# Asymptotic $\int_M \mathrm{exp}[\mathbf{e}\left(n -\frac{t}{2\pi i}\right)] \left( 1 + \frac{5}{6} \mathbf{e}^2 \right)^{1/2}$ on quintic Calabi-Yau

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 $\log n! \approx n \log n$.

Clarifications

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 $\int_M \mathbb{e}^3 = 5$?

• 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? Aug 27 '15 at 12:51
• I fixed a TeX typo. The math content is unchanged. Aug 27 '15 at 15:16
• What is the meaning of $(1+\frac{5}{6}\mathbf{e}^2)^{1/2}$? Aug 27 '15 at 15:18
• 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. Aug 27 '15 at 18:45

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

$$\frac{5A^3}{6}+\frac{25A}{12}$$

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.