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:



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> 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 $\{\cdot,\cdot\}$?
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Can you tell the volume of a symplectic manifold from the Poisson brackets?
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:



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> 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 $\int\_X: \Omega^m \to \mathbb{R}$?
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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](https://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials).) 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](http://en.wikipedia.org/wiki/Serre%25E2%2580%2593Swan_theorem) 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.




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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 $2 \pi \mathbb{Z}$.




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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. 


