A short proof of the Poincare Lemma

[wrobel]

I would like to share a concise proof of the Poincaré Lemma

Theorem (Poincaré Lemma)
Assume that a smooth $k$-form
$$\omega = \sum_{i_1 < \dots < i_k} \omega_{i_1 \dots i_k} (x) dx^{i_1} \wedge \dots \wedge dx^{i_k}$$
is defined in a neighbourhood of the origin in $\mathbb{R}^m$.
If $\omega$ is closed ($d\omega = 0$), then it is exact in some open ball $B$ centered at the origin. That is, there exists a $(k-1)$-form $\Omega$ such that $d\Omega = \omega$.

Proof
Consider the radial vector field $v(x) = x$. The corresponding flow is given by $g^t(x) = e^t x$. The pullback of $\omega$ under this flow is:
$$g^t_* \omega = e^{kt} \sum_{i_1 < \dots < i_k} \omega_{i_1 \dots i_k}(e^t x) dx^{i_1} \wedge \dots \wedge dx^{i_k}.$$
From the definition of the Lie derivative , we have:
$$\frac{d}{dt} g^t_* \omega = g^t_* \mathcal{L}_v \omega.$$
By applying Cartan's magic formula and the fact that the exterior derivative
commutes with the pullback ($d g^t_* = g^t_* d$), we obtain:
$$\frac{d}{dt} g^t_* \omega = g^t_* (d i_v \omega + i_v d \omega) = d(g^t_* i_v \omega)$$
where we used the assumption $d\omega = 0$.

Now, we integrate both sides with respect to $t$ from $-\infty$ to $0$:
$$\int_{-\infty}^0 \frac{d}{dt} (g^t_* \omega) \, dt = \int_{-\infty}^0 d(g^t_* i_v \omega) \, dt.$$
The term $g^t_* \omega$ vanishes as $t \to -\infty$ due to the $e^{kt}$ factor. Thus, the left-hand side evaluates to:
$$g^0_* \omega - \lim_{t \to -\infty} g^t_* \omega = \omega - 0 = \omega.$$
Pulling the exterior derivative out of the integral on the right-hand side, we get:
$$\omega = d \left( \int_{-\infty}^0 g^t_* i_v \omega \, dt \right).$$
Therefore, $\Omega = \int_{-\infty}^0 g^t_* i_v \omega \, dt$ is the required $(k-1)$-form.