# Induced Lie Algebra Representation, Left invariant vector fields and more…

The following is an excerpt from a proof in John Lee's Introduction to Smooth Manifolds I am struggling to understand. I would appreciate if someone was able to help me with whatever it is I am missing!

Theorem: Let $G$ be a Lie group, let $\mathfrak{g}$ be its Lie algebra, and let $\text{Ad}:G\rightarrow GL(\mathfrak{g})$ be the adjoint representation of $G$. The induced Lie algebra representation $\text{Ad}_*:\mathfrak{g}\rightarrow \mathfrak{gl(g)}$ is given by $\text{Ad}_*=\text{ad}$.

Here, $\text{ad}:\mathfrak{g}\rightarrow \mathfrak{gl}(g)$ is the adjoint representation of the Lie algebra $\mathfrak{g}$ defined by $\text{ad}(X):\mathfrak{g}\rightarrow \mathfrak{g}$,$\text{ad}(X)Y=[X,Y]$.

Outline of beginning of proof:

He shows earlier that $\text{Ad}:G\rightarrow GL(\mathfrak{g})$ is a Lie group representation, particularly, it is a Lie group homomorphism. Thus, the quantity $\text{Ad}_* X$ is well defined, being the unique element of $\mathfrak{gl(g)}$ that is $\text{Ad}$-related to $X$. He says (not only here, but uses this terminology a lot) that $\text{Ad}_* X$ is 'determined by its value at the identity'. What I think he means by this is because it's left invariant, its value at any point $g$ is found by computing $(dL_g)_e (\text{Ad}_* X)_e$?? Though I don't fully grasp why this is relevant.

Moving on, he writes, because $t\mapsto \exp tX$ is a smooth curve in $G$ whose velocity vector at $t=0$ is $X_e$, we can compute the action of $\text{Ad}_* X$ on an element $Y\in \mathfrak{g}$ by

\begin{align*} (\text{Ad}_* X)Y&=\left(\frac{d}{dt}\bigg|_{t=0} \text{Ad}(\exp tx)\right)Y. \end{align*}

He then continues the proof. I do not understand how he gets from the left hand side to the right and side in the above equation.

This is what I have got so far, let $\gamma$ be the integral curve generated by $X$,

\begin{align*}\frac{d}{dt}\bigg|_{t=0} \text{Ad}(\exp tX)&=\frac{d}{dt}\bigg|_{t=0}\text{Ad}(\gamma(t))\\ \\ &=(\text{Ad}\circ \gamma)^{\prime}(0)\\ \\ &=d\text{Ad}_e (\gamma^{\prime}(0))\\ \\ &=d\text{Ad}_e(X_e) \end{align*}

This is where I get stuck. I don't see how the above quantity is $\text{Ad}_* X$. As I understand it, the formula for $\text{Ad}_* X$ is given by

\begin{align*} \text{Ad}_* X=(d\text{Ad}_e X_e)^L, \end{align*} where the value at a point $g$ is

\begin{align*} (\text{Ad}_* X)_g=(d\text{Ad}_e X_e)^L_g=(dL_g)_e\left(d\text{Ad}_e X_e\right) \end{align*}

I can see some similar terms lying around but don't understand exactly what is going on here. Is my calculation correct? If it is, it seems he is only evaluating $\text{Ad}_* X$ at the identity? Is he only evaluating at the identity because it is 'determined by its value at the identity' and can somehow conclude later that because equality holds at the identity it holds everywhere? It's clear I am missing something crucial here so if anyone could help it would be much appreciated!