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If $A$ is a vector subspace of $\mathfrak{g}$(Which is a Lie algebra), and $N=\{x\in\mathfrak{g}:[x,A]\subseteq A\}$

So if $N=A$, then $A$ is a subalgebra, and if $N=\mathfrak{g}$ then, $A$ is an ideal, correct?

What can I conclude about $N$?

So many hats
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  • There is a typo, you mean $A\subseteq N$ for $A$ a subalgebra. Also, could you maybe elaborate on your final question? I don't understand what you are after. What you can say in general is that $N$ is a subalgebra. – Hanno Aug 16 '15 at 07:24
  • @Hanno I am not sure what you mean about a typo? As for the final question, I think Dietrich Burde satisfied my curiosity, thanks :). – So many hats Aug 17 '15 at 00:51
  • I only meant that $A$ being a subalgebra is equivalent to $A\subseteq N$, and not to $A=N$ :) – Hanno Aug 17 '15 at 05:05

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The space $N$ is called the normalizer of $A$ in $\mathfrak{g}$, and it is the largest subalgebra of $\mathfrak{g}$ containing $A$ as an ideal. If $A$ is already an ideal in $\mathfrak{g}$, then $N_{\mathfrak{g}}(A)=\mathfrak{g}$ of course. If $N_{\mathfrak{g}}(A)=A$, then we only know that $A$ is self-normalising.

Dietrich Burde
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  • Why is $N$ a subalgebra of $\mathfrak{g}$? So I want $[N,N]\subseteq N$ and we have $[N,A]\subseteq A$, I don't see how we get that it is a subalgebra. – So many hats Aug 16 '15 at 12:17
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    Let $x,y\in N$. Then $[[x,y],A]]=[[y,A],x]+[[A,x],y]\subset A$ by the Jacobi identity. Hence $[x,y] \in N$ by definition of $N$. – Dietrich Burde Aug 16 '15 at 13:00