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On Wikipedia the de Rham cohomology groups are defined to be the cohomology groups of the de Rham cochain complex (equivalence classes of differential $k$-forms).

By this definition the zeroth de Rham cohomology group is the set of all closed differential zero forms modulo all exact $0$-forms (i.e. modulo the image of the exterior derivative). In formula,

$$ H^0_{dR} = {\ker d^{1}\over \mathrm{im } d^0 } = \ker d^{1}$$

Since $d^0: 0 \to \Omega^0$ is the trivial map.

Question1: Am I correct so far?

Using the notation and terminology on Wikipedia $H_0$ is therefore the set of all closed $0$-forms. Since $0$-forms are smooth functions the question arises what it means for a smooth function $f$ to be closed, that is, which $f$ have vanishing exterior derivative $df=0$.

Question2: How to determine whether a smooth function is closed?

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    $\begingroup$ Which book(s) are you using to study de Rham cohomology? Wikipedia is not exactly the best source. From calculus to cohomology? Something else? $\endgroup$ – guest Dec 30 '14 at 2:28
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    $\begingroup$ Do you know what $df$ is in coordinates? $\endgroup$ – user98602 Dec 30 '14 at 2:29
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    $\begingroup$ @guest No, I use Wikipedia at this moment. I don't know of any good sources to learn de Rham theory. $\endgroup$ – a student Dec 30 '14 at 2:42
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    $\begingroup$ @MikeMiller No, what do you mean by "in coordinates"? $\endgroup$ – a student Dec 30 '14 at 2:43
  • $\begingroup$ Wikipedia is perhaps the second worst possible source to learn de Rham theory. Some popular sources are, Lee's book or Warner's book on differentiable manifolds; or (as guest says below) Madsen's "from calculus to cohomology". ("In coordinates" means you take a chart and identify that part of the manifold with $\Bbb R^n$; this is normally how one defines $df$) $\endgroup$ – user98602 Dec 30 '14 at 2:45
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Think about this in the case of $\mathbb{R}^1$. What kind of functions have $\frac{df}{dx} = 0$ everywhere?

After this, what happens in $\mathbb{R}^n$ when all of the partial derivatives vanish?

Now a slightly harder case: In $\mathbb{R}^1-\{0\}$, what kind of functions can have vanishing derivative? Are there more or less such functions than for all of $\mathbb{R}^1$?

Maybe now you are starting to guess that the dimension of $H^0$ counts something. What does it count?

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    $\begingroup$ Thank you for your answer! The first one is easy: constant functions. The second question is already hard for me. In $\mathbb R^3$ it means it is a plane parallel to the $x-y-$axis but it's not clear to me how to generalise to $\mathbb R^n$. Could the answer be that the image of $f$ is an $n-1$-dimensional plane with normal vector $e_n=(0,0,\dots, 0,1)$? $\endgroup$ – a student Dec 31 '14 at 1:25
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    $\begingroup$ @astudent This is connected to a lie we tell calculus students: that an antiderivative looks like $F(x)+C$, where $C$ is a constant. Really $C(x)$ should be a locally constant function! $\endgroup$ – Steven Gubkin Dec 31 '14 at 2:21
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    $\begingroup$ I'm starting to understand, the example $\mathbb R - 0$ really helped. The function with vanishing derivative there are those who equal some constant on $(-\infty,0)$ and some other constant on $(0,\infty)$. So the de Rham cohomology $\mathbb R \oplus \mathbb R$ counts those functions? $\endgroup$ – a student Dec 31 '14 at 3:11
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    $\begingroup$ I'm not sure it's ok to ask you another question in the comments (if not I will post a new question): but now I am wondering about generators of the de Rham cohomology of a connected space. My newly acquired understanding suggest that any constant function should generate $H^0_{dR}$. Is that correct? $\endgroup$ – a student Dec 31 '14 at 3:15
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    $\begingroup$ @astudent Exactly right. Otherwise, it is the locally constant functions, which is isomorphic to $\mathbb{R}^n$, where $n$ is the number of connected components of the space. The next step is to start thinking about $H^1$. For that, I suggest first understanding the Poincare lemma, and then grappling with $H^1$ of $\mathbb{R}^2-\{(0,0)\}$. $\endgroup$ – Steven Gubkin Dec 31 '14 at 3:18

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