# Cohomology Calculation

A couple of days ago I asked this Question on calculating hypercohomology

I tried a similar example for $(\mathbb{C}^*)^2$, and I have a couple of questions.

Here is my calculation:

We have a cochain of $\mathbb{C}[x^{\pm 1}, y^{\pm 1} ]$-modules:

$$0 \longrightarrow \mathbb{C}[x^{\pm 1}, y^{\pm 1} ] \longrightarrow \left< dx,dy \right> \longrightarrow \left< \ dx \wedge dy \ \right> \longrightarrow 0$$

with codifferential maps:

$d (f) = \frac{\partial f}{\partial x} \cdot dx + \frac{\partial f}{\partial y} \cdot dy$

$d (f \cdot dx + g \cdot dy) = \left( \frac{\partial g}{\partial x} - \frac{\partial f}{\partial y} \right) \cdot ( dx \wedge dy) \\$

and we get:

$H^0_{dR} = \mathbb{C} \ \ \ \$ , $\ \ \ \ H^2_{dR} \cong \mathbb{C} \left( \frac{1}{xy} \right)\cdot ( dx \wedge dy) \cong \mathbb{C} \ \ \$ , $\ \ \ H^1_{dR} \cong \mathbb{C} \left(\frac{1}{x} \right)\cdot dx + \mathbb{C} \left(\frac{1}{y} \right)\cdot dx \cong \mathbb{C}^2$.

My Questions:

Firstly, is the above correct?

Secondly, if it is correct then is there a nice way to visualize what I get for $H^1$? I understand the space looks like 4 dimensional euclidean space with two planes removed that touch at a single point?

Here's a bad attempt at trying to imagine it, don't know whether it means anything. I tried to think of squashing the planes at the point so i can see it in 3d, and then use the unseen dimension to move around with one of the paths:

How you visualize what you get for $H^1$ is somewhat clear--and similar to your picture. Namely, you have for each copy of $\mathbb{G}_m$ a generating loop. The loops then fit together to form a pair of independent loops in the product. It's precisely what happens for the torus. This independence is seen by the independence of the differentials.
In fact, this is topologically correct considering the comparison theorem of Grothendieck, telling us that in our case, you're just computing the singular cohomology of $(\mathbb{C}^\times)^2$, which is homotopy equivalent to $S^1\times S^1$.
• Does my calculation above generalize straight away to get something like: $H^p\mathbb{\left( (C^*)^n \right)} = \mathbb{C}^{\binom{n}{p}}$ ? Jul 25 '14 at 11:45
• @JC574 Right :). The comparison theorem is one of the main reasons why algebraic de Rham cohomology is considered a 'nice' cohomology theory, for varieties in characteristic $0$. Jul 26 '14 at 0:31