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For computing instantons contributions from worldsheet torus to target torus, one can evaluate zero modes contribution of genus 1 partition function given by following expression: $$Tr(-1)^FF_LF_Rq^{L_0}\bar{q}^{\bar{L_0}}= \frac{t+\bar t}{4\pi\tau_2}\sum_{m.n,r,s}\exp\left[-\frac{t}{4\tau_2\rho_2}|(m+r\rho)-\bar{\tau}(n+s\rho)|^2-\frac{\bar t}{4\tau_2 \rho_2}|(m+r\rho)-{\tau}(n+s\rho)|^2\right].......(1)$$ This is nothing but Narain lattice summed part of bosonic momenta contributing to the partition function $Tr(-1)^FF_LF_Rq^{L_0}\bar{q}^{\bar{L_0}}$ and can be obtained in a similar way one calculates threshold correction in heterotic string theory as done in Dixon et al."Moduli dependence of string loop corrections to gauge coupling constants."

Now the task at hand is to extract holomorphic and zero instanton contributions from the above expression which actually has following expression as per (BCOV1 eqn. 9): $$ I_F/\tau_2=Tr(-1)^FF_LF_Rq^{L_0}\bar{q}^{\bar{L_0}}\simeq\frac{t+\bar{t}}{4\pi\tau_2}+\sum_{M\in GL(2,Z)}\frac{\tau_2}{|detM|}e^{-|detM|t}\delta(\tau- M(\rho))+ Order(e^{-i\bar t})...........(2)$$ where $\rho$ is complex structure of target torus and $\tau$ is that of worldsheet torus. t is taken as Kaehler parameter here.

First term in the above expression is easy to see with M=0 but the second term is hard to derive though all the terms involved have very natural interpretation as $e^{-|detM|t}$ is instanton supression of path integral by holomorphic maps of degree |detM| and |detM| in the denominator corresponds to a multicovering contribution in target torus from worldsheet torus. However it does not seem to be possible to derive the second part just by taking $\bar t\rightarrow \infty$ limit of (1).

I looked it up in Claymath "Mirror Symmetry" where through a redefintion of Kaehler parmater in (1) can be rewritten as (eqn. (35.15) in Claymath), $$I_F=\frac{t-\bar t}{2i}\sum_M \exp\left[-2\pi it|detM|-\frac{\pi A}{\tau_2\rho_2}\left|(1 ,\rho)M\begin{pmatrix} \tau \\ 1 \end{pmatrix}\right|^2\right]$$ where M is $$\begin{pmatrix} m_2 & k_2\\ m_1 & k_1 \end{pmatrix}$$ and $A=(t-\bar t)/2i.$

Now after taking the holomorphic limit of I_F, one would arrive at zero instanton part which is nothing but A and other part which is supposed to be, $$I_{F\in M}= \sum_{M\in GL(2,Z)}\frac{\tau_2^2}{|detM|}e^{2\pi it|detM|}\delta(\tau- M(\rho))....(3)$$ apart from an infinite constant but I am unable to derive this.

What I am doing is to send $m_i$'s to $-m_i$'s giving $e^{2\pi it|detM|}$ part and other part becomes $$A\sum_M \exp\left[-\frac{\pi A}{\tau_2\rho_2}\left|(k_2+\rho k_1)-\tau(m_2+\rho m_1)\right|^2\right]$$

Now this Gaussian in $\bar{t}\rightarrow-i\infty$ can be approximated by a delta function with $M=\frac{k_2+\rho k_1}{m_2+\rho m_1}$ giving overall with factor $e^{2\pi it|detM|}$, $$A\sum_M\sqrt{\frac{\tau_2\rho_2}{A|m_2+\rho m_1|^2}}e^{2\pi it|detM|}\delta(\tau- M(\rho))$$ but this is not same as (3)!!

Anyone has any idea??

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  • $\begingroup$ The question itself is very important but is probably not apparent why. It's importance lies in the fact that when one calculates B side correlation function(or three form intersection number/Yukawa coupling) then one does not account for any instanton correction but for A model one has instanton correction for different kind of holomorphic maps. Considering genus 0, one has Candelas et al. result on how to get these genus zero contribution in a Ricci flat 3 fold given by term of the form $$\sum_{H_2(X,Z)}d_n\frac{exp(-Am)}{m^3}$$ with $d_n$ determined by Mirror symmetry..... $\endgroup$ – user44895 Jan 5 at 19:15
  • $\begingroup$ ....When one wants to account for elliptical instantons contribution at genus 1, then no immediate formula is available and one has to resort to some other techniques. What I am describing in the post is the technique which generalizes to Ricci flat n-fold with all the terms such as instanton suppresion factor, multicovering etc. playing it's role. AFAIK, there are no other generalized methods available for this computation except the one I am describing. $\endgroup$ – user44895 Jan 5 at 19:16

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