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Is there a good generalisation of Laurent series for several complex variables?

I am interested in generalised power series that have some terms with negative powers, but not too many. In single variable complex analysis, "not too many" means that the (Laurent) series has only a finite number of terms with a negative power of the variable. In several variables, I want that at least something like $\frac1{1- z/w} = \sum_{n\geq 0} z^n w^{-n}$ counts as generalized Laurent series: While the exponent of $w$ may become arbitrarily small, it is at least bounded in terms of the exponent of $z$.

Has someone thought about this kind of series?

rimu
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    One of the approaches to vertex operator algebras is based on the calculus of formal distributions, with the formal delta-function $\delta(z-w)=\cdots+z^{-3}w^2+z^{-2}w+z^{-1}+w^{-1}+zw^{-2}+z^2w^{-3}+\cdots$ having the key property $\operatorname{Res}_{z=0}f(z)\delta(z-w)=f(w)$. See e. g. "Vertex algebras for beginners" by Kac. – მამუკა ჯიბლაძე Sep 26 '22 at 14:47
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    @მამუკაჯიბლაძე: Right, but this delta-function doesn't live in any ring (only in a module over the Laurent polynomial ring). – darij grinberg Sep 27 '22 at 04:03
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    @rimu: Your series is a formal power series in $z/w$ and $w$ (for example). Generally, the ring of formal power series in $x_1/x_2$, $x_2/x_3$, ..., $x_{n-1}/x_n$ and $x_n$ (or Laurent series in these variables) is often used. – darij grinberg Sep 27 '22 at 04:06
  • @darijgrinberg You are right. From "general considerations" (that distributions should live in something dual to where functions live) formal distributions should be equipped with a coalgebra rather than algebra structure (and functions with comodule structure over that coalgebra (?)) – მამუკა ჯიბლაძე Sep 27 '22 at 05:10
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    What about Johnson series? See https://mathoverflow.net/q/430307/16537 (in particular, the 3rd note in that post). These are natural generalizations of Hanh series, mentioned by Ira Gessel in their answer https://mathoverflow.net/a/431192/16537 below. – Salvo Tringali Sep 27 '22 at 05:13
  • Let me just use this question as a pretext to point out the fact that $\mathbb{C}(!(x)!)(!(y)!)$ is not the same as $\mathbb{C}(!(y)!)(!(x)!)$ (nor is either the same as the field of fractions $\mathbb{C}(!(x,y)!)$ of $\mathbb{C}[[x,y]]$). This is certainly obvious to both the asker and the answerers, but some people occasionally get surprised by this, so it's worth making a comment. – Gro-Tsen Sep 27 '22 at 12:22
  • @Gro-Tsen: In a similar spirit, $\mathbb{C}[[x]][y]$ is not the same as $\mathbb{C}[y][[x]]$ – M.G. Sep 27 '22 at 14:24
  • @darijgrinberg Do you have any references that use formal power series in $z/w$ and $w$? It sounds interesting. (You may make it an answer.) – rimu Sep 27 '22 at 18:42
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    @rimu: I wrote a bit about this in https://www.cip.ifi.lmu.de/~grinberg/algebra/va3.pdf , but I don't know a good textbook. – darij grinberg Sep 27 '22 at 19:32
  • @darijgrinberg Many thanks for sharing your notes! – მამუკა ჯიბლაძე Sep 29 '22 at 17:52

2 Answers2

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A good condition is that the exponent vectors lie in a union of finitely many translates of a fixed pointed polyhedral cone $\mathcal{C}$ (with vertex at the origin). ``Pointed'' means that $\mathcal{C}$ does not contain a line (infinite in both directions). With this condition, if we fix the cone $\mathcal{C}$ then the product of two power series is defined formally. Such series appear for instance in Brion's theorem, one reference being Section 9.3 of Beck and Robins, Computing the Continuous Discretely.

Note. The condition that the exponent vectors lie in a finite union of translates of $\mathcal{C}$ is equivalent to saying that they lie in a single translate of $\mathcal{C}$, since a finite union of translates is contained in a single translate.

Richard Stanley
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The easiest way to deal with series like $\sum_{n=0}^\infty z^n w^{-n}$ is with iterated Laurent series. This series is an element of the ring $\mathbb{Z}((w))[[z]]$: power series in $z$ whose coefficients are Laurent series in $w$. (In this case Laurent polynomials in $w$ would suffice.)

A much more general, though more complicated, approach is through Hahn series (also called Mal'cev–Neumann series) in which we have an indeterminate with exponents from an ordered group, with the condition that the exponents corresponding to nonzero terms are well ordered. (The well-ordered condition implies that multiplication of these series is well-defined.) To represent $\sum_{n=0}^\infty z^n w^{-n}$ in this way, we take as our exponent group the additive group $\mathbb{Z}\times\mathbb{Z}$ ordered lexicographically. With $x$ as the indeterminate, the series under consideration are of the form $\sum_{(i,j)\in \mathbb{Z}\times\mathbb{Z}} x^{(i,j)}$. We multiply monomials by $x^{(i_1,j_1)} x^{(i_2,j_2)}=x^{(i_1+i_2, j_1+j_2)}$. We may identify $x^{(i,j)}$ with $z^iw^j$. Then \begin{equation*} \sum_{n=0}^\infty z^n w^{-n}=\sum_{n=0}^\infty x^{(n,-n)} \end{equation*} is allowable since the exponent set $\{(0,0), (1,-1), (2,-2),\dots\}$ contains no infinite decreasing sequence. On the other hand \begin{equation*} \sum_{n=0}^\infty z^{-n} w^{n}=\sum_{n=0}^\infty x^{(-n,n)} \end{equation*} is not allowed since the exponent set contains the infinite decreasing sequence $(0,0)>(-1,1)>(-2,2)>\dots$.

Ira Gessel
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