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I am reading this paper and on pp.19-20 it states the following relation between large gauge transformation and intersection form: for the action on a 4-manifold $M^4$

$$S[A,B] = \int_{M^4}{\sum_{I=1}^s{\frac{N_I}{2\pi} B^I \wedge dA^I} + \sum_{I,J=1}^s{\frac{p_{IJ} N_I N_J}{4\pi N_{IJ}} B^I \wedge B^J}}$$

where $A^I$ and $B^I$ are 1- and 2-form fields respectively and $N_{IJ}$ is the greatest common divisor gcd$(N_I,N_J)$, the gauge transformation reads

$$ A^I \to A^I + dg^I - \sum_J{\frac{p_{IJ}N_J\eta^J}{N_{IJ}}} \qquad B^I \to B^I + d\eta^I $$

If the diagonal elements $p_{II}$ and the integer $N_I$ are odd, $e^{iS}$ is invariant under large gauge transformations only if $M^4$ has even intersection form.

I don't see how the large gauge transformation invariance requires the intersection form $H^{2}(M^4;\mathbb{Z}) \times H^{2}(M^4;\mathbb{Z}) \to \mathbb{Z}$ to be even. Could somebody help to clarify?

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Take $s=1$ for simplicity. Then this is a topological field theory of a 1-form gauge field $A$ and a 2-form gauge field $B$ coupled via

$$S = \frac{n}{2\pi} \int_M B \wedge \mathrm{d}A +\frac{pn}{4\pi} \int_M B \wedge B,$$

where $M$ is a closed 4-manifold.

Clearly the theory is invariant under ordinary gauge transformations of $A$,

$$A \to A + \mathrm{d}\lambda,$$

where $[\mathrm{d}\lambda]/2\pi \in H^1(M,\mathbb{Z})$. When $[\mathrm{d}\lambda]$ is trivial in cohomology (i.e. $\lambda$ is actually a globally defined function), call this a small gauge transformation; when it is non-trivial, call it a large gauge transformation. That is, $$\oint_\Sigma \frac{\mathrm{d}\lambda}{2\pi} \in \mathbb{Z}$$ is zero for a small gauge transformation and non-zero for a large gauge transformation.

The theory is also invariant under "1-form" gauge transformations

$$B \to B + \mathrm{d} \eta\\ A \to A - p \eta,$$

where $\eta$ is a 1-form gauge field (and $p$ must therefore be an integer so that $A-p \eta$ is still a gauge field). Under this transformation the action is deformed by

$$\delta S = \frac{n}{2\pi} \int_M \mathrm{d}\eta \wedge \mathrm{d} A + \frac{pn}{4\pi} \int_M \mathrm{d}\eta \wedge \mathrm{d} \eta,$$

which may be written more suggestively as

$$\delta S = 2\pi n \int_M \frac{\mathrm{d}\eta}{2\pi} \wedge \frac{\mathrm{d} A}{2\pi} + \pi pn \int_M \frac{\mathrm{d}\eta}{2\pi} \wedge \frac{\mathrm{d} \eta}{2\pi}.$$

The integrals are $\mathbb{Z}$-valued, since $A$ and $\eta$ are gauge fields. For small gauge transformations, $\eta$ is globally defined, these integrals are zero, and the action is invariant. Including large gauge transformations, the first term will leave the path integral weight $e^{iS}$ invariant provided $n\in \mathbb{Z}$. For generic $M$ and $n$, the second term leaves the path integral invariant provided $p \in 2 \mathbb{Z}$. However, if $n$ is even, then $p$ may be any integer. Likewise, if it happens that the intersection form on $M$ is even (which will be the case if $M$ is spin),

$$\int_M \frac{\mathrm{d}\eta}{2\pi} \wedge \frac{\mathrm{d} \eta}{2\pi}\in 2\mathbb{Z},$$

then $p$ may again be any integer.

See Kapustin, Seiberg Coupling a QFT to a TQFT and Duality section 6 for more details, (and the rest of the paper for a nice discussion of these kinds of $BF$ theories).

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  • $\begingroup$ Woo~What a clear exposition! Thank you so much! I almost gave up hope after receiving no response for 2 days : ) $\endgroup$ – PhysicsMath Mar 12 '18 at 10:23

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