Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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 $a \in Z(H)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.