MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $(X^{2n},\omega)$ is a compact symplectic manifold. Knowing the algebra $C^\infty(X)$ is equivalent to knowing the manifold $X$, and knowing the Poisson bracket $\{\cdot,\cdot\}:C^\infty(X)\otimes C^\infty(X)\to C^\infty(X)$ is equivalent to knowing $\omega$. Thus it is natural to ask the following question:

Can one describe the functional $f\mapsto\int_Xf\cdot\omega^{\wedge n}$ just in terms of $C^\infty(X)$ and $\{\cdot,\cdot\}$? What about just the special case when $f=1$; that is, does the volume of $X$ admit a description purely in terms of $C^\infty(X)$ and $\{\cdot,\cdot\}$?

share|cite|improve this question
up vote 7 down vote accepted

The Poisson geometry is, I think, a red herring. I will explain in this answer how to construct $\Omega^{2n}(X)$ as a $C^{\infty}(X)$ module, and how to construct the volume form $\omega^{n}$ within it. So I think the right question to ask is:

Let $X^m$ be a smooth, orientable, compact manifold. Suppose that we are given $\Omega^{\bullet}(X)$ with its dga structure. How can we recover the functional $\int_X: \Omega^m \to \mathbb{R}$?

Given $C^{\infty}(X)$, we can construct $T^1(X)$ as the space of $\mathbb{R}$-linear derivations $C^{\infty}(X) \to C^{\infty}(X)$. Then we can construct $\Omega^1(X)$ as $\mathrm{Hom}_{C^{\infty}(X)}(T^1(X), C^{\infty}(X))$. (Attempting to construct $\Omega^{1}(X)$ more directly runs into problems.) We then have the map $d: C^{\infty}(X) \to \Omega^1(X)$ which, given a function $f \in C^{\infty}$, produces the map $Y \mapsto Y(f)$ in $\mathrm{Hom}_{C^{\infty}(X)}(T^1(X), C^{\infty}(X))$. The Serre-Swan correspondence then lets us build the sections of any vector bundle on $X$ which is obtained from tensor powers of $\Omega^1(X)$ by applying the corresponding tensor operations on $C^{\infty}(X)$-modules. For example, $(\bigwedge^2 T^1)(X)$ is the antisymmetric elements of $T^1(X) \otimes_{C^{\infty}(X)} T^1(X)$.

In particular, we can talk about $\bigwedge^2 T^1(X)$, and we have the evaluation pairing $\bigwedge^2 T^1(X) \times C^{\infty}(X) \times C^{\infty}(X) \to C^{\infty}(X)$. So we can talk about the unique bivector $\eta$ in $\bigwedge^2 T^1(X)$ which realizes the Poisson bracket. Also, we can take about the map $(\bigwedge^2 T^1(X))^{\otimes n} \to \bigwedge^{2n}(T^1(X))$. So we can produce the form $\eta^{\otimes n}$.

We can say that the manifold $X$ is symplectic if and only if $\eta^{\otimes n}$ is a generator of the $C^{\infty}(X)$-module $\bigwedge^{2n} T^1(X)$. If so, then $\omega^n$ is the reciprocal element in $\Omega^{2n}(X)$.

All of the above, though a pain, is definitely algebraic. So I claim that we are reduced to the boxed question.

I have a pretty elegant answer to the boxed question for circles, and a not very elegant answer for general $m$. I hope this question will attract some better answers.

Let $X$ be a circle, and let $\eta \in \Omega^1(X)$. We will say that $\eta$ is a resonant class if there are $x$ and $y \in C^{\infty}(X)$ such that $$x^2+y^2=1,\ dx = y \eta,\ dy = - x \eta.$$ The point is that, if these equations are solvable, then we can locally write $(x,y)$ as $(\cos \theta, \sin \theta)$ where $d \theta = \eta$.

So these equations are globally solvable if we can find a global such $\theta$, modulo $2 \pi$. In other words, they are globally solvable if and only if $\int_X \eta \in 2 \pi \mathbb{Z}$. So this gives us a description of the map $\int_X$ up to sign; it takes the resonant classes in $\Omega^1(X)$ to $2 \pi \mathbb{Z}$.

Here is my bad answer to the general case. We will say that $\eta$ "comes from a sphere" if there are functions $x_0$, $x_1$, ..., $x_m$ on $X$ such that $\sum x_i^2=1$ and $\eta = \sum (-1)^r x_r dx_0 \wedge dx_1 \wedge \cdots \widehat{dx_r} \cdots \wedge dx_m$. In other words, $\eta$ is pulled back from the volume form on $S^m$ along a smooth map.

Then we can describe $\int_X$ by the two normalizations that (1) $\int_X$ is $0$ on $d \Omega^{m-1}(X)$ and (2) $\int_X$ takes integer multiples of the area of $S^m$ on those forms that come from a sphere. The normalization is correct because any compact oriented connected $m$-fold has a smooth degree $1$ map to an $m$-sphere. (Proof on request.)

I say this is ugly for two reasons. First, the construction of that degree $1$ map is a partition of unity argument. So I suspect that this description will be completely useless in practice.

Second, although every integral cohomology class is pulled back from a sphere, I don't think that every integral volume form is. So I really need both conditions in the normalization above. That strikes me as inelegant.

share|cite|improve this answer
"I see your question and I raise you one." :) – Gunnar Þór Magnússon Jan 20 '12 at 22:28
@David: to construct a smooth degree 1 map to a sphere, can't you just fix a small ball B in M, and map everything outside the interior of B to a point? This gives a continuous degree 1 map to a sphere, and by perturbing it we obtain a smooth map which is still degree 1. – Tom Church Jan 20 '12 at 22:48
It's that perturbation that I thought required the partition of unity. Maybe I'm wrong. – David Speyer Jan 20 '12 at 23:59

There is a trivial sense in which the Poisson algebra determines the volume, given the initial paragraph. But this is clearly not what you mean. You might look at the section on "symplectic groupoids" in:

A. Weinstein. The volume of a differentiable stack.

My memory is that Alan does say some things about volumes of symplectic manifolds in terms of the corresponding Poisson manifold. But it's still very analytical. Note that the "correct" volume form is not $\omega^{\wedge n}$ but $\omega^{\wedge n}/n!$. I will write $\mu = \omega^{\wedge n}/n!$ for this volume form.

If you suppose that you already know how to compute $\int 1\,\mu$, I think I can describe how to compute $\int f\,\mu$. Let me suppose that your manifold is connected. Then you can check that $\int f\mu = 0$ iff $f = \operatorname{div}_\mu \vec x$ for some vector field $\vec x$ on your manifold. Here $\operatorname{div}_\mu$ is the "divergence with respect to $\mu$"; i.e. $\operatorname{div}_\mu \vec x = (\mathcal L_{\vec x}\mu) / \mu$, where $\mathcal L_{\vec x}$ is the Lie derivative with respect to $\vec x$. I am using the fact that $\mu$ is nowhere vanishing.

The question, then, is to characterize the image of the map $\operatorname{div}_\mu$ from vector fields to functions, because then $\int: C^\infty / \operatorname{im}(\operatorname{div}_\mu) \to \mathbb R$ is an isomorphism taking $1$ to $\int \mu$, which I supposed you already know. But now, let me work temporarily in local coordinates. Then $\mu = \frac1{n!} \det \omega = \frac1{n!} (\det \pi)^{-1}$, and we want the logarithmic derivative (for $\mathcal L_{\vec x}$) of this, but the logarithmic derivative of a determinant is a trace. So if I am not mistaking, $\operatorname{div}_\mu \vec x = -\operatorname{tr}((\mathcal L_{\vec x}\pi)/\pi)$, where here and throughout $\pi$ is the Poisson tensor.

So, modulo arithmetic errors, the point is that the integration you care about is characterized by the volume and by the image of $\vec x \mapsto \operatorname{tr}((\mathcal L_{\vec x}\pi)/\pi)$; namely, it is the unique map vanishing on this image and sending $1$ to the volume. Then I should also mention that $\mathcal L_{\vec x}\pi$ is something you can calculate entirely algebraically. First recall that a "vector field" is exactly a derivation, and so you can talk about $\vec x$ and $\pi$ in terms of (multi)derivations of $C^\infty$. Second, recall that the commutator gives rise to the Schouten—Nijenhuis bracket $[,]$ on multivector fields completely algebraically, and $\mathcal L_{\vec x}\pi = [\vec x,\pi]$.

To invert $\pi$ and take the trace still requires some geometry, of course — you should not be able to do so if your manifold has singular points or if it is Poisson but not symplectic. Then again, to be (connected and) symplectic is an algebraic condition — there is a purely algebraically-defined homology theory that counts symplectic leaves — and so there might be a purely algebraic way to do what you ask.

Hope that helps.

share|cite|improve this answer
what means "correct" ? I mean the sentence about $\omega/n!$ not just $\omega^n$ ? – Alexander Chervov Jan 20 '12 at 12:32
@A.Chervov: let $\omega$ be the standard symplectic form $\sum_{i=1}^n dx_i \wedge dy_i$ on ${\bf R}^{2n}$. The standard volume form $\mu$ on ${\bf R}^{2n}$ is $x_1 \wedge y_1 \wedge x_2 \wedge y_2 \wedge\cdots\wedge x_n \wedge y_n \wedge$, which is $\omega^n/n!$, not $\omega^n$. There might be a more structural reason to prefer in the context of differential geometry to prefer $\mu$ to $n! \mu$; where I come from in number theory, one also needs differential forms in positive characteristic, and there $n! \mu$ is clearly worse in characteristic $p \leq n$. – Noam D. Elkies Jan 20 '12 at 15:41
@Alexander: Of course, for the rest of my "answer", I don't use the normalization of $\mu$. I think that the standard reason in the real case to prefer $\mu$ is that the Poisson tensor defines naturally a pairing of top forms, and $\mu$ is the one that pairs to itself to $1$. In Weinstein's paper cited above, he mentions some other reasons to prefer $\mu$, but they boil down to this one. Anyway, the real reason is philosophical: it is a broad fact that almost always divided powers are preferable to their non-divided cousins. For example, formulas with divided powers tend to be much cleaner. – Theo Johnson-Freyd Jan 20 '12 at 17:41
@Noam, Theo Thank You ! I thought the reason might be related with quantization where the "correct" dimension of the Hilbert space is given by $\int \omega^n/n!$ (actually it is true only when K=0 (canonical class) otherwise Todd(K) enters from Rieman-Roch - this is "half-form" correction). But it seems reasons are more simple. In quantum world this n! is quite usual, but seems always not clear for me. – Alexander Chervov Jan 21 '12 at 14:53

Let me add a few remarks on Theo's answer. For a compact connected symplectic manifold, it is known that the integration with respect to the Liouville volume form (whatever normaliyation you prefer) is the only linear functional on $C^\infty(M) = C^\infty_0(M)$ which vanishes on all Poisson brackets, unique up to rescaling of course. Remarkably, you don't even need to assume that the functional is continuous in any sense: this statement holds in the full algebraic dual of $C^\infty(M)$. There are various proofs of this available. Now on the other hand, you know that the integral of $1$ is the volume (in your prefered normalization) which shows that the constant functions on a compact symplectic manifold are not linear combinations of Poisson brackets (this is of course wrong in the noncompact case). From this you obtain that a function is a linear span of Poisson brackets iff its integral vanishes, a result due to Lichnerowicz (maybe???). This is the space Theo is talking about.

In particular, I'm rather sceptic that one has a simple way of computing the volume in terms of Poisson brackets, as you can rescale the integration functional without destroying the above conclusions about the subspace of linear spans of Poisson brackets. Hmm...

share|cite|improve this answer
I'm not sure I understand the last sentence. Doesn't the Poisson bracket determine the symplectic form completely (and not just up to a constant scalar factor) and therefore the volume? – Deane Yang Jan 20 '12 at 13:56
Hi Deane. Of course, this is true What I mean is that there seems to be not a simple computational formula. If you first recover $\omega$ from the Poisson bracket then, of course, you're done. But it does not seem to be easy to just write down a simple formula (is it?) in terms of Poisson brackets of some functions, in order to get the volume... – Stefan Waldmann Jan 23 '12 at 17:50

There is a different way to state Waldmann's answer which can help in understanding the situation, in my opinion.

His statement depends on the following facts:

  1. The vector space ${\cal C}^\infty(M)$ modulo Poisson commutators is 1-dimensional. This space is what can be called the 0th Poisson homology of $M$. The fact that this space is $1$-dim essentially depends on the fact that 0th Poisson homology here is isomorphic to top de Rham cohomology, via a canonical isomorphism determined by the Liouville volume form.

  2. This 0th Poisson homology coincides with the Corresponding Chevally-Eilenberg homology of the Lie algebra given by Poisson bracket, both if chains are chosen to be differentiable, continuous, algebraic (do you have a ref. for this?).

No doubt that the algebraic CE homology of the Lie algebra defined by the Poisson commutator is a purely algebraic object, in the sense in which the question was formulated.

On the other hand the fact that this algebraic homology is the same as, say, Poisson homology is a result of quote different nature. In general, at the vector fields level, Poisson cohomology is the cohomology of the subcomplex of multidifferential chains inside the Chevalley-Eilenberg complex of the infinite dimensional Lie algebra ${\cal C}^\infty(M)$, and there is no reason why the two should be equal. I've seen, on this point, citations to a paper by Melotte, "cohomologie de Chevalley associée aux variétés de Poisson" Bull. Soc. Royale Sci. Liège 5, 319-413 (1989), but i've never read this paper so I have no idea whether it contains some clear ideas on how and when the two cohomologies should be equal.

This cohomological approach shows that possibly most of this consideration can be carried on for unimodular Poisson manifolds, where an invariant volume form takes the place of Liouville volume.

P.s. the result attributed to Lichnerowicz on the kernel of the integral is, he says, somewhat contained in a paper by Calabi "on the group of automorphisms of a symplectic manifold" 1970 (cited in Brylinski's paper on Poisson homology).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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