# Proving the Poincare Lemma for $1$ forms on $\mathbb{R}^2$

I am trying to prove the Poincare Lemma for $1$ forms on $\mathbb{R^2}$. So I said that if I doing this, I start of with

$$\omega = f_1(x_1,x_2) dx_1 + f_2(x_1,x_2)dx_2.$$

First thing I want to prove is $d\omega = 0$. So, I get

$$d \omega = df_1 \wedge dx_1 + df_2\wedge x_2$$

$$= \left( \frac{\partial f_1}{\partial x_1}dx_1 + \frac{\partial f_1}{\partial x_2} dx_2 \right) \wedge dx_1 + \left(\frac{\partial f_2}{\partial x_1} dx_1 + \frac{\partial f_2}{\partial x_2}dx_2 \right) \wedge dx_2.$$

$dx_1 \wedge dx_1 = 0$ and $dx_2 \wedge dx_2 = 0$ and so we get

$$\frac{\partial f_1}{\partial x_2} dx_2 \wedge dx_1 + \frac{\partial f_2}{\partial x_1}dx_1 \wedge dx_2.$$

From the anti commutatitivty law we have $dx_1 \wedge dx_2 = - dx_2 \wedge dx_2$, and so we put this in and collect like terms and get

$$\left(\frac{\partial f_2}{\partial x_1} - \frac{\partial f_1}{\partial x_2} \right) dx_i \wedge dx_2,$$

which tells me that $d \omega = 0 \iff \frac{\partial f_2}{\partial x_1} = \frac{\partial f_1}{\partial x_2}$, but I'm stuck how to use this to show that $\omega = d \eta$. Can someone help me please?

• You are almost there. – Shuhao Cao Jun 6 '13 at 5:29

Well if $d\omega = 0$ then for any smooth closed curve $\gamma$ in $\mathbb{R}^2$ the area enclosed is a smooth compact manifold $M$ where $\partial M = \gamma$. From stokes theorem we have for any closed curve $\gamma$ $$\int_\gamma \omega = \int_{\partial M} \omega = \int_M d\omega = 0$$ Thus if $\gamma_1$ and $\gamma_2$ are curves with same endpoints then $\int_{\gamma_1}\omega = \int_{\gamma_2}\omega$. So let for $x = (x_1,x_2)$ we define $\gamma_x$ as any curve from origin to $x$ and define $F : \mathbb{R}^2 \rightarrow \mathbb{R}$ as $$F(x) = \int_{\gamma_x}\omega$$ Thus for any curve $\alpha$ from $x$ to $x+he_k$ where $h>0$ and $k\in\{1,2\}$ we have $F(x+he_k) -F(x) = \int_\alpha \omega$. So for $f = (f_1,f_2)$ that are given, if we choose $\alpha$ as the straight line i.e. $\alpha(t) = x + the_k$ then $\alpha (0) = x,\ \alpha(1) = x +he_k$ then we arrive at for $k\in \{1,2\}$ $$F(x+he_k) -F(x) = \int^1_0 (f\circ\alpha)(t).\alpha'(t)dt = \int^1_0 f(x+the_k).he_k dt = h\int^1_0f_k(x+the_k)dt$$ If we define the function $$g_k(t) = \int^t_0f_k(x+ue_k)du$$ then $g_k(0) = 0$ and $g_k'(t) = f_k(x+te_k)$ hence $g_k'(0) = f_k(x)$. Then from above we clearly have $$\frac{F(x+he_k)-F(x)}{h} = \int^1_0f_k(x+the_k)dt = \frac{1}{h}\int^h_0 f_k(x+ue_k)du = \frac{g_k(h)-g_k(0)}{h}$$ Hence taking $h \rightarrow 0$ we have $\partial F/\partial x_k = g_k'(0) = f_k(x)$. Thus $$\omega = f_1dx_1 + f_2dx_2 = \frac{\partial F}{\partial x_1}dx_1 + \frac{\partial F}{\partial x_2}dx_2 = dF$$