If you haven't found something strange: February 2013

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.