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Let $G$ be a connected Lie group with Lie algebra $\mathfrak g$. I'm trying to proof the following assertion:

the kernel of the adjoint representation of $G$ coincides with its center.

What I have done so far:

Let's recall that $Ad: G \to Aut(G)$ is given by $Ad(g) = d(Inn_g)$ where $Inn_g(h) = ghg^{-1}, \, h \in G,$ is the inner automorphism.

For the first one, if $g \in Z(G)$, then $gh = hg$ for all $h \in G$ what implies that $Inn_g (h) = ghg^{-1} = h = id_G(h)$. Hence $$ Ad(g) = d(Inn_g) = d(id_G) = id_G $$ and so $g \in \ker Ad$. I'm stuck in the converse of this assertion. Well, if $g \in \ker Ad$ we have that $Ad(g) = id_G$ and, since $G$ is connected, $G$ is spanned by $\exp [\mathfrak g]$. However I can't proceed from this.

Help?

user 242964
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1 Answers1

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Assume that $Ad_{g}$ is trivial, and let $h$ be an arbitrary element of $G$ of the form $h=exp(v)$. We will first show that $g\cdot h=h\cdot g$.

1) Show that $t\mapsto g\cdot exp(tv)\cdot g^{-1}$ is a one-parameter subgroup of $G$. Its velocity at time zero is given by $Ad_{g}(v)=v$, using that $Ad_{g}=Id_{\mathfrak{g}}$.

2) But also $t\mapsto exp(tv)$ is a one-parameter subgroup of $G$ whose velocity at time zero is $v$. Since one-parameter subgroups are uniquely determined by their initial velocity, we have $$ g\cdot exp(tv)\cdot g^{-1}=exp(tv). $$ In particular, for $t=1$ we get $$g\cdot h \cdot g^{-1}=h,$$ so $$ g\cdot h = h \cdot g $$

Now, since $G$ is connected, any element $h$ of $G$ can be written as $$ h=exp(v_{1}) \cdots exp(v_{k}) $$ for some $v_{1},\ldots,v_{k}\in\mathfrak{g}$. Using what we just proved, we have \begin{align} g\cdot h &= g\cdot exp(v_{1})\cdot exp(v_{2}) \cdots exp(v_{k})\\ &=exp(v_{1})\cdot g \cdot exp(v_{2})\cdots exp(v_{k})\\ &= \cdots\\ &=exp(v_{1})\cdots exp(v_{k})\cdot g\\ &=h\cdot g \end{align}

studiosus
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