understanding $O(n)$ homeomorphic to $SO(n)\times \Bbb Z_2$ proof

Asked Sep 22 '22 at 21:19

Active Sep 23 '22 at 11:14

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I am doing Exercise 4-16 in Armstrong's Basic Topology.

The question is to prove $O(n)$ homeomorphic to $SO(n)\times \Bbb Z_2$. I have some idea on how to approach this, but when I see other people's solution online I was very confused.

The comment suggested a function $$f:O(n)\to SO(n)\times \Bbb Z_2; f(A)=(\det(A)A, \det(A))$$ and claimed that it was a one-one, onto map. I also see this on some greg grant's solution material. But isn't $O(n)$ set of orthogonal matrices, implying the determinant of its element being $\pm1$? How could it map to anything $(..,0)$, then how is it onto?

I understand I can easily map between $\{0,1\}$ and $\{-1, 1\}$. But since it wrote $\Bbb Z_2$, I was wondering if there is some notation or basic things I am not aware of .

Thanks in advance.

edited Sep 23 '22 at 11:14

Kevin.S

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asked Sep 22 '22 at 21:19

fourofour_un

7

You understood everything correctly. It is common practice to identify ${ \pm 1}$ with $\mathbb{Z}_2$. – nilsw Sep 22 '22 at 21:21
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(In particular, by letting the group operation be multiplication.) – fish Sep 22 '22 at 21:23
Ahh, good to know. Thank you both! – fourofour_un Sep 22 '22 at 21:43
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The map is not well-defined if $n$ is even. You need to replace $\det{A}$ in the first coordinate with “a fixed orthogonal matrix with determinant $\det{A}$”. – Aphelli Sep 22 '22 at 21:44
A related question https://math.stackexchange.com/q/1532628/360262 – yi li Sep 25 '22 at 15:21

understanding $O(n)$ homeomorphic to $SO(n)\times \Bbb Z_2$ proof

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