Differential Geometry – Semisimple Lie Groups with Finite Center

differential-geometrylie-algebraslie-groupssemisimple-lie-algebrastopological-groups

Before the problem, here goes the background: let $G$ be a semisimple Lie group with Lie algebra $\mathfrak{g}$. Consider $X_s$ an semisimple element of $\mathfrak{g}$ such that every eigenvalue of $e^{ad(X_s)}$ has norm $1$. By 'semisimple element' we understand that to the complexification $\mathfrak{g}_{\mathbb{C}}$ of $\mathfrak{g}$, the operator $\hbox{ad}X_s$ is conjugated to a diagonal matrix.

Now, consider the set
\begin{equation}
K_0 = \hbox{cl}\{\exp{(tX_s)}: t \in \mathbb{R}\}.
\end{equation}

If $G = KAN$ is the Iwasawa decomposition of the Lie group $G$, by theorem 7.2, pg. 431 of Helgason's book – Differential Geometry, Lie groups and symmetric spaces – the element $e^{ad(X_s)}$ is conjugated to an element of $K$.

My problem is: considering that $G$ has finite center, I'm trying to prove that $K_0$ is compact using the fact that $K$ is compact, but I have no clue about how to do it. I've tried to use open covers, unsuccessfully.

Any help is appreciated.

Best Answer

First recall that an element $g$ of a topological group $G$ is called compact if $g$ is contained in a compact subgroup of $G$. Now let's use the formula ($x=X_s\in\mathfrak{g}$) $$ \operatorname{Ad}(\exp x) = \exp(\operatorname{ad}x) \text{ as operators }\ \mathfrak{g}\to\mathfrak{g} $$ If every eigenvalue of $\exp(\operatorname{ad}x)$ has norm $1$ and $\operatorname{ad}x$ is a semi-simple operator on $\mathfrak{g}$, then $\operatorname{Ad}(\exp x)$ is a compact element in $\operatorname{Aut}(\mathfrak{g})$. Since $G$ is a semi-simple group and $G$ has finite center, it follows that $\exp x$ is a compact element in $G$.

Let $L=\{\exp tx\mid t\in\mathbb{R}\}$. The group $L$ is a one-parameter subgroup of the Lie group $G$. Therefore, by virtue of Exercise C.2 on page 150 of Helgason's book, the subgroup $L$ is either closed or has compact closure. If $L$ is closed and non-compact, then $L\cong\mathbb{R}$ and $L$ has no nontrivial compact elements. This contradicts the fact that $\exp x$ is a compact element.

Consequently $K_0=\operatorname{cl}(L)$ is a compact group.

More detailed answer to the question "Why is $\exp x$ a compact element of $G$ if $\operatorname{Ad}(\exp x)$ is a compact element of $\operatorname{Aut}(\mathfrak{g})$?"

  1. We know that $\operatorname{Ad}:G\to\operatorname{Aut}(\mathfrak{g})$ is a Lie group homomorphism.

  2. Since $G$ is a semisimple connected Lie group, $\operatorname{Ker}(\operatorname{Ad})$ is a discrete subgroup coinciding with the center $Z(G)$ of $G$.

  3. Let $\operatorname{Ad}(\exp x)\in A$, where $A$ is a compact subgroup of the group $\operatorname{Aut}(\mathfrak{g})$.

  4. Since the $Z(G)$ is finite, the preimage $B=\operatorname{Ad}^{-1}(A)$ is a compact subgroup of $G$.

  5. We have $\exp x\in B$. Hence $\exp x$ is a compact element of group $G$.

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