Seifert surfaces

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Pat LeSmithe

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Jun 26, 2009, 11:54:14 PM6/26/09
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According to

http://en.wikipedia.org/wiki/Seifert_surface

every knot or link is the boundary of a compact, connected, and oriented
surface called a Seifert surface.

Recently, I found SeifertView, a Windows application that generates and
displays beautiful representations of such surfaces:

http://www.win.tue.nl/~vanwijk/seifertview/index.htm

Here's a gallery:

http://www.win.tue.nl/~vanwijk/seifertview/knot_gallery.htm

Unfortunately, the source code is not available. However, the author
mentioned that the program is written in Delphi. Perhaps the literature
contains a computational recipe we can adapt for Sage.

Anyway, the binary does run on Linux, in emulation:

wine seifertview.exe


Simon King

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Jun 27, 2009, 1:59:14 AM6/27/09
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Hi Pat!

On 27 Jun., 05:54, Pat LeSmithe <qed...@gmail.com> wrote:
...
> Unfortunately, the source code is not available.  However, the author
> mentioned that the program is written in Delphi.  Perhaps the literature
> contains a computational recipe we can adapt for Sage.

The existence proof for Seifert surfaces is constructive: Given a knot
diagram (i.e., a generic orthogonal projection with over/
undercrossings marked), it is straight forward to construct a Seifert
surface.

Much more complicated would it be to construct a Seifert surface of
*minimal* genus. It is possible, though, based on the theory of normal
surfaces that goes back to Wolfgang Haken. In a nutshell (hope I
remember things correctly):

- Given a knot diagram, construct a triangulation of the knot
complement.

- The triangulation gives rise to a system of linear equations with
integer coefficients, whose non-negative integral solutions with some
extra-condition are in one-one correspondence to Normal Surfaces
(i.e., surfaces in the knot complement that are 'nice' with respect to
the triangulation).

- The cone of non-negative integral solutions is finitely generated,
and it is possible to construct a generating set. As much as I
remember, the size of the generating set is bounded from above by 2 to
the power of (10 times number of tetrahedra). There is a series of
examples of triangulations for which there is now sub-exponential
lower bound for the coefficients occuring in the generating set. Ugly,
but that's life.

- Based on these generators, one can read off minimal Seifert
surfaces.

Note that a while ago I had started a "topology" Wiki:
http://wiki.sagemath.org/topology
Perhaps it could be revived?

In the Wiki, I mention a package t3m, that, as much as I know, could
do the above computation. I don't know about visualisation, though.

Best regards,
Simon

Simon King

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Jun 27, 2009, 2:02:09 AM6/27/09
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PS:

On 27 Jun., 07:59, Simon King <simon.k...@uni-jena.de> wrote:
> The existence proof for Seifert surfaces is constructive: Given a knot
> diagram (i.e., a generic orthogonal projection with over/
> undercrossings marked), it is straight forward to construct a Seifert
> surface.
>
> Much more complicated would it be to construct a Seifert surface of
> *minimal* genus.

That said: The algorithm that comes out of the existence proof is not
bad at all. As much as I remember, for alternating knot diagrams, it
yields a surface of minimal genus. And certainly the algorithm is very
*very* easy. So, the only problem is visualisation.

Cheers,
Simon

Nathan Dunfield

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Jun 27, 2009, 11:38:38 AM6/27/09
to sage-devel
> Much more complicated would it be to construct a Seifert surface of
> *minimal* genus. It is possible, though, based on the theory of normal
> surfaces that goes back to Wolfgang Haken.  In a nutshell (hope I
> remember things correctly):

Yes, you do.

> In the Wiki, I mention a package t3m, that, as much as I know, could
> do the above computation.

It could easily be made to do that, in those small cases where Haken's
algorithm is feasible. However, in practice there are usually better
ways involving quick methods of finding (some) Seifert surfaces
combined with (twisted) Alexander polynomials to give lower bounds on
the genus.

> I don't know about visualisation, though.

No, it doesn't do that.

Best,

Nathan
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