# Deducing the adjoint representation of a Lie algebra from the adjoint representation of the Lie group

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 $$f$$?

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 $$\mathfrak{g}$$.