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This question was previously asked here. I am posting it here also to increase the potential number of people who will see it. I realize that this question might not be entirely in the spirit of mathoverflow. Please bear with me if you could. I am really eager to learn the resolution of the below problem.

Consider the following two form in two complex variables $z_1$ and $z_2$ on $\mathbb{C}\otimes\mathbb{C}$ (where each $\mathbb{C}$ is mapped to a punctured sphere):

$$F(z_1,z_2)=\frac{1}{z_1 z_2}\frac{1}{ f_1(z_1,z_2)}\frac{1}{ f_2(z_1,z_2)}dz_1\wedge dz_2$$

where functions $f_1$ and $f_2$ are defined as:

$$f_1(z_1,z_2)=a+z_1+\frac{z_1}{z_1+z_2}~~~~~,~~~~~f_2(z_1,z_2)=b+z_2+\frac{z_2}{z_1+z_2}$$

and $a,b$ are complex parameters. The global residue theorem states that the sum of all residues gives zero for any function that is holomorphic everywhere up to a finite amount of points (since, by contour deformation on the direct disc product under the generalized Cauchy measure, this corresponds to an integral over a closed contour with no poles enclosed). Let us denote by $R_F([z_1],[z_2])$ the residue obtained from $F(z_1,z_2)$ while localizing to zero the poles $[z_1]$ and $[z_2]$ from the denominator, which are associated with varying $z_1$ and $z_2$ respectively. With this, the global residue theorem yields for $F(z_1,z_2)$:

$$R_F(f_1,f_2)+R_F(z_1,f_2)+R_F(f_1,z_2)+R_F(z_1,z_2)=0$$

In this case there are no poles at infinity. Now it is relatively easy to check that we get:

$$z_1=\frac{-a(1+a+b)}{a+b}~~~,~~~z_2=\frac{-b(1+a+b)}{a+b}~~~\Rightarrow ~~~R_F(f_1,f_2)=\frac{a+b}{ab(1+a+b)}$$ $$z_1=0~~~,~~~z_2=-1-b~~~\Rightarrow ~~~R_F(z_1,f_2)=\frac{-1}{a(1+b)}$$ $$z_1=-1-a~~~,~~~z_2=0~~~\Rightarrow ~~~R_F(f_1,z_2)=\frac{-1}{b(1+a)}$$

The final residue at $z_1=0$ and $z_2=0$ depends on what $a$ and $b$ are. But it can be only one of the two following values:

$$R_F(z_1,z_2)=\frac{1}{a(1+b)}~~~~~~~or~~~~~~~R_F(z_1,z_2)=\frac{1}{b(1+a)}$$

Clearly, the sum of the four terms does not vanish. It is interesting to note that the two possibilities in $R_F(z_1,z_2)$ cancel either $R_F(f_1,z_2)$ or $R_F(z_1,f_2)$, but the first residue $R_F(f_1,f_2)$ is never cancelled by anything. What went wrong?

EDIT

David Speyer asked in the comments what version of the global residue theorem I am using. I am trying to apply the global residue theorem as given in "Principles of Algebraic Geometry" by Griffiths and Harris (in the residues chapter, around page 655 in my version of the book). Also, I found an explicit example of it being applied around equations (3.4) - (3.9) in this paper.

EDIT 2

grghxy pointed out that I should look more carefully into possible contributions from poles at infinity. There are three cases to consider:

1) Only $z_1\rightarrow\infty$

In this case $f_1=z_1+O(z_1^0)$ and $f_2=O(z_1^0)$, therefore

$$F(z_1,z_2)\sim\frac{1}{z_1^2}dz_1\wedge dz_2\rightarrow q_1^2\frac{-1}{q_1^2}dq_1\wedge dz_2 =-dq_1\wedge dz_2$$

so we see that there is no pole and therefore no residue in this case.

2) Only $z_2\rightarrow\infty$

Already checked in 1) by symmetry arguments. No pole and therefore no residue at infinity.

3) $z_1\rightarrow\infty$ and $z_2\rightarrow\infty$

This case is a combination of 1) and 2). So $f_1=z_1+O(z_i^0)$ and $f_2=z_2+O(z_i^0)$, therefore

$$F(z_1,z_2)\sim\frac{1}{z_1^2}\frac{1}{z_2^2}dz_1\wedge dz_2\rightarrow q_1^2q_2^2\frac{-1}{q_1^2}\frac{-1}{q_2^2}dq_1\wedge dq_2 =dq_1\wedge dq_2$$

we see that there is no pole and therefore no residue in this case either.

EDIT 3

grghxy emphasized in the comments that to have a valid setting for the global residue theorem in $\mathbb{C}^2$, one should not only provide the two form $\omega=F(z_1,z_2)$, but also two effective divisors $D_1$ and $D_2$. Then the theorem works if the poles of $\omega$ are "no worse" than $D_1+D_2$ (the exact meaning of this statement is still mostly mysterious to me). I do know that the $D_i$ are constructed from subvarieties of codimension-1, so basically are hypersurfaces defined by vanishing of scalar functions of $z_1$ and $z_2$. However, it is not clear to me which relation and properties in respect to $\omega$ the $D_i$'s should have in order to qualify as input data for the global residue theorem. So I don't know where to start. Any suggestion?

I guess it will be simpler for me to first start with the simplified problem where

$$f_1(z_1,z_2)=1+a_1 z_1+a_2 z_2~~~~~,~~~~~f_2(z_1,z_2)=1+b_1 z_1+b_2 z_2$$

with this the residues obtained from $F(z_1,z_2)$ are:

$$z_1=\frac{b_2-a_2}{a_2 b_1-a_1 b_2}~~~,~~~z_2=\frac{a_1-b_1}{a_2 b_1-a_1 b_2}~~~\Rightarrow ~~~R_F(f_1,f_2)=\frac{a_2 b_1-a_1 b_2}{(a_1-b_1)(a_2-b_2)}$$ $$z_1=0~~~,~~~z_2=\frac{-1}{b_2}~~~\Rightarrow ~~~R_F(z_1,f_2)=\frac{b_2}{a_2-b_2}$$ $$z_1=\frac{-1}{a_1}~~~,~~~z_2=0~~~\Rightarrow ~~~R_F(f_1,z_2)=\frac{a_1}{b_1-a_1}$$ $$z_1=0~~~,~~~z_2=0~~~\Rightarrow ~~~R_F(z_1,z_2)=1$$

These are all the poles and all the residues that appear in this particular $F(z_1,z_2)$. Now one can check straightforwardly that:

$$R_F(f_1,f_2)+R_F(z_1,f_2)+R_F(f_1,z_2)+R_F(z_1,z_2)=\\~~~~~~~~~~=\frac{a_2 b_1-a_1 b_2}{(a_1-b_1)(a_2-b_2)}+\frac{b_2}{a_2-b_2}+\frac{a_1}{b_1-a_1}+1=0$$

so the global residue theorem is satisfied. If I could find the hidden $D_1$ and $D_2$ that are at work here, I might be able to generalize to the harder problem above. Any suggestions on how to go about here?

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    $\begingroup$ Could you provide a link to the multivariate global residue theorem which you are using? Google is pointing me only to this paper by Griffiths projecteuclid.org/euclid.bams/1183544569 , which is about residues of $n$ functions of $n$ variables. $\endgroup$ Aug 14, 2015 at 14:03
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    $\begingroup$ @David Speyer : I am trying to apply the global residue theorem as given in "Principles of Algebraic Geometry" by Griffiths and Harris (in the residues chapter, around page 655 in my version of the book). Also, I found an explicit example of it being applied around equations (3.4) - (3.9) in the paper arxiv.org/pdf/1311.5200v1.pdf $\endgroup$
    – Kagaratsch
    Aug 14, 2015 at 14:17
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    $\begingroup$ Thanks! I think I'll have to wait until I get home to my books to look at this, or wait for someone else to do so. The relevant pages of G&H aren't online. That paper doesn't state the theorem they are using, although it sure looks as if they are factoring their two variable integrand into two factors, which would be consistent with the paper I linked. $\endgroup$ Aug 14, 2015 at 15:05
  • $\begingroup$ @grghxy : Yes, I did mean to multiply $F(z_1,z_2)$ by $dz_1\wedge dz_2$ to make it a two form, I will edit my question to mention it. I was thinking of having each $\mathbb{C}$ be mapped to a punctured sphere, where the puncture corresponds to a point at infinity that can also contain a residue. Since the contour in $z_1$ and $z_2$ is a polydisc I hope that having a direct product of two spheres as the underlying manifold is compact. Not sure if I make any sense or am addressing your question properly? Let me know if I should think again. $\endgroup$
    – Kagaratsch
    Aug 14, 2015 at 15:06
  • $\begingroup$ @grghxy I did consider transformations of the integrand under $z_1\rightarrow 1/q_1$ and/or $z_2\rightarrow 1/q_2$ and it looked like there were no poles at infinity (so at $q_1=0$ and or $q_2=0$) in all the cases. (If I understood the issue you are pointing out correctly.) I will check again! $\endgroup$
    – Kagaratsch
    Aug 14, 2015 at 15:18

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