Consider a (possibly infinite-dimensional) Lie group $\mathcal{G}$ and let $\mathcal{A}$ be an algebra with a product $\cdot$ and the bracket $[u,v]=u\cdot v - v\cdot u$.

The following statement is clear: If $\mathcal{G}\subset \mathcal{A}$, then a trace $\text{Tr}:\mathcal{A}\to\mathbb{R}$ is invariant under conjugation, i.e.

$$\text{Tr}(u\cdot v)=\text{Tr}(v\cdot u) \ \forall u,v\in\mathcal{A}\Rightarrow \text{Tr}(u\cdot v \cdot u^{-1})=\text{Tr}(v)\ \forall v\in\mathcal{A}, \forall u\in\mathcal{G}.$$

If $\mathcal{A}=\text{Lie}(\mathcal{G})$, then also the converse statement holds (i.e. a conjugation-invariant map $\text{Tr}$ is a trace on $\text{Lie}(G)$) -- provided the Lie group $\mathcal{G}$ admits some smooth exponential mapping.

With these two result at hand, we can conclude the following:

If $\mathcal{G}\subset\mathcal{A}$, the Lie group $\mathcal{G}$ admits some exponential mapping and $\mathcal{A}=\text{Lie}(G)$, then there is a bijection of traces and conjugation-invariant maps $\text{Tr}:\mathcal{A}\to\mathbb{R}$.

My question is whether we can have $\mathcal{G}\subset \text{Lie}(\mathcal{G})$?

cf. also Lemma 2, p. 13.

My intuition says no since $\text{Lie}(\mathcal{G})$ is the tangent space at the identity of $\mathcal{G}$; on the other hand, a linear map with trace property should be invariant under conjugation;


If $A$ is a finite-dimensional algebra over ${\mathbb R}$, equipped with its natural topology, and if $G$ is an open subgroup of $(A,\cdot)$ (that is, $(A^{\times})^0\subset G\subset A^{\times}$), then you have natural identifications $\text{Lie}(G)=\text{T}_{1_A}(G)\stackrel{\cong}{\to}\text{T}_{1_A}(A)\cong A$, and in particular, you may naturally identify $G$ with a subset of $\text{Lie}(G)$. Under this identification, the adjoint action of $G$ on $A$ is given by conjugation, while the adjoint action of $A$ on itself is given by the commutator of $A$, which is what you need to deduce part (2) from part (1) in the Lemma you cited.

I suspect that the same works for some classes of infinite-dimensional manifolds and Lie groups modelled on some well-behaved class of topological vector spaces, but I'm not really familiar with that.

  • $\begingroup$ I added a reference where in particular this statement is "shown" for infinite-dimensional Lie groups. There is no hint on the relation between $\text{Lie}(\mathcal{G})$ and the tangent space. $\endgroup$ – mikemike Oct 29 '14 at 12:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.