Is O(2) homeomorphic to anything simplier?

kyam · New Member Offline Joined: 26 Jan 2005 Posts: 5

I'm in the middle of parameterizing the orthogonal group O(2). Is it possible?

O(2) has the form
[ cos(t) , sin(t) ]
[ -sin(t) , cos(t) ].
Is it homemorphic to a circle S^1, and how can you show that it is not
homeomorphic to [0,1), or any simple subset in R? It is classic to show that when
f : [0, 2π) → S^1 is defined by
f(φ) = (cos(φ), sin(φ)),
then f is not the homeomorphism required because inv(f) is not continuous.
But how do you know (or show) that two spaces are not homeomorphic?

Your comments are greatly appreciated.
Thank you very much for consideration in this matter.

Regards,
Kai

**Posted:** Wed Jan 26, 2005 9:09 pm

Kent Merryfield

The set of matrices you wrote down is SO(2), not O(2). SO(2) is indeed homeomorphic to the circle; if you endow the circle with the multiplicative group structure of the complex numbers of modulus 1, that's also a group isomorphism.

O(2) includes reflections, hence matrices of determinant -1 as well as matrices of determinant 1. Topologically, it is disconnected. It has two connected components, each separately homeomorphic to the circle.

How to show the circle is not homeomorphic to a line segment? One way uses the Intermediate Value Theorem. Any continuous mapping from the circle to a segment (or to the whole line) must take on some value at least twice.

**Posted:** Wed Jan 26, 2005 11:01 pm

grobber

More generally, the sphere $\mathbb S^n$ is not homeomorphic to $\mathbb R^n$ for the same reasons (Borsuk Ulam). I think a similar question has been discussed before: how to show that $\mathbb R^3$ is not homeomorhic to $\mathbb R^2$ ? Borsuk Ulam solves this one pretty quickly, but are there any nice proofs that rely less on such advanced stuff?

**Posted:** Thu Jan 27, 2005 5:43 am

kyam · New Member Offline Joined: 26 Jan 2005 Posts: 5

Thank you for mentioning the Borsuk-Ulam theorem. I looked it up and it is very helpful.

**Posted:** Thu Jan 27, 2005 11:28 am