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Projected vector transport on Stiefel and Grassmann

Definition

Let X be a source frame, η∈TXSt⁡(n,r), and Z∈TXSt⁡(n,r). For Stiefel transport let Y=RX(η) be either the QR or polar endpoint. For Grassmann transport, require η to be the unique horizontal representative at [X], so XTη=0, and choose Y=RXpol(η), the polar endpoint; the positive-diagonal QR frame is not used for this representative-based quotient transport. Define the projected Stiefel transport TYSt(Z):=Z−Ysym⁡(YTZ), and, for a horizontal source representative Z, define the projected Grassmann transport representative TYGr(Z):=(I−YYT)Z. The Grassmann expression is interpreted as a tangent representative at [Y]. The polar endpoint is right-equivariant, so the next proposition verifies representative independence, tangency, and identity at a zero step. These projection transports are not claimed to be Levi-Civita parallel transport.

Facts & Assumptions

Given: X∈St⁡(n,r), a tangent step η (horizontal in the Grassmann branch), its Stiefel endpoint Y=RX(η) or its Grassmann polar endpoint Y=RXpol(η), and a source tangent (or horizontal) representative Z.

[F1]

The Stiefel and horizontal projections at a frame Y are the formulas in Tangent spaces and Frobenius projections on Stiefel and Grassmann.

[F2]

QR and polar endpoints are orthonormal frames, and the polar endpoint transforms as RXQpol(ηQ)=RXpol(η)Q (QR and polar retractions on Stiefel and Grassmann).

[F3]

Both retractions fix the zero tangent vector (The QR and polar maps are first-order Stiefel and Grassmann retractions).

Proof

technique · identify each formula as the corresponding orthogonal projection
1.1F1F2given

The first displayed map is exactly the Stiefel tangent projection from [F1], and the second is exactly the horizontal Grassmann projection from [F1]. By [F2], the endpoint is a valid frame; each formula is linear in Z and takes values in the stated destination tangent space.

2.1F1F2F3givenalgebra∎

Under a change (X,η,Z)↦(XQ,ηQ,ZQ), [F2] gives the polar endpoint YQ; hence I−(YQ)(YQ)T=I−YYT and the output changes to VQ. This is exactly the changed horizontal representative, so the Grassmann tangent vector is independent of the frame. At η=0, [F3] gives Y=X; a Stiefel tangent has sym⁡(XTZ)=0, and a Grassmann horizontal representative has XTZ=0, so both transports return Z.

Depends on

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