User:Leo algknt: Difference between revisions

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'''Question 1.'''
'''Question 1.'''


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
''

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
'''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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c, & a\not= b
c, & a\not= b
\end{array}
\end{array}
\right.</math>
\right.</math>,
so that <math>a\wedge b + a + b \equiv 0\mod 3</math>.
so that <math>a\wedge b + a + b \equiv 0\mod 3</math>.


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>
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) =|\mathrm{Null}(M)| = 3^{\dim(\mathrm{Null}(M))}</math>




''B. Prove that <math>\lambda(K)</math> is computable in polynomial time in the number of crossings of K.
Let


''
B. Prove that <math>\lambda(K)</math> is computable in polynomial time in the number of crossings of K.
From part A, we can compute the rank of the <math>M</math> (M is a result of <math>n</math> linear equations coming from <math>n</math> crossings of a knot diagram) using Gaussian elimination and this can be done in polynomial time. From the rank nullity theorm, we get the nullity and hence <math>\lambda(K)</math>.

Revision as of 00:24, 24 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


B. Prove that is computable in polynomial time in the number of crossings of K.

From part A, we can compute the rank of the (M is a result of linear equations coming from crossings of a knot diagram) using Gaussian elimination and this can be done in polynomial time. From the rank nullity theorm, we get the nullity and hence .