# Understanding the global residue theorem

I am studying the global residue theorem which applies for multivariate residues. The theorem is reported in the following references e.g. (1) (Eq.87) , (2) (Eq. 109) in different formulations, that now I would try to understand. It is unnecessary to reveal that my education is not from math studies, so I am sorry if I won't be precise.

• First of all, let us focus on the following definition (1) (see Eq.87).

Theorem 2. (Global residue theorem). Let $$\omega$$ denote a meromorphic $$n$$-form defined on a compact manifold $$M$$. Given an open covering $$\{U_i\}$$, let $$\omega$$ take the local form
$$\omega=\frac{h(z)dz_1 ∧ · · · ∧ dz_n}{f_1(z)· · · f_n(z)}$$ where $$f(z) = (f_1(z), . . . , f_n(z)):\mathbb{C}^n \rightarrow \mathbb{C}^n$$ and $$h(z): \mathbb{C}^n \rightarrow C$$ are holomorphic functions. Let $$D_j= \{z ∈ M : f_j (z) = 0\}$$ with $$j = 1, . . . , n$$ denote the divisors of $$\omega$$, and assume that $$V = D_1 ∩· · ·∩ D_n$$ is a finite set. Then $$\sum_{p∈V} Res_p\omega = 0$$ where each $$Res_p\omega$$ is evaluated locally on a patch $$U_i$$ which contains $$p$$.

Strictly speaking, if we have a form which is defined on $$\mathbb{C}^n$$, the theorem does not apply. This is why in (1) is suggested to compactify $$\mathbb{C}^n$$ into $$\mathbb{CP}^n$$ and then apply the theorem. This is done through the change of coordinate $$z_1 = \frac{w_1}{w_0}\,,\,. . . \,,\,z_n = \frac{w_n}{w_0}$$ and the open covering $$\{U_k\}$$ is defined as

$$U_k = \{(w_0,w_1,...w_n): w_k=1\}\,, \text{for } k=0,1,...n$$

The form $$\omega$$ on the patch $$U_k$$ then takes the expression (see Eq. 92 in (1))

$$\omega|_{U_k} = \frac{(-1)^k\, h(w/w_0)\, dw_0\,\wedge\,...\wedge dw_n}{w_0^{n+1}f_1(w/w_0)...f_n(w/w_0)}$$

Question: Is there a sufficient condition on the polynomials $$h(z),f_i(z)$$ such that the zeros of $$f(z) = (f_1(z),...,f_n(z))$$ are all the points contained in the set $$V$$ in the open covering $$U_0$$?

• According to (2) the theorem can also be stated in the following way

Let $$\omega = h\,dz/f_1 ...f_n$$ be defined by polynomials $$h$$ and $$f_i$$. Let $$F_i = \{z ∈ \mathbb{C}^n : f_i(z) = 0\}$$ be the hypersurface (i.e. $$n − 1$$ dimensional subspace) associated with $$f_i$$ and $$Z = F_1 ∩F_2 ∩...∩F_n$$ be the set of zeroes of $$f$$. Here we assume that $$Z$$ is a discrete set of points. Then one defines the Global residue of $$h$$ with respect to the map $$f$$ as $$Res_f (h) = \sum_{a∈Z} res(ω)_a.$$ Now, the Global Residue Theorem (GRT) states that if $$deg(h) < deg(f_1) + . . . + deg(f_n) − n$$ then $$Res_f (h) = 0$$.

This formulation seems to give an answer to my question about the sufficient condition, i.e. provided $$deg(h) < deg(f_1) + . . . + deg(f_n) − n$$ . However, I don't understand it. For example, this formulation does not talk about compact manifolds and seems to be very general. However, it does not seem quite exact (maybe I am misunderstanding). Consider for example the form $$\omega = \frac{z_2^2 z_1 dz_1\wedge dz_2}{(1-z_2 -z_1 +2z_1 z_2)z_1(z_2-z_1)z_2(z_1-1)}$$ with the map $$f(z) =\left((1-z_2 -z_1 +2z_1 z_2)z_1(z_2-z_1),z_2(z_1-1)\right)$$. The set $$Z$$ is given by discrete points $$Z=\{(0,0),(1,0),(1,1)\}$$. Moreover, $$\text{deg}(h) = 3$$ and $$\text{deg}(f_1)+\text{deg}(f_2) = 5 + 1 = 6$$, then the condition $$3 < 6 - 2 = 4$$ is satisfied and I would expect the theorem holds. Instead, by direct computation (by hand) I get a non-zero global residue. If you don't want to do computations by hand, you can use the Mathematica package MultivariateResidues, the code is shown below. Where am I wrong? Is the definition of $$\text{deg}(...)$$ more complicated?

You can copy and past the following Mathematica Code to reproduce my result

Get["MultivariateResidues.m"];
sols = {(1 - w2 + w1 (-1 + 2 w2)) w1 (w2 - w1) == 0, w2 (w1 - 1) == 0} // Solve;
listResidues = {};
Print[Dynamic[ii], "/", Length[sols]]
For[ii = 1, ii <= Length[sols], ii++,
AppendTo[listResidues, (MultivariateResidue[w2^2 w1, {(1 - 1 w2 + w1 (-1 + 2 w2)) w1 (w2 - w1), w2 (w1 - 1)},sols[[ii]]] // Simplify)]]
listResidues2 /. List -> Plus // Simplify
(* Output: -1 *)


1. Let us start by discussing OP's example. One of the 3 singularities that OP mentions, namely the point $$(0,0)$$, is a removable singularity. Let us for simplicity remove common factors in the numerator and the denominator.

2. OP's 2-form then becomes: $$\omega ~=~ \frac{z_2 \mathrm{d}z_1\wedge \mathrm{d}z_2}{\underbrace{(1-z_2- z_1 +2z_1 z_2)}_{=g_3(z)}\underbrace{(z_2-z_1)}_{=g_2(z)}\underbrace{(z_1-1)}_{=g_1(z)}}. \tag{1}$$

3. We are interested in points where precisely 2 of the denominator parenthesis vanish$$^1$$ \begin{align} g_1(z)~=~0 ~\wedge~ g_2(z)~=~0:& \qquad z~=~(1,1), \cr g_1(z)~=~0 ~\wedge~ g_3(z)~=~0:& \qquad z~=~(1,0), \cr g_2(z)~=~0 ~\wedge~ g_3(z)~=~0:& \qquad z~=~(b_{\pm},b_{\pm}), \qquad b_{\pm}~:=~\frac{1\pm i}{2}.\end{align}\tag{2} The first 2 poles $$(1,1)$$ and $$(1,0)$$ were already mentioned by OP. OP did not mention the last 2 poles $$(b_{\pm},b_{\pm})$$.

4. The issue is that we might have to change how we factorize into 2 factors $$f_1(z)$$ and $$f_2(z)$$ in the denominator $$f_1(z)f_2(z)=g_3(z)g_2(z)g_1(z)$$ to get to all pertinent poles. Apart from this point, which isn't really spelled out in Ref. 2, the non-compact version of the global residue theorem (GRT) is correct.

5. Let us calculate all the 4 residues: \begin{align} \oint\oint_{(1,1)} \omega ~=~& \oint\oint_{(0,0)}\frac{ \mathrm{d}h_1\wedge \mathrm{d}h_2}{(h_2-h_1)h_1} ~=~(2\pi i)^2, \cr \oint\oint_{(1,0)} \omega ~=~& \oint\oint_{(0,0)}\frac{h_2 \mathrm{d}h_1\wedge \mathrm{d}h_2}{(h_2- h_1)(-1)h_1} \cr ~=~& \oint\oint_{(0,0)}\frac{(k_2+h_1) \mathrm{d}h_1\wedge \mathrm{d}k_2}{k_2(-1)h_1}~=~0, \cr \oint\oint_{(b_{\pm},b_{\pm})} \omega ~=~& \oint\oint_{(0,0)}\frac{b_{\pm} \mathrm{d}h_1\wedge \mathrm{d}h_2}{(h_1+h_2)(h_2-h_1)(b_{\pm}-1)^2}\cr ~=~& \oint\oint_{(0,0)}\frac{b_{\pm} \mathrm{d}k_1\wedge \mathrm{d}k_2}{k_1k_2(\mp i)}~=~\frac{-1\pm i}{2}(2\pi i)^2.\end{align}\tag{3} Note that the residues indeed sums up to zero as they should according to the non-compact version of the global residue theorem (GRT) in Ref. 2.

References:

1. K.J. Larsen & R. Rietkerk, MultivariateResidues: a Mathematica package for computing multivariate residues, arXiv:1701.01040; eq. (87).

2. N. Arkani-Hamed, F. Cachazo, C. Cheung, & J. Kaplan, A Duality For The S-Matrix, arXiv:0907.5418; eq. (109).

--

$$^1$$ Another potential candidate is if the second-order polynomial $$g_3(z)$$ simultaneously has a simple zero wrt. both the variables $$z_1$$ and $$z_2$$. However, this is not the case since the differential $$\mathrm{d}g_3~=~ (2z_2-1)\mathrm{d}z_1+(2z_1-1)\mathrm{d}z_2\tag{4}$$ does not vanish on the curve $$0~=~g_3(z)~\equiv~ 2\left(z_1-\frac{1}{2}\right)\left(z_2-\frac{1}{2}\right)+\frac{1}{2}.\tag{5}$$

• Thank you for the answer. The Theorem of Ref 2. refers to a specific map $f$ in terms of which we should compute the residues, right? I was choosing the map $f= (g_1(z),g_2(z)g_3(z))$ which indeed gives only the poles $(1,1)$ and $(1,0)$. So, given this map $f$, why should I sum up the residues wrt the map $f'= (g_1(z)g_2(z),g_3(z))$? This is not clear to me.. I understood that the the order of the factorisation should be fixed. Oct 5, 2019 at 16:53
• No, it should not. We need to take all singularities into account. Oct 6, 2019 at 14:32
• I have accepted the answer because the bounty was expiring and the answer is partially useful. If you like, you can improve the answer, for example by defining the "Global residue of ℎ with respect to the map 𝑓 ". From the way the theorem is written, it is not clear to me that you have to consider all permutations of denominators. Could you expand your answer on this point? thank you. Oct 6, 2019 at 17:40
• I updated the answer. Oct 6, 2019 at 18:30