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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?

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  • $\begingroup$ You are almost there. $\endgroup$
    – Shuhao Cao
    Jun 6, 2013 at 5:29

1 Answer 1

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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 $$ \omega = f_1dx_1 + f_2dx_2 = \frac{\partial F}{\partial x_1}dx_1 + \frac{\partial F}{\partial x_2}dx_2 = dF $$

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  • $\begingroup$ Why nobody vote up this nice answer? +1. $\endgroup$
    – Shuhao Cao
    Jun 6, 2013 at 5:32
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    $\begingroup$ Everyone knows that if you want upvotes on MSE you have to include a graph of the Batman logo. $\endgroup$
    – treble
    Jun 6, 2013 at 5:44
  • $\begingroup$ @ treble: So true. $\endgroup$ Nov 14, 2015 at 18:50
  • $\begingroup$ Wow, finally! have been searching for two weeks now without knowing the lemma's name and now found it. Reading about Poincare's Lemma makes me actually think and feel, that it's actually a very powerful/strong and lemma, similar to the Cauchy–Goursat (integral) theorem. I am so happy now! $\endgroup$
    – Imago
    Mar 7, 2016 at 17:08

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