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If we have that $G$ is a connected linear group and $H<G$, where $H$ is also connected, with $\mathfrak{h}$ the lie algebra of $H$ and we define the centralizers of the elements in the following way:

$Z(H):=\{a\in G| aha^{-1}=h\forall h\in H\}$ and $Z(\mathfrak{h})=\{a\in G|ad(a)Y=Y\forall Y\in \mathfrak{h}\}$

I have a question which is asking me to show that these two are the same but I am unsure how to go about doing this, is there some property of the adjoint that I am not seeing here that may be useful?

Thanks for any help

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Is $H$ connected? –  Eric O. Korman May 18 '13 at 18:21
    
@EricO.Korman yes sorry I had not realised I has missed that out I will edit my question now thanks –  hmmmm May 19 '13 at 9:50

1 Answer 1

up vote 1 down vote accepted

For these sorts of questions, you want to transition between Lie algebras and Lie groups using the exponential map. The useful facts for this problem are:

  1. $\exp(Ad_a Y) = a (\exp Y) a^{-1}$. This follows from $Ad_a$ being the derivative of conjugation and the naturality of $\exp$.
  2. If $X \in \mathfrak g$ then $X \in \mathfrak h \subset\mathfrak g$ if and only if $\exp X \in H$.
  3. A connected Lie group is generated by the image of the exponential map.

So for this problem, suppose first $a \in Z(H)$ and let $Y \in \mathfrak h$. Then by 1., $$ \exp(Ad_a Y) = a \exp(Y) a^{-1} \in H $$ so that $Ad_a Y \in \mathfrak h$ by 2. Thus $a \in Z(\mathfrak h)$.

Conversely, suppose $a \in Z(\mathfrak h)$ and let $h \in H$. Then since $H$ is connected, by 3. we can write $h = \exp(Y_1) \cdots \exp(Y_n)$ for $Y_j \in \mathfrak h$, giving us $$ a h a^{-1} = a\exp(Y_1) \cdots \exp(Y_n)a^{-1} = (a \exp(Y_1) a^{-1})(a \exp(Y_2) a^{-1}) \cdots (a \exp(Y_n) a^{-1}) $$ which lies in $H$ since each $a \exp(Y_j) a^{-1} \in H$. Thus $a \in Z(H)$.

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