# How to evaluate this integral: $\oint dx$?

I am trying to understand differential forms. Now I tried to evaluate

$$\oint_{S^1}dx$$

I should get anything non-zero but I don't know how to do it (even though I know the result).

If $S^1$ in the integral is one of the circles that goes around the torus,

how do I go about actually calculating the integral?

It's clear to me that the function I am integrating is $f(x,y)=1$. Now it's not clear to me whether I can just choose any parametrisatioin of $S^1$ and then integrate, like for example

$$\int_0^{2\pi} 1 \cdot |r'(t)| dt = 2 \pi$$

where $r = (\cos t, \sin t)$.

The reason why I think that this might be wrong is that it does not take into account any information about the space in which $S^1$ lies. But the shape of the space should determine whether a given $1$-form is exact or not.

Edit

In response to the comment by jflipp:

If I use $d \theta = {x dy - y dx \over x^2 + y^2}$ then I can do it: I see that the integral is $2\pi$. Now I want to use either $dx$ or $dy$, the differential $1$-forms in the expression $d \theta = {x dy - y dx \over x^2 + y^2}$.

• What exactly is the $x$ in $\oint_{S^1} dx$? Is this a coordinate function on $\mathbb R^3,$ in which the torus is embedded? Is it something else? – jflipp Jan 22 '15 at 8:52
• @jflipp I don't know, I suspect this is what I'm not clear about. What else can it be? – a student Jan 22 '15 at 8:56
• @jflipp I can answer your question... I edited my question. – a student Jan 23 '15 at 0:24

You write that the function $f$ is on a form $f(x,y)=1,$ so I assume that $S^1$ lies on a plane and $x$ in $dx$ stands for first coordinate on a plane. To calculate your integral, you have to specify orientation of $S^1$. Hence I additionaly assume that the orientation of the circle is standard, i.e. counterclockwise. Now take an arbitrary orientation preserving parametrization $r:[0,2\pi]\rightarrow S^1.$ Let $r(t)=(\cos t,\sin t).$ The integral by DEFINITION (see Spivak - Calculus on Manifolds) is equal $$\oint_{S^1}dx=\int_0^{2\pi} r^*(dx)=\int_0^{2\pi} d(r^*(x))=\int_0^{2\pi} d(x\circ r)=\int_0^{2\pi} d(\cos t)=\int_0^{2\pi} -\sin t dt=0.$$ You can also use Stoke's theorem. Since $S^1$ has an empty boundary we get that $$\oint_{S^1}dx=\oint_\emptyset x=0.$$

Edit

But if you are willing to integrate anything of the form $d(something),$ you are still allowed to use Stoke's theorem.

Edit $2$

I believe I know the answer if we want to calculate the integral $$\oint_{S^1}d\theta,$$ where $d\theta$ is just a SYMBOL (it is not an exact form) for $\frac{xdx-ydy}{x^2+y^2}.$ Now we have to choose circle that we are interested in. Let it be a circle glued along A oriented as in picture. Then the parametrization is given by $r:[0,1]\rightarrow S^1$ with formula $r(t)=(t,1).$ With that, the integral is equal $$\oint_{S^1}\frac{xdx-ydy}{x^2+y^2}=\int_0^1r^*(\frac{xdx-ydy}{x^2+y^2})=\int_0^1(\frac{x\circ r}{x^2 \circ r +y^2 \circ r})d(x\circ r)-0=\int_0^1\frac{t}{t^2+1}dt=\log(2)/2$$

$dx$ is exact by definition, provided that $x$ is a globally well-defined function on your space.
So if I understood your question correctly you're asking about an abstract torus that is considered to be the quotient space of $R^2$, on which the function $x$ is not globally defined.
In this case, you'll need to compute your integral like $\oint_{S^1}dx=x(2\pi)-x(0)=2\pi$

• Which on $S^1$, it isn't. – user147263 Jan 22 '15 at 23:04
• Why did you say $x$ is not globally defined on $S^1$? – Xipan Xiao Jan 22 '15 at 23:44
• OK I got your point. If the torus is not embedded in $R^3$, but only considered a quotient space of $R^2$, then right $x$ is not well-defined. – Xipan Xiao Jan 22 '15 at 23:46
• Thank you. You understand my question correctly. How did you go from $\oint_{S^1}dx$ to $x(2\pi) - x(0)$? Do you just integrate the constant $f(x,y) = 1$ function to get $f(x,y)=x + g(y)$? I am unsure how to evaluate $x+g(y)$... – a student Jan 23 '15 at 0:17
• @FallenApart Xipan is assuming that $dx$ is a differential form on $\mathbb{R}$ which then decends to the quotient space $\mathbb{R}/(2\pi)$, which is a circle. In other words, his $dx$ is what someone might normally call $d\theta$. So both of your answers are correct with a different interpretation of what OP meant by $dx$. – Steven Gubkin Jan 23 '15 at 0:21