BPS

Does not actually stand for brane preserving some supersymmetry, but that's what its doing

Remember the Reissner-Nordstrom metric around a charged black hole. It has two horizons at




Of course the case M

and



Doing a change of basis to this gives instead.





Of course the RHS should be positive. So this gives a bound banning M < Z.

To be continued with actual branes preserving supersymmetry.

Hopf Algebras

Take the diagrams of associativity and unital and flip all the arrows. These are below.

This is the comultiplication and counit of a coalgebra.

Let's put both an algebra and a coalgebra structure at the same time. And I'm going to make them get along. I will also provide an antihomomorphism S. If S^2=Id, then I would have a conjugation operation, but I'm not going to require that.

And this diagram commutes of course. It is just evil to draw non-commuting diagrams.

Let's use the example of the group algebra KG for some K and some G. The coproduct will be



The antipode map will be inversion of all group elements and the counit will send all group elements to 1 in K.

Let's check if this coalgebra structure behaves. We check it on a basis of KG namely g for all g in the group.

For the top path



The bottom path



The middle way



Success!

Another example: This one will generalize in a very rich manner.

Universal enveloping algebra of some Lie algebra.

Δ(x) = x ⊗ 1 + 1 ⊗ x for all x in the Lie algebra.
ε(x) = 0
S(x) = -x


These all respect commutators so they respect the ideal we mod out by in defining the universal enveloping algebra. Therefore they extend throughout the algebra. 

We are now ready to mess this structure by looking at the specific example of SL(2,C)

Start with M(2,C) it has a comultiplication given by



First of all remember these a's are actually functions on M(2,C) that read off the right coefficient. They are not complex numbers. That is why later we will be able to mess them up. This notation is a mishmash of tensor product and matrix multiplication but just replace multiplication by tensoring, but otherwise do matrix multiplication. Then you can just read off the coproducts of the a's which generate the regular functions on 2x2 matrices.

The counit is



Does this coproduct descend to SL(2,C)? Does the ideal generated by ad-bc-1 go both ways?





It is fine with comultiplication, it's also ok with counits. It's a coideal too.

That means we can use these coproducts and counits on SL(2,C)'s regular function space. We find a antipode and it is a Hopf algebra.

The antipode is given by S(f)(x)=f(x^-1). So the antipode of the function that reads the ij component is the function that reads off the ij component of the inverse matrix. This is why we concentrate on SL(2,C). There is a nice way to express the ij component of the inverse in terms of the original matrix components easily.

Well now we could suppose the a_ij's don't commute. Instead of commutativity, we put some different relations. (I am going to change my mind and call them a,b,c and d now)

ba=qab, db=qbd, ca=qac, dc=qcd, bc=cb, ad-da=(q-q^-1)bc

But we'll put the same comultiplication and counit. Again just like ad-bc, you just check that comultiplication and counit respect all the relations.

To get deformed SL(2,C) instead of deformed M(2,C) we see that the analog to ad-bc is now da-qbc. Apply comultiplication and counit and you'll see it is a coideal. It's also in the center of the algebra. It's okay to localize this quantum determinant to 1 and get to SL.

So we can get a Hopf algebra on the functions on any matrix group which are generated by the entry functions. Of course, this group may actually start off as a arbitrary topological group that was turned into a matrix group by some representation.

See http://en.wikipedia.org/wiki/Compact_quantum_group

Atiyah Singer strikes back

We will introduce another parameter t to help us out as a regulator.



This holds only for nonzero t, but we want the t=0 case so we will calculate at small t then take the limit as t goes to 0.



Time for some perturbation theory. Define V(t) to be V conjugated by the free Hamiltonian exponentiated.



Which of these terms will contribute? It will be the one with enough gamma matrices so that the trace does not vanish. Remember V has a so each power of V gives two gamma's. To get the 4 we need for nonvanishing trace we need two powers of V. That is the third term above. We need to take the trace of this:



What do about those propagators from x to x' etc.? Well we know that unperturbed hamiltonian pretty well. It's the free particle which we know.



Notice that when x and y are even a little bit separated the exponential becomes tiny. So we might as well take x' and x'' to be basically x. The corrections to those propagators will be exponentially suppressed.







Success. We have gotten from the index of that elliptic operator to integrating the second Chern class. This is just a chiral anomaly. Indexes could screw our theory in other ways too. A gauge or gravitational anomaly would be more evil.

ch is the Chern character and Td is the Todd, but you can get them both from chern classes. This is a big gun., but no kill like overkill.

Partitions

A partition splits up a natural number into a sum of nonincreasing positive naturals. In the modular forms post we discussed the Dedekind function.

Of course that is an infinite product that is hard to compute. It is hard to read off the coefficients especially for large n. Imagine asking specifically how many different partitions there are for 1000. It blows up pretty fast.



The above gives a pretty good approximation to its growth. For 1000 it is 1.4% off.

Still that's an estimate and exactly how good it is is questionable. There are also some infinite convergent sums, but that again has the same issues as infinite products.

In January, Ono said

"I can take any number, plug it into P, and instantly calculate the partitions of that number. P does not return gruesome numbers with infinitely many decimal places. It’s the finite, algebraic formula that we have all been looking for."

Apparently it is done with a "self similarity" of some specific subsequences. I'll just refer to the paper, but it is pretty clearly epic.

Edit: November 10, 2011

He gave a colloquium today. He explained the connection with Maass forms and mock theta functions.
http://en.wikipedia.org/wiki/Mock_modular_form

Atiyah Singer

Let's start from the beginning.

Take an order N operator which in local coordinates is given as follows.



M is a multi-index which indicates which coordinates to differentiate again. For example if M=(1,2,0,0,1) which is of order N would correspond to the operator



Now we might as well take the term with the most derivatives. After all that is the most relevant operator by dimensional analysis. Now Fourier transform to get the symbol.

For the Laplacian it is



This is only 0 at the origin. This is the definition of elliptic. We have the operator, we can also take the adjoint because the fields live in spaces with inner products. For example, fermions live in the spinor bundle and there is an inner product on spinors. We can ask that the kernel of the operator and its adjoint are finite dimensional. This operator is Fredholm and it is made of win.

Now what is the difference between the kernel's size and cokernel's size. This measures the mismatch between zero modes. This is where we see the notion of the concept of invariance of measure.



Here a and b's are the coefficients of the eigenfunction expansion. The indices that correspond to nonzero eigenvalues match up perfectly.



Of course we see a pairing between n and n' if and only if the eigenvalue is nonzero. Otherwise we don't know how much they mismatch. We also notice that here n and n' correspond to different chiralities because there are an odd number of gamma matrices in the operation from one to another.

We want the difference between left and right chiralities. How do we calculate this? Next time on If you haven't found something strange.

AdS/Awesome

The title is taken from a Sean Carroll tweet

We define a space with p spatial directions and q time directions. Think of a space with signature (p,q+1) $ds^2 = \sum_{i=1}^p dx_i^2 - \sum_{j=1}^{q+1} dy_j^2$ and then take a pseudosphere inside given by $\sum_{i=1}^p x_i^2 - \sum{j=1}^{q+1} y_j^2 = -L^2$. Changing this minus sign will change it to (p-1,q+1) deSitter space instead. (The hyperboloid of one sheet vs two sheets.)

If you choose q=0, this gives a model for hyperbolic p space. To make it more reasonable to handle, we might as well do a Weyl transformation to bring it to a ball model or an upper half space model, but nonetheless it is still a pseudosphere in Minkowski p+1 space.

The restriction to the pseudosphere gets rid of one of the timelike directions and makes the metric of signature (p,q). So in addition to the hyperbolic geometry of (p,0) we like (p,1) as well. We like time, we could use more time directions.

Do some conformal transformations and get the metric to a form as follows

$ds^2 = \frac{1}{y^2} (-dt^2+dy^2+\sum dx_i^2)$ where y is a radial coordinate and y=0 corresponds to the conformal boundary that you never can quite get to.

We can see that if we take constant time slices we get hyperbolic spaces in the upper half space coordinate system. We also see that as you go to the conformal boundary it looks like Minkowski p space. So if we use AdS (4,1) we get our familiar spacetime at the boundary. We could put a quantum field theory on this boundary like $\mathcal{N} =4$ Super Yang-Mills or any other theory we like. Alternatively we could put a quantum gravity theory inside the AdS. It'll mess up the AdS geometry a little but not too much on the boundary. After all the conformal boundary is really far away.

This sets up a correspondence between relevant operators in the QFT on the left and the quantum gravity theory on the right. The right hand side is not the full partition function it is the partition function where we only count fields that match the boundary condition at the boundary. Of course to take care of the fact that this is a conformal boundary not a real boundary, it is regulated by cutting of at $y=\epsilon$ and extrapolating to $y=0$

In the QFT picture the y coordinate is the renormalization scale. That is why CFT's are nicer in this correspondence. Renormalization is easy if you are at a CFT.

Modular Group

The modular group is PSL(2,Z) which naturally acts on the upper half plane/the Poincare unit disk.

The standard generators are S and T. S sends z to -1/z and T sends z to z+1. The relations are S^2=I and (ST)^3=I. Calling U=ST we see it is generated by S and U which have no relations between them. Therefore the group is the free product of S's and U's.
Now for something totally different. Braids on 3 strands. Define a=s1*s2*s1 and b=s1*s2 where si is take the ith strand over the i+1st. Using the braid relations we get a^2=b^3. This element is in the center. Just try taking the commutator with s1 and s2. Quotienting by this cyclic subgroup gives a group with a and b with relations a^2=I and b^3=I. Precisely the same one we had before. We also see there is nothing else in the center. So B3 mod center = PSL(2,Z).

The PSL(2,R) can be easily identified by looking at the defining representation of these groups and then centrally extending both top and bottom.

So we know it acts on points of the upper half plane. Therefore it can act on functions on that space too. Just replace all the z's with gz for some group element g.

Some of these functions are particularly recurrent like the eta function which rears it head all the gorram time. Raising it to the 24th power (for the 24 transverse directions to the bosonic string) and calling that delta gives.



Replace that 12 with arbitrary k and you get general modular forms. If you chose k=0 you get nothing but the constants. There aren't any other functions that are invariant under the group. You have to pay the cost of that (cz+d) factor, but at least it's not that costly.

Some k's have modular forms some don't. They must be even natural numbers and at least 4. In fact even cooler is the fact that you only need two modular forms and you can get the others by polynomials. These two are the first two of the Eisenstein series.

You get a lot of identities with these See here

The key way to work with these functions is to Fourier transform from the tau variable to q



This turns them into Taylor series in q which converge because tau being in the upper half plane means q will be inside the unit disk. The numbers that appear as coefficients are spectacular.

The ones in the Eisenstein series give the summatory function which adds up powers of the divisors of n.



The ones for the inverse eta function after you get rid of that 24th root of q is the partition function. The partition function is cool. Partitions label representations of SU(N) and the symmetric group. So this function counts representations with a fixed number of boxes in the Young Tableaux.

Classifying Spaces

Start off with your base space M and gauge group G. You will get a principal bundle P that is the total space for this.

The question is how to classify the possible P's up to isomorphism. Let's pick another G bundle EG over BG. It will have the property that every G bundle is determined by the homotopy class of a map p from M to BG. Then the bundle over M becomes the pullback of this model bundle.

Why should this special bundle even exist? We have a contravariant functor from the homotopy category to set which takes my space X to all G-bundles on it up to isomorphism. The fact that we pull back bundles not push them indicates the contravariance. The question then becomes is this functor representable by some object in the homotopy category. This is the object that we will call BG. Or at least it is one of the possible choices all of which are equally good.

For a moment let us specialize to U(1) gauge theories. We will then be looking at line bundles. What is BU(1)? Let's use the fact that the infinite sphere is contractible. To see this think about the sphere as the normalized wavefunctions on the interval then push everything to the end so that eventually get the constant function. Now mod out by the U(1) of phases. This gives BU(1)=normalized pure states and EU(1)=all normalized states. Since EU(1) is contractible, this is actually a good classifying space that has the universal property/represents the functor. This is because contractibility of the total space means that the loop space of U(1) and BU(1) are homotopy equivalent via the long exact sequence having a whole bunch of zeroes. This allows us to see isomorphisms between each homotopy group at all n. Now we need to figure out how to classify maps from M to BU(1)=pure states. This is where we need to know about spectra for generalized cohomology.

Define K(Z,0)=Z and K(Z,n)=BK(Z,n-1). This gives K(Z,1)=BZ a space which has Z as its space of loops. Well that's U(1) because we can homotope any loop to a standard form z^n by gradually getting rid of those lower terms. Taking the next step is K(Z,2)=BU(1)=pure states. So to classify U(1) bundles we need the classes of maps from X to K(Z,2) but K(Z,2) is in the spectrum for plain old cohomology and so [X to K(Z,2)] is the second integral cohomology. This is clearly the easiest so far to calculate and is equivalent. We have our way to classify all U(1) bundles. Just compute the second integral cohomology. There will be a bundle associated with each class. This is the chern class of the line bundle. We have lot's of ways of seeing that chern classes classify line bundles. We could also use the long exact sequence in sheaf cohomology associated to transition maps or connections and curvature.

All line bundles have some chern class and the universal one EU(1) over BU(1) is no different so we have to go from the class c in the cohomology of BU(1) to the class in the cohomology of M. But remember we have that map p from X to BU(1), and this is co so we can just pull back c to M and changing p to another homotopic one will not make a bit of difference.

Let's give some more examples of the classifying spaces for some other gauge groups.

For U(n) start with the total space of orthonormal n frames sitting inside countable Hilbert space.

Modding out by U(n) makes all the n frames in the same n plane the same. Therefore BU(n) is the set of n-planes sitting inside Hilbert space. This checks out with n=1 too because we can associate to each pure state the line in Hilbert space that when properly normalized and phased out would give this state.