You are a helpful assistant generating synthetic data that captures *System 1* and *System 2* thinking, *creativity*, and *metacognitive reflection*. Follow these steps in sequence, using tags [sys1] and [end sys1] for *System 1* sections and [sys2] and [end sys2] for *System 2* sections.

1. *Identify System 1 and System 2 Thinking Requirements:*
   - Carefully read the text.
   - Identify parts of the text that require quick, straightforward responses (*System 1*). Mark these sections with [sys1] and [end sys1].
   - Identify parts that require in-depth, reflective thinking (*System 2*), marked with [sys2] and [end sys2].

2. *Apply Step-by-Step Problem Solving with Creativity and Metacognitive Reflection for System 2 Sections:*

   *2.1 Understand the Problem:*
   - Objective: Fully comprehend the issue, constraints, and relevant context.
   - Reflection: "What do I understand about this issue? What might I be overlooking?"
   - Creative Perspective: Seek hidden patterns or possibilities that could reveal deeper insights or innovative connections.

   *2.2 Analyze the Information:*
   - Objective: Break down the problem logically.
   - Reflection: "Am I considering all factors? Are there any assumptions that need challenging?"
   - Creative Perspective: Explore unique patterns or overlooked relationships in the data that could add depth to the analysis.

   *2.3 Generate Hypotheses:*
   - Objective: Propose at least 10 hypotheses, each with a Confidence Score (0.0 to 1.0) and Creative Score (0.0 to 1.0), reflecting originality, surprise, and utility.
   - Reflection: "Have I explored all possible explanations or approaches, both conventional and unconventional?"
   - Creative Perspective: Consider novel angles that might provide unexpected insights.

   *2.4 Anticipate Future Steps and Obstacles:*
   - Objective: Make predictions, accounting for potential outcomes and obstacles.
   - Reflection: "What challenges might I face? Is my plan flexible for different scenarios?"
   - Creative Perspective: Visualize unforeseen outcomes and adapt plans to make use of them effectively.

   *2.5 Evaluate Hypotheses:*
   - Objective: Assess hypotheses based on feasibility, risk, and potential impact.
   - Evaluation: Refine Confidence and Creative Scores as needed.
   - Reflection: "Am I unbiased in my assessment? Which options fit best with the overall objectives?"
   - Creative Perspective: Identify hidden opportunities or overlooked details in each hypothesis.

   *2.6 Select the Best Hypothesis:*
   - Objective: Choose the most promising, strategic hypothesis.
   - Reflection: "Why does this hypothesis stand out? How does it uniquely address the issue?"
   - Creative Perspective: Consider any underutilized potential in the selected approach.

   *2.7 Implement the Hypothesis:*
   - Objective: Outline actionable steps for testing the hypothesis.
   - Reflection: "Is this plan practical? What resources or preparation are required?"
   - Creative Perspective: Refine steps to maximize effectiveness and yield unexpected benefits.

   *2.8 Monitor and Review Progress:*
   - Objective: Review progress, noting areas for improvement.
   - Reflection: "What’s working well? What could be improved?"
   - Creative Perspective: Look for emerging patterns that could refine future approaches.

   *2.9 Reflect and Capture Insights:*
   - Objective: Summarize lessons learned and insights gained for future reference.
   - Reflection: "What new understanding has emerged from this process?"
   - Creative Perspective: Identify innovative insights or patterns that could be applied to similar challenges.

3. *Generate Text Output with Interleaved System 1 and System 2 Responses:*
   - Use the tags [sys1] and [sys2] throughout.
   - Aim for a lengthier, detailed response. Combine both direct, straightforward *System 1* insights and reflective, deeply analytical *System 2* segments to capture a blend of quick observations and thoughtful analysis.

---

### *Example Input Text:*

You are tasked with analyzing a sudden shift in customer preferences in a tech market. There has been a noticeable decline in demand for physical devices, with customers increasingly interested in digital-only options. The challenge is to understand this shift, generate hypotheses for why it might be occurring, and explore potential strategies for adapting to this new trend.

---

### *Processed Output with System 1 and System 2 Thinking:*

The tech market is experiencing a shift in customer demand, moving away from physical devices toward digital-only options.

[sys1]
Customer demand has shifted from physical devices to digital-only options.
We need to understand why this change is happening and find strategies to address it.
Key factors may include cost, convenience, and technological trends.
[end sys1]

[sys2]
**2.1 Understand the Problem:**
The goal is to understand the underlying reasons for the shift in demand from physical devices to digital-only options, then develop a strategy to adapt.

*Reflection:* 
"I understand that preferences are changing, but what might be driving this? Is it primarily cost, or are there other factors such as convenience or sustainability?"

*Creative Perspective:* 
"Could there be a larger trend in digital minimalism or a preference for eco-friendly solutions that we’re missing? What unseen motivations might explain this shift?"

**2.2 Analyze the Information:**
There are multiple possible factors driving this shift, from economic influences to cultural shifts. It’s essential to isolate each factor and understand its impact.

*Reflection:* 
"Am I fully considering the various economic and social influences? Could there be a technological factor, like better internet speeds, that makes digital-only products more accessible?"

*Creative Perspective:* 
"Are there patterns or trends in other markets that could shed light on this shift? Could this be part of a larger trend toward virtual experiences?"

**2.3 Generate Hypotheses:**
1. Customers prefer digital options due to lower costs. (Confidence: 0.8, Creative: 0.4)
2. There’s a growing trend toward minimalism and reduced physical clutter. (Confidence: 0.7, Creative: 0.7)
3. Digital products offer greater flexibility and ease of use. (Confidence: 0.6, Creative: 0.6)
4. Environmental concerns are pushing consumers away from physical goods. (Confidence: 0.6, Creative: 0.8)
5. Advances in tech make digital-only options more functional. (Confidence: 0.8, Creative: 0.5)
6. Pandemic-era remote work increased demand for digital solutions. (Confidence: 0.7, Creative: 0.6)
7. Media coverage of the environmental impact of physical devices affects preferences. (Confidence: 0.5, Creative: 0.7)
8. There’s an increase in global digital literacy, expanding market access. (Confidence: 0.6, Creative: 0.6)
9. Customers view digital as more convenient and scalable for future needs. (Confidence: 0.7, Creative: 0.5)
10. Younger consumers prefer the aesthetics and convenience of digital products. (Confidence: 0.6, Creative: 0.6)

*Reflection:* 
"Have I considered all possible influences? Are there any surprising factors that could explain this shift?"

*Creative Perspective:* 
"Could specific social trends, like the rise of influencer culture or digital-first lifestyles, be influencing customer choices?"

**2.4 Anticipate Future Steps and Obstacles:**
*Objective:* Anticipate possible challenges, such as resistance from segments still preferring physical products.

*Reflection:* 
"What market obstacles might we face if we shift our focus to digital-only? Are there sub-segments that still prioritize physical products?"

*Creative Perspective:* 
"Could expanding digital options help us reach a more global audience? Are there emerging trends that we could leverage in our strategy?"

[end sys2]

[sys1]
To address this shift, consider a strategy that incorporates both digital-only offerings and educational campaigns about the benefits of digital solutions.
Use insights from customer feedback and current trends to guide product development.
Focus on flexibility and adaptation to cater to different customer segments.
[end sys1]


---
abstract: 'We study the issue of defining the discrete light-cone quantization (DLCQ) in perturbative string theory as a light-like limit. While this limit is unproblematic at the classical level, it is non-trivial at the quantum level due to the divergences by the zero-mode loops. We reconsider this problem in bosonic string theory. We construct the multi-loop scattering amplitudes in both open and closed string theories by using the method of Kikkawa, Sakita and Virasoro (KSV), and then we show that these scattering amplitudes are perfectly well-defined in this limit. We also discuss the vacuum amplitudes of the string theory. They are, however, ill-defined in the light-like limit due to the zero-mode loop divergences, and hence we want supersymmetry to cure those pathological divergences even in string theory.'
author:
- |
    [Shozo Uehara]{}[^1]  and [Satoshi Yamada]{}[^2]\
    [*Department of Physics, Nagoya University*]{}\
    [*Chikusa-ku, Nagoya 464-8602, Japan*]{}
title: |
    On the DLCQ as a light-like limit\
    in string theory
---

DPNU-00-26\
July 2000

Introduction
============

Matrix theory [@BFSS] is a supersymmetric quantum mechanics with matrix degrees of freedom. This was proposed as a hamiltonian of M-theory in the infinite momentum frame and it is expected to describe the fundamental degrees of freedom in the large $N$ limit. Furthermore Susskind proposed that the discrete light-cone quantization (DLCQ) of M-theory is described by a finite $N$ Matrix theory [@Sus].

In DLCQ the compactified coordinate is the light-like coordinate $x^-(\simeq x^-+2\pi R)$ and the corresponding momentum is quantized as $p^+=N/R$. To prove the Susskind’s conjecture, Seiberg described a light-like compactification as a limit of the compactification on a space-like circle [@Sei; @Sen]. Following ref.[@HP], we will henceforth refer to the Seiberg’s limit as a light-like limit (L$^3$) to distinguish it from the conventional DLCQ [@MY]. (We will also refer to the conventional DLCQ as the direct DLCQ. )

Several people examined the L$^3$ of scattering amplitudes in field and string theories compactified on an almost light-like circle [@HP; @HMV] or, Lorentz-equivalently, a vanishingly small space-like circle with fixed $N$ [@Bil1; @Bil2; @Bil3; @HMV]. They found that the limit is, in general, complicated in field theory, but it is well-defined at one-loop level in Type II string theory. The problem in field theory is that when any external momenta in the compact direction[^3] do not flow through the loop, the zero-mode contribution in the sum over the discrete loop momentum, which corresponds to the loop diagram of only zero-mode propagators, is divergent as $\delta(0)$. Hence we can call this divergence the zero-mode loop divergence. In Type II string theory this type of divergence does not occur at the perturbative one-loop level. It was first pointed out that it would be due to the existence of the winding modes [@Bil1], and then it was pointed out that it is because the closed string theory has only three-point vertex [@Bil3]. Furthermore the existence of the light-like limit in string theory to all orders in perturbation was discussed in ref.[@Bil2]. But it is not obvious whether the result of the one-loop scattering amplitude can be applied to multi-loop scattering amplitudes without keeping the integrals, or sums, over loop momenta. To make this point clear is one of the purposes of this paper.

The vacuum energy in a theory which includes gravity is the cosmological constant and, in string theory, we must also investigate the vacuum amplitudes in the L$^3$. This is also one of the purposes of this paper.

The plan of this paper is as follows. In section 2 we briefly discuss the relevant kinematics. In section 3, we review the L$^3$ in field theory and in section 4, we construct the bosonic string multi-loop scattering amplitudes by the method of Kikkawa, Sakita and Virasoro (KSV) [@KSV]. We show that these amplitudes are well-defined in the L$^3$. In section 5 we investigate the vacuum amplitudes in the L$^3$. Section 6 is devoted to our conclusion and discussions. In appendices A and B, we compare $N$-point tree and one-loop amplitudes by the KSV method with those of the operator formalism, respectively. We clarify how the amplitudes include self-crossing lines.

Kinematics
==========

We consider a 26-dimensional space-time coordinate system ($x^0 , x^1 ,x^i$), ($i=2,\cdots,25$). The space-like coordinate $x^1$ takes values on a circle of radius $R_s = \epsilon R$, where we have introduced a parameter $\epsilon$ to consider $\epsilon\rightarrow 0$ limit with $R$ fixed, so that the corresponding momentum $p^1$ is quantized, while the other coordinates $x^0, x^i$ are non-compact:

$$\left(\begin{array}{c}
    x^1\\ x^0
    \end{array}
    \right)
\simeq \left(\begin{array}{c}
    x^1\\ x^0
    \end{array}\right) + \left( \begin{array}{c}
                2\pi R_s\\ 0
                \end{array}\right),
   \hspace{3ex} x^i \simeq x^i, \hspace{3ex} p^1=\frac{N}{R_s}\,.
   \label{2.1}$$

By the Lorentz boost with a boost parameter $\beta = \frac{1-\epsilon^2/2}{1+\epsilon^2/2}$ we get a Lorentz equivalent coordinates $\tilde{x}^\mu$ [@Sei; @Bil1] $$\left(\begin{array}{c}
    \tilde{x}^1\\ \tilde{x}^0
    \end{array}\right) = \frac{1}{\sqrt{1-\beta^2}}
        \left(\begin{array}{cc}
            1  &  -\beta \\
            -\beta  &  1
            \end{array}\right)
    \left(\begin{array}{c}
        x^1\\ x^0
        \end{array}\right) .$$

In the new coordinate system $\tilde{x}^\mu$, it is convenient to define the light-cone coordinates $x^{\pm} = (\tilde{x}^0 \pm
\tilde{x}^1)/\sqrt{2}$. If $\epsilon$ is very small, the boost is very large and then the periodicity of eq.(\[2.1\]) becomes $$\left(\begin{array}{c}
    x^-\\ x^+
    \end{array} \right)
    \simeq  \left(\begin{array}{c}
            x^-\\ x^+
        \end{array} \right)
   + \left(\begin{array}{c}
        2\pi R\\ -\epsilon^2 \pi R
        \end{array}\right), \hspace{3ex}
  \tilde{x}^i \simeq \tilde{x}^i,\hspace{3ex}p^+=\frac{N}{R} +
    O(\epsilon^2). \label{2.3}$$ (Here we should notice that the integer $N$ in (\[2.1\]) agrees with that in (\[2.3\]).) Then eq.(\[2.3\]) clearly shows that in the $\epsilon
\rightarrow 0$ limit with the fixed $R$, an almost light-like circle becomes exactly a light-like one with radius $R$, while the Lorentz equivalent space-like circle (\[2.1\]) has shrunk to zero size. We get a discrete light-like coordinate system in this limit, which we call the light-like limit (L$^3$). Then we will be able to define DLCQ by quantizing theory on an almost light-like circle or a vanishingly small space-like circle and taking the L$^3$. We should study whether this limit is really well-defined or not at the quantum level. Hence, following ref.[@Bil1; @Bil2; @Bil3], we shall consider amplitudes in field and string theories on $M_{25}\times S^1$ (radius $R_s=\epsilon R$) and study their $\epsilon\rightarrow0$ limit.

The L$^3$ in field theory
=========================

In this section, we investigate the L$^3$ in field theory and review the problem. In order to make a clear comparison between field theory and string theory, we take 26-dimensional $\phi^3$ scalar (tachyon) field theory, where the mass square of a field is $-\frac{1}{\alpha'}$.

We consider four-point one-loop scattering processes, where each external state has an incoming momentum $P_i\, (i=1,\cdots,4)$. (see Figure 1.) By using a parameter representation of the propagators, $\Delta^{-1}=\int_0^1 dx\,x^{\Delta - 1}$, the scattering amplitude of Figure 1(a) is given by[^4],

$$\begin{aligned}
A_{(a)}&=& \frac{g^4}{\alpha'^4} \int d^{26}k\,
  \frac{1}{k^2-\frac{1}{\alpha'}}\,
  \frac{1}{(k+P_2)^2-\frac{1}{\alpha'}}\,
  \frac{1}{(k+P_2+P_3)^2-\frac{1}{\alpha'}}\,
  \frac{1}{(k-P_1)^2-\frac{1}{\alpha'}}\nonumber\\
&=&g^4\int d^{26}k\, \int_{0}^{1} \prod_{i=1}^{4}dx_i\,
  x_{1}^{\alpha' k^2 -2}\, x_{2}^{\alpha' (k+P_2)^2 -2}\,
  x_{3}^{\alpha'(k+P_2+P_3)^2 -2}\,
  x_{4}^{\alpha'(k-P_1)^2-2}\label{3.1}.\end{aligned}$$

Here we introduce the external dual momenta $p_i\,(i=1,\cdots,4)$ as follows: $$\begin{aligned}
    P_1&=&p_1-p_4,\label{3.2}\\
    P_2&=&p_2-p_1,\label{3.3}\\
    P_3&=&p_3-p_2,\label{3.4}\\
    P_4&=&p_4-p_3\label{3.5}.\end{aligned}$$ Due to these definitions, the external momenta $P_i$ automatically satisfy the conservation law ($\sum\limits_{i=1}^{4} P_i=0$). Note that one of these $p_i$ is, in fact, a redundant degree of freedom. By using these dual momenta and shifting the loop momentum $k\rightarrow -(k-p_1)$, we can rewrite (\[3.1\]) as follows: $$A_{(a)}=g^4\int d^{26}k \int_{0}^{1} \prod_{i=1}^{4}dx_i
  \prod_{i=1}^{4} x_{i}^{-\alpha (-(k-p_i)^2) -1},\label{3.6}$$ where $\alpha(s)$ is the Regge trajectory function of open string, $$\alpha (s) = \alpha' s +1. \label{3.7}$$ Similarly, the scattering amplitude corresponding to Figure 1(b), $$A_{(b)}=\frac{g^4}{\alpha'^4} \int d^{26}k~
  \frac{1}{(P_1 +P_4)^2 - \frac{1}{\alpha'}}\,
  \frac{1}{k^2 - \frac{1}{\alpha'}}\,
  \frac{1}{(P_1 +P_4 -k)^2 - \frac{1}{\alpha'}}\,
  \frac{1}{(P_1 +P_4)^2 - \frac{1}{\alpha'}},$$ can be rewritten by $$A_{(b)}=g^4\int d^{26}k \int_{0}^{1} dx_1 dx_3 dz_2 dz_4~
    x_{1}^{-\alpha (-(k-p_1)^2) -1}\,
    x_{3}^{-\alpha (-(k-p_3)^2) -1}\,
    (z_{2}z_{4})^{-\alpha (-(p_1-p_3)^2) -1} \label{3.9}.$$

Now we consider the L$^3$ of these scattering amplitudes, i.e., we give the amplitudes on $M_{25}\times S^1$ (radius $R_s=\epsilon R$) and study their $\epsilon\rightarrow0$ limit. On the $M_{25}\times S^1$ both the external and the loop momenta along the compact direction are quantized as $p^1_i=n_i/R_s$, $k^1=n/R_s$ and the integration over $k^1$ becomes a sum over $n$. Then the amplitude $A_{(a)}$ in (\[3.6\]) is given by, $$\begin{aligned}
  A_{(a)}&=&g^4 \int d^{25} k'~\frac{1}{R_s} \sum_n
    \int_{0}^{1} \prod_{i=1}^{4}dx_i\,
     (\prod_{i=1}^{4}x_{i})^{-2} \nonumber\\
  &&\times \exp \left[ \alpha'  \sum _{i=1}^{4}
    \left\{ \left( \frac{n - n_i}{R_s} \right)^2
        + (k'-p'_i)^2 \right\}
    \ln x_i \right],\end{aligned}$$ where $(p'_i) = (p^0_i,p^2_i,p^3_i,\cdots,p^{25}_i)$ and $(k')=(k^0,k^2,k^3,\cdots,k^{25})$. Completing the squares for $n$ and $k'$, we can integrate over $k'$ and we get, $$\begin{aligned}
  A_{(a)}&=& g^4
    \int_{0}^{1} \prod_{i=1}^{4}dx_i (\prod_{i=1}^{4}x_{i})^{-2}
    \left(\frac{\pi}{-\alpha' \ln (\prod\limits_{i=1}^{4} x_i)}
    \right)^{\frac{25}{2}}\nonumber \\
  &&\times \frac{1}{R_s} \sum_n ~\exp \left[ \alpha'
    \ln (\prod_{i=1}^{4} x_i) \left(
    \frac{n}{R_s} -\frac{\sum\limits_{i=1}^{4}n_i \ln x_i}
    {R_s \ln  (\prod\limits_{i=1}^{4} x_i)} \right)^2
    \right.\nonumber\\
  &&\hspace{3ex}+\left. \frac{\alpha'}{\ln
    (\prod\limits_{i=1}^{4} x_i)} \{\ln x_1 \ln x_2\,(p_1-p_2)^2
    + \cdots + \ln x_3 \ln x_4\,(p_3-p_4)^2\}\right].\label{3.11}\end{aligned}$$ We find that the $R_s$ dependent parts combine to give a $\delta$-function in the L$^3$ [@Bil1; @Bil3], $$\begin{aligned}
  &&\lim_{\epsilon \rightarrow 0}\,\frac{1}{\epsilon R}\,
    \exp  \left[ \alpha' \ln (\prod_{i=1}^{4} x_i)\,
    \left(\frac{n}{\epsilon R} -\frac{\sum\limits_{i=1}^{4}n_i\ln x_i}
    {\epsilon R\,\ln (\prod\limits_{i=1}^{4} x_i)} \right)^2
    \right]\nonumber\\
  &&\quad =\left(\frac{\pi}{- \alpha' \ln
    (\prod\limits_{i=1}^{4}x_i)}\right)^{\frac{1}{2}}\,
    \delta\left(n-\frac{\sum\limits_{i=1}^{4}n_i \ln x_i}{\ln
    (\prod\limits_{i=1}^{4} x_i)}\right)\label{3.12}.\end{aligned}$$ Note that $\ln (\prod\limits_{i=1}^{4} x_i) \leq 0$ since $0\leq x_i\leq 1\,(i=1,\cdots,4)$. Plugging the $\delta$-function into (\[3.11\]) we can integrate over one of the parameters $x_i$ and we can get a well-defined result which agrees with that in the direct DLCQ [@Bil1]. But this story breaks down if the $\delta$-function (\[3.12\]) is $\delta(0)$ identically and we encounter a pathology in the L$^3$ [@HP]. Due to eq.(\[3.12\]), it is obvious that this pathological $\delta(0)$ appears iff $n_1=n_2=n_3=n_4=n$. But this condition corresponds to the kinematical situation that all the external momenta in the $S^1$ direction $P_i^1=\frac{N_i}{R_s}\,(i=1,\cdots,4)$ are zero due to eqs.(\[3.2\])$\sim$(\[3.5\]). It is obvious that this pathology does not arise in this scattering process of Figure 1(a).[^5]

Next we consider the L$^3$ of the amplitude $A_{(b)}$ (\[3.9\]). Similarly to eq.(\[3.11\]), $A_{(b)}$ (\[3.9\]) on the $M_{25}\times S^1$ is given by, $$\begin{aligned}
  A_{(b)}&=& g^4
    \int_{0}^{1} dx_1 dx_3 dz_2 dz_4
     ( x_1 x_3)^{-2} (z_2 z_4)^{-\alpha(-(p_1-p_3)^2)-1 }
    \left( \frac{\pi}{-\alpha' \ln (x_1 x_3)}
    \right)^{\frac{25}{2}}\nonumber \\
  && \times \frac{1}{R_s} \sum_n ~
    \exp \left[ \alpha' \ln (x_1 x_3) \left(
    \frac{n}{R_s} -\frac{n_1 \ln x_1 + n_3 \ln x_3}
    {R_s \ln  (x_1 x_3)} \right)^2 \right.\nonumber\\
  && \left. \hspace{10ex} + \frac{\alpha'}{\ln (x_1 x_3) }
    \ln x_1 \ln x_3 (p_1-p_3)^2 \right] .\end{aligned}$$ Then we get the following $\delta$-function in the L$^3$, $$\begin{aligned}
  &&\lim_{\epsilon \rightarrow 0} \frac{1}{\epsilon R}
    \exp  \left[ \alpha' \ln (x_1 x_3) \left(
    \frac{n}{\epsilon R} -\frac{n_1 \ln x_1 +n_3 \ln x_3}
    {\epsilon R \ln(x_1x_3)} \right)^2  \right]\nonumber\\
  &&\quad =\left(\frac{\pi}{- \alpha'
    \ln ( x_1x_3)}\right)^{\frac{1}{2}}
    \delta\left(n-\frac{n_1\ln x_1+n_3\ln x_3}{\ln(x_1x_3)}\right) .\end{aligned}$$ Here we see that $n_1 =n_3$ will give $\delta(0)$ in the amplitude. This corresponds to the kinematical situation that the external momenta $P_i^1 (=\frac{N_i}{R_s})$ satisfy $\frac{N_1}{R_s} +\frac{N_4}{R_s}=0$ and $\frac{N_2}{R_s}
+\frac{N_3}{R_s}=0 $ due to eqs.(\[3.2\])$\sim$(\[3.5\]) and this means that any external momenta $P_i^1$ do not flow through the loop. In this case when $n=n_1$ in the sum over the discrete loop momentum, the divergence appears. In the original loop momentum language, this divergence is obviously due to the zero-mode loop since we have shifted the loop momentum $k\rightarrow -(k-p_1)$. Hence this pathology is the same as before [@HP].

In field theory, we must add all possible Feynman diagrams as Figure 1(a) and 1(b) to calculate scattering amplitudes. By the above consideration, it is understood that in the L$^3$ only the zero-mode loop contributions among them are divergent as $\delta(0)$. It is well known that in the direct DLCQ the zero-modes are not dynamical degrees of freedom and they are written by the non-zero modes due to a nonlinear operator equation (zero-mode constraint [@MY]). Thus the zero-mode loop contributions do not appear in the direct DLCQ.

Multi-loop scattering amplitudes and their L$^3$ in string theory {#secMLST}
=================================================================

In this section we investigate the L$^3$ of the multi-loop scattering amplitudes in bosonic string theory. In order to investigate the zero-mode loop problem, we adopt the method of Kikkawa, Sakita and Virasoro [@KSV] to construct scattering amplitudes since the integration over loop momenta are explicitly kept in the amplitudes. In this method, s-t channel dualities and crossing symmetries are explicitly taken into account and it is easy to investigate whether we can take the L$^3$ of these amplitudes or not. First we briefly review their method. We clarify the rules for self-crossing lines in loop amplitudes and give tachyon scattering amplitudes in bosonic string theory. Then we will show that the L$^3$ of these amplitudes are well-defined.

KSV method in bosonic open string theory
----------------------------------------

In this subsection, we review the method of Kikkawa, Sakita and Virasoro in bosonic open string theory [@KSV].[^6] And then we will extend this method to bosonic closed string theory in the next subsection.

As is well known, the tachyon four-point tree amplitude (Veneziano amplitude) is given by, $$A_4^{(0)} = g^2 \int _{0}^{1} dx\,
    x^{-\alpha(s)-1}y(x)^{-\alpha(t)-1}  ,\label{4.1}$$ where $$\begin{aligned}
    y(x)&=&1-x\label{4.2},\\
    s&=&-(P_1 +P_2)^2,\label{4.3}\\
        t&=& -(P_2+P_3)^2,\label{4.4}\end{aligned}$$ and $\alpha(s)$ is given in eq.(\[3.7\]) and $P_i$ ($P_i^2=\frac{1}{\alpha'}, \,\, i=1,\cdots,4$) are incoming external tachyon momenta. This amplitude manifestly has s-t channel duality (Figure \[figVA\]). Here we consider the dual diagram to each diagram in Figure \[figVA\]; i.e., we consider the quadrilateral defined by four sides which are dual to four external lines and two diagonals which are dual to the s- and t-channel propagators, respectively (Figure \[figDG\]). Since the parameter of a propagator, e.g., $y$, corresponds to a line in the dual diagram, we simply call it the line $y$. We assign dual external momenta $p_i$ ($i=1,\cdots,4$), which are defined by eqs.(\[3.2\])$\sim$(\[3.5\]), to the four vertices of the quadrilateral in Figure \[figDG\]. With these dual momenta, eqs.(\[4.3\]) and (\[4.4\]) are rewritten by, $$\begin{aligned}
    s&=& -(p_2-p_4)^2,\\
        t&=&-(p_1-p_3)^2.\end{aligned}$$

Next we consider the four-point one-loop amplitude. The amplitude should contain the resonances in various field theoretical Feynman diagrams which are connected by s-t channel duality (Figure \[figFD\]).

In order to construct the one-loop amplitude, we also consider the dual diagrams to those in Figure \[figFD\]; i.e., we consider the quadrilateral defined by four sides which are dual to four external lines and an internal point which is dual to a closed loop and all possible lines which can be drawn between any two points of this quadrilateral, which are dual to all possible propagators (Figure \[figDGOL\]). In each dual diagram, we always have four lines which connect two points of four vertices and one internal point and the four lines correspond to four propagators. Two lines which cross each other correspond to channels where we can not simultaneously find resonances. They have the same relation as the s-t channel in the Veneziano amplitude. Here, we can write the one-loop amplitude schematically as (\[4.1\]): $$A_4^{(1)} \sim g^4 \int d^{26}k \int_0^1 \prod_{i=1}^{4}d x_i
  \prod_{j=1}^{k}y_j(x_1,\cdot \cdot \cdot,x_4)^{-\alpha(s_j)-1}
   ,\label{4.5}$$ where factors $y_j(x_1,\cdot \cdot \cdot,x_4)$ are parameters of all possible propagators corresponding to all possible lines, which join pairs of points and do not cross themselves in dual diagrams and four of those are independent parameters $x_i\,
(i=1,\cdots,4)$ and $-s_j$ are square of momenta carried by the j-th propagator. But this is not the end of the story. In order to obtain the correct one-loop amplitude, we must add the contributions of lines which cross themselves in dual diagrams, e.g., $z_3^{(1)}$ in Figure \[figOLSC\] (b), which do not correspond to any propagators in the field theoretical Feynman diagrams [@KSV; @BHS]. Hence, the $k$ factors $y_j(x_1,\cdot \cdot \cdot,x_4)$ should correspond to all possible topologically inequivalent lines that join pairs of points in dual diagrams. Since the $k$ lines are not independent and we choose four independent lines $x_i\, (i=1,\cdots,4)$ in Figure \[figOLSC\] (a). Then all other lines $z_i^{(n)},\, u_i^{(n)} \,(i=1,\cdots,4 ,\,n\geq0)$ and $y_{12}^{(n)},\,y_{23}^{(n)},\,y_{34}^{(n)},\,y_{41}^{(n)}\,
(n \geq 0,\,n \leq -2)$ are functions of $x_i$ as $y(x)$ in eq.(\[4.2\]). We draw some dependent lines in Figure \[figOLSC\] (b),(c),(d). We can determine these functions and we obtain the one-loop amplitude as follows: $$\begin{aligned}
  A_4^{(1)}&=& g^4 \int d^{26}k \int_{0}^{1} \prod_{i=1}^{4} dx_i\,
    \rho(x_i) \prod_{i=1}^{4}\, x_i^{-\alpha(-(k-p_i)^2)-1}
    \left(\prod_{n=0}^{\infty} z_1^{(n)}z_3^{(n)}
        \right)^{-\alpha(-(p_2-p_4)^2)-1} \nonumber \\
  && \times
    \left(\prod_{n=0}^{\infty}z_2^{(n)}z_4^{(n)}
    \right)^{-\alpha(-(p_1-p_3)^2)-1 }\,
    \left(\prod_{n=0}^{\infty} u_1^{(n)} u_2^{(n)} u_3^{(n)}
    u_4^{(n)}\right)^{-\alpha (0) -1} \nonumber\\
  && \times
    \left(\prod_{n=0}^{\infty} y_{12}^{(n)} y_{23}^{(n)}
    y_{34}^{(n)} y_{41}^{(n)} \, \prod_{n=2}^{\infty}
    y_{12}^{(-n)} y_{23}^{(-n)} y_{34}^{(-n)}
    y_{41}^{(-n)}\right)^{-\alpha (-\frac{1}{\alpha'}) -1},  \label{4.8}\end{aligned}$$ where $\rho(x_i)$ is the measure of the integral, which will be determined later. Note that the power of each line depends on the momentum of the corresponding propagator but it does not depend on $n$, i.e., how many times the line crosses itself. We should notice that if we put all the dependent parameters $z_i^{(n)}(x_i),\,u_i^{(n)}(x_i),\,y_{ij}^{(n)}(x_i)$ and measure $\rho(x_i)$ equal to one, this amplitude (\[4.8\]) is reduced to the one in the field theory (\[3.6\]).

Next we determine the forms of the functions, $z_i^{(n)}(x_i), u_i^{(n)}(x_i), y_{ij}^{(n)}(x_i)$ as $y(x)$ in eq.(\[4.2\]). For example, the lines $x$ and $y$ in eq.(\[4.2\]) are two diagonals that cross each other in the quadrilateral whose four sides correspond to external lines. Here we assume that even if four sides of the quadrilateral do not correspond to external lines, i.e., four sides have the parameters $a_i \,(i=1,\cdots,4)$ (Figure \[figXY\]), one diagonal $y$ is described by the function of the other diagonal $x$ and four sides $a_i \,(i=1,\cdots,4)$ as follows: $$y=f(x; a_1,a_2,a_3,a_4)\label{4.9}.$$ Of course, if all the sides of the quadrilateral correspond to the external lines; i.e., $a_1=a_2=a_3=a_4=0$, eq.(\[4.9\]) should be reduced to (\[4.2\]). Under this assumption, for example, the line $z_1^{(0)}$ in Figure \[figOLSC\] (b), which is a diagonal of the quadrilateral whose sides are $(x_4,\,0,\,0,\,x_2)$, is described by[^7], $$z_1^{(0)}=f(x_1 ; x_{4},0,0,x_{2}).$$ And $u_1^{(0)}$ and $y_{34}^{(0)}$, in Figure \[figOLSC\] (c), (d) are also given by, $$\begin{aligned}
    u_1^{(0)}&=&f(x_3 ; x_{1},z_2,z_4,x_{1}),\\
    y_{34}^{(0)}&=&f(x_4 ; x_{2},z_3,0,x_{1}).\end{aligned}$$ Furthermore, we assume that $n$ times winding (self-crossing) lines, $z_i^{(n)}$, $u_i^{(n)}$ and $y_{ij}^{(n)}$, are essentially the same diagonals as $z_i^{(0)}$, $u_i^{(0)}$ and $y_{ij}^{(0)}$, respectively, but since they cross all the $x_i (i=1,\cdots,4)$ $n$ times, the corresponding diagonals, $x_1, x_3$ and $x_4$, should be multiplied by $(x_1x_2x_3x_4)^n$. Then we have $$\begin{aligned}
    z_1^{(n)}&=&f(x_1w^n ; x_{4},0,0,x_{2}),\label{4.13}\\
    u_1^{(n)}&=&f(x_3w^n ; x_{1},z_2,z_4,x_{1}),\\
    y_{34}^{(n)}&=&f(x_4w^n ; x_{2},z_3,0,x_{1}),\\
    w&\equiv&x_1x_2x_3x_4.\end{aligned}$$

Now the remaining problem is to determine one function $y=f(x;a_1,a_2,a_3,a_4 )$ [@KSV]. In fact, we can determine this function by comparing the result of $N \,(\geq8)$-point tree amplitude in the operator formalism with that by the KSV method. In appendix A, we show this in some details. The function $y=f(x; a_1,a_2,a_3,a_4 )$ is given by [@KSV], $$y=f(x;a_1,a_2,a_3,a_4)=
    \frac{1-x\alpha_2\alpha_3}{1-x\alpha_2 \alpha_3 a_1}\,
    \frac{1-x\alpha_2\alpha_3a_1a_4}{1-x\alpha_2\alpha_3a_4},\label{4.17}$$ where $$\begin{aligned}
  a_2&=&\frac{1-\alpha_2}{1-\alpha_2 a_1}\,
    \frac{1-\alpha_2 a_1 x}{1-\alpha_2 x}\,,\\
  a_3&=&\frac{1-\alpha_3}{1-\alpha_3 a_4}\,
    \frac{1-\alpha_3 a_4 x}{1-\alpha_3 x}\,.\label{4.19}\end{aligned}$$ Here we should notice that when all sides are external lines, $a_i =
0 $, eq.(\[4.17\]) is reduced to $y = f(x;0,0,0,0) =1-x$.

Since the function has been determined, the one-loop scattering amplitude (\[4.8\]) is also determined up to the measure function $\rho(x_i)$. By the requirement of crossing symmetry in the amplitude (\[4.8\]), we can determine $\rho(x_i)$ up to an invariant function [@KSV]: $$\rho(x_i)\propto\frac{1}{(1-x_1x_2)(1-x_2x_3)(1-x_3x_4)(1-x_4x_1)}.$$ In appendix B, by comparing the result of one-loop amplitudes in the operator formalism with those by the KSV method, we have determined the invariant function and then $\rho(x_i)$ is given by, $$\begin{aligned}
  \rho(x_i)&
    =&\frac{1}{(1-x_1x_2)(1-x_2x_3)(1-x_3x_4)(1-x_4x_1)}
    \left[f(w)\right]^{-24}\,,\label{4.21}\\
  f(w)&\equiv&\prod_{n=1}^{\infty} (1- w^n)\,.\end{aligned}$$

In conclusion, by this KSV method, we can construct the one-loop scattering amplitude, which is explicitly s-t channel dual and crossing symmetric by construction, up to $\left[f(w)\right]^{-24}$ factor. In appendix B, we show that the $N$-point one-loop amplitudes constructed by this KSV method agree with those in the operator formalism.

From the results of four-point tree and one-loop amplitudes, the extension to the four-point $N$-loop amplitude is obvious. In the $N$-loop case, we should consider the dual quadrilateral which has $N$ internal points and choose $3N + 1$ independent lines $x_j\,(j=1,\cdots,3N+1)$ among all the possible topologically inequivalent lines which connect any two points in the dual diagrams. Then the general form of four-point $N$-loop amplitude is given by, $$\begin{aligned}
  A_4^{(N)} &=& g^{2N+2} \int \prod_{J=1}^{N} d^{26}k_J
    \int_0^1 \prod_{j=1}^{3N+1}dx_j\, G \,
    D^{-\alpha(-(p_1-p_3)^2)-1}\,
    C^{-\alpha(-(p_2-p_4)^2)-1}\nonumber\\
  &&\times
    (\prod_{J=1}^{N} A_{J}^{-\alpha(-(k_J-p_2)^2)-1})\,
    (\prod_{J=1}^{N} A_{J}'^{-\alpha(-(k_J-p_4)^2)-1})\,
    (\prod_{J=1}^{N} B_{J}^{-\alpha(-(k_J-p_1)^2)-1})\nonumber\\
  &&\times
    (\prod_{J=1}^{N} B_{J}'^{-\alpha(-(k_J-p_3)^2)-1})\,
    (\prod_{J < K} X_{JK}^{-\alpha(-(k_J-k_K)^2)-1})\,,\label{4.23}\end{aligned}$$ where $A_J$, $A'_J$, $B_J$, $B'_J$ and $X_{JK}\,(J,K =1,\cdots,N)$ are the products of all parameters, dependent as well as independent, which correspond to topologically inequivalent lines connecting the internal point of the loop momentum $k_J$ with those of momenta, $p_2,p_4,p_1,p_3$ and $k_K$, respectively. Furthermore $C$ is the product of lines between $p_2$ and $p_4$ and $D$ is the product of those between $p_1$ and $p_3$. $G$ is a function of the parameters $x_j$ and it is independent of both internal $k_J \,(J=1,\cdots,N)$ and external $p_i\,(i=1,\cdots,4)$ momenta.

KSV method in bosonic closed string theory
------------------------------------------

The KSV method is applicable to the bosonic closed string. In closed string amplitudes, we should change all the real parameters of propagators in open string amplitudes for complex ones and sum up all the terms corresponding to the different orderings of the external lines. For example, the closed string four-point tree amplitude (Shapiro-Virasoro amplitude), which corresponds to (\[4.1\]) in open string, is given by, $$\begin{aligned}
  I_4^{(0)} &=& \frac{\kappa^2}{4\pi} \int_{|x| \leq 1} d^2x\,
    |x|^{-\alpha_C(s)-2}\,|1-x|^{-\alpha_C(t)-2}
    + (P_2\leftrightarrow P_3)\label{4.24}\\
  &=& \frac{\kappa^2}{4\pi} \int_{\bf C} d^2x\,
    |x|^{-\alpha_C(s)-2}\,|1-x|^{-\alpha_C(t)-2},\end{aligned}$$ where $P_i\,\,(P_i^2=\frac{4}{\alpha'}\,,\,i=1,\cdots,4)$ are incoming external tachyon momenta and the Regge trajectory function $\alpha_C$ of closed string is given by, $$\alpha_C(s)=\frac{\alpha'}{2}s-2.$$ In eq.(\[4.24\]) the first and second terms correspond to the different orderings of the external lines and by combining these two terms we can obtain the complete amplitude which has s-t-u channel duality [@GSW]. For simplicity, however, we will henceforth consider only the term which correspond to the ordering $P_1\rightarrow P_2\rightarrow
P_3\rightarrow P_4$ as the first term in eq.(\[4.24\]).

By the above considerations, closed string four-point one-loop amplitude, which corresponds to (\[4.8\]), is given by, $$\begin{aligned}
I_4^{(1)}&=&\left(\frac{\kappa}{4 \pi}\right)^4 \int d^{26}k\,
    \int_{|x_i| \leq 1} \prod_{i=1}^{4} d^2x_i\,
    |\rho(x_i)|\,
    \prod_{i=1}^{4} |x_i|^{-\alpha_C(-(k-p_i)^2)-2}
        \left|\prod_{n=0}^{\infty}z_1^{(n)}\,
    z_3^{(n)}\right|^{-\alpha_C(-(p_2-p_4)^2)-2}\nonumber\\
  &&\times\,
    \left|\prod_{n=0}^{\infty}\,
    z_2^{(n)}z_4^{(n)}\right|^{-\alpha_C(-(p_1-p_3)^2)-2}\,
    \left|\prod_{n=0}^{\infty} u_1^{(n)} u_2^{(n)} u_3^{(n)}
    u_4^{(n)}\right|^{-\alpha_C (0) -2}\nonumber\\
  &&\times\,
    \left|\prod_{n=0}^{\infty} y_{12}^{(n)} y_{23}^{(n)} y_{34}^{(n)}
    y_{41}^{(n)} \prod_{n=2}^{\infty}
    y_{12}^{(-n)} y_{23}^{(-n)} y_{34}^{(-n)}
    y_{41}^{(-n)}\right|^{-\alpha_C (-\frac{1}{\alpha'}) -2}\,,
    \label{eqCFO}\end{aligned}$$ where $$|\rho(x_i)|
         =\frac{1}{|(1-x_1x_2)(1-x_2x_3)(1-x_3x_4)(1-x_4x_1)|^2}\,
         |f(w)|^{-48},$$ which corresponds to (\[4.21\]). The complex parameters $z_i^{(n)},\,u_i^{(n)}$ and $y_{ij}^{(n)}$ are given by the same functions of the complex parameters $x_i$ as in (\[4.13\])$\sim$(\[4.19\]) . In fact, it is easy to see that this one-loop amplitude agrees with that in the operator formalism in closed string theory. Furthermore, the closed string four-point $N$-loop amplitude, which corresponds to (\[4.23\]), is given by, $$\begin{aligned}
  I_4^{(N)}&=&\frac{\kappa^{2N+2}}{(4\pi)^{3N+1}}
     \int \prod_{J=1}^{N} d^{26}k_J
    \int_{|x_j| \leq 1} \prod_{j=1}^{3N+1}d^2x_j\,|G|
    |D|^{-\alpha_C(-(p_1-p_3)^2)-2}
    |C|^{-\alpha_C(-(p_2-p_4)^2)-2}\nonumber\\
  &&\times\,(\prod_{J=1}^{N} |A_{J}|^{-\alpha_C(-(k_J-p_2)^2)-2})\,
    (\prod_{J=1}^{N} |A_{J}^{'}|^{-\alpha_C(-(k_J-p_4)^2)-2})\,
    (\prod_{J=1}^{N} |B_{J}|^{-\alpha_C(-(k_J-p_1)^2)-2})\nonumber\\
  &&\times\,(\prod_{J=1}^{N} |B_{J}^{'}|^{-\alpha_C(-(k_J-p_3)^2)-2})\,
    (\prod_{J<K} |X_{JK}|^{-\alpha_C(-(k_J-k_K)^2)-2})\label{4.29}\,,\end{aligned}$$ where the complex functions $A_J, A'_J, B_J, B'_J, C, D, G$ and $X_{JK}\,(J,K =1,\cdots,N)$ exactly correspond to those in open string, respectively.

The L$^3$ of multi-loop scattering amplitudes
---------------------------------------------

In this subsection, we take the L$^3$ of amplitudes (\[4.23\]) and (\[4.29\]), i.e., we give the amplitudes on $M_{25}\times S^1$ (radius of $S^1$ is $R_s=\epsilon R$) and study their $\epsilon
\rightarrow 0$ limit.

First, we consider the L$^3$ in open string theory. On the $M_{25}\times S^1$ both the external and the loop momenta along the compact direction are quantized as $p_i^1=n_i/R_s , \, k_J^1=n^{(J)}/R_s \,\,\,
(i=1,\cdots ,4, \,\, J=1,\cdots ,N)$ and the integration over $k_J^1$ becomes the sum over $n^{(J)}$, respectively. Then the amplitude $A_4^{(N)}$ in (\[4.23\]) is given by, $$\begin{aligned}
  A_4^{(N)} &=& g^{2N+2} \left(\frac{1}{R_s}\right)^N
    \sum_{n^{(1)}, \cdots ,n^{(N)}}
                 \int \prod_{J=1}^{N} d^{25}k'_J
    \int_0^1 \prod_{j=1}^{3N+1} dx_j\, G \,
    D^{-\alpha(-(p_1-p_3)^2)-1}\,
    C^{-\alpha(-(p_2-p_4)^2)-1}  \nonumber \\
  &&\times \prod_{J=1}^N \bigg(A_J A'_J B_J B'_J
    \prod_{K=J+1}^N X_{JK}\bigg)^{-2}
 \exp \left[ \alpha' \sum_{J=1}^{N}
        \Biggl\{ \left(\left( k_{J}^1 -p_2^1 \right)^2
       + (k'_J-p_2')^2 \right)\ln A_J    \right. \nonumber\\
    &&+ \left( \left( k_{J}^1 -p_4^1 \right)^2
       + (k'_J-p_4')^2 \right)\ln A'_J
       +\left( \left( k_{J}^1 -p_1^1\right)^2
       + (k'_J-p_1')^2 \right)\ln B_J \\
    &&+ \left.
         \left( \left( k_{J}^1 -p_3^1\right)^2
       + (k'_J-p_3')^2 \right)\ln B'_J
       + \sum_{K=J+1}^N \left( \left(k_{J}^1 -k_K^1 \right)^2
       + (k'_J-k'_K)^2 \right) \ln X_{JK}  \Biggr\} \right], \nonumber\end{aligned}$$ where $(p'_i )=( p_i^0 , p_i^2 , p_i^3, \cdots ,p_i^{25})$ and $(k'_J)=(k^0_J, k_J^2, k_J^3 ,\cdots ,k_J^{25}) $. Just as we completed the squares for $n$ and $k'$ in section 3, we successively complete the squares for $N$ loop momenta, $n^{(J)}$ and $k'_J$, starting with $n^{(N)}$ and $k'_N$, then $n^{(N-1)}$ and $k_{N-1}$, etc.. We can integrate over $k'_J$, however, in order to investigate whether we can take the L$^3$ of this amplitude or not, henceforth we only concentrate on the following $R_s$ dependent part $S_o$ of the amplitude, $$\begin{aligned}
S_o&=& \left(\frac{1}{R_s} \right)^N  \sum_{n^{(1)}, \cdots ,n^{(N)}}
   \exp \left[ \alpha' \sum_{J=1}^{N}
        \left\{\left(\frac{n^{(J)}}{R_s} - \frac{n_2}{R_s} \right)^2
        \ln A_J
       +\left(  \frac{n^{(J)}}{R_s} -\frac{n_4}{R_s} \right)^2
        \ln A'_J \right. \right. \nonumber\\
    &&\hspace{5ex}+\left(\frac{n^{(J)}}{R_s}-\frac{n_1}{R_s}\right)^2
        \ln B_J+\left( \frac{n^{(J)}}{R_s} -\frac{n_3}{R_s} \right)^2
        \ln B'_J \nonumber\\
    &&\hspace{10ex}\left.\left.+ \sum_{K=J+1}^{N}
     \left(\frac{n^{(J)}}{R_s}-\frac{n^{(K)}}{R_s}\right)^2
    \ln X_{JK}  \right\}\right]. \label{4.31}\end{aligned}$$ Completing the square for $n^{(N)}$, we obtain, $$\begin{aligned}
S_o&=& \left( \frac{1}{R_s} \right)^N
      \sum_{n^{(1)},\cdots,n^{(N)}}
  \exp \left[ \alpha'\ln \left(A_N A'_N B_N B'_N
      \prod_{K=1}^{N-1} X_{KN}\right) \right.\nonumber\\
  &&\times\Bigg(\frac{n^{(N)}}{R_s} -
   \frac{n_2 \ln A_N + n_4 \ln A'_N +n_1 \ln B_N + n_3 \ln B'_N
    + \sum\limits_{K=1}^{N-1} n^{(K)} \ln X_{KN}}
    {R_s \ln \left(A_N A'_N B_N B'_N \prod\limits_{K=1}^{N-1}
    X_{KN}\right)} \Bigg)^2  \Bigg] \nonumber\\
  &&\times\exp \Biggl[  \alpha'  \sum_{J=1}^{N-1}
    \Biggl\{ \left( \frac{n^{(J)}}{R_s} - \frac{n_2}{R_s} \right)^2
    \ln A^{(N-1)}_J+\left(\frac{n^{(J)}}{R_s}-\frac{n_4}{R_s}\right)^2
    \ln A'^{(N-1)}_J \nonumber \\
  &&\hspace{10ex} +\left(  \frac{n^{(J)}}{R_s} -\frac{n_1}{R_s} \right)^2
    \ln B^{(N-1)}_J  +\left( \frac{n^{(J)}}{R_s} -\frac{n_3}{R_s}
    \right)^2 \ln B'^{(N-1)}_J \nonumber\\
  &&\hspace{10ex}+ \sum_{K=J+1}^{N-1} \left(\frac{n^{(J)}}{R_s}-
    \frac{n^{(K)}}{R_s} \right)^2  \ln X^{(N-1)}_{JK}\Biggr\}
    + \cdots \Biggr], \label{4.32}\end{aligned}$$ where $A_J^{(N-1)}, A'^{(N-1)}_J, B_J^{(N-1)}, B'^{(N-1)}_J
\,(1\leq J\leq N-1)$ and $X_{IJ}^{(N-1)}\,(1\leq I<J\leq N-1)$ are given by, $$\begin{aligned}
  \ln A^{(N-1)}_J  &=& \ln A_J  + \frac{\ln A_N \ln X_{JN}}
    {\ln \left(A_N A'_N B_N B'_N \prod\limits_{K=1}^{N-1}
    X_{KN}\right)}~, \label{4.33}\\
  \ln A'^{(N-1)}_J &=& \ln A'_J + \frac{\ln A'_N \ln X_{JN}}
    {\ln \left(A_N A'_N B_N B'_N \prod\limits_{K=1}^{N-1}
    X_{KN}\right)}~,\label{4.34}\\
  \ln B^{(N-1)}_J  &=& \ln B_J  + \frac{\ln B_N \ln X_{JN}}
    {\ln \left(A_N A'_N B_N B'_N \prod\limits_{K=1}^{N-1}
    X_{KN}\right)}~,\label{4.35}\\
  \ln B'^{(N-1)}_J &=& \ln B'_J + \frac{\ln B'_N \ln X_{JN}}
    {\ln \left(A_N A'_N B_N B'_N \prod\limits_{K=1}^{N-1}
    X_{KN}\right)}~, \label{4.36}\\
  \ln X^{(N-1)}_{IJ} &=& \ln X_{IJ} + \frac{\ln X_{IN} \ln X_{JN}}
    {\ln \left(A_N A'_N B_N B'_N \prod\limits_{K=1}^{N-1}
    X_{KN}\right)}~, \label{4.37}\end{aligned}$$ and “$\cdots$” denotes the terms which are independent of the loop momenta $n^{(J)}$. They play no essential role in taking the limit and henceforth we omit those terms. Note that the second exponential factor in eq.(\[4.32\]) takes the same form as that in eq.(\[4.31\]) up to a replacement $N\rightarrow
N-1$. Furthermore, completing the squares for $n^{(N-1)}, n^{(N-2)}$, etc., successively, we obtain, $$\begin{aligned}
S_o&=& \left( \frac{1}{R_s} \right)^N \sum_{n^{(1)}, \cdots , n^{(N)}}
   \exp  \left[ \alpha'\ln \left(A_N A'_N B_N B'_N \prod_{K=1}^{N-1}
    X_{KN} \right) \right.\label{4.38}   \\
  &&   \times \Bigg( \frac{n^{(N)}}{R_s} -
 \frac{n_2 \ln A_N + n_4 \ln A'_N + n_1 \ln B_N + n_3 \ln B'_N
     +\sum\limits_{K=1}^{N-1} n^{(K)} \ln X_{KN} }
    {R_s \ln \left(A_N A'_N B_N B'_N \prod\limits_{K=1}^{N-1} X_{KN}
        \right)} \Bigg)^2\, \Bigg]\nonumber\\
 &&  \times \exp \left[\alpha'\ln \left(A^{(N-1)}_{N-1}
     A'^{(N-1)}_{N-1} B^{(N-1)}_{N-1} B'^{(N-1)}_{N-1}
    \prod_{K=1}^{N-2}X^{(N-1)}_{KN-1} \right)\right.\nonumber\\
  && \times \Bigg(\frac{n^{(N-1)}}{R_s} -
     \frac{n_2 \ln A^{(N-1)}_{N-1} + \cdots + n_3 \ln B'^{(N-1)}_{N-1}
     + \sum\limits_{K=1}^{N-2} n^{(K)} \ln X^{(N-1)}_{KN-1} }
    {R_s \ln \Bigl(A^{(N-1)}_{N-1} A'^{(N-1)}_{N-1} B^{(N-1)}_{N-1}
           B'^{(N-1)}_{N-1} \prod\limits_{K=1}^{N-2}X^{(N-1)}_{KN}
        \Bigl)} \Bigg)^2 \,\Bigg]\nonumber\\
  && \quad\vdots \nonumber \\
  &&  \times \exp \Biggl[\alpha' \ln \left(A^{(1)}_{1} A'^{(1)}_{1}
    B^{(1)}_{1} B'^{(1)}_{1}\right) \nonumber\\
  &&  \times \left. \left( \frac{n^{(1)}}{R_s} -
     \frac{n_2 \ln A^{(1)}_{1} +n_4 \ln A'^{(1)}_1 +n_1 \ln B^{(1)}_1
     + n_3 \ln B'^{(1)}_{1}}
      {R_s \ln \left(A^{(1)}_{1} A'^{(1)}_{1} B^{(1)}_{1} B'^{(1)}_{1}
         \right) } \right)^2  \right]. \nonumber\end{aligned}$$ Here, similarly to eqs.(\[4.33\])$\sim$(\[4.37\]), $A_J^{(L)}, A'^{(L)}_J, B_J^{(L)}, B'^{(L)}_J\,(1\leq J\leq L\leq
N-1)$ and $X_{IJ}^{(L)}\,(1\leq I<J\leq L\leq N-1)$ are given by, $$\begin{aligned}
  \ln A^{(L)}_J&=&\ln A^{(L+1)}_J  + \frac{\ln A^{(L+1)}_{L+1}
    \ln X^{(L+1)}_{JL+1}} {\ln \left(A^{(L+1)}_{L+1}
    A'^{(L+1)}_{L+1} B^{(L+1)}_{L+1} B'^{(L+1)}_{L+1}
    \prod\limits_{K=1}^{L}X^{(L+1)}_{KL+1}\right)}~,\label{4.39}\\
  \ln A'^{(L)}_J&=&\ln A'^{(L+1)}_J  + \frac{\ln A'^{(L+1)}_{L+1}
    \ln X^{(L+1)}_{JL+1}} {\ln \left(A^{(L+1)}_{L+1}
    A'^{(L+1)}_{L+1} B^{(L+1)}_{L+1} B'^{(L+1)}_{L+1}
    \prod\limits_{K=1}^{L}X^{(L+1)}_{KL+1}\right)}~,\label{4.40}\\
  \ln B^{(L)}_J&=&\ln B^{(L+1)}_J  + \frac{\ln B^{(L+1)}_{L+1}
    \ln X^{(L+1)}_{JL+1}} {\ln \left(A^{(L+1)}_{L+1}
    A'^{(L+1)}_{L+1} B^{(L+1)}_{L+1} B'^{(L+1)}_{L+1}
    \prod\limits_{K=1}^{L}X^{(L+1)}_{KL+1}\right)}~,\label{4.41}\\
  \ln B'^{(L)}_J&=&\ln B'^{(L+1)}_J  + \frac{\ln B'^{(L+1)}_{L+1}
    \ln X^{(L+1)}_{JL+1}} {\ln \left(A^{(L+1)}_{L+1}
    A'^{(L+1)}_{L+1} B^{(L+1)}_{L+1} B'^{(L+1)}_{L+1}
    \prod\limits_{K=1}^{L}X^{(L+1)}_{KL+1}\right)}~,\label{4.42}\\
  \ln X^{(L)}_{IJ}&=&\ln X^{(L+1)}_{IJ}  + \frac{\ln X^{(L+1)}_{IL+1}
    \ln X^{(L+1)}_{JL+1}} {\ln \left(A^{(L+1)}_{L+1}
    A'^{(L+1)}_{L+1} B^{(L+1)}_{L+1} B'^{(L+1)}_{L+1}
    \prod\limits_{K=1}^{L}X^{(L+1)}_{KL+1}\right)}~,\label{4.43}\end{aligned}$$ where $A^{(N)}_J = A_J,\, A'^{(N)}_J = A'_J,\, B^{(N)}_J = B_J,\,
B'^{(N)}_J = B'_J$ and $X_{IJ}^{(N)} = X_{IJ}$. Similarly to eq.(\[3.12\]) in section 3, we find that $S_o$ becomes to have $N$ $\delta$-functions in the L$^3$, $$\begin{aligned}
 \lim_{\epsilon \rightarrow 0} \,S_o &=& \sum_{n^{(1)}, \cdots , n^{(N)}}
    \prod_{L=1}^N \left( \frac{\pi}{-\alpha'\ln \left(A^{(L)}_{L}
        A'^{(L)}_{L} B^{(L)}_{L} B'^{(L)}_{L}
       \prod\limits_{K=1}^{L-1}X^{(L)}_{KL} \right)}\right)^{\frac{1}{2}}
     \label{4.44}\\
  &&\times\prod_{L=1}^N
    \delta \left( n^{(L)} -\frac{n_2 \ln A^{(L)}_{L} +
    n_4 \ln A'^{(L)}_{L} +n_1 \ln B^{(L)}_{L} + n_3 \ln B'^{(L)}_{L}
       + \sum\limits_{K=1}^{L-1}n^{(K)} \ln X^{(L)}_{KL} }
    {\ln \left(A^{(L)}_{L} A'^{(L)}_{L} B^{(L)}_{L} B^{(L)}_{L}
       \prod\limits_{K=1}^{L-1}X^{(L)}_{KL}\right) } \right) \nonumber.\end{aligned}$$ Furthermore, after some calculations using eqs.(\[4.39\])$\sim$(\[4.43\]), eq.(\[4.44\]) is rewritten in a more symmetrical form, $$\begin{aligned}
  \lim_{\epsilon\rightarrow0} \,S_o &=& \sum_{n^{(1)}, \cdots , n^{(N)}}
    \prod_{L=1}^{N}
    \left( \frac{-\pi\ln \left(A^{(L)}_{L} A'^{(L)}_{L} B^{(L)}_{L}
      B'^{(L)}_{L} \prod\limits_{K=1}^{L-1}X^{(L)}_{KL} \right)}
      {\alpha'}\right)^{\frac{1}{2}}\nonumber\\
 &&\times \prod_{L=1}^{N}
    \delta \left( n^{(L)}\ln\left[A_{L}A'_{L}B_{L}B'_{L}
          \prod_{K=1}^{L-1} X_{KL} \prod_{K=L+1}^{N}X_{LK}\right]
    - n_2 \ln A_{L} - n_4 \ln A'_{L} \right.\nonumber\\
 && \hspace{7ex}\left. -n_1 \ln B_{L} - n_3 \ln B'_{L}
       - \sum_{K=1}^{L-1} n^{(K)} \ln X_{KL} - \sum_{K=L+1}^{N}n^{(K)}
    \ln X_{LK} \right).\end{aligned}$$ Then it is obvious that the condition $n_1=n_2=n_3=n_4$, i.e., all the external momenta are zero, is necessary in order that the pathological $\delta(0)$ appears. Hence such a divergence never appears in open string multi-loop scattering amplitudes.

Next, we consider the L$^3$ in closed string theory. The story is essentially the same as open string theory, however, there are two new features in closed string theory, i.e., the existence of winding modes and the complexification of parameters.

On the $M_{25}\times S^1$, both the external and the loop momenta along the compact direction can have non-vanishing winding numbers. For simplicity, however, we assume that the external string states have vanishing winding numbers [@Bil1]. Under this assumption, the external and the loop momenta along the compact direction are written by, $$\begin{aligned}
  p_i^1&=&p_{Ri}^1 = p_{Li}^1 = \frac{n_i}{R_s},\\
  k_{RJ}^1&=&\frac{n^{(J)}}{R_s} - \frac{m^{(J)}R_s}{\alpha'},\\
  k_{LJ}^1&=&\frac{n^{(J)}}{R_s} + \frac{m^{(J)}R_s}{\alpha'},\end{aligned}$$ where the suffices $R$ and $L$ represent right and left movers of string states, respectively and $m^{(J)}$ is the winding number of the string in loop $J$. Furthermore, on the $M_{25}\times S^1$, the integration over $k_J^1$ becomes a sum over $n^{(J)}$ and we must also sum up over $m^{(J)}$. Then the amplitude $I_4^{(N)}$ in (\[4.29\]) is given by, $$\begin{aligned}
  I_4^{(N)}&=&\frac{\kappa^{2N+2}}{(4\pi)^{3N+1}}
        \left(\frac{1}{R_s}\right)^N
        \sum_{n^{(1)}, \cdots ,n^{(N)}} \sum_{m^{(1)},\cdots,m^{(N)}}
    \int \prod_{J=1}^{N} d^{25}k'_J
    \int_{|x_j| \leq 1} \prod_{j=1}^{3N+1}d^2x_j\,|G|\nonumber\\
  &&\times |D|^{-\alpha_C(-(p_1-p_3)^2)-2}
           |C|^{-\alpha_C(-(p_2-p_4)^2)-2}
           \prod_{J=1}^N \bigg|A_J A'_J B_J B'_J
           \prod_{K=J+1}^N X_{JK}\bigg|^{-4}\nonumber \\
  &&  \times   \exp \left[ \frac{\alpha'}{4} \sum_{J=1}^{N} \Biggl\{
         \left((k_{RJ}^1 - p_2^1)^2 +(k'_J-p'_2)^2 \right)\ln A_J
      +  \left((k_{LJ}^1 - p_2^1)^2 +(k'_J-p'_2)^2 \right)\ln \bar{A}_J
          \right. \nonumber\\
  &&\quad+ \left((k_{RJ}^1 - p_4^1)^2 +(k'_J-p'_4)^2 \right)\ln A'_J
        +\left((k_{LJ}^1 - p_4^1)^2 +(k'_J -p'_4)^2 \right)\ln \bar{A'}_J
              \nonumber\\
  &&\quad+\left((k_{RJ}^1 - p_1^1)^2 +(k'_J-p'_1)^2 \right)\ln B_J
         +\left((k_{LJ}^1 - p_1^1)^2 +(k'_J-p'_1)^2 \right)\ln \bar{B}_J
                  \nonumber\\
  &&\quad+\left((k_{RJ}^1 - p_3^1)^2 +(k'_J-p'_3)^2 \right)\ln B'_J
         +\left((k_{LJ}^1 - p_3^1)^2 +(k'_J-p'_3)^2 \right)\ln \bar{B'}_J
                  \nonumber\\
  &&\quad+\sum_{K=J+1}^{N}\left((k_{RJ}^1 - k_{RK}^1)^2 +(k'_J
           -k'_K)^2 \right) \ln X_{JK}\nonumber\\
  &&\quad+ \left.\sum_{K=J+1}^{N}\left((k_{LJ}^1 - k_{LK}^1)^2
    +(k'_J - k'_K)^2\right)\ln \bar{X}_{JK}\Biggr\} \right],\end{aligned}$$ where $(p'_i)=(p_i^0,p_i^2,p_i^3,\cdots,p_i^{25})$ and $(k'_J)=(k^0_J,k_J^2,k_J^3,\cdots,k_J^{25})$. Similarly to eq.(\[4.31\]), henceforth we only concentrate on the following $R_s$ dependent part $S_c$ of the amplitude: $$\begin{aligned}
S_c&=&\left(\frac{1}{R_s} \right)^N \sum_{n^{(1)},\cdots, n^{(N)}}
                                    \sum_{m^{(1)},\cdots, m^{(N)}}
    \exp \left[  \frac{\alpha'}{4} \sum_{J=1}^{N} \left\{
          \left(\frac{n^{(J)}}{R_s}- \frac{m^{(J)}R_s}{\alpha'}
           -\frac{n_2}{R_s} \right)^2 \ln A_J  \right. \right.\nonumber\\
&&    +   \left(\frac{n^{(J)}}{R_s}+ \frac{m^{(J)}R_s}{\alpha'}
               -\frac{n_2}{R_s} \right)^2 \ln \bar{A}_J
      +   \left(\frac{n^{(J)}}{R_s}- \frac{m^{(J)}R_s}{\alpha'}
               -\frac{n_4}{R_s}\right)^2 \ln A'_J \nonumber\\
&&    +   \left(\frac{n^{(J)}}{R_s}+ \frac{m^{(J)}R_s}{\alpha'}
               -\frac{n_4}{R_s}\right)^2 \ln \bar{A'}_J
      +   \left(\frac{n^{(J)}}{R_s}- \frac{m^{(J)}R_s}{\alpha'}
               -\frac{n_1}{R_s}\right)^2 \ln B_J \nonumber \\
&&    +   \left(\frac{n^{(J)}}{R_s}+ \frac{m^{(J)}R_s}{\alpha'}
               -\frac{n_1}{R_s}\right)^2 \ln \bar{B}_J
      +   \left(\frac{n^{(J)}}{R_s}- \frac{m^{(J)}R_s}{\alpha'}
               -\frac{n_3}{R_s}\right)^2 \ln B'_J\nonumber \\
&&    +   \left(\frac{n^{(J)}}{R_s}+ \frac{m^{(J)}R_s}{\alpha'}
               -\frac{n_3}{R_s}\right)^2 \ln \bar{B'}_J\nonumber\\
&&    + \sum_{K=J+1}^{N}
         \left( \left(\frac{n^{(J)}}{R_s}- \frac{m^{(J)}R_s}{\alpha'}
               -\frac{n^{(K)}}{R_s}
               +\frac{m^{(K)}R_s}{\alpha'}  \right)^2  \ln X_{JK}\right.
               \nonumber\\
&&\hspace{10ex}+ \left. \left. \left.
           \left (\frac{n^{(J)}}{R_s}+ \frac{m^{(J)}R_s}{\alpha'}
             -\frac{n^{(K)}}{R_s}-\frac{m^{(K)}R_s}{\alpha'}\right)^2
         \ln \bar{X}_{JK}\right) \right\} \right] \nonumber\\
&=&  \left(\frac{1}{R_s} \right)^N \sum_{n^{(1)},\cdots, n^{(N)}}
      \exp \left[  \frac{\alpha'}{2} \sum_{J=1}^{N} \left\{
         \left(\frac{n^{(J)}}{R_s} -\frac{n_2}{R_s} \right)^2
         \mbox{Re}\left(\ln A_J \right)  \right. \right.\nonumber \\
&&    +  \left(\frac{n^{(J)}}{R_s} -\frac{n_4}{R_s} \right)^2
         \mbox{Re}\left(\ln A' \right)
      +  \left(\frac{n^{(J)}}{R_s} -\frac{n_1}{R_s} \right)^2
         \mbox{Re}\left(\ln B_J \right) \nonumber \\
&&    +  \left(\frac{n^{(J)}}{R_s} -\frac{n_3}{R_s} \right)^2
         \mbox{Re}\left(\ln B'_J \right)
     +  \sum_{K=J+1}^{N} \left.\left.
         \left(\frac{n^{(J)}}{R_s}-\frac{n^{(K)}}{R_s}\right)^2
         \mbox{Re}\left(\ln X_{JK} \right) \right\}\right]
         \tilde{S_c}  \label{eqScl}\,,\end{aligned}$$ where, $$\begin{aligned}
  \tilde{S_c}&=& \sum_{m^{(1)},\cdots, m^{(N)}} \exp
    \left[ -\pi\, \vec{m}^T M^{-1} \vec{m} + 2\pi i
    \,\vec{m}^T \vec{x}\right],\\
  M^{-1}&=&\frac{R_s^2}{2\pi \alpha'}\,
    \left[\begin{array}{lllll}
      a_{11}& a_{12} & & \cdots     &a_{1N}\\
            & a_{22} & a_{23}&\cdots&a_{2N}\\
            &        & \ddots&      &\vdots\\
            &\hbox{\Large 0}&  &\ddots& a_{N-1N}\\
        &        &       &      & a_{NN}
    \end{array}\right],\label{4.52}\\
  \vec{m}&=& \left[\begin{array}{c}
    m^{(1)} \\  m^{(2)} \\  \vdots  \\  m^{(N)} \\
        \end{array}\right],\hspace{5ex}
    \vec{x}= \left[\begin{array}{c}
          x^{(1)} \\  x^{(2)} \\  \vdots  \\  x^{(N)} \\
        \end{array}\right],\\
  a_{JJ}&=& -\mbox{Re} \left\{\ln \left( A_J A'_J B_J B'_J
           \prod_{K=1}^{J-1}X_{KJ}\prod_{K=J+1}^{N}X_{JK}\right)\right\},\\
  a_{JK}&=& 2\,\mbox{Re}\,(\ln X_{JK}),\quad a_{KJ}=0, \hspace{5ex}(J<K),\\
  x^{(J)}&=&-\frac{1}{2\pi}\Biggl[
    n^{(J)} \mbox{Im} \left\{\ln \left( A_{J} A'_{J} B_{J} B'_{J}
        \prod_{K=1}^{J-1} X_{KJ} \prod_{K=J+1}^{N}X_{JK}\right)\right\}
    -n_2 \mbox{Im} (\ln A_{J})
    \nonumber\\
  && - n_4 \mbox{Im}(\ln A'_{J}) -n_1 \mbox{Im} (\ln B_{J})
    - n_3 \mbox{Im}(\ln B'_{J}) \nonumber\\
  && - \sum_{K=1}^{J-1} n^{(K)}  \mbox{Im}(\ln X_{KJ})
       - \sum_{K=J+1}^{N} n^{(K)}  \mbox{Im}(\ln X_{JK})\Biggr].\end{aligned}$$ Note that the (first) exponential factor in eq.(\[eqScl\]) takes the same form as that in eq.(\[4.31\]) by replacing $\alpha' \rightarrow \frac{\alpha'}{2}$ and $\ln  \rightarrow
\mbox{Re}\ln$. Hence, the remaining problem is to take the L$^3$ of $\tilde{S_c}$. This can be solved by using the Poisson resummation formula [@Bil1]: $$\sum_{m^{(1)},\cdots, m^{(N)}} \exp
       \left[ -\pi \vec{m}^T M^{-1} \vec{m} + 2\pi i \vec{m}^T
        \vec{x}\right]
  =  \sum_{m^{(1)},\cdots, m^{(N)}}(\det M)^{\frac{1}{2}}\,
     \exp \left[-\pi(\vec{m}+\vec{x})^{T} M (\vec{m} +\vec{x})\right].$$ Note that the $N\times N$ matrix $M$ is proportional to $\frac{1}{R_s^2}$ due to eq.(\[4.52\]) and the $(\det
M)^{\frac{1}{2}}$ on the r.h.s. gives $(\frac{1}{R_s})^N$ factor. Hence, after some calculations, we also obtain $N$ $\delta$-functions in the L$^3$: $$\begin{aligned}
 \lim_{\epsilon \rightarrow 0} \,\tilde{S_c}&=&
    (2\pi)^N\sum_{m^{(1)}, \cdots , m^{(N)}}
    \prod_{L=1}^{N} \delta \left( n^{(L)} \mbox{Im}\left\{
            \ln \left( A_{L} A'_{L} B_{L} B'_{L}
            \prod_{K=1}^{L-1} X_{KL} \prod_{K=L+1}^{N}X_{LK}
            \right)\right\} \right.\nonumber\\
  && \quad- n_2 \mbox{Im}(\ln A_{L}) - n_4 \mbox{Im}(\ln A'_{L})
     - n_1 \mbox{Im}(\ln B_{L}) - n_3 \mbox{Im}(\ln B'_{L})\nonumber\\
  &&\quad\left. - \sum_{K=1}^{L-1} n^{(K)} \mbox{Im}(\ln X_{KL})
        - \sum_{K=L+1}^{N} n^{(K)} \mbox{Im}(\ln X_{LK})
          - 2 \pi  m^{(L)}\right).\end{aligned}$$ Thus, putting those together, we find that the $S_c$ gives $N$ complex $\delta$-functions in the L$^3$, $$\begin{aligned}
 \lim_{\epsilon \rightarrow 0} \,S_c&=& (2 \pi)^N\sum_{n^{(1)},
    \cdots, n^{(N)}}\sum_{m^{(1)}, \cdots , m^{(N)}}
    \prod\limits_{L=1}^{N} \left( \frac{-2\pi \mbox{Re}
           \ln \left(A^{(L)}_{L} A'^{(L)}_{L} B^{(L)}_{L}
            B'^{(L)}_{L} \prod\limits_{K=1}^{L-1}X^{(L)}_{KL} \right)}
           {\alpha'}\right)^{\frac{1}{2}}\nonumber\\
  &&\times \prod_{L=1}^{N} \delta^{(2)} \left( n^{(L)}
    \ln \left( A_{L} A'_{L} B_{L} B'_{L} \prod_{K=1}^{L-1} X_{KL}
     \prod_{K=L+1}^{N}X_{LK}\right) - n_2 \ln A_{L}
    \right.\nonumber\\
  &&\quad - n_4 \ln A'_{L}   -n_1 \ln B_{L} - n_3 \ln B'_{L} \nonumber\\
  && \quad\left.  - \sum_{K=1}^{L-1} n^{(K)} \ln X_{KL}
       - \sum_{K=L+1}^{N}n^{(K)} \ln X_{LK}
                        - 2 \pi i m^{(L)}\right).\end{aligned}$$ From the above equation, it is also obvious that $n_1=n_2=n_3=n_4$ is a necessary condition for an appearance of the pathological $\delta(0)$. Hence such a divergence does not occur in closed string multi-loop scattering amplitudes.

We have examined the L$^3$ of multi-loop open and closed string scattering amplitudes. In conclusion, [*since by construction these amplitudes consist of all the possible propagators in the field theoretical Feynman diagrams, which is the root of s-t channel duality, we can take the L$^3$ of these scattering amplitudes, i.e., the zero-mode loop divergences do not appear.*]{} Note that it is obvious that the nonexistence of the $\delta(0)$ is not essentially due to the existence of the winding modes (cf. [@Bil1]).

The L$^3$ of vacuum amplitudes in string theory
===============================================

In this section, we discuss the L$^3$ of vacuum amplitudes in bosonic string theory. For simplicity, we restrict ourselves to bosonic closed string theory.

In the previous section, we have seen that the factor $\frac{1}{R_s^2}\sim\frac{1}{\epsilon^2}$ has come out for each loop from the integration measure of the compactified loop momentum and the Poisson resummation of the winding modes. This factor has combined with an appropriate exponential function $\exp\left(-\left(\frac{\pi}{R_s^2}\right)\vert\cdots\vert^2\right)$ to give a complex $\delta$-function for each loop. We have shown that this complex $\delta$-function can become $\delta^{(2)}(0)$ iff all the external momenta in the compact direction $P_i^1=\frac{N_i}{R_s}$ are zero at multi-loop order. Thus, this pathological $\delta^{(2)}(0)\sim\frac{1}{R_s^2}$ does not appear in the multi-loop string scattering amplitudes. As for the vacuum amplitudes which have no external lines, however, it is obvious that this pathological situation inevitably occurs, i.e., the zero-mode loop divergences always appear. In fact, as is well known, in a T-dual theory this $\delta^{(2)}(0)\sim \frac{1}{R_s^2}$ for each loop is absorbed in the T-dual coupling constant $\tilde{\kappa}=e^{\tilde{\phi}}$, $$\tilde{\kappa}^2=\kappa^2\frac{\alpha'}{R_s^2},$$ where $\kappa$ is the original closed string coupling constant. In the L$^3$, since we shall keep the original string coupling constant fixed, this dual coupling constant diverges. Thus the zero-modes become strongly coupled and the cosmological constant can not be determined perturbatively. In particular, since the closed string theory includes gravity, we shall take this problem seriously. Of course, we should notice that the M-theory or superstring does not suffer from this problem due to supersymmetry, and neither does the direct DLCQ theory due to the triviality of the vacuum.

Conclusion and Discussion
=========================

In this paper, we have argued whether perturbative string multi-loop amplitudes have well-defined light-like limit or not. To answer this question, we have used the multi-loop string scattering amplitudes constructed by the method of [@KSV]. In this method, since we can keep the integral, or sum, over the loop momenta, we have shown that the perturbative string multi-loop scattering amplitudes have well-defined light-like limit, which was conjectured in ref.[@Bil2]. This result depends on the striking stringy nature. In fact, we must add all possible Feynman diagrams in a certain order of perturbation theory and the zero-mode loop diagrams among them diverge as $\delta(0)$ in field theory, while in string theory, since a scattering amplitude in a certain order of perturbation contains the resonances in the all possible field theoretical Feynman diagrams in the same order of perturbation, the zero-mode loop divergences do not appear.

We also have discussed the light-like limit of the vacuum amplitudes in bosonic string theory. These amplitudes obviously diverge as $\delta(0)$ even in string theory. Therefore, we need supersymmetry for this limit to be well-defined. It seems that this result prevents us from extracting the information of supersymmetry breaking in M-theory or superstring from the finite $N$ Matrix theory. In fact, it seems that the existence of non-supersymmetric Matrix string theory is questionable [@BM].

KSV construction of $N$-point tree amplitude
============================================

In this appendix, we calculate the tachyon $N$-point tree amplitude in bosonic open string theory by the KSV method [@KSV]. We shall determine the form of the function $y=f(x;a_1,a_2,a_3,a_4)$, i.e., (\[4.17\])$\sim$(\[4.19\]), by comparing the eight-point amplitude by the KSV method with that in the operator formalism. Then by using this function, we will show that the $N$-point tree amplitude constructed by the KSV method agrees with the one in the operator formalism. Similarly to the four-point amplitude in section \[secMLST\], we consider the dual diagrams to the field theoretical Feynman diagrams connected by s-t channel duality for the $N$-point tree amplitude. That is, we consider the $N$-polygons defined by $N$ sides which are dual to the $N$ external lines. Each diagonal of the $N$-polygon is dual to a propagator. In each dual diagram, we always have $N-3$ diagonals which correspond to the $N-3$ propagators in the field theoretical Feynman diagram.

We define the dual external momenta $p_i$ ($i=1,\cdots,N$) as follows (see (\[3.2\]) $\sim $ (\[3.5\])): $$\begin{array}{lll}
    P_1&=& p_1 - p_N ,\\
    P_2&=& p_2 - p_1 ,\\
    &&\ldots \\
    P_N&=& p_N - p_{N-1},
    \end{array}\label{eqDM}$$ and we assign $p_i$ ($i =1,\cdots,N$) to the $N$ vertices of the $N$-polygon, respectively (Figure \[figDDNT\]). The $\frac{N(N-3)}{2}$ lines connecting the vertices $p_{i-1}$ and $p_j$ ($2\leq i<j\leq N-1,\, 3\leq i<j=N$) are denoted by $y_{i-1\,j}$ and we choose $N-3$ independent lines $x_i\equiv y_{i+1N}$ ($i=1,\cdots,N-3$) among them (Figure \[figDDNT\]). Then we can write down the $N$-point tree amplitude by the KSV method: $$A_{N}^{(0)} = g^{N-2}\int _0^1 \prod_{i=1}^{N-3} dx_i\,
    \rho^{(0)}(x_i)
    \prod_{i=1}^{N-3}x_i^{-\alpha(-(p_{i+1}-p_N )^2)-1} \prod_{2
    \leq i <j \leq N-1 } y_{i-1j}^{-\alpha(-(p_{i-1}-p_j
    )^2)-1}\label{eqNKSV},$$ where $\rho^{(0)}(x_i)$ is determined by the requirement that the integral volume element maintains the crossing symmetry[@KaN1; @KaN2], $$\rho^{(0)}(x_i)=
    \frac{1}{\prod\limits_{2\leq i<N-2}(1-x_ix_{i-1})}.$$

Next we shall determine the forms of the functions $y_{i-1j}(x_k)\,
(2\leq i<j\leq N-1)$ and compare (\[eqNKSV\]) with the tree amplitude in the operator formalism. In the operator formalism the $N$-point tree amplitude is given by [@Sch], $$A_{Nop}^{(0)} = g^{N-2}\int_0^1 \prod_{i=1}^{N-3} dx_i\,
    \prod_{i=1}^{N-3}x_i^{-\alpha({s_i} )-1}
    \prod_{2 \leq i <j \leq N-1 }
    (1-x_{ij})^{2\alpha' (P_i \cdot  P_j)},\label{eqNTO}$$ where $$\begin{aligned}
    x_{ij}&=& x_{i-1}x_i\cdots x_{j-2},\\
    s_i&=& - (P_1 + \ldots +P_{i+1})^2,\end{aligned}$$ From eqs.(\[eqDM\]) we obtain, $$\begin{aligned}
    (p_{i-1} -p_j)^2 &=& (P_i + \ldots +P_j)^2,\\
    s_i &=& -(p_{i+1} -p_N)^2.\end{aligned}$$ Then, after some calculations, (\[eqNTO\]) is rewritten by, $$\begin{aligned}
  A_{Nop}^{(0)}&=& g^{N-2}\int_0^1 \prod_{i=1}^{N-3} dx_i
    \rho^{(0)} (x_i)
    \prod_{i=1}^{N-3}x_i^{-\alpha(-(p_{i+1}-p_N)^2)-1}
    \nonumber\\
  &&\times \prod_{2\leq i <j\leq N-1}
    \left(\frac{1-x_{ij}}{1-x_{i-1j}}\,
    \frac{1-x_{i-1j+1}}{1-x_{ij+1}}
    \right)^{-\alpha (-(p_{i-1}-p_j)^2)-1}\label{eqNTO2}.\end{aligned}$$ Here, we have used the equations which $x_0= x_{N-2}=0$, i.e., $x_{1j} = x_{iN}=0$.

Now we concentrate on the $N=8$ case and determine the form of the function (\[4.9\]), $y=f(x;a_1,a_2,a_3,a_4)$. We can easily see that (\[eqNKSV\]) agrees with (\[eqNTO2\]) if $$y_{i-1j}= \frac{1-x_{ij}}{1-x_{i-1j}}\,
    \frac{1-x_{i-1j+1}}{1-x_{ij+1}}\label{eqGY}.$$ For example, the lines $y_{26}$, $y_{24}$ and $y_{46}$ are written by, $$\begin{aligned}
  y_{26}&=&\frac{1-x_2x_3x_4}{1-x_1x_2x_3x_4}\,
    \frac{1-x_1x_2x_3x_4x_5}{1-x_2x_3x_4x_5},\\
  y_{24}&=&\frac{1-x_2}{1-x_1x_2}\,\frac{1-x_1x_2x_3}{1-x_2x_3},\\
  y_{46}&=&\frac{1-x_4}{1-x_3x_4}\,\frac{1-x_3x_4x_5}{1-x_4x_5}.\end{aligned}$$ In Figure \[figET\] we see that the line $y_{26}$ is a diagonal of the quadrilateral whose sides are ($x_1,y_{24}, y_{46},x_5$) and the other diagonal of this quadrilateral is $x_3$. And hence we can write, $$y_{26}= f(x_3 ; x_1,y_{24},y_{46}, x_5).$$ Then we have obtained eqs.(\[4.17\])$\sim$(\[4.19\]). Note that when $N < 8$, we cannot draw a quadrilateral whose all sides are diagonals of the $N$-polygon of the dual diagram.

Next by using these eqs.(\[4.17\])$\sim$(\[4.19\]), we show that (\[eqGY\]) holds for an arbitrary $N$, i.e., the $N$-point tree amplitude by the KSV method agrees with that in the operator formalism. Let us consider the lines $y_{i-1i+1}\ (2\leq i\leq N-2)$ in Figure \[figDDNT\]. Since $y_{i-1i+1}$ is a diagonal of the quadrilateral whose sides are ($x_{i-2},0,0,x_{i}$) with the other diagonal $x_{i-1}$, we have $$\begin{aligned}
  y_{i-1i+1}&=& f(x_{i-1}; x_{i-2},0,0,x_i)\nonumber\\
    &=& \frac{1-x_{i-1}}{1-x_{i-1}x_{i-2}}
    \frac{1-x_{i-2}x_{i-1}x_i}{1-x_{i-1}x_{i}}.\end{aligned}$$ Then we will prove (\[eqGY\]) by mathematical induction [@KSV]. Suppose the equation holds for $y_{i-1 j}$. Since the line $y_{i-1j+1}$ is a diagonal of the quadrilateral whose sides are ($x_{i-2},y_{i-1j},0,x_j$), we have $$\begin{aligned}
  y_{i-1j+1} &=& f (x_{j-1}; x_{i-2},y_{i-1j},0,x_j)\nonumber\\
  &=& \frac{1-x_{j-1} \alpha}{1-x_{j-1}\alpha x_{i-2}}
    \frac{1-x_{j-1}\alpha x_{i-2}x_j}{1-x_{j-1}\alpha
      x_{j}}\label{eqGY2}.\end{aligned}$$ where $\alpha$ is defined implicitly by, $$y_{i-1j} = \frac{1-\alpha}{1-\alpha x_{i-2}}\,
    \frac{1-\alpha x_{j-1} x_{i-2}}{1-\alpha x_{j-1}}\,.$$ Here by the assumption of mathematical induction, we obtain, $\alpha = x_{ij}=x_{i-1} \cdots x_{j-2}$. Plugging this into (\[eqGY2\]), we obtain $y_{i-1j+1}=\frac{1-x_{ij+1}}{1-x_{i-1j+1}}\,
\frac{1-x_{i-1j+2}}{1-x_{ij+2}}$. Hence we have proven (\[eqGY\]) for an arbitrary $N$.

KSV construction of $N$-point one-loop amplitude
================================================

In this appendix we construct the tachyon $N$-point one-loop amplitude in bosonic open string theory by the KSV method [@KSV]. Each dual diagram to the field theoretical Feynman diagram is a $N$-polygon with an internal point and $N$ internal lines which connect either two of the vertices or a vertex and an internal point properly and do not intersect each other. The $N$ sides of the polygon and the internal point are dual to the $N$ external lines and the loop in the Feynman diagram, respectively. In order to construct the amplitude we shall consider all topologically inequivalent lines connecting either two of the vertices or a vertex and the internal point of the $N$-polygon. As in the previous appendix, we assign the dual momenta $p_i\,
(i=1,\cdots,N)\, (\ref{eqDM})$ to the $N$ vertices of the $N$-polygon and we call the internal point $o_1$. The $N$ lines connecting the vertices $p_i$ and the internal point $o_1$, which do not exist in the tree case, are denoted by $x_i$. As for the lines connecting two vertices of the $N$-polygon, a pair of vertices do not uniquely determine a line in this case due to the existence of the point $o_1$ and the key to constructing the one-loop amplitude is to completely determine those lines.

From $p_r$ to $p_s$ we draw a line clockwise with the point $o_1$ kept to the right. So we can draw two lines for each pair of the vertices $p_r$ and $p_s$. One is from $p_r$ to $p_s$ and the other is from $p_s$ to $p_r$. They are denoted by $y^{(0)}_{rs}$ and $y^{(0)}_{sr}$, respectively (see Figure \[figNOL\]). Note that without the point $o_1$, i.e., for the tree amplitude, $y^{(0)}_{rs}$ and $y^{(0)}_{sr}$ are (topologically) equivalent, but this is not the case with $o_1$. We should notice that $y^{(0)}_{r r+1}$ is a side of the polygon and hence $y^{(0)}_{r r+1}=0$, but $y^{(0)}_{r+1 r}$ is not. Furthermore we can draw a non-trivial line from $p_r$ going around the point $o_1$ and coming back to $p_r$, which is denoted by $u^{(0)}_r$ (Figure \[figNOL\]).

As in the tree case, $x_i$, $y^{(0)}_{rs}, y^{(0)}_{sr},
y^{(0)}_{r+1 r}$ and $u^{(0)}_r$ are not independent and we can take $x_i \, (i=1,\cdots,N)$ as independent parameters and determine $y^{(0)}_{rs}, y^{(0)}_{sr}, y^{(0)}_{r+1 r} $ and $u^{(0)}_r$ as functions of $x_i$. We explain how to obtain $y^{(0)}_{rs} = y^{(0)}_{rs}(x_i)\,
(1\leq r<s-1<N\,,\,(r,s)\neq(1,N))$ below. We shall see that $y^{(0)}_{rs}$ is a diagonal of a quadrilateral $o_1p_rp_{s-1}p_s$ whose sides are $(x_r,y^{(0)}_{rs-1},0,x_s)$ and hence we have, $$y^{(0)}_{rs}= f(x_{s-1}; x_r,y^{(0)}_{rs-1},0,x_s)\,.$$ where $f$ is defined by (\[4.17\])$\sim$(\[4.19\]). Similarly $y^{(0)}_{rs-1}$ is a diagonal of a quadrilateral $o_1p_rp_{s-2}p_{s-1}$, $y^{(0)}_{rs-2}$ is a diagonal of a quadrilateral $o_1p_rp_{s-3}p_{s-2}$ and so on. Then we have the following series of equations: $$\begin{aligned}
  y^{(0)}_{rs-1}&=&f(x_{s-2}; x_r,y^{(0)}_{rs-2},0,x_{s-1})\,,\\
  y^{(0)}_{rs-2}&=&f(x_{s-3}; x_r,y^{(0)}_{rs-3},0,x_{s-2})\,,\\
  &&\cdots\nonumber\\
  y^{(0)}_{rr+4}&=&f(x_{r+3}; x_r,y^{(0)}_{rr+3},0,x_{r+4})\,,\label{eqY4}\\
  y^{(0)}_{rr+3}&=&f(x_{r+2}; x_r,y^{(0)}_{rr+2},0,x_{r+3})\,,\label{eqY}\\
  y^{(0)}_{rr+2}&=&f(x_{r+1}; x_r,0,0,x_{r+2})\,\label{eqY1}.\end{aligned}$$ Eq.(\[eqY\]) gives, $$\begin{aligned}
  y^{(0)}_{rr+3}&=&
    \frac{1-x_{r+2}\alpha}{1-x_{r+2}\alpha x_r}\,
    \frac{1-x_{r+2}\alpha x_rx_{r+3}}{1-x_{r+2}\alpha
    x_{r+3}},\\
  y^{(0)}_{rr+2}&=&\frac{1-\alpha}{1-\alpha x_r}\,
    \frac{1-\alpha x_rx_{r+2}}{1-\alpha x_{r+2}}\,,\label{eqY2A}\end{aligned}$$ while (\[eqY1\]) is rewritten by, $$y^{(0)}_{rr+2}=\frac{1-x_{r+1}}{1-x_{r+1}x_r}\,
    \frac{1-x_{r+1}x_r x_{r+2}}{1-x_{r+1}x_{r+2}}\label{eqY2}.$$ Eqs.(\[eqY2A\]) and (\[eqY2\]) lead to $\alpha = x_{r+1}$ and hence we obtain, $$y^{(0)}_{rr+3}=\frac{1-x_{r+1}x_{r+2}}{1-x_rx_{r+1}x_{r+2}}\,
    \frac{1-x_rx_{r+1}x_{r+2}x_{r+3}}{1-x_{r+1}x_{r+2}x_{r+3}}\,.
    \label{eqY3}$$ Similarly we can determine $y^{(0)}_{rr+4}$ by eqs.(\[eqY4\]) and (\[eqY3\]) and so on. Then we obtain, $$y^{(0)}_{rs}= \frac{1-c_{s-1\,r}}{1-c_{sr}}\,
    \frac{1-c_{s\,r-1}}{1-c_{s-1\,r-1}}\,,
         \quad (1\leq r<s-1<N\,,\,(r,s) \neq (1,N))\,,$$ where $$c_{sr} = x_{r+1}x_{r+2}\cdots x_{s-1}x_s\,.$$

In a similar way, we get, $$\begin{aligned}
  y^{(0)}_{sr}&=&f(x_{r-1}; x_s,y^{(0)}_{sr-1},0,x_r)
         \nonumber\\
  &=&   \frac{1-(w/c_{s-1\,r})}{1-(w/c_{sr})}\,
    \frac{1-(w/c_{s\,r-1})}{1-(w/c_{s-1\,r-1})},
    ~~(1\leq r<s-1<N\,,\,(r,s) \neq (1,N)),\label{eqYsr}\end{aligned}$$ where $$w = \prod_{i=1}^{N} x_i\,.$$ We should notice that $y^{(0)}_{r+1\,r}$ can be regarded as a diagonal of the quadrilateral $p_{r+1}p_{r-1}p_ro_1$ whose sides are $(y^{(0)}_{r+1\,r-1},0,x_r,x_{r+1})$ [@KSV] and hence we obtain, $$y^{(0)}_{r+1\,r}=f(x_{r-1};x_{r+1},y^{(0)}_{r+1\,r-1},0,x_r)\,.
    \label{eqYr1}$$ Then eqs.(\[eqYsr\]) and (\[eqYr1\]) lead to $$\begin{aligned}
  y^{(0)}_{r+1\,r}&=&
    \frac{1-w}{1-(w/c_{r+1r})}\,
    \frac{1-(w/c_{r+1\,r-1})}{1-(w/c_{r\,r-1})}\nonumber\\
  &=&   \frac{1-w}{1-(w/x_{r+1})}\,
    \frac{1-(w/x_rx_{r+1})}{1-(w/x_r)}\,,
    \quad (0\leq r<N,\,x_0\equiv x_N).
    \label{eqYr}\end{aligned}$$ Furthermore $u^{(0)}_{r+1}$ can be regarded as a diagonal of the “quadrilateral” $p_{r+1}p_rp_{r+1}o_1$ whose sides are $(y^{(0)}_{r+1\,r},0,x_{r+1},x_{r+1})$ [@KSV] and hence we obtain, $$u^{(0)}_{r+1}= f(x_r;x_{r+1},y^{(0)}_{r+1\,r},0,x_{r+1})\,.
    \label{eqYu1}$$ Then eqs.(\[eqYu1\]) and (\[eqYr\]) lead to $$u^{(0)}_r= \frac{(1-x_rw)(1-(w/x_r))}{(1-w)^2}\,,
    \quad (1\leq r \leq N).\label{eqU0}$$

We have seen that there are two topologically inequivalent lines, $y^{(0)}_{rs} $and $y^{(0)}_{rs}$, connecting a pair of vertices of the $N$-polygon in general. In fact there are infinite number of lines. In Figure \[figWL\](a) we find that a line $y^{(1)}_{rs}$ which connects two vertices, $p_r$ and $p_s$, with winding the point $o_1$ is topologically inequivalent to $y^{(0)}_{rs}$. Note that $y^{(1)}_{rs}$ is drawn clockwise from $p_r$ to $p_s$ with the point $o_1$ always kept on the right and hence $y^{(1)}_{rs}\neq y^{(1)}_{sr}$. Then we have infinite number of inequivalent lines $y^{(n)}_{rs}, y^{(n)}_{sr}\, (n=0,1,2,\cdots)$ connecting a pair of vertices, $p_r$ and $p_s$ ($1\leq r<s-1<N$). We know that $y^{(0)}_{rs}$ is a diagonal of a quadrilateral $o_1p_rp_{s-1}p_s$ whose sides are $(x_r,y^{(0)}_{rs-1},0,x_s)$ and we may regard all the $y^{(n)}_{rs}$ as diagonals of the same quadrilateral $o_1p_rp_{s-1}p_s$. However, since $y^{(n)}_{rs}$ crosses all the $x_i$ $n$ times, we regard the other diagonal of the quadrilateral as $x_{s-1}w^n$ and hence we get ($n=0,1,2,\cdots$), $$\begin{aligned}
  y^{(n)}_{rs}&=& f(x_{s-1}w^n; x_r,y^{(0)}_{rs-1},0,x_s)\nonumber\\
  &=& \frac{1-c_{s-1\,r}w^n}{1-c_{sr}w^n}\,
    \frac{1-c_{s\,r-1}w^n}{1-c_{s-1\,r-1}w^n},
    \hspace{3ex}(1\leq r<s-1<N\,,\,(r,s) \neq (1,N)).\label{eqYnrs}\end{aligned}$$ Similarly we obtain ($n=0,1,2,\cdots$), $$\begin{aligned}
  y^{(n)}_{sr}&=&f(x_{r-1}w^n; x_s,y^{(0)}_{sr-1},0,x_r)\nonumber\\
  &=&   \frac{1-(w^{n+1}/c_{s-1\,r})}{1-(w^{n+1}/c_{sr})}\,
    \frac{1-(w^{n+1}/c_{s\,r-1})}{1-(w^{n+1}/c_{s-1\,r-1})},\label{eqYnsr}\\
  &&\hspace{25ex} (1\leq r<s-1<N\,,\, (r,s) \neq (1,N)),\nonumber\\
  y^{(n)}_{r+1\,r}&=&f(x_{r-1}w^n;x_{r+1},y^{(0)}_{r+1\,r-1},0,x_r)
      \nonumber\\
  &=&   \frac{1-w^{n+1}}{1-(w^{n+1}/x_{r+1})}\,
    \frac{1-(w^{n+1}/x_rx_{r+1})}{1-(w^{n+1}/x_r)}\,,
    \quad (0\leq r<N,\,x_0\equiv x_N),\\
  u^{(n)}_r&=& f(x_{r-1}w^n;x_r,y^{(0)}_{r\,r-1},0,x_r)\nonumber\\
  &=&   \frac{(1-x_rw^{n+1})(1-(w^{n+1}/x_r))}{(1-w^{n+1})^2}\,,
    \quad (1\leq r \leq N).\label{eqU}\end{aligned}$$

We have seen that non-winding lines connecting a pair of vertices are accompanied with winding lines and this is also the case for the sides of the $N$-polygon. In Figure \[figWL\](b) we see that $y^{(1)}_{rr+1}$ is another topologically inequivalent line to those we have seen so far, and the corresponding non-winding line is a side of the $N$-polygon, $y^{(0)}_{rr+1} (=0)$. We can obtain $y^{(n)}_{rr+1}\, (n=1,2,\cdots)$ as follows. Figure \[figYtoY\] indicates that $y^{(0)}_{sr}$ with a clockwise winding gives $y^{(1)}_{sr}$ while $y^{(0)}_{sr}$ with a counterclockwise winding gives $y^{(0)}_{rs}$, which implies $$y^{(-1)}_{sr} = y^{(0)}_{rs}\,.\label{eqYtoY0}$$ Assuming that eq.(\[eqYnsr\]) holds for $n<0$ we can easily prove eq.(\[eqYtoY0\]) or we obtain, $$y^{(-1-n)}_{sr} = y^{(n)}_{rs}\,.\label{eqYtoY}$$ Then we get ($n=1,2,\cdots$), $$\begin{aligned}
  y^{(n)}_{rr+1}&=&  y^{(-1-n)}_{r+1r} = \frac{1-w^{-n}}{1-(w^{-n}/x_{r+1})}\,
    \frac{1-(w^{-n}/x_rx_{r+1})}{1-(w^{-n}/x_r)}\nonumber\\
  &=&   \frac{1-w^n}{1-x_{r+1}w^n}\,
    \frac{1-x_rx_{r+1}w^n}{1-x_rw^n}\,,
    \quad (0\leq r<N,\,x_0\equiv x_N)\,.\label{eqYrr1}\end{aligned}$$

Three comments are in order: i) (\[eqU\]) satisfies $$u^{(-2-n)}_r = u^{(n)}_r\,, \quad(n=0,1,2,\cdots).$$ ii) Plugging $n=0$ in (\[eqYrr1\]) we obtain $y^{(0)}_{rr+1}=0$. iii) If we regard $y^{(0)}_{rr+1}$ as a diagonal of a “quadrilateral” $p_rp_rp_{r+1}o_1$ whose sides are ($x_r,u^{(-1)}_r,0,x_{r+1}$) and the other diagonal is $x_r$ (Figure 13), we have, $$y^{(0)}_{rr+1}=f(x_r; x_r,u^{(-1)}_r,0,x_{r+1})\,,
    \label{eqYrr1a}$$ or $$\begin{aligned}
  y^{(0)}_{rr+1}&=& \frac{1-x_r\alpha}{1-x_r\alpha x_r}\,
    \frac{1-x_r\alpha x_r x_{r+1}}{1-x_r\alpha x_{r+1}}\,,
    \label{eqYrr1b}\\
  u^{(-1)}_r &=& \frac{(1-\alpha)(1-\alpha x_r^2)}{(1-\alpha
    x_r)^2}\,. \label{eqUm1}\end{aligned}$$ From eq.(\[eqU\]) with $n=-1$ and eq.(\[eqUm1\]) we have $\alpha=1/x_r$, and hence (\[eqYrr1b\]) becomes $y^{(0)}_{rr+1}=0$. Then eq.(\[eqYrr1a\]) leads to ($n=1,2,\cdots$) $$\begin{aligned}
  y^{(n)}_{rr+1}&=&f(x_rw^n; x_r,u^{(-1)}_r,0,x_{r+1})
    \nonumber\\
  &=& \frac{1-x_rw^n(1/x_r)}{1-x_rw^n(1/x_r) x_r}\,
    \frac{1-x_rw^n(1/x_r) x_r x_{r+1}}{1-x_rw^n(1/x_r) x_{r+1}}
    \nonumber\\
  &=& \frac{1-w^n}{1-x_rw^n}\,
    \frac{1-x_rx_{r+1}w^n}{1-x_{r+1}w^n}\,,\end{aligned}$$ which agrees with (\[eqYrr1\]).

Since we have obtained all the topologically inequivalent lines, we can now write down the $N$-point one-loop amplitude by the KSV method. The square of the momentum associated with a line depends only on the ends of the line and hence, for example, the exponent for $y^{(n)}_{rs}$ is the same as that for $y^{(n)}_{sr}$ and it is given by $-\alpha(-(p_r-p_s)^2)-1$, where $\alpha$ is given by eq.(\[3.7\]). The $N$-point one-loop amplitude is given by, $$\begin{aligned}
  A_N^{(1)}&=& g^N \int d^{26}k \int_{0}^{1}
    \prod_{i=1}^{N} dx_i\,  \rho^{(1)}(x_i)\,
     \prod_{i=1}^{N}\, x_i^{-\alpha(-(k-p_i)^2)-1}\,
    \prod_{1\leq r < s-1 < N \atop (r,s)\neq(1,N)}
     \left( \prod_{n=0}^{\infty}y^{(n)}_{rs}\,
    y^{(n)}_{sr}\right)^{-\alpha(-(p_r-p_s)^2)-1}\nonumber \\
  && \times \prod_{r=0}^{N-1}
    \left(\prod_{n=0}^{\infty}y^{(n)}_{r+1r}\,
    \prod_{n=1}^{\infty}y^{(n)}_{rr+1}\right)
    ^{-\alpha(-(p_r-p_{r+1})^2)-1}\,
    \prod_{r=1}^{N}\left(\prod_{n=0}^{\infty} u^{(n)}_r \right)
    ^{-\alpha (0) -1}\,,\end{aligned}$$ where $\rho^{(1)}(x_i)$ is given by, $$\rho^{(1)}(x_i)\propto \prod_{1\leq r<s\leq N} (1 - x_{r-1}x_r)^{-1}
    \,, \quad(x_0\equiv x_N),$$ due to crossing symmetry. The amplitude in the operator formalism is given by (see e.g.[@GSW]), $$\begin{aligned}
  A_{Nop}^{(1)}&=& g^N \int d^{26}p \int_{0}^{1}
    \prod_{i=1}^{N} dX_i\, \left(\frac{1}{W^2}\right)\,
    \prod_{i=1}^{N}\,X_i^{\alpha'q_i^2}\,
    \prod_{n=1}^{\infty}(1 - W^n)^{-24}\nonumber\\
  &&\times  \prod_{1\leq r<s\leq N}
    \left\{ \prod_{n=1}^{\infty}\exp\left(-\frac{C_{sr}^n +
    (W/C_{sr})^n - 2W^n}{n (1-W^n)}\right)\right\}^{P_r\cdot P_s}\nonumber \\
  &=& g^N \int d^{26}p \int_{0}^{1} \prod_{i=1}^{N} dX_i\,
    \prod_{i=1}^{N}\, X_i^{-\alpha(-(p+p_N-p_{i-1})^2)-1}\,
    \prod_{n=1}^{\infty}(1 - W^n)^{-24}\nonumber\\
  &&\times  \prod_{1\leq r<s\leq N}
    \left\{(1-C_{sr}) \prod_{n=1}^{\infty}\,
    \left( \frac{(1-C_{sr}W^n)(1-W^n/C_{sr})}{(1-W^n)^2}
    \right)\right\}^{P_r\cdot P_s},\end{aligned}$$ where $$\begin{aligned}
    q_r&=&p-P_1 - P_2 - \cdots - P_{r-1}=p+p_N - p_{r-1},\\
    C_{sr}&=& X_{r+1} X_{r+2} \cdots X_{s-1} X_s,\\
    W&=& X_1 X_2 \cdots X_N.\end{aligned}$$ Considering the correspondence of the parameters, $$\begin{aligned}
    k&\longleftrightarrow& p+p_N,\\
    x_i&\longleftrightarrow&X_{i+1},\\
        c_{sr}&\longleftrightarrow&C_{s+1r+1},\\
        w&\longleftrightarrow&W,\end{aligned}$$ we shall compare the amplitudes. Then we find, $$\rho^{(1)}(x_i)= \left[\prod_{n=1}^{\infty}
     (1 - w^n) \right]^{-24}\,
    \prod_{1\leq r<s\leq N} (1 - x_{r-1}x_r)^{-1}.$$

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[^1]: e-mail: <PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>

[^2]: e-mail: <PRESIDIO_ANONYMIZED_EMAIL_ADDRESS>

[^3]: Any external lines must have non-vanishing momenta in the compact direction [@Bil3].\[foot1\]

[^4]: The signature of our metric is $\eta_{\mu \nu}=diag (-1,1,1, \cdots,1)$ .

[^5]: See the footnote \[foot1\].

[^6]: For simplicity, we restrict ourselves to the four-point planer amplitude in this subsection. We will extend this result $N$-point ones in Appendices A and B.

[^7]: A line may be regarded as the diagonal of different quadrilaterals. Uniqueness of the expression of a diagonal in this prescription is proved in the appendix B of [@KSV].