Quantization of the Kalb-Ramond field In string theories, I have sometimes seen the condition:
$$\frac{1}{4\pi^2}\int_{\Sigma_2} B_2 \in (0,1)  .$$
This is might be obvious, but I don't see why this should be. In this paper (p. 17) they say:

The condition $\frac{1}{4\pi^2 \alpha^\prime}\int_{\Sigma_2} B_2 \in (0,1)$ comes also from the quantization of the string action, $\exp \left( \frac{i}{4\pi^2 \alpha^\prime}\int_{\Sigma_2} B_2 \right)$, as part of the string path integral. This is similar to what happens in quantum mechanics when coupling particles to a gauge field $A_\mu$.

But why should this imply that the integral of $B_2$ is restricted to $(0,1)$?
 A: The restriction of the flux of the Kalb ramond field to the interval $[0, 1)$ is a generalization of the Aharonov-Bohm effect to stringy or surface holonomy.
The Kalb-Ramond field is a type of a higher Abelian gauge connection. Its gauge transformation has the form:
$$\delta B = d\Lambda$$
where $\Lambda$ is a one form.
In the case of the electromagnetism, the gauge potential $A$, minimally couples to a particle moving on a manifold $M$ (it is assumed that $M$ is compact):
$$ S = \frac{q}{\hbar} \int_{\Gamma} A$$
Where $\Gamma$ is a one dimensional closed submanifold of $M$. Let $D$, be a two dimensional surface having $\Gamma$ as a boundary, then by the Stokes theorem:
$$ S = \frac{q}{\hbar} \int_{D} F$$
where, $F$ is the corresponding curvature two-form: $F = dA$.
The minimal coupling has two implications:


*

*When $H^2(M)$ is nonvanishing, then for any closed surface $D$
$$\frac{q}{\hbar} \int_{D} F \in \mathbb{Z}$$
This is the Dirac's quantization condition. The nontrivial field configurations are Dirac magnetic monopoles in this case.


This condition is explained very clearly by Orlando Alvarez in enter Topological quantization and cohomology.


*When $H^2(M) = 0$, but $H^1(M)$  is nonvanishing, the curvature of any gauge potential will be vanishing. In this case the connection is a flat connection. The prototype of this case is the Aharonov-Bohm effect.  


In the Aharonov-Bohm case $M = S^1$ and:
$$A = \frac{q \Phi}{2 \pi r \hbar} d\phi$$
By integration over S^1, the Aharonov-Bohm phase is:
$$\phi_{AB} = \frac{q \Phi}{ r \hbar}$$
Now the physics should not change if we perform a gauge transformation:
$$A\rightarrow A + d\phi$$
Please notice, that $\phi$ is not a single valued function on $S^1$, thus $d\phi$ is closed but not exact, thus we are performing a large gauge transformation on $A$.
This gauge transformation has no effect on the Aharonov-Bohm phase, since it adds $2\pi$ to the phase.
On the other hand, if we increase the flux in the solenoid  by $ \frac{2 \pi r \hbar}{q}$ we will add also exactly $2 \pi$ to the Aharonov-Bohm phase, which is the same effect as the gauge transformation. Thus the physically distinct fluxes lie in the interval:
$$\frac{q \Phi}{ 2 \pi r \hbar} \in [0, 1)$$
For a general manifold, we can take instead of $d\phi$ an integral one form $\omega_1 \in H^1(M, \mathbb{Z})$  and perform the gauge transformation:
$$A\rightarrow A + \omega_1$$
The case of the Kalb-Ramond field is analogous:


*

*It couples minimally to a string:
$$ S = \frac{1}{4 \pi^2\alpha^{\prime}} \int_{\Sigma} B$$
which can be written as:
$$ S = \frac{1}{4 \pi^2\alpha^{\prime}} \int_{D} H$$
Where, $H = dB$ and $D$ is now a three dimensional surface whose boundary is $\Sigma$.

*When $H^3(M)$ is nonvanishing, the Dirac's quantization condition implies that for a closed surface $D$:
$$ \frac{1}{4 \pi^2\alpha^{\prime}} \int_{D} H \in \mathbb{Z}$$
(This case is also described in detail in Alvarez's reference). The nontrivial $B$-fields in this case are types of generalized Dirac monopoles two-forms, explicitly constructed in some cases by Nepomechie and Teitelboim. 

*When $H^3(M) = 0$ and $H^2(M)$ is nonvanishing, the large gauge transformation:
$$B\rightarrow B + \omega_2$$
Where $\omega_2 \in H^2(M, \mathbb{Z})$ does not change the stringy Aharonov-Bohm phase (or surface holonomy):
$$ \frac{1}{4 \pi^2\alpha^{\prime}} \int_{\Sigma} B$$
(Here $\Sigma$ is a closed two dimensional surface)
Which implies, that the fluxes:
$$ \frac{1}{4 \pi\alpha^{\prime}} \int_{\Sigma} B \sim \frac{1}{4 \pi\alpha^{\prime}} \int_{\Sigma} B + 2 \pi $$
are equivalent.
This subject is discussed by Fuchs, Nikolaus, Schweigert and Waldorf
 (page 9).
