Associated Adjoint Bundle

We may let $G$ act as a general linear transformation on its own Lie algebra ${\frak g}$, given a $Y\in{\frak g}$ and $g\in G$ we define

\[\text{Ad}_ {g}(Y) = g Y g^{-1} = L_ {g\ast } \,\circ\,R_ {g^{-1} \ast } Y.\]

Obviously if $G$ is abelian then $\text{Ad}_ {g}$ does not do anything to $Y$. This map $\text{Ad}$ takes an element of the group $G$, then uses it to “distort” a tangent vector in $T_ {e}G$, thus

\[\text{Ad}: G \to GL({\frak g}), \quad g\mapsto \text{Ad}_ {g}\]

where $\text{GL}({\frak g})$ is the group of general linear transformations of ${\frak g}$.

One can check that $\text{Ad}_ {g}$ is an representation of $G$, which is essentially the adjoint representation of $G$. You might have already encountered different definition of adjoint representation of $G$ before, then you should be able to show that these definitions indeed agrees with each other. The subgroup $\text{Ad}_ {G}\subset GL({\frak g})$ is called the adjoint group of $G$.

Furthermore, we can talk about the induced man, or differential, or “push-forward” of $\text{Ad}$. Now, $\text{Ad}_ {\ast}$ would map $TG$ to $TGL({\frak g})$. But how does it work?

Given $g\in G$, we have the one-parameter subgroup $g(t)=\exp(tX)$ where $t$ is the parameter, and the corresponding velocity vector at the identity is

\[X =\frac{d}{d t} g(t) = \frac{d}{d t} \exp(tX) {\Large\mid}_ {t=0}, \quad X \in {\frak g}.\]

Write $g(\varepsilon)$ to denote the group element infinitesimally away from the identity $e$. $\text{Ad}$ maps $e$ to the identity linear transform

\[\text{Ad}_ {e} Y = e Y e^{-1} =Y\]

and maps $g(\varepsilon)$ to some other linear transform

\[\text{Ad}_ {g(\varepsilon)} = g(\varepsilon) Y g(\varepsilon)^{-1} = : Y'\]

which is not the same as $Y$. The induced map $\text{Ad}_ {\ast}$ should give the difference between $Y’$ and $Y$, which are both element of ${\frak g}$, hence so should be their difference,

\[\text{Ad}_ { {\ast } }: X \mapsto \text{some linear transform of } Y.\]

Specific calculation shows that

\[\text{Ad}_ {\ast }(X) Y = \frac{d}{d t} e^{ tX } Y e^{ -tX }{\Large\mid}_ {t=0}=XY-YX = [X,Y].\]

We can also write $\text{Ad}_ {\ast}(X)$ as $\text{Ad}_ {X}$.

Let us write $\text{ad}(X)$ or $\text{ad}_ {X}$ for the above linear transformation. Thus

\[\text{ad}: {\frak g}\to{\frak g},\quad X\mapsto[X,-].\]

In summary,

\[\text{Ad}_ {\exp(tX)} Y =e^{ tX } Y e^{ -tX }\]

is a curve in ${\frak g}$, its velocity is also in ${\frak g}$. For fixed $X$, the one parameter group

\[\text{Ad}_ {\exp(tX)}\]

is generated by $\text{ad}_ {X}$.