Knots On a Cubic Lattice

Let the function f from S^1 to R^3 be defined as follows.

f(2 pi k / N) = (m_k, n_k, l_k)
|m_{k+1}-m_k|+|n_{k+1}-n_k|+|l_{k+1}-l_k|=1
f(2 pi (k+r) / N) = r (m_k, n_k, l_k) + (1-r) (m_{k+1}, n_{k+1}, l_{k+1})

where N is a constant, k is an integer between 0 and N, 0 < r < 1, while m_k,n_k,l_k are integers. Such a function defines a knot if and only if (m_0, n_0, l_0) = (m_N, n_N, l_N) while for any i,j not equal to each other, other than 0 and N, (m_i, n_i, l_i) is not equal to (m_j, n_j, l_j). The first condition implies that N must be even.

Such a function may be denoted as follows. First, one observes that the vectors (m_{k+1}-m_k, n_{k+1}-n_k, l_{k+1}-n_k) are unit vectors and may be along one of the six axes x,y,z,-x,-y,-z. Once these are defined, the function f and thus the knot are fixed. Therefore for a value of N there are at most 6^N such knots. In reality, due to the constraints shown above, this number is much smaller. These knots are denoted through a sequence a_1, a_2, ..., a_N, where a_k takes values in {1,2,3,4,5,6}, and a_k=1,2,3,4,5,6 means that (m_{k+1-m_k, n_{k+1}-n_k, l_{k+1}-l_k) is along the axes x,y,z,-z,-y,-x respectively.

As was the case with the notation based on regular projections, the following questions emerge.

To proceed with answers to these questions, click here

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Charilaos Aneziris, aneziris@hades.ifh.de

Copyright 1996