- #1
Geometry_dude
- 112
- 20
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.
$$U_0= (0, 2/3) \, , \, U_1= (1/3, 1) \, , \, U_2= (2/3, 1] \cup [0, 1/3)$$
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.