If you haven't found something strange

Sunday, July 14, 2013

2 dimensional (T)(C)FT choose at least 1

Theo, Zach and Dan B-E at tea: February 2013

What does a TFT do to every topological surface?

It gives an invariant. Let us say that invariant lives in V.

What does a CFT do to every topological surface?

It gives a function on the moduli space of Riemann surfaces with that genus. The function will take value in some V.

What about a TCFT?

Well now we want a locally constant function on the moduli. In this way a lot of the dependence on conformal structure is lost. But large transformations, namely the mapping class group, can still do stuff. Another way to say that is it is a closed degree 0 form. But V might have a Q on it making it differential graded vector space.
So what we are really looking at is $(\Omega^\bullet , d ) \otimes (V,Q)$

There are maps to TFT and CFT.

If you have a choice of volume form on the moduli stack you can integrate the CFT and TCFT results. This is what you do in perturbative string theory where you need to integrate out all the choices you made about conformal structure on the worldsheet.

Koszul Duality

I have a f.d. semisimple Lie algebra L. What is it's structure? I could give you a basis and structure constants and then you could check that that works.

$[x_i , x_j ] = c_{ij}^k x_k$

Or I could give you this chain complex where the differential is as follows

$0 \to k \to \mathfrak{l^*} \to \wedge^2 \mathfrak{l^*} \to \cdots$

$d (x_i^*) = c_i^{jk} x_j^* \wedge x_k^*$

This chain complex is generated in degree 1 so that is all I had to give you. If you compute

, you get Jacobi Identity. This is Chevalley-Eilenberg.

So the data of a Lie algebra is the same as a semifree differential graded-commutative algebra generated in degree 1.

Aside:
You can change 1 to 1-n to get a Lie n-algebra, but then you would need to specify the higher brackets because of the other generators. If you replace it with just n, you get a n-Lie algebra. Removing all restrictions on generators whatsover lands you in $L_\infty$ .

Now let's flip the arrows and go the other way.

I have a free graded Lie algebra structure on something called SC^*. If you see the condition for it to be a complex, you see it forces C to be a commutative algebra.

Again we could play with where it is generated to get the rest of the E_n type structures instead of just Comm.

Saturday, April 27, 2013

AGT back in 1996

http://arxiv.org/pdf/hep-th/9605193.pdf

Tuesday, March 12, 2013

Atiyah Singer III: The Witten Genus

$\begin{matrix} K(X) & \rightrightarrows & K(pt)\\ \downarrow & & \downarrow\\ H_{dR}^{ev}(X) & \rightrightarrows & H_{dR}^{ev}(pt) \end{matrix}$

The two maps on top are analytic and topological shriek maps.

If you mix up which of the analytic or topological maps you do, you get the index theorem in all it's forms.

Replace K theory with TMF and the A-hat with the Witten genus should give you a beefed up version.
I don't know which of the maps have been constructed, but this is where a punt to Stolz-Teichner and crew goes.

The idea is we know the Spectra which represent K theory in all it's flavors, BU and all that. If that TMF is also a generalized cohomology theory, what is it's spectrum. Insert the suspected one here.

credit talks given by Dan Berwick-Evans

Sunday, February 24, 2013

Doplicher Roberts

The statement in Chari and Pressley Guide to Quantum Groups

Associated to every local quantum field theory in four or more spacetime dimensions is a compact group G whose representation ring is isomorphic to the fusion ring of the theory.

Also the appendix of
http://arxiv.org/pdf/math-ph/0602036v1.pdf

Need to unpack this into a post.

Wednesday, February 6, 2013

Ishibashi (as in without the k)

Consider the annulus.

You can view it as a segment rotated around to meet up with itself. Supposing boundary condition a and b on the outside and inside, what you are doing is calculating the effect of two boundary changing operators at 0 and at infinity.

If you calculate the partition function you get some linear combination of characters over all levels h showing up with some coefficients n.

$\sum_h n^{ab}_h \chi_h (q)$

where $q=e^{-\pi \delta}$ is the

But you could also thing of this same situation as a closed loop moving from the a side to the b side in imaginary time $\delta^{-1}$ . In this picture we need to calculate the matrix element $\langle a \mid e^{-H/\delta} \mid b \rangle$

But what a and b make sense to put there? We have the condition $L_n = \widetilde{L}_{-n}$ in the weak sense on this space.
Look at n=0, this says that the state lives in $\bigoplus V_h \otimes \tilde{V}_h$ with no cross terms.

Keep looking at the others and you see an explicit expression

$\mid h \rangle \rangle = \sum_{N=0}^\infty \sum_{j=1}^{d_h(N)} \mid h,N,j \rangle \otimes \overline{\mid h,N,j \rangle}$

equal combinations over everything allowed with that h.

a and b have to be a linear combination of these.

$\mid a \rangle = \sum \langle \langle h \mid a \rangle \mid h \rangle \rangle$

Now let's get back to that matrix element.

$\langle \langle h' \mid b \rangle \langle a \mid h'' \rangle \rangle \delta_{h',h''} \chi_{h'} (e^{-4\pi/\delta})$

That last two pieces is the matrix element between the Ishibashi states. But we see the argument of the character is the victim of an S transformation, so we can rewrite it in terms of the character we had in the first picture.

$n_{ab}^h = \sum_h S_{h'}^h \langle \langle h' \mid b \rangle \langle a \mid h' \rangle \rangle$

These are called the Cardy conditions. It is an positive integer condition so it is restrictive. The operator content at the boundary is constrained to fit this.

Tuesday, February 5, 2013

nKdV Hierarchies

I haven't had a new post in a while, so let's finish up some of the drafts.

http://math.berkeley.edu/~ilya/software2/tmp/Dickey-classical-W.pdf

Write an nth order differential operator L.

$L=\partial^n + u_{n-2}\partial^{n-2}+u_{n-3}\partial^{n-3}+\cdots+u_0$

Consider the algebra generated by the u symbols and their derivatives A.

Everything in A can be differentiated. Take the quotient by the subspace of closed elements.

Also consider a ring of Laurent polynomials in the derivative symbol but the other way around. Meaning they can go down as far as they need but have to cut out at the top.
This ring splits up into the positive part and negative part.

You can also take residues in the same manner with taking the term that looks like $\partial^{-1}$ .

Because the top coefficient of L is 1 we can actually take it to a fractional power m/n. $L^{m/n}$ which commutes with L.

But let's mess that up by a violent projection to the positive part. Now we have a n-2 order differential operator which we can call the time derivative of L which describes how the u's are changing. Since the first u that shows up is at the right place, we are OK.

If you take different fractions, you see lots of commuting times, so we get lots of conserved quantities which are written via residues of all those $L^{m/n}$ for all m.

Let n=2. Then we only have one u and this becomes KdV classic.

Wednesday, December 12, 2012

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

$d &=& \frac{1}{1-y} a + b^\dagger\\ \bar{d} &=& a^\dagger + \frac{y}{1-y} b\\$

respectively.

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

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

Sunday, December 9, 2012

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

Monday, November 19, 2012

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 $\mathbb{R}^4$ with metric

. Look at the self-dual Yang-Mills equations on it. In these coordinates they read:

$\big[D_x+D_y,D_u\big]&=&0\\ \big[D_x-D_y,D_x+D_y\big]+\big[D_u,D_t\big]&=&0\\ \big[D_x-D_y,D_t\big]&=&0\\$

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.

$(\partial_x - 2A + \lambda B )s=0\\ (\partial_t-C+\lambda \partial_x)s=0\\$

We have put three equations into 2 by inserting a $\lambda$ 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.

$2A &=& \bigl(\begin{smallmatrix} 0 & \psi\\ -\tilde{\psi} & 0 \end{smallmatrix}\bigr)\\ 2\kappa C &=& \bigl(\begin{smallmatrix} \psi\tilde{\psi} & \psi_x\\ -\tilde{\psi}_x & -\psi\tilde{\psi} \end{smallmatrix}\bigr)\\$

and

$2\kappa\psi_t = \psi_{xx}+2\psi^2\tilde{\psi}\\ 2\kappa\tilde{\psi}_t = -\tilde{\psi}_{xx}-2\tilde{\psi}^2\psi\\ 2\kappa=1\; \text{or} \; -i$

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.

Thursday, October 25, 2012

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.

Sunday, October 21, 2012

Youtube videos better than NYAN cat

Cool people saying cool things

Friday, October 19, 2012

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.

$\lambda \wedge d\lambda > 0\\ d\lambda (R,-)=0\\ \lambda (R) =1\\$

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

$\sum a_i \alpha_i\\ \alpha_i = {(R_1,m_1),\cdots (R_n,m_n)}$

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

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.

Tuesday, September 11, 2012

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

Friday, August 24, 2012

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.

Thursday, August 9, 2012

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.

Thursday, August 2, 2012

(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.