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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?

a student
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    Which book(s) are you using to study de Rham cohomology? Wikipedia is not exactly the best source. From calculus to cohomology? Something else? – guest Dec 30 '14 at 02:28
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    Do you know what $df$ is in coordinates? –  Dec 30 '14 at 02:29
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    @guest No, I use Wikipedia at this moment. I don't know of any good sources to learn de Rham theory. – a student Dec 30 '14 at 02:42
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    @MikeMiller No, what do you mean by "in coordinates"? – a student Dec 30 '14 at 02:43
  • 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 $df$) –  Dec 30 '14 at 02:45
  • @astudent Judging from your reply to Mike, you need a good book on de Rham cohomology starting at the beginning and being explicit. I recommend 'from calculus to cohomology'. – guest Dec 30 '14 at 02:45
  • @guest Thanks for the book recommendation. The library is closed until next week. Once it reopens I will check out the book. I hope that in the meantime I can still carry on learning about the de Rham cohomology using what I have. – a student Dec 30 '14 at 02:50
  • I completely support you finding a better source and getting a book from the library. There may be an acceptable online reference (do a search, or if you can't find one, post a question regarding that). Here's a wikipedia page to read for the moment. – aes Dec 30 '14 at 03:00
  • @MikeMiller Oh that kind of coordinates. I do know about charts. But I'm unsure about the relation between charts and the exterior derivative. So, to answer your question, no I don't know how to write $df$ in coordinates. Thank you for the book recommendations. I have to wait until the library reopens next year. – a student Dec 31 '14 at 01:21
  • I looked at Madsen on amazon and the reviews suggest that it is unsuitable for a beginner and hard to read? – a student Dec 31 '14 at 01:26

1 Answers1

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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?

Steven Gubkin
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    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 $e_n=(0,0,\dots, 0,1)$? – a student Dec 31 '14 at 01:25
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    (I know the de Rham theorem and I know cellular homology: $H^0$ counts connected components) – a student Dec 31 '14 at 01:29
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    If $\frac{\partial f}{\partial x_i} = 0$ for some coordinate $x_i$ then $f$ is constant in that direction. – hjhjhj57 Dec 31 '14 at 01:32
  • @astudent when all of the partials vanish in $\mathbb{R}^n$, the answer is the same: the function must be a constant function. Probably the easiest way to see this is to use the one variable result repeatedly.

    Do you see how de Rham cohomology is counting the connected components for the case of $\mathbb{R}-{0}$?

     – Steven Gubkin Dec 31 '14 at 02:20
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    @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! – Steven Gubkin Dec 31 '14 at 02:21
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    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? – a student Dec 31 '14 at 03:11
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    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? – a student Dec 31 '14 at 03:15
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    @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 $\mathbb{R}^2-{(0,0)}$. – Steven Gubkin Dec 31 '14 at 03:18
  • @astudent you may also be interested in my answers here: http://math.stackexchange.com/q/664648/34287, http://math.stackexchange.com/q/975327/34287, http://math.stackexchange.com/q/595657/34287 – Steven Gubkin Dec 31 '14 at 03:23
  • By the way, I really recommend Shiffrin's book "Multivariable mathematics" as a precursor to any study of de Rham cohomology. It is "multivariable calculus done right". You might also be interested in my course here: ximera.osu.edu/course/kisonecat/m2o2c2/course/ which leads to the multivariable taylor theorem, with tensors. This is sort of orthogonal to the differential forms story (symmetric instead of alternating tensors), but pretty cool story which is not often told. – Steven Gubkin Dec 31 '14 at 03:25
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    I have started to work through your comments sequentially. I got stuck after trying to apply the Poincaré lemma to calculate $H^1$. I will post this question in a new post. – a student Dec 31 '14 at 04:15