# $df_1,…,df_k$ linearly independent $\Rightarrow \frac{\omega^n}{n!}=df_1\wedge…\wedge df_k\wedge\sigma$

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 $L_{X_H}\sigma$.

• 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 $X_H$) – rmdmc89 Jul 21 '18 at 12:58
• @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 $M_c$. – Jordan Payette Jul 21 '18 at 14:20