# Adjoint representation and tangent vectors

Let $G$ be a Lie group, $\mathfrak{g}$ its Lie algebra, $\text{Ad}:G\rightarrow GL(\mathfrak{g})$ the adjoint representation of $G$. Then, for $X,Y\in \mathfrak{g}$,

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

I'm just wondering exactly why this equation is true?

All I'm able to say about the situation is that $\text{Ad}(\exp(tX))$ is a curve in $GL(\mathfrak{g})$, thus \begin{align*} \frac{d}{dt}\bigg|_{t=0}\text{Ad}(\exp(tX)) \end{align*} is a tangent vector in $T_{\text{Id}}GL(\mathfrak{g})$.

Any help here would be much appreciated.

## 1 Answer

The point is that "evaluating in $Y$" is (the restriction of) a linear map, so its derivative coincides with the map itself. This is not specific to a Lie algebra being involve. If $V$ is a vector space, $v\in V$ an element and $c:I\to GL(V)$ is a smooth curve, then the derivative of the curve $I\to V$ defined by $t\mapsto c(t)(v)$ is just $(c'(t))(v)$.

• Thanks for your response. I still don't really follow. I understand the derivative of a linear map is the map itself but can't really piece the rest together. Do you know of a resource I could read up on to get more comfortable with this? – beedge89 Apr 29 '15 at 10:00
• I don't know a nice reference. Keep in mind that $GL(\mathfrak g)$ is simply an open subset in the finite dimensional real vector space $L(\mathfrak g,\mathfrak g)$. So you really just use that for a smooth curve $c$ in $\mathbb R^n$ and a linear map $A:\mathbb R^n\to \mathbb R^m$ you have $(A\circ c)'(t)=A(c'(t))$. Viewing $GL(\mathfrak g)$ as a manifold is rather confusing at this point (although of course you have the same argument for the restriction of a linear map to a submnaifold). – Andreas Cap Apr 29 '15 at 11:41