# Gracious Living

Banach-Tarski part 2
November 21, 2010, 23:30
Filed under: Algebra, Math | Tags: , , , , ,

So I sort of left you hanging last time.  We talked about equidecomposability, showed that $F_2$ was paradoxical under its own action on itself, and embedded $F_2$ into $SO(3)$.  From here, it just becomes a matter of putting all the steps together: first the sphere, then the ball minus its center, then the whole ball.

First, let’s look at the sphere.  Here’s a simple fact about group actions: any group action on a set restricts to a group action on a subset of that set if and only if that subset is a union of orbits.  For $SO(3)$, the orbit of a point is the sphere about the origin containing it, since all orthogonal matrices preserve distance.  In particular, $SO(3)$ acts on the unit sphere, and since this is just one orbit, the action is transitive (for any pair of points, an element of $SO(3)$ takes the first to the second).

In addition, group actions restrict to subgroup actions, so the copy of $F_2$ in $SO(3)$ also acts on the sphere.  Recall that this is generated by rotations of $\arccos(3/5)$ around the $x$– and $z$-axes, called $a$ and $b$ respectively.  It’s clear that this is countable: we can just list the words in increasing order of length.  It also partitions the sphere into orbits.

Again, we partition this group into four sets, $W(a),W(a^{-1}),W(b),W(b^{-1})$, such that $aW(a^{-1})\sqcup W(a)=bW(b^{-1})\sqcup W(b)=F_2$.  Then we pick a point out of each orbit in the sphere of $F_2$, calling this set $X$.  Then the sets $W(a)X,W(a^{-1})X,W(b)X,W(b^{-1})X$ are disjoint, partition the sphere (since the action is free), and we also have partitions $aW(a^{-1})X\sqcup W(a)X=bW(b^{-1})\sqcup W(b)X=S^2$!  Geometrically, what we’ve done is split the sphere into these four sets, and each of them can be rotated by a certain angle so that it completely covers two others.  If we rotate $W(a^{-1})X$ by $a$, then together with $W(a)X$, we have a whole sphere — then we just have to translate this to the side, and rotate $W(b^{-1})X$, and get a second sphere.  This paradoxical decomposition of the sphere is called the Hausdorff paradox.

There are three things that need to be said about this proof.  One is a deliberate error that I hope you caught, one isn’t considered an error now but would have been when the proof was published, and one is why this isn’t that strange.

The first thing is: the action of $F_2$ on the sphere is not free.  Since all the elements are rotations, they each fix two points, where their axis of rotation intersects the sphere.  What this means is that the partition of $F_2$ doesn’t necessarily give a partition of $S^2$ when it acts on $X$: rather, points in $X$ that get fixed could wind up sent to the same place twice, meaning that the sets overlap and potentially breaking our proof.

But fortunately, this doesn’t happen.  The reason why goes back all the way to our studies of cardinality: $F_2$, and thus the set of points fixed by any rotation in $F_2$, is countable, and the sphere is not.  If we remove this countable fixed point set (call it $C$) from the sphere, then the action of $F_2$ on $S^2-C$ is free, and so we can paradoxically decompose $S^2-C$.  We just have to show that $S^2-C$ is equidecomposable with $S^2$, which really only requires the same trick we did on the circle last time.  Pick any irrational rotation $r\in SO(3)-F_2$.  Then the axis of rotation of $r$ is disjoint from $C$, so $C,rC,r^2C,\dotsc$ are all disjoint.  Then applying $r$ to $C\cup rC\cup r^2C\cup\dotsb$ gives $rC\cup r^2C\cup\dotsb$, which is the original set minus $C$.  So $S^2\sim S^2-C\sim S^2-C\sqcup S^2-C\sim S^2\sqcup S^2$.

The second thing is the sentence “pick an element out of each orbit of $F_2$.”  Without the Axiom of Choice, there’s no reason we should be able to do this.  Most mathematicians today use choice without reservation, but in Hausdorff, Banach, and Tarski’s day, this raised some eyebrows.  In fact, the Banach-Tarski paradox was one of the main sources of fuel behind the anti-choice (or, as I like to call it, “pro-life”) movement.  Without choice, the paradox disappears, but I feel that some very beautiful mathematics disappears as well.

The third thing is that the Hausdorff paradox really isn’t as bad as it seems.  Why?  Because in $\mathbb{R}^3$, surface area isn’t well-defined.  I find it easier to think about the analogous problem with length in $\mathbb{R}^2$.  Imagine measuring the coastline of Great Britain.  If you’re looking at a map, you can just say “well, it looks like a triangle” and use the perimeter of that triangle as a first approximation.  Then you can zoom in on the map, measure each of the tiny lines, and get a larger approximation.  Then you can actually go there and start measuring it with surveying equipment, and your approximation will be bigger.  If you go back with a yardstick, the approximation will be bigger still.  As you increase your level of detail, you uncover more kinks in the coastline, making your approximation longer, and this sequence won’t, in general, converge!  (Fortunately, this isn’t a problem with differentiable curves — we can just integrate the length of the tangent vector.)

In $\mathbb{R}^2$, we can approximate any area to any level of precision — for example, it must be less than the area of any disk containing it.  But in $\mathbb{R}^3$, we again have the problem of craggy landscapes, and it’s entirely possible that the area of a bounded surface can be infinite!  Since the paradoxical decomposition given in the Hausdorff paradox is really weird (so weird that we need the axiom of choice to prove it exists), it’s not really that surprising that we can double the area of a surface through isometries.

What Banach and Tarski did was extend the Hausdorff Paradox to the ball, which has volume.  We expect to be able to measure volume in $\mathbb{R}^3$.  But if we connect each point on $S^2$ to the origin by a line segment (not including the origin itself), then $F_2$ acts (almost) freely on the set of those segments, whose union is, of course, the ball minus its center.  Repeating the above steps, we get $B^3-\{0\}\sim(B^3-\{0\})\sqcup(B^3-\{0\})$.

Finally, imagine a small circle through the origin and inside the unit ball.  We showed in the previous post that such a circle was equidecomposable with itself minus a point under the group of rotations about its center.  Such rotations are in $SE(3)$ (to rotate about an axis not going through the origin, you just have to conjugate by a translation sending the origin to it in a perpendicular direction — investigate this).  So under the $SE(3)$ action, the ball is equidecomposable to itself minus its center.  Putting this all together, we have that the ball is equidecomposable with two balls.

$\square$, bitches!

The implications of this are extensive.  There are two directions I want to go.  First, in order for this to make sense, at least one of the sets in the paradoxical decomposition must have no definable volume.  This naturally leads us to asking when we can define volume (or length, area, et cetera) at all, a question at the heart of measure theory.  Second, the reason we can do this in $\mathbb{R}^3$ but not $\mathbb{R}^2$ is due to certain abstract group-theoretic properties of $SE(3)$.  We can continue our discussion of group theory by examining these properties, and follow up the Banach-Tarski paradox with von Neumann’s extension to $\mathbb{R}^2$, where a paradoxical decomposition is possible if we extend the acting group to the group of all transformations that preserve area, $SA(2)$.  Of course, I also want to head back to topology at some point — we’ve yet to discuss metric spaces, which are relevant here, and there are a bunch of other cool topics to address.

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[…] of the big lessons learned from the Banach-Tarski paradox is that even in something as simple as a unit ball, we can find sets of impossible or undefinable […]