# Computing Čech cohomology groups

1. Jul 19, 2014

### Geometry_dude

I'm trying to understand Čech cohomology and for this I'm looking at the example of $S^1$ defined as $[0,1]/\sim$ with $0 \sim 1$. To compute everything, I have the cover $\mathcal U$ consisting of the sets
$$U_0= (0, 2/3) \, , \, U_1= (1/3, 1) \, , \, U_2= (2/3, 1] \cup [0, 1/3)$$
Obviously, $\mathcal U$ is a finite, good, open cover.
Since
$\check C ^0 (\mathcal U, \mathbb R) \simeq \mathbb R^{\# \mathcal U}$
for any good, finite, open cover of a topological space, we have in our case
$$\check C ^0 (\mathcal U, \mathbb R) \simeq \mathbb R ^3 \, .$$
Taking intersections, we get the sets
$$U_{20}\equiv U_2 \cap U_0 = (0,1/3) \, ,\, U_{01}=(1/3, 2/3) \, ,\, U_{12}=(2/3,1) \, .$$
Again, we only need to count these and get $\check C ^1 (\mathcal U, \mathbb R) \simeq \mathbb R^{3}$. All these are disjoint and hence $\check C ^2 (\mathcal U, \mathbb R)= \mathbb \lbrace 0 \rbrace$.
Now we want to compute the Čech cohomology groups $\check H ^k (S^1, \mathbb R)$, which are independent of the cover. We take the Čech differential
$$\delta_k \colon \check C^k( \mathcal U, \mathbb R ) \to \check C^{k+1} ( \mathcal U, \mathbb R )$$
defined as usual and compute
$$\delta_0 f _{\alpha_0 \alpha_1} = f_{\alpha_1} - f_{\alpha_0}$$
for $f \in \check C ^0 (\mathcal U, \mathbb R)$ and
$$\delta_1 g _{\alpha_0 \alpha_1 \alpha_2} =g_{ \alpha_1 \alpha_2} - g_{ \alpha_0 \alpha_2} +g_{ \alpha_0 \alpha_1} \, .$$
Since $\check C ^2 (\mathcal U, \mathbb R)= \mathbb \lbrace 0 \rbrace$, we have
$$\delta_1\bigl(\check C ^1 (\mathcal U, \mathbb R)\bigr) = \lbrace 0 \rbrace \quad \implies \quad \ker \delta_1 = \check C ^1 (\mathcal U, \mathbb R) \simeq \mathbb R^3$$
and since $\check C ^{-1} (\mathcal U, \mathbb R) := \mathbb \lbrace 0 \rbrace$ and $\delta$ is linear, we get
$$\delta_{-1}\bigl(\check C ^{-1} (\mathcal U, \mathbb R)\bigr)= \lbrace 0 \rbrace .$$
The latter is true in general. It follows that for a general topological space $M$ we get
$$\check H^0(M,\mathbb R) = \ker \delta_0$$
and by looking at the formula for $\delta_0 f$ to get $\ker \delta_0$ we observe that $f$ has to be constant on each connected component of $M$ and thus we have $1$ degree of freedom for each connected component, which implies
$$\check H^0(M,\mathbb R) \simeq \mathbb R^{\# M} \, ,$$
where $\# M$ is the number of connected components of $M$. Thus
$$\check H^0(S^1,\mathbb R) \simeq \mathbb R \, .$$
Here comes my question: How do I compute the image of $\delta_{0}$ directly? Do I just take $g= \delta f$, look at all the representatives and count degrees of freedom?
If I do that, I get
$$\delta_0 \bigl(\check C ^0 (\mathcal U, \mathbb R)\bigr) \simeq \mathbb R ^3 \, ,$$
which is wrong, since
$$\dim \ker \delta_0 + \dim \delta_0 \bigl( \check C ^0 (\mathcal U, \mathbb R) \bigr) = \dim \check C ^0 (\mathcal U, \mathbb R) = 3 \, .$$
As $$\dim \ker \delta_0 = 1$$ by the previous reasoning, we get
$$\dim \delta_0 \bigl( \check C ^0 (\mathcal U, \mathbb R) \bigr) = 2$$
and hence
$$\check H^1 \simeq \mathbb R^3 / \mathbb R ^2 \simeq \mathbb R \, .$$
Is that correct?
It is easy to see that all the higher cohomology groups have to vanish.

2. Jul 19, 2014

### Geometry_dude

Nevermind, I think I understood my error now. We have
$$\delta_0 \colon \check C^0( \mathcal U, \mathbb R) \to \check C^1( \mathcal U, \mathbb R)$$
and
$$g_{20}= f_0 - f_2 \, , \, g_{01}= f_1 - f_0 \, , \, g_{12}= f_2 - f_1 \, .$$
It's a linear map, so we look at the matrix representation
$$\tilde \delta_0 \colon \mathbb R^3 \to \mathbb R^3$$
$$\tilde \delta_0 = \begin{pmatrix} 1 & 0 & -1 \\ -1 & 1 & 0 \\ 0 & -1 & 1 \end{pmatrix} \, ,$$
whose determinant vanishes. The rank is two and thus everything is correct.
Thus
$$\check H^0 (S^1, \mathbb R) \simeq \mathbb R ^ {\# S^1} = \mathbb R ^1 \quad , \quad \check H^1 (S^1, \mathbb R) \simeq \mathbb R ^ {3} / \mathbb R^2 \simeq \mathbb R^1$$

3. Aug 7, 2014