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How big is the center of an arbitrary orthogonal group $O(m,n)$?

In the special case of the "circle group" $O(2)$, it's clear that $|\zeta O(2)|$ = 1. In the case of $O(3)$, it seems clear that the center has two elements $\zeta O(3) = \lbrace 1, -1 \rbrace$. I can see this by visualizing a sphere in an arbitrary $(i, j, k)$ basis, and observing that both the identity and the "complete" reversal $(i, j, k) \mapsto (-i, -j, -k)$ commute with everthing.

But I'd like a simple way to see how the situation changes for more general orthogonal groups like the (inhomogeneous) Lorentz group $O(3,1)$.

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The centre of $O(m,n)$ for $m+n\ge1$ always as exactly two elements: $I$ and $-I$. – Robin Chapman Jul 16 '10 at 13:20
I'm maybe utterly pedantic, however let's say that the center is trivial in char 2, since $I$ and $-I$ coincide. – Francesco Polizzi Jul 16 '10 at 13:30
Francesco, you're not being utterly pedantic. You could go further: In characteristic $2$ the group $O(2)$ is commutative and does not consist entirely of scalar matrices. – Tom Goodwillie Jul 16 '10 at 14:48
Soulphysics -- You are confusing the centre with the group of components. $O(n)$ has two connected components, $O(3,1)$ has four. – Bruce Westbury Jul 16 '10 at 15:27
Francesco, Tom: in char 2, you use bad notion of center; e.g., ${\rm{SL}}_p$ has scheme-theoretic $\mu_p \ne 1$ in char. $p > 0$. If $(V,q)$ is a non-deg. quad. space of dim $n \ge 3$ then ${\rm{SO}}(q) := {\rm{O}}(q) \cap {\rm{SL}}(V)$ is smooth (so has good notion of scheme-theoretic center) and its scheme-th. center is the "scalar" $\mu_2$ when $n$ is even and trivial when $n$ is odd. Also, ${\rm{O}}(q) = {\rm{SO}}(q)$ when $n$ even and ${\rm{O}}(q) = {\rm{SO}}(q) \times \mu_2$ when $n$ odd. So for char. 2 & $n > 2$, one should say ${\rm{O}}(q)$ has scheme-th. center $\mu_2$ ($ \ne 1$). – BCnrd Jul 16 '10 at 16:18

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