semi-simple and simple lie group,SO(n) for n even

Question

It is stated in the literature that SO(n) for n even is semi-simple or simple and either one of these means that it has no non-identity abelian factor groups but I,-I is an order 2 abelian factor group which is contained in SO(n) which contradicts this. Also if eliminate this one possibility,or say SO(n)/{I,-I} for n even >1, I would say SO(n) would be simple and not semi-simple . Can anyone give example of why these(my) statements are not true. It is stated in the literature that SO(n)/{I,-I} ,n even, is simple except for n=4 which is stated as only semi-simple. How can that be - in particular in the natural 4x4 matrix representation what exactly is a possible proper normal subgroup ? It can't be restricted to the space of any 2 or 3 dimensional submatrix that is where every element of the normal subgroup has a 1 on a certain diagonal(s). Could a normal subgroup be taken as a finite group? No but found an answer in Lie Algebras for Physicists by Ashok Das and Susumo Okubo, kupdf.net_das-amp-okubo-lie-groups-and-lie-algebras-for-physicist.pdf, pages 93-4. Subgroups $\exp(a_1J_1+a_2J_2+a_3J_3)$ for all constants $a_i$ or same but $J_i$ replaced by $K_i$ since $[J_i,K_j]=0$ and SO(4) is direct product of the two SO(3) type groups. Now would like to have a general proof of why the same type of analysis and development could not yield proper normal subgroups in SO(n) for n>4. I think possibly a similar type of analysis holds for the discrete finite alternating permutation group.

Answer 1

The first thing to resolve such a confusion is to know the correct definitions (which can be found for instance here).

Definition. A connected Lie group $G$ is said to be simple if its Lie algebra is simple (the usual definition of the latter includes the nonabelian condition). Equivalently, $G$ is nonabelian and contains no connected closed proper normal subgroups (proper means $\ne G$ and $\ne \{1\}$). A connected Lie group is semisimple if its Lie algebra is semisimple, i.e. is a direct sum of simple Lie algebras.

Caveat: It is sometimes convenient to regard the trivial group $\{1\}$ as a simple Lie group.

Accordingly, the fact that the group $SO(2k)$ contains the normal subgroup $\{\pm 1\}$ does not contradict its simplicity. In fact, $SO(n)$ is simple for all $n\ne 4, n\ge 3$, and $SO(4)$ is semisimple (its Lie algebra is isomorphic to $o(3)\oplus o(3)$.

The confusion comes from the fact that a simple Lie group need not be simple as an abstract group: $SO(2n), n\ge 3$, is simple as a Lie group and is not simple as an abstract group.

Answer 2

to MOishe : I had to put it as an answer because the following was too long for comment. Ok so you are considering the $\{\pm I\}$ subgroup, being finite and discrete is considered not connected I may guess? as far as that internet wikpedia article goes yes i did read that prior and in fact was my reason for saying, as far as technicalities of exact language definition, SO(2k) were not simple nor semi-simple even though practically it does share the same important properties of other simple or semi-simple groups. For i quote part of that article :

"As a counterexample, the general linear group is neither simple, nor semisimple. This is because multiples of the identity form a nontrivial normal subgroup, thus evading the definition. Equivalently, the corresponding Lie algebra has a degenerate Killing form, because multiples of the identity map to the zero element of the algebra. Thus, the corresponding Lie algebra is also neither simple nor semisimple. Another counter-example are the special orthogonal groups in even dimension. These have the matrix ? I {\displaystyle -I} -I in the center, and this element is path-connected to the identity element, and so these groups evade the definition. Both of these are reductive groups. "

so you note esp where he says 'another counter-example' which would logically mean of the same status as the general linear group - being also not simple nor semi-simple since he is implying both of the same category and note his phrase " Both of these are reductive groups. "

But anyway the important is not getting caught up in these exact definitions but the fact that for SO(2k), unlike the general linear group, the Killing metric tensor matrix is already NON-singular for SO(2k) without having to consider SO(2k)/{I,-I} which ofcourse is also NON-singular if the metric tensor has a meaning for SO(2k)/{I,-I}