15.5 The dual of a Lie algebra and symplectic geometry

15.5 The dual of a Lie algebra and symplectic geometry

Concept links · terms present in this machine draft; source roles are unverified: vector space · dual space · Lie algebra · adjoint representation · group action · Lie bracket

We have been careful to keep track of the diference between phase space $M={\mathbf{R}}^{2d}$ and its dual $\mathcal{M}=M^{\ast }$ , even though the symplectic form provides an isomorphism between them (see equation 14.6). One reason for this is that it is $\mathcal{M}=M^{\ast }$ that is related to the Heisenberg Lie algebra by

$${\mathfrak{h}}_{2d+1}=\mathcal{M}\oplus \mathbf{R}$$

with $M$ the linear functions on phase space, $R$ the constant functions, and the Poisson bracket the Lie bracket. It is this Lie algebra that we want to use in chapter 17 when we define the quantization of a classical system.

Another reason to carefully keep track of the diference between M and $M$ is that they carry two diferent actions of the Heisenberg group, coming from the fact that the group acts quite diferently on its Lie algebra (the adjoint action) and on the dual of its Lie algebra (the “co-adjoint” action). On $M$ and M these actions become:

• The Heisenberg group $H_{2d+1}$ acts on its Lie algebra ${\mathfrak{h}}_{2d+1}=\mathcal{M}\oplus \mathbf{R}$ by the adjoint action, with the diferential of this action given as usual by the Lie bracket (see equation 5.4). Here this means

$$\begin{array}{rl}ad\left(\left(\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q}{{\mathbf{c}}_p}\right),c\right)\right)\left(\left(\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q^{\prime }}{{\mathbf{c}}_p^{\prime }}\right),c^{\prime }\right)\right)& =\left[\left(\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q}{{\mathbf{c}}_p}\right),c\right),\left(\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q^{\prime }}{{\mathbf{c}}_p^{\prime }}\right),c^{\prime }\right)\right]\\ & =\left(\left(\genfrac{}{}{0ex}{}{0}{0}\right),\Omega \left(\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q}{{\mathbf{c}}_p}\right),\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q^{\prime }}{{\mathbf{c}}_p^{\prime }}\right)\right)\right)\end{array}$$

This action is trivial on the subspace $\mathcal{M},$ taking

$$\left(\genfrac{}{}{0ex}{}{{\mathbf{c}}_q^{\prime }}{{\mathbf{c}}_p^{\prime }}\right)\to \left(\genfrac{}{}{0ex}{}{0}{0}\right)$$

• The simplest way to define the “co-adjoint” action in this case is to define it as the Hamiltonian action of $H_{2d+1}$ on $M={\mathbf{R}}^{2d}$ such that its moment map ${\mu }_L$ is just the identification of ${\mathfrak{h}}_{2d+1}$ with functions on $M$ . For the case $d=1$ , one has

$${\mu }_L=L=c_{q}q+c_{p}p+c\in {\mathfrak{h}}_3$$

and

$$X_{{\mu }_L}=-c_q\frac{\partial }{\partial p}+c_p\frac{\partial }{\partial q}$$

This is the action described in equations 15.3 and 15.4, satisfying

$$q\left(\left(\left(\genfrac{}{}{0ex}{}{x}{y}\right),z\right)\cdot m\right)=q(m)+y,p\left(\left(\left(\genfrac{}{}{0ex}{}{x}{y}\right),z\right)\cdot m\right)=p(m)-x\text{\hspace{1em}(15.13)}$$

Here the subgroup of elements of $H_3$ with $x=z=0$ acts as the usua translations in position $q$.

It is a general phenomenon that for any Lie algebra $g$, a Poisson bracket on functions on the dual space ${\mathfrak{g}}^{\ast }$ can be defined. This is because the Leibniz property ensures that the Poisson bracket only depends on Ω, its restriction to linear functions, and linear functions on ${\mathfrak{g}}^{\ast }$ are elements of g. So a Poisson bracket on functions on ${\mathfrak{g}}^{\ast }$ is given by first defining

$$\Omega (X,X^{\prime })=[X,X^{\prime }]\text{\hspace{1em}(15.14)}$$

for $X,X^{\prime }\in \mathfrak{g}=({\mathfrak{g}}^{\ast }{)}^{\ast }$ , and then extending this to all functions on ${\mathfrak{g}}^{\ast }$ by the Leibniz property.

Such a Poisson bracket on functions on the vector space ${\mathfrak{g}}^{\ast }$ is said to provide a “Poisson structure” on ${\mathfrak{g}}^{\ast }$ . In general it will not provide a symplectic structure on ${\mathfrak{g}}^{\ast }$ , since it will not be non-degenerate. For example, in the case of the Heisenberg Lie algebra

$${\mathfrak{g}}^{\ast }=M\oplus \mathbf{R}$$

and Ω will be non-degenerate only on the subspace $M,$ the phase space, which it will give a symplectic structure.

Digression. The Poisson structure on ${\mathfrak{g}}^{\ast }$ can often be used to get a symplectic structure on submanifolds of ${\mathfrak{g}}^{\ast }$ . As an example, take $\mathfrak{g}=\mathfrak{so}(3)$ , in which case ${\mathfrak{g}}^{\ast }={\mathbf{R}}^3$ , with antisymmetric bilinear form $\omega$ given by the vector cross-product. In this case it turns out that if one considers spheres of fixed radius in ${\mathbf{R}}^3$ $\omega$ provides a symplectic form proportional to the area two-form, giving such spheres the structure of a symplectic manifold.

This is a special case of a general construction. Taking the dual of the adjoint representation Ad on $g$, there is an action of $g\in G$ on ${\mathfrak{g}}^{\ast }$ by the representation $A{d}^{\ast }$ , satisfying

$$(A{d}^{\ast }(g)\cdot l)(X)=l(Ad(g^{-1})X)$$

This is called the ${}^{\omega }co-adjoin{t}^{\prime \prime }$ action on ${\mathfrak{g}}^{\ast }$ . Picking an element $l\in {\mathfrak{g}}^{\ast }$ , the orbit ${\mathcal{O}}_l$ of the co-adjoint action turns out to be a symplectic manifold. It comes with an action of $G$ preserving the symplectic structure (the restriction of the co-adjoint action on ${\mathfrak{g}}^{\ast }$ to the orbit). In such a case the moment map

$$\mu :{\mathcal{O}}_l\to {\mathfrak{g}}^{\ast }$$

is just the inclusion map. Two simple examples are

• For $\mathfrak{g}={\mathfrak{h}}_3,$ , phase space $M={\mathbf{R}}^2$ with the Heisenberg group action of equation $15.13$ is given by a co-adjoint orbit, taking $l\in {\mathfrak{h}}_3^{\ast }$ to be the dual basis vector to the basis vector of h given by the constant function 1 on ${\mathbf{R}}^2$

• For $\mathfrak{g}=\mathfrak{so}(3)$ the non-zero co-adjoint orbits are spheres, with radius the length of l, the symplectic form described above, and an action of $G=$ $SO(3)$ preserving the symplectic form.

Note that in the second example, the standard inner product on ${\mathbf{R}}^3$ provides an SO(3) invariant identification of the Lie algebra with its dual, and as a result the adjoint and co-adjoint actions are much the same. In the case of $H_3,$ , there is no invariant inner product on ${\mathfrak{h}}_3,$ so the adjoint and co-adjoint actions are rather diferent, explaining the diferent actions of $H_3$ on $M$ and M described earlier.

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