User:Leo algknt: Difference between revisions
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A. Prove that the set of all 3-colourings of a knot diagram is a vector space over <math>{\mathbb Z}_3</math>. Hence <math>\lambda(K)</math> is always a power of 3 |
A. Prove that the set of all 3-colourings of a knot diagram is a vector space over <math>{\mathbb Z}_3</math>. Hence <math>\lambda(K)</math> is always a power of 3 |
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Attempt: Let <math>D</math> be a knot diagram with <math>n</math> crossings. There are <math>n</math> arcs. Let <math>a_1, a_1, \ldots, a_n \in {\mathbb Z}_3</math> represent the arcs. Now let <math>a,b,c \in {\mathbb Z}_3</math>. Define <math>\wedge : {\mathbb Z}_3 \times {\mathbb Z}_3 \rightarrow {\mathbb Z}_3</math> by |
Attempt: Let <math>D</math> be a knot diagram for the knot <math>K</math> with <math>n</math> crossings. There are <math>n</math> arcs. Let <math>a_1, a_1, \ldots, a_n \in {\mathbb Z}_3</math> represent the arcs. Now let <math>a,b,c \in {\mathbb Z}_3</math>. Define <math>\wedge : {\mathbb Z}_3 \times {\mathbb Z}_3 \rightarrow {\mathbb Z}_3</math> by |
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\end{array} |
\end{array} |
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\right.</math> |
\right.</math> |
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so that <math>a\wedge b + a + b \ |
so that <math>a\wedge b + a + b \equiv 0\mod 3</math>. |
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Then, with the above definition, we get a linear equation <math>a_{i_1} + a_{i_2} + a_{i_3} \equiv 0\mod 3</math> for each each of the <math>n</math> crossings, where <math>i_1, i_2, i_3 \in {1, 2, \ldots, n}</math>. Thus we get a system of <math>n</math> linear equation, from which we get a matrix <math>M</math>. The nullspace <math>\mathrm{Null}(M)</math> of <math>M</math> is the solution to this system of equation and this is exactly the set of all 3-colourings of <math>D</math>. This is a vector space of size <math>\lambda(K) = 3^{|\mathrm{Null}(M)|}</math> |
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Revision as of 23:49, 23 May 2018
Home Work 1
Question 1.
A. Prove that the set of all 3-colourings of a knot diagram is a vector space over . Hence is always a power of 3
Attempt: Let be a knot diagram for the knot with crossings. There are arcs. Let represent the arcs. Now let . Define by
so that .
Then, with the above definition, we get a linear equation for each each of the crossings, where . Thus we get a system of linear equation, from which we get a matrix . The nullspace of is the solution to this system of equation and this is exactly the set of all 3-colourings of . This is a vector space of size
Let
B. Prove that is computable in polynomial time in the number of crossings of K.