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I think this question should already be abound in literature but the only place I find is from this article:

http://math.uchicago.edu/~lxiao/files/notes/Fundamental%20Groups.pdf

which seems to be elaborating this definition:

http://en.wikipedia.org/wiki/%C3%89tale_fundamental_group

but unfortunately as I do not understand much algebraic geometry I do not how to make use of this definition.


I am thinking about extending classical Bott periodicity to arbitrary rings that is good enough (UFD, for example). By extending I mean that I want to measure infinite matrices of entires in a ring $R$ with determinant 1 by the "one point compactification" of $R^{n}$ via introducing some topology. Hence in the classical case we can measure $U$ by $S^{n}$. I want to ask:

1): Is this possible? ( I thought about it over a bus trip but do not know how to establish universal bundles if the base ring is discrete, so I am stuck in here).

2): Is there any previous such constructions? What are their properties?

I feel there must be something well-known because Bott-periodicity theorem is a very old theorem. I do not know whether this is more appropriate for MO or for stack exchange, but I decided to put in here.

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I'm really curious as to what the answer is. Did you end up thinking about this some more? Also, what do you mean when you say you want to “measure infinite matrices?” Aren’t you just looking for a result saying something like $\pi_k(U(R)) \cong \pi_{k+2}(U(R))$ for $U_n(R)$ being $n\times n$ unitary matrices over $R$? –  David White Apr 14 '11 at 17:33
    
I’m also curious: how do the links at the top have anything to do with the question on Bott Periodicity? It is clear how to relate Bott Periodicity to etale maps and the algebraic $\pi_1$? –  David White Apr 14 '11 at 17:50
    
I did thought about the situation more and I still feel it is hopeless as I do not know much about general linear groups. I think you know what Bott periodicity is, I mean an analog by using sphere like objects to "measure" the topological quality of such matrices. –  Kerry Apr 22 '11 at 6:13
    
The links sited above introduced fundamental groups to schemes, but I do not understand much algebraic geometry. So I got stuck. –  Kerry Apr 22 '11 at 6:15
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1 Answer

Perhaps an easier approach if you wanted to avoid going into the world of schemes would be to define orthogonal, symplectic, and unitary matrices over $R$ in analogy with the case when $R=\mathbb{C}$. We should be able to at least find conditions on $R$ for which this could work (defining your ``good ring’’ of the title). For example, to define determinant and $GL(R)$, it's sufficient for $R$ to be commutative. To get the infinite unitary group $U(R)$ you need $R$ to have an involution (because of conjugate transpose). A good place to look for more info on this would be The Book of Involutions: http://www.amazon.com/Book-Involutions-Colloquium-Publications-Mathematical/dp/0821809040

We know that when quaternion algebras have involutions. This theory is developed in depth in the book. The book also looks at more general central simple algebras and includes a great example for endomorphism algebras on page 23. They prove that a central simple algebra $A$ over field $F$ has an involution of the first kind (fixing the center elementwise) iff $A\otimes_F A$ splits. If $B$ is a central simple algebra over $K$, a separable quadratic extension of $F$, then $B$ has an involution of the second kind (acts as an order 2 automorphism on the center) which fixes $F$ pointwise iff $N_{K/F}(B)$ splits.

This book also discusses the connection between Clifford algebras and orthogonal involutions, between the discriminant algebra and unitary involutions, and between Tits algebras and irreducible representations of the classical groups. So you might be able to use these connections to at least get some hands-on examples to see if Bott Periodicity extends (although you of course already have it for Clifford algebras).

Anyway, I’d also be very curious to see if the other approach (via schemes and algebraic homotopy groups) would work. But it seems much harder to me.

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Thank you for this comment. It is out of my expectation and very helpful. –  Kerry Apr 22 '11 at 6:16
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