# Properness of moment map

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?

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.