## Reflections on the Letters r,s,t (Groups II)

Continuing the discussion from last week, let’s consider the 3-letter alphabet {r,s,t}. We are allowed to form all possible words in these letters (and their inverses, if you want to), but we agree that rr=ss=tt=1 and (rs)^2=(st)^3=(tr)^6=1. This defines the Coxeter group G(2,3,6). Last time we saw that the very similar group G(2,3,3) is finite, today we will see that G(2,3,6) is infinite. Below is the beginning of its Cayley graph.

We travel from one word to the next by appending r,s, or t. This looks much more complicated than what we saw for G(2,3,3), but things become clearer when we look at another group. Consider a yellow triangle with 30, 60, 90 degree angles (writing this as π/2, π/3,π/6 makes 2-3-6 reappear), and let ρ, σ, τ be the reflections at the lines extending its edges.

These three reflections generate a group Γ(2,3,6) of Euclidean symmetries which has the yellow triangle as its fundamental domain. The clue is that Γ(2,3,6) = G(2,3,6). We can easily map G(2,3,6) to Γ(2,3,6) by sending R to ρ, s to &sigma, t to τ. This works because ρ, σ, τ satisfy the same relations as r, s, t. It doesn’t work as easily the other way because ρ, σ, &tau could also satisfy other, hidden relations.

Let’s look at the word tsrtst. Reading it from left to right gives us a path on the Cayley graph from the initial triangle to a target triangle. Translating from Latin tsrtst to Greek τσρτστ gives a composition of reflections that takes the initial triangle to the same target triangle. This is not completely trivial, you prove it by induction. Remember that the composition is applied from the right to the left, so we also change reading direction.

This observation can be used to show that the translation map G(2,3,6) to Γ(2,3,6) is injective. If a word in G is the identity in \$γ, its path in the Cayley graph must be a closed loop. As the Euclidean plane where the tiling lives is simply connected, we can homotope it to a constant path, using elementary operations: Backtracking an edge, or shrinking a loop around a vertex to a point. The former is the accomplished using the relations rr=ss=tt=1, the latter using the other relations. This shows that the geometric homotopy can be realized using the relations of the group, and thus we can reduce the word to the trivial word 1.

This is essentially the proof of a famous theorem by Walther Franz Anton von Dyck: The group G(a,b,c) is finite if and only if 1/a+1/b+1/c>1. We have seen the relevant examples in the case
1/a+1/b+1/c>1 and 1/a+1/b+1/c=1. If 1/a+1/b+1/c <1, we need hyoperbolic geometry. Above is a picture of the Cayley graph of G(2,3,7) within the tiling of the hyperbolic plane by (π/2,π/3,π/7)-triangles.

## Reflections on the Letter T

Almost a year ago, when there was still hope, I posted a few Fall themed images with the title Yellow. The third image shows a view of McCormick’s Creek with a tree trunk that looked in 2008 like this:

The perspective of the two images is not quite identical, but you will see that in the older image above there are two prominent trees, the right of which has become the trunk in the second image of the image of last year’s post.

Above is another image, from 2009, tree still standing, again from a slightly shifted perspective. The view has always tantalized me, because it looked promising, but I could never turn it into a picture I was happy with. Below is a view from the other side, another year later.

Then, in 2015, this unexpected change:

With the tree reduced to a stump,

the place has become more balanced and serene.

Sometimes it is worth the wait.

## Do and Undo (Groups I)

Groups are mathematical games being played with letters. In the simplest version, we use just one letter (say a), and are allowed to add it to or to remove it from a word. This is the free group of one generator, or the infinite cyclic group.

Clearly, this game of create and destroy needs more rules. A simple rule is to make it truly cyclic and finite by insisting that after using the letter a say 7 times, we are back where we started. This means aaaaaaa=1, which is a relief, but still not very interesting.

With two letters a and b, our game expands.

This Cayley graph is the dual graph of the tiling of the hyperbolic plane by ideal squares, and not accidentally so.

Again we can restrict the rules of the game. Let’s play with the three letters r, s, and t, and insist that rr=ss=tt=1 to avoid any repetitions, and also that rsrs=ststst=trtrtr=1. The result is the Coxeter group G(2,3,3), and after a while playing around with the words, you find its Cayley graph below, neatly laid out.

This is dual to a tiling of the sphere by spherical triangles with angles π/2,π/3,π/3, and this is also not accidentally so.

## Time to Leaf

This year, times seems to be running faster, as if everything wants it to be over.

The freshly fallen leaves already look like they are from last year. And they don’t even read the news.

This is the game I played, called Still Live, ironically: You walk around and take the leaves as they are. You may remove a stem or piece of dirt, but you may not add.

So one fights against the randomness of every appearance, without creative power, only allowed to select.

This makes for a nice morning walk in the woods, despite.

## Always or Never

Take two ellipses, one within the other. Take a point on the outer ellipse, and draw one of the two tangents to the inner ellipse, and find its second intersection with the outer ellipse. Use this point to start the process again, and again. You will get a polygonal path in the ellipse that will most likely not close up. But in case you are lucky, something miraculous happens: If you pick any other point and repeat the game, the polygon will again close up.

This is the content of a famous theorem by Jean-Victor Poncelet.

In spirit, it is similar to a theorem of Jakob Steiner that asserts that a chain of circles in an annulus bounded by two circles either always or never closes up. While Steiner’s theorem follows immediately by inverting the circles into a pair of concentric circles, such a simple proof is not available for Poncelet’s theorem. Until recently, all proofs I know of were, let’s say, advanced.

At the core of a new proof by Lorenz Halbeisen and Norbert Hungerbühler are some fundamental theorems from projective geometry.

Let’s first recall that five points, no three collinear, determine a unique conic.

This is because through four points, you can find two different degenerate conics consisting each of a pair of lines, and by forming linear combinations, accommodate a fifth point. Below we will need the dual theorem: Given five lines, no three concurrent, there is a unique conic tangent to them.

Pascal’s theorem is a condition for six points to lie on a conic: They do if and only if opposite sides intersect in collinear points. Above you see this for six points on the two branches of a hyperbola.

Dual to this is Brianchon’s theorem (illustrated above): The sides of a hexagons are tangent to a conic if and only of its diagonals are concurrent.

As an application, Halbeisen and Hungerbühler show: If the six vertices of two triangles a1,a2,a3 and b1,b2,b3 lie on a conic, than there is a conic tangent to the six sides of the triangles. The proof is easy: Applying Pascal to the hexagon a1,b2,a3,b1,a2,b3 gives us three collinear points c12,c13,c23.

Then applying Brianchon to the hexagon a1,c12,b1,b3,c23,a3 shows that it is tangent to a conic. But the sides of this hexagon are the same as the sides of the two triangles, so we are done

From here, we obtain Poncelet’s theorem for triangles: Suppose you have two ellipses inside each other, and a triangle whose vertices lie on the outer ellipse and whose sides are tangent to the inner. Take another point on the outer ellipse, and form a second triangle by drawing the tangents to the inner ellipse. We have to show that the third side of the triangle is also tangent to the inner ellipse.

By the theorem by Halbeisen and Hungerbühler, the two triangles have an inscribed common ellipse. The given inner ellipse touches five of the same six lines by construction. But a conic is uniquely determined by five tangent lines.

The general case follows of n-gons the same idea, but requires more bookkeeping.

## Too Wide?

When I started using an SLR, I had just two lenses: A 28-85mm zoom, and a 20mm wide angle lens. That was too wide for me, back then,
and it took me a while to appreciate it.

When I moved on to a DSLR, one of the first new lenses I bought was Nikon’s 14-24mm zoom. That was something else, and again it took me many years to make use of the wider end of it.

This year, I decided to push myself again, and I acquired an 11mm lens.

This lens works like the news these days: It shows a distorted reality. If you want the truth, look elsewhere.

But, as with the news these days, the distortion is so extreme, that we are never tricked into believing it is real. It is more a provocation.

The benefit? Maybe we can learn to resist to undergo this distortion ourselves. Or is the remaining path too narrow?

## Odd Angles

For a while, this will be my last post about conformal spiderwebs. Today, we will still look at circular quadrilaterals that are conformal images of squares, but allow the angles to be multiples of 90 degrees. Like so:

Let’s call this a square of type (1,1,3,3). Multiply the numbers by 90 and you get the angles at the vertices. I have again employed Möbius to place three corners at (1,0), (0,1), and (-1,0). The fourth vertex is again moving cautiously along the unit circle. Below is a square of type (1,3,3,3), and here the fourth vertex is on the x-axis, the second possible case we noticed for right angled circular quadrilaterals.

Similarly, here is a square of type (1,1,1,3), also with the fourth vertex on the x-axis.

Missing are squares of type (1,3,1,3). While there are quadrilaterals of this type, all conformally correct squares I could find were only immersed (i.e. overlapping).

Then one can also have squares of type (2,2,2,2), for instance. The circle would be an example, with artificial vertices at (1,0),(0,1),(-1,0) and (0,-1), but there are also bean shaped squares like the one below.

Finally, the square with zero angles, in its most regular form.

## Out of Order

I don’t get often to Nevada. The last time was in March 2015, flying into Las Vegas to drive north to Zion. We spent a night and a day in Moapa Valley to visit the Valley of Fire State Park. Both town and state park completely cured me of all prejudices I had about Nevada. Moapa Valley is a small relaxed community with lots of local artists and friendly people, and the Valley of Fire State Park is an amazing piece of landscape that is on my list of places to return to.

Typical views are like the ones above, and full of interesting details. Can you spot the head below in the image above?

Everything here is created by light and shadow, and changes within minutes.

One can go instantly from harsh contrast to soft pastell. You would think a landscape like this can cure every ailment of the soul.

From there, on our way north, we passed through another little town: Mesquite, Nevada, not even 40 miles northeast. Little did we know about what had taken residence there a few months earlier, brooding, breeding the incomprehensible.

In Memoriam, once more: Las Vegas, October 1, 2017.

## More Spiderwebs: Conformal Squares with Circular Edges

The Schwarzian derivative of a map f(z) from the upper half plane to a right angled circular quadrilateral that sends the points -1, 0, 1 and infinity to the vertices of the quadrilateral (and thus making it conformally a square) is given by the expression

From it, one can find this map by taking the quotient of two solutions of the linear ordinary differential equation u”(z) + 2 Sf(z) u(z)=0. This is one step more complicated that the hypergeometric differential equation needed for triangles.

The parameter “a” is the accessory parameter. We have seen last week that there is a two-parameter family of right angled circular quadrilaterals, and the parameter a singles out the 1-dimensional subfamily of those quadrilaterals that are conformally squares.

In these images I have used a Möbius transformation to move three of the vertices to the points (1,0), (0,1) and (-1,0). The fourth vertex is then somewhere on the lower unit circle.

This is somewhat remarkable: First, it shows that we can find a conformal square for any such choice of four points (the first three normalized, the fourth on the half-circle). Secondly, it appears that the second family of right circular quadrilaterals we found last week where the fourth corner would move on line through (-1,0) and (1,0) does not contain any conformal squares. Thirdly, remember again from last week that for any such choice of four vertices, there is a 1-parameter family of right circular quadrilaterals with these points as vertices, but only one of them is conformally a square.

Of course one can also play with the angles. As a teaser for what’s to come next week, below is an anti-square.