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This is a problem from the book "Introduction to Lie-Algebra"-Erdmann & Wildon, Chapter 5-"Subalgebras of $gl(V)$"

Let $L=gl(n, \mathbb{C})$. Let $A\subset L$ be a subalgebra of $L$, such that $A$ contains all the diagonal matrices. Prove that $N_{L}(A)=A$, where $$N_{L}(A)=\{\ x\in L : [x,a]\in A \forall a\in A \}\ $$, called the normalizer of $A$ in $L$.

Now whenever I take a particular example for $A$, that is , say $A=d(n,\mathbb{C})$, or $A=t(n,\mathbb{C})$, it becomes easy to compute and we can see that they are self-normalizing. But whenever we have to take an arbitrary subalgebra containing $d(n,\mathbb{C})$, I have no clue about how to compute the elements of $N_{L}(A)$.

The book suggests, two ways- one a direct approach , which I tried without succeeding, another a little sophisticated one- by using the "invariance Lemma", which is there in the book in the same chapter.

I want to have a clue for both the suggested ways. I will really appreciate some help. Thanks in advance!

Riju
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1 Answers1

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A subalgebra $\mathfrak{h}$ of $\mathfrak{gl}_n(K)$, $K$ a field of characteristic zero, containing the diagonal matrices $\mathfrak{d}$, is normalized by diagonal matrices. It follows that it is a sum of weight spaces: $\mathfrak{h}=\bigoplus_{(i,j)\in M}\mathfrak{a}_{ij}$, where $\mathfrak{a}$ is the 1-dimensional line generated by the matrix $E_{ij}$, and $M$ is a subset of $\{1,\dots,n\}^2$ containing the diagonal.

Hence the normalizer $\mathfrak{n}$ of $\mathfrak{h}$ has the same form, with $M$ replaced with a possibly larger subset $M'$. If $(i,j)$ belongs to $M'-M$, then $i\neq j$ and $[\mathfrak{n},\mathfrak{h}]\subset\mathfrak{h}$. Since $\mathfrak{n}$ contains $\mathfrak{a}_{ij}$ and $\mathfrak{h}$ contains $\mathfrak{d}$, we deduce $\mathfrak{a}_{ij}=[\mathfrak{a}_{ij},\mathfrak{d}]\subset\mathfrak{h}$, hence $(i,j)\in M$, a contradiction. Thus $M=M'$, that is, $\mathfrak{h}$ is self-normalized.

YCor
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