Moment map of the action of $SU(2)$ on $\mathbb C^{2n}$

Question

Let $SU(2)$ acts on symplectic space $((\mathbb C^2 -\ (0,0))^{n},\omega)$, where

$$\omega=dx_1\wedge dx_2+dx_3\wedge dx_4+\cdots+dx_{4n-3}\wedge dx_{4n-2}+dx_{4n-1}\wedge dx_{4n}$$

as $T_A(z_1,\ldots,z_{n})=(Az_1,\ldots,Az_n)$. I need to check that this action is hamiltonian and find for what $n$ $0$ is a regular value of moment map $\mu$. I verified that this just by coordinate method and find that

$$\mu(x_1,\ldots,x_{4n})=(-1/2(x_1^2+x_2^2-x_3^2-x_4^2+x_5^2+x_6^2-\cdots-x_{4n-1}^2-x_{4n}^2),x_3x_2-x_1x_4+x_7x_6-x_5x_8+\cdots+x_{4n-2}x_{4n-1}-x_{4n}x_{4n-1}, x_1x_3+x_2x_4+\cdots+x_{4n-3}x_{4n-1}+x_{4n-2}x_{4n}).$$

I checked that for $n=1,$ $\mu^{-1}(0)$ is empty so $0$ is regular :). But I don't know how can it be solved for arbitrary $n$, because the equations are huge.

Answer 1

Theorem: Let $(M_i,\omega_i)_{i=1,2}$ be Hamiltonian $G$-spaces with moment maps $\mu_i: M_i \to \mathfrak g^*$. If $\pi_i: M_1 \times M_2 \to M_i$ is the projection map onto the $i^{th}$ factor, then $(M_1 \times M_2, \pi_1^* \omega_1+\pi_2^*\omega_2)$ is a Hamiltonian $G$-space under the diagonal action of $G$.

In your case, to show that the map is Hamiltonian you proceed by induction. You've shown $n=1$ so the base case is done. Now assume the result holds for $n$ and let's show $n+1$. Notice that the $SU(2)$ action on $\mathbb C^{2n+2}$ acts invariantly on each copy of $\mathbb C^2$, and hence we have an $SU(2)$-invariant decomposition of $\mathbb C^{2n+2} \cong \mathbb C^{2n} \times \mathbb C^{2}$. The diagonal action on $\mathbb C^{2n} \times \mathbb C^{2}$ is precisely the one above given, so that also checks out.

Exercise: Show that your $\omega$ is prescribed as the sum of the projection pullbacks on $\mathbb C^{2n}$ and $\mathbb C^2$.

Now apply the theorem, and the result follows.

Since you have that the moment map can be decomposed into a sum of simpler moment maps, try to use this to deduce your regularity result.

Edit: For regularity, notice that for $p \in M$ we have $\operatorname{im}d\mu_p = \mathfrak g_p^\perp$; the annihilator of the Lie algebra of the stabilizer of $p$. At the point $0$, your stabilizer is $$SU(2)_{\vec 0} = \{ A \in SU(2): (A\cdot 0, \ldots, A\cdot 0) = (0,\ldots, 0) \} = SU(2),$$ so that $\mathfrak{su}(2)_{\vec 0} = \mathfrak{su}(2)$. It then follows that $\mathfrak{su}(2)_{\vec 0}^\perp = \mathfrak{su}(2)^\perp = \{ 0 \},$ so $n=0$ is the only $n$ for which $\vec 0$ is a regular point.