Is a finite volume Lie group compact?

Question

I know an example of a finite volume homogeneous space which is not compact, $SL_2(\mathbb(R)) / SL_2(\mathbb{Z})$. But what about a Lie group with this property? Can it happen?

(The Lie group is assumed to have the Haar measure.)

Answer 1

Yes, it is true. Here is a more general statment:

Suppose that $G$ is a locally compact group with Haar measure $\mu$. Then $G$ is compact iff it has finite Haar measure. (A.5.1. In Kazhdan's Property (T) by Bekka, de la Harpe, Valette)

Proof that finite Haar measure implies $G$ is compact:

1. Assume $G$ is not compact.
2. Let $U$ be a compact neighborhood of $e$.
3. By 1, we can inductively build a sequence of $g_i \in G$ so that $g_{n + 1} \not \in \cup_{i = 1}^n g_i U$.
4. Letting $V$ be a neighborhood of $e$ so that $V^{-1} = V$ and $V^2 \subset U$, then $g_n V \cap g_m V = \emptyset$ if $n \not = m$. (Follows from $3$.)
5. Then $\mu(G) \geq \mu ( \bigcup g_n V) = \Sigma \mu(g_n V) = \Sigma \mu(V) = \infty$. QED

We use that open sets have positive measure in Haar measure. The converse is because compact sets have finite Haar measure.