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This is about the theory of Borel-$\acute{\textrm{E}}$calle re-summation and resurgence, see Refs below.

This states that the perturbative series (say of the vacuum expectation value of an operator $\mathcal{O}$ in quantum field theory) should be thought of as part of the so-called trans-series expansion, containing both perturbative and non-perturbative corrections: $$\langle \mathcal{O} \rangle = \sum_{k=0}^{\infty} c_{0,k} g^k+ \sum_{I} e^{-\frac{S_I}{g^2}} \left( \sum_{k=0}^{\infty} c_{I,k} g^k \right)+ \dots \;, $$ where $I$ is a label for the saddle points of the action $S$, and $S_I$ denotes the value of the action at the $I$-th saddle point; the coefficients $c_{I,k}$ represent the perturbative expansions around the $I$-th saddle point. Assuming we know all the saddle points contributing to the path integral, all the coefficients $c_{0,k}$ and $c_{I,k}$ can be computed by perturbative methods, and we obtain a trans-series. Now we hope to turn the trans-series into a well-defined function $\langle \mathcal{O} \rangle(g)$ as a function of the coupling constant with the help of the resurgence theory. If that function is resurgent, we may adiabatically continue back to the large value of the coupling constant along a suitable choice of path in the complex plane.

Questions:

(1). What are the precise mathematical conditions/criteria to turn the trans-series above into a well-defined function as a function of the coupling constant with the help of the resurgence theory?

When we say that the function is resurgent, does it require infinite differentiable or an infinite continuable?

p.s. The resurgence in the physics literature may refer to a stronger statement. A large-order asymptotic growth (as $k$ large) of the perturbative coefficients $c_{0,k}$ around the trivial saddle point contains the information of the non-perturbative saddle points within the same topological sector.

(2). What well-controlled mathematical properties can we state precisely for the above function $\langle \mathcal{O} \rangle$? What are the mathematical rigorous results for the re-summation and resurgence?

The context is partially inspired by this post.


  1. J. E ́calle, Les fonctions r ́esurgentes. Tome I, vol. 5 of Publications Math ́ematiques d’Orsay 81 [Mathematical Publications of Orsay 81]. Universit ́e de Paris-Sud, D ́epartement de Math ́ematique, Orsay, 1981. Les alg`ebres de fonctions r ́esurgentes. [The algebras of resurgent functions], With an English foreword.

  2. J. E ́calle, Les fonctions r ́esurgentes. Tome II, vol. 6 of Publications Math ́ematiques d’Orsay 81 [Mathematical Publications of Orsay 81]. Universit ́e de Paris-Sud, D ́epartement de Math ́ematique, Orsay, 1981. Les fonctions r ́esurgentes appliqu ́ees `a l’it ́eration. [Resurgent functions applied to iteration].

  3. O. Costin, Asymptotics and Borel summability, vol. 141 of Chapman & Hall/CRC Monographs and Surveys in Pure and Applied Mathematics. CRC Press, Boca Raton, FL, 2009.

  4. C. Mitschi and D. Sauzin, Divergent series, summability and resurgence. I, vol. 2153 of Lecture Notes in Mathematics. Springer, [Cham], 2016. Monodromy and resurgence, With a foreword by Jean-Pierre Ramis and a preface by E ́ric Delabaere, Mich`ele Loday-Richaud, Claude Mitschi and David Sauzin.

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I can only offer a partial answer here. I don't really know how resurgence is used in QFT, so I am going to talk about the general principles.

First thing first. The definition of an asymptotic series does not need complex analysis. However an asyptotic series in a real setting is much less useful. Lets say that a function $f$ admits $x$ as an asymptotic series to $+\infty$. Now define the function $g$ to be $1$ on $[n,n+1)$ if $n$ is even and $0$ if $n$ is odd. Then the function $f\cdot g$ also admits $x$ as an asymptotic serries. So in the real setting, we do not have any form of uniquness and we cannot extract any information (well maybe the word any is too strong here).

So when we talk about resurgent functions we are talking necessarily about complex analytic functions. Of course in the complex setting, having one derivative is equivalent to having all of them and a bit more. I will come back to endlessly continuability later.

So now let's look at a somewhat trivial example: the ODE $-x''(t)+x(t)=1/t$.

If we assume that the solution is an analytic function in a neighbourhood of infinity we can substitute $\tilde x(t)=\sum_{n\ge0}\frac{c_n}{t^n}$ in the ODE and collect terms to find eventually that $$\tilde x(t)=\sum_{n\ge0}\frac{(2n)!}{t^{2n+1}}.$$ And we realise that our assumption was wrong because this series has $0$ radius of convergence. So let's resum it. Let's us $s$ as the dual variable of $t$. The Borel transform of this is $$\hat x(s)=\sum_{n\ge0}s^{2n}.$$ This series converge in the unit disc. Then because we were "lucky", we can easily extent it beyond the unit disc: $$ \hat x(s) = \frac{1}{1-s^2}. $$ Then the Laplace transform of this is a solution of the ODE. But which Laplace transform?

We can define 2 here. One by integrating on the positive real axis and the other by integrating on the negative real axis. In practice we can move the contour of integration freely as long as we don't hit a singularity of the function. The singularities are of course $\pm i$, this is the reason that we have 2 Laplace transforms.

So the solutions of the ODE are $$ x^\pm(t) = \int_0^{\pm\infty}\frac{e^{-ts}}{1-s^2}ds. $$ These are two particular solutions, i.e. $x^\pm(t)$ tends to $0$ as $t$ tends to $\pm\infty$.

OK, let's talk about the resurgence here. Unfortunately, the term resurgent function is used liberally to describe 3 different objects. In the above example we can call a resurgent function $\tilde x$, $\hat x$ or $x^\pm$. I think that it is more appropriate to call $\hat x$ a resurgent function and nothing else, so let's go with that.

So ok, what do we get from what we did in the example? We know that we have 2 solution and we know that these 2 solutions are not the same. But how different are they? First of all what is the domain of definition of these solutions? I will not go into details here but for $x^+$ it is $\mathbb C$ minus the negative real axis with the origin. If this is not clear to you why, you should look into the literature for the Laplace transform. The domain of definition for $x^-$ is the reflection with respect to the imaginary axis.

Then we can take the difference $x^+(t)-x^-(t)$ for non-real $t$ and this is well defined. In order to calculate this difference we notice that since $$ x^\pm(t) = \int_0^{\pm\infty}\frac{e^{-ts}}{1-s^2}ds, $$ we have $$ x^+(t)-x^-(t) = \int_{-\infty}^\infty\frac{e^{-ts}}{1-s^2}ds. $$ However as it is written this integral does not converge, we have to deform the contour of integration. So let's choose $t$ with negative imaginary part. This means that we have to tilt the contour of integration "upwards on both sides". Then because the singularities of the function are just poles we get $$ x^+(t)-x^-(t) = \oint_i\frac{e^{-ts}}{1-s^2}ds, $$ which means the closed integral around the complex number $i$. So not only we know that the 2 solutions are different, but also we can explicitely compute their difference.

OK, so what is the big deal with resurgence? In general the situation is as follows. We want an analytic solution, but the solution fails to be analytic and we get an asymptotic series. Then we want to construct a solution and we need to use resummation. The problem we find there is that typically the solution we construct is different in different sections of the complex plane and the way it is different depends on the singularities of the Borel transform, $\hat x$ in our case. The theory of resurgent functions gives a toolset that allows us to analyse the singularities of such functions. Before this theory this was largely science fiction.

So let's go back to the endlessly continuability business. Initially we have $\hat x$ as a convergent series, i.e. it is a function defined in a disc, but we need to take its Laplace transform. The first thing we do is to extend it to (hopefully) its maximal domain, which means that it need to extend to infinity in a sector. If moreover the function is of exponential growth in this sector, then we can define the Laplace transfor. No other conditions are needed.

However, if we want to use the resurgence's toolkit we need to have a functions with isolated singularities. With meromorphic functions what isolated singularities are is straightforward. If the function is defined in a cover of $\mathbb C$, in other words "it has branching singularities", this becomes tricky. Let's say that this means that if we project the singularities of the funcition on $\mathbb C$, they are isolated points. This condition is stricter than what the theory needs, but I hope your case is covered by it.

So what do you need in order to rigorously prove your statement? All the above. You need to have a well defined asymptotic series. You need to prove that it's Borel transform converges. You need to prove that this functions is endlessly continuable. And finally you need to have exponential bounds for the growth of the functions in "most" directions.

I hope this helps. Unfortunately I only touched the surface of this amazing theory. In reality the rabbit hole is too deep. I hope this helps for now. Ask if something is unclear.

ADDED LATER:

I forgot to comment on this transseries business. So going back to the example we had $$ x^+(t)-x^-(t) = \oint_i\frac{e^{-ts}}{1-s^2}ds. $$ Let's move this our the origin: $$ x^+(t)-x^-(t) = \oint_0\frac{e^{-t(s+i)}}{1-(s+i)^2}ds = e^{-it}\oint_0\frac{e^{-ts}}{1-(s+i)^2}ds. $$ So you see that we get exponential in front of a "series" (in this case it is not really a series).

In the general case we get a function that has many singularities that are ramification points, let's look at another simple example: the finite difference equation $x(t+1)-x(t)=1/t^2$.

The Borel transform of this equation is $e^{-s} \hat x(s)-\hat x(s) = s$, so we get $$ \hat x(s) = \frac{s}{1-e^{-s}}. $$ Now this equation has a pole on $2\pi i n$ for all integers $n$ but the origin.

We can do basically the same as above, i.e. define 2 solutions (that are of course not general) and take their difference. But now the difference of the solutions is: $$ x^+(t)-x^-(t) = \sum_{n=1}^\infty\oint_{2\pi i n}\frac{s\,e^{-ts}}{1-e^{-s}}ds $$ and if we move all the integrals to the origin we get $$ x^+(t)-x^-(t) = \sum_{n=1}^\infty e^{-2\pi i n t} \oint_{0}\frac{(s+2\pi i n)\,e^{-ts}}{1-e^{-s-2\pi i n}}ds. $$ Of course these integrals do not give us series because the function is simple. However in the general case, if for example add a non-linear term in the example, we will get ramification points instead of ajust a pole and then we will get an actual transseries.

There is one last thing that I was wondering if I should write. I decided to do so, but it may be confusing at least in the beginning. OK here we go:

Now that you read all the previous, I have to tell you that I lied. A lot. So the classical Borel transform is defined as the formal inverse of the Laplace transform and it devides by a factorial. This is the reason that we can get a convergent series. However, in resurgence we need a generalization of this.

I will just give you some references because this is too big to be written here.

So the proper way to define a resurgent function is through hyperfunction, which are a generalisation distributions. I think the easiest textbook on the subject is "Introduction to Hyperfunctions and Their Integral Transforms" by Urs Graf.

Resurgent functions can be thought as a very special class of hyperfunctions. The paper enter link description here gives an introduction but I think it is not so easy to follow if you don't know already this. I looked around a bit and I found this PhD thesis enter link description here that explains it a bit. It is rather dense, but potentially easier. Read Chapter 2.

Also I don't know if you know them already but Sternin and Shatalov are probably more relevant to you. They have a book titled "Borel-Laplace Transform and Asymptotic Theory: Introduction to Resurgent Analysis" which for some reason is very expensive and not all libraries have it. Take a look also at their papers.

A word of caution about the book though. Their definition of resurgent functions is not the same as Ecalle's. They define a smaller class that has some extra properties, which means that probably they are the only ones using this definition of resurgence. Nevertheless it can help you understand the theory.

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  • $\begingroup$ thanks, very nice +1 $\endgroup$ – wonderich Mar 7 at 18:01
  • $\begingroup$ @wonderich I added abit more. Take a look $\endgroup$ – tst Mar 8 at 9:56

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