Lie groups in Harish-Chandra's class A Lie groups in Harish-Chandra's class satisfies the following:

*

*$G$ is a real Lie group and its Lie algebra $\mathfrak g$ is reductive;

*$G$ has only finitely many connected components;

*$\mathrm{Ad}(G)$ is contained in $\mathrm{Int}(\mathfrak g_\mathbb C)$;

*The center of the Lie subgroup $G_{ss}$ is finite, where $G_{ss}$ is generated by the Lie subalgebra $[\mathfrak g,\mathfrak g]$.

I don't quite understand the motivation of the third and fourth requirements. Is there any concrete example of Lie groups satisfying this class, particularly the last two requirements? I also wonder whether all compact matrix Lie groups are in this class.
It will be also helpful if you could provide any further references than the original literature in explaining more about its motivation and examples.
 A: Here are some examples of groups that violate condition #3 or #4:

*

*$O(2n)$ does not satisfy #3. (Found in Knapp's Lie groups beyond an introduction, Chapter VII §2.)

*The universal cover of $\operatorname{SL}(2, \mathbb R)$ does not satisfy #4 because it has infinite center.

*Here is an example where $Z(G)$ has finitely many components but $Z(G_{ss})$ is infinite: Let $G_1 = \operatorname{SL}(2, \mathbb R)$, $\widetilde G_1$ its universal cover, $K = \operatorname{SO}(2, \mathbb R)$ and $\widetilde K$ its preimage in $\widetilde G_1$. The center of $\widetilde G_1$ is infinite cyclic and contained in $\widetilde K$. Define
$$G = (\widetilde K \times \widetilde G_1)/Z(\widetilde G_1)$$
where the action on the direct product is the diagonal one. Then $Z(G) = (\widetilde K \times Z(\widetilde G_1))/Z(\widetilde G_1)$ is connected even, yet $G_{ss} =  \widetilde G_1$ which embeds naturally in $G$ and it has infinite center. In this case, $G_{ss}$ is closed.

*Here is an example where $G_{ss}$ is not closed. The construction is similar. Let $\alpha$ be an irrational number and define
$$G = (K \times \widetilde G_1)/Z(\widetilde G_1)$$
where the action is the diagonal one and $Z(\widetilde G_1)$ is viewed as a subgroup of $K$ via $\exp_K(\alpha \cdot \log_{\widetilde K}(\cdot))$. Then the subgroup $\exp_G(\mathfrak k)$ is dense in $(K \times \widetilde K)/Z(\widetilde G_1)$ and the closure of $G_{ss}$ is all of $G$.

As to the motivation of #3 and #4, I can think of the following:
#3 Implies (but is not equivalent to) the condition that $\operatorname{Ad}(G)$ acts trivially on the center of $\mathfrak g$. That is, that $Z(G)$ does not have smaller dimension than $Z(G^0)$, $G^0$ being the identity component. This I think is completely natural. It is not equivalent, because there are outer automorphisms that fix the center of $\mathfrak g$ and you can make groups out of those using semidirect products. Those automorphisms are precisely the outer automorphisms of $\mathfrak g_{ss}$ and there are finitely many of them. (A Cartan involution is sometimes outer.) Most of the theory actually goes through when we assume only that $\operatorname{Ad}(G)$ acts trivially on $Z(\mathfrak g)$. But it is convenient to impose that all those automorphisms are inner, because then it makes sense to speak of regular elements of $\operatorname{Ad}(G)$ and of regular elements of $G$.
#4 Is useful because it implies that the $K_{ss}$ in the Cartan decomposition of $G_{ss}$ is compact, and that the same is true for every (connected) semisimple subgroup of $G$. And compact means that you can easily do analysis: you can integrate on a compact group without worrying about convergence and you can sometimes prove the existence of elements satisfying a certain smooth equation by studying the points where something is minimized or maximized. Compactness is just nice to have.
