# About the hessian of a hamiltonian function of a hamiltonian torus action

I am trying to understand several claims about the hessian of a hamiltonian function associated to a fundamental field, appearing in Lemma 5.54 of McDuff-Salamon's "Intro to symplectic topology".

(Summary of context: The torus $$\mathbb{T}^m$$ acts on a symplectic manifold $$(M, \omega)$$ and the action be hamiltonian (in the sense of McDuff - Salamon, i.e. the map sending elements in the Lie algebra of the torus to their corresponding Hamiltonian functions with the Poisson bracket is a Lie algebra homomorphism). Let $$H$$ be a hamiltonian function associated to some fixed $$\theta \in \mathfrak{t} = Lie(\mathbb{T}^m)$$ i.e. $$dH = i_{X_H} \omega$$, where $$X_H$$ is the fundamental vector field associated to $$\theta$$. Also, we have $$J$$ an almost complex structure compatible with $$\omega$$ and invariant under the action.)

If $$x$$ a fixed point of the action (so a critical point of $$H$$ (proven earlier)) they look at $$d_x X_H:T_xM \rightarrow T_{X_H(x)} (T_xM)$$ which they claim is $$-J_x \nabla_x^2 H$$. Not only can't I prove this calim, but it doesn't even make sense to me, because $$-J_x \nabla_x^2 H \in End(T_xM)$$, so an identification of $$T_{X_H(x)}(T_xM)$$ with $$T_xM$$ would be needed, but they don't specify a canonical one and I can't see one. Then, in what sense is $$d_x X_H$$ even a vector field, given that it doesn't associate to a point $$v \in T_x M$$ a vector in $$T_v (T_x M)$$ but in $$T_{X_H(x)} (T_x M)$$?

So, condensed, this is my main question:

1. What does it mean that $$d_x X_H = - J_x \nabla_x^2 H$$ and why is this so? (more on this one down below, in the paragraph about the Guillemin-Sternberg approach)

Furthermore, denoting the action as $$\tau \cdot x =: \psi_{\tau}(x)$$, they next claim

1. that the $$1$$-parameter group of $$d_x X_H$$ is $$d_x \psi_{t \theta}$$
2. that it is also $$exp(-tJ_xS_x)$$ (which I'm assuming is matrix exponentiation, not the geodesic exponential w.r.t. the induced metric).

Why do these hold?

Guillemin-Sternberg follow a similar path in Theorem 32.6 of "Symplectic techniques in physics", but there they don't really look at $$d_x X_H$$ but they firstly take $$\phi_t$$ the flow of $$X_H$$ and consider $$L:= \frac{d}{dt} |_{t=0} (\phi_t(x)): T_xM \rightarrow T_xM$$ (this again makes sense because $$x$$ is fixed). This eliminates the problem of the needed identification at least. Then they claim that $$(H_{**}(x))(v, v) = \omega_x (Lv, v)$$, $$H_{**}$$ being the hessian as a bilinear form. I can easily find that an $$L$$ satisfying this equality must be precisely $$-J_x\nabla_x^2 H$$, using the definition of $$\nabla_x^2H$$ via the invariant metric defined by $$\omega$$ and $$J$$. But why is the $$L$$ defined as before precisely the $$L$$ satisfying the last equality? I suspect a good expression of $$H_{**}(x)(v,v)$$ is needed. Of course, $$H_{**}(x)(v,v)= v(i_{X_H} \omega (\tilde{v}))$$, with $$\tilde{v}$$ a vector field extending $$v$$, but does this help in any way?

I also think this good expression is needed to see that $$J_x \nabla_x^2 H = \nabla_x^2H J_x$$, which is another point I'm unable to prove further on in the McDuff-Salamon lemma.

The notation in both of your sources is somewhat mysterious to me, but it seems clear that $$d_xX_H$$ is not referring to the differential of $$X_H:M\to TM$$. Instead, it is referring to the $$(1,1)$$ tensor $$\nabla X_H(x)$$. The two are only tenuously related, in that for $$v\in T_xM$$, $$\nabla_vX_H$$ can be identified with the vertical part of $$d_xX_H(v)$$ through various canonical maps. Note that $$\nabla X_H(x)$$ is independent of the choice of metric, at least, since $$x$$ is a zero of $$X_H$$.
Also, the Hessian $$\nabla^2H(x)$$ is generally interpreted as a $$(0,2)$$ tensor rather than a $$(1,1)$$ tensor. It seems the notation $$\nabla^2_xH$$ is instead referring to the hessian with an index raised using the metric $$g$$. $$\nabla^2H(x)$$ is also independent of the choice of metric, since $$x$$ is a critical point of $$H$$.
To your second question, $$\nabla X_H(x)$$ is indeed the generator of the group action $$d_x\psi_t:T_xM\to T_xM$$. Since this is a linear $$\mathbb{R}$$-action, it suffices to show that $$\nabla_vX_H=\frac{d}{dt}(d_x\varphi_t(v))_{t=0}$$ for all $$v\in T_xM$$. To show this, we can define a vector field $$V$$ with $$V(p)=v$$ and use Lie derivatives: \begin{align} \frac{d}{dt}(d_x\varphi_t(v))_{t=0}=&-\frac{d}{dt}(d_x\varphi_{-t}(v)-v)_{t=0} \\ =&-\frac{d}{dt}(d_x\varphi_{-t}(V)-V)_{t=0}(x) \\ =&-(\mathcal{L}_{X_H}V)(x) \\ =&-[X_H,V](x) \\ =&-\nabla_{X_H}V(x)+\nabla_VX_H(x) \\ =&\nabla_vX_H \end{align} If you're familiar with abstract index notation, it might be useful in working with these tensor identities. For instance, the equality in the first question, as I've interpreted it, is $$\nabla_bX_H^a=J^a{}_cg^{cd}\nabla_d\nabla_bH$$. To show this, let $$\omega^{ab}$$ denote the inverse of $$\omega_{ab}$$, satisfying $$\omega^{ab}\omega_{bc}=\delta^a_c$$. We can rewrite the definition of the Hamiltonian vector field $$X_H^a\omega_{ab}=\nabla_bH$$ as $$X_H^a=-\omega^{ab}\nabla_bH$$ and rewrite the compatibility condition $$g_{ab}=\omega_{ac}J^c{}_b$$ as $$\omega^{ab}=J^a{}_cg^{cb}$$. $$X_H^a=-\omega^{ac}\nabla_cH \\ \implies\nabla_bX^H_a=-\omega^{ac}\nabla_b\nabla_cH-\nabla_b\omega^{ac}\nabla_cH \\ =-\omega^{ac}\nabla_c\nabla_bH-\nabla_b\omega^{ac}\nabla_cH \\ =-J^a{}_d(g^{dc}\nabla_c\nabla_bH)-\nabla_b\omega^{ac}\nabla_cH$$ The first term gives the desired equality, since the second term vanishes at $$x$$.
• Indeed the hessian is defined via $g(\nabla^2H(x)(v), w) = v(w(H))$ w/ $w$ extended to a v.f. So in $dX_H(x) = \nabla X_H(x)$, is $\nabla$ the Levi-Civita connection assoc. to $g$? If so, I can follow your computation for the second question, using that $\nabla$ is torsion free. But I still don't see why $\nabla_* X_H(x)= -J_x \nabla^2_x H (*)$, which is what I'm now understanding the first claim to be. Commented Sep 8, 2021 at 8:58
• Yes, $\nabla$ is the Levy-Civita connection of $g$ throughout. I've added a bit more detail on showing the equality $\nabla X_H=-J\nabla_x^2H$. Commented Sep 8, 2021 at 15:41