The String Coffee Table

Nonabelian B-field equations of motion

Posted by urs

[Update 07/15/04: The issue discussed below can now be found in hep-th/0407122.]

In the previous entry I discussed the boundary state which should describe superstrings in the background of a nonabelian 2-form field. Consistency requires that $F_A+B=0$ and hence the boundary state reads

i.e. it is obtained from the boundary state $\mid \text{D9}\rangle$ of the bare D9 brane by acting on it with the untraced Wilson loop of the nonabelian 1-form gauge field $A$.

This is essentially (up to the $B$-contribution which follows from hep-th/0401175) a straightforward generalization of the technique described in

Koji Hashimoto: Generalized supersymmetric boundary state (2000)

for nonabelian $A$, nonvanishing $B$ field and the special case $F_A+B=0$. What I wrote previously serves to demonstrate that this boundary state has indeed the intended physical interpretation by showing that pulling the worldsheet supercurrent through the above Wilson line does indeed deform the exterior derivative in loop space in a way expected from 2-form gauge theory.

The next logical step is to check for the equations of motion of the background fields $A$ and $B$. In the above paper Koji Hashimoto demonstrated - for the case of abelian $A$ and $B$ = 0 - that these can be read off from the divergeces of the deformed boundary state itself.

Here I want to check for these divergences for the case of the nonabelian deformation given above. It turns out that in the nonabelian case this is actually much simpler than in the abelian case. For nonabelian $A$ we can, without loosing information, restrict attention to constant gauge fields, ${\partial }_{\mu }A=0$, for convenience. This removes a lot of worldsheet fields whose contractions would mess up the calculation.

Expanding the above $\mid \text{D9}(A,B)\rangle$ in $A$ yields

Divergences in these expressions appear as follows:

The bare boundary state $\mid \text{D9}\rangle$ is a Fock state with lots of worldsheet oscillators acting on the worldsheet vacuum and designed in such a way that

On this we are acting with $X^{\prime }$ which has the oscillator expansion

By using the above boundary condition on the oscillators annihilators can be turned into creators when applied to the boundary state, so that

So when acting on this again with $X^{\prime \nu }(\kappa )$ there is a divergence from pulling the annihilators at $\kappa$ through the creators at $\sigma$:

The sum over cosines is what has to be inserted into the above path-ordered integral in prder to compute the divergences. Obviously the first two terms have trivially no divergence and in the third they also obviously vanish. So the first nontrivial contribution comes at order $𝒪(A^3)$.

I have taken the liberty to simply feed the relevant mixed ordered integrals over these cosines into Mathematica. This yields for the term at $𝒪(A^3)$ a contraction proportional to

So the divergences vanish iff

Assuming that this result is independent of our choice to work with constant $A$ this means that

i.e. that $A$ satisfies the equations of motion of nonabelian YM theory, up to higher order corrections.

This does certainly not look surprising, even though I think the derivation is pretty nice (assuming that I didn’t screw up somewhere). But I think, since this is in the context of nonabelian 2-form gauge theory, this result is not uninteresting.

To recall, what it really says (all assuming that I am not confused) is that the background field equations for strings in a nonablian 1-form background $A$ and a nonabelian 2-form background $B$ are

and

This is curious from the point of view of 2-group gauge theory. The reason is that these equations are not invariant under the 2nd order gauge transformation

But I think the discussion from the previous entry clarifies why this second transformation cannot be expected to be true in the boundary state formalism.

Namely there I had argued that the above gauge transformation is really one of the closed string, and it is still in the nonabelian case. But the closed string does not really couple to the nonabelian $A$ and $B$-fields, so that’s not of much physical importance. What does couple to these backgrounds is the open string. And here it matters which value $A$ has and so the above shift $A\mapsto A+\Lambda$ should not be a symmetry.

Or so I think. Comments are welcome.

Posted at July 13, 2004 12:23 PM UTC