# Multivariate residues in simple cases

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!

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.
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.