# The norm-squared of a moment map behaves like a Morse-Bott function

Let $$G$$ be a compact Lie group with Lie algebra $$\mathfrak{g}$$. Let $$<.,>$$ denote a $$G$$-invariant inner product on $$\mathfrak{g}$$.

Let $$(M,\omega)$$ be a symplectic compact manifold endowed with a hamiltonian action of $$G$$, and let $$\mu : M \longrightarrow \mathfrak{g}^*,$$ be a moment map associated to this action. We fix a Riemannian metric $$g$$ on $$M$$.

I have read somewhere that every gradient flow line $$\phi_x(t)$$ of the norm-squared of the moment map $$\mu$$ begins and ends at a critical point, i.e $$\lim_{t \rightarrow + \infty} \phi_t(x)$$ and $$\lim_{t \rightarrow - \infty} \phi_t(x)$$ exist, and they are both critical points of the norm-squared of $$\mu$$. but I couldn't find a proof of this fact anywhere, Hopefully someone can help.

(This is a follow-up question to This one: The negative gradient flow of a Morse-Bott function on a compact manifold converges to a critical point?)

• For a Hamiltonian $S^1$-action with Hamiltonian $H$, the square norm of the Hamiltonian may not be differentiable on the set $\{H=0\}$. For example when $\{H=0\}$ is disjoint from the fixed point set, then the square norm is not differentiable anywhere along it. So please clarify what you mean by critical points. Aug 25 at 11:46
• Also it may help if you could link to the place where you read it (if possible). Aug 25 at 11:53
• @NickL, the definition of critical points that I'm using is the following: $\textbf{Def}$: we say that $x \in M$ is a critical point of $\vert \vert \mu \vert \vert ^2$ if $d_x \vert \vert \mu \vert \vert ^2 =0$.
– asma
Aug 26 at 0:34
• Ah ok, that is usually referred to as the norm squared. Square norm usually means $||\mu||$. Indeed the norm squared is differentiable. Aug 26 at 10:45
• This is proved in Gradient flow of the norm squared of a moment map by Eugene Lerman who attributes the proof to Duistermaat. Aug 28 at 20:03