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I have look here in order to find an answer why the orthogonal group $O(n)$ has two path-components, given by $\det =1$ and $\det =-1$.

But what is a possible proof for that fact? I can easily show that there are at least two components (not path-components) by using the determinant-function.

  • No not really since the answer shows that there are at least two components (not path-components). This I already know. But thanks for the link anyway! –  May 12 '17 at 19:02
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    Every matrix Lie group is a smooth manifold, hence it is path-connected if and only if it is connected. – Dietrich Burde May 12 '17 at 19:04
  • Alright, but we don't know yet that this two sets are connected. Besides I don't know anything about matrix Lie groups und you statemant which you have cited, sorry! –  May 12 '17 at 19:07
  • How does it even make sense to talk about $O(n)$ having path components, without first construing it as a Lie group? – MJD May 12 '17 at 19:08

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$O(n)$ is not connected, hence not path-connected, which implies it has at least two path components, or else it would be path-connected.

Now say there are at least 3 path-connected components of $O(n)$, call them $A$,$B$ and $C$.

So the image of any continuous function, say the determinant function $f$ from $A\rightarrow \mathbb{R}$ is path-connected, so range of $f$ is a singleton otherwise range contains at least $2$ points hence range of $f$ is exactly $\{-1,1\}$, as determinant of an orthogonal matrix cannot be anything else, hence it would be not path-connected(using the intermediate value property of continuous functions).

Similarly, if we take the determinant of the elements of $B$ it would either all be $1$ or $-1$ , same is the case for the set $C$.

Now by pigeon-hole principle either all elements of $A$ and $B$ have the same determinant or all elements of $A$ and $C$ have the same determinant or all elements of $B$ and $C$ have the same determinant, hence union of the two of the the $3$ sets $A$,$B$,$C$ is path-connected(By the logic of existence of continuous functions), hence a contradiction to our hypothesis that $A$,$B$,$C$ were different path components!

Hence $O(n)$ has exactly $2$ path-connected components.

Arpan1729
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  • But how do you know that it has only two path components? " I can easily show that there are at least* two components (not path-components) by using the determinant-function"* – Ben Grossmann May 12 '17 at 22:07