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In $\mathbb{R}^2$, differential forms are as follows:

  • 0-forms: $f:\mathbb{R}^2\longrightarrow\mathbb{R}$ differentiable
  • 1-forms: $g\mathrm{d} x+h\mathrm{d} y$, $g$, $h:\mathbb{R}^2\longrightarrow\mathbb{R}$ differentiable
  • 2-forms: $k\mathrm{d} x\wedge\mathrm{d} y$, $k:\mathbb{R}^2\longrightarrow\mathbb{R}$ differentiable

and the exterior derivative acts on them:

  • $\mathrm{d}(f)=\mathrm{d} f=\dfrac{\partial f}{\partial x}\mathrm{d} x+\dfrac{\partial f}{\partial y}\mathrm{d} y$
  • $\mathrm{d}(g\mathrm{d} x+h\mathrm{d} y)= \mathrm{d} g\wedge\mathrm{d} x+\mathrm{d} h\wedge\mathrm{d} y= \left(\dfrac{\partial g}{\partial x}\mathrm{d} x+\dfrac{\partial g}{\partial y}\mathrm{d} y\right)\wedge\mathrm{d} x+\left(\dfrac{\partial h}{\partial x}\mathrm{d} x+\dfrac{\partial h}{\partial y}\mathrm{d} y\right)\wedge\mathrm{d} y\\ \phantom{\mathrm{d}(g\mathrm{d} x+h\mathrm{d} y)}= \dfrac{\partial g}{\partial x}\mathrm{d} x\wedge\mathrm{d} x+\dfrac{\partial g}{\partial y}\mathrm{d} y\wedge\mathrm{d} x+\dfrac{\partial h}{\partial x}\mathrm{d} x\wedge\mathrm{d} y+\dfrac{\partial h}{\partial y}\mathrm{d} y\wedge\mathrm{d} y= \left(\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}\right)\mathrm{d} x\wedge\mathrm{d} y$
  • $\mathrm{d}(k\mathrm{d} x\wedge\mathrm{d} y)= \mathrm{d} k\wedge\mathrm{d} x\wedge\mathrm{d} y= \left(\dfrac{\partial k}{\partial x}\mathrm{d} x+\dfrac{\partial k}{\partial y}\mathrm{d} y\right)\wedge\mathrm{d} x\wedge\mathrm{d} y= \dfrac{\partial k}{\partial x}\mathrm{d} x\wedge\mathrm{d} x\wedge\mathrm{d} y+\dfrac{\partial k}{\partial y}\mathrm{d} y\wedge\mathrm{d} x\wedge\mathrm{d} y=0$

Now we may compute the cohomology groups


  • $\text{ker }\mathrm{d}_0=\left\{f\middle|\dfrac{\partial f}{\partial x}=\dfrac{\partial f}{\partial y}=0\right\}=\{f|f=cte\in\mathbb{R}\}$
  • $\text{im }\mathrm{d}_{-1}=\{0\}$ (we take $\mathrm{d}_{-1}:0\longrightarrow\Omega^0(\mathbb{R}^2)$ the zero map)

$$H^0(\mathbb{R}^2)=\dfrac{\{f=cte\in\mathbb{R}\}}{0}\simeq\mathbb{R}$$


  • $\text{ker }\mathrm{d}_1=\left\{g\mathrm{d} x+h\mathrm{d} y\middle|\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}=0\right\}$
  • $\text{im }\mathrm{d}_0=\left\{\dfrac{\partial f}{\partial x}\mathrm{d} x+\dfrac{\partial f}{\partial y}\mathrm{d} y\right\}$

Obviously, $\text{im }\mathrm{d}_0\subset\text{ker }\mathrm{d}_1$, because $\dfrac{\partial (\partial f/\partial y)}{\partial x}-\dfrac{\partial (\partial f/\partial x)}{\partial y}=0$. Now we wonder whether equality holds. Let $g$, $h$ with $\dfrac{\partial h}{\partial x}=\dfrac{\partial g}{\partial y}$; there will exist some $f$ with $\dfrac{\partial f}{\partial x}=g$, $\dfrac{\partial f}{\partial y}=h$? If so, this function could be found by integration, moving along the $x$-axis and then vertically; let $f(0,0)=0$:

$$f(x,y)=\int_0^x f_x(t,0)\mathrm{d} t+\int_0^y f_y(x,t)\mathrm{d} t=\int_0^x g(t,0)\mathrm{d} t+\int_0^y h(x,t)\mathrm{d} t$$

Does this function meet our requirements?

$$\dfrac{\partial f}{\partial x}=g(x,0)+\int_0^y \dfrac{\partial h}{\partial x}(x,t)\mathrm{d} t=g(x,0)+\int_0^y \dfrac{\partial g}{\partial y}(x,t)\mathrm{d} t=g(x,0)+ [g(x,t)]_0^y=g(x,0)+g(x,y)-g(x,0)=g(x,y)$$

$$\dfrac{\partial f}{\partial y}=h(x,y)$$

so $f$ is our desired function. Therefore there is equality in our previous sets and

$$H^1(\mathbb{R}^2)=\dfrac{\left\{g\mathrm{d} x+h\mathrm{d} y\middle|\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}=0\right\}}{\left\{\dfrac{\partial f}{\partial x}\mathrm{d} x+\dfrac{\partial f}{\partial y}\mathrm{d} y\right\}}=0$$


  • $\text{ker }\mathrm{d}_2=\Omega^2(\mathbb{R}^2)=\{k\mathrm{d} x\wedge\mathrm{d} y\}$
  • $\text{im }\mathrm{d}_1=\left\{\left(\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}\right)\mathrm{d} x\wedge\mathrm{d} y\right\}$

Given some fixed function $k$, there will exist $g$, $h$ with $\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}=k$? Now is very easy: we may take $g=0$ and $h(x,y)=\int_0^x k(t,y)\mathrm{d} t$, and everything is ok. So again kernel and image coincide and

$$H^2(\mathbb{R}^2)=\dfrac{\{k\mathrm{d} x\wedge\mathrm{d} y\}}{\left\{\left(\dfrac{\partial h}{\partial x}-\dfrac{\partial g}{\partial y}\right)\mathrm{d} x\wedge\mathrm{d} y\right\}}=0$$