Correspondence between ideals of Lie algebra and connected normal subgroups of connected algebraic group

Question

Let $G$ be a connected linear algebraic group (LAG) and $H\leq G$ a closed connected subgroup such that $\operatorname{Lie}H\subseteq\operatorname{Lie}G$ is an ideal. I want to show that $H$ is a normal subgroup of $G$.

To do so the main result I have is :

1. For $\phi,\psi:G\to H$ morphisms of LAG, then : $(d\phi)_e=(d\psi)_e\implies \phi=\psi$
2. If $H_1,H_2\leq G$ are closed, $\operatorname{Lie}(H_1\cap H_2)=\operatorname{Lie}H_2\cap\operatorname{Lie}H_2$, in particular $\operatorname{Lie}H_1=\operatorname{Lie}H_2\implies H_1^{o}=H_2^{o}$ (connected components)
3. For $\phi:G\to H$, $H'\leq H$ closed, we have that $\operatorname{Lie}(\phi(G))=(d\phi)_e(\operatorname{Lie}G)$ and $\operatorname{Lie}(\phi^{-1}(H'))=(d\phi)_e^{-1}(\operatorname{Lie}H')$. In particular $\operatorname{Lie}(\operatorname{Ker}\phi)=\operatorname{Ker}((d\phi)_e)$.

So my idea is to consider $\phi_g:H\to G$ the conjugation by $g$ and show that $\operatorname{Lie}H=\operatorname{Lie}\phi_g(H)$ which would lead to $H=\phi_g(H)$ since they are connected. We have $\operatorname{Lie}(\phi_g(H))=(d\phi_g)_e(\operatorname{Lie}H)$ and I want to show its equal to $\operatorname{Lie}H$. But $(d\phi_g)_e:\operatorname{Lie}H\to\operatorname{Lie}G$ is defined by $A\mapsto A\circ \phi_g^*$ where $\phi_g^*:\mathcal O(G)\to \mathcal O(H)$ is the induced algebra morphism between coordinate rings. From this I'm not sure on how to proceed to show that the image of this map is in $\operatorname{Lie}H$.

Am I on the right way ?

Answer 1

Let $G$ be a smooth connected (affine) algebraic group over some ground field $k$ and write $\mathfrak{g}$ for its Lie algebra. As was already pointed out in Armado's answer, we have the adjoint representation of $G$ $$\operatorname{Ad} \colon G \to \operatorname{GL}_{\mathfrak{g}}$$ and its derivative $$\operatorname{ad} = d(\operatorname{Ad}) \colon \mathfrak{g} \to \mathfrak{gl}_{\mathfrak{g}}$$ and the Lie bracket on $\mathfrak{g}$ is basically defined by $[x, y] = \operatorname{ad}(x)(y)$.

Now suppose we are given a smooth connected algebraic subgroup $H \subseteq G$ such that its Lie algebra $\mathfrak{h}$ is an ideal in $\mathfrak{g}$. Then we claim that $H \subseteq G$ is actually normal. To prove this we proceed in two steps:

• Consider the algebraic subgroup $P \subseteq \operatorname{GL}_{\mathfrak{g}}$ given as the stabilizer of the subspace $\mathfrak{h} \subseteq \mathfrak{g}$. Its Lie algebra $\mathfrak{p} \subseteq \mathfrak{gl}_{\mathfrak{g}}$ is then given as the set of those endomorphisms of $\mathfrak{g}$ that stabilize $\mathfrak{h}$. The condition that $\mathfrak{h} \subseteq \mathfrak{g}$ is an ideal thus translates into saying that $\operatorname{ad}(\mathfrak{g}) \subseteq \mathfrak{p}$, which implies that $\operatorname{Ad}(G) \subseteq P$.

• Now suppose that $g \in G(k)$ and consider the conjugation map $\phi_g \colon G \to G$. By definition of the adjoint representation we have $\operatorname{Ad}(g) = d(\phi_g)$. Thus the previous bullet point implies that $d(\phi_g)(\mathfrak{h}) \subseteq \mathfrak{h}$ which implies that $\phi_g(H) \subseteq H$. As $g \in G(k)$ was arbitrary, this shows that $H \subseteq G$ is normal.

Answer 2

Given the representation of Lie groups induced by the conjugation:

$\text{Ad}:G \rightarrow \text{GL}(\text{Lie}(G))$,

we can differentiate it and obtain:

$d(\text{Ad}):\text{Lie}(G) \rightarrow \mathfrak{gl}( \text{Lie}(G))$

It is easy to see that this map is extacly the $\text{ad}$ of Lie algebras (to compute this remember that $\frac{d}{dt}|_{t=0} e^{tX}=X $).

So by the universal property of the differential we obtain that $\text{Ad}(e^X)=e^{\text{ad}(X)}$ $\forall X \in \text{Lie}(G)$.

Since $G$ is connected it is generated by $\text{exp}(\text{Lie}(G))$ so a generic $g\in G$ is of the form $e^{X_1}…e^{X_t}$, we can assume $g=e^X$ (the computation is similar).

We can now see that $\text{Ad}(g)(H)\subseteq H$, let $h \in H$:

$\text{Ad}(e^X)(h)=e^{\text{ad}(X)}(h)=\sum_n \frac{\text{ad}(X)^n}{n!} (h) $,

now since also $h$ is of the form $h=e^{Y}=\sum_m \frac{Y^m}{m!}$ and since $\text{Lie}(H)$ is and ideal of $\text{Lie}(G)$ we can easily conclude the proof.