$Knot Theory, 3-colourbility of knots$

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$Knot Theory, 3-colourbility of knots$

Decide the 3-colourability of each of the knots 7_5, 7_6, 7_7: In each case, either give a valid
3-colouring or prove that no 3-colouring exists.

Use the Rolfsen Knot Table

Topology Algebraic Topology

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To check if a knot is 3-colorable, we proceed as follows. We associate to each strand a color $a \in \mathbb{F}_3$: then a coloring is valid if all three colors are used and, for each crossing between the strands $i, j, k$, we have $a_i + a_j + a_k = 0 \in \mathbb{F}_3$, as this is equivalent to saying that $a_i, a_j, a_k$ are either all equal or all different.

Each crossing gives us a linear equation, so if we put those all together in a matrix, finding a valid 3-coloring is the same as finding a solution to the linear system that is not a multiple of $(1,1,1)$, which exists if and only if the kernel of the matrix has dimension at least $2$.

With reference to the attached picture, $7_5$ corresponds to the matrix \[ \begin{pmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 1 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 1 & 1 & 0 \\ 0 & 1 & 1 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & 1 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 \end{pmatrix} \] which has rank $6$, and $7_6$ corresponds to the matrix
\[ \begin{pmatrix} 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 1 & 0 \\ 1 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix} \] which also has rank $6$. It follows that neither of these knots is 3-colorable.

On the other hand, the picture shows a valid 3-coloring of $7_7$, so that one is 3-colorable.

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Alessandro Iraci

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Matchmaticians Alumnus

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Thanks I understand 7_7 however what I did for 7_5 and 7_6 is that I proved it by trying to draw the strands with only 3 colours however I said that since the under/over crossing strands must be different it must use more than 3 colours, do u think that is still a valid proof ?
Alessandro Iraci

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If you formalize it correctly, it is, but I think that formalizing it correctly is essentially equivalent to doing what I did here. You can't just say "I tried and it didn't work", you have to do a case-by-case analysis, explain why forced choices are actually forced, and then make sure that your cases are exhaustive. It is doable, but you need to be careful. I find this linear algebra approach more streamlined.
Alessandro Iraci

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Also under/over crossings don't have to be different, they can be the same color if the third strand in the crossing also has the same color, that's the tricky part I think.
- Matchmaticians Alumnus
  
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  on wikipedia it says 7_5 is tricolorable?
Matchmaticians Alumnus

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Ok makse sense I will look over what I did!
- Matchmaticians Alumnus
  
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  on wikipedia it says 7_5 is tricolorable?
Matchmaticians Alumnus

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oops i misread it sorry

$Knot Theory, 3-colourbility of knots$

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