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In a previous blog post, I discussed the recent result of Guth and Katz obtaining a near-optimal bound on the Erdos distance problem. One of the tools used in the proof (building upon the earlier work of Elekes and Sharir) was the observation that the incidence geometry of the Euclidean group ${SE(2)}$ of rigid motions of the plane was almost identical to that of lines in the Euclidean space ${{\bf R}^3}$:

Proposition 1 One can identify a (Zariski-)dense portion of ${SE(2)}$ with ${{\bf R}^3}$, in such a way that for any two points ${A, B}$ in the plane ${{\bf R}^2}$, the set ${l_{AB} := \{ R \in SE(2): RA = B \}}$ of rigid motions mapping ${A}$ to ${B}$ forms a line in ${{\bf R}^3}$.

Proof: A rigid motion is either a translation or a rotation, with the latter forming a Zariski-dense subset of ${SE(2)}$. Identify a rotation ${R}$ in ${SE(2)}$ by an angle ${\theta}$ with ${|\theta| < \pi}$ around a point ${P}$ with the element ${(P, \cot \frac{\theta}{2})}$ in ${{\bf R}^3}$. (Note that such rotations also form a Zariski-dense subset of ${SE(2)}$.) Elementary trigonometry then reveals that if ${R}$ maps ${A}$ to ${B}$, then ${P}$ lies on the perpendicular bisector of ${AB}$, and depends in a linear fashion on ${\cot \frac{\theta}{2}}$ (for fixed ${A,B}$). The claim follows. $\Box$

As seen from the proof, this proposition is an easy (though ad hoc) application of elementary trigonometry, but it was still puzzling to me why such a simple parameterisation of the incidence structure of ${SE(2)}$ was possible. Certainly it was clear from general algebraic geometry considerations that some bounded-degree algebraic description was available, but why would the ${l_{AB}}$ be expressible as lines and not as, say, quadratic or cubic curves?

In this post I would like to record some observations arising from discussions with Jordan Ellenberg, Jozsef Solymosi, and Josh Zahl which give a more conceptual (but less elementary) derivation of the above proposition that avoids the use of ad hoc coordinate transformations such as ${R \mapsto (P, \cot\frac{\theta}{2})}$. The starting point is to view the Euclidean plane ${{\bf R}^2}$ as the scaling limit of the sphere ${S^2}$ (a fact which is familiar to all of us through the geometry of the Earth), which makes the Euclidean group ${SE(2)}$ a scaling limit of the rotation group ${SO(3)}$. The latter can then be lifted to a double cover, namely the spin group ${Spin(3)}$. This group has a natural interpretation as the unit quaternions, which is isometric to the unit sphere ${S^3}$. The analogue of the lines ${l_{AB}}$ in this setting become great circles on this sphere; applying a projective transformation, one can map ${S^3}$ to ${{\bf R}^3}$ (or more precisely to the projective space ${{\bf P}^3}$), at whichi point the great circles become lines. This gives a proof of Proposition 1.

Details of the correspondence are provided below the fold. One by-product of this analysis, incidentally, is the observation that the Guth-Katz bound ${g(N) \gg N / \log N}$ for the Erdos distance problem in the plane ${{\bf R}^2}$, immediately extends with almost no modification to the sphere ${S^2}$ as well (i.e. any ${N}$ points in ${S^2}$ determine ${\gg N/\log N}$ distances), as well as to the hyperbolic plane ${H^2}$.

My colleague Ricardo Pérez-Marco showed me a very cute proof of Pythagoras’ theorem, which I thought I would share here; it’s not particularly earth-shattering, but it is perhaps the most intuitive proof of the theorem that I have seen yet.

In the above diagram, a, b, c are the lengths BC, CA, and AB of the right-angled triangle ACB, while x and y are the areas of the right-angled triangles CDB and ADC respectively. Thus the whole triangle ACB has area x+y.

Now observe that the right-angled triangles CDB, ADC, and ACB are all similar (because of all the common angles), and thus their areas are proportional to the square of their respective hypotenuses. In other words, (x,y,x+y) is proportional to $(a^2, b^2, c^2)$. Pythagoras’ theorem follows.