Show that the form $w$ is closed but not exact

Question

Let $$w~=~\dfrac{-y}{x^2+y^2}dx+\dfrac{x}{x^2+y^2}dy, \qquad (x,y)~\in\mathbb{R}^2\backslash \{(0,0)\}.$$

Showing that $w$ is closed is easy. Just calculate $dw$ and you'll get 0.

But how do I show that $w$ is not exact?

In other words, I need to prove that there is no form $\lambda$ such that $w=d \lambda$

Should I assume that $w=d \lambda$ and try to get to a contradiction?

Answer 1

I think that perhaps one of the most illuminating ways to look at this is to transform $w$ to polar coordinates. Since

$w = -\dfrac{y}{x^2 + y^2}dx + \dfrac{x}{x^2 +y^2}dy, \tag{1}$

with

$x = r\cos \theta, \tag{2}$

$y = r\sin \theta, \tag{3}$

we immediately see that

$r^2 = x^2 + y^2, \tag{4}$

leading to

$w = -\dfrac{\sin \theta}{r} dx + \dfrac{\cos \theta}{r} dy; \tag{5}$

we also have, from (2) and (3), that

$dx = (\cos \theta) dr - r(\sin \theta)d\theta, \tag{6}$

$dy = (\sin \theta) dr + r(\cos \theta)d \theta, \tag{7}$

and plugging (6) and (7) into (5) yields, after a little algebraic maneuvering,

$w = d\theta. \tag{8}$

Of course in performing the above calculations, we need to remember one little caveat: we must stay away from the point $(x, y) = 0$, that is $r = 0$, where in fact $w$ isn't even defined; we are in "point" of fact operating in the punctured plane $\Bbb R^2 \setminus \{ 0 \}$. And though (8) gives the superficial impression that $w$ is exact, this only appears to be the case, since $\theta$ is not in fact definable as a function on $\Bbb R^2 \setminus \{ 0\}$. This of course may be concluded from the fact that in traversing a circular path centered at the origin the value if $\theta$ will have increased by $2\pi$ when the starting point is first re-visited; we can in fact express this observation in integral form by computing the line integral of $w = d\theta$ around a circle of radius $R$ centered at the origin. Let then the circle be given parametrically by $c(t) = (R\cos t, R \sin t)$, $0 \le t \le 2 \pi$; we have

$\displaystyle \int_c w = \int_0^{2\pi} d\theta(c(t))(\dot c(t))dt$ $= \displaystyle \int_0^{2\pi} d\theta(c(t))((-R\sin t, R\cos t)^T)dt, \tag{9}$

and if we combine (5) and (8) with the definition of $c(t)$ we see that

$d\theta(c(t)) = -\dfrac{\sin t}{R} dx + \dfrac{\cos t}{R} dy, \tag{10}$

and thus

$d\theta(c(t))(\dot c(t)) = \sin^2 t + \cos^2 t = 1, \tag{11}$

and the integral becomes

$\displaystyle \int_c w = \int_0^{2\pi} dt = 2\pi. \tag{12}$

(12) shows that: i.) $w = d\theta$ is not exact in $\Bbb R^2 \setminus \{ 0 \}$; and ii.) $\theta$ cannot really be defined as a function on $\Bbb R^2 \setminus \{ 0 \}$, since we obtain multiple values by integrating $d\theta$ over a path such as $c(t)$. But I guess the main point here is that $w$ is indeed not exact, and this is how it is shown.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

Answer 2

Let $\alpha :[0,2\pi]\to \mathbb{R}^2\setminus\{0\}$ with $\alpha(t)=(cos(t),sin(t))$. Remark that $\alpha$ is a closed curve (i.e., $\alpha(0)=\alpha(2\pi)$). Thus, if $\omega$ is exact, then $$ \int_{\alpha}\omega=0. $$ But $$ \int_{\alpha}\omega=\int_0^{2\pi}\omega(\alpha(t)).\alpha'(t)dt=\int_0^{2\pi}(-sin(t),cos(t)).(-sin(t),cos(t))dt=2\pi $$ Then $\omega$ is not exact.

Answer 3

The origin of the monster: Complex Analysis. The function $$z\longmapsto\frac1z$$ is holomorphic in $\Bbb C\setminus\{0\}$ without primitive in $\Bbb C\setminus\{0\}$. Integrating along a path that surrounds the zero: $$\int_{|z|=1}\frac1z\,dz=2\pi i.$$ The field $w$ will appear while doing the calculations in this line integral.

Answer 4

If $w$ is exact, its integral along any contour $C$ should be equal to zero. But, plotting the vector field $\mathbf{E}=(E_x,E_y)$ with $$E_x=-\frac{y}{x^2+y^2},\qquad E_y=\frac{x}{x^2+y^2},$$ we will get something like this:

and it is clear that the integral $\int \mathbf{E}\cdot d\mathbf{r}$ along the circles centered at the origin does not vanish.

Answer 5

Another take on this (albeit obviously similar to the above solutions) is:

Note that $\omega|_{\delta B_1(0)}=-ydx+xdy$. but this is already its own Pullback onto $\delta B_1(0)$, as can be for example verified using polar coordinates.

Therefore $\omega|_{\delta B_1(0)}=\iota_{\delta B_1(0)}^*(\omega)=-ydx+xdy$.

Now obviously $-ydx+xdy\neq 0$ on $\delta B_1(0)$, because else $ydx=xdy$, wich cannot be because $dx$ and $dy$ are linearly independant and on $\delta B_1(0)$ $x$ and $y$ are not zero at the same time.

But then $-ydx+xdy$ is a nowhere vanishing top degree form on $\delta B_1(0)$, and therefore $\int_{\delta B_1(0)} -ydx+xdy\neq 0$. Actually $\iota_{\delta B_1(0)}^*\omega$ can be reckoned to be up to sign the "canonical" volume form of $\delta B_1(0)$, and therefore cant have Integral 0, but has the 1-volume of $\delta B_1(0)$, up to sign, as integral.

But if $\iota_{\delta B_1(0)}^*\omega$ was exact one would have $\iota_{\delta B_1(0)}^*\omega=d\omega'$ for some $\omega'$ on $\delta B_1(0)$, because $\delta B_1(0)$ is compact and without boundary, stokes theorem applies to $\iota_{\delta B_1(0)}^*\omega$ on $\delta B_1(0)$, yielding:

$\int_{\delta B_1(0)}\iota_{\delta B_1(0)}^* \omega=\int_{\delta B_1(0)}d\omega'=\int_{\delta\delta B_1(0)=\emptyset}\omega'=0$, contradiction to $\int_{\delta B_1(0)}\iota_{\delta B_1(0)}^*\omega\neq 0$ as established earlier.

But therefore $\iota_{\delta B_1(0)}^*\omega$ not exact.

But now suppose $\omega=d\omega''$, so $\omega$ exact, then because pullbacks commute with $d$ we have:

$\iota_{\delta B_1(0)}^* \omega=\iota_{\delta B_1(0)}^* d\omega''=d\iota_{\delta B_1(0)}^*\omega''=d\omega'$ for $\omega':=\iota_{\delta B_1(0)}^*\omega''$, so $\iota_{\delta B_1(0)}^*\omega$ would be exact after all, contradiciton.

The upshot of the argument is: Any differential form whose Pullback to some compact submanifold with positive dimension is a nonvanishing top degree form on the submanifold is not exact due to such a contradiction involving stokes.