Real form of $\mathfrak{sl}_2 \mathbb{C}$ in Fulton-Harris Representation Theory

Question

In Fulton-Harris' Representation Theory on page 431 there is an analysis of the structure of a real form $\mathfrak{g}_0$ if complex Lie algebra $ \mathfrak{g} := \mathfrak{sl}_2 \mathbb{C}$ I not understand:

To get the idea of what to expect, let us work out real forms of $\mathfrak{sl}_2 \mathbb{C}$ in detail. To do this, suppose $\mathfrak{g}_0$ is any real Lie subalgebra of $\mathfrak{sl}_2 \mathbb{C}$, with $\mathfrak{g}_0 \otimes \mathbb{C} = \mathfrak{sl}_2 \mathbb{C}$. The natural thing to do is to try to carry out our analysis of semisimple Lie algebras for the real Lie algebra $\mathfrak{g}_0$ that is, find an element $H \in \mathfrak{g}_0$ such that $H$ under the adjoint rep $ad(H)$ acts semisimply on $\mathfrak{g}_0$ decompose $\mathfrak{g}_0$ into eigenspaces, and so on. The first part of this presents no problem: since the subset of $\mathfrak{sl}_2 \mathbb{C}$ of non-semisimple matrices is a proper algebraic subvariety, it cannot contain the real subspace $\mathfrak{g}_0 \subset\mathfrak{sl}_2 \mathbb{C}$, so that we can certainly find a semisimple $ H \in \mathfrak{g}_0$.

Questions:

Why should the subset of $\mathfrak{sl}_2 \mathbb{C} - (\mathfrak{sl}_2 \mathbb{C})_s \subset \mathfrak{sl}_2 \mathbb{C}$ of non-semisimple matrices have structure of a proper algebraic subvariety? In other words: Is beeing semisimple (=diagonalizable since we are working over $\mathbb{C}$) an open condition? And how general is this statement? Is it for example true that for any Lie algebra $\mathfrak{g}$ contained in a matrix algebra $\mathcal{M}$ the subset $\mathfrak{g} - \mathfrak{g}_s \subset \mathfrak{g}$ of it's non-semisimple elements have structure of a proper algebraic subvariety?

Secondly, assuming that the subset of non-semisimple matrices is indeed a proper algebraic subvariety of $\mathfrak{sl}_2 \mathbb{C}$, why we can conclude that it NOT contains the real form $\mathfrak{g}_0$? Does the condition $\mathfrak{g}_0 \otimes \mathbb{C} = \mathfrak{sl}_2 \mathbb{C}$ imply that $\mathfrak{g}_0$ is dense in $\mathfrak{sl}_2 \mathbb{C}$?

Answer 1

1. I don't believe this claim about semisimplicity being a (Zariski) open condition, but a closely related statement is true and suffices here: having distinct eigenvalues is an open condition, because its complement is cut out by the discriminant of the characteristic polynomial. So we can work instead with the open subset of matrices with distinct eigenvalues.

2. Yes. It suffices to show that $\mathbb{R}^n$ is (Zariski) dense in $\mathbb{C}^n$, or equivalently that any polynomial $p \in \mathbb{C}[x_1, \dots x_n]$ vanishing on $\mathbb{R}^n$ vanishes identically. Here is sort of a fun way to do it: if $p(\mathbb{R}^n) = 0$ then it follows that the same is true for all partial derivatives $\frac{\partial p}{\partial x_i}$. We can repeatedly take partial derivatives until only a constant is left (given by the leading term with respect to lexicographic order, say), but this constant must be zero, from which it follows that $p$ can't have any nonzero terms, so $p = 0$ identically.

However, stronger statements are true: for example it's also true that $\mathbb{Z}^n$ is Zariski dense in $\mathbb{C}^n$, and the above argument is unavailable although I guess one could make a similar argument using finite differences. More generally it's true that if $S_1, \dots S_n$ are any collection of infinite subsets of $\mathbb{C}$ then $\prod_{i=1}^n S_i$ is Zariski dense in $\mathbb{C}^n$, which can be proven by induction on $n$ or using the combinatorial Nullstellensatz.

Answer 2

For the record, here is an arguably simpler argument showing that $\mathfrak g_0$ contains at least one ad-semisimple element $\neq 0$.

Given $X \in \mathfrak g_0$, let $\lambda_1, \lambda_2, \lambda_3$ be the (possibly complex, possibly repeated) roots of the characteristic polynomial of $ad_{\mathfrak{g}_0}(X)$. Since the trace of this operator is $0$, $\sum \lambda_i=0$, and since $[X,X]=0$, w.l.o.g. $\lambda_1=0$, i.e. $\lambda_2 = -\lambda_3$. This means that for each $X$, either all $\lambda_i=0$ (which means $X$ is ad-nilpotent), or the $\lambda_i$ are mutually distinct, i.e. $X$ is ad-semisimple.

But now: If all $X \in \mathfrak g_0$ were ad-nilpotent, then by Engel's Theorem $\mathfrak g_0$ would be a nilpotent Lie algebra, a fortiori so would be $\mathfrak g_0 \otimes \mathbb C \simeq \mathfrak{sl}_2(\mathbb C)$, which it is not. QED.

Admittedly this argument is tailor-made to the case at hand, as it uses the relative smallness of the number $3$; in higher dimensions, we cannot make such strong statements about the $\lambda$'s, and many elements are neither ad-nilpotent nor ad-semisimple, in that their adjoint operators can contain proper Jordan blocks for non-zero eigenvalues.