I am trying to work out how one obtains the adjoint representation of a Lie algebra from the adjoint representation of the Lie group. I apologise if this is a trivial question but I am a physicist so I am rusty.

Consider a Lie group $G$ and its corresponding Lie algebra $\mathfrak{g}$. The adjoint representation of $G$ is given by

$$ \operatorname{Ad}: G \rightarrow \operatorname{Aut}(\mathfrak{g}).$$

According to the Wiki article, I can obtain the adjoint representation $\operatorname{ad}$ of the Lie algebra from the pushforward of $\operatorname{Ad}$:

$$ \operatorname{ad} : = \operatorname{Ad}_* : \mathfrak{g} \rightarrow \operatorname{Der}(\mathfrak{g}), $$

where $\operatorname{Der}(g)$ is the derivation algebra.

Now, from my knowledge of the pushforward, I would say that the pushforward maps vectors in one tangent space, to vectors in another. So I would expect $\operatorname{Ad}_*$ to act as

$$ \operatorname{Ad}_* : \mathfrak{g} \rightarrow T_I(\operatorname{Aut}(\mathfrak{g})), $$

where $I$ is the identity of $\operatorname{Aut}(\mathfrak{g})$. What I do not understand is why the elements of $T_I(\operatorname{Aut}(\mathfrak{g}))$ could be interpreted as operators on $\mathfrak{g}$. In other words, I do not understand why $T_I(\operatorname{Aut}(\mathfrak{g})) \cong \operatorname{Der}(\mathfrak{g})$. From a differential geometry perspective, tangent vectors are objects which act on functions $f : M \rightarrow \Bbb R$, so where do we deduce that these tangent vectors of $T_I(\operatorname{Aut}(\mathfrak{g}))$ are in fact operators on $\mathfrak{g}$?

Theorem 2.3 of this states that $ A \in T_I(\operatorname{Aut}(\mathfrak{g})) $ then for $Y \in \mathfrak{g}$ it acts as

$$ A(Y) = \frac{\Bbb d}{\Bbb dt} \operatorname{Ad}(\exp tX) Y \bigg|_{t=0},$$

where $A = \operatorname{ad}(X)$. I do not understand whether this is a definition or can be deduced purely from $\operatorname{ad}$, but I do not know why this is how $\operatorname{ad}$ should act. It also seems like an abuse of notation here too, should this not be acting on some test function $f$?


1 Answer 1


A canonical identification is being made between a vector space and one of its tangent spaces.

Let $V$ be a finite dimensional vector space, and $M_V=\operatorname{End}(V)$ be the algebra of linear maps $V\to V$. $M_V$ is itself a finite dimensional vector space, so we have a canonical isomorphism $T_AM_V\cong M_V$ for each $A\in M_V$. $GL(V)$ and its Lie subgroups are submanifolds of $M_V$, so we may identify their Lie algebras with subspaces of $M_V$. Under this identification, the Lie bracket can be written in terms of multiplication in $M_V$ by $[A,B]=AB-BA$. This is why we can write the Lie algebras of linear algebraic groups ($\mathfrak{gl}(n)$, $\mathfrak{so}(n)$, etc.) as sets of matrices.

$\operatorname{Aut}(\mathfrak{g})\subset GL(\mathfrak{g})$ is a linear algebraic group, so, as above, we may identify $T_I\operatorname{Aut}(\mathfrak{g})$ with a subspace of $M_\mathfrak{g}$, the set of linear maps $\mathfrak{g}\to\mathfrak{g}$. With a bit of computation, one can show that these linear maps are in fact derivations on $\mathfrak{g}$.


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