers (Stiefel spaces, Grassmannians, and tautological bundles).

Proof

technique · differentiate the Hermitian constraint and orthogonally decompose into vertical, horizontal, and normal components
1.1F1F2givenalgebra

Regard the Hermitian r×r matrices as a real vector space. The derivative of F(X)=X∗X−Ir is DFX[Z]=X∗Z+Z∗X. It is onto that space: for any Hermitian H, Z=XH/2 gives DFX[Z]=H. Hence the level is regular, and [F2] gives the displayed tangent kernel.

2.1givenstep 1.1algebra

Set S=Herm⁡(X∗G). Then X∗(G−XS)+(G−XS)∗X=0, so G−XS is tangent. For tangent Z, put K=X∗Z, which is skew-Hermitian. For Hermitian S, tr⁡(SK)‾=−tr⁡(SK) by adjoint and cyclic-trace identities; thus Re⁡tr⁡(SK)=0. Therefore XS is normal in the real Frobenius metric, and the stated formula is the Stiefel orthogonal projection.

2.2F1F3givenstep 1.1algebra

A curve Q(t) in U⁡(r) through Ir has Ω=Q′(0) skew-Hermitian by differentiating Q(t)∗Q(t)=Ir. Conversely, for any skew-Hermitian Ω, the Cayley curve Q(t)=(Ir−t2Ω)−1(Ir+t2Ω) is defined near zero, satisfies Q(t)∗Q(t)=Ir, and has Q(t)=Ir+tΩ+O(t2); hence the vertical tangent directions are exactly XΩ. Every Stiefel tangent decomposes as Z=XΩ+(I−XX∗)Z with Ω=X∗Z. The second summand is horizontal. For any vertical XΩ and horizontal W, ⟨XΩ,W⟩R=Re⁡tr⁡(Ω∗X∗W)=0, so the summands are orthogonal. The decomposition is unique, and the graph-chart quotient identifies its horizontal component with the quotient tangent.

3.1F1givenstep 2.2algebra∎

The matrix (I−XX∗)G is horizontal. For every horizontal Z, ⟨G−(I−XX∗)G,Z⟩R=Re⁡tr⁡((XX∗G)∗Z)=Re⁡tr⁡(G∗XX∗Z)=0 since X∗Z=0. Thus it is the horizontal orthogonal projection. Under (X,G)↦(XQ,GQ) for unitary Q, it transforms to (I−XX∗)GQ, the same tangent vector in the changed frame.

Depends on

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Sources