Background: suppose $F:G\to H$ is a homomorphism of Lie groups. Then, there is an induced homomorphism $F_*:\mathfrak g\to \mathfrak h$ defined by

$Y_h:=(F_*X)_h=d(L_h\circ F_e)X_e \tag1\label{eq:1}$

This definition is the one it makes sense to use because it makes $X$ and $Y\ F$-related.

Now, if $X\in \mathfrak g,$ we have $X_e=\gamma'(0)$ where $\gamma(t)=\exp tX$, which means that

$(F_*X)_e=dF_e(X_e)=dF_e(\gamma'(0))=\frac{dF(\exp tX)}{dt}|_{t=0}\tag{2}\label{eq:2}$

Now, apply \eqref{eq:1} and \eqref{eq:2} to the following data: fix $g\in G$ and set $C_g:G\to G: x\mapsto gxg^{-1}$ and $\text{Ad}:G\to GL(\mathfrak g):g\mapsto (C_g)_*.$ Then, $\text{Ad}_*:\mathfrak g\to \mathfrak gl(\mathfrak g)$ satisfies

$(\text{Ad}_*X)_Y=d(L_Y\circ \text{Ad}_e)X_e \tag{3}\label{eq:3}$

$(\text{Ad}_*X)_e=\frac{d\text{Ad}(\exp tX)}{dt}|_{t=0}\tag{4}\label{eq:4}$

My question is simply this: in the course of his proof, Lee has instead of \eqref{eq:4} the following equation

$\text{Ad}_*X=\frac{d\text{Ad}(\exp tX)}{dt}|_{t=0}\tag{5}\label{eq:5}$

which does not make sense to me. Now, if we take $X$ to be a vector in $T_eG$, then we can find a curve $c$ in $G$ such that $c'(0)=X$ and so we get


and so if we take $c=\gamma$ where $\gamma(t)=\exp tX$, we get \eqref{eq:5}. But then, we are considering $X$ to be a left-invariant vector field again. Is this type of argument legitimate? If so, why? If not, why not?

I believe my confusion is that although the definition of the induced homomorphism $F_*X$ on a Lie algebra is obviously not the same as $(F_*X)_e$ and yet in defining the adjoint representation, Lee uses the former, in which case, I do not see clearly how $(5)$ obtains, whereas in another proof I have seen, one defines $\text{ad}(X) = (d \text{Ad})_e(X)$ (and then proves that $\text{ad}(X)Y=[X,Y]$) and then $(5)$ is just a consequence of the definiton. But of course, $(d \text{Ad})_e(X)\neq \text{Ad}_*X$.


1 Answer 1


There's an implicit identification going on here. For any finite-dimensional vector space $V$, let $GL(V)$ denote the Lie group of invertible linear transformations of $V$. There's a canonical Lie algebra isomorphism between the Lie algebra $\operatorname{Lie}(GL(V))$ of left-invariant vector fields on $GL(V)$ and the Lie algebra $\mathfrak g\mathfrak l(V)$ of endomorphisms of $V$ (under commutator bracket). (See Corollary 8.42 in my Introduction to Smooth Manifolds, 2nd ed.)

The equation you labeled as (5) above is using this identification for the Lie algebra of $GL(\mathfrak g)$. The right hand side is the initial velocity of a smooth curve in $GL(\mathfrak g)$. Considering the latter as an open subset of the vector space $\mathfrak g\mathfrak l(\mathfrak g)$, the velocity vector can be regarded unambiguously as an element of the vector space $\mathfrak g\mathfrak l(\mathfrak g)$ itself.

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    $\begingroup$ Got it! Thanks for taking the time to explain this in detail. I am self-studying this material and I have learned a great deal from your book. It is a real gem. $\endgroup$ Commented Mar 6, 2020 at 3:46

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