Grassmannians from unitary symplectic reduction

For $\xi \in \mathfrak{u}(k)$ and $X\in T_{A}M=M$, using [F3] and $(\xi A{)}^{\ast }=A^{\ast }{\xi }^{\ast }=-A^{\ast }\xi$, $\omega ({\xi }_M(A),X)=\mathrm{Im}\mathrm{tr}((-\xi A{)}^{\ast }X)=\mathrm{Im}\mathrm{tr}(A^{\ast }\xi X)\mathrm{.}$

With $\mu (A)=-\frac{i}{2}A{A}^{\ast }+\lambda iI$ we have $\mu (A{)}^{\ast }=-\mu (A)$, so $\mu (A)\in \mathfrak{u}(k)$ and the component is ${\mu }^{\xi }(A)=⟨\mu (A),\xi ⟩=\mathrm{Re}\mathrm{tr}(\mu (A{)}^{\ast }\xi )$; its derivative in the direction $X$ picks up $\frac{i}{2}(X{A}^{\ast }+A{X}^{\ast })$ and, using $\mathrm{tr}(X^{\ast }\xi A)=-\overline{\mathrm{tr}(A^{\ast }\xi X)}$ together with cyclicity, equals $d{\mu }_A^{\xi }(X)=-\mathrm{Im}\mathrm{tr}(A^{\ast }\xi X)=-\omega ({\xi }_M(A),X)\mathrm{.}$ Hence the component moment equations hold.

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

The zero level is ${\mu }^{-1}(0)=\{A:A{A}^{\ast }=2\lambda I\}$: it is nonempty because $k\le n$ and contains $A=\sqrt{2\lambda }(I_{k}0)$. If $gA=A$ on this level, then $g=gA{A}^{\ast }(2\lambda {)}^{-1}=A{A}^{\ast }(2\lambda {)}^{-1}=I$, 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.

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

Dimension check: $\dim M=2kn$ and $\dim G=\dim G_0=k^2$ because $0$ is a central coadjoint value, so by [F5] $\dim M_0=2kn-2k^2=2k(n-k)$, the dimension of $\mathrm{Gr}(k,n)$. For the form, let $S_{\lambda }(A)=\sqrt{2\lambda }A$ carry the unit-frame level $A{A}^{\ast }=I$ onto the level $A{A}^{\ast }=2\lambda I$. This map is $U(k)$-equivariant, preserves row spaces, and satisfies $S_{\lambda }^{\ast }\omega =2\lambda \omega$. Pulling the two reduction identities back along $S_{\lambda }$ therefore gives ${\bar{S}}_{\lambda }^{\ast }{\omega }_{\lambda }^{\mathrm{r}\mathrm{e}\mathrm{d}}=2\lambda {\omega }_{1\mathrm{/}2}^{\mathrm{r}\mathrm{e}\mathrm{d}}$ on the common Grassmannian quotient. Thus, with ${\omega }_{\mathrm{Gr}}:={\omega }_{1\mathrm{/}2}^{\mathrm{r}\mathrm{e}\mathrm{d}}$, the reduced form at level $\lambda$ is $2\lambda {\omega }_{\mathrm{Gr}}$, rather than one fixed form for every $\lambda$.