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Lemma: If $\mathfrak{k}$ is a subalgebra of $\mathfrak{g}$ that contains an Engel subalgebra, then $\mathfrak{k}$ is self-normalizing.

Proof: Suppose $\mathfrak{k}\supset \mathfrak{g}_0(ad\; x)$ for some $x\in \mathfrak{g}$, then $x\in\mathfrak{k}\subset N_{\mathfrak{g}}(\mathfrak{k})$ and $[x\mathfrak{k}]\subset \mathfrak{k}$. $ad\; x$ acts on $N_{\mathfrak{g}}(\mathfrak{k})/\mathfrak{k}$ and without eigenvalue 0 (?!!!). However $[x,N_{\mathfrak{g}}(\mathfrak{k})]\subset \mathfrak{k}$, so action is trivial. This means $N_{\mathfrak{g}}(\mathfrak{k}) = \mathfrak{k}$.

I understand everything from the proof except the part that ad $x$ has no 0 eigenvalue on $N_{\mathfrak{g}}(\mathfrak{k})/\mathfrak{k}$. Can someone please explain why?

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$\mathfrak{g}_0(ad\ x)$ contains all $y$ such that $ad\ x (y)=0$. So the zero eigenspace of $ad\ x$ in $N_g(\mathfrak{k})$ is contained in $\mathfrak{k}$. Thus there is no zero eigenspace in the factor $N_g(\mathfrak{k})/\mathfrak{k}$

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  • $\begingroup$ Why does the second sentence of your argument imply the third sentence ? $\endgroup$ – Amr Apr 21 at 1:40
  • $\begingroup$ Since $\mathfrak{l}$ is stable under $ad \, x$, the generalized eigenspaces of $ad\, x$ on $\mathfrak{g}/\mathfrak{\ell}$ are exactly $(\mathfrak{p} + \mathfrak{l})/\mathfrak{l}$ for $\mathfrak{p}\subseteq \mathfrak{g}$ a generalized eigenspace of $ad \, x$. But $\mathfrak{l}$ contains the generalized $0$-eigenspace of $ad \, x$, so there is no eigenvalue zero on $\mathfrak{g}/\mathfrak{l}$. $\endgroup$ – Joshua Mundinger Apr 21 at 14:47
  • $\begingroup$ @JoshuaMundinger Thank you a lot for your comment and your time. I m still not convinced, probably I m missing something too obvious. I don't know what you mean by a generalized eigenspace, but I suppose you want to make that statement: Let $T:V \rightarrow V$ be a linear transformation that stabilizes a subspace $W$. One gets naturally another map $\overline{T}:V/W\rightarrow V/W$. Then the kernel (or $0$ eigenspace) of $\overline{T}$ equals $(ker T+W)/W$. If yes, then I disagree with this statement as $ker(\overline{T})=f^{-1}[W]/W$ instead $\endgroup$ – Amr Apr 21 at 16:43
  • $\begingroup$ @Amr The generalized $\lambda$-eigenspace of $T: V \to V$ is $\cup_{k\geq 1}ker ( (T - \lambda)^k)$. If $B$ is a basis such that $T$ is in Jordan form with respect to $B$, then the generalized $\lambda$-eigenspace is exactly the span of the vectors in $B$ corresponding to $\lambda$-blocks. Since $ad \, x$ is not necessarily diagonalizable, we must use this instead of a decomposition of $\mathfrak{g}$ into eigenspaces. $\endgroup$ – Joshua Mundinger Apr 22 at 17:20

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