DefinitionDefinition: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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Polar retraction on the complex Grassmannian

Definition

For $X \in {St}_{C} (n, r)$ and horizontal $η \in T_{[X]} {Gr}_{C} (r, n)$ , so $X^{*} η = 0$ , define $R_{X}^{p o l} (η) = (X + η) ((X + η)^{*} (X + η))^{- 1 / 2}$ and $R_{[X]}^{p o l} (η) = range (R_{X}^{p o l} (η))$ . The inverse square root is the inverse of the unique positive-definite Hermitian square root. The horizontal condition gives $(X + η)^{*} (X + η) = I + η^{*} η ≻ 0$ , so this map is defined for every horizontal tangent vector, and its frame representative is unchanged as a Grassmann point when $(X, η)$ is replaced by $(X Q, η Q)$ with unitary $Q$ .

Facts & Assumptions

Given: $X \in {St}_{C} (n, r)$ and horizontal $η$ as in Complex Stiefel and Grassmann tangent projections.

[F1]

A horizontal representative satisfies $X^{*} η = 0$ (Complex Stiefel and Grassmann tangent projections).

[F2]

On a finite-dimensional real or complex inner-product space, a nonnegative operator has a unique nonnegative square root; for a positive-definite input its eigenvalues are the positive square roots of the input eigenvalues, so the root is positive definite and invertible (A non-negative operator has a unique non-negative square root).

[F3]

The complex Grassmannian is the quotient of complex orthonormal frames by right unitary changes of frame (Stiefel spaces, Grassmannians, and tautological bundles).

Proof

technique · verify full rank, orthonormality, and unitary equivariance of the polar formula

1.1F1F2givenalgebra

By [F1], $(X + η)^{*} (X + η) = I + η^{*} η$ . For every nonzero $v \in C^{r}$ , $v^{*} (I + η^{*} η) v = ∥ v ∥_{2}^{2} + ∥ η v ∥_{2}^{2} > 0$ , so the Gram matrix is Hermitian positive definite. Apply [F2] on $C^{r}$ : its unique nonnegative Hermitian square root has strictly positive eigenvalues and is invertible, which defines the inverse root in the formula.

2.1F2givenstep 1.1algebra

Write $C = (X + η)^{*} (X + η)$ and $Y = (X + η) C^{- 1 / 2}$ . Since $C^{- 1 / 2}$ is Hermitian and $C^{- 1 / 2} C C^{- 1 / 2} = I_{r}$ , we have $Y^{*} Y = I_{r}$ . Thus $Y$ is a complex orthonormal frame and its range is a point of the stated Grassmannian.

3.1F1F3step 2.1algebra∎

By the quotient convention [F3], frames related by a unitary matrix represent the same point. For unitary $Q$ , the Gram matrix of $(X Q, η Q)$ is $Q^{*} C Q$ . Uniqueness of the positive square root gives $(Q^{*} C Q)^{- 1 / 2} = Q^{*} C^{- 1 / 2} Q$ . Hence the resulting frame is $Y Q$ , with the same column space as $Y$ .

Depends on

Used by

Finite-budget Riemannian trust-region method Algorithm
Complex finite-basis PC-VAR optimization on Grassmann Example
The complex Grassmann polar map is a first-order retraction Theorem

Dependency tree · two levels

12 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Peter A. Absil, Robert Mahony, and Rodolphe Sepulchre, Optimization Algorithms on Matrix Manifolds, Chapter 4 (2008) (standard reference, not scraped)
Nicholas D. S. Manton, Optimization algorithms exploiting unitary constraints, IEEE Transactions on Signal Processing 50 (2002), §§IV and VI (standard reference, not scraped)
Nicolas Boumal, An Introduction to Optimization on Smooth Manifolds (2023 prepublication) (standard reference, not scraped)