1

I want to show that if $G$ is a connected Lie group, then its centre, $Z(G)$ equals $Ker \ Ad$, where Ad is the adjoint function define by:

$$\forall X \in G, Y \in \mathfrak{g}: Ad(X)(Y) = XYX^{-1}$$

Now, I know also the next few relations:

$$ X \in \mathfrak{g}: \ Ad(e^X) = e^{ad(X)}$$

$$ ad(X)(Y)= [X,Y]$$

Now, from the fact that $G$ is connected I know that each $A \in G$ can be written as:

$$\exists X_1,\cdots,X_m \in \mathfrak{g}: \ A=e^{X_1}\cdots e^{X_m}$$

So now I want to show that for every $B \in G, \ A\in \mathfrak{g}: \ AB=BA$.

Now I see that: $Ad(AB)=Ad(BA)$ cause $Ad$ is homomorphism and $Ad(A)=Id$.

But I don't see how does this imply that $AB=BA$.

Any hints or advice is much helpful.

Thanks in advance.

MathematicalPhysicist
  • 4,191

1 Answers1

3

Let $A\in G$, $X\in \frak{g}$. If $Ad(A)=Id$, then $X = Ad(A)(X) =AXA^{-1}$. Now consider

\begin{align*} e^{AXA^{-1}} &= 1 + AXA^{-1} + \frac{1}{2}(AXA^{-1})^2 + \cdots \\ &= 1 + AXA^{-1} + \frac{1}{2}AX^2A^{-1} + \frac{1}{3!}AX^3A^{-1} + \cdots\\ &= A (1+ X + \frac{1}{2}X^2 + \frac{1}{3!}X^3+\cdots)A^{-1} = Ae^XA^{-1} \end{align*}

So if $Ad(A)=Id$, then $e^X = Ae^XA^{-1}$, so $Ae^X = e^XA$ for any $X\in \frak{g}$. Since any element $B\in G$ can be written as $e^{X_1}e^{X_2}\cdots e^{X_n}$, and $A$ commutes with each $e^{X_k}$, then $A$ commutes with their product: $AB = BA$, and since $B$ was arbitrary, we have $A\in Z(G)$.

Now suppose $A\in Z(G)$, so $AB=BA$ for all $B\in G$. In particular $Ae^X=e^XA$ for every $X\in \frak{g}$ so that $e^X = Ae^X A^{-1}$. Introducing a real parameter $t$, we have $e^{tX} = Ae^{tX} A^{-1}$. Taking the derivative of both sides with with respect to $t$: $Xe^{tX} = AXe^{tX}A^{-1}$. Setting $t=0$ we get $X = AXA^{-1}$, so $Ad(A)(X)=A$. Since $X$ was arbitrary, $Ad(A)=Id$ and so $A\in \operatorname{Ker}(Ad)$.

Zavosh
  • 5,966