I'm working through Problem 4.16 in Armstrong's Basic Topology, which has the following questions:
- Prove that $O(n)$ is homeomorphic to $SO(n) \times Z_2$.
- Are these two isomorphic as topological groups?
Some preliminaries:
Let $\mathbb{M_n}$ denote the set of $n\times n$ matrices with real entries. We identify each matrix $A=(a_{ij}) \in \mathbb{M_n}$ with the corresponding point $(a_{11},a_{12},…,a_{1n},a_{21},a_{22}…,a_{2n},…,a_{n1},a_{n2},…,a_{nn}) \in \mathbb{E}^{n^2}$, thus giving $\mathbb{M_n}$ the subspace topology.
The orthogonal group $O(n)$ denotes the group of orthogonal $n \times n$ matrices $A \in \mathbb{M_n}$, i.e. with $det(A)=\pm{1}$.
The special orthogonal group $SO(n)$ denotes the subgroup of $O(n)$ with $det(A)=1$.
$Z_2=\{-1, 1\}$ denotes the multiplicative group of order 2.
My attempt
For odd $n$, the answer to both questions is yes, as we verify below. Consider the mapping
$f:O(n)\to SO(n)\times Z_2, A \mapsto(det(A)\cdot A, det(A))$.
We have the following facts about $f$:
-
It is injective. If $f(A)=f(B)$ then $(det(A)\cdot A, det(A))=(det(B)\cdot B, det(B))$.
Therefore, $det(A)=det(B) \neq 0$ so $A=B$. -
It is surjective. For $(D,d) \in SO(n) \times Z_2$, we can take $dD \in O(n)$,
giving $f(dD)=(det(dD)\cdot dD, det(dD))=(d^n\cdot det(D) \cdot dD,d^n \cdot det(D))=(d^{n+1}D, d^n)=(D,d)$, since $n$ is odd. -
It is a homomorphism. $f(AB)=(det(AB)\cdot AB, det(AB))=(det(A)det(B)\cdot AB, det(A)det(B))$
$=((det(A)\cdot A)(det(B)\cdot B), det(A)det(B))=f(A)f(B)$. -
It is continuous. Let $\mathcal{O} \in SO(n) \times Z_2$ be open. Then $\mathcal{O}=U \times V$ for $U$ open in $SO(n)$ and $V$ open in $Z_2$.
Since $SO(n)$ is open in $O(n)$, $U$ is therefore open in $O(n)$. $-U=\{-A\mid A\in U\}$ is also open in $O(n)$. But $f^{-1}(\mathcal{O})=f^{-1}(U\times V)=U\cup -U$.
Since $O(n)$ is compact and $SO(n)\times Z_2$ is Hausdorff, we therefore have that $f$ is a homeomorphism. Thus, they are isomorphic as topological groups.
For even $n$, this mapping is not well-defined: if $A \in O(n)$ with $det(A)=-1$ then, $det(det(A)\cdot A)=(det(A))^{n+1}=-1$, so $det(A)\cdot A \notin SO(n)$.
My question then is are they homeomorphic as topological spaces if $n$ is even?
From the related questions, it seems like for even $n$, the two groups cannot be isomorphic due to one being abelian while the other is not and them having different centers and derived subgroups (I don't fully understand these arguments but I will brush up on them). So they cannot be isomorphic as topological groups. But can they be homeomorphic as topological spaces?
Related questions:
Are $SO(n)\times Z_2$ and $O(n)$ isomorphic as topological groups?
Two topological groups $\mathrm{O}(n)$ (orthogonal group) and $\mathrm{SO}(n)\times \mathbb{Z}_2$
understanding $O(n)$ homeomorphic to $SO(n)\times \Bbb Z_2$ proof
Best Answer
Consider the map $\varphi\colon SO(n) \times \mathbb{Z}/2 \to O(n)$, $\varphi(M, s) = I_s M$ where $I_0 = \mathrm{Id}_n$ is the identity matrix and $I_1 = \operatorname{diag}(-1, 1, \ldots, 1)$ is the diagonal matrix with a $-1$ as the first entry and all other diagonal entries equal to 1. Clearly $I_s \in O(n)$ for both $s = 0$ and $s = 1$, so the map is well-defined and evidently continuous. Since the image of $SO(n) \times \{0\}$ under $\varphi$ consists exclusively of matrices with determinant 1 and that of $SO(n) \times \{1\}$ exclusively of matrices of determinant $-1$, the map is injective (since on each component it is just given by multiplication with a group element), and surjectivity is also not difficult to see: If $M \in O(n)$ has $\det(M) = -1$, then $\det(I_1 M) = 1$ and consequently $I_1 M \in SO(n$), whence $M = \varphi(I_1 M)$ lies in the image. Thus, we conclude (by the compact-Hausdorff lemma) that $\varphi$ is a homeomorphism.