# Global Residue Theorem in $\mathbb{CP}^2$.

Consider the following meromorphic form defined on $\mathbb{CP}^2$: Be $\phi_{1,2}$ the chart given by $\phi_{1,2}(Z_1,Z_2) = (1,Z_1,Z_2)$, in these coordinates define $\omega= \frac{1}{Z_1 Z_2} dZ_1 \wedge dZ_2$. Then the form can be given as $\bigwedge_i \frac{dZ_i}{(Z_i)}$ in any of the coordinate charts where one of the $z$'s is $1$ and the others are free (e.g. $(Z_1,1,Z_2)$) a part from a possible sign that will be irrelevant in the following.

Now, it seems to me that such form has exactly three poles, $(1,0,0)$,$(0,1,0)$,$(0,0,1)$. The residues can be calculated as in pag. $650$ of Griffiths & Harris and one obtains either $1$ or $-1$.

Now, since $\mathbb{CP}^2$ is compact and with no boundary, the sum of the residues of $\omega$ should be zero, but this is impossible (no matter what the actual signs are, $\pm1\pm1\pm1$ cannot be zero).

So where is the problem?

• In dimensions $> 1$ there are no isolated singularities. The pole set is a (complex) one-dimensional subspace. – Daniel Fischer Jul 24 '15 at 14:16
• By "pole" i meant the common zeroes of the functions appearing in the denominator, using the notation of pag 649 f_i = Z_i, so they should be just (1,0,0), ... and not the points of the form (1,z,0), ... (is this the 1-dim subspace you had in mind?) – giulio bullsaver Jul 24 '15 at 15:22
• You seem to believe that there is a theorem which says that, if $X$ is a compact complex surface, and $\omega$ is a meromorphic $2$-form with normal crossing poles, then the sums of the residues of $\omega$ at the zero-strata of the pole divisor should be $0$. There is no such theorem; you have just given a counter-example. If you could tell us why you believe this, we could maybe say more about why you are wrong. – David E Speyer Jul 30 '15 at 19:39

Ah, I have just looked up what the global residue theorem says. According to Griffiths, the statement is the following: Let $X$ be a compact connected complex $n$-fold and let $D_1$, ..., $D_n$ be $n$ divisors such that $D_1 \cap \cdots \cap D_n$ is a discrete set $Z$. Let $\omega$ be a meromorphic form with poles along $\bigcup D_i$. Then $\sum_{p \in Z} \mathrm{res}_p \omega = 0$.
In your setting, you want $X = \mathbb{P}^2$, and $\omega = \frac{d (x_1/x_3) \wedge d (x_2/x_3)}{(x_1/x_3) (x_2/x_3)}$. The pole locus of $\omega$ is $\{ (x_1: x_2: x_3) : x_1 x_2 x_3=0 \}$. This divisor has three components, but in order to apply Griffiths formulation I must write it as a union of two components: Say $D_1 = \{ x_1 x_2 =0 \}$ and $D_2 = \{ x_3 = 0 \}$. Then $D_1 \cap D_2$ is just two points, consistent with Griffiths formulation. This explains your confusion: The third $0$-stratum of your boundary is not a point of $D_1 \cap D_2$.
I guess this reflects my lack of classical training, but I had never seen the result expressed this way. The statement I knew was the following: Suppose we are giving $n+1$ divisors, $D_0$, $D_1$, ..., $D_n$ such that any $n$-fold intersection is finite and any $(n+1)$-fold intersection is empty. Let $\omega$ be a meromorphic differential form with poles restricted to $\bigcup D_i$. Define $$R_j(\omega) = (-1)^{j} \sum_{p \in D_0 \cap \cdots \widehat{D_j} \cdots \cap D_n} \mathrm{Res}_p \omega.$$ (We compute the residue with respect to the ordering $(D_0, \cdots \widehat{D_j}, \cdots, D_n)$ of the coordinates.) Then $R_j$ is independent of $j$.
Presumably, some easy linear algebra will deduce this formulation from Griffiths' formulation. To deduce Griffiths from this one, just take $D_0 = \emptyset$.
The reason this result is important is the following: Set $U_i = X \setminus D_i$, so $(U_0, U_1, \ldots, U_n)$ is a cover of $X$. Then $\omega$ is a Cech cocycle for $H^n(X, \Omega^n)$, and $(-1)^j R_j$ is (more or less) the Serre duality isomorphism $H^n(X, \Omega^n) \cong \mathbb{C}$.