Ytukyg

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 !

We want to show that on $mathbb R^n$, all closed forms of degree $pgeq 1$ 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. $$dalpha+alpha d=1.$$

Let $omega$ be a closed $p$-form. Then, for any $xinmathbb R^n$ we define

$$(alpha omega)(x):=int_0^1 t^{p-1}i_xomega(tx)dt, $$

where $i_x$ is the interior product operator.
Then (by Cartan's magic formula)

$$begin{aligned}
((dalpha + alpha d)omega)(x) & = int_0^1 t^pmathcal L_xomega(tx)dt \
& =(text{use chain rule and pull-back definition of Lie derivative}) \
& = int_0^1frac{d}{dt}(t^pomega(tx))dt=omega(x) end{aligned} $$

and we are done. The diff. form $theta:=t(omega)$ 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. $t^p omega(tx) = M_t^* omega (x)$.

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 $I in I_2$. $alpha (fdx_I) = left(int_0^{x_1} f dx_1right) dx_{i_2} wedge cdots wedge dx_{i_p}$ 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.

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_Xomega(tx)dt
end{eqnarray}
where $X=sum_ix^ipartial_i$, this generates the one parameter group of diffeomorphisms $Phi_s:x^imapsto e^sx^i=:y^i$. 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_Xomega)+i_X(domega)Big)(tx)dt\
&=&int_0^1t^{p-1}(mathcal{L}_Xomega)(tx)dt
end{eqnarray}
Then from the definition of the Lie derivative in terms of the pullback,
begin{eqnarray}
(mathcal{L}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_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}_Xomega)_x=Big(frac{d}{ds}Big|_{s=0}(Phi_s)^*omegaBig)_x=pomega_x+x^ifrac{partial f}{partial x^i}(x)dx^I
end{eqnarray}
Therefore,
begin{eqnarray}
t^{p-1}(mathcal{L}_Xomega)_{tx}&=&t^{p-1}Big(pomega_{tx}+tx^ifrac{partial f}{partial x^i}(tx)dx^IBig)\
&=&frac{d}{dt}Big(t^pf(tx)dx^IBig)=frac{d}{dt}(t^pomega(tx))
end{eqnarray}
Hence,
begin{eqnarray}
((h_{k+1}circ d_k+d_{k-1}circ h_k)omega)(x)=t^pomega(tx)|_{t=0}^1=omega(x)
end{eqnarray}
Therefore since $omega$ is closed, $domega=0$ so $mathcal{L}_Xomega=d(i_Xomega)$ and hence $d_{k-1}h_komega=omega$, i.e. $omega$ is exact.