Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have tried to prove the following problem:

Suppose $G$ is a connected matrix Lie group. Then the center of $\mathfrak{g}$ which is $\ker \textrm{ad}$ is equal to the Lie algebra of $Z(G)$.

enter image description here

Recall that $\textrm{ad} : \mathfrak{g} \to \text{gl}(\mathfrak{g})$ is the map that sends $X \in \mathfrak{g}$ to $\textrm{ad}_X$, where $\textrm{ad}_X$ is defined by

$$\textrm{ad}_X(Y) = [X,Y].$$ Similarly $\textrm{Ad}_A$ is the map such that for all $X \in \mathfrak{g}$, we have $\textrm{Ad}_A(X) = AXA^{-1}$. Now I have proven the problem above as follows: If $Y \in \ker \textrm{ad}$, then certainly for all $t \in \Bbb{R}$ we have $\textrm{ad}_{tY} = 0$. Now we have given any $x \in G$ that

$$\begin{eqnarray*} e^{tY}xe^{-tY} &=& \textrm{Ad}_{e^{tY}}(x) \\ &=& e^{\textrm{ad}_{tY}}(x) \\ &=& x \end{eqnarray*}$$

by assumption and since this holds for all $x$, we conclude that $e^{tY} \in Z(G)$ for all $t$ and so $Y$ is in the Lie algebra of $Z(G)$.

For the converse, if $Y$ is in the Lie algebra of $Z(G)$, then $e^{tY} \in Z(G)$ for all $t$ and so given any $x \in G$, $e^{tY}xe^{-tY} = x$. We now have $e^{\textrm{ad}_{tY}}x =x$ for all $x$, which implies that $e^{\textrm{ad}_{tY}} = I \in \textrm{GL}(\mathfrak{g})$. Taking the derivative at $t = 0$ implies that $\textrm{ad}_Y = 0$ and so $Y \in \ker \textrm{ad}$.

However: I have nowhere used the assumption of connectedness of $G$. What have I done wrong?

share|improve this question
add comment

2 Answers

up vote 4 down vote accepted

Since $\operatorname{ad}_{tY}$ acts on the Lie algebra of $G$ (only), you cannot expect $e^{\operatorname{ad}_{tY}}=\operatorname{Ad}_{e^{tY}}$ to hold as identity of functions defined beyond the connected component $G_0$ of the identity of $G$. Indeed if you think of $G=O(2,\mathbf R)$ where $G_0$ is the commutative group of $SO(2,\mathbf R)$, you can see that $\operatorname{ad}_{tY}=0$ for any $Y$, but $\operatorname{Ad}_{e^{tY}}$ is conjugation by a rotation, which does not fix reflections (unless $e^{tY}=\pm \mathrm{Id}$).

In fact now that I reread your question carefully, I see that you defined $\operatorname{Ad}_A$ to be a function defined on the Lie algebra $\mathfrak{g}$, not on $G$, so this makes your argument that applies it to $x\in G$ invalid in the first place. But my point is that although you can define $\operatorname{Ad}_g$ (maybe with a different notation) to be a conjugation by $g$ in $G$, you cannot expect $\operatorname{ad}_{tY}$ to tell you what any such $\operatorname{Ad}_g$ does beyond $G_0$.

share|improve this answer
    
Thanks for your answer. Does this mean that in my argument above, everywhere where I equated $e^{\textrm{ad}_{tY}}$ with $\textrm{Ad}_{e^{tY}}$ is not valid unless $G$ is connected? –  fpqc Sep 6 '12 at 10:07
    
I see the problem now. I have applied that identity to group elements and not to elements of the Lie algebra. This is where my mistake lies. –  fpqc Sep 6 '12 at 10:11
    
Thanks Marc for your answer. I have now gone to recheck my proof of another fact that I proved earlier, namely that $Z(G) = \ker \textrm{Ad}$. –  fpqc Sep 6 '12 at 10:17
    
I have posted an answer below, correcting my proof above. It relies on a lemma that uses connectedness of $G$, hooray! –  fpqc Sep 6 '12 at 10:48
add comment

As pointed out by Marc, I applied a certain identity that held valid only for elements of the Lie algebra to all of the Lie group. I have now reproduced a new proof that relies on the following lemma:

$\textbf{Lemma:}$ Suppose $G$ is a connected matrix Lie group. Let $\textrm{Ad}$, $\textrm{ad}$ be defined as above and let $\mathfrak{g}$ be the Lie algebra of $G$. Then $Z(G) = \ker \textrm{Ad}$.

I will write lower case letters for elements of $G$ and capitals for elements of the Lie algebra.

Proof: Suppose $x \in \ker \textrm{Ad}$. Then connectedness of $G$ implies that I can write any $y \in G$ as $e^{Y_1}\ldots e^{Y_n}$ for some $Y_1,\ldots,Y_n \in \mathfrak{g}$. Then

$$\begin{eqnarray*} xyx^{-1} &=& xe^{Y_1}\ldots e^{Y_n}x^{-1} \\ &=& xe^{Y_1}x^{-1}\ldots xe^{Y_n}x^{-1} \\ &=& e^{xY_1x^{-1}}\ldots e^{xY_nx^{-1}}\\ &=& e^{Y_1}\ldots e^{Y_n}\\ &=& y \end{eqnarray*}$$

proving that $x \in Z(G)$. Conversely, suppose that $y \in Z(G)$. Then I can define a group homomorphism

$$\begin{eqnarray*} \psi_y :& G& \to G \\ &x&\mapsto yxy^{-1}. \end{eqnarray*} $$

Now if $\phi_y$ is the induced map on Lie algebras, we see given any $X \in \mathfrak{g}$ that

$$\begin{eqnarray*} e^{t\phi_y(X)} &=& e^{\phi_y(tx)} \\ &=& \psi_y(e^{tX}) \\ &=& ye^{tX}y^{-1} \\ &=& e^{t(yXy^{-1})}. \end{eqnarray*}$$

Now if I differentiate both sides at $t = 0$, I see that $\phi_y(X) = yXy^{-1}$, so that $\phi_y$ is actually $\textrm{Ad}_y$. However from the second last step, $ye^{tX}y^{-1} = e^{tX}$ by definition of $y \in Z(G)$. It follows that $\phi_y(X) =\textrm{Ad}_y(X) = X$ for all $X \in \mathfrak{g}$, from which it follows that $\textrm{Ad}_y = I \in \textrm{GL}(\mathfrak{g})$. It now follows that $y \in \ker \textrm{Ad}$, completing the proof of the lemma.

$$\hspace{6in} \square$$

Having completed the proof of the lemma above, we can now prove our main result:

The Lie algebra of $Z(G)$ is equal to $\ker \textrm{ad}$.

If $Y $ is in the Lie algebra of $Z(G)$, then $e^{tY} \in Z(G)$ for all $t$. By the lemma, this means that given any $X \in \mathfrak{g}$,

$$e^{t \textrm{ad}_Y}= e^{\textrm{ad}_{tY}} (X) = \textrm{Ad}_{e^{tY}}(X) = X.$$

Differentiating at $t = 0$ now gives that $\textrm{ad}_Y = 0$, proving that $Y \in \ker \textrm{ad}$.

Reverse inclusion: If $Y \in \ker \textrm{ad}$, by the lemma we just need to prove that $e^{tY} \in \ker \textrm{Ad}$ for all $t$. Now for all $X \in \mathfrak{g}$, we have

$$\begin{eqnarray*} X &=& e^{\textrm{ad}_{tY}}(X) \\ &=& \textrm{Ad}_{e^{tY}}(X) \end{eqnarray*}$$

implying that $\textrm{Ad}_{e^{tY}} = I \in \textrm{GL}(\mathfrak{g})$ for all $t$, from which it follows that $e^{tY}$ is in $\ker \textrm{Ad}$.

$$\hspace{6in} \square$$

share|improve this answer
add comment

Your Answer

 
discard

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.