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I need a reference for the proof that the complex orthogonal group $SO_{2n+1}($ℂ$) = \{A\in SL_{2n+1}($ℂ$): A^TA = Id\}$ is simple in a group theoretical sense (if it is true). How about the simplicity of $SO_{2n+1}(K)$ in general (i.e. $K$ an arbitrary infinite field)? It there any criterion? It seems that if $K$ is "big", then $SO_{2n+1}(K)$ is not simple, e.g. if $K$ has a valuation, then (I think) one can find somehow a proper normal subgroup.

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That's $O_n(\mathbb{C})$ you've defined. You should write $A \in SL_n(\mathbb{C})$ (otherwise there's an obvious normal subgroup :)). –  Ryan Reich Jun 21 '10 at 19:42
(I presume you mean $A\in\mathrm{SL}_n(\mathbb C)$.) When $n$ is even the group has non-trivial centre and hence is not simple but that is the only problem. Generally results on normal subgroups of semi-simple groups is due to Tits but this particular case may be earlier. –  Torsten Ekedahl Jun 21 '10 at 19:43
Thanks! I changed to odd dimension. –  Jakub Gismatullin Jun 21 '10 at 20:05
Another problem with the even case is that $SO_4$ is not simple as an algebraic group: it is isogenuous to $SO_3\times SO_3.$ –  Victor Protsak Jun 22 '10 at 0:50
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4 Answers

up vote 3 down vote accepted

The structure of classical groups goes back a long way and has been treated in a number of books, but in varying generality (arbitrary fields, various commutative rings, etc.). One older source in French is J.A. Dieudonne's concise Springer Ergebnisse volume La geometrie des groupes classiques (1963). A probably more readable modern textbook with limited aims is Larry Grove's Classical Groups and Geometric Algebra (AMS, 2002), just cited by Skip. There is also Emil Artin's old book Geometric Algebra and a much larger book by Hahn-O'Meara oriented more to algebraic K-theory. Anyway your group is simple both as an algebraic and as an abstract group (special orthogonal groups in odd dimension are also adjoint groups). There's no real need to get into algebraic groups, BN-pairs, or the like, though this is the "correct" general setting as Tits showed.

Actually, simplicity of various classical groups is proved sometimes in graduate algebra textbooks (which I don't have at hand). It depends how far you want to go. Over more general fields, especially of characteristic 2, a little more care is needed but these groups are still simple or very close to it even over most finite fields.

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In many places they assume "isotropicity" of the group (i.e. Witt index > 0) to get simplicity of the projectivisation of the derived subgroup (L. Grove, thm 6.31, page 58). In the case of usual SO_n, the Witt index is 0, so the simplicity is not clear. –  Jakub Gismatullin Jun 21 '10 at 21:33
Yes, I was following your formulation at first over $\mathbb{C}$. Once you get into the forms of Witt index 0 it gets more complicated. This is why the theory got developed further around isotropic and anisotropic forms over a given field, which gets reformulated in appropriate generality in the general theory of algebraic groups. After a while that setting seems more natural for simplicity questions. –  Jim Humphreys Jun 21 '10 at 22:00
To reinforce what Jim wrote in the preceding comment, if you assume the group is isotropic, you can typically prove results that depend only on the type of the group and not on the arithmetic of the ground field $K$. But if you allow the group to be isotropic, results typically involve heavily the arithmetic of $K$. To underline this distinction, one can compare the Kneser-Tits Problem (for isotropic groups) with the Margulis-Platonov Conjecture (same issue, but for anisotropic groups over number fields). These are the modern context for Jakub's question. –  Skip Jun 22 '10 at 0:52
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You have to be kind of careful. For the complex numbers, there is no problem as BCnrd wrote above. For a general field $K$, you are asking a basic question from the field of geometric algebra (there's an AMS GTM volume by Grove on the topic, and Dieudonne's books on classical groups).

If you take a nondegenerate quadratic form $q$, then we know:

  • If $q$ is isotropic over $K$, then $Spin(q)(K)$ is "projectively simple" (i.e., the quotient by its finite center is simple)
  • If $q$ is anisotropic, then $Spin(q)(K)$ can be far from simple. You can construct examples using valuations, as you suggest.

But you asked about $SO(q)(K)$. Then you use Galois cohomology (where I assume that $K$ has characteristic different from 2 for simplicity):

$1 \to \mu_2(K) \to Spin(q)(K) \to SO(q)(K) \to K^{\times}/K^{\times 2}$

where the last map is the spinor norm. You should think of the spinor norm as usually having a big image, so $SO(q)(K)$ will be pretty far from simple.

Your specific group

You asked specifically about $SO(q)$ where $q$ is a sum of squares. Then the spinor norm map has image products of sums of squares (typically not actually squares themselves), so you should expect the image of $Spin(q)(K)$ to be a normal subgroup in $SO(q)(K)$ of large index. (That is all very imprecise, but a precise answer depends on the arithmetic of $K$ and the dimension of $q$.) But now you can ask: Is $Spin(q)(K)$ modulo its center a simple group?

Here you have a strong advantage. If $q$ is isotropic over $K$ (i.e., you can write $0$ as a sum of a small enough number of nonzero squares), then you know from classical results that the answer is "yes".

If $q$ is not isotropic over $K$, then $Spin(q)$ is anisotropic (as an algebraic group) but it is split by the quadratic extension $K(\sqrt{-1})$. For such groups, under some hypotheses on $K$ (maybe as weak as characteristic zero), you know that the answer is again "yes" by Chernousov. See Theorem 9.7 on page 514 of the book "Algebra & Number Theory" by Platonov and Rapinchuk. You will have to inspect the proof (which starts on p.546) to see the precise hypotheses they need on $K$, or you can consult some of the original papers, where the proof is slightly different.

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A nice way to interpret Skip's example is that there's an exact sequence of $k$-gps $$1 \rightarrow \mu \rightarrow \widetilde{G} \rightarrow G \rightarrow 1$$ where $\widetilde{G}$ is s.c. central cover and $\mu$ is the mult. type finite center. Passing to $k$-points gives an exact sequence of pointed sets $$1 \rightarrow \widetilde{G}(k)/\mu(k) \rightarrow G(k) \rightarrow {\rm{H}}^1(k,\mu) \rightarrow {\rm{H}}^1(k,\widetilde{G})$$ with connected map a homomorphism. In many cases, final term is 1 and the 2nd-to-last term is computed by Hilbert 90 to be big (e.g., ${\rm{H}}^1(k,\mu_2)$). –  BCnrd Jun 21 '10 at 20:42
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A reference with a small amount of patching to do is Bourbakie: Groupes et algèbres de Lie, Chap IV, 2.7, it uses the theory of BN-pairs (which there are called Tits systems). It is shows that the only non-trivial normal subgroup is that consisting of the scalar matrices (which is non-trivial exactly when $n$ is even). As I said above this is probably overkill for this classical situation.

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You need to be more careful, or else there may be a non-trivial central element. Note that for $n$ even you have $-I$ in the special orthogonal group.

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How it works in odd dimensions? –  Jakub Gismatullin Jun 21 '10 at 20:05
Use Ch. IV, sec. 2.7, Cor. to Thm. 5 in Bourbaki LIE. If $G$ is split conn'd ss gp over field $k$ s.t. $G(k)$ is own commutator subgp and gen'td by $U(k)$'s for unip. radicals $U$ of Borel $k$-subgps $B$ of $G$ then structure theory for ss gps provides Tits system, so Bourbaki ref. implies $G(k)$ mod center is simple abstract gp. If $G$ is abs. simple and s.c. then these hypotheses on $G$ hold, by using "big cell" and structure of Weyl group (gen'td by simple reflections!) to reduce to case of ${\rm{SL}}_2(k)$ (win if $|k| > 3$!). So for $k$ alg. closed, $G(k)$ simple if $G$ is adjoint. –  BCnrd Jun 21 '10 at 20:23
I know this simplicity criterion (due to Tit's) with BN pair buissness, but the main point is the perfectness of the group. I don't know how to show that SO_n's are perfect. –  Jakub Gismatullin Jun 21 '10 at 21:04
Jakub, I had in mind the case of split groups (since I usually write ${\rm{SO}}_n$ to denote the special orthogonal group associated to the split quadratic form). Now I see that you specifically defined it relative to the "sum of squares" form, so then there will indeed be problems if the form is anisotropic (for example). Not sure what to suggest in that case, though over the real numbers there must be classical facts from compact groups which clarify the matter. Sorry! –  BCnrd Jun 21 '10 at 22:06
Jakub, then just keep reading. Basically, it's a refined group version of the Lie algebra fact that pairs of opposite root spaces generate $\mathfrak{sl}_2$'s. (The "refined" aspect is Bruhat decomposition using the Weyl group to nail down the entire group, not just some open around the identity.) At least over an alg. closed field, the version in Jim's book will give you the group version which I was invoking above. –  BCnrd Jun 21 '10 at 23:18
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