TITLE: Is the Poisson bracket uniquely determined by its properties?
QUESTION [0 upvotes]: Let $D\subset \mathbb{R}^2$ be open, as $C^{\omega}(D)$ I denote the space of analytic real-valued functions on $D$. The poisson bracket is an alternating bilinear map: $[ \cdot, \cdot ]: C^{\omega}(D) \times C^{\omega}(D) \rightarrow C^{\omega}(D)$. So that for analytic functions $f_1(x,p)$, $f_2(x,p)$, $g(x,p)$ we have:
$$
[f_1f_2,g]= f_1[f_2,g]+ [f_1,g]f_2
$$
and:
$$
[x,p]=1, \ [x,x]=[p,p]=0
$$
My physics prof said that these properties would determine the poisson bracket uniquely, indeed I could proof inductively:
$$
[x^mp^n,x^{i}p^j]=(mj-ni)x^{m+i-1}p^{n+j-1}
$$
Therefore the poisson bracket acts on polynomials as:
$$
[f,g]=\frac{\partial f}{\partial x} \frac{\partial g}{\partial p}-\frac{\partial f}{\partial p} \frac{\partial g}{\partial x}
$$
My prof now said that since any analytic function can, by definition, be expressed as a Taylor series, the function above is the only function with said properties on the space of all analytic functions.
I don’t get why this would be so clear, could maybe someone help me to understand why we can make this step from polynomials to all analytic functions? Because if I take an analytic function that let’s say only depends on x: $f(x)=\sum_{i=1}^{\infty} a_i x^{i}$, and I’d like to show how $\frac{\partial f}{\partial x} \frac{\partial g}{\partial p}-\frac{\partial f}{\partial p} \frac{\partial g}{\partial x} $ acts on it, I wouldn’t know if I could swap derivative and $\lim$, since Taylor series don’t need to converge uniformly, do they?

REPLY [1 votes]: Here is what I would try: Let us assume that $[\cdot, \cdot] : C^{\infty}(D) \times C^{\infty}(D) \to C^{\infty}(D)$ and that the derivation identity
$$[f_1f_2, g] = f_1[f_2, g] + [f_1, g]f_2$$
holds for all $f_1, f_2, g \in C^{\infty}(D)$. First using bump functions and the derivation identity, I think we can show that $[\cdot, \cdot]$ acts locally, meaning that if $f, g \in C^{\infty}(D)$ and $x \in D$ and $\tilde{f} = f$ and $\tilde{g} = g$ on a neighborhood of $x$ in $D$, then $[f, g](x) = [\tilde{f}, \tilde{g}](x)$. This means that for any open $U \subset D$, $[\cdot, \cdot] : C^{\infty}(U) \times C^{\infty}(U) \to C^{\infty}(U)$ is well defined and satisfies the two identites.
Then if $f, g$ are analytic on $D$, given $x \in D$, there is a neighborhood $U$ of $x$ such that the Taylor series of $f$ and $g$ and the Taylor series for all the derivatives converge absolutely and uniformly to $f$ and $g$. So we get $[f_n,g_n](x) =\frac{\partial f_n}{\partial x}(x) \frac{\partial g_n}{\partial p}(x) - \frac{\partial f_n}{\partial p}(x) \frac{\partial g_n}{\partial x}(x)$ for $f_n$ and $g_n$ being the $n$th partial sums of the Taylor series of $f$ and $g$. Let's assume that $[f_n, g_n] \to [f, g]$, essentially that $[\cdot, \cdot]$ is continuous wrt the $C^\infty(D)$ topology. Taking $n \to \infty$ and using uniform convergence and boundedness of derivatives (so that the product also converges uniformly), we get $[f,g](x) =\frac{\partial f}{\partial x}(x) \frac{\partial g}{\partial p}(x) - \frac{\partial f}{\partial p}(x) \frac{\partial g}{\partial x}(x)$.
{"set_name": "stack_exchange", "score": 0, "question_id": 4564763}