Clavelli-Shapiro Trick

Computing the expected energy for a harmonic oscillator in a heat bath isn't bad. Other observables aren't too bad either, but here is a trick that will generalize.

First introduce new b operators obeying the same commutation relations.

To compute the thermal expectation value with parameter y ( fugacity goes with chemical potential, I mean the one that corresponds to temperature), you take the regular expectation value of your operator, but replace all the a's and their adjoints by

respectively.

3 Proofs in A.6 at least in a useful special case.

This seems insanely useful depending on how far it is pushed.

Relative QFT

http://arxiv.org/pdf/1212.1692.pdf

Expectation 5.3: Given a real Lie algebra, invariant inner product such that all coroots have length square 2, and a full lattice in the Cartan such that on the lattice the inner product is integral and even, there should exist a 7 dimensional topological quantum field theory and a 6 dimensional theory relative to it.

This is supposed to be the big goal of the (2,0) siX theory.

A relative QFT is an extended field theory in one higher dimension. To be continued...

Decay

MOA2DIS mother of integrability

Why self dual yang mills in 4 dimensions is the Mother Of All 2d Integrable Systems. At least that is what Ketov says in his last chapter.

Found the sources
http://arxiv.org/pdf/hep-th/9307021v2.pdf
http://ac.els-cdn.com/037596018990964X/1-s2.0-037596018990964X-main.pdf?_tid=410488d6-3295-11e2-8bd0-00000aab0f6b&acdnat=1353362920_03278f52f496a8df2a191a8dae74b67f

Let's work on (2,2) signature  with metric  . Look at the self-dual Yang-Mills equations on it. In these coordinates they read:




Calling the components of the gauge field A,B,C,D respectively. We can ask that they don't depend on u or y. We can also ask that A=D, but I don't know why you would ask such a thing.




We have put three equations into 2 by inserting a  and asking to read of the coefficients of the commutator of the above operators.

The quadratic piece demands that B not depend on x. There are three possibilities for B after gauge transforming. Let us pick the one where B is proportional to the generator for the Cartan in it's usual form multiplied by some t dependent function.

The equations are solved if we write the following for A and C.




and



The second choice above gives the nonlinear Schrodinger equation with attractive self interaction.

We made several choices along the way, other choices would have resulted in NLS with repulsive interaction or KdV.

If that is all there is to it, the title MOA2DIS seems to be excessive.

Enriching some categories

Start with a plain old category C. You have some objects A and B and Hom(A,B) to describe all the arrows from A to B.

What structure does Hom(A,B) have?

Not much. Ok, now I'm stuck. I can't say any more.

If the Hom(A,B) were just sets, it would be locally small. Not good enough.

If I somehow knew they were abelian groups, then I would be in pre-additive categories, and I would start to have some hope about doing all the abelian category chases.

What are the structures I could possibly ask Hom(A,B) to have? I need to have a well defined map Hom(B,C) x Hom(A,B) to Hom(A,C) so that tells me that it has to be monoidal in order for the left hand side to make sense.

And there should be the identity in Hom(A,A).  That means there should be a morphism from the monoidal unit to Hom(A,A) in this monoidal category.

AND coherence relations time:

Now another example, what if I enrich over chain complexes of complex vector spaces. Each hom is now differential graded. So this is a dg-category.

I can relax conditions a little bit more and give you an example of an A_\infty category. The specific one I have in mind is the Fukaya category of some symplectic manifold.

Picture time

Draw more holomorphic polygons to give the higher maps that mess up the dg-category structure.

Youtube videos better than NYAN cat

Cool people saying cool things

Contact Homology

GRASP usually gives me things I can post about, and today's talk was embedded contact homology.

Start with a contact 3 manifold Y. Find it's Reeb vector field R.



Look for where this vector field makes circular orbits. Those are Reeb orbits.

Take the module generated by all unions of them.

An arbitrary element will look like

 

where R lists all the orbits and m's are the multiplicities in  .

You say the boundary of a particular union of Reeb circles is the sum over all output union of Reeb circles such that each one is counted by the size of the moduli space of holomorphic curves connecting the first set of inputs to the second set of outputs in the symplectic manifold associated to Y.

The picture looks like

The top and bottom pieces are such holomorphic curves. This therefore is showing a contribution to d^2 which we hope all cancel out.

Of course you need a condition on indices. It is so much easier to explain the index in Lagrangian Floer where it is the number of times the tangent planes turn. These indices save the day and give some control of what the end curves can be.

I don't know the index in ECH, but since he didn't say what it was in the talk, it sounds profoundly ugly. This also means I have no hope for explaining why this is a chain complex. These all have that same Morse homology flavor of trajectories breaking like shown above. In the Morse case, I can visualize all the pieces cancelling and that is where mileage can be drawn from.

Fivebrane structure

So you know how you need the second Stieffel Whitney class of spacetime to vanish for fucking fermions to make sense? Why is that?

First you have an SO(n) bundle. But you want it to be an Spin(n) bundle so that the sections on the associated vector bundle will become fermion fields.
Viewing the bundles as homotopy classes of maps instead you are looking at a lift of the map from X to BSO(n) to a map from X to BSpin(n).

There is a map from BSO(n) to BBZ_2 given by the special element 1 of [BSO(n), BBZ_2]=H^2 (BSO(n),Z_2)=Z. This is the second Stieffel-Whitney class. Pullback the universal bundle gets you BSpin(n). So in order for the map to lift, the diagram must commute. This means that the map X to BSO(n) to BBZ_2 has to vanish all the time. That only works when this class vanishes for the tangent bundle.

Now to get from Spin(n) to String(n). Repeat the construction.

There is a map from BSpin(n) to BBBU(1) given by H^3 (BSpin(n) , U(1))=Z. Pullback the universal bundle to get BString(n). The diagram commutes when the composite X to BBBU(1) vanishes. This class is half the first Pontryagin class in H^4(X,Z).

Repeat again from String(n) to Fivebrane(n). This time replace BBBU(1) with 7 applications of delooping. The obstruction class is now one sixth the second Pontryagin class in H^8(X,Z).

http://arxiv.org/pdf/0805.0564v3.pdf

Goodbye Thurston

Thurston has left :( so I figure a geometerization of 3 manifolds post is in order.

This is all in his book Three Dimensional Geometry and Topology.

For surfaces, uniformization is prehistoric. We have the easy invariant of the genus to tell us the covering space and it gets a constant curvature metric.

For 3d manifolds, the situation is trickier. We again want to find model manifolds along with their Lie groups of isometries. We also want these actions to be transitive. None of the points should get any special treatment.

The connected component of the point stabilizers are going to be Lie subgroups of SO(3).

If they are the full SO(3), then all directions you stare into look the same so we have a symmetric space. It must be one of euclidean, spherical or hyperbolic depending on the curvature.

If the component of the stabilizer is SO(2), then look in the leftover direction at each point to construct a vector field. The flow lines of the vector field give a foliation of the manifold where every leaf is 1D. The leaves are either circles or lines.
The easiest example of such would be surface cross R. This gives two new geometeries and euclidean 3-space over again.
There are two more of these, the cover of SL2 and the Heisenberg group.

This leaves when the point stabilizer is 1D. The only possibility is Sol geometry.

Don't let go of money you don't have to

http://www.chrisstucchio.com/blog/2012/hft_apology.html

I would suspect a high frequency trader to be with the Alliance.

(2,0)

“The relation between 4D N=4 SYM and the 6D (2, 0) theory is just like that between Darth Vader and the Emperor. You see Darth Vader and you think “Isn’t he just great? How can anyone be greater than that? No way’.Then you meet the Emperor”. - Nima

It all comes down to the distinction between string theory and M theory. Instead of thinking about the worldvolume theory of a short stack of D3 branes, you know think about stacked M5 branes compactified on a torus. Since we are compactifying on a torus, that most BORING of all the Calabi-Yaus, we keep all that nice supersymmetry. It explains S duality nicely as the mapping class group of the torus. If we compactify on a general Riemann surface, we get other S-duality groups corresponding to that surface.

5 SIGMA!

http://havewefoundthehiggsbosonyet.org/

And ATLAS comes in with 5 Sigma

Rumor Has It

http://physicsworld.com/blog/2012/06/cern_calls_press_conference_fo.html

Excitement!

Keep an eye on the Jackie and Toves tumblr for liveblogging the revelation.
http://physicistsofdiscerningtaste.tumblr.com/

AdS Integrability

The idea behind this is that in the N=4 SYM on the boundary at large N, only the single trace operators will really matter.

They will be like Tr(abcdef...) so for example if they were a bunch of those scalars coming from the transverse motion of the D3 branes you would get a SO(6) spin chain. This you can then solve by Bethe Ansatz.

It's all thanks to planarity.

Of Monsters and Mathematicians

I did a moonshining post about Matheiu, but not about it's daddy the Monster. That oversight will be fixed now.

Suppose you have a discrete subgroup of PSL(2,R) call it G. The quotient of the hyperbolic plane by G will be some surface possibly with bad stuff. When the surface is genus 0 call that a genus 0 group.

We also want two more conditions on our group. The group should contain all of the T duality subgroup. T just translates the complexified coupling by 1. Shifting the action by multiples of 2 pi would make sense to include. Let us also have it contain the the congruence subgroup for some N.

Because the group contains those translations, we can Fourier expand on the quotient in powers of q.

The field of functions on this surface is the rational functions of a single function J called the Hauptmodul.

If you do this for SL(2,Z) you get the familiar one.


Finally the monster raises it's head.

The Hauptomodul looks like



where V is a infinite dimensional graded module for the Monster. This was originally constructed from looking at those numbers in the SL(2,Z) J function and noticing the fact that 196884=1+196883 and those were known irreps of the Monster. The fact that you could break up those numbers in such combinations of the dimensions of Monster irreps was the distilling information illegally from character tables.

You can apply Hecke operator like constructions to those weighted characters.



This is modular and can only blow up at the cusps so it is a polynomial in the original Tg.

These are called replication formulae because they relate repeated applications of g.

We're still relying on coincidences between the numbers to establish connection between these two series. We need to construct the representation V.

This is done by compactifying the 26 dimensional bosonic string target space on the 24 dimensional Leech lattice and also quotienting by the -1 involution of this lattice. Recall the heterotic string construction.

I enjoy our Little Talks

Terry Fucking Tao!

http://www.nature.com/news/mathematicians-come-closer-to-solving-goldbach-s-weak-conjecture-1.10636

Spooky Action from the FUTURE!

I have 2 machines that produce entangled pairs of photons. Call them thing 1 and thing 2.

Thing 1 sends a photon to Dick. Thing 2 sends a photon to Sally. They both send the other photons to the cat.

Dick and Sally measure their photons in whatever basis they choose.

Later, the cat decides whether or not to entangle the two photons it got.

If the cat did nothing, Dick and Sally's measurements have nothing to do with each other. There is no way they could talk to each other.

But if the cat did entangle them, the two measurements are correlated. That's weird, the cat acted in the future and influenced what Dick and Sally saw before.

An actual experiment

http://arxiv.org/ftp/arxiv/papers/1203/1203.4834.pdf

It reminds me of those time travelling neutrinos that came back and loosened that cable so they wouldn't be found out. (Credit Kate)

Some cute integrals

Normally I would hate evaluating integrals with trig functions with the white hot intensity of a thousand suns, but these are so cute they can get away with it.



Everybody get  . Good!
Next one



Everybody get  . Good!
Next one



Keep doing these all the way until



Big surprise!  What did you get?

Before doing the next one, guess the next one.

If you guessed   , you would be wrong. It is actually about 2.31e-11 lower than that. Still a rational multiple of   though. If you managed to guess that I tip my hat to you.

You can't possibly be satisfied with this. Something has to explain why the jump suddenly happened so far into the sequence.

Well, to do these better double the integral by going over the entire real line. Now use the property that integrating in real space and in momentum space gives the same results. The products of sincs becomes convolutions of box functions. In fact you might as well do this with a general sequence like so:

 instead of the reciprocals of the odd integers like we did before.

Look at the running sums of the alphas we had before. You see that



The step where things mess up is when the sum jumps above 2. This is generally true as well. To show why this is true you need to look at the widths of the convolutions and how that grows.

Next time someone tells you to fill in the pattern



you can confidently tell them you have no fucking clue.

Plant a seed and grow a cluster

As the title says start with a seed.

It is a n tuple of variables.

Also give an n by n skew symmetric matrix that is called the exchange matrix.

For each i between 1 and n, there is a mutation.

The action on the exchange matrix is given as follows.

View the exchange matrix as the adjacency matrix for a quiver. Now flip the arrows that touch vertex i. Also create shortcuts whenever you flip two arrows that are lined up. Cancel any loops you might have created and then give back the adjacency matrix.

The action on the seed is as follows.

Leave everything except the ith in the tuple alone. Call that one y and replace it with w. The other elements in the tuple are t's.

Now we have an n valence tree where each vertex has a seed and an exchange matrix and traversing an edge tells you to mutate in that direction.

A cluster algebra is called finite type if you end up only getting a finite number of seeds. It has been shown that the classification of finite type cluster algebras is the same as the classification of semisimple Lie algebras and finite root systems.

Baker, baker give me recurrence as fast as you can

The Baker's map is a simple dynamical system that has deterministic chaos. The map is defined on a square. Cut the dough in half vertically then stack them on top of each other after kneading each piece. It is illustrated below.

The exponential growth between two different points is especially easy to see for two points that are just separated in the horizontal direction and not in the vertical direction at all. The distance between them doubles every time, at least until they end up on different sides of the vertical cut, then its not so easy to say what happens.

It doubles the horizontal direction and halves the vertical direction, so it will preserve area for every rectangle, which is a basis. So extending this preservation to the sigma algebra they generate, you see that this map preserves Lebesgue measure.

A picture of the chaotic nature.

By composition, we get an action of this map on functions on the square. It is unitary on the Hilbert space of square integrable ones. Once we figure out this operator's spectrum, we will be able to exactly say what this map does to arbitrary square integrable functions. Express the function in the diagonalized basis then use the eigenvalues to determine the action on each summand.

Each eigenvalue will have some power N that takes it as close to 1 as you want. If you want an entire square integrable function to come back to look like itself, take the least common multiple of each N of the summands that contribute a significant amount. Some analysis will give you exactly which of the components you'll need to consider. Let's call that applying the baker map M times.

The punchline to the above is you can get a function that is arbitrary close after M iterations. This demonstrates Poincare recurrence for this dynamical system. So it starts out looking more and more disordered, but then it comes back to the original state or at least close enough.

If you deleted a bit when it was in it's chaotic state, then you would be screwed. Think like if you forgot the states in the Black Hole and didn't remember to keep track of all the Hawking radiation.

Virtually Haken

Apparently every compact, orientible, irreducible 3 manifold with infinite fundamental group has a finite sheeted cover that is Haken. Will keep an eye out to see Agol's proof.

What is Haken?
It means it should have an orientable incompressible surface.

So what to do with this orientable incompressible surface? Thicken it up a little bit and then remove that from the 3-manifold. Now you have 3-manifold(s) that have a boundary that is that surface. But if that surface isn't a sphere, then the 3-manifold we are talking about has infinite first homology. This gives a properly embedded 2-sided nonseparating incompressible surface in each of the pieces. So the result after the cut is still Haken. Repeat until the only surface you can find is a sphere. But they are irreducible, so these spheres bound 3-balls. This gives a decomposition of the manifold into 3-balls.

3-balls are easy, and the gluing is doable so we have some control. A nice theorem is that homotopy equivalence is homotopic to an honest homeomorphism. This means we only need to be concerned with fundamental groups. It is also true that these fundamental groups have a solvable word problem. We can actually tell when a complicated curve given as a big product of nasty group generators is actually trivial. This is also true for virtually Haken.

What are some examples of Haken manifolds? There is everybody's favorite type of 3-manifold, link complements. There are also surface bundles over the circle.

Now that we have the definitions written, we play the waiting game for when the preprint goes up to see how this is proved.

http://en.wikipedia.org/wiki/Virtually_Haken_conjecture

http://en.wikipedia.org/wiki/Haken_manifold

Being Judged for ALL THE LIES

I am amused. :)

More seriously, Strominger talked about dS/CFT as analytically continuing AdS/CFT. This is not normally kosher, but in if your bulk theory is higher spin you don't run into the problems you do when your bulk is say type IIB. So with that you end up getting 3d CFT's that have SO(-N) symmetry.

What the fuck does that even mean? Novikov asked the same question in 1980. It means Sp(N). Why? Because the change in the quadratic forms transforms the scalars from commuting to anticommuting. That means that the loops get factors of -N instead of N. 

But that is the liar's (read physicist's) explanation. Maybe you should just call me a liar.

Steenrod Algebra, Lie Groups and Formal Series! Oh My!

http://math.berkeley.edu/~reb/courses/261/all.pdf
Page 24

Disclaimer: I don't know the difference between 1 and -1.

Recall the operations Sq which act on the cohomology as degree i maps and looking at the right domain    gives the cup square.

These operations have the ugly relations
 

In addition there is a coproduct on this algebra



so the Steenrod algebra is a cocommutative Hopf algebra over the 2 element field.

But we know where cocommutative Hopf algebras come up a lot. Universal enveloping algebras of some Lie group.

This "lie group" will be like the automorphism group of something. What is that something?

Here is where we get to formal power series. Infinitesimal transformations of the real line fixing 0 are given by  formal power series with 0 constant term and invertible linear term. Look at the composition of such power series and you end up finding a complicated product or in other words a complicated coproduct on

Now we really want the additive homomorphisms. There aren't any besides multiplication by constants. But if we reduce mod p we are ok. We do the same process of formal power series and figure out the coproducts.

So our formal power series are


Try composing one with coefficients a's and one with b's. You get



What are the coproducts from this?

 

Set p=2, treat the 0th term as 1 and dualize to get the relations we had before.

Motives and Deformation Quantization

3/3/12

http://arxiv.org/pdf/math/9904055v1.pdf

I heard Nima has been thinking about actions of the Motivic Galois Group lately.
What's up with that???

4/29/13

http://www.its.caltech.edu/~matilde/ObiMotivesSurveyFinal.pdf

How do YOU quantize???

I think I'll fill out one of the four sections for this post at a time.

We aren't lucky enough to get strong quantization and that sucks! This was proven by Groenwald and van Hove.

Weyl
Geometric
Brane

Sources: (in order)
Weinstein's Notes
Weyl
John Baez
Gukov

Why can we solve some QM problems?

Here is some old stuff back from Paulo Bedaque's 622.

Write the two operators



They are H.C.s of each other so lets call them A and  respectively. Write two Hamiltonians




If you have an eigenstate   of the second, then you automatically get an eigenstate of the first by taking  . Therefore the two Hamiltonians have the same spectrum except possibly at the bottom rung.

Expanding out the Hamiltonians gives the two potentials as .

Playing with the ground state of the Hamiltonian gives a formula for the superpotential, or the other way around if you were given the ground state instead.

Why is this supersymmetric? Take the direct sum of the two Hilbert spaces. A and  placed in blocks of a 2 by 2 in order to switch between the two pieces. If we call those two spaces, the fermionic and bosonic pieces then we just got the supersymmetry exchanging the two. The superalgebra here is SL(1,1)

This can be done in higher dimensions after passing to the effective 1 dimensional problem after including the angular momentum barrier pushing away from the origin.

Now we can get to the title. Why are certain potentials solvable?

This happens when you have the following condition on the partner potentials.



where the a's are parameters. If we have the above condition, you can construct a ground state for the first Hamiltonian and shape invariance guarantees that the spectra are just shifts of each other. You can build up the entire bound state spectrum from this by applying raising operators and bouncing back and forth between Hamiltonians.

Your favorite potentials that you have solved like harmonic oscillators and the hydrogen atom and their partners form shape invariant pairs. So they are exactly solvable as you knew because you solved them back in your misspent youth.
Shape invariance is a sufficient but not necessary condition for solvability.

For more:
http://arxiv.org/pdf/hep-th/9405029.pdf