# Analytic continuation of the double sum $\sum_{n,m\ge0}x^ny^mt^{nm}$

Define function $f(x,y,t)$ as the analytic continuation of the series $$f(x,y,t)=\sum_{n,m\ge0}x^ny^mt^{nm}$$ This series definitely converges when all the arguments are small enough. I would like to understand the global properties of this function: presence of zeros, of poles or other singularities etc. In particular, I would like to locate the poles of this function w.r.t. to $y$ and find the corresponding residues.

Let us do a partial resummation $$f(x,y,t)=\sum_{n\ge0}\frac{x^n}{1-yt^n}$$ which is possible when say $|y|,|t|<1$.

Naively, from the above expression I expect that $f(x,y,t)$ has poles in $y$ for $y=t^{-n}, n\ge0$ with residues $$\operatorname{Res}_{y=t^{-n}}f(x,y,t)=-x^n/t^n$$

However, there is a problem here for me. I will first formulate is loosely and then give more accurate description.

## The problem

I. Loose formulation. Locations of poles of $f(x,y,t)$ are independent of $x$. Therefore, we don't expect them to change if $x$ changes. However, one can prove formula $f(x^{-1},y,t)=f(x,y^{-1},t)$ (modulo not-so-interesting terms). In the lhs we've only changed $x$, and hence we don't expect locations of poles to change. On the other hand in the rhs we've changed $y\to y^{-1}$ which causes poles to invert!

II. More accurate description. By formally operating with the series one can deduce three following properties

1. $f(x,y,t)=f(y,x,t)$, obvious symmetry.
2. $f(x,y^{-1},t)=-yf(xt^{-1},y,t^{-1})$ rule to invert an argument.
3. $f(xt,yt,t)=(xyt)^{-1}(f(x,y,t)-(1-x)^{-1}-(1-y)^{-1}+1)$ scaling rule.

Using their combination one shows that

$$f(x^{-1},y,t)=f(x,y^{-1},t)+(1-y)^{-1}-(1-x)^{-1}$$

Again, as stated in the loose formulation, the poles in $y$ do not seem to agree between the lhs and rhs: in the lhs they are at the points $y=t^{-n}, n\ge0$ while at the rhs at the points $y=t^n, n\ge0$.

## Summary

I understand that most likely I messed up with divergent series somewhere, but I can't exactly find where. I would be grateful if somebody explained me where, but my main question is the following: what are locations and residues of poles for function $f(x,y,t)$ wrt $y$-variable? Do they depend on $x$? If yes, how?

## Appendix

Here I present formal derivation of properties 2, 3.

$$f(x, y^{-1},t)=\sum_{n\ge0}\frac{x^n}{1-y^{-1}t^n}=\sum_{n\ge0}\frac{x^nt^{-n}y}{yt^{-n}-1}=-yf(xt^{-1}, y, t^{-1})$$

And

$$f(xt, yt,t)=\sum_{n,m\ge0}{x^ny^mt^{nm+n+m}}=(xyt)^{-1}\sum_{n,m\ge0}x^{n+1}y^{m+1}t^{(n+1)(m+1)}=(xyt)^{-1}\left(\sum_{n,m\ge0}-\sum_{n\ge0,m=0}-\sum_{n=0,m\ge0}+\sum_{n=0,m=0}\right)x^{n}y^{m}t^{nm}=(xyt)^{-1}(f(x,y,t)-(1-x)^{-1}-(1-y)^{-1}+1)$$

As far as I can the the factorization procedure suggested in this answer is not actually valid since the two factors has non-overlapping region of convergence. So, I'm still looking for a solution.

• There seems to be a typo: the exponent of $y$ is supposed to be $m$, right? – user1688 Feb 3 '16 at 10:43
• Could you make precise, in which domains the formulas 2. and 3. hold? – user1688 Feb 3 '16 at 11:17
• Let me present a formal derivation and leave it up to reader to judge. I am afraid to confuse things. – Weather Report Feb 3 '16 at 11:26
• Here's a guess: Are you sure that formula 3 is actually valid beyond formally? Swapping $x \iff x^{-1}$ (and similarly for $y$) may result in a sum that doesn't actually converge... – Simon Rose Feb 3 '16 at 12:55
• I think that if you instead consider the sum for $n\in \mathbb Z$, then you get an easy to write elliptic function. But I am not sure what to do with $n\geq 0$. – Lev Borisov Feb 4 '16 at 0:38

I think you can do the following (I haven't checked convergence carefully, so don't trust me too much). Write $a = \frac{m+n}{2}, b = \frac{m-n}{2}.$ Then $mn = a^2-b^2.$ Now (let's ignore indeterminacy in the square root for now), $\ x^m y^n = (xy)^a(x/y)^b$. The advantage of this sketchy manipulation is that your sum can be expressed, formally, as $$\sum_{a, b\in \frac{\mathbb{N}}{2}a\equiv b\text{ mod }1} \left((xy)^a t^{a^2}\right)\left((x/y)^b t^{-b^2}\right),$$ which is almost a product of two theta functions. A little more rigorously, if you choose square roots $\alpha^2 = xy, \beta^2 = x/y$ and a fourth root $\tau^4 = t$ and write $A = 2a, B = 2b$ then under a suitable convergence hypothesis, $$f(x,y,t) = \frac{G(\alpha,\beta,\tau)+G(-\alpha, -\beta, \tau)}{2},$$ where $$G(\alpha, \beta, \tau) = \sum_{A, B} \alpha^{A}\beta^{B}\tau^{A^2-B^2} = \left(\sum_{A} \alpha^A\tau^{A^2}\right)\left(\sum_{B}\beta^B\tau^{-B^2}\right),$$ which is actually a product of two theta functions. Of course this product cannot converge either when $|\tau|>1$ or $|\tau|<1,$ but convergence on the circle $|\tau| = 1$ with small $\alpha,\beta$ (corresponding to small $x$ compared to $y, y^{-1}$) should be enough to get your analytic continuation. From this it should not be very hard to deduce the poles. I wonder if such a formula can also be obtained from an appropriate version of the triple product identity. A little bit more context for how this problem came up could be helpful to see if this is the case.

Edit: As Lev Borisov points out, this only works when the sum is taken over $n\in \mathbb{Z}$ (equivalently, when studying $f(x,y,t) + f(x,y^{-1},t^{-1})$). As such, it is not an answer to the OP. -DV

• Thank you for the answer! I know very little about elliptic functions and now I will try to fill this gap. Meanwhile I can mention that this problem came as a simplified version of a physics-inspired problem (modular transformations of conformal blocks in CFT). In fact, for this problem I only need the symmetric combination $f(x,y,t)+f(-x,y,t)$. As Lev Borisov suggested in the comments this could simplify things. Any comments on that? Thanks! – Weather Report Feb 4 '16 at 9:20
• Typo. I meant $f(x,y,t)+f(x^{-1},y,t)$. – Weather Report Feb 4 '16 at 9:55
• You seem to be missing the inequalities $m,n\geq 0$. This will lead to weird inequalities for $m+n,m-n$, and I don't see how you would get products. – Lev Borisov Feb 4 '16 at 14:40
• That is a very good point. @WeatherReport, if you were interested in $f(x,y,t) + f(x^{-1}, y, t^{-1})$ then the above derivation should work (with $x$ replaced by $y$). Are you sure that what you want is $f(x,y,t) + f(x^{-1}, y, t)$? – Dmitry Vaintrob Feb 4 '16 at 14:56
• @DmitryVaintrob Well, I'm not really sure how this toy example relates to my actual problem. However, I feel like this is the part of the research I should carry on myself:) Your trick to factorize series into a product and relation to elliptic functions looks very promising. I will probably soon mark this as an accepted answer, unless somebody will appear with a full solution to the problem formulated originally in the OP. Thank you very much, once again. – Weather Report Feb 4 '16 at 15:31

I will give an alternative expression for $f(x,y,t)$ that may be helpful for others trying to solve this problem. Initially suppose $|x|, |y|< 1$, and $0 < t < 1$. Then $t^{mn} = \exp(-mn \log(1/t))$, and so by the Mellin formula for $e^{-x}$, we have $$t^{mn} = \frac{1}{2\pi i} \int_{(\sigma)} \Gamma(s) (mn \log(1/t))^{-s} ds,$$ where $\sigma > 0$.

By a simple rearrangement, we have $$f(x,y,t) = -1 + \frac{1}{1-x} + \frac{1}{1-y} + \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} x^m y^n t^{mn}.$$ The infinite double sum above can then be expressed as $$\frac{1}{2 \pi i} \int_{(\sigma)} \Gamma(s) (\log(1/t))^{-s} Li_s(x) Li_s(y) ds,$$ where $Li_s(z) = \sum_{n=1}^{\infty} x^n n^{-s}$ is the polylogarithm.

I haven't thought much about if this integral representation can lead to a meromorphic continuation. The polylogarithm does have analytic continuation but the above integral may not converge absolutely if $|x| > 1$ or $|y| > 1$.

• You don't need absolute convergence, locally uniform convergence is sufficient. – GH from MO Feb 10 '16 at 18:09

Suppose that $|x|<1$ and $|t|<1$. Then the power series $\sum_{n\geq 1}\frac {x^n}{1-yt^n}$, considered as a function of $y$, converges uniformly on compact sets disjoint from $y=t^{-k}$. Indeed, the denominators approach $1$ uniformly as $n\to\infty$ so you can bound by the geometric series.

Moreover, the same argument shows that in a neighborhood of $y=t^{-k}$ the sum $\sum_{n\geq 1, n\neq k}\frac {x^n}{1-yt^n}$ converges uniformly. This shows that the poles are indeed simple with residues $-\frac {x^n}{t^n}$.

If you try to analytically continue beyond $|x|,|t|<1$ region, then you may run into trouble. There is still convergence if $1<|x|<|t|$, so I think that in this regime the pole locations and residues are the same. Note that you will now have an essential singularity at $y=0$.

It is less clear what happens if you try to analytically extend away from these regions. Moreover, I am not at all sure whether such extension would be possible, and whether there would be some monodromy. My guess is that you would need to avoid the $x=t^{-n}$ locations, and there may be monodromy associated to loops around these.

• Your arguments look reasonable. However, I remain in some confusion. Concerning my self-contradictory conclusion: I think something is wrong with the property 2 ($f(x,y^{-1},t)=-yf(xt^{-1},y,t^{-1})$). It should also work if the summation is over $n\in\mathbb{Z}$. But equation (11) from your paper does not behave in this way. However, what exactly is wrong I do not understand. It seems enough for convergence of $\sum_{n\in\mathbb{Z}}\frac{x^n}{1-yt^n}$ that $|t|<|x|<1$, irrelevant of $y$. Not sure how can one run into trouble by replacing $y\to y^{-1}$ and making identical transformations. – Weather Report Feb 12 '16 at 16:46
• Not sure what the problem is: you have poles at $y^{-1}=t^{-n}$ on one side and at $y=(t^{-1})^{-n}$ on the other side, which is the same thing (as viewed as a function of $y$, in appropriate ranges for $|x|$ and $|t|$). – Lev Borisov Feb 12 '16 at 19:25
• I am saying the following: take function $\phi(x,y,t)=\sum_{n\in\mathbb{Z}}\frac{x^n}{1-yt^{n}}$. Now it seems that $\phi(x,y^{-1},t)=\sum_{n\in\mathbb{Z}}\frac{x^n}{1-y^{-1}t^{n}}=-y \sum_{n\in\mathbb{Z}} \frac{(xt^{-1})^n}{1-yt^{-n}}=-y\phi(xt^{-1},y,t^{-1})$. This is in contradiction with the explicit formula from your paper. The poles do agree, but the zeros do not: $(1-xyt^{k-1})(1-x^{-1}y^{-1}t^{k})$ vs $(1-xy^{-1}t^{-k})(1-x^{-1}yt^{-k+1})$. – Weather Report Feb 13 '16 at 9:06
• In my paper, I only consider $|t|<1$. The products would diverge, badly, if $|t|>1$. So the product formula is only valid in this regime. – Lev Borisov Feb 13 '16 at 12:42

I though it is worth pointing out a closely related case which appears to be solvable. The following formula is due to paper http://arxiv.org/abs/math/9904126 (formula (11) with slight relabeling and rearrangement)

$$\sum_{n\in\mathbb{Z}}\frac{x^n}{1-yt^n}=\prod_{k\ge1}\frac{(1-xyt^{k-1})(1-x^{-1}y^{-1}t^{k})(1-t^k)^2}{(1-xt^{k-1})(1-x^{-1}t^{k})(1-yt^{k-1})(1-y^{-1}t^{k})}$$

Series in the l.h.s. converges for $|t|<|x|<1$ (probably in some other region too). If $|t|<|x|,|x|^{-1},|y|,|y|^{-1}$ it can be presented as a more symmetric double expansion

$$\sum_{n\in\mathbb{Z}}\frac{x^n}{1-yt^n}=-1+\frac1{1-x}+\frac1{1-y}+\sum_{n,m>0}\left(x^ny^m-x^{-n}y^{-m}\right)t^{nm}$$