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Jason
Apr 13, 2008, 10:10:59 PM4/13/08
to knot theory reu
Click on http://groups.google.com/group/knottheoryreu/web/arc-presentation-and-arc-index?hl=en
- or copy & paste it into your browser's address bar if that doesn't
work.
quiddit...@gmail.com
Apr 15, 2008, 1:55:51 AM4/15/08
to knot theory reu
I am most interested in the pairing of the permutations and whether or
not such a pairing of permutations constitutes sufficient information
to determine a knot uniquely. I would think the easiest and best shot
for exposing the inability to completely uniquely determine a knot by
its page permutation pair would be to see if the left and right handed
trefoil can give the same permutations. I would really like to know
what the permutation pairs of some knots are and especially if you
come by a proof that such a notation uniquely determines a knot or a
counter example I would like to know.
not such a pairing of permutations constitutes sufficient information
to determine a knot uniquely. I would think the easiest and best shot
for exposing the inability to completely uniquely determine a knot by
its page permutation pair would be to see if the left and right handed
trefoil can give the same permutations. I would really like to know
what the permutation pairs of some knots are and especially if you
come by a proof that such a notation uniquely determines a knot or a
counter example I would like to know.
Jason
Apr 20, 2008, 9:16:11 PM4/20/08
to knot theory reu
I will definitely look into this. As I think about it though, at least
for the trefoil, mirroring the crossings doesn't change the ordering
of the permutations.
for the trefoil, mirroring the crossings doesn't change the ordering
of the permutations.
marg...@math.utah.edu
Apr 21, 2008, 11:04:57 AM4/21/08
to knot theory reu
Hi Jason,
Great pictures!
How do we know the smallest number of pages needed for the trefoil or
the figure 8? Can we argue that the unknot is the only knot on 4
pages, and the trefoil knot is the only knot on 5 pages?
How does the arc index relate to the stick number, that is, the
smallest number of straight lines needed to draw a given knot?
Is there an alternate way to compute the Alexander polynomial by
looking at these pictures?
Dan
Great pictures!
How do we know the smallest number of pages needed for the trefoil or
the figure 8? Can we argue that the unknot is the only knot on 4
pages, and the trefoil knot is the only knot on 5 pages?
How does the arc index relate to the stick number, that is, the
smallest number of straight lines needed to draw a given knot?
Is there an alternate way to compute the Alexander polynomial by
looking at these pictures?
Dan
Jason
Apr 25, 2008, 2:53:21 PM4/25/08
to knot theory reu
this is a proof that for 2,3,and 4 pages the only knot that can be
represented is the unknot and that two unlinks and the hopf link can
be represented on four pages.
given the definition of an arc, a curve that doesn't cross itself and
intersects the binding at two points called vertices, and the fact
thatt we are limiting ourselves to one arc per page, for any given
number of pages we have a finite number of arcs that can be
represented. This implies that for n pages we can have at most n*2
vertices. Also in order to have closed simple loops we can have at
most two arcs intersecting at one vertex. For one page we have one arc
and at most two vertices, for two vertices we have no knot because we
have no closed loop. On two pages we can have at most two arcs and
four vertices. If the four vertices are disjoint then we don't have a
closed loop, if we have three vertices we have two arcs with one
vertex in common but the other two are disjoint so we don't have a
closed loop. If we have two vertices and two arcs we have a closed
loop, this loop is the unknot, therefore there are no other knots that
can be drawn on two pages.
Given three pages and three arcs we again state that we can have at
most 6 vertices. To get 6 vertices our 3 arcs are disjoint so we have
no knot. With 5 vertices and 3 arcs two arcs meet at one vertex and
have disjoint two distinct vertices at their other intersections with
the binding, third is arc is then disjoint from the first two this
gives no closed loop there is no knot with three arcs and five
vertices. With 3 arcs and four vertices we two possible cases to
check, the first is were all three arcs share a vertex but thi s is
not allowed. The second case is to have two arcs meet at two vertices
and a third arc that is disjoint this gives one unknot and a disjoint
arc. We are not allowed to have 3 arcs with 3 vertices since this
violates the condition that our curve is simple, the same argument
rules out 3 arcs with 2 vertices.
For 4 pages with 4 arcs we have at most 8 vertices. The 8 vertice case
gives us disjoint arcs. The 7 vertice case gives us no closed loops.
The 6 vertice case gives us at most one closed loop, an unkot, and two
disjoint arcs. The 5 vertice case gives at most one closed loop and
two arcs that have one common vertex. The 4 vertice case has at most
two closed loops, these loops are disjoint unknots, they can be chosen
in three ways see figure, four arc pages.jpg in the files. All other
cases violate the simple curve condition.
We can generalize this and say that for n pages any presentation with
n-1 vertices violates the condition that at most 2 arcs meet at one
vertex and that any presentation with more than n vertices has at lest
one arc that is not part of a closed loop.
represented is the unknot and that two unlinks and the hopf link can
be represented on four pages.
given the definition of an arc, a curve that doesn't cross itself and
intersects the binding at two points called vertices, and the fact
thatt we are limiting ourselves to one arc per page, for any given
number of pages we have a finite number of arcs that can be
represented. This implies that for n pages we can have at most n*2
vertices. Also in order to have closed simple loops we can have at
most two arcs intersecting at one vertex. For one page we have one arc
and at most two vertices, for two vertices we have no knot because we
have no closed loop. On two pages we can have at most two arcs and
four vertices. If the four vertices are disjoint then we don't have a
closed loop, if we have three vertices we have two arcs with one
vertex in common but the other two are disjoint so we don't have a
closed loop. If we have two vertices and two arcs we have a closed
loop, this loop is the unknot, therefore there are no other knots that
can be drawn on two pages.
Given three pages and three arcs we again state that we can have at
most 6 vertices. To get 6 vertices our 3 arcs are disjoint so we have
no knot. With 5 vertices and 3 arcs two arcs meet at one vertex and
have disjoint two distinct vertices at their other intersections with
the binding, third is arc is then disjoint from the first two this
gives no closed loop there is no knot with three arcs and five
vertices. With 3 arcs and four vertices we two possible cases to
check, the first is were all three arcs share a vertex but thi s is
not allowed. The second case is to have two arcs meet at two vertices
and a third arc that is disjoint this gives one unknot and a disjoint
arc. We are not allowed to have 3 arcs with 3 vertices since this
violates the condition that our curve is simple, the same argument
rules out 3 arcs with 2 vertices.
For 4 pages with 4 arcs we have at most 8 vertices. The 8 vertice case
gives us disjoint arcs. The 7 vertice case gives us no closed loops.
The 6 vertice case gives us at most one closed loop, an unkot, and two
disjoint arcs. The 5 vertice case gives at most one closed loop and
two arcs that have one common vertex. The 4 vertice case has at most
two closed loops, these loops are disjoint unknots, they can be chosen
in three ways see figure, four arc pages.jpg in the files. All other
cases violate the simple curve condition.
We can generalize this and say that for n pages any presentation with
n-1 vertices violates the condition that at most 2 arcs meet at one
vertex and that any presentation with more than n vertices has at lest
one arc that is not part of a closed loop.
Jason
Apr 25, 2008, 3:01:03 PM4/25/08
to knot theory reu
There is probably a simpler way to say this using the two permutations
of an arc representation. but I felt that the above post was a little
more rigorous. If you would like to see that one I could work it out
and post it.
of an arc representation. but I felt that the above post was a little
more rigorous. If you would like to see that one I could work it out
and post it.
Jason
Apr 25, 2008, 3:26:29 PM4/25/08
to knot theory reu
marg...@math.utah.edu
Apr 26, 2008, 1:15:12 PM4/26/08
to knot theory reu
Hi Jason,
Great!
Any thoughts on how arc index behaves under connect sum? Can we show
that ArcIndex(J#K) is at most ArcIndex(J)+ArcIndex(K)?
Dan
> > Dan- Hide quoted text -
>
> - Show quoted text -
Great!
Any thoughts on how arc index behaves under connect sum? Can we show
that ArcIndex(J#K) is at most ArcIndex(J)+ArcIndex(K)?
Dan
>
> - Show quoted text -
marg...@math.utah.edu
Apr 26, 2008, 1:23:08 PM4/26/08
to knot theory reu
Hi Jason,
Great pictures for the spanning tree for the trefoil. It looks like
this is very similar to the spanning trees you get for the (1,1,1)
pretzel knot (a.k.a the trefoil). In what sense are they different?
Is there an example of a pretzel knot where your and Onye's trees are
fundamentally different?
Dan
On Apr 25, 1:26 pm, Jason <Jasmea...@gmail.com> wrote:
> Click onhttp://groups.google.com/group/knottheoryreu/web/arc-presentation-and...
Great pictures for the spanning tree for the trefoil. It looks like
this is very similar to the spanning trees you get for the (1,1,1)
pretzel knot (a.k.a the trefoil). In what sense are they different?
Is there an example of a pretzel knot where your and Onye's trees are
fundamentally different?
Dan
On Apr 25, 1:26 pm, Jason <Jasmea...@gmail.com> wrote:
> Click onhttp://groups.google.com/group/knottheoryreu/web/arc-presentation-and...
Jason
Apr 26, 2008, 1:34:06 PM4/26/08
to knot theory reu
claim: In order to get only closed loops the permutation labeling of
our presentation on n pages can only have n vertices.
since every vertex has one arc coming out it must have another coming
in in order to close off that end point.
a crossing in a typical knot diagram is depicted by either interleaved
arcs and vertices
Using this claim we see we only need to check what are the closed
loops produced by the permutations of n vertices on n pages. To see
that the only knot obtained on 2,3,4 is the unknot we just check the
cases. We ignore the case of one page since there are no arcs with
only one vertex by our definition. for two pages we have two vertices
and two arcs this can be described as the permutations vertex(a,b),
pages(1,2). For three arcs we have three vertices a,b,c and three
pages 1,2,3 there are two permutations that give closed loops, (a,b,c)
(1,2,3) and (a,c,b) (1,2,3) both of these are the unknot as the arcs
and vertices are not interleaved. For four arcs we have four pages
1,2,3,4 and four vertices a,b,c,d we get the following permutations
with closed loops (a,b)(c,d) (1,2)(3,4), (a,b)(c,d) (1,3)(2,4), (a,b)
(c,d) (1,4)(2,3), (a,d)(c,b) (1,2)(3,4), (a,d)(c,b) (1,3)(2,4), (a,d)
(c,b) (1,4)(3,2), (a,c)(b,d) (1,3)(2,4). These last four are all
combinations of two unknots as seen in the other description two posts
above and in the added figure (four page arcs.jpg).
to check that a knot is presentable on n pages one would have to check
all permutations on those pages until it is presented or ruled out.
This is rather inefficient and tedious and one will notice that for
the hopf link the picture permutations are the same for both the cases
mentioned in the figure (four page arcs.jpg) in files.
our presentation on n pages can only have n vertices.
since every vertex has one arc coming out it must have another coming
in in order to close off that end point.
a crossing in a typical knot diagram is depicted by either interleaved
arcs and vertices
Using this claim we see we only need to check what are the closed
loops produced by the permutations of n vertices on n pages. To see
that the only knot obtained on 2,3,4 is the unknot we just check the
cases. We ignore the case of one page since there are no arcs with
only one vertex by our definition. for two pages we have two vertices
and two arcs this can be described as the permutations vertex(a,b),
pages(1,2). For three arcs we have three vertices a,b,c and three
pages 1,2,3 there are two permutations that give closed loops, (a,b,c)
(1,2,3) and (a,c,b) (1,2,3) both of these are the unknot as the arcs
and vertices are not interleaved. For four arcs we have four pages
1,2,3,4 and four vertices a,b,c,d we get the following permutations
with closed loops (a,b)(c,d) (1,2)(3,4), (a,b)(c,d) (1,3)(2,4), (a,b)
(c,d) (1,4)(2,3), (a,d)(c,b) (1,2)(3,4), (a,d)(c,b) (1,3)(2,4), (a,d)
(c,b) (1,4)(3,2), (a,c)(b,d) (1,3)(2,4). These last four are all
combinations of two unknots as seen in the other description two posts
above and in the added figure (four page arcs.jpg).
to check that a knot is presentable on n pages one would have to check
all permutations on those pages until it is presented or ruled out.
This is rather inefficient and tedious and one will notice that for
the hopf link the picture permutations are the same for both the cases
mentioned in the figure (four page arcs.jpg) in files.
Jason
Apr 26, 2008, 1:51:49 PM4/26/08
to knot theory reu
marg...@math.utah.edu
Apr 28, 2008, 12:05:51 AM4/28/08
to knot theory reu
Hi Jason,
You are right that this method seems much more efficient.
Is there a way to look at a permutation and see that you have the
unknot, without drawing the picture?
Dan
> > and post it.- Hide quoted text -
You are right that this method seems much more efficient.
Is there a way to look at a permutation and see that you have the
unknot, without drawing the picture?
Dan
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