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Let $(\mathcal M,g)$ be a smooth Riemannian manifold and $\Delta$ be the standard (positive) Laplace operator given in coordinates by the usual $$ \Delta=-\vert g\vert^{-1/2}\partial_j(\vert g\vert^{1/2} g^{jk}\partial_k(\cdot)). $$ The cotangent bundle of $\mathcal M$ has a standard symplectic structure, given in coordinates by $ \omega =d(\xi\cdot dx). $ The principal symbol of $\Delta$ is $ p(x,\xi)=g^{jk}(x)\xi_k\xi_j. $ Let $\phi$ be a smooth function defined on the base $\mathcal M$. I can calculate the Poisson bracket $$ \{p,\phi\}=2g(X,\nabla \phi),\quad X=g^{-1}\xi, $$ but I am in trouble with the calculation of $\{p,\{p,\phi\}\}$. I think that $$ \frac14 \{p,\{p,\phi\}\}=D_XD_X \phi-\frac12 D_{\nabla g(X,X)}\phi=(\nabla^2\phi)(X,X), $$ but I am not able to prove the second equality. Maybe a simpler question would be about the proof (if correct !) of $$ \nabla \bigl(g(X,X)\bigr)=2D_X X. $$ Thanks in advance for your help.

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Your simpler relation is not correct for the following reason : if $X$ has constant norm, then all integral curves of $X$ would be geodesics ! –  Thomas Richard Jul 5 '13 at 17:16
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This is a case, where I think it's easiest to do the calculation initially in local co-ordinates and using Christoffel symbols. You can turn it into a more elegant co-ordinate-free proof after it all becomes clear. –  Deane Yang Jul 5 '13 at 18:59
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@ThomasRichard: $X$ does not have constant norm and is not a vector field: it is a section of the pull-back of $T\mathcal{M}$ to $T^*\mathcal{M}$. That said, using the Levi-Civita connection, $X$ can be viewed as a horizontal vector field on $T^*\mathcal{M}$ and then (up to signs and factors of $2$) it is the Hamiltonian vector field corresponding to the OP's function $p$ and the integral curves are geodesics. –  Fran Burstall Jul 6 '13 at 19:54
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1 Answer 1

I am going to shoot for coordinate-free. First, some notation: let $\pi:T^*\mathcal{M}\to\mathcal{M}$ be the bundle projection, set $E:=\pi^{-1}T\mathcal{M}$, the pullback of the tangent bundle of $\mathcal{M}$, so that $E^*=\pi^{-1}T^*\mathcal{M}$ is the bundle along the fibres. Equip both with the pullback $D$ of the Levi-Civita connection of $\mathcal{M}$.

Now $E^*$ has a tautological section $\xi$ given by $\xi(\alpha)=\alpha$ and then the Liouville form is $\langle\xi,d\pi\rangle$. Thus the symplectic form is $$ \omega=d\langle\xi,d\pi\rangle=\langle D\xi\wedge d\pi\rangle+\langle \xi,d^Dd\pi\rangle=\langle D\xi\wedge d\pi\rangle $$ since $d^Dd\pi=0$, $D$ being the pullback of a torsion-free connection.

The connection gives us a splitting into vertical and horizontal bundles $T(T^*\mathcal{M})=\mathcal{V}\oplus\mathcal{H}$ where $\mathcal{V}=\ker d\pi$ and $\mathcal{H}=\ker D\xi:T(T^*\mathcal{M})\to E^*$. Clearly $d\pi:\mathcal{H}\cong E$. Note that both $\mathcal{V}$ and $\mathcal{H}$ are Lagrangian subbundles. Let $X$ be the section of $\mathcal{H}$ corresponding to $g^{-1}\xi\in\Gamma E$. Thus $X$ is the vector field on $T^*\mathcal{M}$ satisfying $$ D_X\xi=0,\qquad d\pi(X)=g^{-1}\xi. $$

Now let's get to work: the function $p=g(\xi,\xi)$ and, since we want to compute Poisson brackets with $p$, we should compute the Hamiltonian vector field of $p$: $dp=2g(D\xi,\xi)$ so that $$dp(Y)=2g(D_Y\xi,\xi)=2\langle D_Y\xi,g^{-1}\xi\rangle=2\langle D_Y\xi,d\pi(X)\rangle=2\omega(Y,X).$$ Thus $p$ has Hamiltonian vector field $2X$.

Now let $\phi$ be a function on $\mathcal{M}$ then $$ \{p,\phi\circ\pi\}=d(\phi\circ\pi)(2X)=2d\phi(g^{-1}\xi)=2g(\xi,d\phi), $$ in agreement, modulo the violence I have done to their notation, with OP.

Now $$ \tfrac14\{p,\{p,d(\phi\circ\pi)\}\}=d_Xg(\xi,d\phi)=g(D_X\xi,d\phi)+g(\xi,D_X(d\phi))=g(\xi,D_X(d\phi)), $$ since $X$ is horizontal. This pretty much bakes the cake: we are viewing $d\phi$ is a section of $E^*$ (so we should really have written $(d\phi)\circ\pi$) and $D$ is the pullback of the Levi-Civita $\nabla$ so the defining property of pullback connections tells us that $D_X((d\phi)\circ\pi)=(\nabla_{d\pi(X)}d\phi)\circ\pi$. Putting it all together now yields $$ \tfrac14\{p,\{p,d(\phi\circ\pi)\}\}=\pi^*(\nabla^2\phi)(X,X)=\nabla^2\phi(g^{-1}\xi,g^{-1}\xi). $$

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Great! Thanks a lot. –  Bazin Jul 10 '13 at 12:39
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