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I find it intuitive enough that the radical of $\mathfrak{gl}_n\mathbb F$ is the scalar matrices, but I have trouble finding an easy, but complete proof:

Proof. Let $\mathfrak s$ denote the scalar matrices. Clearly $\mathfrak s\subset\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$. Suppose that $\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$ is generated by more than one element, so that $X\in\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$, but $X\notin\mathfrak s$. We can change basis such that $\mathrm{rad}(\mathfrak{gl}_n\mathbb F)$ is upper-triangular. (The change of basis leaves the scalar matrices invariant.) Then $X$ is upper-triangular.

I want to conclude that there exists a $Y\in\mathfrak{gl}_n\mathbb F$ s.t. $[X,Y]$ is not upper-triangular. How can I see that $Y$ always exists, except by waving hands?

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  • $\begingroup$ Try looking at proofs of the fact that $\mathfrak{sl}_n$ is simple. $\endgroup$ – Alistair Savage May 5 '13 at 20:53
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If $I$ is a solvable ideal of $\mathfrak{gl}_n(\mathbb{F})$ then $I\cap \mathfrak{sl}_n(\mathbb{F})$ is a solvable ideal of $\mathfrak{sl}_n(\mathbb{F})$ which is semisimple (in fact simple) as long as the characteristic of $\mathbb{F}$ does not divide $n$, so the radical of $\mathfrak{gl}_n(\mathbb{F})$ intersects trivially with $\mathfrak{sl}_n(\mathbb{F})$, and this forces it to have dimension $1$.

On the other hand, if we take the example $\mathfrak{gl}_2(\mathbb{F}_2)$ then one can check that this is actually solvable, so in this case the radical does not just consist of the scalar matrices (the derived subalgebra is contained in $\mathfrak{sl}_2(\mathbb{F}_2)$ which contains the scalar matrices. The quotient of $\mathfrak{sl}_2(\mathbb{F}_2)$ by the scalar matrices has dimension $2$ and is thus solvable).

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  • $\begingroup$ Thank you for your answer. This is exactly the reasoning I had in mind. My professor, however, argued that I should not assume the fact that $\mathfrak{sl}_n$ is (semi)simple. In particular, completing the proof in my question would give a proof that $\mathfrak{sl}_n$ is semisimple. The argument you outline does it "the wrong way around"... $\endgroup$ – Earthliŋ May 21 '13 at 19:35
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    $\begingroup$ That $\mathfrak{sl}_n(\mathbb{F})$ is simple when $n$ is not divisible by the characteristic of $\mathbb{F}$ (this condition is important!) can be done directly with some work (and I guess that work will also usually essentially include a proof that the radical of the general linear Lie algebra consists of the scalar matrices). Alternatively, one could use a sledgehammer and provide a basis that satisfies the Serre relations for type $A_n$ which then guarantees that it is simple at least in characteristic $0$. $\endgroup$ – Tobias Kildetoft May 21 '13 at 19:41
  • $\begingroup$ From Serre's lecture notes, a rapid way to prove sl_n is semisimple (in characteristic 0 let's say), is to use (a corollary of) Theorem 5.1 in Part I Chapter 6: The existence of an irreducible faithful representation of a Lie algebra g whose center is trivial is equivalent to g being semisimple. Since sl_n is easily seen to have trivial center and acts faithfully irreducibly on k^n, we are done. This is Serre's own justification at the start of Part I Chapter VII. $\endgroup$ – Doug Aug 20 '18 at 11:41

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