Moment maps unitary group acting on matrices

Question

I am reading the fifth chapter of "An introduction to extremal Kaehler metrics" by Gabor Szekelyhidi. At the very beginning of that chapter, the author describes moment maps and Hamiltonian action. Here is a short description: Suppose that a connected Lie group $G$ acts on a Kaehler manifold $M$ preserving the Kaehler form $\omega.$ The derivative of the action gives rise to a Lie algebra map

$$ \rho: \mathfrak{g} \rightarrow \mathfrak{X}(M)$$

Where $\mathfrak{g}$ is the Lie algebra of $G$ and $\mathfrak{X}$ is the algebra of vector fields on $M.$ Then the action of $G$ on $M$ is said to be Hamiltonian if there exists a $G-$equivariant map

$$ \mu: M \rightarrow \mathfrak{g}^{*} $$

such that for any $\xi \in \mathfrak{g}$ the function $\langle\mu, \xi\rangle$ is a hamiltonian function for the vector field $\rho(\xi):$

$$ d\langle\mu, \xi\rangle = -\iota_{\rho(\xi)} \omega. $$

I got stuck with the following exercise: Let $M_n$ be the set of $n \times n$ complex matrices equipped with the Euclidean metric under the identification $M_n=\mathbb{C}^{n^2}.$ The unitary matrices $U(n)$ act on $M_n$ by conjugation, preserving this metric. I.e. $A \in U(n)$ acts by $M \mapsto A^{-1}MA.$ Find a moment map $\mu: M_n \rightarrow \mathfrak{u}(n)^{*}.$ So far, I've just noticed that if $n=1$ then the action of $U(1)$ by conjugation on complex number is the trivial action, what will be then the moment map? I am trying to see what happens also in dimension 2. I know that $\mathfrak{u}(2)$ is generated by skew-Hermitian matrices, so every skew-hermitian matrix can be diagonalized and the eigenvalues must be pure imaginary. How to find a moment map in this case?

Answer 1

Ok, so first, you should know what is the Kähler metric that you are defining on the space $M_n$. Since it is not specified, I will assume it is the induced by the standard hermitian product $$h(A,B)=Tr(AB^\dagger)$$, where $B^\dagger$ denotes the adjoint (that is the transpose conjugate) of $B$. This of course has associated a scalar product $$ g(A,B)=Re(Tr(AB^\dagger))$$ and a symplectic form $$ \omega(A,B)=-Im(Tr(AB^\dagger))$$ (the minus sign here is just a convention).

As you noted, there $U(n)$ acts symplectically on $M_n$ via $A\mapsto U^\dagger A U$. Now, to see how the infinitesimal associated action $\mathfrak{u}(n)\rightarrow M_n$ is, just compute $$ \rho_X(A)=\left.\frac{d}{dt}\right|_{t=0}(e^{-Xt}Ae^{Xt})=-XA+AX=[A,X]. $$

Now we want to find a moment map $\mu:M_n \rightarrow \mathfrak{u}(n)^{\ast}$ for this action. To do this we will identify $\mathfrak{u}(n)$ with its dual using the pairing given by the hermitian metric, that is $$ \langle X,Y \rangle = Tr(XY^\dagger). $$ Therefore, I claim that the momentum map is given by the mapping $\mu:M_n \rightarrow \mathfrak{u}$ such that $$\mu(A)=\tfrac{i}{2}[A,A^\dagger].$$

Now note that $$ f_X(A)=\langle \mu(A),X \rangle = \tfrac{i}{2} Tr([A,A^\dagger]X), $$ so (check this out) $$ df_{X,A}(Y)=\left.\frac{d}{dt}\right|_{t=0}f_X(A+tY)= \tfrac{i}{2}Tr([Y,A^\dagger]X+[A,Y^\dagger]X). $$ Now, using the cyclic property of the trace ($Tr(ABC)=Tr(BCA)=Tr(CAB)$), is easy to check that this is indeed $$ df_{X,A}(Y)=-Im(Tr([X,A]Y^\dagger))=\omega([X,A],Y)=-\omega(\rho_X(A),Y). $$ So $\mu$ is indeed a moment map for this action.

This example is indeed very useful, for example, to find the moment map that gives Hitchin equations for Higgs Bundles, check out Section 4 on Hitchin's paper "The self-duality equations on a Riemann surface" if you want to read more about this.

Answer 2

For the Hermitian case, there is a more direct formula which can avoid using the dual $\mathfrak{g}*$, and the tangent map.

Let's assume $(V,h)$ is an Hermitian vector space (hence $\mathrm{Im} h$ is a symplectic form) which endows with an Hermitian $G$-action (hence also Hamiltonian), pick $\langle\cdot,\cdot\rangle$ be an Ad-invariant Euclidean metric on $\mathfrak{g}$, then the moment map computes as $$\langle\mu(x),\xi\rangle=\frac{i}{2}h\left(\underline{\xi}(x),x\right)$$ where $\underline{\xi}(x)$ means the fundamental vector field associated to the $G$-action.

Notice that this formula makes sense, as the RHS is indeed taking values in $\Bbb{R}$, because the action is $G\longrightarrow U(V)$, thus the infinitesimal action is $\mathfrak{u}(V)$, which makes $\underline{\xi}(\cdot)$ a skew-Hermitian operator.

The proof of above formula is a straightforward computation.

Now back to our case, the $V$ is $\text{End}(\Bbb{C}^n)$, the Hermitian product is given by $$h(X,Y)=\text{Tr}(XY^*)$$ the Lie group $G$ is $U(n)$, the Ad-invariant metric on its Lie algebra is $$\langle\xi,\zeta\rangle=\text{Tr}(\xi\zeta^*)$$ the fundamental vector field is $$\underline{\xi}(X)=[\xi,X]$$ therefore, the moment should satisfy $$\begin{aligned}\text{Tr}(\mu(X)\xi^*)&=\frac{i}{2}\mathrm{Tr}\left([\xi,X]X^*\right)\\ &=-\frac{i}{2}\text{Tr}([X,X^*]\xi^*)\end{aligned}$$ for all $\xi$. So $$\mu(X)=-\frac{i}{2}[X,X^*]$$

Answer 3

It is easier to compute the moment map of the left and right action $(A,M)\to M.A$ and $(A,M)\to A^{-1}M$, and one find respectively ${i\over 2} M M^*, {-i\over 2}M^*M)$, so that the moment map is ${i\over 2}(MM^*-M^*M)$.