First proof of Poincaré Lemma I know that a way of proving Poincare lemma is to use the homotopy invariance and contractibility of the Euclidean space. Is there is a way of doing it directly (without using the contractibility of $\mathbb{R}^n$)?
What was the first proof of this statement ? I wish to know all the different ways of proving this lemma. Please provide references. Thanks !
 A: We describe a linear functional $\alpha : \Omega^p \mathbb R^n \to \Omega^{p-1} \mathbb R^n$. The space alternating $p$-linear functions on $\mathbb R^n$ has dimension $n \choose p$, and you can write the basis as $dx_{i_1} \wedge dx_{i_2} \wedge \cdots \wedge dx_{i_p}$ where $1 \leq i_1 < i_2 < \cdots < i_p \leq n$. 
If $I = (i_1, i_2, \cdots, i_p)$ is such a multi-index let $dx_I = dx_{i_1} \wedge dx_{i_2} \wedge \cdots \wedge dx_{i_p}$. Let $I_1$ be the collection of multi-indices with $i_1=1$ and let $I_2$ be the collection of multi-indices with $i_1 > 1$.  
Given a $p$-form $f dx_I$ with $f: \mathbb R^n \to \mathbb R$ we define $\alpha$ linearly, by $\alpha (f dx_I) = 0$ if $I \in I_2$.  $\alpha (fdx_I) = \left(\int_0^{x_1} f dx_1\right) dx_{i_2} \wedge \cdots \wedge dx_{i_p}$ if $I \in I_1$. You can think of $\alpha$ is a type of `total contraction' of the form in the coordinate $x_1$-direction.
It's fairly direct to check that 
$$ d(\alpha(\omega)) + \alpha(d \omega) = \omega - \pi^*(i^* \omega)$$
for every $p$-form $\omega$.  Here $i : \mathbb R^{n-1} \to \mathbb R^n$ is
the inclusion $i(x_2,\cdots,x_n) = (0,x_2,\cdots,x_n)$ and $\pi : \mathbb R^n \to \mathbb R^{n-1}$ is projection $\pi(x_1,x_2,\cdots,x_n) = (x_2,\cdots,x_n)$. 
So this is a less technically-sophisticated argument than Avitus's but it might be a little simpler to follow conceptually. In the end you reduce showing a closed $p$-form on $\mathbb R^n$ is exact to solving the problem for one smaller $n$.  Dimension $p=n$ is the base case, where the above formula starts the induction. 
A: The above calculation in more detail. Define $h_k:\Omega^p(\mathbb{R}^n)\rightarrow \Omega^{p-1}(\mathbb{R}^n)$ by,
\begin{eqnarray}
h_k(\omega)(x)=\int_0^1t^{p-1}i_X\omega(tx)dt
\end{eqnarray}
where $X=\sum_ix^i\partial_i$, this generates the one parameter group of diffeomorphisms $\Phi_s:x^i\mapsto e^sx^i=:y^i$. To see this note that $X^i(t)=\frac{dx^i(t)}{dt}=x^i(t)$, hence we solve this ODE to find the stated solution. Now, 
\begin{eqnarray}
((h_{k+1}\circ d_k+d_{k-1}\circ h_k)\omega)(x)&=&\int_0^1t^{p-1}\Big(d(i_X\omega)+i_X(d\omega)\Big)(tx)dt\\
&=&\int_0^1t^{p-1}(\mathcal{L}_X\omega)(tx)dt
\end{eqnarray}
Then from the definition of the Lie derivative in terms of the pullback,
\begin{eqnarray}
(\mathcal{L}_X\omega)_x=\Big(\frac{d}{ds}\Big|_{s=0}(\Phi_s)^*\omega\Big)_x
\end{eqnarray}
Now we can compute the pullback explicitly,
\begin{eqnarray}
[((\Phi_s)^*\omega)_x]_{k_1...k_p}=\frac{\partial y^{i_1}}{\partial x^{k_1}}...\frac{\partial y^{i_p}}{\partial x^{k_p}}[\omega_{\Phi_s(x)}]_{i_1...i_p}=e^{ps}[\omega_{e^sx}]_{k_1...k_p}
\end{eqnarray}
Due to linearity we can restrict to the simple case where $\omega(x)=f(x)dx^I$. Therefore, 
\begin{eqnarray}
(\mathcal{L}_X\omega)_x=\Big(\frac{d}{ds}\Big|_{s=0}(\Phi_s)^*\omega\Big)_x=p\omega_x+x^i\frac{\partial f}{\partial x^i}(x)dx^I
\end{eqnarray}
Therefore, 
\begin{eqnarray}
t^{p-1}(\mathcal{L}_X\omega)_{tx}&=&t^{p-1}\Big(p\omega_{tx}+tx^i\frac{\partial f}{\partial x^i}(tx)dx^I\Big)\\
&=&\frac{d}{dt}\Big(t^pf(tx)dx^I\Big)=\frac{d}{dt}(t^p\omega(tx))
\end{eqnarray}
Hence, 
\begin{eqnarray}
((h_{k+1}\circ d_k+d_{k-1}\circ h_k)\omega)(x)=t^p\omega(tx)|_{t=0}^1=\omega(x)
\end{eqnarray}
Therefore since $\omega$ is closed, $d\omega=0$ so $\mathcal{L}_X\omega=d(i_X\omega)$ and hence $d_{k-1}h_k\omega=\omega$, i.e. $\omega$ is exact. 
A: We want to show that on  $\mathbb R^n$, all closed forms of degree $p\geq 1$ are exact. To do so we construct a linear operator
$$\alpha:\Omega^p(\mathbb R^n) \rightarrow \Omega^{p-1}(\mathbb R^n) $$
s.t.  $$d\alpha+\alpha d=1.$$
Let $\omega$ be a closed $p$-form. Then, for any $x\in\mathbb R^n$ we define
$$(\alpha \omega)(x):=\int_0^1 t^{p-1}i_x\omega(tx)dt, $$
where $i_x$ is the interior product operator.
Then  (by Cartan's magic formula)
$$\begin{aligned} 
((d\alpha + \alpha d)\omega)(x) & = \int_0^1 t^p\mathcal L_x\omega(tx)dt \\
 & =(\text{use chain rule and pull-back definition of Lie derivative}) \\
 & = \int_0^1\frac{d}{dt}(t^p\omega(tx))dt=\omega(x) \end{aligned} $$
and we are done. The diff. form  $\theta:=\alpha(\omega)$ is the exact form we need.
edit: For the chain rule step one wants to consider the pull-back $M_t^* \omega$ where $M_t : \mathbb R^n \to \mathbb R^n$ is scalar multiplication by $t$. i.e. $t^p \omega(tx) = M_t^* \omega (x)$.
