Suppose that a torus $T$ acts on a non-compact symplectic manifold $M$. Assume that this action is Hamiltonian and that the fixed point set of $T$ is compact. Let $\mu:M\to\mathfrak{t}^{*}$ denote the moment map of the action, where $\mathfrak{t}$ denotes the Lie algebra of $T$.

If there exists an $X\in\mathfrak{t}$ such that $\mu(X):M\to\mathbb{R}$ is a proper function that is bounded below, why is $\mu:M\to\mathfrak{t}^{*}$ necessarily proper?


1 Answer 1


This follows from a point set topology argument. The assumptions that the fixed point set is compact, or that $\mu(X)$ is bounded below, are not necessary.

Consider the linear map $F: \mathfrak{t}^{*} \rightarrow \mathbb{R}$ given by $L \mapsto L(X)$. Note that $\mu(X) = F \circ \mu$. Suppose $C \subset \mathfrak{t}^{*}$ is compact, then $F(C)$ is compact since $F$ is continuous. Now, $\mu^{-1}(C) \subset \mu(X)^{-1}(F(C))$, which is compact since $\mu(X)$ is proper by assumption.

$\mathfrak{t}^{*}$ is homeomorphic to a Euclidean space, so $C$ is closed and bounded in $\mathfrak{t}^{*}$. Since $\mu$ is continuous, $\mu^{-1}(C)$ is closed. On the other hand, we saw in the previous paragraph that it was contained in the compact subset $\mu(X)^{-1}(F(C))$, so it is itself compact. Hence, $\mu$ is proper.


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