\wedge d\psi\end{aligned}$$

where we have introduced $dx_{\parallel}^2 = -dx_0^2 + dx_1^2
+ dx_2^2 + dx_3^2 + dx_4^2 + dx_5^2 + dx_6^2$ and $dx_\perp^2 = dx_7^2 + dx_8^2 + dx_9^2$, along with $V = 1 + \frac{1}{2r}$, $r =x_7^2 + x_8^2 + x_9^2$, $\theta = \cos^{-1} \left( \frac{-x_9}{r} \right)$, and $\psi = \tan^{-1}\left( \frac{x_8}{x_7} \right)$. Here, to simplify the formulas we have set the radius $R_{10}$ of the M-theory circle to one.

If the D2-brane is extended in two directions (say, $x_7,x_8$) orthogonal to the D6-brane, then it will capture half of the flux from the D6-brane and must therefore have half of a fundamental string ending on the D2. In [@half], this question was studied using the method of [@NOYY]; i.e., by considering test M2-branes in the Kaluza-Klein monopole background. There certainly do exist configurations in which the D2-brane is extended orthogonally to the D6-brane, and for these cases the issue is merely one of proper accounting. In particular, while there is indeed 1/2 unit of fundamental string ‘brane source’ charge in this system (in particular, this charge can be shown to flow along the D2-brane world-volume to infinity), brane-source charge is not in general quantized (see [@3Q]). Instead, the 