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Constructing Poisson algebras

(Commutative) Poisson algebras are clearly very interesting, so it would be nice to have ways of constructing examples. We know that $k[x, p]$ is a Poisson algebra with bracket uniquely defined by $\{ x, p \} = 1$; this describes a classical particle in one dimension, and is the classical limit of a quantum particle in one dimension (essentially the Weyl algebra).

More generally, if $A, B$ are Poisson algebras, then the tensor product $A \otimes_k B$ can be given a Poisson bracket given by extending

$\displaystyle \{ a_1 \otimes b_1, a_2 \otimes b_2 \} = \{ a_1, a_2 \} \otimes b_1 b_2 + a_2 a_1 \otimes \{ b_1, b_2 \}$

linearly. At least when $A, B$ are unital, this Poisson algebra is the universal Poisson algebra with Poisson maps from $A, B$ such that the images of elements of $A$ Poisson-commute with the images of elements of $B$. In particular, it follows that $k[x_1, p_1, ..., x_n, p_n]$ is a Poisson algebra with the bracket

$\{ x_i, x_j \} = \{ p_i, p_j \} = 0, \{ x_i, p_j \} = \delta_{ij}$.

This describes a classical particle in $n$ dimensions, or $n$ different classical particles in one dimension, and it is the classical limit of a quantum particle in $n$ dimensions, or $n$ different quantum particles in one dimension.

Today we’ll discuss the question of how one might go about constructing Poisson brackets more generally.

Alternating biderivations

Recall that an alternating biderivation $\{ -, - \}$ on an algebra $A$ is an alternating bilinear map which is a derivation in each variable. Furthermore, recall that if $D_1, D_2$ are derivations on $A$, then $\{ a, b \} = D_1(a) D_2(b) - D_1(b) D_2(a)$ is an alternating biderivation. Thus we get a natural alternating map $\text{Der}(A, A) \times \text{Der}(A, A) \to \text{ABDer}(A, A)$ which factors through the exterior square

$\displaystyle \Lambda^2(\text{Der}(A, A)) \to \text{ABDer}$.

This map is injective since $D_1(a) D_2(b) - D_1(b) D_2(a) = 0$ for all $a, b \in A$ if and only if $D_1, D_2$ are scalar multiples of each other. A natural question is when this map is also surjective.

Proposition: The above map is surjective when $A = k[x_1, ... x_n]$ is a polynomial algebra.

Proof. Any alternating biderivation $\{ -, - \}$ on $A$ is determined by $\{ x_i, x_j \} = c_{ij}, i < j$. Furthermore,

$\displaystyle \{ a, b \} = \frac{\partial a}{\partial x_i} \frac{\partial b}{\partial x_j} - \frac{\partial b}{\partial x_i} \frac{\partial a}{\partial x_j}$

is an alternating biderivation such that $\{ x_i, x_j \} = 1$ and such that $\{ x_k, - \} = 0$ for all $k \neq i, j$. It follows that for any choice of elements $c_{ij} \in A, i < j$ there exists a unique alternating biderivation satisfying $\{ x_i, x_j \} = c_{ij}$ given by

$\displaystyle \{ a, b \} = \sum_{i < j} c_{ij} \left( \frac{\partial a}{\partial x_i} \frac{\partial b}{\partial x_j} - \frac{\partial b}{\partial x_i} \frac{\partial a}{\partial x_j} \right)$.

This is a very special case of the Hochschild-Kostant-Rosenberg theorem, at least once you know that the Harrison cohomology $H^2_s(A, A)$ of a polynomial algebra vanishes. But this isn’t difficult to see: any commutative first-order deformation is necessarily itself a polynomial algebra because no relations can exist between any lifts of the generators.

The following alternate perspective on the above result may be enlightening. Note that any derivation $D : A \to A$ factors through the $A$-module generated by formal symbols of the form $da, a \in A$ subject to the following relations:

1. $dc = 0$ whenever $c \in k$.
2. $d(a + b) = d(a) + d(b)$.
3. $d(ab) = a d(b) + d(a) b$.

This $A$-module is denoted $\Omega^1_{A/k}$ and known as the space of Kähler differentials. Since all of the above axioms are satisfied by the exterior derivative of a function on a smooth manifold, the intuition here is that $\Omega^1_{A/k}$ is analogous to the space of differential $1$-forms on $\text{Spec } A$. This fits in nicely with the intuition that derivations are analogous to vector fields on $\text{Spec } A$, since by definition a derivation $A \to A$ is an $A$-module morphism $\Omega^1_{A/k} \to A$, so there is a natural pairing between the two.

But now it’s clear that an alternating biderivation on $A$ is nothing more than an $A$-module morphism

$\displaystyle \{ -, - \} : \Lambda^2(\Omega^1_{A/k}) \to A$.

Thus alternating biderivations are dual to $2$-forms. Intuitively, they are therefore bivector fields on $\text{Spec } A$. If $\Omega^1_{A/k}$, as an $A$-module, behaves sufficiently similar to a finite-dimensional vector space over a field, then it follows that the dual of its exterior square ought to be isomorphic to the exterior square of its dual, which is what we showed above when $A$ is a polynomial algebra. In this case, $\Omega^1_{A/k}$ is in fact a free $A$-module on generators $dx_1, dx_2, ... dx_n$, which is what makes the above argument work abstractly. More generally I think the above argument goes through whenever $\Omega^1_{A/k}$ is finitely-generated and projective.

(Above I am glossing over the distinction between the exterior square and the space of alternating $2$-tensors. I haven’t yet made up my mind about when this distinction is worth making.)

The Jacobi identity

Now that we have a reasonably good grasp of alternating biderivations, what can we say about the ones that satisfy the Jacobi identity (and therefore are Poisson brackets)?

Proposition: An alternating biderivation on an algebra $A$ satisfies the Jacobi identity if and only if the Jacobi identity is satisfied when a set of generators of $A$ is plugged in.

Proof. It suffices to observe that the Jacobiator

$\displaystyle \{ a, \{ b, c \} \} + \{ b, \{ c, a \} \} + \{ c, \{ a, b \} \}$

is a triderivation: it is trilinear and satisfies the Leibniz rule in each of its components separately. (This is a straightforward computation using the Leibniz rule for $\{ a, b \}$.) Thus it is determined linearly by its values on a set of generators of $A$.

We record the following two immediate corollaries.

First, if $V$ is a vector space equipped with an alternating bilinear form $\omega : V \times V \to k$, then $\omega$ extends to a Poisson bracket on the symmetric algebra $S(V)$. (The Jacobi identity is clearly satisfied on generators since $\{ v, w \} \in k$ is a scalar for any $v, w \in V$.) These are precisely the polynomial Poisson algebras for which the Poisson bracket is graded with degree $-2$. In particular, if $V$ has a basis $x_1, ... x_n, p_1, ... p_n$ such that $\omega$ is given by

$\displaystyle \omega(x_i, x_j) = \omega(p_i, p_j) = 0, \omega(x_i, p_j) = \delta_{ij}$

then we get precisely the algebra of observables on $n$ classical particles as described earlier.

Second, if $\mathfrak{g}$ is a Lie algebra, then the Lie bracket $[-, -]$ extends to a Poisson bracket on the symmetric algebra $S(\mathfrak{g})$. These are precisely the polynomial Poisson algebras for which the Poisson bracket is graded with degree $-1$.

In both of these cases we can explicitly find a deformation quantization: that is, we can identify a formal deformation from which we get the above Poisson algebras as classical limits. This will be expanded on in later posts.

One Response

1. Great post. Can you recommend any references for this material?