Maximum number of independent constants of motion in involution

Question

Consider a holonomic system. Let $M$ be the phase space $T^{\ast}Q$ where $Q$ is the configuration space with $\dim(Q)=n$. Two functions $f,g: M \to \mathbb{R}$ are in involution if they Poisson-commute ($\{f,g\}=0$). I was told that the maximum number of (functionally) independent constanst of motion in involution is $n$, but I can't prove that (or even see why). If you know about this stuff and you're able to prove this fact, thank you so much.

My idea is that it's probably related to the fact that the symplectic form $\omega$ is not-degenerate, but I couldn't get anything from this.

Answer 1

You can start noticing that $\omega(X_f,X_g)=\{f,g\}$, where $X_h$ is the Hamiltonian vector field of $h:T^*Q\rightarrow \mathbb{R}$. At this point assume to have $f_1,...,f_n:Q\rightarrow \mathbb{R}$ which are pairwise in involution functionally independent functions, i.e. $df_i\wedge df_j\neq 0$, $\{f_i,f_j\}=0$ for any $i\neq j$.

Assume now that there exists another function $\bar{f}: Q\rightarrow \mathbb{R}$ which is in involution with all the other $n$s and does not depend on them.

You then get that $\omega(X_{\bar{f}},X_{f_i})=0$ for any $i=1,...,n$. However this gives an immediate contradiction. Indeed, let us take an arbitrary point $q\in Q$. The tangent vectors $X_{f_1}(q),...,X_{f_n}(q)\in T_qQ$ form a basis of this linear space. Therefore, any $v\in T_q Q$ can be written as $$v = \sum_{i=1}^n \lambda_i X_{f_i}(q)$$ and hence, by bilinearity of the symplectic form, you get that $$ \omega(X_{\bar{f}},v) = \sum_{i=1}^n\lambda_i\omega(X_{\bar{f}},X_{f_i})=0.$$

This is a contradiction since the symplectic form is non-degerate by construction, which means if $\forall v\in T_qQ$ $\omega(X,v)=0$, then necessarily $X=0$. However, we assumed $\bar{f}$ to be functionally independent from the other functions, which is not the case when $\bar{f}=0$. We can then conclude the maximum number of independent functions in involution on $Q$ is $n$.

The same can be said in terms of constants of motion, for example when you talk about Liouville-Arnold's integrability theorem you assume to have the maximum number of functionally independent first integrals of the Hamiltonian vector fields you are considering.

I am sure this is not the shortest proof of this fact, however, at this moment this seemed more clear to me than other possibilities.

Answer 2

The essence is the following simple fact.

Suppose we have a symplectic form $\omega $ on a $2n$-dim vector space $V$. By Darboux's theorem, we can find a basis of $V$, $\{x_i,y_i | 1\leq i \leq n \}$, such that

$$\omega(x_i,y_j) = \delta_{ij}, \quad \omega(x_i,x_j) =\omega(y_i,y_j )=0. $$

We have actually a direct decomposition of $V$, $V= \oplus_{i=1}^n V_i $, with $V_i$ the 2-dim subspace spanned by $(x_i,y_i )$.

Now consider a subspace $W$ which is of at least $n+1$ dim. We have

$$W = \oplus_{i=1}^n ( W \cap V_i ). $$

Counting the dimension, we know that at least for some $i$, the intersection $W\cap V_i $ is of dim 2. That is, $V_i \subset W$. By $\omega $ is nonzero on $V_i $.