5

On page 35 in his book Riemannian Geometry, Manfredo do Carmo states the following:

Giving a surface $S \subset \Bbb{R}^{3}$, we have a natural way of measuring the lengths of vectors tangent to $S$, namely: the inner product $\langle v, w\rangle$ of two vectors tangent to $S$ at a point $p$ of $S$ is simply the inner product of these vectors in $\Bbb{R}^{3}$. The way to compute the length of a curve is, by definition, to integrate the length of its velocity vector. The definition $\langle \phantom{X} , \phantom{X}\rangle$ permits us to measure not only the lengths of the curves in $S$ but also the area of domains in $S$, as well as the angle between two curves, and all the other "metric" ideas used in geometry.

I don't understand

(1) what we are integrating

(2) how the lengths of the tangent vectors can be used to define the length of a curve

Can someone address those two points?

Note: I mean, visually, I could imagine stringing a bunch of little arrows together like the pictures of path integral arrows in physics, but I assume nothing like that is going on here. Its not like we are putting tangent vectors tip to tail and summing over their lengths is it?

Stan Shunpike
  • 4,921

1 Answers1

6

"but I assume nothing like that is going on here. Its not like we are putting tangent vectors tip to tail and summing over their lengths is it? " -- in a sense it is.

(1) we are integrating $\sqrt {\langle \gamma\prime, \gamma\prime\rangle}$ if $\gamma$ is a parametrical $C^1$ curve with values in $M$, i.e. if $\gamma:[a,b]\rightarrow M$ is $C^1$ you calculate $$\int_a^b \sqrt {\langle \gamma^\prime(t), \gamma^\prime(t)\rangle} dt$$

(2) this is the limit of an approximation procedure. If the curve is contained in some Euclidean vector space (like $\mathbb{R}^3$) it turns out that the length of approximating linear polygons converges to exactly the integral in question. Actually the tangents to such approximating polygons tend locally to the tangent vectors of the curve.

Thomas
  • 22,361
  • That was very helpful. One question: I am new to applying calculus to manifolds. And this integral is defined on $[a,b]$. So does this mean $\gamma'(t)$ is integrable over both $[a,b]$ and the corresponding values in$M$ or just $[a,b]$? – Stan Shunpike Jan 02 '15 at 04:38
  • 2
    @StanShunpike A parametric curve is defined on a real interval. The integrand is, for each $t$, just a real number. So the particular integral we are looking at is an integral of a real valued function on a real interval. The geometric meaning of the integral is, however, coming from the geometry of $M$ and coded in the metric. In particular the parametrization of the curve does not matter, if you reparametrize you get the same length. – Thomas Jan 02 '15 at 04:47