### Not in Kyoto

Scheduling prevents me from being at the Strings Conference in Kyoto this week. But I can live vicariously over the web.

Here are some of the talks that struck me.

#### Strominger

Andy Strominger’s been thinking about spacelike branes. One version of the story is to introduce a boundary interaction on the string worldsheet,

Following Sen, one notes that this is the analytic continuation of Boundary Sine-Gordon theory, which is exactly soluble. For $\lambda=\pm 1/2$, the theory is particularly interesting, as the boundary interaction is equivalent to setting Dirichlet boundary conditions for $x^0$ (an “SD-brane”).

The boundary state for $x^0$ is

For $\lambda\to\pm 1/2$, this has the limit

For $\lambda\to - 1/2$, this is indeed sets $x^0=0$ at the boundary. But the physics is a little awkward. It corresponds to a rolling tachyon which rolls up from $- \infty$ towards the local maximum of the tachyon potential and then rolls back down. This has no analogue in the superstring. $\lambda=+1/2$ seems to be even more boring. It looks like the boundary state vanishes (for real $x^0$), and Sen has interpreted this as the closed string vacuum.

“Not so!” says Strominger. The closed string field configuration which is sourced by the brane satisfies

Clearly, $|\phi\rangle$ is *a* solution, but it is not, according to Strominger, the one dictated by the usual rules of string perturbation theory. Rather, one should take the analytic continuation from Euclidean signature. Equivalently, one should solve

using the Feynman propagator and analytically-continue to imaginary $t_i$. This gives

This, indeed, vanishes near the origin, but is nonetheless nontrivial.

This is a really interesting subtlety. For static branes, finding the closed string field configuration sourced by the brane involves solving an elliptic equation, and demanding that the solution die off at infinity gives a unique solution. In the present situation, one has a hyperbolic equation to solve, and there is some extra physical input that is required. The proposal for SD branes is very nice, but I wonder what the general prescription might be like.

#### Bousso

Raphael Bousso talked about proving Bekenstein’s original Entropy Bound for weakly-gravitating systems. The Bound states that the entropy of a system of total mass, $M$, contained in as sphere of radius $R$ is bounded by

Since $M\ll M_{bh} = R/2G$, the mass of a black hole whose Schwarzschild radius is $R$, this is much more stringent than the Holographic Bound

of 't Hooft and Susskind.

To prove Bekenstein’s bound, Raphael invokes a version of the Covariant Entropy Bound due to Flanagan, Marolf and Wald. Let $A_1$ and $A_2$ be spacelike hypersurfaces, and $L$ be a lightsheet whose boundary is $A_1-A_2$, which is non-expanding with respect to $A_1$ and expanding with respect to $A_2$. Then the entropy crossing $L$ is bounded by

What’s cool about this version of the Covariant Entropy Bound is that the surfaces $A_i$ don’t need to be closed. Indeed, Raphael takes them to be disks, such that the lightsheet emanating from $A_1$ is just *barely* nonexpanding. The lightrays passing through our weakly-gravitating system are focussed, so $A_2$ is a slightly smaller disk than $A_1$. The radius of $A_1$ is roughly $\pi R^2$. Because of the focussing, $A_2$ is smaller by a fraction $2 M/R$. So, sloughing over all the technical details, one obtains something like Bekenstein’s bound.

Theoretically, the bound you obtain this way *could* actually be tighter than Bekenstein’s, but it’s not clear that, in practice, a self-gravitating system can give a tighter bound.

#### Aganagic

Mina Aganagic gave a real tour de force talk about recent work on the open topological A-Model. In the A-Model, one has a Calabi-Yau manifold $X$, and a Riemann surface $\Sigma$ with boundaries. One is interested in counting holomorphic maps from $\Sigma$ into $X$ such that the boundaries of $\Sigma$ get mapped into Special Lagrangian submanifolds of $X$. In the case of $X$ a local Calabi-Yau, with special Lagrangian $T^2\times\mathbb{R}$ fibers over $\mathbb{R}^3$, these guys have an incredibly elegant solution which, alas, defies my humble attempts at a simple summary.

More next time …