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?
[Math] how to compute the de Rham cohomology of the punctured plane just by Calculus
algebraic-topologycalculusdifferential-geometryhomology-cohomologysmooth-manifolds
Related Solutions
If you remove a point from $P^2$ you are left with something which looks like a Moebius band. You can use this to compute $H^\bullet(P^2)$.
Let $p\in P^2$, let $U$ be a small open neighborhood of $p$ in $P^2$ diffeomorphic to an open disc centered at $p$, and let $V=P^2\setminus\{p\}$. Now use Mayer-Vietoris.
The cohomology of $U$ you know. The open set $V$ is diffeomorphic to an open moebious band, so that tells you the cohomology; alternatively, you can check that it deformation-retracts to the $P^1\subseteq P^2$ consiting of all lines orthogonal to the line corresponding to $p$ (with respect to any inner product in the vector space $\mathbb R^3$ you used to construct $P^2$), and the intersection $U\cap V$ has also the homotopy type of a circle. The maps in the M-V long exact sequence are not hard to make explicit; it does help to keep in mind the geometric interpretation of $U$ and $V$.
Later: alternatively, one can do a bit of magic. Since there is a covering $S^2\to P^2$ with $2$ sheets, we know that the Euler characteristics of $S^2$ and $P^2$ are related by $\chi(S^2)=2\chi(P^2)$. Since $\chi(S^2)=2$, we conclude that $\chi(P^2)=1$. Since $P^2$ is of dimension $2$, we have $\dim H^p(P^2)=0$ if $p>2$; since $P^2$ is non-orientable, $H^2(P^2)=0$; finally, since $P^2$ is connected, $H^0(P^2)\cong\mathbb R$. It follows that $1=\chi(P^2)=\dim H^0(P^2)-\dim H^1(P^2)=1-\dim H^1(P^2)$, so that $H^1(P^2)=0$.
Even later: if one is willing to use magic, there is lot of fun one can have. For example: if a finite group $G$ acts properly discontinuously on a manifold $M$, then the cohomology of the quotient $M/G$ is the subset $H^\bullet(M)^G$ of the cohomology $H^\bullet(M)$ fixed by the natural action of $G$. In this case, if we set $M=S^2$, $G=\mathbb Z_2$ acting on $M$ so that the non-identity element is the antipodal map, so that $M/G=P^2$: we get that $H^\bullet(P^2)=H^\bullet(S^2)^G$.
We have to compute the fixed spaces:
$H^0(S^2)$ is one dimensional, spanned by the constant function $1$, which is obviously fixed by $G$, so $H^0(P^2)\cong H^0(S^2)^G=H^0(S^2)=\mathbb R$.
On the other hand, $H^2(S^2)\cong\mathbb R$, spanned by any volume form on the sphere; since the action of the non-trivial element of $G$ reverses the orientation, we see that it acts as multiplication by $-1$ on $H^2(S^2)$ and therefore $H^2(P^2)\cong H^2(S^2)^G=0$.
Finally, if $p\not\in\{0,2\}$, then $H^p(S^2)=0$, so that obviously $H^p(P^2)\cong H^p(S^2)^G=0$.
Luckily, this agrees with the previous two computations.
Best Answer
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,
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 $\frac{\partial v}{\partial x} - \frac{\partial u }{\partial y} = 0$.
Proof: This is just Green's theorem (or Stokes' theorem) (in its most simple setting, where it's not hard to show directly).
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.
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.
Proof: This is a direct computation.
Now're done:
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
Note that the "smooth" condition can be weakened.