show/hide this revision's text 3 Removed wrong idea

Remove denominators in a differential relationde Rham cohomology

Let $\omega = \mathrm d \eta$ be an exact rational $n$-form on $\Bbb P^n$.

It may happen that the polar locus of $\eta$ is not included in the polar locus of $\omega$. But is it true that $\omega = \mathrm d \eta_0$, where $\eta_0$ is a $(n-1)$-form which is regular where $\omega$ is ?

In other words (and considering the affine case), assume that $$ F = \partial_1 G_1 + \dotsb + \partial_n G_n $$ with $F$ and $G_i$'s rational functions in $x_1,\dotsc,x_n$. Is it possible to choose the $G_i$ such that there denominators are a power of the denominator of $F$ ?

I've been unable to provide a counter example to this question whereas the following formulation seems to indicate that there exists a one.

Let $X$ the open set of $\Bbb P^n$ where $\omega$ is regular, and $Z$ the polar locus, in $X$, of $\eta$, so that $Z$ is a hypersurface of $X$. If I'm not mistaken, a positive answer to my question is equivalent to the injectivity of the restriction map in the de Rham cohomology $$H^n(X)\to H^n(X\setminus Z).$$

And following Hartshorne's On the De Rham cohomology of algebraic varieties, we have an exact sequence $$ \dotsb \to H^{n-1}(X\setminus Z) \to H^{n-2}(Z) \to H^n(X)\to H^n(X\setminus Z), $$ so that the injectivity of the last arrow is equivalent to the nullity of the previous one, or the sujectivity of the one before, the Poincaré residue map. Is there any reason for the this residue map to be surjective ? What if $Z$ is an smooth ? or is a hyperplane ?

Yet another formulation uses homology, thanks to Poincaré duality, the question boils down

For $n=2$ we have to check that $H_n(X\setminus $0\to H^1(X) \to H^1(X\setminus Z) \to H_n(X)$ H^0(Z) \to 0$$ is surjectiveexact. My intuition tells me it's true when $n = 2$ : if we have a $2$-cycle in $X$ my guess is that it can be deformed in order Which seems to avoid $Z$ since both object (the cycle and $Z$) will have real dimension 2 in a space of real dimension 4. For $n$ larger, it may be falsetrue... Although I don't know why.

Any thought will be much appreciated!, including on the specific case $n=2$.
show/hide this revision's text 2 Improve the question

Injectivity of the restriction map Remove denominators in de Rham cohomologya differential relation

Let $\omega = \mathrm d \eta$ be an exact rational $n$-form on $\Bbb P^n$.

It may happen that the polar locus of $\eta$ is not included in the polar locus of $\omega$. But is it true that $\omega = \mathrm d \eta_0$, where $\eta_0$ is a $(n-1)$-form which is regular where $\omega$ is ?

In other words (and considering the affine case), assume that $$ F = \partial_1 G_1 + \dotsb + \partial_n G_n $$ with $F$ and $G_i$'s rational functions in $x_1,\dotsc,x_n$. Is it possible to choose the $G_i$ such that there denominators are a power of the denominator of $F$ ?

I've been unable to provide a counter example to this question whereas the following formulation seems to indicate that I'm wrongthere exists a one.

Let $X$ the open set of $\Bbb P^n$ where $\omega$ is regular, and $Z$ the polar locus, in $X$, of $\eta$, so that $Z$ is a hypersurface of $X$. If I'm not mistaken, a positive answer to my question is equivalent to the injectivity of the restriction map in the de Rham cohomology $$H^n(X)\to H^n(X\setminus Z).$$

And following Hartshorne's On the De Rham cohomology of algebraic varieties, we have an exact sequence $$ \dotsb \to H^{n-1}(X\setminus Z) \to H^{n-2}(Z) \to H^n(X)\to H^n(X\setminus Z), $$ so that the injectivity of the last arrow is equivalent to the nullity of the previous one, or the sujectivity of the one before, the Poincaré residue map. Is there any reason for the this residue map to be surjective ? What if $Z$ is an smooth ? or is a hyperplane ?

Yet another formulation uses homology, thanks to Poincaré duality, the question boils down to check that $H_n(X\setminus Z) \to H_n(X)$ is surjective. My intuition tells me it's true when $n = 2$ : if we have a $2$-cycle in $X$ my guess is that it can be deformed in order to avoid $Z$ since both object (the cycle and $Z$) will have real dimension 2 in a space of real dimension 4. For $n$ larger, it may be false.

Any thought will be much appreciated !
show/hide this revision's text 1

Injectivity of the restriction map in de Rham cohomology

Let $\omega = \mathrm d \eta$ be an exact rational $n$-form on $\Bbb P^n$.

It may happen that the polar locus of $\eta$ is not included in the polar locus of $\omega$. But is it true that $\omega = \mathrm d \eta_0$, where $\eta_0$ is a $(n-1)$-form which is regular where $\omega$ is ?

I've been unable to provide a counter example to this question whereas the following formulation seems to indicate that I'm wrong.

Let $X$ the open set of $\Bbb P^n$ where $\omega$ is regular, and $Z$ the polar locus, in $X$, of $\eta$, so that $Z$ is a hypersurface of $X$. If I'm not mistaken, a positive answer to my question is equivalent to the injectivity of the restriction map in the de Rham cohomology $$H^n(X)\to H^n(X\setminus Z).$$

And following Hartshorne's On the De Rham cohomology of algebraic varieties, we have an exact sequence $$ \dotsb \to H^{n-1}(X\setminus Z) \to H^{n-2}(Z) \to H^n(X)\to H^n(X\setminus Z), $$ so that the injectivity of the last arrow is equivalent to the nullity of the previous one, or the sujectivity of the one before, the Poincaré residue map. Is there any reason for the this residue map to be surjective ? What if $Z$ is an smooth ? or is a hyperplane ?