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## Poisson algebras and the classical limit

In the previous post we described the Heisenberg picture of quantum mechanics, which can be phrased quite generally as follows: given a noncommutative algebra $A$ (the algebra of observables of some quantum system) and a Hamiltonian $H \in A$, we obtain a derivation $[-, H]$, which is (up to some scalar multiple) the infinitesimal generator of time evolution. This is a natural and general way to start with an algebra and an energy function and get a notion of time evolution which automatically satisfies conservation of energy.

However, if $A$ is commutative, all commutators are trivial, and yet classical mechanics somehow takes a Hamiltonian $H \in A$ and produces a notion of time evolution. How does that work? It turns out that for algebras of observables $A$ of a classical system, we can think of $A$ as the classical limit $\hbar \to 0$ of a family $A_{\hbar}$ of noncommutative algebras. While $A$ is commutative, the noncommutativity of the family $A_{\hbar}$ equips $A$ with the extra structure of a Poisson bracket, and it is this Poisson bracket which allows us to describe time evolution.

Today we’ll describe one way to formalize the notion of taking the classical limit using the deformation theory of algebras. We’ll see how Poisson brackets pop out along the way, as well as the relevance of the lower Hochschild cohomology groups.

Hochschild cohomology

In this post we consider associative, but not necessarily unital, algebras over a fixed field $k$ of characteristic not equal to $2$.

Let $A$ be such an algebra. A formal deformation of $A$ is an associative $k[\epsilon]$-bilinear product $\star$ on $A[[\epsilon]]$ (when $A$ is noncommutative, $\epsilon$ should be central) such that for $a, b \in A$ we have

$\displaystyle a \star b = ab + \epsilon (\text{other terms})$.

In other words, quotienting by $\epsilon$ we obtain the original product on $A$. For the purposes of understanding the classical limit, $A$ should be thought of as the classical algebra of observables and $A[[\epsilon]]$, together with the deformed product $\star$, should be thought of as the quantum one. Roughly speaking, formal deformations of an algebra $A$ describe the “formal neighborhood” of $A$ in the moduli space of algebras.

Formal deformations turn out to be difficult to construct in general. As a first approximation, instead of quotienting by $\epsilon$ we can quotient by $\epsilon^2$. A first-order deformation of $A$ is an associative $k[\epsilon]/\epsilon^2$-bilinear product on $A[\epsilon]/\epsilon^2$ satisfying the same condition as above. Explicitly, it is given by a product

$\displaystyle a \star b = ab + \epsilon f(a, b)$

where $f(a, b) : A^2 \to A$ is a $k$-bilinear map such that $(a \star b) \star c = a \star (b \star c)$, or equivalently

$\displaystyle f(ab, c) + f(a, b) c = f(a, bc) + a f(b, c)$.

The $k$-vector space of all such maps is denoted $Z^2(A, A)$ and called the space of Hochschild $2$-cocycles of $A$ with coefficients in $A$. Roughly speaking, first-order deformations of $A$ describe the “tangent space” to $A$ in the moduli space of algebras.

Of course, this is not really the space we’re interested in. If we actually want to understand first-order deformations of $A$, then it would be a good idea to identify $2$-cocycles that give isomorphic deformations. Hence suppose $f_1, f_2$ are $2$-cocycles describing deformations $\star_1, \star_2$ which are isomorphic in the sense that there is a $k[\epsilon]$-linear map $\phi : A[\epsilon]/\epsilon^2 \to A[\epsilon]/\epsilon^2$ sending one to the other which reduces to the identity $\bmod \epsilon$. Writing $\phi = I + \epsilon T$, we want

$\displaystyle (a + \epsilon Ta) \star_1 (b + \epsilon Tb) = a \star_2 b + \epsilon T(a \star_2 b)$

which after simplification gives

$\displaystyle f_2(a, b) - f_1(a, b) = (Ta) b + a (Tb) - T(ab)$.

In other words, $f_1, f_2$ give isomorphic deformations if and only if $f_2 - f_1$ lies in the subspace of $2$-cocycles of the form $(Ta) b + a (Tb) - T(ab)$ for some $k$-linear map $T : A \to A$. These $2$-cocycles are precisely the $2$-coboundaries $B^2(A, A)$; as $2$-cocycles, they describe the deformations isomorphic to the trivial deformation given by $A[\epsilon]/\epsilon^2$ with its usual product. The quotient space $Z^2(A, A) / B^2(A, A)$, the second Hochschild cohomology $H^2(A, A)$, is the desired space which parameterizes first-order deformations of $A$.

It’s worth mentioning at this point that Hochschild cohomology can be computed from a cochain complex beginning

$0 \to A \xrightarrow{d_1} \text{Hom}_k(A, A) \xrightarrow{d_2} \text{Hom}_k(A \otimes A, A) \xrightarrow{d_3} ...$

where $d_1$ sends an element $a \in A$ to a map $b \mapsto ab - ba$ and $d_2$ sends a map $f : A \to A$ to a map $a \otimes b \mapsto f(a) b + a f(b) - f(ab)$ (at least up to a sign I may have wrong). The kernel of $d_3$ is the space of Hochschild $2$-cocycles, and its quotient by the image of $d_2$ gives $H^2(A, A)$ as above.

But there are two smaller cohomology groups we can describe reasonably concretely as well. The kernel of $d_1$ defines zeroth Hochschild cohomology

$\displaystyle H^0(A, A) \cong Z(A)$

which is precisely the center of $A$. The kernel of $d_2$ defines the space of Hochschild $1$-cocycles $Z^1(A, A) \cong \text{Der}(A, A)$, which is precisely the space of derivations $A \to A$. The image of $d_1$ defines the space of Hochschild $1$-coboundaries $B^1(A, A)$, which are given by inner derivations $b \mapsto [a, b]$ (named by analogy with inner automorphisms). The quotient $Z^1(A, A)/B^1(A, A)$ defines the first Hochschild cohomology

$\displaystyle H^1(A, A) \cong \text{Der}(A, A)/\text{Inn}(A, A)$

which is precisely the space of outer derivations of $A$.

Introducing the lower Hochschild cohomology groups gives Hochschild cohomology extra structure. For reasons I don’t particularly understand, if $D_1, D_2$ are two derivations, then

$f(a, b) = D_1(a) D_2(b)$

is a Hochschild $2$-cocycle. Moreover, if either of the derivations $D_i$ is inner, then $f$ turns out to be a $2$-coboundary. So the above map descends to a “cup product”

$H^1(A, A) \otimes H^1(A, A) \to H^2(A, A)$.

Hodge decomposition

Now suppose $A$ is commutative. Taking the opposite algebra of a first-order deformation gives another first-order deformation; concretely, if $f(a, b)$ is a $2$-cocycle, then $f(b, a)$ is also a $2$-cocycle. Then (and this is where the hypothesis that $\text{char}(k) \neq 2$ is important) every $2$-cocycle admits a canonical decomposition

$\displaystyle f(a, b) = \frac{f(a, b) + f(b, a)}{2} + \frac{f(a, b) - f(b, a)}{2}$

into symmetric and alternating $2$-cocycles. Moreover, since $A$ is commutative, trivial deformations of $A$ are commutative, so the $2$-coboundaries all lie in the symmetric subspace. Hence we can identify a direct summand $H^2_s(A, A)$ of the second Hochschild cohomology consisting of the quotient of the symmetric $2$-cocycles by the $2$-coboundaries. This is known as second Harrison cohomology, and describes commutative deformations of $A$.

On the other hand, since all $2$-coboundaries are symmetric, it follows that the alternating $2$-cocycles map injectively into Hochschild cohomology. Moreover, since

$\displaystyle a \star b - b \star a = \epsilon (f(a, b) - f(b, a))$

is a commutator, it follows that alternating $2$-cocycles are biderivations (derivations in each variable separately). On the other hand, it is straightforward to verify that any alternating biderivation is a $2$-cocycle. It follows that we have a “Hodge decomposition”

$\displaystyle H^2(A, A) \cong \text{ABDer}(A, A) \oplus H^2_s(A, A)$.

Notably, $\text{ABDer}(A, A)$ is itself a space of functions rather than a quotient of a space of functions by another space of functions, so it is a little more concrete to work with than Hochschild or Harrison cohomology. In addition, biderivations are uniquely determined by what they do to generators of $A$, so it is easier to write down biderivations than general $2$-cocycles.

Example. If $D_1, D_2$ are two derivations $A \to A$, then $\frac{D_1(a) D_2(b) - D_1(b) D_2(a)}{2}$ is an alternating biderivation. Note that this is just the image of the cup product in $\text{ABDer}(A, A)$.

As it turns out, a large class of commutative algebras, the “smooth algebras,” have the property that $H^2_s(A, A) = 0$ (so have no nontrivial commutative first-order deformations). For such algebras, $H^2(A, A)$ can be understood very concretely as the space of alternating biderivations. I believe such algebras include the algebras of functions on a smooth affine variety, but I don’t understand these issues well yet.

Poisson algebras

A second-order deformation of $A$ is an associative $k[\epsilon]/\epsilon^3$-bilinear product on $A[\epsilon]/\epsilon^3$ such that quotienting by $\epsilon$ we obtain the original product on $A$. Writing such a deformation as

$\displaystyle a \star b = ab + \epsilon f_1(a, b) + \epsilon^2 f_2(a, b)$

its first-order part $f_1(a, b)$ is necessarily a Hochschild $2$-cocycle, but one which satisfies an additional condition: the second-order part of the identity is $(a \star b) \star c - a \star (b \star c) = 0$ gives

$\displaystyle f_1(f_1(a, b), c) - f_1(a, f_1(b, c)) = f_2(a, bc) + a f_2(b, c) - f_2(ab, c) - f_2(a, b) c$.

Functions $A^{\otimes 3} \to A$ of the form on the RHS are Hochschild $3$-coboundaries, and in fact this condition can be interpreted to mean that the associator of $f_1$ must be zero in $H^3(A, A)$ (which it would take us too far afield to define here). In particular, if $H^3(A, A) = 0$, then every first-order deformation of $A$ extends to a second-order deformation.

The above condition on $f_1(a, b)$ implies another condition which we are more interested in for the time being. Namely, the commutator

$\displaystyle [a, b] = \epsilon (f_1(a, b) - f_1(b, a)) + \epsilon^2 (f_2(a, b) - f_2(b, a))$

has first-order part $\epsilon (f_1(a, b) - f_1(b, a)) = \epsilon \{ a, b \}$, and after a quick calculation, it turns out that the fact that $[a, b]$ satisfies the Jacobi identity implies that $\{ a, b \}$ does as well.

This suggests the following definition. A Poisson bracket on $A$ is a Lie bracket $\{ a, b \}$ which is also a biderivation. In formulas, $\{ a, b \} : A \otimes A \to A$ is a $k$-bilinear map satisfying

1. $\{ a, a \} = 0$
2. $\{ a, bc \} = \{ a, b \} c + b \{ a, c \}$
3. $\{ a, \{ b, c \} \} = \{ \{ a, b \}, c \} + \{ b, \{ a, c \} \}$.

A Poisson algebra is an algebra equipped with a Poisson bracket. Any noncommutative algebra has a canonical Poisson bracket given by the commutator $ab - ba$, but significantly, even if $A$ is commutative, it can still admit a nontrivial Poisson bracket if it admits any second-order deformations which are noncommutative to first order.

Poisson algebras are the natural setting for the most general form of the Heisenberg picture. Given a Poisson algebra $A$ and a Hamiltonian $H$, the derivation $\{ -, H \}$ is now the infinitesimal generator of time evolution. The procedure above, where we took the first-order part of a noncommutative deformation, also elegantly explains how Poisson brackets on a quantum system descend to Poisson brackets on a classical system.

Hamiltonian mechanics

Let’s see how this works more explicitly. We return to the setting of a quantum particle in one dimension in a potential. Recall that in this case the Hamiltonian is given by

$\displaystyle H = \frac{p^2}{2m} + V(x)$

where $x, V(x)$ are multiplication operators and $p = - i \hbar \frac{d}{dx}$. Suppose for simplicity that $V(x)$ is a polynomial in $x$. Then all of the operators we care about are more or less contained in the algebra of operators generated by $x, p$ subject to the relation

$\displaystyle [x, p] = i \hbar$.

Taking the classical limit as above with $\epsilon = i \hbar$ we obtain the commutative algebra $\mathbb{C}[x, p]$ with Poisson bracket uniquely defined by $\{ x, p \} = 1$ and extended via the Leibniz rule. Recall that $\{ a, b \} = D_1(a) D_2(b) - D_1(b) D_2(a)$ is a Poisson bracket for any two derivations $D_1, D_2$; this particular Poisson bracket can be written

$\displaystyle \{ a, b \} = \frac{\partial a}{\partial x} \frac{\partial b}{\partial p} - \frac{\partial b}{\partial x} \frac{\partial a}{\partial p}$

for all $a, b \in \mathbb{C}[x, p]$. We find that

$\displaystyle \{ x, H \} = \frac{\partial H}{\partial p} = \frac{p}{m}$

$\displaystyle \{ p, H \} = - \frac{\partial H}{\partial x} = -V'(x)$.

Since we divided by $i \hbar$, the infinitesimal generator of time evolution is now given by $\{ -, H \}$, and so we have recovered Hamilton’s equations (as well as Newton’s second law, again) from the Heisenberg picture.

Note that the Poisson bracket defined above naturally extends to the algebra $C^{\infty}(\mathbb{R}^2)$ of smooth functions on the classical phase space $\{ (x, p) \in \mathbb{R}^2 \}$ of a classical particle in one dimension. More generally, given a smooth manifold $M$, there is a natural Poisson algebra structure on the algebra $C^{\infty}(T^{\ast}(M))$ of smooth functions on the cotangent bundle of $M$ induced by a canonical symplectic form. In other words, the cotangent bundle is a Poisson manifold.

A natural question to ask about Poisson manifolds, and more generally about commutative Poisson algebras, is whether they admit a unique formal deformation whose first-order part gives the corresponding Poisson bracket; this is then the unique deformation quantization of the classical system described by the Poisson algebra. That this is true for Poisson manifolds is a deep result of Kontsevich.