# De Rham cohomology groups of $\mathbb{R}^n$

I want to show that for each $$1 \le k\le n$$ we have $$H_{dR}^k(\mathbb{R}^n)=0$$

The strategy is to construct for each $$k$$ a linear map $$h_k:\Omega^k(\mathbb{R}^n)\to \Omega^{k-1}(\mathbb{R}^n)$$ such that $$d_{k-1}\circ h_k + h_{k+1}\circ d_{k}=Id_{\Omega^k(\mathbb{R}^n)} \qquad\qquad[1]$$ where $$d_k:\Omega^{k}(\mathbb{R}^n)\to \Omega^{k+1}(\mathbb{R}^n)$$ is the exterior derivative.

I have the following definition: for each $$\omega=\omega_{i_1\dots i_k}dx^{i_1}\wedge\dots\wedge dx^{i_k}$$ in $$\Omega^k(\mathbb{R}^n)$$ we define $$h_k(\omega)(x)=k \left[ \int_0^1t^{k-1} \omega_{i_1\dots i_k}(tx) \;dt\right]x^{i1} dx^{i_2}\wedge\dots\wedge dx^{i_k}$$

Can you help me to show that this definition satisfies $$[1]$$?

For example I have $$d_k(w)=\sum_j \frac{\partial(\omega_{i_1\dots i_k})}{\partial x^j} dx^j\wedge dx^{i_1}\wedge\dots\wedge dx^{i_k}$$ and

$$d_{k-1}( h_k (w))(x)=k\sum_j \frac{\partial(\left[ \int_0^1t^{k-1} \omega_{i_1\dots i_k}(tx) \;dt\right]x^{i1})}{\partial x^j} dx^j\wedge dx^{i_2}\wedge\dots\wedge dx^{i_k}$$ and

$$h_{k+1}(d_k(\omega))(x)=(k+1)\left[ \int_0^1t^{k} \frac{\partial(\omega_{i_1\dots i_k})}{\partial x^j} (tx) \;dt\right]x^{j}dx^{i_1}\wedge\dots\wedge dx^{i_k}$$

I don't know if my calculations above are correct, and even if they are, I can't go on.

Please give most details you can if you do sofisticated manipulations/calculatons.

• This sounds a lot like a homework problem, and you're probably expected to provide the sophisticated manipulations and calculations on your own. – Matthew Leingang Oct 30 '18 at 17:47
• @MatthewLeingang This is homework I gave myself, since I'm self-studing this subject. – Minato Oct 30 '18 at 17:49
• – Tsemo Aristide Oct 30 '18 at 17:51
• @Minato: OK, good. What if you try it in low dimensions and low degrees? Like $n=3$ and $k=1$? That might help you see what's going on. – Matthew Leingang Oct 30 '18 at 17:52
• @TsemoAristide I already looked that, but i don't know the notions of Lie derivative, interior multiplication and Cartan's formula. – Minato Oct 30 '18 at 17:54