How may we determine if a pretzel knot is actually mod 3 colorable?
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onye
Apr 21, 2008, 4:29:46 PM4/21/08
to knot theory reu
We can generalize the algorithm pq+pr+qr=k for any (p,q,r) pretzel
knot, not necessarily alternating pretzel knots. If k is such that k
is a multiple of 3, then we may say that such a pretzel knot is
colorable mod 3. For example, the pretzel knots (7,-2,7) and (-7,8,5)
are both tri-colorable. This is because ((7*7)+(-2*7)+(-2*7)=21) and
((-7*5)+(-7*8)+(8*5)=-51). Both 21 and -51 are multiples of 3.
Therefore, we may conclude that these two knots are tri-colorable.
However, the knot (5,-3,7) is not tri-colorable because ((-3*5)+
(-3*7)+(5*7)=-1) and -1 is not a multiple of 3.
By this algorithm, we may also generalize that if pq+pr+qr=k, such
that k is a multiple of z, then the pretzel knot (p,q,r) is mod z
colorable.
I will need to set up the proof in my next posting. I don't know if
somebody else had discovered this algorithm before. But I actually did
figure this out myself!
Nice isn't it?
knot, not necessarily alternating pretzel knots. If k is such that k
is a multiple of 3, then we may say that such a pretzel knot is
colorable mod 3. For example, the pretzel knots (7,-2,7) and (-7,8,5)
are both tri-colorable. This is because ((7*7)+(-2*7)+(-2*7)=21) and
((-7*5)+(-7*8)+(8*5)=-51). Both 21 and -51 are multiples of 3.
Therefore, we may conclude that these two knots are tri-colorable.
However, the knot (5,-3,7) is not tri-colorable because ((-3*5)+
(-3*7)+(5*7)=-1) and -1 is not a multiple of 3.
By this algorithm, we may also generalize that if pq+pr+qr=k, such
that k is a multiple of z, then the pretzel knot (p,q,r) is mod z
colorable.
I will need to set up the proof in my next posting. I don't know if
somebody else had discovered this algorithm before. But I actually did
figure this out myself!
Nice isn't it?
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