Multivariate residues in simple cases

Question

I have been looking at residues of multivariate functions and found there are quite a few difficulties (see e.g. Multivariate Residue Theorem? or Multivariate/multidimensional residues). In the literature, this is discussed in the context of manifolds, 1-forms and currents. Unfortunately, I am not an expert on manifolds.

Question: Are there "simple rules" deriving from the general treatment that can be applied in more basic cases.
I am thinking of multivariate functions $f(x,y,z, ...)$ with simple poles at equal points $x=y,~ x=z, ...$, where I would like to evaluate the residue at multiple, coinciding points $x=y=z$ as consecutive residues $\text{Res}_{x=y} \text{Res}_{y=z} \cdots$ in a consistent way.

Example: Consider the function $f(x,y,z) = \frac{1}{(x-y)(x-z)}$ defined on $\mathbb{C}^3$. It has singularities on and the 1-dimensional subspaces $\{(x,y,z) | x=y \}$ and $\{ (x,y,z) | x=z\}$ which intersect at $x=y=z$. Computing the residue on the intersection can be done through consecutive application of residues: $$ \text{Res}_{y=z} \text{Res}_{x=y} \frac{1}{(x-y)(x-z)} = \text{Res}_{y=z} \frac{1}{(y-z)} = 1. $$ However, exchanging the residues leads to a wrong result: $$ \text{Res}_{x=y} \text{Res}_{y=z} \frac{1}{(x-y)(x-z)} = 0. $$

Is there a procedure that tells me how to correctly take certain residues or at least relate different combinations of residues which give the same result? (For example $\text{Res}_{y=z} \text{Res}_{x=y}$ and $\text{Res}_{z=x} \text{Res}_{y=x}$ in the previous example)

Background: In quantum field theory, amplitudes (vacuum expectation values of time-ordered products of fields) are meromorphic functions in $\mathbb{C}^k$. Poles correspond to the temporary fusion of particles and higher-order poles at the intersection of more than two points (which I would like to evaluate as consecutive residues) appear when there are more complicated composite particles.

I appreciate any help or literature recommendation!

Answer 1

If think I have an idea which works in some cases at least. The idea is to cancel all poles and consider residue as limits of this (now) analytic function. This only works for simple poles of course. Here is what I mean:

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Assuming that $f(x,y,z)$ has simple poles at $x=z$ and $y=z$. It follows that $(x-z)(y-z) f(x,y,z)$ is analytic around $x=z$ and $y=z$.
Therefore, it is possible to take limits of this continous function \begin{align} &\lim_{x\to z} \lim_{y\to z} (x-z)(y-z) f(x,y,z) = \lim_{y\to z} \lim_{x\to z} (x-z)(y-z) f(x,y,z) \\ \Leftrightarrow& \lim_{x\to z} (x-z) \lim_{y\to z} (y-z) f(x,y,z) = \lim_{y\to z} (y-z) \lim_{x\to z} (x-z) f(x,y,z) \end{align} The limits commute of course, because for an analytic function it does not matter from which direction we approach the point $x=y=z$.
These limits yield the residues at $x=z$ and $y=z$, so it follows that: \begin{align} &\text{Res}_{x=z} \text{Res}_{y=z} = \lim_{x\to z}(x-z) \lim_{y\to z} (y-z) = \lim_{y\to z} (y-z) \lim_{x\to z}(x-z) = \text{Res}_{y=z} \text{Res}_{x=z}. \end{align} This means: If one can exchange the limits consistently, one can exchange the residues.

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In the example I gave, the function $f(x,y,z) = \frac{1}{(x-y)(x-z)}$ has poles at $y=x$ and $z=x$. Looking at the residue $\text{Res}_{y=z} \text{Res}_{x=y}$ means to consider $(y-z) (x-z) f(x,y,z)$. This is not analytic because the pole at $x=y$ is not cancelled. We can approach $x=y=z$ only by going first to $x=z$ and then $y=x$, not the other way around.

But $\text{Res}_{z=x} \text{Res}_{y=x} = \text{Res}_{y=x} \text{Res}_{z=x}$ can be exchanged consistently here.

Answer 2

Multidimensional residues generally require isolated singularities, just as in the $C^1$ case. In the example you propose, $\frac{1}{(x-y)(x-z)}$, situated in $C^3$, the function is in some sense "underdetermined" to produce an isolated singularity: you have a "line of singularities" of a sort, along $x=y=z$. Since there is no isolated pole / singularity, the residue will always be $0$. (A similar thing occurs the post you cited MSE 1599033.)

It may help to see this by considering the function $1/(xy)$ in $C^3$. Without going into too many details (polydiscs and the ability to approach this as an iterated integral, which I think helps avoid the complications of differential forms), consider the iterated integral, $$ \frac{1}{(2\pi i)^3} \int_{D_z} \int_{D_y} \int_{D_x} \frac{1}{xy} dxdydz $$ (w/ disc domains $D_w \equiv \{w\in C^1 \mid |w|<r\}$). The inner $dx$ produces residue $1/y$; next, the $dy$ produces residue $1$ (so we have the function $g(x,y,z)=1$), then $dz$ will produce zero no matter what closed contour you consider (the function $g=1$ has no poles, anywhere). The result will always be $0$ (and would have been in your example above if you had used the full multidimensional formalism and included the "other" dimension--because there is technically no isolated singularity in $C^3$ for your given function, $\frac{1}{(x-y)(x-z)}$).

The form $\frac{g}{f_1...f_n}$, $g,f_i:C^n->C$, where $g$ is holomorphic in the domain of interest (has no poles) and where there are one or more points in $C^n$ such that $f_1=f_2=...=f_n=0$ is common in multidimensional residues (I believe this falls under the term "complete intersection"--a particularly simple type of multidimensional complex pole as your cite MSE 481656 mentions), this being related to sums over multiple / systems roots and associated resultants (see MO 365866).

It is easy to place your example in such a form by fully "pinning down" a pole--e.g. $\frac{1}{(x-y)(x-z)(z-1)}$--then this has a "multidimensional pole" in $C^3$ at $(1,1,1)$. Methods like Grothendieck residues can be used to approach this (with these linear cases--ie where each $f_i$ is linear, and the pole of multiplicity $1$--being particularly easy). (To my current knowledge in fact, you can take the transforms, $t_1=x-y$, $t_2=x-z$, $t_3=z-1$, and, provided you account for the altered $dxdydz$ element via the Jacobian $\implies |J|dt_1 dt_2 dt_3$, as is standard for change of variables on volume elements, this can be recast as a "simple" iterated integral over function $1/(t_1t_2t_3)$ "around" the point $(0,0,0)$--but that may be better for a separate question.)