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I would like to share a concise proof of the Poincaré Lemma
Theorem (Poincaré Lemma)
Assume that a smooth ##k##-form (##k\in\mathbb{N}##)
$$\omega = \sum_{i_1 < \dots < i_k} \omega_{i_1 \dots i_k} (x) dx^{i_1} \wedge \dots \wedge dx^{i_k}$$
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.
Theorem (Poincaré Lemma)
Assume that a smooth ##k##-form (##k\in\mathbb{N}##)
$$\omega = \sum_{i_1 < \dots < i_k} \omega_{i_1 \dots i_k} (x) dx^{i_1} \wedge \dots \wedge dx^{i_k}$$
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.
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