Tag: Schwarz Christoffel Map

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:

2quarter a 01

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.

Mixed31b 0

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

1quarter a 01

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.

Bean0

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

Zero 4 01

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

Schwarzian

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.

Square 0

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.

Square 1

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.

Square 0 5

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

3quarter b

Safely Footed Spiderwebs

… et je continue encore de fouler le parvis sacré de votre temple solennel…

Tri10 01

Let’s talk again a little about triangles. The last time I wrote about triangles is not quite a year ago, and it didn’t help.

Tri90 01

What you see in this post are all triangle that have their vertices at the same place, the third roots of unity, to be precise. They are, however, not Euclidean triangles with straight edges, but curved ones, with circular edges.

Tri240b 01

The first three are equiangular still, making angles of 10, 90 and 240 degrees at the corners, respectively. The spiderwebs are conformal images of polar coordinates on the disk, thus illustrating the Schwarz-Christoffel formula for circular polygons. The bat down below is a neat optical illusion, too: Would you think that the vertices are at the corners of an equilateral triangle?

Tri 1 1 3 01

The theory behind this is based on Schwarzian derivatives and the Schwarz reflection principle, so clearly Hermann Amandus Schwarz owns all this.
It is also intimately connected to hypergeometric functions and much more recent mathematics.

Bird 01

And there is some mystery, still. While circular triangles are safe (they are determined by their angles, up to a Möbius transformation, and the Schwarz-Christoffel formula will deliver), quadrilaterals are not. Even Euclidean straight & right quadrilaterals can be differently shaped rectangles, and things get worse with circular ones. In this case, the Schwarz-Christoffel formula will have some extra parameters, the so-called accessory parameters. Changing the will not change the conformal nature of the quadrilateral nor its angles, but its “bulge”. More about this later.

Flat is Beautiful

Every flat 2-dimensional torus can be obtained by identifying opposite edges of a parallelogram.

Flatmarked 01

Each such torus has an involution that fixes four points, marked in four colors above. We can visualize the quotient as a tetrahedron with 180 degree angles at every corner by taking the triangle consisting of the lower left half of the parallelogram, and folding it together.

Tetra

So the space of all tori is nothing but the space of tetrahedra. Each such tetrahedron determines a unique point on the thrice punctured sphere. This can be seen by constructing the elliptic function on the torus determined by sending red to infinity, yellow to 0, and green to 1. The unique point is then the value of blue, and is called the modular invariant of the torus. To go backwards, take a point in the thrice punctured sphere and compute the quotient of elliptic integrals (using independent cycles)

Latex image 1

This complex number is the ratio of the two edges of the parallelogram that defines the torus.
This map is a Schwarz-Christoffel map: It maps the upper half plane to a circular triangle with all angles 0.
Restricting it to the upper half disk has as its image one half of such a triangle, namely

A 0 0 x

Let’s repeat all this starting again with a parallelogram, which now has been removed from the plane.

Complement 01

Identifying again opposite edges defines a translation structure on a punctured torus that corresponds to a meromorphic 1-form with a double order zero at red (because the cone angle there is 6π), and a double order pole at infinity (yellow) (because the holomorphic 1-form dz has a double order pole at infinity). For a given modular invariant, we can determine the parallelogram to use using another Schwarz-Christoffel map

Latex image 3

which maps the upper half disk to a different circular polygon:

A 1 0 x

Curiously, we see that the ratio of the edge vectors of the parallelogram can now also lie in the lower half plane or even be real, in which case the parallelogram degenerates. For instance, we can construct a torus that corresponds to the quotient -1 by slitting the complex plane from -1 to 1, and identifying the top of the slit from -1 to 0 (resp. 0 to 1) with the bottom of the slit from 0 to 1 (resp. -1 to 0).

Slit70 01

This corresponds to an ordinary torus whose parallelogram is a rhombs with angle 70.7083 degrees. Next time we will see what this torus is good for.

Squaring the Circle

Squaring the circle is easy, you just need to know what you want to do. My personal favorite method is to use elliptic functions defined on rectangular tori to map rectangles to disks, as shown below for a square. These maps don’t preserve area (which is what the Greeks had wanted), but they preserve angles.

Circlesquare

I had some leftover architecture images from Columbus and wanted to see how they look when made circular. Here, for instance, is the AT&T building

DSC 5596

and this is a circular version:

Ex5

There are three degrees of freedom one can play with (the dimension of the automorphism group of the hyperbolic plane), which means that one can squeeze parts of the image towards the boundary cirle. Here are two other versions of the same image.

Att2

Another favorite of mine is the atrium of the Cummins office building with its wonderfully intricate play with straight lines and black and white.

DSC 5562

Now we only have to find architects and builders who create buildings that have these curves in reality.

Cummins

2-3-6

I like it when apparently simple things evolve all by themselves into complex objects. Like watching cactus seeds grow into cacti. That was a distraction, but I do like it. Below is a left over piece of mathematics that would have fit nicely into a paper I wrote with Shoichi about triply periodic minimal surfaces.

M236

It is, evidently, quite complicated. To unravel it, here is a smaller portion of it, its seed, so to speek.

Prism

That is a minimal surface inside a prism over a 2-3-6 triangle (which has a right angle, a 30 degree angle, and a 60 degree angle).
The curves in the vertical faces of the prism are symmetry curves of the surface, and reflecting at these faces of the prism extends the surface. The two curves in the bottom and top face of the prism are not symmetry curves, but when you place two prisms on top of each other (by translation), the curves will fit. The pattern the curves make on the prism determines the surface almost completely, there is just one degree of freedom. Here is another, equally pretty, version, using a different parameter.

M236b

Another way to seed these surfaces is through conformal geometry. Below is the conformal image of a circular annulus onto a polygonal annulus bounded by two nested 2-3-6 triangles. The parameter lines are images of radii and concentric circles, respectively. This map is the main ingredient in the Weierstrass representation of all these surfaces. Simple, isn’t it?

Tri