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(adsbygoogle = window.adsbygoogle || []).push({});

$$U_0= (0, 2/3) \, , \, U_1= (1/3, 1) \, , \, U_2= (2/3, 1] \cup [0, 1/3)$$

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.

**Physics Forums - The Fusion of Science and Community**

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Computing Čech cohomology groups

Loading...

Similar Threads - Computing Čech cohomology | Date |
---|---|

A How to compute limit of removable singularity? | Jan 15, 2017 |

References for Self Study in de Rham Cohomology | Dec 2, 2015 |

2nd order pole while computing residue in a complex integral | Apr 16, 2015 |

Computing a Generating Set in Cohomology | May 18, 2014 |

How to compute monodromy of a particular algebraic function | Nov 19, 2013 |

**Physics Forums - The Fusion of Science and Community**