MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been trying to learn about snappy's method for encoding once-punctured torus bundles ( As you can see from the link, they are imported via the Manifold() function.

I have looked through documentation on the snappy website and much of the twister website. I am having trouble finding a reference for the encoding ‘b++LLR’, ‘b+-llR’, ‘bo-RRL’, ‘bn+LRLR’.

Following the literature (see for example, I am guessing that $L$ and $R$ correspond to writing an element in $SL(2,\mathbb{Z})$ as a word in

$L = \begin{pmatrix} 1 & 0 \\\\ 1 & 1 \end{pmatrix}$ and $R = \begin{pmatrix} 1 & 1 \\\\ 0 & 1 \end{pmatrix}$ with lower case elements used for inverses.

However, what to the b++, b+-, bn+ and bo- denote?

share|cite|improve this question
I haven't seen how Twister has been implemented in SnapPy but I suspect that prefix "b++" or "b+-" probably indicates whether or not you're dealing with an orientable bundle or not -- whether or not the resulting manifold is orientable. The punctured torus has an automorphism group that's technically a little bigger than $SL_2(\mathbb Z)$ since you don't have to be the identity in a neighbourhood of the puncture. The automorphism group is an extension of $SL_2(\mathbb Z)$. The "b+-" prefix probably indicates composition by an orientation-refersing involution. – Ryan Budney Mar 6 '13 at 16:47
Every Anosov element of $SL_2(\mathbb{Z})$ is conjugate to a product of $L$ and $R$'s and $\pm I$ (depending on the sign of the trace). This is what SnapPea allows one to do. Elements of $GL_2(\mathbb{Z})$ with negative determinant may be obtained from $SL_2(\mathbb{Z})$ elements by multiplying by a diagonal matrix with $\pm 1$ on the diagonal. So I suspect the notation is indicating a product of $L$'s and $R$'s like you say, together with an adjustment for the sign of the trace and determinant. But it's hard to guess from the notation which corresponds to which. – Ian Agol Mar 6 '13 at 17:19
I think Schleimer has documentation for Twister on his webpage. Right, here it is. – Ryan Budney Mar 6 '13 at 18:34
I believe the twister page is not relevant -- the b++ etc is "classic" SnapPea notation. – Sam Nead Mar 6 '13 at 21:51
up vote 3 down vote accepted

I believe that your question is answered by section (3e) at this page at the Geometry Center. Note that b+ and bo are equivalent as are b- and bn (o and n stand for orientation preserving and reversing, respectively). You can check a few examples using the is_isometric method of Manifold.

share|cite|improve this answer
Thank you. This is exactly what I was looking for. – Neil Hoffman Mar 6 '13 at 23:02

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.