It is known that the sphere $\mathbb S^3$ may be decomposed as two solid tori. It may be done with the following "parametrization" $f:(\mathbb S^1\times \mathbb S^1 \times [0,1])_{/\sim}\to \mathbb S^3\subseteq \mathbb R^4$: $$f(\alpha, \beta , t) = (\cos \frac{t\pi}{2} \cdot \alpha,\ \sin \frac{t\pi}{2} \cdot \beta).$$
There must be a mistake somewhere, but I can't find it: I parametrized $\textrm{SO}(3)$ (which is the same as $\mathbb R\textrm P^3\neq \mathbb S^3$) in the same way.
Any matrix from $\textrm{SO}(3)$ is uniquely determined by its two first columns.
Assume that the first column $A$ is different from the poles $\pm(0,0,1)$. Then the second column $B$ can be described in terms of angle. Let's look at $B$ as a vector tangent to the sphere $\mathbb S^2$ at point $A$. Then, $0$-angle corresponds to vector $B$ pointing east, $\pi/2$ corresponds to $B$ pointing north and so on.
So this part of $\textrm{SO}(3)$ may be paramtrized by $\mathbb S^2\setminus \{\pm(0,0,1)\} \times \mathbb S^1 \simeq (-1,1) \times \mathbb S^1 \times \mathbb S^1$ (with coordinate functions: $t,\varphi, \psi$).
Now assume that $A=\pm(0,0,1)$. Parametrize $B$ with angle in the following way: $B(\phi)=(0,\cos(\phi+\frac{\pi}{2}), \sin(\phi+\frac{\pi}{2}))$. How to glue it with the previous one? Let:
$$ B(-1,\varphi, \psi) = B(\varphi+\psi),\\ B(1, \varphi, \psi) = B(\varphi-\psi).$$
If we reparametrize the torus $\mathbb S^1\times \mathbb S^1$ with $\alpha = \varphi + \psi$ and $\beta = \varphi - \psi$, we can see that function $g=(A,B)(t,\alpha,\beta):[-1,1] \times \mathbb S^1 \times \mathbb S^1 \to \textrm{SO}(3)$ identifies the same points as $f$ did (with obvious corrections on order of the factors and length of the interval).
The proof relies on the three: 1) any continuous map from a compact space $X$ onto a Hausdorff space $Y$ induces a homeo between the obvious quotient $X_{/\sim}$ and $Y$, 2) $B$ is continuous, 3) the identifications are the same.
What am I missing?
Answer
Since savick01 already wrote a self-answer describing where the proof goes wrong, I'll concentrate in this answer on a more intuitive picture of what happens. Therefore I'll also not directly start at the mapping, but first describe some equivalences which help to visualize what happens
Mathematically, this identification is trivial. We use the standard embedding of $S^3$ into $\mathbb R^4$ as the unit sphere around the origin, given by the equation $$x^2 + y^2 + z^2 + w^2 = 1\ ,$$ where a point in $x^4$ is given by the coordinates $(x,y,z,w)$. Solving this equation for $w$ gives the solutions $$w = \pm\sqrt{1-x^2-y^2-z^2}\ .$$ Thus the equation has two solutions for $x^2 + y^2 + z^2 < 1$, one for $x^2+y^2+z^2=1$, and none for $x^2+y^2+z^2>1$. Now the equation $x^2+y^2+z^2\le 1$ describes the closed unit ball in $\mathbb R^3$. So we have two unit balls, one for positive and one for negative $w$. The exception is at the border, where $w=0$ and therefore the points of both unit balls with equal coordinates describe the very same point of $S^3$.
So far for the description. But what does it mean? To see that, we go to one dimension less and consider the sphere $S^2$, embedded in $\mathbb R^3$. Now consider the parallel projection when looking for example from positive $z$ direction (that is, from above). This projection means basically removing the $z$ coordinate. The image of the sphere is, of course, the unit disk. Now if a point on the sphere has a positive $z$ value, we can see it from above, otherwise we can't. It is customary to draw "unseen" lines on a sphere dashed in the projection. Another option is to use grey instead of black, which has the advantage that it also works for points. See the following image for an example (although you've probably already seen tons of such images):
This shows the disk (actually the circle which is the border of the disk), and the image of a great circle, where the front part ($z>0$) is drawn in black and the back part in grey. Now what this actually means is that there are two disks drawn on top of each other, where one disk is drawn in black, and the other in grey:
The circle which forms the border of the disk of course only exists once, and only there we can get from one of the disks to the other.
Those two "front" and "back" disks for the 2-sphere are the exact equivalents to the two balls for the 3-sphere. Indeed, we also can combine those two balls into one and use different colours to mark "front" and "back". This gives a visual mental model for $S^3$.
To see what the parametrization $f$ does, we expand the circles $S^1$ into two coordinates each using the standard embedding, so we get the coordinates $$f(\alpha, \beta, t) = (\cos(t\pi/2)\cos\alpha, \cos(t\pi/2)\sin\alpha, \sin(t\pi/2)\cos\beta, \sin(t\pi/2)\sin\beta)$$
Now if we project it into our two balls, we first see that for $t=\mathrm{const}$ and $\beta=\mathrm{const}$ we get a circle with radius $\cos(t\pi/2)$ parallel to the $x$-$y$ plane with center on the $z$ axis, an height $z=\sin(t\pi/2)\cos\beta$ on one of the two spheres.
Now we also let the parameter $\beta$ run, to get the full torus. By doing so, we find that the circle just moves along the $z$ axis, until it reaches the border of the ball where it "jumps" to the other ball and moves again along the $z$ axis. That is, we get a pair of cylinders, one on each sphere, both equal. Note that each pair of corresponding vertical lines on both spheres give a circle on $S^3$; this is the sane as the latitude circles look like lines when looking at earth from the side, and the longitude circles do so when looking from a pole.
The toruses for different $t$ correspond to cylinders of different radius, just as if we had used apple corers of different size to cut through the spheres.
There are two special cases: $t=0$ and $t=1$. For $t=0$, we get the equator of the balls (the equator is shared because it's at the border). That is, the torus degenerated to a great circle. For $t=1$, we get the z axis for both balls. Note that this also is a great circle of $S^3$.
Now having explored $S^3$, let's explore the manifold corresponding to $SO(3)$, the group of rotations in the three-dimensional space. As everyone knows, a rotation can be described by its rotation axis and the rotation angle. Since an axis gives a direction and an angle gives a magnitude, one can combine both into a vector. The zero vector is then the identity, and any other vector describes a rotation of the angle given by its length around the corresponding axis. One can show that this is a continuous mapping, that is, only slightly different rotations lead to only slightly different vectors. However, a rotation of $\pi$ around an axis and a rotation of $-\pi$ around that same axis are the same transformation, therefore one has to identify both. This of course means that we can restrict ourselves to vectors of length $\le\pi$ because any longer vector would "wrap around" to the other side. Therefore the rotations are described by a ball of radius $\pi$, where antipodal points on the border sphere are identified.
Of course one can scale that ball to an unit ball.
The first vector of the rotation matrix describes the direction into which the unit vector $e_x$ is rotated. So if we define Euler angles using the $x$ direction instead of the $z$ direction and replace the angle for the $y$ rotation (which goes from $0$ to $\pi$) by its cosine, we get exactly the parametrization from the question: $A$ is determined by two angles describing a point on $S^2$, and $B$ is described by the third angle.
(To be continued; it's now 1am here :-))
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