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I have an exercise of Lie group as follows: "Let $G,H$ be closed connected subgroup of $GL_n(\mathbb{R})$, and $H$ be subgoup of $G$. Suppose that $Lie(H)$ is an ideal of $Lie(G)$. Prove that $H$ is a normal subgroup of $G$." I get stuck to solve this problem. Also I have no idea to use the connectedness of $G$ and $H$. Some one can help me? Thanks a lot!

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This question has an open bounty worth +50 reputation from Tim kinsella ending in 3 days.

The current answers do not contain enough detail.

This has been asked at least twice on this site, so I don't want to post it again. There doesn't seem to be a full proof in either post of this extremely natural question.

2 Answers 2

This is essentially an application of the Lie subalgebra-subgroup correspondence:

Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. Suppose that $\mathfrak{h}$ is a Lie subalgebra of $\mathfrak{g}$. Then there is a unique connected immersed Lie subgroup $H\subseteq G$ whose Lie algebra corresponds to $\mathfrak{h}$.

You are given $H$ a closed connected Lie subgroup of the Lie group $G$. Choose $g\in G$, and let $H' = gHg^{-1}$. Then $H'$ is a Lie group with corresponding Lie algebra $Lie(H')\subseteq Lie(G)$. However, the assumption that $Lie(H)$ is an ideal of $Lie(G)$ says exactly that $Lie(H') = Lie(H)$. You then have that $H$ and $H'$ are two connected Lie subgroups of $G$ with the same Lie algebra. By the uniqueness in the above theorem, it follows that $H = H'$, and hence that $H$ is normal.

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Thank you so much. But I don't know about the theorem you said. Can you tell me the name of the theorem? By chance, if the Lie group $G$ is linear, i.e a closed subgroup of $GL_n{\mathbb{R}}$, do we have simple proof for this theorem? –  mapping Nov 3 '12 at 15:11
    
@mapping: I don't know if the theorem has a name. It is essentially the Frobenius Theorem‌​. The idea is as follows. Take a basis $X_1,\ldots, X_r$ for $\mathfrak{h}$. These are invariant vector fields. Since $\mathfrak{h}$ is a subalgebra, these vector fields involutive. Frobenius' Theorem says there is a unique connected integral submanifold $H\subseteq G$ through the identity element. This is the subgroup $H$. Of course, you have to prove it is a Lie subgroup. –  froggie Nov 3 '12 at 16:44
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Thanks a lot! I will try to prove by this way. However I have trouble to prove $Lie(H')=Lie(H)$: If I take $X\in Lie{H'}$ then we have $\exp(tX)\in H'\forall t$. Then I can deduce that $g^{-1}Xg=Z\in Lie(H)$. So I need to prove $gZg^{-1}\in Lie(H)$. Using the condition $Lie(H)$ is an ideal of $Lie(G)$, we obtain $gZ-Zg\in Lie(H)$, hence $[gZ-Zg,g^{-1}]\in Lie(H)$. Finally I get $gZg^{-1}+g^{-1}Zg\in Lie(H)$. Until here, I don't know how to continue. Can you help me? –  mapping Nov 3 '12 at 17:53

There seems to be a full proof contained in lemma 0.1 here http://math.berkeley.edu/~ianagol/261A.F09/Simplegroups.pdf

Here is the argument:

Claim: For $X \in \frak{g}$, $Y \in \frak{h}$, we have $e^Xe^Ye^{-X}\in H$.

Proof of claim: Denote by $exp$ the exponential map $End(\frak{g}) \rightarrow $ $Aut(\frak{g}) $ (where Aut and End are the vector space automorphisms and endomorphisms) . Since $\frak{h}$ is an ideal, $ad_X^n$ preserves $\frak{h}$ for all $n \in \mathbb{N}$, and therefore so does $exp(ad_X)$. We therefore have $$e^Xe^Ye^{-X}=e^{Ad_{e^X}(Y)}=e^{exp(ad_X)Y} \in e^{\frak{h}} $$

This establishes the claim.

Now the fact that $G$ normalizes $H$ follows from the fact that $$H= \bigcup_{n\in \mathbb{N}} (e^\frak{h})^n$$ $$G= \bigcup_{n \in \mathbb{N}}(e^\frak{g})^n$$ Since $G$ and $H$ are connected.

In particular, this doesn't seem to use the fact that $G$ is a closed subgroup of $GL(n, \mathbb{R})$, although this is a hypothesis of lemma 0.1. Nor does it require even that $H$ be closed in $G$.

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