# 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

Staff Emeritus
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

Staff Emeritus
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.