Payton Rasmussen

2022-10-09

(1) Prove that O(n) is homeomorphic to $\mathrm{S}\mathrm{O}(n)\times {\mathbb{Z}}_{2}$. (2) Are these two isomorphic as topological groups?

(O(n) : Orthogonal group, SO(n) : Special orthogonal group)

My attempt for (1) :Let's pick any element C from $\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)$, and let

$$\begin{array}{rl}& f:\mathrm{S}\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n);\phantom{\rule{thickmathspace}{0ex}}f(A)=A\\ & g:\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n);\phantom{\rule{thickmathspace}{0ex}}g(A)=CA\end{array}$$

Then f and g are homeomorphisms. Since SO(n) is compact, ${g}^{-1}(\mathrm{S}\mathrm{O}(n))=\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)$ is also compact. So, they are closed in the subspace topology on O(n). Thus, by the glueing lemma, the function

$$\phi :\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n)\times {\mathbb{Z}}_{2};\phantom{\rule{thickmathspace}{0ex}}\phi (A)=\{\begin{array}{ll}(f(A),0)\phantom{\rule{1em}{0ex}}& (\text{when}A\in \mathrm{S}\mathrm{O}(n))\\ (g(A),1)& (\text{otherwise})\end{array}$$

becomes continuous. We can easily confirm that $\phi $ is bijective and ${\phi}^{-1}$ is continuous.

Questions:

(1) I'm not sure I'm going in the right direction (in proving (1)). I used glueing lemma to justifying continuousness of the function $\phi $, but I doubt whether I'm using that lemma in the right place.

(2) I tried to prove that two groups are NOT isomorphic, but I can't find any key for that. I tried to compare orders of elements in two groups, but I couldn't find some good elements to compare... Is there any simple way to check whether two groups are isomorphic?

(O(n) : Orthogonal group, SO(n) : Special orthogonal group)

My attempt for (1) :Let's pick any element C from $\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)$, and let

$$\begin{array}{rl}& f:\mathrm{S}\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n);\phantom{\rule{thickmathspace}{0ex}}f(A)=A\\ & g:\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n);\phantom{\rule{thickmathspace}{0ex}}g(A)=CA\end{array}$$

Then f and g are homeomorphisms. Since SO(n) is compact, ${g}^{-1}(\mathrm{S}\mathrm{O}(n))=\mathrm{O}(n)-\mathrm{S}\mathrm{O}(n)$ is also compact. So, they are closed in the subspace topology on O(n). Thus, by the glueing lemma, the function

$$\phi :\mathrm{O}(n)\to \mathrm{S}\mathrm{O}(n)\times {\mathbb{Z}}_{2};\phantom{\rule{thickmathspace}{0ex}}\phi (A)=\{\begin{array}{ll}(f(A),0)\phantom{\rule{1em}{0ex}}& (\text{when}A\in \mathrm{S}\mathrm{O}(n))\\ (g(A),1)& (\text{otherwise})\end{array}$$

becomes continuous. We can easily confirm that $\phi $ is bijective and ${\phi}^{-1}$ is continuous.

Questions:

(1) I'm not sure I'm going in the right direction (in proving (1)). I used glueing lemma to justifying continuousness of the function $\phi $, but I doubt whether I'm using that lemma in the right place.

(2) I tried to prove that two groups are NOT isomorphic, but I can't find any key for that. I tried to compare orders of elements in two groups, but I couldn't find some good elements to compare... Is there any simple way to check whether two groups are isomorphic?

antidootnw

Beginner2022-10-10Added 10 answers

Step 1

Consider the mapping $f:O(n)\to SO(n)\times {\mathbb{Z}}_{2}$ given by

$$K\to \{\text{det}(M)\cdot M,\text{det}M\}$$

This is a well defined one to one and onto map. You can check that the map above is continuous.

Step 2

Now since we know that O(n) is compact and $SO(n)\times {\mathbb{Z}}_{2}$ is Hausdorff , the following map is a homeomorphism.

Consider the mapping $f:O(n)\to SO(n)\times {\mathbb{Z}}_{2}$ given by

$$K\to \{\text{det}(M)\cdot M,\text{det}M\}$$

This is a well defined one to one and onto map. You can check that the map above is continuous.

Step 2

Now since we know that O(n) is compact and $SO(n)\times {\mathbb{Z}}_{2}$ is Hausdorff , the following map is a homeomorphism.

Riya Andrews

Beginner2022-10-11Added 4 answers

Step 1

Consider the mapping $f:O(n)\to SO(n)\times {Z}_{2}2$ given by $M\to (det(M)\cdot M,detM)$

Step 2

I think your answer is not right. Check det $[det(M)M]=[det(M){]}^{n+1}$. Thus if n is even, it will be wrong.

Consider the mapping $f:O(n)\to SO(n)\times {Z}_{2}2$ given by $M\to (det(M)\cdot M,detM)$

Step 2

I think your answer is not right. Check det $[det(M)M]=[det(M){]}^{n+1}$. Thus if n is even, it will be wrong.

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