# The Unapologetic Mathematician

## The Fundamental Theorem of Line Integrals

At last we can state one of our special cases of Stokes’ theorem. We don’t need to prove it, since we already did in far more generality, but it should help put our feet on some familiar ground.

So, we take an oriented curve $c$ in a manifold $M$. The image of $c$ is an oriented submanifold of $M$, where the “orientation” means picking one of the two possible tangent directions at each point along the image of the curve. As a high-level view, we can characterize the orientation as the direction that we traverse the curve, from one “starting” endpoint to the other “ending” endpoint.

Given any $1$-form $\alpha$ on the image of $c$ — in particular, given an $\alpha$ defined on $M$ — we can define the line integral of $\alpha$ over $c$. We already have a way of evaluating line integrals: pull the $1$-form back to the parameter interval of $c$ and integrate there as usual. But now we want to use Stokes’ theorem to come up with another way. Let’s write down what it will look like:

$\displaystyle\int\limits_cd\omega=\int\limits_{\partial c}\omega$

where $\omega$ is some $0$-form. That is: a function. This tells us that we can only make this work for “exact” $1$-forms $\alpha$, which can be written in the form $\alpha=df$ for some function $f$.

But if this is the case, then life is beautiful. The (oriented) boundary of $c$ is easy: it consists of two $0$-faces corresponding to the two endpoints. The starting point gets a negative orientation while the ending point gets a positive orientation. And so we write

$\displaystyle\int\limits_cdf=\int\limits_{\partial c}f=\sum\limits_{j=0,1}(-1)^{1+j}f(c_{1,j})=f(c(1))-f(c(0))$

That is, we just evaluate $f$ at the two endpoints and subtract the value at the start from the value at the end!

What does this look like when we have a metric and we can rewrite the $1$-form $\alpha$ as a vector field $F$? In this case, $\alpha$ is exact if and only if $F$ is “conservative”, which just means that $F=\nabla f$ for some function $f$. Then we can write

$\displaystyle\int\limits_c\nabla f\cdot ds=f(c(1))-f(c(0))$

which should look very familiar from multivariable calculus.

We call this the fundamental theorem of line integrals by analogy with the fundamental theorem of calculus. Indeed, if we set this up in the manifold $\mathbb{R}$, we get back exactly the second part of the fundamental theorem of calculus back again.

October 24, 2011 - Posted by | Differential Geometry, Geometry

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3. Excellent site; nice job!.
I hope this is not too simple-minded. The last part of this segment seems to say , with the expression f(c(1))-f(c(0)) , that the line
integral of f along c, is independent of the path joining c(1) with c(0) in the manifold. But I believe this independence of path is true
only when the manifold M is simply-connected, right? Equivalently, the integral depends only on the homotopy class ( with fixed
endpoints c(1),c(0)) of the curve joining the endpoints, so that, if M is simply-connected, any two curves are homotopic, but this is
not so if M is not simply-connected. Could you please tell me where the assumption of simple-connectedness is made in your post–
if it is made at all?
Thanks, and keep up with the nice site.

Comment by genios | May 10, 2013 | Reply

4. @genios:

The extra assumption that makes path-independence work is not (in this case) about the space (i.e. simple-connectedness), but about the thing being integrated – it must be the gradient of something.

(I hope I’m not wrong.)

Comment by Andrei | May 10, 2013 | Reply

5. @Andrei: I think you’re right. My comment comes from something I remember about homology, re closed curves defined in a region R of the complex plane, where C~C’ (homologous) , if , given f analytic in R, we had Int_C fdx =Int_C’fdx if there is only one class, then R is homologically-trivial. I think if R is not simply-connected, then there is at least one form w defined in R that is not exact. So maybe I should
change my statement to saying that if R is simply-connected, then not all forms are exact.

Comment by genios | May 10, 2013 | Reply

6. Yes, exactly. It’s more about the homology than the homotopy type of the region, but nontrivial first homotopy does imply nontrivial first homology.

So, where does the assumption of trivial homology come in? It’s actually sort of deep, and comes from the relationship between first homology defined in terms of curves in the region and first cohomology defined in terms of 1-forms. The key assumption is that in our region we can write our 1-form $\alpha$ as $df$ for some function $f$.

To see how this matters, think about the real plane with the origin removed. We can define a vector field wrapping around the origin like

$\displaystyle\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$

which, locally, is the gradient of the “angle function” $\theta$. We can integrate this around the unit circle and get a nonzero answer when we get back to the place where we started, seemingly in violation of the fundamental theorem of line integrals.

But the problem is that we can’t define the angle function consistently on our whole region! There is no single differentiable function on the region whose gradient is our given vector field. And the ability to define such a vector field is deeply connected to the fact that it has a hole around which we can wrap a closed curve.

Comment by John Armstrong | May 10, 2013 | Reply