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Complex projector tangents are skew-Hermitian commutator directions

Statement

Let $P \in P_{r}$ and define the tangent space by velocities of differentiable curves in $P_{r}$ . Then $T_{P} P_{r} = {[K, P] : K^{*} = - K}, [K, P] : = K P - P K .$ For each skew-Hermitian $K$ , the curve can be realized near $t = 0$ as $P_{K} (t) = Q_{K} (t) P Q_{K} (t)^{*}$ , where $Q_{K} (t) : = (I + t K / 2) (I - t K / 2)^{- 1} .$

Facts & Assumptions

Given: $P \in P_{r}$ in a finite-dimensional complex inner-product space and a skew-Hermitian matrix $K^{*} = - K$ .

[F1]

The rank- $r$ projector domain consists of Hermitian idempotents with trace $r$ (Rank-r orthogonal projector domain).

[F2]

At $P$ , the tangent directions are exactly the Hermitian off-diagonal blocks between $range P$ and its orthogonal complement; every such block is realized by the graph chart (Graph charts and tangent directions for complex rank-r projectors).

[F3]

The inner product is linear in its first argument, conjugate-linear in its second, and positive definite; the adjoint satisfies $⟨ A x, y ⟩ = ⟨ x, A^{*} y ⟩$ (Real and complex inner-product spaces and their induced length, The norm $∥ v ∥ = \sqrt{⟨ v, v ⟩}$ induced by a real or complex inner product, The adjoint $T^{*} : W \to V$ is characterised by $⟨ T v, w ⟩_{W} = ⟨ v, T^{*} w ⟩_{V}$ ).

[F4]

For a linear map on an $n$ -dimensional space, $\dim \ker T + \dim im T = n$ (Rank-nullity: $\dim_{F} V = nullity T + rank T$ ). A subspace of an $n$ -dimensional space having dimension $n$ equals the whole space (If $\dim_{F} V = n$ and $U$ is a linear subspace of $V$ , then $U$ is finite-dimensional, $\dim_{F} U \leq n$ , and $\dim_{F} U = n$ if and only if $U = V$ , claim 2).

[F5]

Matrix inversion has derivative $D inv (A) [H] = - A^{- 1} H A^{- 1}$ , and matrix differentials obey product and adjoint rules (On the invertible locus, $D inv (A) [H] = - A^{- 1} H A^{- 1}$ , Matrix differentials obey the sum rule, product rule, and adjoint rule).

[F6]

Matrix multiplication, identity, and trace use their finite-coordinate definitions (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes, The trace $tr (A)$ as the sum of the diagonal entries).

Proof

1.1F3F4givenalgebra

Put $M_{t} = I - t K / 2$ . If $t \neq 0$ and $M_{t} x = 0$ for a nonzero $x$ , then $K x = (2 / t) x$ . By [F3], $Re ⟨ K x, x ⟩ = 0$ , while this equation gives $⟨ K x, x ⟩ = (2 / t) ∥ x ∥^{2}$ , a nonzero real number. This contradiction shows $\ker M_{t} = {0}$ ; $M_{0} = I$ is also injective. By [F4], $\dim im M_{t} = n$ , hence $im M_{t} = C^{n}$ . Thus each vector has a unique preimage. This inverse map is linear: if $M_{t} x = u$ and $M_{t} y = v$ , then $M_{t} (a x + b y) = a u + b v$ , so uniqueness identifies the inverse of $a u + b v$ with $a x + b y$ . Therefore $M_{t}$ is invertible for every real $t$ .

2.1F5F6step 1.1algebra

Put $N_{t} = I + t K / 2$ , so $M_{t}^{*} = N_{t}$ . Since $M_{t}$ and $N_{t}$ are polynomials in $K$ , they commute, and the same holds for $M_{t}^{- 1}$ . For $Q_{K} (t) = N_{t} M_{t}^{- 1}$ , the adjoint rules in [F5] give $Q_{K} (t)^{*} = N_{t}^{- 1} M_{t}$ . Thus $Q_{K} (t)^{*} Q_{K} (t) = N_{t}^{- 1} M_{t} N_{t} M_{t}^{- 1} = I$ . In finite dimension this makes $Q_{K} (t)$ unitary.

3.1F1F5F6step 2.1algebra

We have $Q_{K} (0) = I$ . From $M_{0}^{'} = - K / 2$ and [F5], $(M_{t}^{- 1})^{'} ∣_{t = 0} = K / 2$ , hence the product rule gives $Q_{K}^{'} (0) = K$ . The curve $P_{K} (t) = Q_{K} (t) P Q_{K} (t)^{*}$ lies in $P_{r}$ : unitarity preserves self-adjointness and idempotence, and $tr (Q_{K} (t) P Q_{K} (t)^{*}) = tr (P Q_{K} (t)^{*} Q_{K} (t)) = tr P$ by the finite trace identity $tr (A B) = tr (B A)$ . Its derivative at zero is $P_{K}^{'} (0) = K P - P K = [K, P]$ , so every such commutator is tangent.

4.1F2F6algebra∎

Conversely, by [F2] any tangent at $P$ has block form $H_{X} = (\begin{array}{cc} 0 & X^{*} \\ X & 0 \end{array})$ in $range P \oplus (range P)^{⊥}$ . Set $K_{X} = (\begin{array}{cc} 0 & - X^{*} \\ X & 0 \end{array})$ , which satisfies $K_{X}^{*} = - K_{X}$ . With $P = (\begin{array}{cc} I & 0 \\ 0 & 0 \end{array})$ , block multiplication gives $[K_{X}, P] = H_{X}$ . Thus every tangent is a skew-Hermitian commutator direction, proving the equality.

Depends on

Used by

Projector-energy stationarity is equivalent to Fock commutation Theorem

Dependency tree · two levels

59 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

Alan Edelman, Tomás A. Arias, and Steven T. Smith, The Geometry of Algorithms with Orthogonality Constraints, SIAM Journal on Matrix Analysis and Applications 20 (1998), §2.5 (standard reference, not scraped)
P.-A. Absil, R. Mahony, and R. Sepulchre, Optimization Algorithms on Matrix Manifolds, Chapter 3, Princeton University Press (2008), §§3.4.4 and 3.5.7 (standard reference, not scraped)