I am confused about two definitions for the notion of a semisimple Lie group i found. Lets say for simplicity i am only interested in matrix groups. In this case, do the following two object-classes coincide? 1) A connected Lie group that does not contain non-trivial connected solvable (equivalently: ...connected Abelian) normal subgroups. 2) A linear connected reductive group with finite center, where a linear connected reductive group is a closed connected group of real or complex matrices that is stable under conjugate transpose.
Remark: if i only use definition 1) the definition implies that in every connected ss. Liegroup the center must be discrete, but i cannot follow that it must even be finite.
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$\begingroup$For simplicity, all Lie groups below are connected.
My preferred definition is that a Lie group $G$ is semisimple if its Lie algebra $\mathfrak{g}$ is semisimple. This is equivalent to definition 1 since solvable ideals in $\mathfrak{g}$ exponentiate to connected solvable normal subgroups of $G$ and vice versa.
Proposition: Definition 2 implies definition 1.
Sketch. Recall that one definition of a reductive Lie algebra $\mathfrak{g}$ is that it is a Lie algebra whose adjoint representation is semisimple. Any Lie algebra of real or complex matrices closed under conjugate transpose $X \mapsto X^{\dagger}$ (this is equivalent to $G$ being closed under conjugate transpose) has this property because the form
$$X, Y \mapsto \text{Re}(\text{tr}(X^{\dagger} Y))$$
is a positive-definite (and in particular nondegenerate) invariant symmetric bilinear form on such matrices, so we can take the complement of a subrepresentation of the adjoint representation and get another subrepresentation.
A reductive Lie algebra is the direct sum of an abelian subalgebra (their center) and a semisimple algebra (their commutator), and hence is semisimple iff it has trivial center, which is equivalent to $G$ having discrete center. $\Box$
In general, definition 1 does not imply definition 2. For example, the universal cover $\widetilde{SL}_2(\mathbb{R})$ of $SL_2(\mathbb{R})$ is semisimple in the sense that its Lie algebra $\mathfrak{sl}_2(\mathbb{R})$ is semisimple, but its center is $\mathbb{Z}$.
However, one can say the following. If $\mathfrak{g}$ is a semisimple Lie algebra, then it admits a Cartan involution, namely an involution $\theta : \mathfrak{g} \to \mathfrak{g}$ such that
$$B_{\theta}(X, Y) = B(X, \theta Y)$$
is negative definite, where $B$ is the Killing form. If $\text{ad}_X : \mathfrak{g} \to \mathfrak{g}$ denotes the adjoint action of $X \in \mathfrak{g}$, then the adjoint (!) of $\text{ad}_X$ with respect to $B_{\theta}$ turns out to be $-\text{ad}_{\theta X}$ by direct calculation, so the adjoint representation exhibits $\mathfrak{g}$ as a Lie algebra of real matrices closed under transpose, and hence exhibits $G/Z(G)$ as a Lie group of real matrices closed under transpose. So definition 1 implies definition 2 up to taking covers, or equivalently up to discrete center. I don't know off the top of my head if requiring $G$ to be a matrix group fixes this.
$\endgroup$ 1 $\begingroup$In case of linear connected Lie groups, and only in this case as the counterexample of Qiaochu Yuan shows, the two definitions are actually equivalent, since the center of a connected linear Lie group whose Lie algebra is semisimple is always finite
You can see that by noticing that for a connected semisimple Lie group, the derived group is identical to the group. So a faithful representation of the lie group (that you can take irreducible by complete reducibility) will be inside $\text{SL}(V)$ for $V$ a complex vector space. By Schur lemma, the center will then necessarily be made of scalar matrices of determinant $1$. Since it is discrete, it is then finite.
If you go along the line of the Cartan decomposition, to follow the suggestion of Qiaochu Yaun, but at the level of the group, you will also obtain that if the Lie algebra (real or complex) is semisimple then the full group of automorphisms of this algebra has a finite center, even if it is not connected.
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