I have a classmate learning algebra.He ask me how to compute the de Rham cohomology of the punctured plane $\mathbb{R}^2\setminus\{0\}$ by an elementary way,without homotopy type,without Mayer-Vietoris,just by Calculus. I have tried and failed.Is it possible to compute the de Rham cohomology just by Calculus?

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    $\begingroup$ Maybe using the winding number?! $\endgroup$
    – Sigur
    Dec 19, 2013 at 11:37

2 Answers 2


Let $M = \mathbb{R}^2 \setminus \{0\}$. We have a global coordinates $(x, y)$ on $M$. We want to compute the cohomology of the complex $$0 \to \Omega^0(M)\ \stackrel{d}{\to}\ \Omega^1(M)\ \stackrel{d}{\to}\ \Omega^2(M)\ \to 0$$ where $\Omega^k(M)$ is the space of smooth $k$-forms on $M$. So we want to compute $H^k(M) = Z^k(M)/B^k(M)$ where $Z^k(M) = \{\alpha \in \Omega^k(M),\, d\alpha = 0\}$ (closed $k$-forms) and $B^k(M) = \{d\beta,\, \beta \in \Omega^{k-1}(M)\}$ (exact $k$-forms). In this situation,

  • Elements of $\Omega^0(M)$ are just smooth functions $f(x,y)$ on $M$
  • Elements of $\Omega^1(M)$ can be written $u(x,y) dx + v(x,y) dy$ (where $u$ and $v$ are smooth functions on $M$)
  • Elements of $\Omega^2(M)$ can be written $g(x,y) dx \wedge dy$ (where $g$ is a smooth function on $M$)

Compute $H^0(M)$:

$B^0(M) = \{0\}$ and $Z^0(M)$ consists of functions $f(x,y)$ such that $df = 0$. Since $M$ is connected, this implies that $f$ is constant. It follows that $H^0(M)$ is isomorphic to $\mathbb{R}$.

Compute $H^1(M)$:

This is the where the all the fun happens :)

Let $\alpha \in Z^1(M)$, this means that $\alpha = u(x,y) dx + v(x,y) dy$ with $\frac{\partial v}{\partial x} - \frac{\partial u }{\partial y} = 0$.

Lemma 1: Let $R$ be a closed rectangle in $\mathbb{R}^2$ which does not contain the origin. Then $\int_{\partial R} \alpha = 0$.

Proof: This is just Green's theorem (or Stokes' theorem) (in its most simple setting, where it's not hard to show directly).

Lemma 2: Let $R$ and $R'$ be closed rectangles in $\mathbb{R}^2$ whose interiors contain the origin. Then $\int_{\partial R} \alpha = \int_{\partial R'} \alpha$.

Proof: This is a consequence of Lemma 1. It might be a bit tedious to write down (several cases need to be addressed, according to the configuration of the two rectangles), but it's fairly easy. You need to cut and rearrange integrals along a bunch of rectangles so that the two initial integrals agree up to integrals along rectangles who do not contain the origin.

Let's denote by $\lambda(\alpha)$ the common value of all integrals $\int_{\partial R} \alpha$ when $R$ is a closed rectangle whose interior contains the origin.

Lemma 3: If $\alpha$ is exact iff $\lambda(\alpha) = 0$.

Proof: "$\Rightarrow$" is trivial1, let's prove the converse. Fix permanently some point $m_0 \in M$, whichever you like best. For any $m\in M$, consider a rectangle $R$ in $\mathbb{R}^2$ whose boundary contains $m_0$ and $m$ but avoids the origin. Let $\gamma$ be one of the two paths joining $m_0$ and $m$ along $\partial R$. Let $f(m) = \int_\gamma \alpha$. Since $\lambda(\alpha) = 0$, this definition does not depend on the choice of the rectangle or the path. In other words $f: M \rightarrow \mathbb{R}$ is well defined. Let's see that $df = \alpha$. Check that $\frac{f(x+h, y) - f(x,y)}{h} = \frac{1}{h}\int_x^{x+h} u(t,y) dt$, so that taking the limit when $h\rightarrow 0$ yields $\frac{\partial f}{\partial x} = u$. Same for $v$.

Finally let's consider the $1$-form $d\theta = \frac{-ydx + xdy}{x^2 + y^2}$. NB: Be well aware that $d\theta$ is a misleading (but standard) notation: it is not an exact form.

Lemma 4: $d\theta$ is a closed $1$-form and $\lambda(d\theta) = 2\pi$.

Proof: This is a direct computation.

Now're done:

Proposition: $H^1(M)$ is the one-dimensional vector space spanned by $[d\theta]$.

where $[d\theta]$ denotes the class of $d\theta$ in $H^1(M)$. Note that $[d\theta] \neq 0$: see Lemma 4 and 3.

Proof: Let $\alpha$ be a closed $1$-form. Consider $\beta = \alpha - \frac{\lambda(\alpha)}{2 \pi} d\theta$. We have $\lambda(\beta) = 0$ so $\beta$ is exact by Lemma 3. This proves that $[\alpha] = \frac{\lambda(\alpha)}{2 \pi} [d\theta]$.

Compute $H^2(M)$:

Let's show that $H^2(M) = 0$, in other words every closed $2$-form on $M$ is exact. This solution is taken from Ted Shifrin in the comments below.

Here's the idea: in polar coordinates, a $2$-form $\omega$ can be written $\omega = f(r,\theta) dr \wedge d\theta$. Then $\eta = (\int_1^r f(\rho, \theta) d\rho)\, d\theta$ is a primitive of $\omega$.

Although it's not very insightful, this can be checked by a direct computation without refering to a change of variables.

Let $\omega(x,y) = g(x,y) dx \wedge dy$. Define $$h(x,y) = \int_1^{\sqrt{x^2+y^2}} t\, g\left(\frac{tx}{\sqrt{x^2+y^2}}, \frac{ty}{\sqrt{x^2+y^2}}\right) \,dt$$ Check that $h(x,y) d\theta$ is a primitive of $\omega$.

NB: Any course / book / notes on de Rham cohomology will show that if $M$ is a connected compact orientable manifold, $H^n(M) \approx \mathbb{R}$. However, I don't think I've read anywhere that when $M$ is not compact, $H^n(M) = 0$. I wonder if there is an "elementary" proof.

1 I'll explain this by request of OP. It is a general fact that if $\gamma : [a, b] \rightarrow M$ is a ${\cal C}^1$ path and $\alpha$ is a smooth exact one-form i.e. $\alpha = df$ where $f$ is a smooth function, then $\int_\gamma \alpha = f(\gamma(b)) - f(\gamma(a))$. This is because $\int_\gamma \alpha = \int_a^b \alpha(\gamma(t))(\gamma'(t))\,dt$ (by definition) and here $\alpha(\gamma(t))(\gamma'(t)) = df(\gamma(t))(\gamma'(t)) = f'(\gamma(t)) \gamma'(t) = \frac{d}{dt}\left(f(\gamma(t))\right)$.

This fact extends to piecewise ${\cal C}^1$ maps by cutting the integral. In particular, is $\gamma$ is a closed piecewise ${\cal C}^1$ path and $\alpha$ is exact, then $\int_\gamma \alpha = 0$.

In fact, it is useful (e.g. for Cauchy theory in complex analysis) to know that

  • A one-form is closed iff its integral along any homotopically trivial loop is zero (any boundary of a rectangle contained in the open set is enough).
  • A one-form is exact iff its integral along any loop is zero.

Note that the "smooth" condition can be weakened.

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    $\begingroup$ You mean $M=\Bbb R^2-\{0\}$. :) For $H^2(M)$, write a $2$-form as $\omega = f(r,\theta)\,dr\wedge d\theta$. Since $\omega$ is well-defined, we have $\int_0^{2\pi} f(r,\theta)d\theta = 0$, and so you can set $\eta = -\big(\int_0^\theta f(r,t)\,dt\big)dr$. $\endgroup$ Dec 19, 2013 at 16:31
  • $\begingroup$ Pas de quoi. :) $\endgroup$ Dec 19, 2013 at 17:02
  • $\begingroup$ Actually, what you wrote leaves me very skeptical :) $\omega$ being well-defined just tells me that $f(r,0) = f(r,2\pi)$. For instance what do you do with $\omega = dr \wedge d\theta$ ($f$ = 1)? $\endgroup$
    – Seub
    Dec 19, 2013 at 17:05
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    $\begingroup$ Oops. You're right. Easy, $\omega = d(r\,d\theta)$. That's a smooth $1$-form on $M$. So, I made it too hard. Set $\eta=\big(\int_1^r f(u,\theta)du\big)d\theta$. $\endgroup$ Dec 19, 2013 at 17:10
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    $\begingroup$ @Eric Hi Eric, to be honest I don't remember if I checked the computation in (x,y)-coordinates, it's been a while. As I wrote in my answer, is the idea is to first write the solution in polar coordinates, and then translate it to (x,y)-coordinates. Checking the answer by direct computation should work though. Let me know if you figure it out. $\endgroup$
    – Seub
    Nov 17, 2021 at 18:04

Here's a fairly detailed sketch to get your friend started:

Let $D = \mathbf{R}^2\setminus\{0\}$ denote the punctured plane, and cover $D$ by two slit planes, say $D_{-} = \mathbf{R}^2\setminus(-\infty, 0]$ and $D_{+} = \mathbf{R}^2\setminus[0,\infty)$. Each set is star-shaped, so if $\eta$ is a closed $1$-form in $D$, then $\eta$ is exact in $D_{-}$ and in $D_{+}$.

The $1$-form $$ \omega = \frac{-y\, dx + x\, dy}{2\pi(x^2 + y^2)}\qquad \left(= \frac{d\theta}{2\pi} \right) $$ is closed (easy computation) and non-exact (the integral over the unit circle is $1$) in $D$.

Let $\alpha$ be an arbitrary closed smooth $1$-form in $D$, and let $c$ be the value of the integral of $\alpha$ over the unit circle. To achieve your friend's goal, it suffices to prove that the closed $1$-form $\beta := \alpha - c\omega$ is exact in $D$.

By the second paragraph above, there exist smooth functions $f_{-}$ in $U_{-}$ and $f_{+}$ in $U_{+}$ such that $df_{\pm} = \beta$ in $U_{\pm}$. Particularly, $df_{-} = df_{+}$ off the horizontal axis; $f_{+} - f_{-}$ is constant in the upper half plane and in the lower half plane. Without loss of generality, $f_{+} - f_{-} = 0$ in the upper half plane, and $f_{+} - f_{-} = C$ in the lower half plane. Argue that $C$ is the integral of $\beta$ around the unit circle, i.e., $C = 0$.

Extracting the deRham cohomology (including handling compact supports as needed) is left to your friend.

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    $\begingroup$ Thanks Georges! Fixed the first and was editing by the time you posted the second. I'll leave the $D$/$U$ idiosyncrasy. :) $\endgroup$ Jun 4, 2021 at 16:16
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    $\begingroup$ Dear Andrew, those were benevolent comments, which I have now deleted. Congratulations for your lightning fast reaction to my comments and +1 for your fine answer. Also, I'm happy that you are still on this site: we complex geometers should try to imperialistically dominate this site :-) $\endgroup$ Jun 4, 2021 at 16:35

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