Sunday, October 21, 2012
Friday, October 19, 2012
Contact Homology
GRASP usually gives me things I can post about, and today's talk was embedded contact homology.
where R lists all the orbits and m's are the multiplicities in
.
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.
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.
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
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.
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.
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.
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.
Wednesday, July 4, 2012
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