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!
Normal subgroup and Lie algebra
3 Answers
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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1Thanks 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? – 2012-11-03
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 $G$.
I will two theorems that can be found in Lie groups and Algebraic group book by Vinberg.
$Theorem:1$ A a homomorphism from a connected lie group H to a lie group G is uniquely determined by tangent algebra homomorphism.
$Theorem:2$ Let $f:H \rightarrow G$ where is H is connected.If $G_1\subset G$ such that $df(Lie(H))\subset Lie(G_1)$. Then $f(H)\subset G_1$.
Let take arbitary $g\in G$ then $a(g):x\rightarrow gxg^{-1}$. The differential of this map say Ad(g) coincide with the adjoint representation of lie algebra hence this map Lie(H) to Lie(H). So $gHg^{-1}\subset H$ for all g hence H is normal.// You can find all the proofs in the book I mentioned in chapter 1 section 2 I hope.
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0@Timkinsella To make GGT's argument precise, observe that since $\mathfrak{h}$ is an invariant subspace of the Lie algebra rep $\mathrm{ad} \colon \mathfrak{g} \to \mathfrak{gl}(\mathfrak{g})$, it follows that $\mathfrak{h}$ is also an invariant subspace of the Lie group rep $\mathrm{Ad} \colon G \to \mathrm{GL}(\mathfrak{g})$. So $\mathrm{Ad}(g)$ does send $\mathfrak{h}$ to itself. – 2018-04-05