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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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    $\begingroup$ The centre of $O(m,n)$ for $m+n\ge1$ always as exactly two elements: $I$ and $-I$. $\endgroup$
    – Robin Chapman
    Commented Jul 16, 2010 at 13:20
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    $\begingroup$ I'm maybe utterly pedantic, however let's say that the center is trivial in char 2, since $I$ and $-I$ coincide. $\endgroup$
    – Francesco Polizzi
    Commented Jul 16, 2010 at 13:30
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    $\begingroup$ 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. $\endgroup$
    – Tom Goodwillie
    Commented Jul 16, 2010 at 14:48
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    $\begingroup$ 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$). $\endgroup$
    – BCnrd
    Commented Jul 16, 2010 at 16:18
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    $\begingroup$ I'm reluctant to add yet another comment, but the original question was too imprecisely formulated to have a clear answer. The structure of classical groups depends heavily on the underlying field, apparently meant to be the real field here; but also the notation used gets somewhat out of control from a mathematical viewpoint. There is definitely a communication barrier between mathematics and physics even in the most classical areas. $\endgroup$
    – Jim Humphreys
    Commented Jul 18, 2010 at 22:19

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