# For intersecting branes, are we allowed to compactify on a torus such that one of the branes becomes dense in it? What is the result?

The story of how to get chiral fermions in the low-energy effective theory of a string theory with intersecting branes goes something like this:

At the point of intersection of two branes, a direct evaluation of the allowed string states shows massless chiral (or anti-chiral, depending on the "direction" of the intersection) fermionic states arising which will then be captured by the low-energy effective QFT. A toy model of how compactifications then produce 4D chiral fermions is compactifying the $\mathbb{R}^{1,9}$ with 2 stacks of $D_6$-branes in it to $\mathbb{R}^{1,3}\times T^2\times T^2\times T^2$, where the brane worldvolumes fill all of $\mathbb{R}^{1,3}$ and intersect in each of the tori.

In this model, it is then that "obviously" the branes wrap multiple times around the torus (this is indeed obvious). It is then said that we can count these wrapping numbers in terms of the basic (homology) cycles on the torus, so each brane gets a wrapping number $(n,m)\in\mathbb{Z}^2$ on each torus. Computing the topological intersection number of the two branes on $T^2\times T^2\times T^2$ then counts the difference between chiral and anti-chiral modes of the theory, and this number is stable against smooth deformations of the branes on the tori.

Now, the statement that the branes get wrapping numbers is what irks me: Yes, I can write down cases where this happens. But there is another case, which, for randomly chosen radii/angles of the tori (or rather the $\mathbb{Z}^2\times\mathbb{Z}^2\times\mathbb{Z}^2$ lattice we quotient out of the $\mathbb{R}^6$ where the branes intersect), seems to happen far more often: If the brane divides one of the basis vectors in the lattice such that the two parts have an irrational ratio, the brane will not wrap around the torus a finite number of times and then return to its starting point, but instead never return to its starting point and only get arbitrarily close to it, and be dense in the torus. (This is a variant of the argument in classical Hamiltonian mechanics why orbits are either closed or dense in a certain subset of the phase space.)

Since the irrational numbers are dense in the reals (and the rationals merely countable, hence a zero measure set), this means that for almost all possible compactification radii/angles, we don't get branes that have a well-defined homology class because they are not closed, and there is no "topological intersection number". Usually, one would want the radii of such compactifications to be determined dynamically, i.e. the radii should be allowed to vary smoothly. But we cannot look at the effective theory for smoothly varying radii because we only know the effective theory for the cases where the branes are closed in the tori.

Does this mean that the radii must be chosen such that both branes become closed in the compacification, and are not allowed to vary, i.e. they cannot be dynamical themselves? Or is there an effective theory for the cases where the branes are not closed, and if yes, how does this theory arise?

• The branes aren't for free. They have a nonzero tension – mass/energy per unit volume. The addition of the "irrationally wrapped branes" is mass-wise equivalent to the additional of infinitely many parallel, normally finitely or rationally wrapped branes. So such a state has a higher energy density shifted by an infinite amount relatively to the rational or finite-brane case. You may say that it's not a problem - there may still be lots of SUSY and zero cosmological constant. But I am afraid the compactification - as bad as $U(\infty)$ on the branes - wouldn't be consistent, after all. – Luboš Motl May 31 '16 at 13:25
• You know, with the weak coupling branes, you are normally neglecting the gravitational backreaction of the branes which is OK because their gravitational field goes like $G_N\sim g_s^2$ but the tension is just $1/g_s$, so the Newton's constant wins and makes the backcreation small for $g_s\to 0$. But for any finite nonzero $g_s$, you may only have something like $N=1/g_s$ branes so that the backreaction is neglected. You effectively have infinitely many parallel branes within a string separation, so the backreaction can't be neglected, and you shouldn't trust the pert. ST in flat space. – Luboš Motl May 31 '16 at 13:29