Conference rusure::math

    -t^2 ( 11 t^5 - 21 t^4 + t^3 + 11 t^2 - 11 t + 11 )
    
    
    				-- edp
    
    
Public key fingerprint:  8e ad 63 61 ba 0c 26 86  32 0a 7d 28 db e7 6f 75
To find PGP, read note 2688.4 in Humane::IBMPC_Shareware.

This sounds like one of those problems where if the higher level problem
were approached in a different way, perhaps the ugly determinant could be
avoided completely ?

/Eric

    re .0
    
    Just curious but when you say that "you are free to choose which ones
    to remove" are you saying that the determinant is the same regardless
    of which two columns are removed? If not, perhaps the result is "nicer"
    for some pairs of columns than for others.

        So, what's the connection to knots? :-)
        
        Dan

    I felt a bit guilty entering this since it's on behalf of my son. He is
    doing a thesis on knots - in particular (as far as I can understand it,
    which isn't very far :-) ambiguities in the representation of knots by
    (Alexander) polynomials. This is supposedly the polynomial for an
    extension of the granny knot.
    
    He's at home at the moment, and therefore Maple-less, so my note was a
    result of a phone call from him.
    
    I think he thinks that throwing away any 2 columns only affects the
    determinant by a factor of t^n. He thought that this polynomial would
    factorise as +/-t^n times the square of a quadratic, so either he's
    wrong or one of us has made a transcription error. Throwing away
    columns 1 and 5 (to get rid of most t's) and doing it by hand I got
    -t((1-t+t^2)^2).
    
    A reference work is "On knots" by Kauffman.
    
    I think it's about time I got one of the symbolic maths packages on my
    home pc. I'll follow up the references in this notesfile.
    
    Dick
    
    And yes, I agree that having to evaluate a determinant is often a red
    herring, but I'm not so sure in this case.

Has he seen Martin Gardner's book, I think it's called "the Knotted Doughnut" ?

I've got it at home, some fun puzzles in there, including of course a
fascinating section on knots.

/Eric

    Yes, I've got the Knotted Doughnut, but I haven't looked at it for a
    while. I'll get it out again.
    
    On the matrix issue, it turns out that only certain pairs of columns
    can be discarded, for example 1/2, 1/4, 1/5, 1/6, 1/8, 1/10... I was
    surprised by his initial statement that you could throw _any_ 2 away.
    
    My vague understanding is that each column represents a segment of rope
    (delimited by crossings) and each row represents a region contained by
    segments of rope. You can only throw away columns representing adjacent
    regions to square the matrix. 
    
    Thanks again, especially to edp for the quick response.
    
    Dick

    Interesting.
    
    The description in -.1 is very informative. I think there is a minor
    glitch in rows vs columns.
    
    Each row corresponds to an edge, terminated where it is crossed.
    
    Each column corresponds to a region.
    
    A non-zero entry corresponds to where row i borders column j. So each
    row has four non-zero entries. Each column will have a number of
    entries corresponding to the number of sides it's got.
    
    It's an immediate consequence of the ubiquitous Euler's formula that
    the number of rows + 2 = number of columns.
    
    I have some problems in the detailed labelling. Each edges has four
    incident regions, labelled in order t,-t,-1,1. But where do I start the
    labelling? I can't see any obvious rule.
    
    I remember reading in a SciAm article years ago that there any
    rearrangement of a closed string can be achieved by composing three
    atomic operations. One is slipping a string under another. A second is
    twisting a string to put a loop in it. What's the third?
    
    I drew the granny knot that the base note refers to. Pretty. If these
    Alexander polynomials are a knot invariant as claimed, then one would
    expect the poly here to factorize into a square, because the granny is
    a double trefoil. The reef knot is a trefoil composed with its mirror
    image.
    
    What I would like to see is that the poly for trefoil is not preserved
    under mirror image. Because if not, Alexander polys can't distinguish
    between granny and reef knot.
    
    So tell me how the labelling is determined, and I can press on.
    
    Thanks,
    Andrew.

    Andrew
    
    I'm beginning to feel like a go-between. Also I'm still reeling from
    Hardy's comment I just read in "A Mathematician's Apology" to the
    effect that the ability to pursue maths disappears at 50!
    
    Anyway, Ed (my son) wrote out a reply which I'm having trouble reading,
    so I'll just include the main bits. In essence he was impressed (and so
    was I) that you had managed to reconstruct knot theory from the scant
    info in the previous replies. Meanwhile I have found a very readable
    introduction in http://sgi.ith.de/~aneziris which I am trying to find
    time to work through (in spite of Hardy).

    Dick
    
    ------------
    
    Extracts from Ed's reply:

    ...
    Yes, each row is a crossing and each column is a region.
    ...
    The knot must then be orientated and, as you travel around it, every
    time you get to a crossing on the underpass the corresponding matrix
    entries are:

               ^
            t  |  -1
          -----------
               ^
           -t  |   1
    ...
    (atomic moves)
    R1 is the twist, putting in a half twist.
    R2 is the pull, pulling a loop from under another strand:

         ---|       ---  |
            |\         \ |
            |/         / |
         ---|       ---  |

    R3 is the swap, pulling a strand from one side of a crossing to the
    other.
    ...
    (granny/reef) The Alexander poly cannot distinguish between the two
    trefoils (LH & RH) and, as knot addition translates to Alex poly
    multiplication, neither can it distinguish between grannies and reefs.
    ...
    For a better knot invariant either the Jones or HOMFLY polys will
    suffice, but still hit problems around knots of 10 or more
    crossings.
    ...
    Please keep setting my dad problems as it keeps a smile on his face.
    ...
    Ed


    

re .9
    
>    I remember reading in a SciAm article years ago that there any
>    rearrangement of a closed string can be achieved by composing three
>    atomic operations.
    
    Presumably the Reidemeister moves.
    
>   One is slipping a string under another. A second is twisting a string to
>   put a loop in it. What's the third?
    
    I'll look it up if noone beats me to it.