Darboux theorem for $2$-dimensional manifolds

Question

Let $M$ be a $2$-dimensional manifold. Using the fact that every non-vanishing $\alpha\in\Omega^1(M)$ can be written as $\alpha=fdg$ locally for convenient smooth functions $f,g$, prove Darboux's theorem for $2$ dimension:

If $(M^2,\omega)$ is a symplectic manifold, there are local coordinates $(x,y)$ such that $\omega=dx\wedge dy$.

I guess the only way to use the given result is by considering a field $X$ such that $X\neq 0$ in some neighbourhood and the non-vanishing $1$-form $i_X\omega$. Then we would have $f,g$ with $i_X\omega=fdg$, but I don't know how to deal with that. I tried to prove that $\omega=df\wedge dg$, but I think that's not true.

Answer 1

Here's a proof along the lines that you are looking for and which is valid only in dimension 2.

Consider a smoothly embedded open ball $B \subset M$. Since $B$ is contractible and the restriction of the symplectic form $\omega$ to $B$ is still closed, this restriction is exact. Thus there exists a 1-form $\lambda$ on $B$ such that $\omega = d \lambda$; by adding to $\lambda$ a closed 1-form and considering if necessary a smaller ball, we can assume without loss of generality that $\lambda$ does not vanish on $B$. From the result mentioned in the question, $\lambda$ can be written as $\lambda = f dg$ for some smooth functions $f, g \in C^{\infty}(B)$. We thus deduce that $\omega = d\lambda = d(fdg) = df \wedge dg$. Hence, considering $\mathbb{R}^2$ equipped with coordinates $(x,y)$ and the standard symplectic form $\omega_0 = dx \wedge dy$, we get that the map $\Phi : B \to \mathbb{R}^2$ given by $\Phi(p) = (f(p), g(p))$ is symplectic, i.e. $\Phi^* \omega_0 = \omega$. It follows that $\Phi$ is an immersion, and it is also a submersion since $B$ and $\mathbb{R}^2$ have the same dimension. This map is thus a local diffeomorphism; Upon shrinking $B$ again if necessary, we can assume that $\Phi$ is a symplectic diffeomorphism onto its image i.e. a Darboux chart.

Answer 2

$\textbf{Hint}$: Use Moser's Theorem and let $X = \{p\}$. If you want to see the proof, look here at Ana's notes: https://people.math.ethz.ch/~acannas/Papers/lsg.pdf. (pg.46)

Answer 3

Notice that since $\dim M = 2$, a symplectic $2$-form $\omega$ on $M$ is nothing but a nonzero volume form. If $(U, \varphi)$ is a chart on $M$ with a parametrization $\varphi : \Bbb R^2\to U$, then the restriction $\omega|_U$ can be pulled back to a volume form $\varphi^*(\omega|_U)$ on $\Bbb R^2$.

Recall that if $M$ is an $n$-manifold in general, then a volume form is a section of the top exterior power $\text{Sym}^n (TM)^*$ of the cotangent bundle $(TM)^*$. Since $\dim T_pM = n$, the fibers of this bundle are isomorphic (but not canonically in general) to $\text{Sym}^n(\Bbb R^n)$ which is a $1$-dimensional $\Bbb R$-vector space since for any two sets $\{v_1, \cdots, v_n\}$ and $\{w_1, \cdots, w_n\}$ of linearly independent vectors, $v_1 \wedge \cdots \wedge v_n = \text{det}(A) w_1 \wedge \cdots \wedge w_n$ where $A$ is the basechange matrix between those two basis (if $\{v_1, \cdots, v_n\}$ does not consistute a basis of $\Bbb R^n$ then $v_1 \wedge \cdots \wedge v_n = 0$). If $\mathbf{e}_1, \cdots, \mathbf{e}_n$ are the standard elementary basis vectors of $\Bbb R^n$ then every element of $\text{Sym}^n(\Bbb R^n)$ is a scalar multiple of $\mathbf{e}_1 \wedge \cdots\wedge \mathbf{e}_n$, where the scalar is the determinant of the matrix of $n$ vectors in the wedge.

This implies that $\text{Sym}^n(TM)^*$ is a line bundle over $M$, and therefore is trivial if $M$ is orientable (which is equivalent to asking for an existence of a nonzero volume form, which gives a section hence triviality of the said bundle). If $\omega$ and $\eta$ are two nonzero volume forms on $M$ then they constitute sections of this bundle. Define $f : M \to \Bbb R$ to be the fiberwise scale factor, i.e., $\omega_p = f(p) \eta_p$. Then globally, $\omega = f \eta$ - any two nonzero volume forms on a manifold are $C^\infty(M)$-multiples of one another.

On $\Bbb R^2$, the standard coordinates $(x, y)$ gives a canonical volume form $dx \wedge dy$. Therefore there is a nonzero function $f$ on $\Bbb R^2$ such that $\varphi^*(\omega|_U) = fdx \wedge dy$. Finally consider a diffeomorphism $\psi : B \subset \Bbb R^2 \to \Bbb R^2$ from some ball $B \subset \Bbb R^2$ (this ball might have finite radius because $\psi$ is a volume-preserving diffeomorphism, and $f dx\wedge dy$ might induce finite area on the codomain $\Bbb R^2$, see Jordan Payette's comments below) such that $\psi^*(f dx \wedge dy) = dx \wedge dy$ (exercise: construct one). Then $\phi = \varphi \circ \psi$ is a parametrization of $U$ such that $\phi^* \omega = dx \wedge dy$. The desired local coordinates on $M$ are then $(\mathbf{x}, \mathbf{y}) = (\pi_1 \circ \phi^{-1}, \pi_2 \circ \phi^{-1})$, in which $\omega = d\mathbf{x} \wedge d\mathbf{y}$.

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This part of the answer is tangential to the original question, and is speculative thoughts on my part. I'd really appreciate if someone can indicate if this sketch can be made into a proof of the general Darboux theorem.

Suppose $\omega$ is a symplectic $2$-form on a $2n$-manifold $M$ in general. Darboux's theorem in general says that there exists local coordinates $(\mathbf{x}^1, \cdots, \mathbf{x}^n, \mathbf{y}^1, \cdots, \mathbf{y}^n)$ such that $\omega = \sum d\mathbf{x}^i \wedge d\mathbf{y}^i$ There's a proof of this using the Moser's trick (see Faraad's answer), but the geometry of this statement was never clear to me in that proof. I'll try to sketch a different perspective on this, partially related to the first part of the answer.

$\omega$ is a linear symplectic form on the fibers of the tangent bundle $\pi : TM \to M$, so we can choose a fiberwise basis $(a_1(p), \cdots, a_n(p), b_1(p), \cdots, b_n(p))$ for each of the tangent spaces $T_p M$ such that $\omega_p = dx^1(p) \wedge dy^1(p) + \cdots + dx^n(p) \wedge dy^n(p)$ where $(x^1(p), \cdots, x^n(p), y^1(p), \cdots, y^n(p))$ is the dual basis of $(T_pM)^*$. This gives a section $s : M \to F(TM)$ of the frame bundle such "$\omega$ is standard along $s$". Darboux's theorem is equivalent to claiming locally $s$ is integrable/holonomic, i.e., $s = (\partial_{x_1} \phi, \cdots, \partial_{x_n} \phi)$ for some parametrization $\phi : B \subset \Bbb R^n \to U$ on $M$. The condition that $\omega$ is closed ($d\omega = 0$) should precisely be the condition for integrability of this moving frame.

I wonder if one can prove this by considering the plane fields $\xi^k = \text{Span}(a_k, b_k)$ on $M$ for $k = 1, \cdots, n$. These are "jointly transverse" in the sense that $\xi^1 + \cdots + \xi^n = TM$, and the symplectic form $\omega$ on $M$ decomposes as a sum of the "volume forms $a_k \wedge b_k$ on $\xi^k$". If $\xi^k$'s are Frobenius-integrable subbundles of $TM$, then they would produce $n$ jointly transverse 2-dimensional foliations $\mathcal{F}_1, \cdots, \mathcal{F}_n$ on $M$ and $\omega$, since it is a volume form on each leave of $\mathcal{F}_k$, has the standard local form on each leaf. Perhaps one can produce a multifoliated chart $U \to \Bbb R^{2n}$ with $\Bbb R^{2n}$ (say with standard coordinates $x_1, \cdots, x_n, y_1, \cdots, y_n$) foliated by the affine $x_ky_k$-planes that takes $\omega$ to the standard symplectic form by preserving the tangential decomposition of $\omega$?