# Computing a Generating Set in Cohomology

1. May 18, 2014

### WWGD

Hi, this issue came up in another site:

We want to compute ( not just ) the deRham cohomology of $X=\mathbb R^2 -${p,q} , but also

an explicit generating set for $H^1 (X) = \mathbb Z (+) \mathbb Z$ in deRham cohomology . Only explicit
generating set I can see here is {(0, +/- 1),(+/-1,0)}. How to map this into a pair of 1-forms that generate the first deRham cohomology?

We can get the actual cohomology using , e.g., Mayer Vietoris, but , without an explicit isomorphism, I don't see how to get a generating set. Maybe we can use the explicit maps in the
MV sequence to get some generators (in the deRham chain complex) ?

I was thinking of this: we compute $H^1(X)$, then we know all theories are equivalent, so this gives us singular cohomology , and then we use the explicit isomorphism in deRham's theorem to get some generators? We would see where the generators are sent.

Last edited: May 18, 2014
2. May 18, 2014

### micromass

A standard example is the 1-form

$$\omega = \frac{ydx - xdy}{x^2 + y^2}$$

This defines a closed form on $\mathbb{R}^2\setminus \{(0,0)\}$ that is not exact. So it serves as a basis for the cohomology group $H^1(\mathbb{R}^2\setminus\{(0,0)\)$.

Then consider the diffeomorphisms $F,G:\mathbb{R}^2\rightarrow \mathbb{R}^2$ by $F(x,y) = (x,y) - p$ and $G(x,y) = (x,y) - q$, where $p,q\in\mathbb{R}^2$.

Are then $F^*\omega$ and $G^*\omega$ not generating elements of $H^1(X)$

3. May 18, 2014

### WWGD

Thanks; if I understood you correctly, I think you need to define maps F,G on or into $\mathbb R^2 -p$ to be able to pull back the information, because w is not defined in the entire plane.

I was trying something similar, but the problem is that $\mathbb R^2 -{p,q}$ retracts into a wedge of circles, but is not homeomorphic to it, and I don't see how to use a homotopy equivalence to pullback a form.

I still think deRham's isomorphism , if we can use singular, will send generators to generators. But, again, we would need to know an explicit generating set for singular. Maybe the actual maps in Mayer-Vietoris can help get the explicit generating set. Thanks.

4. May 18, 2014

### micromass

Yes, you can just restrict domain and codomain to something suitable.

Sure.

Finding an explicit generating set for singular homology doesn't seem all that difficult. The intuition is that a circle around one of the points cannot be deformed to a point. So the idea is just to take two cycles, one surrounding $p$ and one surrounding $q$. Then (in theory) we can find 1-forms associated with these cycles which form a basis for the deRham groups, but this seems difficult in practice.

5. May 23, 2014

### WWGD

Never mind, thanks, I found the answer. It is simpler than I thought.