The Argoknot

Names and notations

Rolfsen notation

$7_2$

Dowker-Thistlethwaite notation

12 10 8 14 4 2 6

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So the full correspondence is

1	3	5	7	9	11	13
12	10	8	14	4	2	6

Which gives a final notation of 12 10 8 14 4 2 6

Conway notation

-5 -2

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It is quite easy to move this to the normal projection using only ambient isotopy, so "-5 -2" is a
Conway notation for the Argoknot.

Braid notation

TBD

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Type justification here

Put a ">(space)" in front of each new line to create the box (even on empty lines and in front of images).

You can type mathematics inline, $f(x)=\sin x$, but not in equation mode.

Signed Planar Graph

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An image of the knot with the vertices drawn on and the connected marked and oriented:

Which of course begets the following graph:

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Knot properties

Alternating knot

Yes

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This projection is alternating. Ergo, the knot is an alternating knot.

Tricolorable

No

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Since tricolorability is independent of projection, we will only consider the standard projection:

Let black be "uncolored;" our three colors will be red, green, and blue. If the projection can be tricolored, one of the following cases is certainly true. Either (1) the top two strands will be the same color, say, red, or (2) the top two strands will be different colors.

Case 1: The top two strands are the same color.

One of the intersections now has two of the three relevant strands colored (both red), so the other strand is completely determined. This determines another intersection and another strand and in fact continues through the entire knot in the order indicated.

The end result of this situation is that every strand is colored red, which is not a valid tricoloring. Therefore, if it is possible to tricolor this projection, the top two strands must be different colors.

Case 2: The top two strands are different colors.

Without loss of generality we label these strands red and blue. As before, one uncolored strand is completely determined, so we color it, determining another strand, and so on in the order shown below.

The end result is a problem intersection, where exactly two colors meet. This is an invalid tricoloring, but is the result of our original assumption (that the top two strands must be different colors). Therefore, the top two strands must be the same color. However, by (1), this is also impossible, so the premise is false: the project is not tricolorable.
Since the projection is not tricolorable, no other projection of this knot could be tricolorable (as that would imply the tricolorability of this projection through a sequence of Reidemeister moves), so this knot is not tricolorable.

Amphicheiral

Yes/No

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Type justification here

Put a ">(space)" in front of each new line to create the box (even on empty lines and in front of images).

You can type mathematics inline, $f(x)=\sin x$, but not in equation mode.

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Numerical knot invariants

Unknotting number

1

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The following series of pictures will show that $7_2$ has an unknotting number of at most 1:
First we reverse the circled crossing to get the next projection:

With a simple Reidemeister II-Move, we get the following projection:

We can clearly unwrap this loop with a series of Reidemeister I-Moves, resulting in the unknot.
Therefore, with only one crossing reversed, we get a projection of the unknot. Thus $U(7_2) \leq 1$. Since $7_2$ is not the unknot, $U(7_2) \geq 1$, which forces the unknotting number to be 1 exactly.

Bridge number

2

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The following (poorly drawn) projection of $7_2$ has two (maximal) overpasses and two underpasses.

Therefore, $B(7_2) \leq 2$. Since $7_2$ is not the unknot, $B(7_2) > 1$, so $B(7_2) = 2$.
Note: the poorly drawn projection was constructed using the plane-division / post-untangling method used in class.

Crossing number

7

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The following is the standard projection of $7_2$.

Since this projection is alternating and reduced, the number of crossings here is the minimum number. Ergo, $c(7_2)=7$.
As a small corollary, $7_2 \neq 0$.

Genus

1

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The following is an oriented version of the standard projection:

Using Seifert's procedure, the separation into Seifert circles results in the following:

The next step is to attach twisted bands at each crossing point; however, this picture provides all the information we need. We can clearly see there are $s=6$ Seifert circles and $c=7$ crossing bands to be added. By a previous result, we know the Euler Characteristic $\chi = s-c = 6-7 = -1$. Since there is only one boundary component, we also know $\chi = 2 - 2g - 1$, so $g=\frac{1-\chi}{2}=\frac{1-(-1)}{2}=1$. Therefore this projection proves that $g(7_2) \leq 1$. Since $7_2$ is not the unknot, we also know $g(7_2) \geq 1$, and this forces $g(7_2)=1$.

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Knot polynomials

Jones polynomial

$t^{-1} - t^{-2} + 2t^{-3} - 2t^{-4} + 2t^{-5} - t^{-6} + t^{-7} - t^{-8}$

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Polynomial Invariants for 7.2

Alexander polynomial

$3t -5 + 3t^{-1}$

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A knot table for 7_2

HOMFLY polynomial

$−L^8 + M^2L^6 + L^6 + M^2L^4 + M^2L^2 + L^2$

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A knot table for 7_2