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In this question, I recently asked if there were free subgroups of rank 3 or higher of the group of rotations in $\mathbb{R}^3$. From the answers, it follows that any free subgroup of rank 2 admits subgroups of arbitrary countable rank.

My question now is whether this can be used to extend the Banach-Tarski Paradox to show that the sphere cannot only be duplicated (leveraging the subgroup of rotations of rank 2), but may be done in a way to produce $n$ copies using a finite number of disjoint subsets of the original unit sphere (leveraging a subgroup of rotations of rank $n$). All I have seen in this vein is re-applying the original statement $n$ times in order to create $n$ spheres, but it seems they could be created all at once using a free subgroup of rank $n$, correct?

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    Great to know!! Thank you!2011-08-04

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This question is answered in the affirmative in Chapter 6 of Stan Wagon's book on the Banach-Tarski Paradox, "Free Groups of Large Rank: Getting a Continuum of Spheres from One."

Using the free subgroups of rank $n$ (of a free group of rank 2), one can extend BT to $n$ copies at once.

Furthermore, as stated in the comments of the above-referenced question, one can use the fact that a free group of rank 2 has a subgroup of countably infinite rank to produce a countably infinite number of copies of the original sphere.

Moreover, it is possible to produce copies of the ball with cardinality of the continuum, though I do not understand the development presented in the text well enough to summarize here.

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    @Pierre, thank you! The link you reference is precisely the talk I gave at MathFest in Kentucky last week, using Stan Wagon's book as a guide (though I only looked at chapters 1-3). The motivation for my question above was from a question posed tome by another student, asking why it only deals with 2 balls at the final state, which prompted me to think about methods of extending this.2011-08-11