Torus and circle action

Question

Consider a Hamiltonian action of torus on a symplectic 4_manifold M with the moment map $\mu \to R^2$ , $\mu=(\mu_1,\mu_2)$. Consider circle action on M induced by the homomorphism $S^1 \to T^2$ by $\theta \to (2\theta, 3\theta)$. How the moment map $\phi$ for this action is expressed in terms of $\mu_1$ and $\mu_2$.

Answer 1

Let $H$ denote a Lie group, $\mathfrak{h}$ denote its Lie algebra and $\mathfrak{h}^*$ denote its dual. A Hamiltonian action of $H$ on $(M, \omega)$ determines (and is determined by) a moment map $\mu : M \to \mathfrak{h}^*$. A Lie group morphism $F : G \to H$ induces a Lie algebra morphism $f : \mathfrak{g} \to \mathfrak{h}$ with transpose $f^* : \mathfrak{h}^* \to \mathfrak{g}^*$. It turns out that $\mu_G := f^* \circ \mu : M \to \mathfrak{g}^*$ is the moment map of the induced Hamiltonian action of $G$ on $(M, \omega)$.

In your case, let $T = T^2$ and $S = S^1$; you chose an identification $S \simeq \mathbb{R}/p\mathbb{Z}$ (where $p$ is the 'period' of the circle) and a decomposition $T \simeq S \times S$ which induce identifications $\mathfrak{t} \simeq \mathfrak{s} \oplus \mathfrak{s} \simeq \mathbb{R} \oplus \mathbb{R}$ (seen as column vectors) and $\mathfrak{t}^* \simeq \mathfrak{s}^* \oplus \mathfrak{s}^* \simeq \mathbb{R} \oplus \mathbb{R}$ (seen as row vectors). Since $F : S \to T : \theta \mapsto (2 \theta \, \mathrm{mod} \, p, 3\theta \, \mathrm{mod} \, p )^T$, you have $f : \theta \mapsto (2 \theta, 3 \theta)^T$, hence $f^* : (x, y) \mapsto 2x + 3y$. Indeed, $[f^*(x,y)](\theta) = (x,y)f_*(\theta) = (x,y)(2\theta, 3\theta)^T = x2\theta + y 3\theta = (2x + 3y)\theta$.

Consequently, $\mu_S : M \to \mathfrak{s}^* \simeq \mathbb{R} : p \mapsto 2 \mu_1(p) + 3 \mu_2(p)$.