Everyday I'm $\mathbb{RP}^{n+1}$

When I was just a little boy and my dad was tucking me into bed one night, we (somehow) got to talking about the overall shape of the universe.  It just sort of goes on forever... doesn't it?  According to my old man, there were some people who thought that space didn't exactly go on forever.  "When you travel far enough in that direction," he said, pointing to one corner of the room, "you eventually come back to the place you started, but you'll approach it from there," he said as he pointed to the opposite corner.

At the time,  I had imagined that he was describing something like a 3-torus, i.e. $S^1 \times S^1 \times S^1$.  Nowadays, I'm certain he was talking about $\mathbb{RP}^3$ instead.

Hrm... I should ask him about this stuff the next time I see him...

Pinkie Pie's parents must have taught her how to appreciate abstract notions of space as well.

Today we'll construct $\mathbb{RP}^3$ and see why it's homeomorphic to $SO(3)$, a subset of $O(3)$.

Don't fret if you don't know what it means for two spaces to be 'homeomorphic'.  We're just going to visualize each one and convince ourselves that they're both the same shape.

First: that stuff about $\mathbb{RP}^2$ should be somewhat fresh in your mind.  There's yet another way to build this space, which we'll be using today:

Take a solid circle, $\{\vec{x} \in \mathbb{R}^2 : ||\vec{x}|| \leq 1\}$, and associate
each boundary point (i.e. $||\vec{x}||=1$) with its antipode.

While this is much easier to visualize than equivalence classes of lines in $\mathbb{R}^3$, we don't get the nice ways of turning geometric objects in $\mathbb{RP}^2$ (points, lines, and conic sections) into familiar geometric objects in $\mathbb{R}^3$ (lines through the origin, planes through the origin, and cones whose apex is the origin, respectively).  Cramming real affine $2$-space into a unit disc doesn't create any problems from the topological perspective, but all the familiar algebraic structures get trampled.

Anyway, since this post is going to be a lot of topological hand-waving, you should take a moment to convince yourself that the procedure above really does give you the same $\mathbb{RP}^2$ as in the last post.  Because we'll be building $\mathbb{RP}^3$ in a similar fashion:

Take a solid sphere, $\{\vec{x} \in \mathbb{R}^3 : ||\vec{x}|| \leq 1\}$, and associate
each boundary point (i.e. $||\vec{x}||=1$) with its antipode.

You should recognize this boundary as a copy of $\mathbb{RP}^2$: that is, $\mathbb{RP}^3$ has a "plane at infinity."  The lines at infinity are simply the 'great circles' or diameters of this boundary sphere.  Every 'affine line' in this configuration touches a point at infinity, (and therefore must connect one pole to the other within this closed sphere).  Note that when we consider the word 'straight line' in the sense of the copy of $\mathbb{R}^3$ within which this solid sphere exists, (and not in terms of what is a straight line in the copy of $\mathbb{R}^3$ being embedded in the solid sphere) there is only one 'straight line' between these antipodes of the boundary circle.

Remember how I said this sort of construction destroys the familiar algebraic structures?  The 'affine lines' belonging to the embedded $\mathbb{R}^3$ that don't go through $\vec{0}$ have to 'bend' around the origin in a certain way.  The way this bending happens is dependent on what function we used to compress $\mathbb{R}^3$ to an open sphere.  Again: the algebraic/geometric picture is a bit ugly, but this is the most straightforward approach from the topological angle.

It's simple, it works, just roll with it.

Now to talk about $SO(3)$...

You may recall that $O(3)$ is the set of $3 \times 3$ real-valued orthogonal matrices, and that the columns of every orthogonal matrix form a basis of $\mathbb{R}^n$.  In case you weren't aware of it before, an ordered orthonormal basis of $\mathbb{R}^3$ can be right handed or left handed.  Because the world is run by left-brained people, the canonical basis for $\mathbb{R}^3$ of $\{ \vec{x}, \vec{y}, \vec{z} \}$ is right handed.  Why does this handedness matter? It is the basis upon which $O(3)$ splits into two connected components

In one component of $O(3)$, the columns of the matrix form a right-handed system.  Thus by changing our basis to the columns of such a matrix, we preserve the 'handedness' of our basis.  Such a transformation can always be expressed as a rotation, and the determinant of such a matrix is always $1$.  These matrices form a normal subgroup of $O(3)$, and they are denoted $SO(3)$, or the special orthogonal group.

Therefore the other component, sometimes denoted $o(3)$, corresponds to the matrices that reverse the handedness of the basis.  These matrices have determinant $-1$ and correspond to a reflection of the canonical basis (plus or minus some rotations).

How do we build a 'space' out of a set of rotations?  For any elements $A, B \in SO(3)$, let the distance from $A$ to $B$ be the minimum number of degrees (or radians) one must rotate the basis given by $A$ to turn it into the basis given by $B$.

Exercise: Find your favorite criteria for a distance metric, and show that this satisfies all of them.

Now let's visualize stuff.  There are three 'dimensions' you can rotate yourself in: there is roll, pitch, and yaw.  These rotations don't commute though.  If we let the rotations occur relative to the axes we start with (instead of Rainbow Dash's current heading) rolling $90^\circ$ degrees clockwise is equivalent to 'yawing' $90^\circ$ degrees to the right, 'pitching' $90^\circ$ degrees down, and then 'yawing' $90^\circ$ degrees to the left.  Still, this nonabelian reduction tends to give us angles much bigger than the ones we started with, so if I want to answer 'local' questions like what Rainbow Dash's angular velocity is, I'll still need three dimensions.

Topology is now 20% cooler

Now let's get back to that visualization stuff...

We won't prove it here, but every $A \in SO(3)$ can be written as a rotation around some axis $\vec{v} \in \mathbb{R}$ (see "$SO(3)$ What").  But we can do even better: if $A$ is a counterclockwise rotation around $\vec{v}$, it is a clockwise rotation around $-\vec{v}$;  Thus each $A$ is a clockwise rotation around a unit vector $\vec{v}$.

We can now arrange these points in $\mathbb{R}^3$, based on their 'distance' from our starting position: We can map each $n^\circ$ rotation clockwise around the unit vector $\vec{v}$ to the point $n\vec{v}$.  All these distinct orientations for Rainbow Dash sort of 'bloom' in $\mathbb{R}^3$, forming a solid ball that grows until we start considering rotations of $180^\circ$ or more.  Rolling $180^\circ$ clockwise is equivalent to rolling $180^\circ$ counterclockwise, and the same goes for 'pitching', 'yawing', or rotation about any other axis.  So we must associate each point on the boundary of this solid sphere with its antipode.

Sound familiar?

So to this very low level of rigor, $SO(3) \cong \mathbb{RP}^3$.  This is actually one of the routes through which some abstract math can apply to reality.  If you make two clockwise twists in a belt, the resulting 'kink' is something you can work out by just translating one end of the belt around the other without introducing any sort of counterclockwise twisting.

This belt-twisting phenomenon is, in the proper jargon, a demonstration of a homotopy between the path in $SO(3)$ that makes two full rotations and the 'identity' (or do-nothing) path.  This phenomenon exists, in some sense, because of what the first homotopy group of $SO(3)$ is: $\pi_1\left(SO(3)\right) \cong \pi_1\left(\mathbb{RP}^3\right) \cong \mathbb{Z}_2$.

My apologies for dumping that on you without explaining what 'homotopy' is.  But this is one of the more surprising and sort of magical results out there.