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Let $G$ be a Lie group and let $\mathfrak{g}$ be its Lie algbera. Let $E$ be a subspace of $\mathfrak{g}$. Define the normalizer of $E$ in $G$ as:

$$ N_{G}(E) = \{ g \in G \;\; | \;\; Ad(g)E = E\}$$

Let the normalizer of $E$ in $\mathfrak{g}$ be defined as:

$$ \mathfrak{n}_{\mathfrak{g}} = \{ x \in \mathfrak{g} \;\; | \;\; [x,E] \subseteq E \}$$

Then I must show that $N_{G}(E)$ is closed and its Lie algebra is $\mathfrak{n}_{\mathfrak{g}}$.


I could prove that it is closed as the adjoint representation is continuous. For the Lie algbera, this is what I did:

For $x \in L(N_{G}(E))$ (the Lie algebra of $N_{G}(E)$), and $z \in E$, we have

$$ [x,z] = ad(x)(z) = \left(\frac{d}{dt}\bigg|_{t=0}e^{ad(tx)}\right)(z) = \left(\frac{d}{dt}\bigg|_{t=0}Ad(exp_{G}(tx)\right)z$$

where $e^{x}$ is the matrix exponentiation and $exp_{G}$ is the Lie groups exponential map. Now I know that $Ad(exp_{G}(tx))z \in E$ by various definitions, but I do not know how to pull the $z$ inside the derivative.

I think that since $Ad(exp_{G}(tx)) \in GL_{E}$ ($GL_{E}$ is the subset of $GL(\mathfrak{g})$ consisting of linear maps that preserve $E$) for every $t$, so is its derivative. However, I am not sure about this? Could some one help me in this particular respect?

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1 Answer 1

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The derivative of $\mathrm{Ad}(\mathrm{exp}_G(tx))$ is not in $\mathrm{GL}(\mathfrak{g})$ since as you note it is $\mathrm{ad}(tx)$, which is in general not invertible. However it does preserve $E$ which is seen for example by pulling $z$ inside the derivative as you suggested:

$$ \left(\frac{d}{dt}\Bigg|_{t=0} \mathrm{Ad}(\mathrm{exp}_G(tx))\right)z = \left(\lim_{t\to 0} \frac{\mathrm{Ad}(\mathrm{exp}_G(tx)) - I}{t} \right)z = \lim_{t\to 0}\frac{\mathrm{Ad}(\mathrm{exp}_G(tx))z - z}{t} = \gamma'(0) $$

Where $\gamma(t) = \mathrm{Ad}(\mathrm{exp}_G(tx))z$, this gives you that $[x, z] \in E$ since $\gamma(t) \in E$.

You still need to prove that every element of $\mathfrak{n}_\mathfrak{g}$ is an element of $L(N_G(E))$. To prove that you once again use that $\mathrm{Ad}(\mathrm{exp}_G(tx)) = e^{\mathrm{ad}(tx)}$ and note that since $\mathrm{ad}(x)$ preserves $E$ then $\mathrm{ad}(tx)$ does too and also $e^{\mathrm{ad}(tx)}(E) \subset E$, you can see this for example by writing the matrix exponential as an infinite sum. All in all this gives that $\mathrm{exp}_G(tx) \in N_G(E)$ and hence $x \in L(N_G(E))$.

I hope this answers your question.

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  • $\begingroup$ But my point is precisely that why can I pull $z$ inside? $\endgroup$ Aug 30, 2013 at 1:44
  • $\begingroup$ Since $(\lim_{t \to 0} f(t))x = f(0)x = \lim_{t \to 0} (f(t)x)$ for $f$ a map into $\mathbb{C}$ continuous at $0$ and $x \in \mathbb{C}$. Then the matrix argument case above is just multiplying out and taking the limit coordinate-wise since limits of vectors and matrices are just the coordinate-wise limit. Note that $z$ is constant in $t$. Is that a sufficiently clear explanation? $\endgroup$ Aug 30, 2013 at 7:09
  • $\begingroup$ Yeah. Thanks for the help. $\endgroup$ Jan 16, 2014 at 4:18

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