Laurent series in several complex variables

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 $$z$$.

• 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. Sep 26 at 14:47
• @მამუკაჯიბლაძე: Right, but this delta-function doesn't live in any ring (only in a module over the Laurent polynomial ring). Sep 27 at 4:03
• @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. Sep 27 at 4:06
• What about Johnson series? See 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 mathoverflow.net/a/431192/16537 below. Sep 27 at 5:13
• @rimu: I wrote a bit about this in cip.ifi.lmu.de/~grinberg/algebra/va3.pdf , but I don't know a good textbook. Sep 27 at 19:32

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 $$(0,0)>(-1,1)>(-2,2)>\dots$$.
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.