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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;

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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.

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  • $\begingroup$ 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

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