Help me! I'm trying to understand Bakalov and Kirillov

This is a page dedicated to helping all those who’re struggling with understanding the classic book:

- B. Bakalov & A. Kirillov,
*Lectures on tensor categories and modular functors*AMS, University Lecture Series, (2000) (web)

The procedure is: go to the nForum Help me! I’m trying to understand Bakalov and Kirillov page, or to Math Overflow. Type in your question and get someone to answer it. Then put down the answer here in final form for future generations.

- pg 112, bottom of proof of Theorem 5.3.8, “the Dehn twist axiom MF7” should read MS7, I believe. Bruce Bartlett

Bruce Bartlett: I don’t understand the $s$ map in Example 5.1.11 in the online version of the book. It’s supposed to be the automorphism

(1)$s : \Gamma_{1,1} \rightarrow \Gamma_{1,1}$

of the torus with one boundary circle which is the analogue of a map which I *do* understand, the automorphism

(2)$s_t : \Gamma_{1,0} \rightarrow \Gamma_{1,0}$

of the ordinary torus (no boundary circle) to itself, which sends

(3)$(\theta, \phi) \mapsto (\phi, -\theta).$

But the map $s_t$ doesn’t seem to work if there is a boundary circle involved. Because an automorphism of $\Gamma_{1,1}$ is supposed to be the identity on the boundary circle, but the map $s_t$ isn’t: it rotates the boundary circle a quarter rotation (right?). The reason I say so is the following picture: we think of the torus as $\mathbb{R}^2 / \mathbb{Z}^2$, as Bakalov and Kirillov encourage us to do. Then the presence of the boundary circle can be thought of as having a little tangent vector pointing to the right at all the lattice positions (this uses their alternative Definition 5.1.10 the extended surface category). The map $(x,y) \mapsto (y, -x)$ rotates the plane by a quarter rotation and takes the lattice to itself, hence it descends to the quotient. But this map rotates the tangent vector by a quarter rotation, and we’re supposed to fix the tangent vectors!

So how does the map $s$ work?

Answer: Domenico Fiorenza explained it as follows: consider the torus with one boundary circle as a square made of elastic material with with opposite sides identified, with an open disc missing in the center. Then turn the square through one quarter of a rotation, while keeping the boundary circle fixed. That’s the $s$ map.

If you picture it at the level of the plane (i.e. before we mod $R^2$ out by $Z^2$), then you need to imagine a whole bunch of circles (“knobs”) located at the integer lattice points of the plane. Get a whole bunch of friends (one for each knob), and then count down “3,2,1,turn!” and turn the knobs through a quarter revolution clockwise (this distorts the elastic material from which the plane is made). Then rotate the whole thing rigidly one quarter of a revolution counterclockwise. The resultant map is the $s$-map at the level of the plane.

category: reference

Revised on September 4, 2010 21:50:36
by Toby Bartels
(173.190.156.19)