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

[redtree]

TL;DR

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.

 

[George Jones]

$Spin\left(n\right)$ 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

 

[redtree]

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?

 

[George Jones]

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?

 

[redtree]

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?