# Prove Poincaré Lemma for $1$-form

Let $$U\subseteq\mathbb{R}^n$$ be an open set that contains $$0$$, and for all $$t\in[0,1]$$ and $$x\in U$$, $$tx\in\mathbb{R}^n$$. Show that every closed differentiable 1-form $$w$$, (i.e. $$dw=0$$) is an exact form (i.e. there exists a function $$f:U\to\mathbb{R}^n$$ such that $$w=df$$).

Hint: Present $$w=\sum_{i=1}^nw_idx_i$$ and define $$f=\sum_{i=1}^n\int _0^1w_i(tx)x_idt$$.

I have a start but I don't know how to keep on: \begin{align} df&=d(\sum_{i=1}^n\int_0^11w_i(tx)x_idt) \text{ =[by Leibnitz]}\\ &= \sum_{i=1}^n\int_0^1d(w_i(tx)x_idt) \\&= \sum_{i=1}^n\int_0^1\sum_{j=1}^n{\partial\over\partial x_j}(w_i(tx_i)x_i)dx_j\wedge dt \\&= \sum_{i=1}^n\int_0^1( \sum_{j\ne i} {\partial w_i\over\partial x_j}(tx_i)\cdot0\cdot x_i)dx_j\wedge dt \\ &+( x_it{\partial w_i\over \partial x_i}(tx_i)+w_i(tx_i))dx_i\wedge dt) \\&= \sum_{i=1}^n\int_0^1{\partial w_i\over \partial x_i}(tx_i)\cdot tx_i dx_i\wedge dt+\sum_{i=1}^n\int _0^1w_i(tx_i)dx_i\wedge dt \end{align}

$$df=d(\int_0^1\sum w_i(tx)dx_idt)=\int_0^1\sum d(w_i(tx)x_i)dt$$.

We have $$d(w_i(tx)x_i)=\sum_{j=1}^{j=n}{\partial \over{\partial x_j}}(w_i(tx)x_i)dx_j$$ $$=\sum_{j=1}^{j=n}{\partial \over{\partial x_j}}w_i(tx)tx_i +w_i(tx)dx_i$$,

since $$dw=0$$, we have $${\partial \over{\partial x_j}}w_i={\partial \over{\partial x_i}}w_j$$,

we deduce that $$df=\int_0^1\sum_{i=1}^{i=n}(\sum_{j=1}^{j=n}({\partial \over{\partial x_j}}w_i(tx))(tx_j)dx_i+w_i(tx)dx_i$$

Remark that $${\partial \over{\partial t}}w_i(tx)=\sum_{j=1}^{j=n}{\partial \over{\partial x_j}}w_i(tx)x_j$$, we can integrate by parts and obtain:

$$\int_0^1(\sum_{j=1}^{j=n}({\partial \over{\partial x_j}}w_i(tx))(tx_j)dx_i+w_i(tx)dx_i)dt =(w_i(tx)]_0^1-w_i(tx)+w_i(tx))dx_i$$.

This implies that $$d(f+\sum_{i=1}^{i=n}w_i(0)x_i)=\sum_{i=1}^{j=n}w_i(x)dx_i$$.

• 1. In the second paragraph, why don't you have $dx_j$ after $\sum_{j=1}^n{\partial w_i\over \partial x_j}(tx)tx_i \boldsymbol{dx_j}+w_i(tx)dx_j$? 2. I don't understand why if $0=dw=\sum_{i=1}^n{\partial w\over \partial x_i} dx_i$ then ${\partial w_i\over \partial x_j}={\partial w_j \over \partial x_i}$? Thanks! – J. Doe Feb 13 at 15:05