# Poincaré residue

In mathematics, the Poincaré residue is a generalization, to several complex variables and complex manifold theory, of the residue at a pole of complex function theory. It is just one of a number of such possible extensions.
Given a hypersurface $X\subset \mathbb {P} ^{n}$ defined by a degree $d$ polynomial $F$ and a rational $n$-form $\omega$ on $\mathbb {P} ^{n}$ with a pole of order $k>0$ on $X$, then we can construct a cohomology class $\operatorname {Res} (\omega )\in H^{n-1}(X;\mathbb {C} )$. If $n=1$ we recover the classical residue construction.

## Historical construction

When Poincaré first introduced residues he was studying period integrals of the form
   ${\underset {\Gamma }{\iint }}\omega$ for $\Gamma \in H_{2}(\mathbb {P} ^{2}-D)$
where $\omega$ was a rational differential form with poles along a divisor $D$. He was able to make the reduction of this integral to an integral of the form
   $\int _{\gamma }{\text{Res}}(\omega )$ for $\gamma \in H_{1}(D)$
where $\Gamma =T(\gamma )$, sending $\gamma$ to the boundary of a solid $\varepsilon$-tube around $\gamma$ on the smooth locus $D^{*}$of the divisor. If
   $\omega ={\frac {q(x,y)dx\wedge dy}{p(x,y)}}$
on an affine chart where $p(x,y)$ is irreducible of degree $N$ and $\deg q(x,y)\leq N-3$ (so there is no poles on the line at infinity $^{page150}$). Then, he gave a formula for computing this residue as
   ${\text{Res}}(\omega )=-{\frac {qdx}{\partial p/\partial y}}={\frac {qdy}{\partial p/\partial x}}$
which are both cohomologous forms.

## Construction
### Preliminary definition

Given the setup in the introduction, let $A_{k}^{p}(X)$ be the space of meromorphic $p$-forms on $\mathbb {P} ^{n}$ which have poles of order up to $k$. Notice that the standard differential $d$ sends
    $d:A_{k-1}^{p-1}(X)\to A_{k}^{p}(X)$
Define
    ${\mathcal {K}}_{k}(X)={\frac {A_{k}^{p}(X)}{dA_{k-1}^{p-1}(X)}}$
as the rational de-Rham cohomology groups. They form a filtration
   ${\mathcal {K}}_{1}(X)\subset {\mathcal {K}}_{2}(X)\subset \cdots \subset {\mathcal {K}}_{n}(X)=H^{n+1}(\mathbb {P} ^{n+1}-X)$
corresponding to the Hodge filtration.

### Definition of residue

Consider an $(n-1)$-cycle $\gamma \in H_{n-1}(X;\mathbb {C} )$. We take a tube $T(\gamma )$ around $\gamma$ (which is locally isomorphic to $\gamma \times S^{1}$) that lies within the complement of $X$. Since this is an $n$-cycle, we can integrate a rational $n$-form $\omega$ and get a number. If we write this as
    $\int _{T(-)}\omega :H_{n-1}(X;\mathbb {C} )\to \mathbb {C}$
then we get a linear transformation on the homology classes. Homology/cohomology duality implies that this is a cohomology class
    $\operatorname {Res} (\omega )\in H^{n-1}(X;\mathbb {C} )$
which we call the residue. Notice if we restrict to the case $n=1$, this is just the standard residue from complex analysis (although we extend our meromorphic $1$-form to all of $\mathbb {P} ^{1}$. This definition can be summarized as the map
   ${\text{Res}}:H^{n}(\mathbb {P} ^{n}\setminus X)\to H^{n-1}(X)$

### Algorithm for computing this class

There is a simple recursive method for computing the residues which reduces to the classical case of $n=1$. Recall that the residue of a $1$-form
    $\operatorname {Res} \left({\frac {dz}{z}}+a\right)=1$
If we consider a chart containing $X$ where it is the vanishing locus of $w$, we can write a meromorphic $n$-form with pole on $X$ as
    ${\frac {dw}{w^{k}}}\wedge \rho$
Then we can write it out as
    ${\frac {1}{(k-1)}}\left({\frac {d\rho }{w^{k-1}}}+d\left({\frac {\rho }{w^{k-1}}}\right)\right)$
This shows that the two cohomology classes
    $\left[{\frac {dw}{w^{k}}}\wedge \rho \right]=\left[{\frac {d\rho }{(k-1)w^{k-1}}}\right]$
are equal. We have thus reduced the order of the pole hence we can use recursion to get a pole of order $1$ and define the residue of $\omega$ as
    $\operatorname {Res} \left(\alpha \wedge {\frac {dw}{w}}+\beta \right)=\alpha |_{X}$

## Example

For example, consider the curve $X\subset \mathbb {P} ^{2}$ defined by the polynomial
    $F_{t}(x,y,z)=t(x^{3}+y^{3}+z^{3})-3xyz$
Then, we can apply the previous algorithm to compute the residue of
    $\omega ={\frac {\Omega }{F_{t}}}={\frac {x\,dy\wedge dz-y\,dx\wedge dz+z\,dx\wedge dy}{t(x^{3}+y^{3}+z^{3})-3xyz}}$
Since
    ${\begin{aligned}-z\,dy\wedge \left({\frac {\partial F_{t}}{\partial x}}\,dx+{\frac {\partial F_{t}}{\partial y}}\,dy+{\frac {\partial F_{t}}{\partial z}}\,dz\right)&=z{\frac {\partial F_{t}}{\partial x}}\,dx\wedge dy-z{\frac {\partial F_{t}}{\partial z}}\,dy\wedge dz\\y\,dz\wedge \left({\frac {\partial F_{t}}{\partial x}}\,dx+{\frac {\partial F_{t}}{\partial y}}\,dy+{\frac {\partial F_{t}}{\partial z}}\,dz\right)&=-y{\frac {\partial F_{t}}{\partial x}}\,dx\wedge dz-y{\frac {\partial F_{t}}{\partial y}}\,dy\wedge dz\end{aligned}}$
and
    $3F_{t}-z{\frac {\partial F_{t}}{\partial x}}-y{\frac {\partial F_{t}}{\partial y}}=x{\frac {\partial F_{t}}{\partial x}}$
we have that
    $\omega ={\frac {y\,dz-z\,dy}{\partial F_{t}/\partial x}}\wedge {\frac {dF_{t}}{F_{t}}}+{\frac {3\,dy\wedge dz}{\partial F_{t}/\partial x}}$
This implies that
    $\operatorname {Res} (\omega )={\frac {y\,dz-z\,dy}{\partial F_{t}/\partial x}}$