Technical issue in the approach to Lie groups taken in a book

Question

I'm teaching Lie groups and Lie Algebras out of Brian C. Hall's book (Lie Groups, Lie Algebras, and Representations: An Elementary Introduction, Springer), which I've enjoyed using. I'm confused about a technical hitch though that I'm not sure how to avoid.

The approach taken in this book has two notable simplifying features:

1. Studying "matrix Lie groups" (closed subgroups of $\mathrm{GL}_n$) instead of general Lie groups.
2. Defining reductive as "is the complexification of the Lie algebra of a compact matrix Lie group" and semisimple as "reductive and center-free." I'll call these compact-reductive and compact-semisimple.

Both of these are nice simplifications that allow one to avoid some technical issues while still getting at the most important material for an introductory course. The latter is very nicely motivated by the question of trying to classify compact Lie groups via Lie algebra theory.

However, one of my students pointed out today a possible problem with combining these two simplifications. It is easy to see that any compact-semisimple Lie algebra is a direct sum of simple Lie algebras, but it is not obvious that these simple summands are themselves compact-semisimple! This means you can't just reduce the classification of compact-semisimple Lie algebras to compact-simple Lie algebras. It seems to me (and this may be my error) that Hall never deals with this issue and tacitly assumes that the simple summands are compact-simple

The obvious theorem to use to get around this issue is the following (Hall Thm 5.11):

Suppose that G is a simply connected matrix Lie group and the Lie algebra $\mathfrak{g}$ of G decomposes as $\mathfrak{g} \cong \mathfrak{h}_1 \oplus \mathfrak{h}_2$. Then there exists closed, simply connected subgroups $H_1$ and $H_2$ of $G$ with Lie algebras $\mathfrak{h}_1$ and $\mathfrak{h}_2$ such that $G = H_1 \times H_2$.

However, there's now a problem, because the compact group $K$ whose complixified Lie algebra of $\mathfrak{g}$ need not have a matrix universal cover! (e.g. $\mathrm{SL}_2(\mathbb{R})$). So this argument won't work in Hall's setting.

Is there some simple way to avoid this issue so that one can use both simplifying approaches 1 and 2 simultaneously?

Answer 1

Thanks for the question. It is true that I never proved that the simple summands in Theorem 7.8 are themselves semisimple in the sense that I define "semisimple." I would not characterize this omission as a problem, however, since I don't claim they are semisimple. Furthermore, I do not prove the classification of semisimple algebras. (And in any case, the classification would go through the classification of root systems, which decompose in terms of irreducibles.)

Nevertheless, it is an interesting question, and I believe we can prove that each $\mathfrak{g_j}$ is semisimple, using only the methods in Chapter 7 of my book. There are actually two issues, which I am not sure have been clearly separated in the discussion so far. First, we need to show that the decomposition of $\mathfrak{g}$ in Theorem 7.8 actually comes from a decomposition of $\mathfrak{k}$. To establish this, we first observe that if $\mathfrak{g}$ is center-free, then $\mathfrak{k}$ must also be center-free. Thus, $\mathfrak{k}$ will decompose as a direct sum of real, simple algebras $\mathfrak{k_j}$, by the same argument as the proof of Theorem 7.8. But since $\mathfrak{k_j}$ admits an Ad-invariant inner product, the proof of Theorem 7.32 shows that the complexification $\mathfrak{g_j}$ of $\mathfrak{k_j}$ is also simple as a complex Lie algebra. Then by the uniqueness result in Proposition 7.9, these $\mathfrak{g_j}$'s must be actually be the ones in Theorem 7.8.

Second, we need to show that each $\mathfrak{k_j}$ is the Lie algebra of a compact group, which would show that $\mathfrak{g_j}$ is semisimple in the sense used in the book. For this, the argument of suggested by Noah Snyder seems best: Since $\mathfrak{k_j}$ is an ideal, it is invariant under the adjoint action of $K$.The image $K_j$ of $K$ inside $GL(\mathfrak{k_j})$ is compact since it's the continuous image of the compact group $K$. Since $\mathfrak{k_j}$ is center-free, the associated Lie algebra homomorphism (the map sending $X$ to the restriction of $ad_X$ to $\mathfrak{k_j}$) will be zero on each $\mathfrak{k_l}$ with $l\neq j$ but will be injective on $\mathfrak{k_j}$. Thus, the image of the associated Lie algebra map will be isomorphic to $\mathfrak{k_j}$.

Answer 2

Here's a much easier argument more in the spirit of Hall's book.

If K is a compact matrix Lie group with discrete center and its Lie algebra decomposes as $\mathfrak{k}=\mathfrak{a}\oplus \mathfrak{b}$ then look at the image of K under its adjoint action on $\mathfrak{a}$. This is a continuous image of a compact set and so is compact (and closed!) so it gives a compact matrix group A. Since the derivative of Ad is ad, the Lie algebra of A is $\mathfrak{a}$.

Answer 3

As suggested by Vít Tuček, we can take the following approach. Since $K$ is compact and has no center, the Killing form on $\mathfrak{k}$ is negative definite. Thus the restriction of the Killing form to each of the $\mathfrak{k}_i$ is negative definite. We now want to exhibit compact matrix groups $K_i$ with Lie algebra $\mathfrak{k}_i$ using the negative definiteness of the Killing form.

There are many arguments in the literature that negative definiteness implies the existence of a compact group (though about half the ones I found seemed to have a gap). Here's one argument (which I got from Hsiang) which doesn't leave the world of matrix groups.

Suppose $\mathfrak{g}$ is a Lie algebra with negative definite Killing form, and consider $\mathfrak{D}$ its Lie algebra of derivations. A short calculation shows that the Killing form on $\mathfrak{D}$ restricts to the Killing form on $\mathfrak{g}$ and by considering the perpendicular compliment of $\mathfrak{g}$ in $\mathfrak{D}$ it follows that $\mathfrak{g} = \mathfrak{D}$. Hence, $\mathfrak{g}$ is the Lie algebra of the matrix Lie group $\mathrm{Aut}(\mathfrak{g}) \subset \mathrm{GL}(\mathfrak{g})$ (this is clearly a closed subgroup because it is cut out by polynomial equations). This subgroup is compact because it preserves the negative definite Killing form.

Answer 4

Another way to define a (complex) reductive Lie algebra ${\mathfrak g} \subset {\mathfrak g}{ \mathfrak l}_n({\mathbb C})$ is to assume that (after a conjugation by an element of $GL_n({\mathbb C})$ it is closed under the "Cartan involution " $x\mapsto ^t {\overline x}$. Then $\mathfrak g=\mathfrak k\oplus i \mathfrak k$, where $\mathfrak k$ is the (real) subalgebra of $\mathfrak k$ consisting of skew hermitian matrices. The same arguments as in one of the answers shows that the real analytic subgroup of $G(\mathbb C)$ with $\mathfrak k$ as Lie algebra is compact (if we also assume $\mathfrak k$ is semi-simple, not just reductive).

A decomposition of $\mathfrak k$ into real simple Lie algebras now gives an analogous decomposition of $\mathfrak g$ as complex simple Lie algebras.