MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
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
@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

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 $$ \tfrac14\{p,\{p,d(\phi\circ\pi)\}\}=\pi^*(\nabla^2\phi)(X,X)=\nabla^2\phi(g^{-1}\xi,g^{-1}\xi). $$

share|cite|improve this answer
Great! Thanks a lot. – Bazin Jul 10 '13 at 12:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.