why a particular map $f: \mathbb{S}^2 \times \mathbb{S}^1 \to SO(3)$ is not a homeomorphism?

Question

Let $SO(3)$ denote group of rotations of the unit sphere $\mathbb{S}^2$.

It is a well known fact that the fundamental groups $$ \pi_1 \Big( SO(3)\Big) \cong \frac{\mathbb{Z}}{2\mathbb{Z}} \not\cong \mathbb{Z} \cong \ \pi_1(\mathbb{S}^2 \times \mathbb{S}^1) $$

do not agree, hence two spaces $SO(3)$ and $\mathbb{S}^2 \times \mathbb{S}^2$ are not homeomorphic.

On the other hand, I can provide this map $f_{p,q}: \mathbb{S}^2\times\mathbb{S}^1 \to SO(3)$ which seems to be bijection and I don't see any points of discontinuity for it and its inverse.

Explain to me why particular $f_{p,q}$ is not a homeomorphism. It must either be non-injective, or non-sujecticive, or the map itself or its inverse must fail to be continuous, or fail to have a continuous inverse.

In order to construct $f_{p,q}$ choose two points $p,q \in \mathbb{S^2}$, such that $0 <d(p,q) = r < 2$ or equivalently $q \neq p \neq -q$. Let $$F_p(x) = \{ T \in SO(3) : Tp = x \}$$ It is easy to see that $F_p(x)$ is non-empty, select unit vector $v$ from $(\mathrm{span}(p,x))^\bot \cap \mathbb{S}^2$, then rotation by suitable angle $\alpha$ around $v$, namely $T_{v,\alpha} \in F_p(x)$. Select $y \in \mathbb{S}(x, r)$, where $\mathbb{S}(x,r) \subset \mathbb{S}^2$ is the circle of the radius $r$ around the point $x$.

I claim that the set $$ F_{p,q}(x,y) = \{ T \in F_p(x) : Tq= y \} \tilde{=} \{ f'_{p,q}(x,y) \} $$ is a singleton.

To see that $F_{p,q}(x,y)$ is nonempty take any map $A \in F_p(x) $ and compose it with rotaion around $x$ by the suitable anlge $\beta$, wich must exist as $A(q) \in \mathbb{S}(x,r)$ ($A$ is an isometry). By construction

$$T_{x,\beta}A(p) = T_{x,\beta}x = x \: \: \text{and} \: \: T_{x,\beta}A(q) = y $$

and whence $T_{x,\beta}A \in F_{p,q}(x,y)$.

If both $A,B \in F_{p,q}(x,y)$ then the map $B^{-1}A$ fixes the plane $\mathrm{span}(p,q)$ which must be 2-dimensional by selection of $p$ and $q$. The only rotation in $SO(3)$ which fixes the plane is the trivial one, which means that $B^{-1}A = I$ or just $A = B$.

At this point we have a map $f'_{p,q}$ with a wierd signature. But for each $x \in \mathbb{S}^2$ it is possible to find a homeomorphisms $\varphi_x:\mathbb{S}(x,r) \to \mathbb{S}^1$ so new map $f_{p,q}(x,y) = f'_{p,q}(x,\varphi_x^{-1}(y))$ is continuous. This must be possible as circles slide continuously across the sphere with their centers.

For every $A,B \in \mathrm{SO}(3)$ if $A(p) \neq B(p)$ or $A(q) \neq B(q)$ than $A \neq B$, hence $f_{p,q}$ is injective.

As every $A \in \mathrm{SO}(3)$ is uniquely characterized by its values $A(p)$ and $A(q)$ the map $f_{p,q}$ must be surjective.

The inverse must be continuous as both sets are Hausdorff and compact and at this point we have a continuous bijection.

Some values:

$f_{p,q}(p, 1 ) = I$

$f_{p,q}(p, e^{\mathrm{i} \alpha}) = T_{p,\alpha}$ is a rotation around $p$ by an angle $\alpha$.

having $x \neq p$

$f_{p,q}(x,1) \neq I$ is not an identity because $f_{p,q}(x,0)(p) = x \neq p$.

Answer 1

For each ${\def \S{\mathbb S}}x$ in $\S^2$ it must be possible to find a homeomorphism $\varphi_x:\S(x,r)\to \S^1$ so the new map $f_{p,q}(x,y)= f'_{p,q}(x,\varphi_x^{-1}(y))$ is continuous.

I think that this is where your error is. Supposing you have chosen these maps $\varphi_x$ in a continuous way, consider the vector field $\vec F$ on $\S^2$ where $\vec F(x)$ is unit magnitude pointing along the great circle from $x$ to $\varphi_x^{-1}(1)$. This is a non-vanishing continuous vector field on the sphere, contradicting the Hairy Ball Theorem.

Answer 2

I guess Mike pointed out a specific mistake in your construction, but I will try to show the "moral" reasons why you cannot do what you are trying to do.

I did not read your construction carefully, but I believe it is more or less the following: an element of $SO(3)$ is determined by what it does to first two vectors in the standard (or any fixed) basis: the third is completely determined because $SO(3)$ acts by isometries and preserves orientation. So we can just look at where the first basis vector goes (which can be anywhere in $S^2$), and then the second vector has to go somewhere on the circle orthogonal to the the image of the first vector.

In other words, we have a transitive action of $SO(3)$ on $S^2$, and the stabiliser of any point is an embedded copy of $SO(2)$, which is homeomorphic to $S^1$. In this way, $SO(3)$ is homeomorphic to a (locally trivial) bundle of circles over $S^2$. What you try to show is that this is a trivial bundle.

You can certainly define a bijection with the trivial bundle, but the problem is, you cannot make it continuous (if it was continuous, the inverse would automatically be continuous as well, by compactness). It is hard to visualise, but we can think of a simpler example of a non-trivial bundle, like the Mobius bundle.

(Note: I think one can also make an argument against triviality using the symmetry in choosing the order of the basis vectors: we could start with the second one and then choose where to put the first one. This would correspond to a homeomorphism $S^2\times S^1\to S^1\times S^2$, which you can probably show does not exist.)