What is Topology, and what's the big deal?

Sorry, that notation in the first line should be \( i,j \in \{1,2\} \).

So let's write out a few things:

If we "run over" the values of the index set in increasing order as follows:

\( \delta^a_b:=\; \delta^1_1, \delta^1_2, \delta^2_1, \delta^2_2 \)

Then \( \delta^a_c \delta^b_d:=\; \delta^1_1 \delta^b_d, \delta^1_2\delta^b_d, \delta^2_1 \delta^b_d, \delta^2_2\delta^b_d \;=\; (1)\delta^b_d, (0)\delta^b_d, (0)\delta^b_d, (1)\delta^b_d \)

Expanding the b,d indicies means you get the identity where the first a,b indicies have a delta of 1. That is, where a 1 appears in the first sequence you replace it with the 2x2 identity matrix and where a 0 appears replace it with the 2x2 0-matrix, this is a matrix product also called the Kronecker product.

What you end up describing is \( I_2 \otimes I_2 \;=\; I_4 \), but the two identity matrices on the left each have a pair of indicies, and the right hand matrix has four, it's a 2x2 matrix of 2x2 matrices

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So then the crossed strands will look like:

\( \delta^a_d \delta^b_c := \delta^1_d \delta^b_1, \delta^1_d \delta^b_2, \delta^2_d \delta^b_1, \delta^2_d \delta^b_2 \), which will be a permutation.
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There is a much easier and quicker way to represent all this notation, which is with abstract tensor diagrams . . .

 

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m'kay, let's introduce another tensor-like object, define:

\( \epsilon_{ij}\;=\; \begin{cases} 1, & \text{if } i < j \\ -1, & \text{if } i > j \\ 0, & \text{otherwise} \end{cases} \)

Now we can show that \( \epsilon^{ab} \epsilon_{cd} \;=\; \delta^a_c \delta^b_d\; - \; \delta^a_d \delta^b_c \). A useful identity according to Kauffman.

Note that the two terms on the right are isomorphic to \( S_2 \), and Kauffman mentions that three strands with crossings (transpositions) represent \( S_3 \), hence n strands will represent \( S_n \). A strand is also then (abstractly) one of n "letters" or symbols acted on by elements of \( S_n \).​

 

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Then Kauffman is saying the Kronecker delta is a single letter or symbol, and therefore is an element of \( S_1 \), which has only one permutation: the identity.

Also, as a function that has two arguments, the order of these "inputs" to the Kronecker delta doesn't matter (here, I don't mean inputs and outputs like {a,b} and {c,d}, which are only relevant with two strands). So that:

\( \delta^a_b \;=\; \delta^b_a \)
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Further, an index (or both indices), can be raised or lowered without changing anything:

\( \delta^{ab} \;=\; \delta^{ba} \;=\; \delta_{ba} \;=\; \delta_{ab} \)​

In other words, the single strand with labeled ends isn't oriented as such, or at least the labels aren't; there is no difference between a vertical and a horizontal strand.
Aha! but when you introduce a second labelled strand plus a way to cross one strand over the other, you do get an orientation, as previously described. This isn't an orientation that defines a direction for both strands, since you move towards a crossing in one of the two directions in \( \mathbb R^1 \), roughly, you don't choose or fix a global direction, because you can choose to move "towards" the crossing in either direction, if you start somewhere on the knot that leads to an underscrossing . . . (heh).

We need a way to define the 90° "between" A-type and B-type switches, or crossings, and this is exactly what a matrix representation (with extra indices) does. It also suggests that the bracket notation will be useful, which it is; with a substitution of parameters, Kauffman's bracket polynomials become Jones polynomials (just sayin').

 

Oops, another mistake: in post 161, "Expanding the b,d indicies means you get the identity where the first a,b indicies have a delta of 1 " should read: "Expanding the b,d indicies means you get the identity where the first a,c indicies have a delta of 1."

Perhaps I can present an example of a knot (diagram!) as a tensor.
The trefoil knot's diagram represents a tensor diagram when it's been labelled (decorated) a certain way.

Since tensors are objects with indices, the decorations will include index variables. These are placed on the edges of the "graph". The trefoil is oriented when a direction is defined everywhere; draw an arrow somewhere, move along the curve and draw another arrow, repeat until there are arrows "going in" and "coming out" of each crossing.

The last type of decoration is identifying each crossing as positive or negative, which is done by drawing a small circle around each crossing and coloring it one of two colors.

Now write out the Kronecker deltas with upper indices as inputs (arrows going in) from left to right, i.e. clockwise around the crossing, and with lower indices (I've changed my mind about the spelling) as outputs (arrows coming out), but in the opposite sense.

So the particular labelling of edges (some permutation of index variables like a, b, c, . . .) defines how these labels are ordered over pairs of Kronecker deltas, which multiplied together determine the entries in a 0-1 matrix, Kauffman's "R-matrix" which then has two upper and two lower indices (two inputs, two outputs).

Then since the R-matrices are connected (there are no free stands), this decorated diagram represents a contracted tensor; matrix multiplication can be defined as a sum (of products) over a repeated index. The R-matrices multiply together this way, as a sum over repeated idices.

 

On a sad note, I just learned of the passing of Prof. Tim Cochran on December 16, 2014.

 

Ok, I've managed to cobble together some Tex symbols which should suffice.

The bracket states of a particular knot are independent of the knot's (or link's) global orientation, so we don't need any arrows anywhere.

We consider instead pairs of labeled arcs as vertical or horizontal (if they aren't, we can deform the boundary until they are); these arcs are segments or sections of the boundary.

So we have a matrix described by these labels as its indices over some set \( \mathcal I\,=\, \{1, 2, 3, ..., n\} \). We can suppose that n = 2 since this makes it easier to keep track of all the matrix elements.

So we have: \(R^{ab}_{cd}\,=\, A\,{}^a_c)({}^b_d \;+\; B\,{}^a_c { \overset { \smile } \frown}{}^b_d \).
Here, I've used parentheses with labels for the vertical pairs, and smile over frown for the horizontal pairs. But that isn't important right now.

We want to show that rotating the index set by 90° is equivalent to changing the switch from a B-type to an A-type (and vice-versa).

That is we want: \(R^{bd}_{ac}\,=\, B\,{}^a_c)({}^b_d \;+\;A\,{}^a_c { \overset { \smile } \frown}{}^b_d \).

 

Dammit, don't know why I typed "index set" when I meant "indices". The second equation clearly has R with a different permutation of the indices a, b, c, d, however.

Furthermore, it's now straightforward to show that \( R^{ab}_{cd} \) and \( R^{bd}_{ac} \) are inverses.

 

If we stay with the notion that a Kronecker delta represents a labelled vertical or horizontal strand, and that the labels as indices of the delta "function" can be raised or lowered, then clearly \( \delta^a_b\,=\, \delta^{ab} \).

So there is no distinction between horizontal and vertical (as arrangements or configurations of strands or arcs) except for how the indices are written; mathematically the previous equality says that both configurations define a set of elements with indices which define where they "belong" in a matrix--the identity matrix. My previous mathematical description of the elements you get when the indices are enumerated or "run over", the order isn't important since the indices uniquely identify a place in the matrix for each element.

So we then have:

 

Sorry, I've got Internet access problems at the moment.

Anyway, this method of using Kronecker deltas and the order and position of their indices to denote (notate) vertical or horizontal pairs of strands seems a bit unsatisfactory.

If you define \( \delta^a_c \delta^b_d \) as a pair of vertical strands, you know that this is mathematically just \((I_4)^{ab}_{cd} \), the identity with the same indices.
But \( \delta^{ab} \delta_{cd} \) is the same identity matrix. You can rewrite \( \delta^a_c \delta^b_d \) as \( \delta^{ac} \delta_{bd} \), a pair of horizontal strands, and clearly \( \delta^a_d \delta^b_c \), is not the same as the first vertical pair, it's a representation of the first pair as crossed over each other (without describing which strand crosses which). Kauffman mentions a "useful" identity, but I haven't seen it being used again in his book (or it's well buried), 'sigh'.

Also quite obvious is that a pair of strands, in terms of a braid group, has only two permutations but three distinct 'states': two uncrossed strands (= the identity operation), and two distinct crossed strands (left over right or right over left). The last two operations are the same permutation; repeating either operation is again the identity permutation but the braid then has two crossings in the same direction, clockwise or counter-clockwise. This is the permutation group \( S_2 \) as the braid group \( B_2 \) quotiented by this equivalence between the "squares" of braid operations (there is only one of these in \( B_2 \) plus its inverse) and permutations over two strands.

All this is really just my sketch of how the background mathematics hangs together, Kauffman takes it a lot further and for me to try to go there I need to get my head around a whole lot more.

 

I never understood why mathematics is necessary to describe nature as opposed to a non-mathematical approach to things like the non-mathematical approach taken in biology, chemistry, psychology, cognitive psychology, cognitive science, philosophy, linguistics and neuroscience for example.

For all we know nature might turn out to be non-mathematical at all.

 

I can assure you from personal and professional experience that the first 2 on this list (at least) do make use of mathematics.Others may have insights into the remainder of your list.

It is very hard to attach any meaning to this supposition. Who ever said "nature is mathematical"?

Would you not rather say that, at the least, mathematics is a useful tool for describing nature, and at most has proved to be "unreasonably effective" to quote Eugene Wigner?

 

Topology is the study of three dimensional surfaces

 

Topology in simple words is the study of geometric properties and spatial relations unaffected by the continuous change of shape or size of figures.

 

The Wikipedia article on Topology lists four subfields: general, algebraic, geometric, and differential topology.

A topological invariant is something that doesn't change when the space is "deformed" (topologically!). If it was only about geometry that would limit the subject somewhat.

 

One thing I haven't looked at very hard is the loop value.

Initially, this is assumed to be 1, but the type I and type II moves are satisfied if the "operator" d has a value of \( -A^2 - B^2 \).

A,B, and d are 'algebraic variables', in that, they are assumed to be placeholders for some value we aren't concerned about. Recall that we have a bracket polynomial for K, a general knot diagram (which need not be knotted, meh):

\( \langle K \rangle\,=\, \sum_{\sigma} \langle K | \sigma \rangle d^{||\sigma||} \)
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Suppose \( \langle K \rangle\,=\, \langle \bigcirc \rangle \), then obviously there is only one state. Note though, there is some ambiguity between the notion of different states and the number of states; really we sum over the number of them, and we don't exactly "sum over states" unless we can do \( \sigma_i,\, i \in \{1, 2, ..., n\} \).

\( \langle \bigcirc \rangle\,=\, \sum_{\sigma}\langle \bigcirc | \sigma \rangle d^{||\sigma||}\)
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Since \( ||\sigma||\) is defined as the number of loops (disjoint circles) in K minus 1, \( d^{||\sigma||}\,=\, 1\). So

\( \langle \bigcirc \rangle\,=\, \sum_{\sigma}\langle \bigcirc | \sigma \rangle \)
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But \( \sigma\,=\, \langle \bigcirc \rangle\,=\, K \), and since there is only one state, we can drop the "sum over sigma" notation, and we get something that looks superficially like bra-ket notation:

\( \langle \bigcirc \rangle\,=\, \langle \bigcirc | \bigcirc \rangle \).​

The expression on the right just says that a single loop has only one state, itself, so it should "sum to 1" . . .

 

I'm going to use a png file of a figure-8 loop, the arrows and the yellow region can be ignored; we focus on the single crossing at the top.

\( \langle \rangle \,=\, (A\,+\,Bd) \langle \bigcirc \rangle \)

(Ok, can someone please tell me how to insert a .png file into tex, it looks like this

Please Register or Log in to view the hidden image!

? thankyou)
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Suppose \( d\,=\, -A^2 - B^2 \), then:

\( A + Bd\,=\, A + (-A^2 - B^2)B\,=\, A + (-BA^2 - B^3) \)​

If AB = BA = 1, then \( A + (-BA^2 - B^3)\,=\, A - A + B^3\,=\, B^3 = A^{-3} \).

Suppose also that \( d\,=\, -A^2 - B^2\,=\, 1 \), then \( -A^2 - B^2 - 1\,=\, 0 \), so we have \( A^2 + B^2 + 1\,=\, A^2 + B^2 + AB\,=\, A^2 + AB + B^2 \).

Which looks superficially like a binary quadratic (just sayin').

 

Actually, I used the wrong diagram. And I figured out that if I can post the .png, I can do this:

\( \Biggr \langle \)

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\( \Biggr \rangle \,=\, (Ad\,+\,B) \langle \bigcirc \rangle \)

(shame about the size)

​

=>

\( Ad + B\,=\, A (-A^2 - B^2) + B\,=\, -A^3 - AB^2 + B\,=\, -A^3 - B + B\,=\, -A^3 \)​

Underlining that you need to be careful about crossings and their states.

 

Kauffman mentions, when discussing his abstract tensor model (crossings with index variables), that the loop value in this model is the trace of a Kronecker delta: \( \delta^i_i\,=\, n \).

So we have a vertical (or horizontal, or maybe neither) strand with the same index variable at "both" ends. If this is a loop though, then the strand must be looped from i to i, type of thing; there is only one index after all; a loop with an index is like a graph with one vertex which has a loop on it connecting the vertex to itself.

We also require the notion of implied summation, or: \( \sum_{i=1}^{n} \delta^i_i\,=\, n \). The (upper and lower) indices are repeated for this Kronecker delta, this implies the summation and so the trace.

What if you try something like \( \delta^i_i \delta^j_j \)? This will just be n × n. How about \( \delta^i_j \delta^j_i \)?

Maybe to hammer home that you need four indices to be able to permute them such that you don't get the identity matrix, the following paper exercise:

Write out the 4 x 4 identity matrix, then label each element (0 or 1) with four indices; first divide the matrix into four equal blocks. Each block will be a 2 x 2 matrix; draw an upper and a lower "1" next to each element in the upper left block, then an upper "1" and a lower "2" on the elements in the upper right block, an upper "2" and lower "1" on elements in the left lower block, and finally an upper and lower "2" on elements in the lower right block. Next to the first pair of numbers on each element, draw a second pair according to the row and column they occupy in each block 2 x 2 submatrix.

Over each element, the left pair of indices describes the "outer structure" or block matrix, the right pair describes the "inner structure" or each submatrix, in terms of rows and columns (resp. subrows and subcolumns).

So you have four index variables which describe elements: \( \delta^a_c \delta^b_d \), say. You can swap these around so the outer pair of indices is on the right instead of the left and nothing will change.

But if you permute the order of the indices, say you swap b and c around; looking at the 4 x 4 identity block matrix with indices identified, you can see that this will be a permutation which is not an identity, it swaps the middle pair of columns (resp. rows). Really all you need to do is consider when \( \delta^a_c \delta^b_d \) is not zero, and then when it will be nonzero if you swap b and c.

 

Agreed

Geometry would definitely limit the subject

Because geometry is not three dimensional