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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?

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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

1 Answer 1

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.

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Thank you. This is exactly what I was looking for. – Neil Hoffman Mar 6 '13 at 23:02

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