This question concerns the boundary conditions that A-branes solve.

Consider the following problem: Suppose that an A-brane wraps a submanifold $Y$ of $X$. Let $\mathcal{L} \rightarrow Y$ be a rank one Chan-Paton bundle over $Y$ equipped with a unitary connection. What are the allowed boundary conditions on the endpoints of an open string with embeding functions $\Phi$, worldsheet fermions $\psi$ and whose endpoints lie on $Y$?.

According to the paper "Remarks on A-branes, Mirror Symmetry, and the Fukaya category" (equation 4, page 8) the answer is that at $z=\bar{z}$ (if the open string worldsheet is the upper-half complex plane), $\Phi$ and the fermions $\psi, \bar{\psi}$ should obey $$\partial_{z}\Phi = R[\partial_{\bar{z}}\Phi] \ \ \ , \ \ \ \psi= R[\bar{\psi}].$$ Where the operator $R$ is defined as $$R=(-1)_{NY}\oplus (g-F)^{-1}(g+F).$$

and a local decomposition of $TX = NY \oplus TY$ is assumed, g is the restriction of the metric of $X$ to $Y$, and $F$ is the curvature tensor of the unitary connection on $\mathcal{L} \rightarrow Y$.

Question: How can I derive that answer?


I noticed that in fact my question was straightforward, and not specific to topological string theory. It solves the general problem of an open string (with the worldsheet topology of a disk) with endpoints on a stack of branes.

Sketch of solution:

Impose Dirichlet boundary conditions for $\Phi$ on $NY$. The computation is developed in the texbook "String theory on a nutshell" (equation 4.16.3); up to a constant it reads $$(\partial-\bar{\partial})\Phi=0,$$ on $z=\bar{z}.$

The correct boundary conditions for the directions $TY$ are $$[g(\partial-\bar{\partial})-F(\partial+\bar{\partial})]\Phi=0.$$

That expression can be found in String Theory and Noncommutative Geometry (equations 2.2 and 2.3); or explicitly worked on the textbook "A first course in string theory" ( second edition, chapter 16).

Then the operator that impose the correct boundary conditions on the whole $TX$ is $$ [ (\partial-\bar{\partial}) \oplus (g(\partial-\bar{\partial})-F(\partial+\bar{\partial})) ] \Phi=0,$$

the required solution.


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