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:
- 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
- that the $1$-parameter group of $d_x X_H$ is $d_x \psi_{t \theta}$
- 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 $x$.