Differential Forms - Key Proof

We prove the Poincaré lemma using an explicit homotopy operator, demonstrating the power of geometric constructions in solving PDEs.

ProofPoincaré Lemma

Statement: Let $U \subset \mathbb{R}^n$ be a star-shaped domain (say, star-shaped with respect to the origin). If $\omega \in \Omega^k(U)$ with $k \geq 1$ and $d\omega = 0$ , then there exists $\eta \in \Omega^{k-1}(U)$ with $d\eta = \omega$ .

Proof: Define the homotopy operator $K: \Omega^k(U) \to \Omega^{k-1}(U)$ by

(K\omega)_x(v_1, \ldots, v_{k-1}) = \int_0^1 t^{k-1} \omega_{tx}(x, v_1, \ldots, v_{k-1}) dt

This integrates $\omega$ along straight lines from the origin to each point. We claim that $dK\omega + K(d\omega) = \omega$ .

Step 1: Verify the claim for 0-forms (functions). For $f \in \Omega^0(U)$ :

(Kf)_x = \int_0^1 f(tx) dt

Then

d(Kf)_x(v) = \frac{d}{ds}\bigg|_{s=0} \int_0^1 f(tx + sv) dt = \int_0^1 df_{tx}(v) dt

By the fundamental theorem of calculus applied to $g(t) = f(tx)$ :

f(x) - f(0) = \int_0^1 \frac{d}{dt}f(tx) dt = \int_0^1 df_{tx}(x) dt

Since $U$ is star-shaped from the origin and $\omega$ is defined on $U$ , we can extend to get $(dK + Kd)f = f - f(0)$ . For functions, $df = 0$ means $f$ is constant, so $f = d(0) + c$ is exact (trivially).

Step 2: For $k$ -forms with $k \geq 1$ , write locally $\omega = \sum_I f_I dx^I$ where $I$ ranges over multi-indices. Applying $K$ component-wise and using the fundamental theorem, we compute:

(dK + Kd)\omega = \omega - \omega|_0

But $\omega|_0 = 0$ when $k \geq 1$ because at the origin, $dx^I|_0 = 0$ for products of differentials.

Step 3: If $d\omega = 0$ , then

\omega = dK\omega + K(d\omega) = d(K\omega) + 0 = d(K\omega)

Thus $\eta = K\omega$ is the required primitive. $\square$

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Remark

The homotopy operator $K$ provides an explicit formula for the primitive, though it's typically not the most natural one. The proof works because star-shaped domains are contractible to a point via the homotopy $h_t(x) = tx$ .

Proof$d^2 = 0$ - Coordinate Computation

We verify that $d^2 = 0$ by direct computation in coordinates. For a $k$ -form

\omega = \sum_I f_I dx^{i_1} \wedge \cdots \wedge dx^{i_k}

we have

d\omega = \sum_{I,j} \frac{\partial f_I}{\partial x^j} dx^j \wedge dx^{i_1} \wedge \cdots \wedge dx^{i_k}

Applying $d$ again:

d^2\omega = \sum_{I,j,\ell} \frac{\partial^2 f_I}{\partial x^\ell \partial x^j} dx^\ell \wedge dx^j \wedge dx^{i_1} \wedge \cdots \wedge dx^{i_k}

For each $I$ and pair $j < \ell$ , the terms with $(\ell, j)$ and $(j, \ell)$ appear:

\frac{\partial^2 f_I}{\partial x^\ell \partial x^j} dx^\ell \wedge dx^j + \frac{\partial^2 f_I}{\partial x^j \partial x^\ell} dx^j \wedge dx^\ell

By equality of mixed partials and anticommutativity of wedge products:

= \frac{\partial^2 f_I}{\partial x^\ell \partial x^j} (dx^\ell \wedge dx^j - dx^\ell \wedge dx^j) = 0

Summing over all indices gives $d^2\omega = 0$ . $\square$

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This algebraic proof complements the geometric interpretation: $d^2 = 0$ reflects the topological fact that the boundary of a boundary is empty.