Coadjoint action of Lie group on dual of its Lie algebra

From Marsden and Ratiu's $$\textit{Introduction to Mechanics and Symmetry}$$:

The $$\textbf{coadjoint representation}$$ of $$G$$ on $$\mathfrak{g}^*$$, the dual of the Lie algebra $$\mathfrak{g}$$ is defined as follows. Let $$\textrm{Ad}_g^*:\mathfrak{g}^* \rightarrow \mathfrak{g}^*$$ be the dual of $$\textrm{Ad}_g$$ defined by $$\langle \textrm{Ad}_g^*\alpha, \xi\rangle = \langle \alpha, \textrm{Ad}_g \xi \rangle$$ for $$\alpha \in \mathfrak{g}^*$$ and $$\xi \in \mathfrak{g}$$. Then the map $$\Phi^*:G\rightarrow GL(\mathfrak{g}^*,\mathfrak{g}^*), \:\:\:\:\:\:\textrm{Ad}_{g^{-1}}^*=(T_e(R_g \circ L_{g^{-1}}))^*$$

$$\textbf{Question}:$$ How are the natural pairings given above related to the map $$\Phi^*$$? What is being taken in and what is being produced by the mapping? Also, in the definition of $$\Phi^*$$, why is there a $$g^{-1}$$ written in $$\textrm{Ad}_{g^{-1}}^*$$. Also, what is the actual representation of $$g$$ on $$\mathfrak{g}^*$$? It should be a matrix, yes? I don't see it. My apologies if the answers to these questions seem obvious to some but to me, this definition seems almost deliberately opaque and any clarification will be much appreciated.

• Check that $\mathrm{Ad}^*_g\circ \mathrm{Ad}^*_h= \mathrm{Ad}^*_{hg}$, but that $\mathrm{Ad}^*_{g^{-1}}\circ \mathrm{Ad}^*_{h^{-1}}= \mathrm{Ad}^*_{(gh)^{-1}}$. So without the $^{-1}$ you do not have a group homomorphism. I don't understand what you mean with "It should be a matrix", $GL(V,V)$ is the space of invertible linear maps $V\to V$, if you choose a basis of $V$ you may identify this space with matrices, but this identification depends on the choice of basis. As to the relation of $\Phi^*$ to the pairing$$\langle \Phi^*(g) \alpha, \xi\rangle = \langle \alpha, Ad_{g^{-1}} \xi\rangle$$ Aug 19, 2019 at 10:45

1 Answer

This definition is compressing different steps together and IMO it's cleaner to separate them, especially if e.g. taking duals of representations is unfamiliar.

The dual linear map

In general, if $$V$$ is a finite-dimensional vector space over a field $$k$$, we can take its dual to get the dual vector space $$V^{\ast} = \text{Hom}(V, k)$$. This operation is a contravariant functor, meaning it also acts on linear maps but in a way that reverses the order of composition, as follows: if $$f : V \to W$$ is a linear map then it induces a dual linear map

$$f^{\ast} : W^{\ast} \ni w^{\ast} \mapsto \left( v \mapsto w^{\ast}(f(v)) \right) \in V^{\ast}.$$

Equivalently, $$f^{\ast}$$ is defined by the adjoint-like property that

$$f^{\ast}(w^{\ast})(v) = w^{\ast}(f(v))$$

where I've written the dual pairing between vectors in $$V$$ and dual vectors in $$V^{\ast}$$ as composition; if you write the dual pairing as something more like $$\langle -, - \rangle$$ (which I would personally avoid to avoid confusion with inner products), then this condition can be rewritten

$$\langle f^{\ast}(w^{\ast}), v \rangle = \langle w^{\ast}, f(v) \rangle$$

which is how it's written in Marsden and Ratiu.

This duality operation is a coordinate-free version of taking transposes; if you work with a basis of $$V$$ and a basis of $$W$$ and the corresponding dual bases of $$V^{\ast}$$ and $$W^{\ast}$$, you'll find that the matrix of $$f^{\ast}$$ with respect to the dual bases is precisely the transpose of the matrix of $$f$$ with respect to the bases.

In particular, like taking transposes, it reverses the order of composition: we have $$(fg)^{\ast} = g^{\ast} f^{\ast}$$.

Dual linear representations

Now we generalize to actions and representations. If $$V$$ is a finite-dimensional vector space over $$k$$ and $$M$$ is a monoid acting on it (the point of generalizing to this case is to clarify a point which is easy to ignore if we jump straight to groups), with action $$\rho(m) : V \to V$$, then the dual vector space $$V^{\ast} = \text{Hom}(V, k)$$ canonically acquires an action of the opposite monoid $$M^{op}$$ ($$M$$ with the opposite multiplication), by taking duals:

$$M^{op} \ni m \mapsto \rho(m)^{\ast} \in \text{End}(V^{\ast}).$$

This reverses the order of multiplication, which is why it's only an action if we take the opposite monoid.

If $$M$$ is a group $$G$$, then a funny thing happens, which is that every group $$G$$ is canonically isomorphic to its opposite group $$G^{op}$$, via the map $$g \mapsto g^{-1}$$. So the standard convention is that $$V^{\ast}$$ is a representation of $$G$$, rather than $$G^{op}$$, defined now by

$$G \ni g \mapsto \rho(g^{-1})^{\ast} \in \text{GL}(V^{\ast}).$$

This is called the dual representation of $$V$$. Very concretely, the idea is that we need to both take inverses and take duals (two operations which reverse the order of multiplication) in order to make sure that this is a group homomorphism.

The coadjoint representation

Now let's separate the construction of the coadjoint representation into two conceptually independent steps:

1. Construct the adjoint representation $$\text{Ad} : G \to \text{GL}(\mathfrak{g})$$.
2. Take its dual representation $$\text{Ad}^{\ast} : G \to \text{GL}(\mathfrak{g}^{\ast})$$.