The Lantern Relation

(Edited… this now has the true lantern relation. Thanks Saul.)

Look at the four punctured sphere with the three annuli on it.

Picture 20

The lantern relation says that the composition of right handed Dehn twists about the red, blue, then green annuli is the same as right handed Dehn twists about each of the boundary components. Or instead, right handed Dehn twists about the red, blue, then green annuli followed by left handed Dehn twists about the boundary components is the identity.

Let’s see this in action…. We’ll also see that the identity can look pretty messed up. And it can look pretty when messed up.

Hmmm… troubles embedding the video… watch it here.

Continue on for stills of each stage.

Picture 10 Picture 9
Picture 8 Picture 7
Picture 6 Picture 5
Picture 4 Picture 3

Picture 2
(I’m trying out a screen capture program called Jing.  This was motivated by some previous conversations with Jesse, Ben’s first vlog, and the bit of discussion at God Plays Dice.)

~ by Ken Baker on June 1, 2008.

7 Responses to “The Lantern Relation”

  1. That’s really cool. What program are you using to draw these pictures?

  2. Thanks. These are done with Rhino3D. The operation I used to do the Dehn twists is called Maelstrom. You can get an overview of this one and some of the other operations here.

    Also, I’m working with the pre-beta port to Mac… Thus it’s free, but not fully functional.

  3. It would be cool to use a curve shortening method to see that the arcs isotope to be standard. Maybe the Hass-Scott method:

  4. btw, there’s a story (possibly apocryphal) that Thurston was trying to understand surface homeomorphisms. He wrote a program to see how curves were taken under a particular diffeomorphism (maybe similar to what you’re doing), and printed them out on an inkjet printer. The resulting images looked like train tracks, and thus the theory was born.

  5. Thanks Ian. That would be a great function to have in these modelers. The full version of Rhino allows for scripting… I’ll have to look into Vaughn’s implementation of Hass-Scott and see if (one of these days) I can adapt it for use here.

  6. If I remember correctly, some care is required here: suppose that A, B, C, D are left twists about the boundaries and x, y, z are right twists about the equators. Then only one of xyzABCD and xzyABCD is isotopic to the identity. (At least, that is what Twister thinks!)

  7. Yeah, Saul, I think I’ve got this one backwards. Dang. Checking the reference I pulled the relation from, there’s the little issue of order of composition. Fun stuff. Should’ve checked it more carefully. I’ll have to show off the difference later.

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