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I am trying to prove that $O(2)$ has two connected components. This is what I have done:
Suppose $A \in O(2)$. Then $A^tA=I$, where $A^t$ is the transpose of $A$ and $I$ is the identity. Taking the determinant of both sides of this equation, we get $\det(A^t)\det(A) = (det(A))^2$

which implies $\det(A) = \pm 1$.

And since $\det$ is continuous (which means that the inverse image of an open set is open), we have $O(2)$ is the disjoint union of $O^+(2)$ and $O^-(2)$. Is this right? Would this be sufficient or would this only prove that there are at least two components?

Andres Mejia
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Bob Joe
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  • That's right. A continuous function like the determinant must map connected sets to connected sets, so a disconnected set like ${-1, 1}$ is not possible. – Theo Bendit May 10 '19 at 03:21
  • So would this be sufficient or would this only prove that there are at least two components? – Bob Joe May 10 '19 at 03:26
  • You're right: you would need to prove that $O^+(2)$ and $O^-(2)$ are both connected. – Theo Bendit May 10 '19 at 03:28

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Actually, as written this is still not enough to show that there are at least two components. You would need to exhibit a matrix achieving both determinants (easy.)

To show that $O^+(2)$ is connected, you need to exhibit a path that is inside of the orthogonal group from one element to another. This can be done by showing that there is always a path to the identity. here is one such way.

Luckily for you $SO(2)$ has a very simple structure. Namely, take $S^1 \subset \mathbb C$ and try the map $S^1 \to SO(2)$ given by

$$t \mapsto \begin{pmatrix}\cos t& -\sin t \\ \sin t & \cos t \end{pmatrix}$$

where $t \in [0,2 \pi]$.

You can easily check that this map is well defined and injective. The only thing left to see is that this map is surjective. This can be done by noting that any columns in a member of $SO(2)$ are two orthonormal vectors sitting inside of $S^1$.

Why is this a homeomorphism?

Andres Mejia
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