15.3 Group actions on M and the moment map

15.3 Group actions on M and the moment map

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

Our fundamental interest is in studying the implications of Lie group actions on physical systems. In classical Hamiltonian mechanics, with a Lie group $G$ acting on phase space $M$, such actions are characterized by their derivative, which takes elements of the Lie algebra to vector fields on $M$. When these are Hamiltonian vector fields, equation 15.6 can often be used to instead take elements of the Lie algebra to functions on $M$. This is known as the moment map of the group action, and such functions on $M$ will provide our central tool to understand the implications of a Lie group action on a physical system. Quantization then takes such functions to operators which will turn out to be the important observables of the quantum theory.

Given an action of a Lie group $G$ on a space $M$ , there is a map

$$L\in \mathfrak{g}\to X_L$$

from g to vector fields on $M$. This takes L to the vector field $X_L$ which acts on functions on $M$ by

$$X_{L}F(m)=\frac{d}{dt}F(e^{tL}\cdot m{)}_{|t=0}\text{\hspace{1em}(15.10)}$$

This map however is not a homomorphism (for the Lie bracket 15.1 on vector fields), but an antihomomorphism. To see why this is, recall that when a group $G$ acts on a space, we get a representation $\pi$ on functions $F$ on the space by

$$\pi (g)F(m)=F(g^{-1}\cdot m)$$

The derivative of this representation will be the Lie algebra representation

$${\pi }^{\prime }(L)F(m)=\frac{d}{dt}F(e^{-tL}\cdot m{)}_{|t=0}=-X_{L}F(m)$$

so we see that it is the map

$$L\to {\pi }^{\prime }(L)=-X_L$$

that will be a homomorphism.

When the vector field $X_L$ is a Hamiltonian vector field, we can define:

Definition (Moment map)

Given an action of $G$ on phase space $M$, a Lie algebra homomorphism

$$L\to {\mu }_L$$

from g to functions on $M$ is said to be a moment map if

$$X_L=X_{{\mu }_L}$$

Equivalently, for functions $F$ on $M$, ${\mu }_L$ satisfies

$$\{{\mu }_L,F\}(m)=-X_{L}F=\frac{d}{dt}F(e^{-tL}\cdot m{)}_{|t=0}\text{\hspace{1em}(15.11)}$$

This is sometimes called a “co-moment map”, with the term “moment map” referring to a repackaged form of the same information, the map

$$\mu :M\to {\mathfrak{g}}^{\ast }$$

where

$$(\mu (m))(L)={\mu }_L(m)$$

A conventional physical terminology for the statement 15.11 is that “the function ${\mu }_L$ generates the symmetry $L^{\text{dprime}}$ , giving its infinitesimal action on functions.

Only for certain actions of $G$ on $M$ will the $X_L$ be Hamiltonian vector fields and an identity $X_L=X_{{\mu }_L}$ possible. A necessary condition is that $X_L$ satisfy equation 15.7

$$X_L\{g_1,g_2\}=\{X_L{g}_1,g_2\}+\{g_1,X_L{g}_2\}$$

Even when a function ${\mu }_L$ exists such that $X_{{\mu }_L}=X_L$ , it is only unique up to a constant, since ${\mu }_L$ and ${\mu }_L+C$ will give the same vector field. To get a moment map, we need to be able to choose these constants in such a way that the map

$$L\to {\mu }_L$$

is a Lie algebra homomorphism from g to the Lie algebra of functions on $M$. When this is possible, the G-action is said to be a Hamiltonian G-action. When such a choice of constants is not possible, the G-action on the classical phase space is said to have an “anomaly”.

Digression. For the case of M a general symplectic manifold, the moment map can still be defined, whenever one has a Lie group $G$ acting on $M$, preserving the symplectic form $\omega$. The infinitesimal condition for such a $G$ action is (see equation 15.9)

$${\mathcal{L}}_X\omega =0$$

Using the formula

$${\mathcal{L}}_X=(d+i_X{)}^2=d{i}_X+i_{X}d$$

for the Lie derivative acting on diferential forms $(i_X$ is interior product with the vector field $X$), one has

$$(d{i}_X+i_{X}d)\omega =0$$

and since $d\omega =0$ we have

$$d{i}_X\omega =0$$

When $M$ is simply-connected, one-forms $i_{X\omega }$ whose diferential is 0 (called “closed”) will be the diferentials of a function (and called ${}^{\prime \prime }exac{t}^{\prime \prime })$ . So there will be a function µ such that

$$i_X\omega (\cdot )=\omega (X,\cdot )=d\mu (\cdot )$$

although such a µ is only unique up to a constant.

Given an element $L\in \mathfrak{g}.$ , a G action on $M$ gives a vector field $X_L$ by equation $\mathrm{15.10.}$ When we can choose the constants appropriately and find functions ${\mu }_L$ satisfying

$$i_{X_L}\omega (\cdot )=d{\mu }_L(\cdot )$$

such that the map

$$L\to {\mu }_L$$

taking Lie algebra elements to functions on $M$ (with Lie bracket the Poisson bracket) is a Lie algebra homomorphism, then this is called the moment map. One can equivalently work with

$$\mu :M\to {\mathfrak{g}}^{\ast }$$

by defining

$$(\mu (m))(L)={\mu }_L(m)$$

Previous section: 15.2 Hamiltonian vector fields and canonical transformations · Next section: 15.4 Examples of Hamiltonian group actions · 中文版本

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