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Computing Čech cohomology groups

  1. Jul 19, 2014 #1
    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. jcsd
  3. Jul 19, 2014 #2
    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
    $$\check H^0 (S^1, \mathbb R) \simeq \mathbb R ^ {\# S^1} = \mathbb R ^1 \quad , \quad
    \check H^1 (S^1, \mathbb R) \simeq \mathbb R ^ {3} / \mathbb R^2 \simeq \mathbb R^1$$
     
  4. Aug 7, 2014 #3

    mathwonk

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