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.
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.