Example of a non-compact semisimple Lie algebra

Question

I apologize if this question is elementary - I'm coming from a physics background. For reasons that aren't important, physicists like their gauge group Lie algebras to have Killing forms that are negative definite. The textbooks often say something along the lines of "we require the gauge group's Lie algebra to be compact, so that the Killing form is negative semidefinite, and also semisimple, so that the Killing form is strictly negative definite." This kind of makes it sound like a semisimple Lie algebra is a special case of a compact Lie algebra, although I don't think this is the case.

Just so that I get a mental picture of every possible combination of choices for compactness and semisimplicity: is there a simple (in the colloquial sense of the word!) example of a Lie algebra that is semisimple but non-compact? What does its Killing form look like?

Note that this question is the same as Is there an example of a non compact, semisimple, amenable Lie group?, but without the requirement of amenability, which I don't know or care about.

Edit: As levap points out, there are two inequivalent definitions of "compact," which correspond to the red zone and the intersection of the red and green zones of the following Euler map:

Answer 1

There are two common but inequivalent definitions of a compact lie algebra: A compact lie algebra can either mean a lie algebra of a compact lie group or a lie algebra whose killing form is negative definite (see here for a comparison of the definitions). I'll assume you work with the first definition (and then the killing form of a compact lie algebra is negative semidefinite).

Consider the lie algebra $\mathfrak{sl}(n,\mathbb{R})$ which is the lie algebra of the group $\operatorname{SL}(n,\mathbb{R})$ which is not compact. This is a semisimple lie algebra whose Killing is given by

$$ K(X,Y) := 2n \operatorname{tr}(XY). $$

The form $K$ is a non-degenerate quadratic form (as it should, being the Killing form of a semisimple lie algebra) which has a mixed signature. To calculate it, write

$$ \mathfrak{sl}(n,\mathbb{R}) = V_{+} \oplus V_{-}, \\ V_{+} := \{ X \in M_n(\mathbb{R}) \, | \, X^T = X, \operatorname{tr}(X) = 0 \}, V_{-} := \{ X \in M_n(\mathbb{R}) \, | \, X^T = -X \}. $$

Note that on $V_{+}$, we have

$$ K(X,X) = 2n \operatorname{tr}(X^2) \geq 0 $$

because $X^2 = XX^T$ is positive semi-definite and $K(X,X) = 0$ iff $X = 0$. That is, $K|_{V_{+}}$ is positive-definite.

On the other hand, on $V_{-}$, we have $K(X,X) \leq 0$ because $X^2 = -XX^T$ is negative semi-definite and $K(X,X) = 0$ iff $X = 0$. That is, $K|_{V_{-}}$ is negative-definite. Hence, $K$ is of mixed signature

$$ \left( \frac{n(n+1)}{2} - 1, \frac{n(n-1)}{2} \right). $$

For example, if $n = 2$ then the signature of $K$ is $(2,1)$.

Answer 2

The short answer is: $\mathfrak{sl}(n,\mathbb{R})$ is semisimple (hence non-degenerate Killing form), but non compact (hence not the Lie algebra of any compact Lie group).

Now, to understand better the statement you quoted, you need to understand this:

• Compact Lie algebra $\implies$ negative semidefinite Killing form.
• Semisimple Lie algebra $\implies$ non-degenerate Killing form.
• Negative semidefinite, non-degenerate, symmetric bilinear form $\implies$ negative definite form.

The third point is just an application of the Cauchy–Schwarz inequality.

The first point is with the definition that a compact Lie algebra is the Lie algebra of a compact Lie group.