Let $(M,\omega)$ be a symplectic manifold, $H,f_1,...,f_k\in C^\infty(M)$ non-zero functions with $\{H,f_i\}=0$. If $c\in\mathbb{R}^k$ is a regular value of $F:=(f_1,...,f_k):M\to \mathbb{R}^k$, consider the submanifold $M_c:=F^{-1}(c)$.

a) Let $U$ be a neighbourhood of $M_c$ in which $df_1,...,df_k$ are linearly independent. Show that $\Lambda_\omega:=\frac{\omega^n}{n!}$ can be written as $\Lambda_\omega=df_1\wedge...\wedge df_k\wedge \sigma$ for some $\sigma\in\Omega^{2n-k}(M)$. [hint: find $\sigma$ locally and use partitions of unity]

b) Show that $df_1\wedge...\wedge df_k\wedge L_{X_H}(\sigma)=0$ and use this fact to see that $L_{X_H}(\sigma)$ can be written as $L_{X_H}(\sigma)=\sum_{i=1}^kdf_i\wedge \rho_i$. Conclude that $\Lambda_c:=i^*\sigma$ is invariant by the flow of $H$ (where $i:M_c\hookrightarrow M$ is the inclusion).

c) Show that $\Lambda_c$ does not depend on the choice of $\sigma$.

Here is where I'm at:

a) Taking Darboux coordinates $(x_1,...,x_n,y_1,...,y_n)$ and considering $x_{n+i}:=y_i$, we have $\Lambda_\omega=dx_1\wedge...\wedge dx_{2n}$, $df_i=\sum_{j=1}^{2n}\frac{\partial f_i}{\partial x_j}dx_j$. Consequently $$df_1\wedge...\wedge df_k=\sum_{1\leq j_1<...<j_k\leq 2n}\det(M_{j_1,...,j_k})\,dx_{j_1}\wedge...\wedge dx_{j_k} $$ where $M_{j_1,...,j_k}$ are $k\times k$ minors of the matrix $\left(\frac{\partial f_i}{\partial x_j}\right)_{i=1,...,k,j=1,...,2n}$ given by the columns $j_1,...,j_k$. Taking $p$ with $(df_1)_p,...,(df_k)_p$ linearly independent, we may assume $\det(M_{1,...,k})\neq 0$, so $$\sigma:=\frac{1}{\det(M_{1,...,k})}dx_{k+1}\wedge...\wedge dx_{2n}$$ is such that $df_1\wedge...\wedge df_k\wedge \sigma=dx_1\wedge...\wedge dx_{2n}=\Lambda_\omega$. The problem is that this works for a neighbourhood $V\subset M$ with $V\cap U\neq\emptyset$, but I don't know how to extend it for the whole $U$. I don't get the hint, because introducing a partition $\{U_\alpha, \rho_\alpha\}$ and defining $\sigma$ as a $\rho_\alpha$-linear combination may break the equality with $\Lambda_\omega$.

b) I can show that $df_1\wedge...\wedge df_k\wedge L_{X_H}(\sigma)=0$, but I don't see how to use this to prove $\sigma$ can be written that way. Besides I don't know what it means for $\Lambda_c$ to be invariant by the flow of $H$.

c) Taking another $\sigma'$ with the same property, we need to prove $$i^*(df_1)\wedge...\wedge i^*(df_k)\wedge i^*(\sigma-\sigma')=0$$ but I don't know how to deal with the pullbacks.


Here are some hints.

(a) This is not a symplectic problem, as $H$ is not involved and since $\Lambda_{\omega}$ could be replaced by any volume form $\Lambda$ on $M$. This suggests that the use of Darboux coordinates might be unnatural, perhaps even irrelevant. Moreover, the $f_i$'s might be very 'unrelated' to any Darboux chart. Consequently it might be best to look for another argument.

One could use the constant rank theorem in order to work in a local chart $V$ with coordinates $(f_1, \dots, f_k, g_1, \dots, g_{2n-k})$ where the $g_i$'s are (morally speaking) coordinates on $M_c$. It shouldn't be difficult to conclude to the existence of $\sigma_V \in \Omega^{2n-k}(V)$ such that $\left. \Lambda \right|_V = df_1 \wedge \dots \wedge df_k \wedge \sigma_V$. Covering $U$ by charts $V_j$ and using a partition of unity $\chi_j$, it shouldn't be difficult to see that $\sigma := \sigma_U := \sum_j \chi_j \sigma_{V_j}$ does the job.

(b) Read $L_{X_H}\sigma$ in any of the charts $V_j$ and express it as a linear combination of wedges of the 1-forms $df_i$ and $dg_l$. Argue that each primitive summand of $L_{X_H}\sigma$ has to contain at least one $df_j$.

Next, for a form $F$ to be invariant under (the flow of) a vector field $X$ means that $L_X F = 0$ where $L_X$ is the Lie derivative. By extension, $F$ is invariant under the flow of a Hamiltonian function $H$ if it is invariant under that of $X_H$.

Here it is useful to recall that pullbacks and Lie derivatives commute, for example $L_{X_H} \circ \iota^* = \iota^* \circ L_{X_H}$. Apply both sides to $\sigma$; you are interested in proving that the left-hand side vanishes, so compute the right-hand side by using the fact that the $f_j$'s are constant on $M_c$.

(c) If $\sigma$ and $\sigma'$ both do the job in part (a), you wish to prove that $\iota^*(\sigma - \sigma') = 0$; so study the difference $\sigma - \sigma'$ in a similar way to what I suggested in part (b) for $L_{X_H}\sigma$.

| cite | improve this answer | |
  • $\begingroup$ I have a question about the meaning of $L_{X_H}\circ \iota^*$. If $\eta\in\Omega^k(M)$, then $\iota^*\eta\in\Omega^k(M_c)$. In order to apply $L_{X_H}$ to $\iota^*\eta$, it would be necessary that $\left.X_H\right|_{M_c}\subset \mathfrak{X}(M_c)$, right? Otherwise how could we guarantee $\phi^*(\iota^*\eta)$ is a form in $M_c$? (here $\phi$ is the flow of $X_H$) $\endgroup$ – rmdmc89 Jul 21 '18 at 12:58
  • 1
    $\begingroup$ @rmdmc89 You are absolutely right about this, and we do have here that $\left. X_H \right|_{M_c} \subset \mathcal{X}(M_c)$. Indeed, we have in general $\{H, f_j\} = \pm df_j(X_H)$ (where $\pm$ is a matter of convention); since we assume $\{H, f_j\} = 0$, it follows that $X_H$ is everywhere tangent to the level sets of $f_j$. As this is true for all $j=1, \dots, k$, it follows that $X_H$ is tangent to $M_c$. $\endgroup$ – Jordan Payette Jul 21 '18 at 14:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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