I am currently working through J-P. Serre's Complex Semisimple Lie Algebras. Progress is slow but steady — I'd never heard of a Lie algebra before starting on this book.

On the second page, the following example of a Nilpotent Algebra is given:

Let $V$ be a vector space of finite dimension $n$. A flag $D = (D_i)$ of $V$ is a descending series of vector subspaces

$$V = D_0 \supset D_1 \supset \cdots \supset D_n = 0$$

of $V$ such that $\operatorname{codim}(D_i) = i$.

Let $D$ be a flag, and let $n(D)$ be the Lie subalgebra of $\operatorname{End} (V) = \mathfrak{gl}(V)$ consisting of the elements $x$ such that $x(D_i) \subset D_{i+1}$. One can verify that $n(D)$ is a nilpotent Lie algebra of class $n-1$.

I'm struggling on base 1 with this one — I can't work out why this should be a subalgebra at all. I understand that what I'm trying to show is that $[n(D),n(D)] \subset n(D)$, because that's what I understand to be the definition of a subalgebra with the usual Lie bracket, but I can't see how this is immediately true.

  • $\begingroup$ If you never had a look at Lie algebras maybe you want to start with something more user friendly such as "Introduction to Lie Algebras" of Erdmann, or "Lie Groups, Lie Algebras, and Representations" from Hall. Serre is great but really suppose you already know everything and he's just summirizing things... $\endgroup$ – Dac0 Oct 12 '17 at 18:58

If $X,Y\in n(D)$, then, for each $i$, $X(D_i)\subset D_{i+1}$ and $Y(D_i)\subset D_{i+1}$. But then $X\bigl(Y(D_i)\bigr)\subset D_{i+2}$ and $Y\bigl(X(D_i)\bigr)\subset D_{i+2}$. Therefore, $(XY-YX)(D_i)\subset D_{i+2}$.

  • $\begingroup$ So this is using the fact that the Lie bracket is represented by the commutator for the Endomorphisms, yes? Do I need to know it's the commutator for the statement to work? I was wondering initially how the statement holds without knowing that, (instead just using the properties of a general Lie bracket) but I assume I need to use the fact that its represented by the commutator? $\endgroup$ – Matt Oct 12 '17 at 14:12
  • $\begingroup$ @MattS Yes, that is what I did (and that is the only way of doing it). $\endgroup$ – José Carlos Santos Oct 12 '17 at 14:25

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