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?

  • $\begingroup$ In dimensions $> 1$ there are no isolated singularities. The pole set is a (complex) one-dimensional subspace. $\endgroup$ – Daniel Fischer Jul 24 '15 at 14:16
  • $\begingroup$ 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?) $\endgroup$ – giulio bullsaver Jul 24 '15 at 15:22
  • $\begingroup$ 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. $\endgroup$ – 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 $H^n(X, \Omega^n) \cong \mathbb{C}$.

| cite | improve this answer | |
  • $\begingroup$ Can you suggest (maybe a reference) some explicit example of the statement you knew? Just to train with the theorem in this form. Thanks $\endgroup$ – apt45 Sep 28 '19 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.