Suppose your manifold is just M = R^2 with the standard differential structure (so the atlas is {(R^2, id, R^2)}). Suppose we have a 1-form [tex]\omega[/tex] on M. Then ofcourse [tex]\omega = a_1 dx_1 + a_2 dx_2[/tex], where the a_i are just c-infinity functions from R^2 to R. Suppose we have a function f on M defined as follows:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]f(x) = x_1 \int_0^1 a_1(tx) dt + x_2 \int_0^1 a_2(tx) dt[/tex]

Show that df = w when dw = 0. Use the fact that:

[tex]\frac{d}{dt}(t a(tx)) = a(tx) + tx_1 \frac{\partial a}{\partial x_1}(tx) + tx_2 \frac{\partial a}{\partial x_2}(tx)[/tex]

Using this hint i rewrote f(x):

[tex]f(x) = a_1 (x)x_1 + a_2 (x)x_2 - x_1^2 \int_0^1 t\frac{\partial a_1}{\partial x_1}dt - x_2^2 \int_0^1 t \frac{\partial a_2}{\partial x_2}dt - x_1 x_2 \int_0^1 t\left(\frac{\partial a_1}{\partial x_2} + \frac{\partial a_2}{\partial x_1}\right) dt[/tex]

Now I want to use the fact that [tex]d\omega = 0[/tex]

This means that:

[tex]d\omega = d(a_1 dx_1 + a_2 dx_2) = d(a_1)\wedge dx_1 + d(a_2)\wedge dx_2 = \left(\frac{\partial a_2}{\partial x_1} - \frac{\partial a_1}{\partial x_2}\right) dx_1 \wedge dx_2

[/tex]

so [tex]\frac{\partial a_2}{\partial x_1} = \frac{\partial a_1}{\partial x_2}[/tex]

If somehow df = w, the differential of the integral terms in f(x) should be zero. But I don't see how i could show that.

Could someone give me a hint?

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# Homework Help: Exactness of a differential form

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