Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I have heard that Polylogarithms are very interesting things. The wikipedia page shows a lot of interesting identities. These functions are indeed supposed to have caught the attention of Ramanujan. Moreover they seem to be important in physics for various purposes like Bose-Einstein integrals, which I am not really knowledgeable enough to understand. These are all things I have heard from people after I queried "why polylogarithms are interesting".

So this function

$$Li_s (z) = \sum_{k=1}^\infty \frac{z^k}{k^s}$$

is very interesting and has a lot of useful properties as I can see. Especially for integral values of $s$.

What is bothering me is the following.

Let $f(z) = \sum a_n z^n$ be analytic inside the unit disc. That is, the radius of convergence at least $1$. For simplicity, we assume it is $1$, i.e.,

$$r = \frac{1}{\limsup_{n\rightarrow\infty}\sqrt[n]{|a_n|}} = 1$$

implying that inside the unit disc, term-by-term differentiation and integration are possible. Since $\left\( a_n \right)^{1/n}$ goes to $1$ as $n$ goes to $\infty$, it is also true that $\left\(na_n \right)^{1/n}$ goes to $1$. Similarly, $\left\(\frac{a_n}{n} \right)^{1/n}$ also goes to $1$. Now, defining

$$f(z,s) = \sum_{n=1}^\infty \frac{a_n z^n}{n^s}$$

it is the case that $f_{2}(z) = f(z,2)$ and in general iterating, $f_n(z) = f(z, n)$ are analytic inside the unit disc. In fact, uniform bounds hold true for compact sets contained in the domain $|z| \leq 1$ and $|s| \leq K$ for arbitrary $K$, and so $f(z,s)$ is analytic in the domain $D^\circ \times \mathbb C$, where $z \in D^\circ$, the open unit disc, and $s \in \mathbb C$.

So any analytic function in the unit disc can be extended just like the logarithm is extended to polylogarithms. But there is a rich theory about polylogarithms, and I haven't heard of any such theories about other functions analytic inside the unit disc.

So, what makes polylogarithm so amenable to an extended theory of polylogarithms yielding so many results? Why is this not giving such an interesting theory for just any other complex functions?

share|improve this question

3 Answers 3

up vote 42 down vote accepted

The reason why polylogarithm are so important/interesting/ubiquitous is they are the simplest non trivial examples of analytical functions underlying variations of mixed Hodge structure. This goes back to Beilinson and Deligne.

A variation of mixed Hodge structure is a very sophisticated gadget. You can think of it as

  1. a nice differential equation (the underlying connexion)
  2. its solutions (the underlying local system of horizontal sections) with a $\mathbb{Q}$-structure
  3. some additional data that make the structure very rigid.

Typical examples of VMHS on $X$ come from the cohomology of families of varieties parametrized by $X$. They can be used to encode the interaction between between topological and arithmetical properties of $X$.

For example, there is a rank 2 variation of mixed Hodge structure $K \in Ext^1_{VMHS(\mathbb{C}^\times)}(\mathbb{Q},\mathbb{Q}(1))$ over $\mathbb{C}^\times$ known as the Kummer sheaf. The underlying local system $K_{\mathbb{Q}}$ has fiber $H_1(\mathbb{C}^\times,\{1,z\};\mathbb{Q})$ at $z$. The underlying connexion is a trivial vector bundle of rank 2 with nilpotent connexion $$ \nabla = d - \begin{pmatrix} 0 & 0 \cr % \frac{dt}{t} & 0 \end{pmatrix} $$ The "periods" are obtained by integrating the coefficient of the matrix over the paths $\gamma \in H_1(\mathbb{C}^\times,\{1,z\};\mathbb{Q})$. So we get a the non trivial period by integrating $\frac{dt}{t}$ over paths $[1,z]$, i.e. we get determinations of $\log(z)$. Conclusion: we have an object $K$ in $VMHS(\mathbb{C}^\times)$ "categorizing" the classical logarithm function. On the arithmetic side, the transcendance of $\log(z)$ for generic $z$ mirrors the fact that $H_1(\mathbb{C}^\times) = \mathbb{Z}$.

The same can be done for polylogarithm functions. The Logarithmic sheaf is the symmetric algebra $Log := Sym(K)$ (it corresponds to the whole family of $\log^n(z)$, $n\in \mathbb{N}$). The Polylogarithm sheaf is a canonical extension $Pol$ of $\mathbb{Q}$ by the restriction of $Log(1)$ to $\mathbb{P}^1\setminus \{0,1,\infty\}$. Its periods encodes the monodromy of the polylogarithm functions in the same way the Kummer sheaf does for the logarithm function. These are the most elementary unipotent variations of mixed Hodge structures.

Now we have "categorized" the classical polylogarithm functions. In fact, this can actually be done on a more fundamental level using only algebraic cycles defined over $\mathbb{Z}$ (this is the motivic story, the variation of mixed Hodge structure being just a realization of the motivic object). This has very interisting arithmetic consequences. For example, specializing to 1, this implies that we have motivic cohomology classes in $H^1(\mathbb{Z},\mathbb{Q}(n))$ whose images under the Hodge regulator corresponds to $\zeta(n)$. Using this picture, the period conjecture then implies that the $\zeta(2n+1)$ are algebraicly independent over $\mathbb{Q}$. To give you an idea of how powerful this intuition is: we can't even prove $\zeta(5)$ is irrational!

In conclusion, the polylogarithm functions are interesting because they correspond to non trivial algebraic cycles. This leads to interactions between the analytical properties of the functions, arithmetics of special values and algebraic geometry. Lots of classical functions should have similar interpretations. For example, there is a similar picture for Euler's Beta and Gamma functions.

share|improve this answer
    
I took the liberty of editing your matrix so that it would render properly (at least for me). Apparently \\ does not work and one has to use \cr instead. –  José Figueroa-O'Farrill May 21 '10 at 9:40
    
Thank you for the great post! –  Mariano Suárez-Alvarez Feb 7 '11 at 20:40
    
YBL: Do you have some reference for this beautiful topics? –  Filippo Alberto Edoardo Sep 19 '12 at 4:38
    
I mentionned Beilinson and Deligne's original paper "Motivic Polylogarithm and Zagier Conjecture". I think Hain's survey "Classical polylogarithm" or Goncharov's "Multiple polylogarithms and mixed Tate motives" should be a good place to start. –  YBL Sep 19 '12 at 18:29

One of the reasons is that it satisfies $$\Theta_z f = T_s f $$ where $\Theta_z = z\frac{\partial}{\partial z}$ and $T_s g(s) = g(s-1)$ (where you take $f(s,z) = Li_s(z)$, and is in some sense the simplest non-trivial function which satisfies this. This first-order mixed (Euler) differential-shift equation was studied some in the 1950s, but then ignored for a long time. That, of course, is not the only reason, but a number of 'weird' identities have simpler proofs when looked at this way.

share|improve this answer

Polylogarithms have interesting connections in the Theory of Partitions. This is mainly because a class of generating functions for bivariate partition statistics can be approximated by polylogarithms.

Consider for example the class of generating functions with $s>1,$

$\displaystyle \sum_{n=0}^\infty \sum_{m=0}^\infty p(n,m)x^n u^m =\prod_{k=1}^\infty \frac{1}{(1-xu^k)^{k^{s-2}}}. $

When $s=2,$ $p(n,m)$ is the number of partitions of $m$ with $n$ parts. When $s=3,$ $p(n,m)$ represents the number of plane partitions of $m$ with trace $n.$

For simplicity assume $1>x>0,$ as $u\to ^-1,$

$\displaystyle \prod_{k=1}^\infty \frac{1}{(1-xu^k)^{k^{s-2}}} \thicksim\sqrt[ - \frac{1}{\zeta(2-s)}]{1-x}\exp(\Gamma(s-1)Li_{s}(x)/(\ln u^{-1})^{s-1}). $

When used in conjunction with other techniques like the Circle Method, this estimate serves as a foundation for many facts about $p(n,m).$

Off the top of my head here are three papers that use this kind of estimate: http://www.math.drexel.edu/~rboyer/papers/partitions_experimental.pdf http://plms.oxfordjournals.org/content/s2-36/1/117.full.pdf http://www.springerlink.com/content/d65348128235x7m1/

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.