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