ukmuiik

Question

Let $M$ be a $2$-dimensional manifold. Using the fact that every non-vanishing $alphainOmega^1(M)$ 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 $omega=dxwedge dy$.

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

Jordan Payette 2,681147 · Accepted Answer · 2018-07-17 19:49:59Z

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 lost 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 $omega = dlambda = d(fdg) = df wedge dg$. Since $omega neq 0$, we deduce that the functions $f$ and $g$ are linearly independent on $B$, hence shrinking $B$ again if necessary we see that the map $(f, g) : B to mathbbR^2$ is a Darboux chart.

just to be sure, when you say "an embedded open ball", you mean an open set homeomorphic to an open ball in $mathbbR^n$, or is it something else? — Jul 17 at 22:55
@rmdmc89 I meant exactly what you understood (with 'diffeomorphic to the usual ball' instead of 'homeomorphic to...', but that this distinction is not really relevant here). — Jul 18 at 1:20

score 2 · Answer 2 · 2018-07-19 12:04:45Z

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 $varphi : Bbb R^2to U$, 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 $textSym^n (TM)^*$ of the cotangent bundle $(TM)^*$. Since $dim T_pM = n$, the fibers of this bundle are isomorphic (but not canonically in general) to $textSym^n(Bbb R^n)$ 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, $v_1 wedge cdots wedge v_n = textdet(A) w_1 wedge cdots wedge w_n$ 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 $mathbfe_1, cdots, mathbfe_n$ are the standard elementary basis vectors of $Bbb R^n$ then every element of $textSym^n(Bbb R^n)$ is a scalar multiple of $mathbfe_1 wedge cdotswedge mathbfe_n$, where the scalar is the determinant of the matrix of $n$ vectors in the wedge.

This implies that $textSym^n(TM)^*$ 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 $f dxwedge dy$ 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 $(mathbfx, mathbfy) = (pi_1 circ phi^-1, pi_2 circ phi^-1)$, in which $omega = dmathbfx wedge dmathbfy$.

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 $(mathbfx^1, cdots, mathbfx^n, mathbfy^1, cdots, mathbfy^n)$ such that $omega = sum dmathbfx^i wedge dmathbfy^i$ 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 ($domega = 0$) should precisely be the condition for integrability of this moving frame.

I wonder if one can prove this by considering the plane fields $xi^k = textSpan(a_k, b_k)$ 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 $mathcalF_1, cdots, mathcalF_n$ on $M$ and $omega$, since it is a volume form on each leave of $mathcalF_k$, 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 $omega$?

In the 1st part of your answer, it might be that $omega$ gives a bounded area to $U$, whereas $dx wedge dy$ gives an infinite area to $mathbbR^2$. In that case, your diffeomorphism $psi$ doesn't exist; rather a diffeomorphism $psi' : V to mathbbR^2$ would exist for some open set $V subset mathbbR^2$. The main difficulty of the problem is then to find $V$ and $psi'$, a nontrivial exercise (a solution to which is given by Moser's trick). — Jul 18 at 12:55
In the 2nd part of your answer, $phi$ is subject to a similar remark to the one in my previous comment. As for the suggestion in the last paragraph, it has the flavour of a $h$-principle result; I don't doubt that such a proof of Darboux' theorem exists, but the condition $domega = 0$ would play a crucial and nontrivial role, and the proof would be relatively sophisticated (cf the work of Datta & Islam to deform just one distribution $xi$ into a symplectic foliation). It's best to use integrability from the outset to construct the Darboux chart, as in Arnol'd's book on mechanics. — Jul 18 at 13:05
Thanks a lot for your comments on the second half. I'll look into the reference you suggested (coincidentally I have been looking at h-principles for a while, so this might be a good learning project for me). I have been told that Arnold constructs the Darboux charts by hand, I'll look into his text. — Jul 19 at 3:48
@JordanPayette Regarding the first half: Thanks, that's a subtle point I completely missed. I am not entirely sure yet that I would need Moser's trick to accomplish this; say $f dx wedge dy$ is my volume form on $Bbb R^2$. I can locally solve the first order differential equation $partial h/partial x = f$ on some ball $B subset Bbb R^2$. On $B$, this form then is $dh wedge dy$. Consider $F : B to Bbb R^2$, $F(x, y) = (h(x, y), y)$. Then $F^*(dx wedge dy) = f dx wedge dy$. $F$ is a diffeomorphism onto image because $textdet(dF) = h_x = f neq 0$ on $B$. — Jul 19 at 4:25
(or at least locally so by inverse function theorem, which is enough) — Jul 19 at 4:41

Faraad Armwood 7,4292619 · Answer 3 · 2018-07-17 04:27:50Z

$textbfHint$: 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)

Jordan Payette 2,681147 · Accepted Answer · 2018-07-17 19:49:59Z

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 lost 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 $omega = dlambda = d(fdg) = df wedge dg$. Since $omega neq 0$, we deduce that the functions $f$ and $g$ are linearly independent on $B$, hence shrinking $B$ again if necessary we see that the map $(f, g) : B to mathbbR^2$ is a Darboux chart.

just to be sure, when you say "an embedded open ball", you mean an open set homeomorphic to an open ball in $mathbbR^n$, or is it something else? — Jul 17 at 22:55
@rmdmc89 I meant exactly what you understood (with 'diffeomorphic to the usual ball' instead of 'homeomorphic to...', but that this distinction is not really relevant here). — Jul 18 at 1:20

score 2 · Answer 5 · 2018-07-19 12:04:45Z