Incidences

Why do we still teach geometry?  The constructions with ruler and compass were essential for the Egyptians and Greeks in order to accurately lay out large scale buildings with the only tool available (the rope). But today we have many other tools available, so there is no reason to confine ourselves to ruler and compass, unless we want to use them as a vehicle to teach the concept of proofs. That, however, is also in low demand, to the extent that graduating majors in mathematics are neither able to prove Pythagoras’ theorem nor to compute the distance of a point from a line.

Central

I have been teaching a Geometry class twice now, and I am releasing the notes I wrote for the first part of the class into the wild. For this first part, I had set myself a few goals: I wanted to use only the most fundamental notions of geometry,  I wanted a plethora of interesting examples, and I wanted to be able to prove a substantial theorem. Finally, I needed to be able to give homework problems. 

Parallel

The solution was to study incidence geometries, specializing pretty quickly to affine and projective spaces over arbitrary fields. So I did not develop axiomatic projective geometry, but rather taught the computational skills needed to quickly get to the geometry of conics in projective planes. This provided plenty of  exercises. Affine and projective transformations are intensely used in order to prove theorems or to simplify computations. The big theorem I prove at the end is Poncelet’s theorem.

Hyperrat

The second half of the course? This deals with the two-dimensional geometries where we have circles: Möbius, Euclidean, spherical, hyperbolic. Emphasis is again on groups, and here in particular on reflection groups, proving Dyck’s theorem at the end. But I am not quite happy with the notes yet, so this part will have to wait.

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So here is part I. Enjoy.

Notes on Geometry – Part I: Incidences

 

 

 

 

 

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Quartet (DePauw Nature Park IV)

I like the days in late fall when Nature has gone to rest, but winter hasn’t arrived yet.

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We should do the same. Instead of denying the approaching darkness by putting up silly lights on dead trees, we should hesitate and contemplate the state of everything around us. 

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So I will conclude this year with posts and images that have more the character of still lives.

 

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It is time to pay tribute to what we will use for building: tree and stone.

 

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And to be thankful that time is still passing.

Six by Six

This puzzle consists of a 6×6 square board and 36 arrow cards with 6 arrows in 6 colors. The goal is to place all arrow cards on the board such that

  • the entire board is covered with cards;
  • each arrow points to exactly one arrow of the same color in the same row or column.

Badmoves

The example above shows incorrect placements of arrows: The yellow arrow cannot point to another arrow, the left blue arrow points to two different blue arrows, and it is not possible to add to the four purple arrows, while correctly placed, two more purple arrows without violating the rules.

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Indeed, if the six arrows of one color are correctly placed, the arrows can be connected to a closed circuit. The figure above shows two closed circuits of six arrow cards. Note that the circuit paths may well cross. The following puzzles have a few arrows already placed, and need to be completed. Below to the left is simple example that shows how to find the solution.

Ex 1

First consider the green arrows. The arrow at e4 points to the right and therefore we must have another green arrow at f4, either pointing up or down. As there is already a green arrow at f6, the new arrow at f4 must point up. This leaves us with two more green arrows to place. Because we already have three horizontal arrows, the remaining arrows must be both vertical, and point to existing horizontal arrows. The only possibility is to place the new green arrows at c6 and e2, as shown above to the right.

Now let’s consider the red arrows. The one at e3 points down, leaving no choice but to place a red arrow pointing left at e1. There are two more horizontal arrows to place, and the only possibilities are b5 and f3, see below to the left.

Ex 2

Turning to blue, there clearly needs to be a left pointing arrow at f1, and the two remaining vertical arrows need to go to a1 and b6.

For yellow we have only two arrows given, but there are not many free spots available. We first are forced to put a left pointing arrow at e6 and then a down pointing arrow at d6. The remaining horizontal arrows go to c5 and d4.

Ex 3

The purple arrow at d4 can only be reached by a right pointing arrow at a5, and the one at c3 only by an up pointing arrow at c1. Then the arrow at c1 requires a left pointing arrow at d1, and the final purple arrow goes to a3.

Filling the remaining spots with cyan arrows is now easy, giving the solution.

Below is a new puzzle to warm up:

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And here is a more difficult one. In both cases, there is only one solution.

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Everyday is silent and grey

Driving from Bloomington to Chicago takes four to five hours and is a pretty boring drive. This year’s highlight was a lone sign saying “Boycott Fox News”, a stark contrast to what we had last year at around this time.

There are many reasons to go to Chicago. This time I had the privilege to be needed as company, because my daughter wanted to see Morrissey.

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I am not very literate when it comes to popular music. The concerts I look back to with fondness had “stars” like Luigi Nono, Karlheinz Stockhausen, or Bhimsen Joshi present. No, I didn’t listen to The Smiths. But this is what our kids are for: To educate us.

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And Morrissey impresses. He says what he thinks not to be a clown, but to be himself, and he is still good at it. His band was dressed in T-shirts saying “This country makes me sick”. Morrissey was obviously very angry.

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The theater was sold out, with a very diverse crowd. A few rows away a group started smoking weed early on. Chicago. But they all had a good time.

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It was good to see this kind of solidarity across generations. It is the kind of angriness that brings people together.

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Ends and Handles

Mathematics gets most exciting when new connections between different areas are discovered. In the theory of minimal surfaces, maybe the most fruitful discovery of this sort was Robert Osserman’s theorem from 1964 that a complete minimal surface of finite total curvature has the conformal type of a compact Riemann surface punctured at finitely many points, with the Weierstrass data extending meromorphically to the compact surface. This was a giant step from the much earlier discovery of the Weierstrass representation, that provided a first link between minimal surfaces and complex analysis.

Symfinite

However, at that time, the only known examples where of genus 0, i.e., punctured spheres, and of those only the catenoid (and the plane) were embedded. In fact, it is pretty easy to make examples like the one above: punctured spheres with many ends that will intersect.

This changed dramatically in 1982 when Celso José da Costa constructed a minimal torus with three ends that was proven to be embedded by David Hoffman and Bill Meeks in 1985. Examples with more ends and of higher genus followed rapidly, all nicely embedded. But there was a pattern: It looked like that if you wanted n ends, you needed to have genus at least n-2. This is the Hoffman-Meeks conjecture. For n=2 this follows from a theorem of Rick Schoen. But why can’t we have (say) a torus with four ends?

Fourends

All attempts to produce such an example have failed. In the image above, the ends will intersect, eventually. On the other hand, there is a fair amount of evidence that there are examples of genus g with precisely g+2 ends. Below is such a surface of genus 3 with 5 ends.

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That such examples should exist for any genus is supported by the Callahan-Hoffman-Meeks surface, a periodic version of the Costa surface. One just needs to chop it into pieces and put catenoidal ends at the top and bottom…

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By recent work of Bill Meeks, Joaquin Perez, and Antonio Ros, the number of ends of a complete, embedded minimal surface of finite total curvature is bounded above by some constant (which is not explicit).

But even the case of tori remains very much open.

Writing on Water

This year I tried to explore a few new places that are not more than an hour’s drive away, at the cost of neglecting a few places that are really close, like the Beanblossom Bottoms Nature Preserve.

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This is wetland, made accessible through slippery boardwalks. The grasses here make even more short-lived art than the sand art made by dune grasses.

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Either from the treacherous safety of the board walks, or right from the center of things,

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the grasses are at work, through gentle dipping of their stalks, or mere reflection. This is like writing blog posts. Words and images not to be bound between covers and shelved, nor streamed into instant oblivion, but just left there for a little while to wither.

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Then, finding the path back out of the forests becomes a possibility, because it is, after all, also only written on water.

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Eraserhead (Games on Circles I)

This is not about the film by David Lynch. Eraserhead is played by two players on finite graph with some vertices occupied by erasers. At each turn, the player chooses one eraser, moves it to an unoccupied adjacent edge, thereby erasing the connecting edge. The player who moves last wins.

Let’s analyze this for cycle graphs.
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If played with just one eraser, the first player chooses the direction, and all other moves are forced, until the eraser has erased all edges. So the first player wins if the cycle graph has an odd number of vertices. That was boring.

For two erasers, it gets slightly more interesting, but the outcome is the same: The first player can win if the number of vertices is odd. The two erasers divide the graph into two components whose number of vertices have different parity (3 and 4 in the example below).
Moving an eraser into the even component would be a mistake, as the second player will then move into the same component with the other eraser. This leaves an even number of moves for both players, and the second player will win.

But a winning move for the first player consists of moving one eraser into the odd component. This leaves the second player with two choices: Moving the other eraser either way leaves an odd number of moves for both players, and the first player wins. If the second player moves the first eraser again, the first player can move either way and will win.

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So while a little harder, this was still easy. Does this even/odd pattern for winning and losing persist with more erasers? At first, it seems so. For any number of erasers (except the cases of no erasers and erasers everywhere, which leave no moves to begin with), the first (resp. second) player can always win with if the cycle graph has an odd (resp. even) number of vertices — up to cycle graphs with 8 vertices. For the following position with three erasers, the first player has no winning move. There are three essentially different moves, and the diagram give winning responses for the second player.

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The first line is interesting. With the first player to move again, there are still 8 edges to delete (possibly). If the first player moves with the topmost eraser right, this will actually happen, and the first player will loose. To prevent this, s(he) must move that eraser to the left. This leaves all erasers in one component with a total of four edges left. This looks like a good thing, but no matter how they play, only three edges will be deleted, and the second player must win.

So the game does get complicated. What helps is to know the nimbers for simple subpositions, like chains with or without erasers at the end. For instance, the line graph with n vertices and one eraser at the end has nimber 1 for odd even n and 0 for odd n. Similarly, the line graph with n vertices and erasers at both end has nimber 0 for odd even n and 1 for odd n. This is quite trivial, of course, as all moves are forced. More surprisingly, for the line graph with two erasers at the end and a third eraser in between, the nimbers are again 1 for n odd, and 0 for n even, except when n=3 or n=4. Then the nimbers are 0 and 2, respectively. So things start off with promising simple patterns, but then deviate. Below I have, for line graphs of length 12 and 13, respectively a unit cube at coordinate (x,y,z) if the position with erasers at x, y, and z is a lost position for the first player.

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Finally, below, the corresponding images for cycle graphs, viewed diagonally to emphasize the symmetry. The discrepancy is baffling.

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