Are $O(n)$ and $SO(n)\times Z_2$ homeomorphic as topological spaces?

Question

I'm working through Problem 4.16 in Armstrong's Basic Topology, which has the following questions:

1. Prove that $O(n)$ is homeomorphic to $SO(n) \times Z_2$.
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.

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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?

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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

Why is the orthogonal group $\operatorname{O}(2n,\mathbb R)$ not the direct product of $\operatorname{SO}(2n, \mathbb R)$ and $\mathbb Z_2$?

Answer 1

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.

Answer 2

Let
$$ \mathrm{O}(n) = \{A \in \mathrm{Mat}_n(\mathbb R): A^\intercal A = I_n\}, \quad C_2 = \{\pm 1\} $$ be, respectively, the orthogonal group and the group with two elements. Since $\det\colon \mathrm{Mat}_n(\mathbb R)\to \mathbb R$, is multiplicative and $\det(A^\intercal) = \det(A)$, if $A \in \mathrm{O}(n)$ then $1=\det(A^\intercal)\det(A) = \det(A)^2$, so that $\det(A)\in C_2$. Moreover, if $s\colon C_2\to \mathrm{O}(n)$ is given by $s(\epsilon) = \mathrm{diag}(\epsilon,1,\ldots,1)$, then $s$ is a group homomorphism and $\det(s(\epsilon))=\epsilon$. Since by definition $\mathrm{SO}(n):=\det_{|\mathrm{O}(n)}^{-1}(\{1\})$ it follows that $\mathrm{O}(n)= \mathrm{SO}(n).H$ where $H = s(C_2)$.

An internal product like this can also be described as a semi-direct product. In this case, as a set the semidirect product is $\mathrm{SO}(n)\times C_2$, but the group operation is not simply componentwise, rather it is twisted by an action of $C_2$ by automorphisms on $\mathrm{SO}(n)$: for a pair of elements $(g_1,\epsilon_1),(g_2,\epsilon_2) \in \mathrm{SO}(n)\rtimes C_2$ their product is defined to be $$ (g_1,\epsilon_1).(g_2,\epsilon_2) = (g_1 s(\epsilon_1) g_2 s(\epsilon_1)^{-1}, \epsilon_1\epsilon_2), $$ or, if we write $c_g(h) = ghg^{-1}$ for $g,h \in \mathrm{SO}(n)$, then $(g_1,\epsilon_1).(g_2,\epsilon_2) = (g_1c_{s(\epsilon_1)}(g_2), \epsilon_1\epsilon_2)$. The fact that $\mathrm{O}(n) = \mathrm{SO}(n)\times H$ can then be rephrased as saying that the map

$$ m \colon \mathrm{SO}(n) \rtimes C_2 \to \mathrm{O}(n), \quad m(g,\epsilon) = g.s(\epsilon), (\forall g \in \mathrm{SO}(n), \epsilon \in C_2), $$ is an isomorphism of groups.

We may view $C_2$ as a topological group by equipping it with the discrete topology (so that any subset of $C_2$ is open in $C_2$) so that $\mathrm{SO}(n)\rtimes C_2$ is a product of topological groups and hence a topological group itself. Thus it makes sense to ask if $m$ is also homeomorphism.

Now since $\det$ is continuous and the image of $\det_{|\mathrm{O}(n)}$ is the discrete set $\{1,-1\}$, it follows that $\mathrm{SO}(n)$ and $\mathrm{SO}(n).s(-1)$ are both open in $\mathrm{O}(n)$. Now if $g \in \mathrm{SO}(n)$ then $m^{-1}(g) = (g,+1)$, which is clearly continuous since its components $g \mapsto g$ and $g\mapsto 1$ are, and similarly if $g\in \mathrm{SO}(n)s(-1)$, then $m^{-1}(g) = (gs(-1),-1)$ so that the restriction of $m^{-1}$ to any coset of $\mathrm{SO}(n)$ in $\mathrm{O}(n)$ is continuous. Since these cosets form an open cover of $\mathrm{O}(n)$, it follows that $m^{-1}$ is continuous, and hence $m$ is an isomorphism of topological groups, i.e. an isomorphism which is also a homeomorphism.