Witten and Knots

November 16, 2011

If you are at all interested in mathematical physics you will want to watch Ed Witten’s recent talk on his work in knot theory that he gave at the IAS. Witten gives a general overview of how he discovered that the Jones polynomial used to classify knots turns out to be “explained” as a path integral using Cherns-Simon theory in 3D. More recently the Jones Polynomial was generalised to Khovanov homology which describes a knotted membrane in 4D and Witten wanted to find a similar explanation. He was stuck until some work he did on Geometric Langlands gave him the tools to solve (or partially solve) the riddle.

Geometric Langlands was devised as a simpler variation on the original Langlands program that is a wide-ranging set of ideas trying to unify concepts in number theory. Witten makes some interesting comments during the question time. He says that one of the main reasons that physicists (such as himself) are able to use string theory to answer questions in mathematics is that string theory is not properly understood. If it was then the mathematicians would be able to use it in this way themselves, he says. Referring to the deeper relationship between string theory and Langlands he said.

“I had in mind something a little bit more ambitious like whether physics could affect number theory at a really serious structural level like shedding light on the Langlands program. I’m only going to give you a physicists answer but personally I think it is unlikely that it is an accident that Geometric Langlands has a natural description in terms of quantum physics, and I am confident that that description is natural even though I think it mught take a long time for the math world to properly understand it. So I think there is a very large gap between these fields of maths and physics. I think if anything the gap is larger than most people appreciate and therefore I think that the pieces we actually see are only fragments of a much bigger totality.”

See also NEW

It’s 11/11/11 11:11:11.111111… UTC

November 11, 2011

In honour of this auspicious moment here is some fun binary maths.

What is this sequence 2,3,7,11,13,19,25,… ?

It is actually the sequence of binary dropped primes. If you want a longer list it is Sloane sequence A014580 These are the primes when you multiply in binary but always drop the carry bits. For example 3 x 3 = 112 x 112 = 1012 = 5. I dropped a carry that would normally add 4 to the answer so 5 is not a binary dropped prime. 9 is not prime (I will drop the “dropped” from now on) either, because it is 9 = 3 x 7.

A curious feature of these primes is that if you reverse the digits in binary you always get another prime so 11 = 10112 and 13 = 11012 are both primes. All primes except 3 must have an odd number of non-zero binary digits because otherwise they are a multiple of 3.

In honour of the date and time it seems appropriate to factorise some numbers that consist of all ones in binary

1111 = 11 x11 x 11

111111 = 11 x 111 x 111

1111111 = 1011 x 1101

11111111 = 11 x 11 x 11 x 11 x 11 x 11 x 11 x 11

Others that I missed are prime. You can probably see that a number in this form can only be a binary dropped prime it the number of digits is a normal prime, but the converse is not true because 127 = 11 x 13 in decimal terms

As well as multiplication you can also do binary dropped addition and this gives you a ring which is a unique factorization domain. Don’t worry about subtraction because it is the same as addition. You can also do modulo arithmetic with the rule that a = b (mod n) iff a+b is a multiple of n.

Are these useful or are they just a curiosity for setting math puzzles? In fact they can be used to generate interesting cyclic binary linear codes . Binary linear codes are sets of codewords consisting of sequences of bits that are closed under binary dropped addition. There are always 2k codewords for some k in such a binary linear code and if the block length is n it can be used to code k bits into n bits in a way that is easy to reverse and very efficient for error-correcting. Cyclic codes have the extra property that you can cycle the digits of the code by shifting them left and moving the last digit to the first. In a cyclic code, when you cycle a codeword in this way you always get another codeword in the set.

The simplest non-trivial cyclic codes are parity codes where the number of non-zero bits in a codeword is always even, In this case n = k+1 so encoding consists of adding a parity bit to a k-bit string. Now remember that a number with an even number of non-zero bits is always a multiple of 3 in binary dropped arithmetic, so can we generate other cyclic codes by using multiples of other numbers in binary dropped arithmetic?

(11 minutes and 11 .111 seconds to go, now observing a minutes silence…)

The answer is yes, but it only works when the numbers are factors of the numbers which are sequences of ones. This is because cycling the numbers is equivalent to working modulo such a number. So we can have codes with a block length that are a multiple of three where all codewords are multiples of 111. More interesting examples are codes with 7 bits using multiples of 1011 or 1101. This is the familiar Hamming code Ham(7,4) which is linked to the Fano plane and the octonions. Even better a string of 23 ones must factorize in binary dropped arithmetic into two number of 12 binary digits, because this would be needed for the Golay code, but it is 11.11.1111… so I will have to check the factorization another time.

International Maths Olympiad 2011

July 19, 2011

The International maths olympiad is an annual competition for high school mathematcains who compete in national teams. This year the event is in the Netherlands. The contestants have now completed their 6 question test and can relax while they wait for the results.

Terrance Tao who is a Fields medalist and expert solver of IMO type problems has been running a mini-polymath feature on his blog each year and the third one starts today, in case you want to join in the fun. He will choose the hardest problem and invite you to find solutions, so being less ambitious I will take a look at the easiest, which is problem number one.

Given any set  A = {a1,a2,a3,a4} of four distinct positive integers, we denote the sum a1+a2+a3+a4 by sA. Let nA denote the number of pairs ( i, ) with 1 ≤ i < j ≤ 4  for which aaj divides sA. Find all sets A of four distinct positive integers which achieve the largest possible value of nA.

You might want to try to solve this before you read any further.

Here is a curious geometric way to construct the sets with maximum nA

Take a platonic solid with V vertices, E edges and F faces, take  A = {1, F-1, V-1, E-1}. Then sA = V+E+F -2 = 2E by Euler’s polyhedron formula.

a1+a2 = F  but for a regular solid the number of faces multiplies by the number of sides of each face must equal twice the number of edges, so  a1+a2 divides sA.

a1+a3 = V  but for a regular solid the number of vertices times the number of edges that meet at each vertex is also twice the number of edges so a1+a3 divides sA.

a1+a4 = a2+a3 = E which obvious divides sA = 2E

So for these sets nA ≤ 4

from the cube and octahedron we get A = {1,5,7,11} and for the dodecahedron and icosahedron we get  A = {1,11,19,29}. The tetradron gives  A = {1,3,3,5} but this is disallowed because the numbers must be distinct. You can check that nA = 4 in each case.

Other sets with  nA = 4 can be constructed by multiplying these two solutions through by a common factor. Your job is to show that this provides all solutions, and that nA = 4 is the maximum.

Update 20-July-2011: Tao choose problem number 2 for the mini-polymath. It was a geometric/combinatoric question about a “windmill” (very appropriate for Holland of course). The solution requires an observation that there is an invariant that is not immediately obvious to everyone. Meanwhile Lubos who took part in IMO 1992 has looked at problem number 5 , see his solution. (My olympiads were 1977 and 1978 by the way) If you are really interested in this type of problem you should visit the mathlinks forums where all the questions were already posted and solved long before any of us bloggers had looked at them.


The 24-cell and 4 qubits

February 28, 2011

A few days ago Lubos reported on an intriguing new paper by Volker Braun describing how to construct a Calabi-Yau manifold with 6 real dimensions and minimal Hodge numbers using the 24-cell. Such manifolds can be applied to the compactification of superstring theory down to our familiar 4 dimensional spacetime. The predictions for physics based on this particular manifold would be unrealistic but its discovery is an important step towards understanding the fuller range of possibilities. It is also of considerable mathematical interest in its own right.

I’m not going to say anything more about that paper but I do want to say something about the 24-cell and its curious relation to 4 qubits as well as a surprising relationship between the invariants of 4 qubits and the platonic solids. I found out about these things after talking to Mike Duff and his coworkers on the qubit/black-hole correspondence (Borsten, Dahanayke, Duff,  Marrani, Rubens). I think this is a bit more specialized than the kind of stuff I usually report on here, but some of our regular commenters expressed an interest and I am always happy to oblige. For anyone who does not understand any of the mathematical terms here they are all explained in wikipedia.

The 24-cell is a very special regular polytope in 4-dimensional space. It has the special property of being self dual in the same sense as the tetrahedron is self-dual in 3 dimensions. It can also be tessellated to fill 4 dimensional space just as a cube can tessellate  to fill 3d space. in fact the 24-cell is the only regular polytope in more than 2 dimensions that has both of these properties. The only comparable shapes in this sense are the triangle, square and hexagon in two dimensions.

The vertices of the 24-cell can be plotted in 4D co-ordinates at the 24 points given by

(±1,0,0,0), (0,±1,0,0),(0,0,±1,0),(0,0,0,±1),(±½,±½,±½,±½)

It’s dual can be plotted at the points,


As many readers of viXra Log are undoubtedly aware, there are many connected mysteries surrounding the number 24 in mathematics and the 24-cell is one of the more enigmatic. 24 is famous as the dimension of the Leech lattice which is connected to the significance of the number in the theory of finite simple groups, especially examples such as the Mathieu groups, the Conway groups and the Fisher groups. The existence of the Leech lattice can be explained in terms of the 24-bit Golay code which can in turn be constucted using special properties of quadratic residues in Z24. Alternatively the Leech lattice is a reduction of an alternating lattice in 25+1 dimensions using a null vector relying on the fact that  the sum of the first 24 square numbers is 70^2. This closely connects together one set of circumstances where the number 24 appears in mathematics.

Then there is a second interlinked set of places where the number 24 shows up in number theory and the theory of special functions. This includes the  Ramanujan discriminant function, a modular form that is the 24th power of the Dedekind eta function. This can be connected to the fact that the value of zeta(-1) is 1/12. It has implications in bosonic string theory where it is linked to the critical dimension where the two dimensional worldsheet vibrates in the remaining 24 dimensions.

These two sets of places where the number 24 appears, one in group theory and the other in number theory, do not seem to have a causal link. You cannot reason that one implies the other. Yet you can combine the two by compactifying bosonic string theory over the Leech lattice. This was a realisation that led to the famous proof of the monstrous moonshine conjectures and a Fields medal for Richard Borcherds. This much will be familiar to anyone who follows related discussions on the internet and especially if they have read John Baez’s lecture on the number 24. As far as I know there is nothing else that clarifies the mystery of this connection. For example there is nothing that directly links the Golay Code to the Ramanujan Discriminant function except Moonshine.

What about the 24 cell, where does that fit in? According to Baez it relates to the number theory side via elliptic curves, but in a way that involves a group of order 24. Perhaps this is a clue to a missing direct link between between number theory and group theory. The connection described by Baez points to the fact that the moduli space of elliptic curves is given by modding out the group SL(2,3). The vertices of the 24-cell when plotted as quaternions (Hurwitz quaternions)  also form a group, and it is the same one, also known as the ditetrahedral group because it is the double cover of the rotation group of the tetrahedron. This seems very nice but actually there are only 15 groups of order 24 and only seven that are not direct products of smaller groups, so saying that two structures form the same group of order 24 is only a small factor better than saying that they have the same size. What we really need to find is a more direct way in which the 24-cell relates to elliptic curves.

This is where the 4-qubit system comes in. The wavefunction of 4 qubits is represented by a 2x2x2x2 hypermatrix of 16 complex numbers. Local transformation on these qubits take the form of SL(2,C) transformations applied to each qubit independently so the overall symmetry group of the system is SL(2,C) 4 . To understand the entanglement possibilities for 4 qubits the first step is to find the polynomial invariants under this group. This is a non-trivial computation but  it can be shown that there are 4 independent invariants of degree 2, 4, 4 and 6 in the 16 components of the hypermatrix (see e.g. http://arxiv.org/abs/quant-ph/0212069 for a construction.) However, there is a special invariant that is a combination of these known as the hyperdeterminant which is of degree 24. The hyperdeterminant is a discriminant for the hypermatrix that is zero iff the quadriliear form constructed from the hypermatrix has singular points where all derivatives vanish. You don’t have to understand the details, just notice that this is another structure where the number 24 has special significance.

It turns out that the 4 qubit hypermatrix is related in a fundamental way to elliptic curves with the hyperdeterminant being related to the Ramanujan discriminant modular form mentioned above. I have described this relationship in detail at http://arxiv.org/abs/1010.4219 so I won’t repeat it here. This makes a direct link between the number 24 that appears as the degree of the hyperdeterminant and its appearance in the theory of modular forms linked to bosonic string theory. After discussing this with Mike Duff I was also able to link the 4 qubit system directly to bosonic strings and I used this in my FQXi essay.

The classification of 4-qubit entanglement is a tricky business. The SL(2,C) 4 transformation group has 12 independent paprameters so it should be possible to use these transformations to reduce any state with its 16 components to representative states parameterised by just 16-12=4 variables.  A clean solution was provided by Verstraete et al in http://arxiv.org/abs/quant-ph/0109033 . They found nine perameterised classes of states where the largest class known as Gabcd has 4 parameters and includes all states whose hyperdeterminant is non-zero. For present purposes I am only interested in this class. It takes a form that can be written in qubit terms as

Φ =  x (|0000> +|1111>) + y (|0011>+|1100>) + z (|0110> + |1001>) + t (|1010> + |0101>)

For this class of states, we can work out any of the invariants including the hyperdeterminant which is going to be a polynomial of degree 24 in the four variables xyz and t. This has the potential to be a complicated expression, after all the full hyperdeterminant in 16 variables is an expression with 2894276 terms. In practice for the reduced state the hyperdeterminant simplifies and when you work it out you will notice that the result factorised into 24 simple factors

Det(Φ) = x2y2z2t2(x+y+z+t)2(x+y+z-t)2(x+y-z+t)2(x+y-z-t)2(x-y+z+t)2(x-y+z-t)2(x-y-z+t)2(x-y-z-t)2

These factors correspond in an obvious way to the Hurwitz quaternions and therefore the vertices of the 24-cell. This provides a direct link between the number of vertices in the 24-cell and the degree of the hyperdeterminant for 4-qubits which in turn is linked to the exponents in modular forms and the critical dimension of bosonic string theory, just as we wanted.

Is there a better way to understand why the hyperdeterminant factorises so conveniently? Yes there is. Although all the 12 dimensions of the group SL(2,C) 4 were used to reduce the hypermatrix to the class Gabcd , there remains a discrete subgroup that maps states of Gabcd , (in the form above) onto themselves, so this discrete subgroup provides a group of linear transformations on the 4D space parameterised by xy, z and t. This subgroup turns out to be the Weyl group of D4 whose system of root vectors is the 24-cell. The polynomial invariants of this reflection group as functions of the four parameters xyz and t are also of degree 2, 4, 4 and 6 and correspond to the 4 qubit invariants. The Weyl group is the symmetry group of the root system so it just permutes the 24 factors in the hyperdeterminant making it an obvious invariant. Notice that D4 as a Lie-algebra is SO(8) or its split form SO(4,4) which is the group used to construct the 4-qubit/black-hole correspondence. This was what Borsten et al used to classify the entanglement of four qubits using a classification of nilpotent orbits that had already been worked out for black holes in M-theory. Their answer matches the Verstraete classification nicely.

We can go one step further and extend the group of transformations to include permutations of the 4 qubits. This gives a larger discrete group acting on Gabcd ,which can be identified as the Weyl group of F4. The corresponding root system is now the 48 vertices of a 24-cell combined with its dual. The polynomial invariants of this group are of degree 2,6,8 and 12 and they correspond to the primitive invariants of the hypercube that is symmetric under the permutations of the qubits as well as the usual  SL(2,C) 4 transformations.

This leads to one last curious correspondence that I want to point out. The degrees of the primitive invariants of the hypercube (2,4,4,6) are not trivial to work out, but can you see what they are related to? Think of a tetrahedron with 4 vertices, 6 edges and 4 faces. The remaining number 2 corresponds to the inside and outside of the tetrahedron which can be regarded as the two three dimensional parts which cover space in combination with the vertices, edges and faces. Remember that the 24-cell as a group is the double cover of the rotation group of the tetrahedron so there is a connection. For the invariants that are symmetric under permutations, the larger root system of 48 vectors from the two 24-cells combined also forms a group when the root vectors are regarded as unit quaternions. The full set of unit quaternions form the group SU(2) which is the double cover of SO(3) so any finite subgroup must be the double cover of some rotation group in 3D. In this case it is the rotation group of the cube or octahedron. This corresponds to the fact that the symmetric invariants for four qubits are of degrees (2,6,8,12) because the cube has 6 faces, 8 vertices and 12 edges (or you can use the dual octahedron).

So despite the fact that the invariants of the 4 qubit system are non-trivial to construct, their polynomial degrees correspond to the geometric elements of three of the platonic solids. What about the other two regular solids, the dodecahedron and icosahedron? There is another reflection group H4 whose root system corresponds to these solids and it therefore has invariants of degree (2,12,20,30) . Since this group acts on the same 4D sapce you can use it to construct four invariants of the 4 qubit system with these degrees, but there is no lie algebra corresponding to H4 and its significance is not so obvious. However, these three cases are part of a system of mysterious “trinities” as noted by the mathematician Vladimir Arnold. This means that there must be a lot more going on that we don’t really understand yet.

String Theory and Partition Numbers

February 12, 2011

Recently there was an intriguing breakthrough in the study of partition numbers. partitions numbers P(n) count the number of ways of expressing n as a sum of positive integers. E.g. P(4) = 5 because 4 = 1+1+1+1 = 1+1+2 = 1+3 = 2+2 = 4. The sequence of partition numbers goes 1,1,2,3,5,7,11,15,22, … It starts off wanting to be the Fibonacci numbers but then after the first five it changes its mind and gives the first five prime numbers before degenerating into a rapidly increasing sequence of interesting numbers.

There are lots of well known things that can be said about the partition numbers but I’ll leave that to Ken Ono who is the main figure in the new discovery, actually two discoveries. The first discovery is a new finite formula for partition functions and the second is a new explanation for some congruence relations discovered by Rananujan. I highly recommend this low level lecture by Ken Ono as an introduction to the new finds.

One thing that is not mentioned in all the recent news coverage is the important connection between partition numbers and string theory that is very easy to see even a very basic level. From the theory of musical harmonics you know that a string has vibration modes labelled by integers k, whose frequency is ωk = kα for some constant α that depends on things like the tension in the string. When a string is quantized it can be treated like a set of decoupled harmonic oscillators with energy levels Ek = (1/2 + mk)ħωk where the sequnece of non-negative integers mk labels the eignestates of the oscillators.  So the total energy is given by

E = Σk (1/2 + mkkα

The zero-point energy is E0 = 1/2 ħkα(1+2+3+4+…) . We can either ignore this as an irrelevant constant while pretending not to notice that it is infinite, or we can use zeta regulation to deduce that 1+2+3+4+ … = ζ(-1) = -1/12. In any case, what we are really interested in is the rest of the sum and to understand it we just need a simple trick. Write mk k = (  k+k+…+k ) (mk times) Then you will immediately notice that the number of states with an energy En = E0nħα is exactly P(n), the partition number of n. This is also valid for the relativistic bosonic string in 26 dimensional spacetime, except that then you need to multiply by 24 because the one dimensional string can vibrate independently in any of the 24 space dimensions transverse to the string.

The partition function for bosonic string theory is therefore given by

Z = Σn P(n) exp( – (24n-1)ħα)

Perhaps that’s why they call it the partition function 🙂

A Pint With Mike Duff

September 16, 2010

In a recent post I cheekily suggested that I could tell Mike Duff some interesting things about hyperdeterminants and elliptic curves over a pint. Mike runs one of the world’s top string theory groups at Imperial College which itself is ranked as the UK’s top research institution according to the Times list, so it was a quite an honour and delight that he took me up on my offer.

Imperial College is based in London and is only about an hours train ride from where I live so yesterday I spent some time with, Mike and two of his students Duminda Dahanayake and Leron Borsten partly in the pub and partly in his office. On Mike’s blackboard were some equations left after a recent visit by Witten that could have held the key to understanding M-theory and therefore the laws of the universe, but Mike had to rub off them off so that I could explain my work.

In the pub we talked about many things including M-Theory, viXra, anti-crackpots, and mostly hyperdeterminants. I learnt some fascinating things and I did indeed explain my number theory stuff and how it might connect to quantum gravity. It turned out that Leron had come up against some number theory problems he needs to solve while working on the classification of black holes, so it was easy to find common ground and discuss some useful things.

They were very welcoming and I got a lot out of the visit. To prove it here are some pictures taken after we had finished off a few pints.