The Question:
The Lie algebra of functions on phase space (under the Poisson bracket) is a central extension of the Lie algebra of Hamiltonian vector fields on phase space (under the vector field Poisson bracket). This is because constant functions have trivial Hamiltonian vector fields. What is the explicit map $c: \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}$ that makes this central extension? (Here $\mathfrak{g}$ is the Lie algebra of vector fields.)
Note that this central extension is exactly the one that occurs during prequantization, as explained below.
Background:
In Hamiltonian mechanics, you usually have some phase space, let's just say $\mathbb{R}^{2N}$, labelled by coordinate functions $q_i, p_i$ for $i = 1 \ldots N$, which comes with a Poisson bracket. Functions on phase space comprise a Lie algebra where the "Lie bracket" is
$$
\{ f, g\}.
$$
You can also construct vector fields on phase space which are "generated" by these functions. These are called "Hamiltonian vector fields."
$$
X_h \equiv \{ h, \cdot \}
$$
From the Jacobi identity,
$$
\{ \{g, h\}, f \} + \{ \{h, f\}, g \} + \{ \{f, g\}, h \} = 0
$$
which can be rearranged to be
$$
\{ \{g, h\}, f \} = \{g, \{h, f\} \} – \{ h, \{g, f\} \}
$$
$$
X_{ \{g, h\} } (f) = X_g (X_h (f) ) – X_h (X_g (f))
$$
$$
X_{ \{g, h\} } = [X_g, X_h].
$$
So $h \mapsto X_h$ is a Lie algebra homomorphism where the Lie bracket is just the Lie bracket of vector fields.
However, this map is not injective. If you add a constant to a function, it will map to the same vector field. That is, for any $a \in \mathbb{R}$,
$$
X_{h + a} = X_h.
$$
A classic example is that, while
$$
\{q_i, p_j\} = \delta_{ij}
$$
we have
$$
X_{q_i} = \frac{\partial}{\partial p_i} \hspace{1 cm} X_{p_j} = -\frac{\partial}{\partial q_i}
$$
$$
[X_{q_i}, X_{p_j} ] = 0.
$$
This shows that the Lie algebra of functions with the Poisson bracket is a central extension of the Lie algebra of Hamiltonian vector fields.
Let me quickly review central extensions. If you have a Lie algebra $\mathfrak{g}$, a central extension of $\mathfrak{g}$ is a map
$$
c : \mathfrak{g} \times \mathfrak{g} \to \mathbb{R}
$$
which is anti symmetric, bilinear, and satisfied the Jacobi identity
$$
c (X, [Y, Z]) + c (Y, [Z, X]) + c (Z, [X, Y]) = 0
$$
for all $X, Y, Z \in \mathfrak{g}$. The idea is that, if you have a Lie algebra $\mathfrak{g}$ with a commutation relation between $X$ and $Y$, we add a central element (a multiple of the identity) to our Lie algebra and declare the new commutation relation to be
$$
[X, Y] \mapsto [X, Y] + c(X, Y) \mathbf{1}.
$$
(Note that you can always construct a central extension by choosing an arbitrary linear function $b : \mathfrak{g} \to \mathbb{R}$ and setting $c(X, Y) = b([X, Y])$. However, not all central extensions are of this trivial form.)
That concludes my review of central extensions. Let's get back to Hamiltonian mechanics. It turns out that there is a way to "extend" the algebra of Hamiltonian vector fields so that it recreates the algebra of functions on phase space. This is what is done in the "prequantization" procedure during geometric quantization.
First, recall that phase space has a symplectic form
$$
\omega = d q_i \wedge d p_i
$$
(where the summation of $i = 1 \ldots N$ is implied)
which is the exterior derivative of
$$
\theta = -p_i dq_i.
$$
In particular,
$$
\theta(X_f) = p_i \frac{\partial f}{\partial p_i}
$$
Taking a cue from prequantization, we can define
$$
Q(f) \equiv X_f + \frac{i}{\hbar} \big(f – \theta(X_f) \big)
$$
which, happily, (assuming I didn't make a sign error) satisfies
$$
[Q(f), Q(g)] = Q(\{f, g\} )
$$
which could be shown after a bit of work. So these operators comprise a Lie algebra homomorphism. What's more, we can see that constants are not annihilated by $Q$, so this is a central extension of the Lie algebra of vector fields.
To reiterate, my question is, how can I write the commutation relations of these operators $Q(f)$ in terms of a simple central extension $c(X_f, X_g)$? What is $c$ explicitly?
Best Answer
My definition of a central extension is slightly different to yours. I would say a central extension $\widetilde{\mathfrak{g}}$ of $\mathfrak{g}$ by an abelian Lie algebra $\mathfrak{a}$ is an exact sequence $$ 0 \to \mathfrak{a} \to \widetilde{\mathfrak{g}}\to \mathfrak{g}\to 0. $$ Then given a splitting $s:\mathfrak{g}\to\widetilde{\mathfrak{g}}$ of this exact sequence, there is a corresponding 2-cocycle $c:\mathfrak{g}\times\mathfrak{g}\to \mathfrak{a}$, given by $$ c(\xi,\eta) := [s(\xi),s(\eta)]_{\widetilde{\mathfrak{g}}} - s([\xi,\eta]_{\mathfrak{g}}) $$ (which is easily shown to lie in the image of $\mathfrak{a}$). Note $c$ depends on the splitting $s$. It can be shown that a different choice of splitting changes $c$ by the boundary of a 1-cocycle.
In this case, we are considering a central extension $$ 0 \to \mathbb{R} \to C^\infty(M) \to \mathfrak{ham}(M,\omega) \to 0, $$ where $(M,\omega)$ is the symplectic manifold, and $\mathfrak{ham}(M,\omega)$ denotes the set of Hamiltonian vector fields on $M$. Suppose for simplicity that $M$ is connected (which is true in your case of $\mathbb{R}^{2n}$). Then taking any $m_0\in M$, a splitting $s:\mathfrak{ham}(M,\omega)\to C^\infty(M)$ can be defined by $$ s(X_f) := f - f(m_0). $$ Note this is well-defined: if $X_f = X_g$, then $f-g$ is a constant function on $M$ (using here the connectedness of $M$), and so $$ s(X_f)-s(X_g) = (f-f(m_0)) - (g-g(m_0)) = (f-g) - (f-g)(m_0) = 0. $$ The corresponding 2-cocycle is \begin{align*} c(X_f,X_g) &= \lbrace s(X_f),s(X_g)\rbrace - s([X_f,X_g]) = \lbrace f-f(m_0), g-g(m_0)\rbrace - s(X_{\lbrace f,g\rbrace}) \\ &= \lbrace f,g\rbrace - (\lbrace f,g\rbrace - \lbrace f, g\rbrace (m_0)) \\ &= \lbrace f, g\rbrace(m_0). \end{align*}