Let $M = \{ G(x) = 0 \} \subseteq \mathbb{P}^4$ be a quintic Calabi-Yau and $\mathbf{e} \in H^2(M, \mathbb{Z})$ such that $\int_M \mathbf{e}^3 = 5$. Then as $t \gg 1$:

$$ \int_M e^{n \mathbf{e}} e^{-\frac{t\mathbf{e}}{2\pi i}} \left( 1 + \frac{5}{6} \mathbf{e}^2 \right)^{1/2} = - \frac{5}{12} \left( \frac{t}{2\pi i} - n\right) \left( 2\left( \frac{t}{2\pi i} - n\right)^2 - 5 \right) \tag{$\ast$}$$

How does a high-frequency integral like this get computed? This could be steepest descent or stationary phase. In that case what are the Leftschetz thimble and Morse flow? Perhaps set $G$ bo the the Fermat quintic or something for an example.

Question Because there are so many new elements for me here: the surface in $\mathbb{P}^4$, digesting what it means $\int_M \mathbf{e}^3 =5$. In a deep way, this is hardly more than an Laplace transform. I would like help seeing how steepest descent / stationary phase is implemented and computed here.

The original paper talks about the pullback of the line bundle $\mathcal{O}_{\mathbb{P}^4}(n)$ to $\mathcal{O}_M(n)$ and details of $\hat{A}$-genus that I am relegating to a separate question.

Withholding the physics context, the left side looks like particular case of a more general result, but this is already an involved computation. I am struggling to see even this much and connections to more classical mathematics.

Tongue-in-cheek remark if the moments on the left side could be computed, then we get some analogue of Stirling formula $\log n! \approx n \log n$.


The notation $(1 + \frac{5}{6} \mathbf{e}^2)^{1/2}$ is taken from an arXiv.hep-th paper from 2013 which in turn cites others. One possible interpretation is to use the Taylor series:

$$(1 + \frac{5}{6} \mathbf{e}^2)^{1/2} = 1 + \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^1 \mathbf{e}^2 + \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^2 \mathbf{e}^4 + \binom{\tfrac{1}{2}}{1} \left(\tfrac{5}{6}\right)^3 \mathbf{e}^6 + \dots $$

The remaining terms are $0$ since $M$ is 6-dimensional. At least that's the gist of one of the comments.

  • Another comment suggests the integral $\ast$ is exact: an equality for all values of $t,n$.

  • Also $\mathbb{e}$ is just a 2-form like $x \, dx \vee dy$ restricted to the manifold $M$. How are we assured these exists one with $\int_M \mathbb{e}^3 = 5$?

  • 3
    $\begingroup$ It looks like you just expand out the integrand as a power series in $\mathbf e$, set $\mathbf e^3=5$ and kill everything else. Am I missing something? $\endgroup$ Aug 27, 2015 at 12:51
  • $\begingroup$ I fixed a TeX typo. The math content is unchanged. $\endgroup$ Aug 27, 2015 at 15:16
  • $\begingroup$ What is the meaning of $(1+\frac{5}{6}\mathbf{e}^2)^{1/2}$? $\endgroup$ Aug 27, 2015 at 15:18
  • $\begingroup$ I got something nearly identical to the RHS by just expanding the integrand as suggested, and I probably made a small mistake somewhere. Given that your manifold is Calabi-Yau, an example of such an $\mathbf{e}$ is a multiple of the Kahler form itself, that's how you're assured there is one. $\endgroup$ Aug 27, 2015 at 18:45

1 Answer 1


OK, let's see if I can put my money where my commenting mouth is. Let me say at the outset that I have no idea where such an integral comes from, but I claim that it doesn't matter to answer the question.

First of all, $\mathbf e$ is nothing mysterious: it is the 2-form dual to a hyperplane section of $M$. The equation $\int_M \mathbf e^3=5$ is just a fancy way of writing that the intersection of 3 hypeplanes in $\mathbf P^4$ (i.e. a line) intersects $M$ in 5 points.

Next, the coefficient $\left(n -\frac{t}{2\pi i}\right)$ never gets modified in any way, so let's just write it as $A$ for simplicity.

Finally, everything in the integrand is supposed to be an element of $H^*(M)$, so we should expand things as power series in $\mathbf e$. Since $\mathbf e \in H^2(M)$ we already have $\mathbf e^4=0$ (contrary to what was written in the edit). So we get

$$ (1+\frac{5}{6}\mathbf e^2)^{\frac12} = 1 + \frac{5}{12} \mathbf e^2 \\ \operatorname{exp} (A \mathbf e) = 1 + A \mathbf e + \frac{A^2}{2} \mathbf e^2 + \frac{A^3}{6} \mathbf e^3$$.

Now multiply these together: the result is

$$ \left( \frac{A^3}{6} + \frac{5A}{12} \right) \mathbf e^3 + \text{lower order terms}. $$

Applying $\int_M$ kills the lower order terms and turns the $\mathbf e^3$ into a 5, so we end up with


The sign of the last term is opposite to what you wrote above, but agrees with the formula on p.37 of the paper you linked.

By the way, I think you misinterpreted what the paper is saying about taking $t \rightarrow + \infty$. In the paper, they are saying that as $t \rightarrow +\infty$, some other integral expression (higher up the page) reduces to the left-hand side of your equation. They are then saying that the left-hand side equals the right-hand side, but that equality is valid for all $t$, as we just saw.


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