2
$\begingroup$

I have been trying to learn about snappy's method for encoding once-punctured torus bundles (http://www.math.uic.edu/t3m/SnapPy/manifold.html#snappy.Manifold). 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 http://arxiv.org/pdf/math/0406242.pdf), 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?

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ Mar 6, 2013 at 16:47
  • $\begingroup$ 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. $\endgroup$
    – Ian Agol
    Mar 6, 2013 at 17:19
  • $\begingroup$ I think Schleimer has documentation for Twister on his webpage. Right, here it is. homepages.warwick.ac.uk/~masgar/Maths/twister.html $\endgroup$ Mar 6, 2013 at 18:34
  • 2
    $\begingroup$ I believe the twister page is not relevant -- the b++ etc is "classic" SnapPea notation. $\endgroup$
    – Sam Nead
    Mar 6, 2013 at 21:51

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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