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Grassmannians from unitary symplectic reduction

Example

Let $1 \leq k \leq n$ and let $M = C^{k \times n}$ with the real inner product $⟨ X, Y ⟩ = Re tr (X^{*} Y)$ and the symplectic form

$ω (X, Y) = Im tr (X^{*} Y),$

and let $U (k)$ act on the left by matrix multiplication, $g \cdot A = g A$ . Then, with $u (k)^{*}$ identified with $u (k)$ through the inner product,

$μ (A) = - \frac{i}{2} A A^{*} + λ i I (λ > 0)$

is an equivariant moment map. Its zero level is the scaled Stiefel manifold ${A : A A^{*} = 2 λ I}$ , on which $U (k)$ acts freely and properly, and the reduction is the Grassmannian

$M_{0} = {A : A A^{*} = 2 λ I} / U (k) = Gr (k, n),$

of dimension $2 k (n - k)$ , carrying the reduced form characterised by $π^{*} ω_{λ}^{r e d} = ι^{*} ω$ . Under the common row-space identification of all these quotients with $Gr (k, n)$ , the form depends linearly on the level: if $ω_{Gr} : = ω_{1 / 2}^{r e d}$ is the unit-frame normalization, then $ω_{λ}^{r e d} = 2 λ ω_{Gr}$ .

Facts & Assumptions

Given: ${A C}_{ω}$ , integers $1 \leq k \leq n$ , the space $M = C^{k \times n}$ with the forms above, the left action of $U (k)$ , and $λ > 0$ .

[F1]

$U (k)$ is a Lie group with Lie algebra $u (k) = {X : X^{*} + X = 0}$ by Unitary and special unitary Lie groups. It is compact: inside $M_{k} (C) ≅ R^{2 k^{2}}$ the equation $A^{*} A = I$ defines a closed set, and it is bounded because $\sum_{i, j} ∣ A_{i j} ∣^{2} = tr (A^{*} A) = k$ ; Heine--Borel now applies (Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line).

[F2]

$ω (X, Y) = Im tr (X^{*} Y)$ is a symplectic form on the real vector space $M$ , since it is an alternating bilinear form with $ω (X, i X) = ∥ X ∥^{2} > 0$ . [algebra]

[F3]

The fundamental field of $ξ \in u (k)$ is $ξ_{M} (A) = {\frac{d}{d t} ∣}_{0} e^{- t ξ} A = - ξ A$ . Fundamental vector fields for a left action.

[F4]

The coadjoint action of $U (k)$ on $u (k)^{*}$ corresponds under the invariant inner product to $H \mapsto g H g^{- 1}$ , so central elements are fixed. The coadjoint representation, action and orbits.

[F5]

The reduction theorem applies when the value is regular and the stabilizer acts freely and properly; regularity is equivalent to local freeness on the level, and the reduced dimension is $\dim M - \dim G - \dim G_{α}$ . Marsden--Weinstein--Meyer symplectic reduction, Regularity of a moment map is equivalent to local freeness, The dimension of a regular reduced space at a nonzero value.

Verification

technique · direct

1.1F2F3

For $ξ \in u (k)$ and $X \in T_{A} M = M$ , using [F3] and $(ξ A)^{*} = A^{*} ξ^{*} = - A^{*} ξ$ , $ω (ξ_{M} (A), X) = Im tr ((- ξ A)^{*} X) = Im tr (A^{*} ξ X) .$

2.1step 1.1F1F2

With $μ (A) = - \frac{i}{2} A A^{*} + λ i I$ we have $μ (A)^{*} = - μ (A)$ , so $μ (A) \in u (k)$ and the component is $μ^{ξ} (A) = ⟨ μ (A), ξ ⟩ = Re tr (μ (A)^{*} ξ)$ ; its derivative in the direction $X$ picks up $\frac{i}{2} (X A^{*} + A X^{*})$ and, using $tr (X^{*} ξ A) = - \overline{tr (A^{*} ξ X)}$ together with cyclicity, equals $d μ_{A}^{ξ} (X) = - Im tr (A^{*} ξ X) = - ω (ξ_{M} (A), X) .$ Hence the component moment equations hold.

3.1step 2.1F4

Equivariance: $μ (g A) = - \frac{i}{2} g A A^{*} g^{- 1} + λ i I = g μ (A) g^{- 1}$ , which is the coadjoint action by [F4]; the added central term $λ i I$ is fixed. Hence $μ$ is an equivariant moment map.

4.1step 3.1F1F5

The zero level is $μ^{- 1} (0) = {A : A A^{*} = 2 λ I}$ : it is nonempty because $k \leq n$ and contains $A = \sqrt{2 λ} (I_{k} 0)$ . If $g A = A$ on this level, then $g = g A A^{*} (2 λ)^{- 1} = A A^{*} (2 λ)^{- 1} = I$ , so the action is free; [F5] therefore makes $0$ a regular value. The action is proper because $U (k)$ is compact, so the zero level is an embedded submanifold and reduction applies.

5.1step 4.1

Quotient: two frames $A, A^{'}$ with $A A^{*} = A^{'} A^{*} = 2 λ I$ lie in the same $U (k)$ -orbit exactly when their rows span the same $k$ -plane, so the quotient is the Grassmannian $Gr (k, n)$ of $k$ -planes in $C^{n}$ .

6.1step 5.1F2F5algebra∎

Dimension check: $\dim M = 2 k n$ and $\dim G = \dim G_{0} = k^{2}$ because $0$ is a central coadjoint value, so by [F5] $\dim M_{0} = 2 k n - 2 k^{2} = 2 k (n - k)$ , the dimension of $Gr (k, n)$ . For the form, let $S_{λ} (A) = \sqrt{2 λ} A$ carry the unit-frame level $A A^{*} = I$ onto the level $A A^{*} = 2 λ I$ . This map is $U (k)$ -equivariant, preserves row spaces, and satisfies $S_{λ}^{*} ω = 2 λ ω$ . Pulling the two reduction identities back along $S_{λ}$ therefore gives ${\overset{ˉ}{S}}_{λ}^{*} ω_{λ}^{r e d} = 2 λ ω_{1 / 2}^{r e d}$ on the common Grassmannian quotient. Thus, with $ω_{Gr} : = ω_{1 / 2}^{r e d}$ , the reduced form at level $λ$ is $2 λ ω_{Gr}$ , rather than one fixed form for every $λ$ .

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Sources

Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)
Ana Cannas da Silva, Lectures on Symplectic Geometry (standard reference, not scraped)