I Is the spin group spin(n) a double cover for O(n)?

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The elements of the orthogonal group O(n) may not be continuously connected as in SO(n), but it seems that they should bijectively map to spin(n) for ##n\geq 3##
Every rotation in O(n) seems like it should map to spin(n) even if some rotations are not continuously connected to the identity.
 
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##Spin\left(n\right)## is the double cover of ##SO\left(n\right)##, while ##Pin\left(n\right)## is the double cover of ##O\left(n\right)##. The name is a pun. See (and references therein)

https://en.wikipedia.org/wiki/Pin_group
 
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Thanks! I read the link (and much else).

From the Wikipedia page: "The pin group of a definite form maps onto the orthogonal group, and each component is simply connected: it double covers the orthogonal group. The pin groups for a positive definite quadratic form Q and for its negative −Q are not isomorphic, but the orthogonal groups are."

Why doesn't the same hold true for semidefinite quadratic forms?
 
redtree said:
From the Wikipedia page: "The pin group of a definite form maps onto the orthogonal group, and each component is simply connected: it double covers the orthogonal group. The pin groups for a positive definite quadratic form Q and for its negative −Q are not isomorphic, but the orthogonal groups are."

Why doesn't the same hold true for semidefinite quadratic forms?
##\mathrm{Pin}\left(p.q\right)## is a double cover of ##\mathrm{O}\left(p.q\right)##. Is this what you mean?
 
I apologize. My question was poorly worded.

Pin##(p,q)## is the double cover of ##O(p,q)##. For a definite form, each element in Pin##(p,q)## is simply connected to ##O(p,q)##.

My question: for a semidefinite form, Pin##(p,q)## is still the double cover of ##O(p,q)##, but each element in Pin##(p,q)## is NOT simply connected to ##O(p,q)##. Is my understanding correct?
 
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