In a physics paper entitled "Five-Dimensional Supersymmetric Gauge Theories and Degenerations of Calabi-Yau Spaces" by Intriligator, Morrison and Seiberg, the authors write an equation

$$2\pi \frac{c_{ijk}}{6}\int_{Y}\left(\frac{F^i}{2\pi}\right)\wedge\left(\frac{F^j}{2\pi}\right)\wedge\left(\frac{F^k}{2\pi}\right) \equiv 2\pi i \frac{c_{ijk}}{6} c_1(L^i)c_1(L^j)c_1(L^k)$$

(This is equation 2.7 on page 5 of the paper.)

This is the extension of the Chern Simons term on a 5-manifold $W$ which is the boundary of a 6-manifold $Y$ (so $W = \partial Y$).

How does one arrive at this equality? The left-hand side is the integral of a 6-form, and the right-hand side has a product of three first Chern classes. What are the underlying mathematical theorems being used here?

When can the triple intersection number be written as a product of Chern classes over (complex) line bundles in this form?

EDIT: Following some statements on page 8 of a related paper by Witten entitled "Phase Transitions in M- and F- Theory", I can say that $L^i$'s are complex line bundles over the oriented six-manifold $Y$. But the equality above is still unclear to me.


I don't think there's anything underhanded going on here. The Chern class of a line bundle $L$ is represented by the form $\dfrac{\sqrt{-1}}{2\pi} F^L$. Wedge product of closed forms corresponds to cup product of the corresponding cohomology classes. So you need to interpret the right-hand side as a product in cohomology of $Y$ (rel boundary). Algebro-geometrically, you can think of the Chern classes as divisors (complex codimension-1 submanifolds) and then intersect the three. I don't know much about this theory, but I assume that these are going to be compact and away from $\partial Y$.

You should let me know if I've missed a subtlety.

  • $\begingroup$ Thank you @Ted! I'm somewhat familiar with the first and second Chern class being the integrands $Tr(F/2\pi i)$ and $Tr((F/2\pi i)\wedge(F/2\pi i))$, so equating the integral on the left to the product of first Chern classes (unintegrated) seemed a bit of a mystery due to my lack of knowledge about the underlying math. $\endgroup$ – leastaction Oct 17 '17 at 6:18
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    $\begingroup$ To belabor, if $\alpha$, $\beta$, $\gamma \in H^2(Y,\Bbb Z)$ are Poincaré dual to $\sigma, \tau, \eta\in H_4(Y,\Bbb Z)$, then topologists and algebraic geometers often write (omitting product) $\alpha\beta\gamma \in H^6(Y,\Bbb Z)$ to represent the integer $\alpha\beta\gamma([Y])$. So there's your "integration." ... Alternatively, think of intersecting the transverse cycles $\sigma,\tau,\eta$ and getting an integer. Ordinarily one would write $\sigma\cdot\tau\cdot\eta$. Does this explain the mystery a bit more clearly? :) $\endgroup$ – Ted Shifrin Oct 17 '17 at 6:42
  • $\begingroup$ That helps a lot; thank you! $\endgroup$ – leastaction Oct 17 '17 at 11:55

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