Definition of arc length on manifolds without parametrization

[mma]

Curves are functions from an interval of the real numbers to a differentiable manifold.
Given a metric on the manifold, arc length is a property of the image of the curves, not of the curves itself. In other word, it is independent of the parametrization of the curve. In the case of the Euclidean space, arc length can be defined without using any parametrization of the curve, namely by the limit of the length of the approximating polygons. My question is whether could we give similar definition of the arc length in the case of manifolds? The standard definition is $$l(\gamma)=\int_{0}^{1} {|\dot\gamma(t)|} dt\ ,$$ and unlike in the case of the definition with polygons, this definition involves the parametrization of the curve, in spite that all parametrizations give the same result. How could we eliminate the unnecessary parametrization from the definition?

 

[evouga]

But how do you define a "curve" that doesn't have a parametrization? For instance, if you say a "curve" is a connected compact 1-dimensional submanifold with boundary, it's easy to show by a gluing argument that this "curve" is just a piecewise regular curve in the ordinary sense, and you can use your formula to get the arc length.

 

[mma]

But how do you define a "curve" that doesn't have a parametrization? For instance, if you say a "curve" is a connected compact 1-dimensional submanifold with boundary, it's easy to show by a gluing argument that this "curve" is just a piecewise regular curve in the ordinary sense, and you can use your formula to get the arc length.

I say an analogy of what I mean.
The definition of the tangent vector of a curve also uses the parametrization of the curve (and what is more, depends on in). The tangent space over a point of the manifold consinsts of the tangent vectors of the curves at this point. Even so, tangent space can be defined without mentioning curves: it is the space of derivations. Derivation is a linear and Leibniz-rule compliant real-valued map from the space of the smooth functions around the point.

 

[Doodle Bob]

the answer to your question is "no, there is no canonical method of defining the arc length of a curve on a general manifold." The reason is quite simple: defining arc length on a curve amounts to locally defining a (not necessarily Euclidean) arc length of a curve in R^n, i.e., on the coordinate patches of the manifold. But the local coordinate systems of the manifold are related to each other by diffeomorphisms, which in general will not preserve a distance-function or arc length function.

In order for your specific scheme to work, the transition mappings of the local coordinate systems of the manifold would have to be elements of SO(n,R), i.e., it would be an affine manifold. And, in general, for any method of assigning arc length to curves, the transition mappings would have to somehow preserve this method.

I imagine, though, that there are plenty of particular manifolds with particular diffeomorphic structures on which an arc length function can be defined parametrization-free on its smooth curves.

 

[evouga]

Why not? The coordinate charts don't have to be isometric to R^n with the standard inner product.

It seems like you can generalize the argument given by the OP for standard Euclidean space. Obviously you need a metric (but Euclidean space has one too). Then cover your curve in finitely many totally normal neighborhoods of radius < epsilon, and construct a piecewise-geodesic curve such that each piece is a minimizing geodesic of arc length < epsilon, and each point between two pieces lies on your original curve. Take epsilon -> 0.

 

[Doodle Bob]

It seems like you can generalize the argument given by the OP for standard Euclidean space. Obviously you need a metric (but Euclidean space has one too). Then cover your curve in finitely many totally normal neighborhoods of radius < epsilon, and construct a piecewise-geodesic curve such that each piece is a minimizing geodesic of arc length < epsilon, and each point between two pieces lies on your original curve. Take epsilon -> 0.

I somewhat misunderstood the original question. I didn't realize that he was assuming a fixed metric, in which case, your method works fine. Of course, in order for this to be easier than the standard parametrization method of calculating arc length, we need a manifold for which it is relatively easy to calculate the distance between two given points, e.g., the hyperbolic plane.

Off-hand, this is a nice way to intuitively envision arc length.

 

[mma]

It seems like you can generalize the argument given by the OP for standard Euclidean space. Obviously you need a metric (but Euclidean space has one too). Then cover your curve in finitely many totally normal neighborhoods of radius < epsilon, and construct a piecewise-geodesic curve such that each piece is a minimizing geodesic of arc length < epsilon, and each point between two pieces lies on your original curve. Take epsilon -> 0.

I am afraid that I don't understand this. It doesn't look like a definition of arc length.

 

[Doodle Bob]

I am afraid that I don't understand this. It doesn't look like a definition of arc length.

It essentially amounts to the polygonal situation on the plane that you described.

You're given a smooth curve whose arc length you want to determine from point a to point b. For every positive integer n, choose n points on the curve between points a and b such that any two consecutive points on the curve are connected by a unique shortest-length geodesic (we are assuming that a and b are among these points). Then connect the points by those geodesics and sum up their lengths. If we do this so that the longest length of the geodesics for each partitioning corresponding to n goes to zero as n goes to infinity, then the length of the arc will be the limit of the sums of the lengths of the geodesics as n goes to infinity.

Of course, even in the Euclidean case, this only works with a smooth curve (at least C^1), since we all know that the coast of England is infinitely long.

 

[mma]

the length of the arc will be the limit of the sums of the lengths of the geodesics as n goes to infinity.

OK, but then how is the length of a geodesic defined?

 

[Doodle Bob]

OK, but then how is the length of a geodesic defined?

Well, actually, through the use of parametrizations. I'm afraid there's just no getting around it. See Do Carmo's Riemannian Geometry for details.

 

[mma]

Well, actually, through the use of parametrizations. I'm afraid there's just no getting around it. See Do Carmo's Riemannian Geometry for details.

It's clear that parametrization is superfluous in the definition because the value of the arc length does not depend on it. But I don't know how can we get rid of it.:(

 

[evouga]

On a Lie group can you use the inverse of the exponential map to locally compute the arc length of geodesics?

 

[mma]

On a Lie group can you use the inverse of the exponential map to locally compute the arc length of geodesics?

Do you mean of curves t->exp(tv) where |v|=1?. On the one hand, such curves are only very special ones among all curves on the Lie-group, and on the other hand this works only for Lie-groups. Otherwise good idea.

 

[evouga]

Yes, they are the geodesics, but then you can find the arc length of any arbitrary curve by using the argument above.

You're right about it only working on Lie groups, of course.

 

[mma]

Yes, they are the geodesics, but then you can find the arc length of any arbitrary curve by using the argument above

I am not too familiar in Lie groups, so I don't know how are geodesics deined on Lie groups. According your remark it seems to me that all t-> exp(tv)g is a geodesic where g is an arbitrary elemt of the group (perhaps this is the definition itself), because the above mentioned t-> exp(tv) curves are only those curves that pass across the unit element of the group; and of course we need geodesics across arbitrary points in arbitrary direction if we want do define the arc length of any curves by them. And it means of course that Lie groups are always equipped with this unique metric determined by the group itself. Am I right?

 

[Doodle Bob]

It's clear that parametrization is superfluous in the definition because the value of the arc length does not depend on it.

Actually, that's not very clear at all. Just because the arc length doesn't depend on it, does not mean that it's superfluous. There might be other ways to define arc length, but in order to find them you should first understand why parametrization is used in the definition in the first place:

If you're given two points and a smooth curve connecting them, this is an immersed submanifold of the overall manifold. A parametrization of this curve can be then thought of as a generalized coordinate patch of this submanifold. The Riemannian metric on the ambient space induces a Riemannian metric onto the curve, which in turn induces a "volume form" on the curve. Arc length is then the measure of the entire curve with respect to this volume form, which is in fact a 1-form on the curve. If you can find some parametrization-free way to redefine the integral of this 1-form along the curve, then you've done it.

The idea of splitting the curve up into partitions and taking the limit as the number of points go to infinity and the distances between these points go to 0, etc., is essentially calculating some particular Riemann sums of the integral described above. Conceptually a nice idea, but calculationwise quite prohibitive in specific cases. For example, try using this method to calculate the length of a lemniscate on the plane.

 

[mma]

Yes, integration of forms on manifolds also suffers from the same thing. Perhaps it is unavoidable to use sometimes some inert objects in the definitions.

 

[gel]

How about the following. If C(M) is the space of smooth functions M->R, then you can define a distance between points as
$$d(P,Q)=\sup\left\{ |f(P)-f(Q)| : f\in C(M), \Vert \nabla f \Vert \le 1\right\}$$

You can then define the length of a curve (even a non-smooth one) in the same way as for Rⁿ. I've seen this idea used in noncommutative geometry, where you don't have coordinate charts.

 

[mma]

How about the following. If C(M) is the space of smooth functions M->R, then you can define a distance between points as
$$d(P,Q)=\sup\left\{ |f(P)-f(Q)| : f\in C(M), \Vert \nabla f \Vert \le 1\right\}$$

You can then define the length of a curve (even a non-smooth one) in the same way as for Rⁿ. I've seen this idea used in noncommutative geometry, where you don't have coordinate charts.

What does here $$\Vert \nabla f \Vert$$ mean?

 

[gel]

What does here $$\Vert \nabla f \Vert$$ mean?

By $\nabla f$ I meant the differential of f, maybe df would be better notation. and ||df|| is the norm under the metric, $\sqrt{g(df,df)}$.

 

[Doodle Bob]

How about the following. If C(M) is the space of smooth functions M->R, then you can define a distance between points as
$$d(P,Q)=\sup\left\{ |f(P)-f(Q)| : f\in C(M), \Vert \nabla f \Vert \le 1\right\}$$

You can then define the length of a curve (even a non-smooth one) in the same way as for Rⁿ. I've seen this idea used in noncommutative geometry, where you don't have coordinate charts.

Where have you seen this? It's a rather interesting idea. This does assume a Riemannian metric on M, doesn't it? Since otherwise you would not be able to calculate $$\Vert \nabla f \Vert.$$

 

[gel]

Where have you seen this? It's a rather interesting idea. This does assume a Riemannian metric on M, doesn't it? Since otherwise you would not be able to calculate $$\Vert \nabla f \Vert.$$

Not sure where I saw this originally, but google came up with the following - see equation 3.5 on page 34 of http://ncg.mimuw.edu.pl/index.php?option=com_content&task=view&id=148&Itemid=98". This assumes you have a spinor structure on the manifold, which does define a Riemannian metric.

edit: in fact, see the equation right in the middle of the very first page of the introduction.

 

[Doodle Bob]

Thanks. It cracks me up that the abstract describes Clifford algebra, spin structures, etc. as "conventional differential geometry." I guess, compared to what he's describing in the lecture, that's true, but I'm much more of a Riemannian curvature kind of guy.

 

[mma]

Not sure where I saw this originally, but google came up with the following - see equation 3.5 on page 34 of http://ncg.mimuw.edu.pl/index.php?option=com_content&task=view&id=148&Itemid=98". This assumes you have a spinor structure on the manifold, which does define a Riemannian metric.

edit: in fact, see the equation right in the middle of the very first page of the introduction.

Sorry for replaying so late, but I took only now the necessary time and effort to understand this definition. Now I understand the derivation of eq. (3.4) in the article you referred, i.e.
$$\sup\{|a(y)-a(x)| : a \in C(M), \Vert{\mathrm{grad} a\Vert _\infty \leq 1\} \leq d(x,y)$$.
But I don't understand the derivation of (3.5) (i.e. the equation above with equality instead of the inequality between tha sides). The triangle inequality $$\vert a_x(y) - a_x(z)\vert \leq d(y,z)$$ has the same direction as (3.4), so it doesn't make equality from it. We need a reverse inequality for proving the equation, isn't it?

 

[mma]

Sorry, I'm silly. I misunderstood the logic of the proof. As now I see, the proof simply shows that the supremum is a maximum at a = a_x, because

|a_x(y) - a_x(x)| = |d(x,y) - d(x,x)| = d(x,y).

But I still don't see the role of the triangle inequality

|a_x(y) - a_x(z)| ≤ d(x,z)

here.

 

[gel]

You need the triangle inequality to show that if a_x(y) = d(x,y)
then
|a_x(y)-a_x(z)| = |d(x,y) - d(x,z)| <= d(y,z)
so a_x is Lischitz, with Lipschitz constant one. The supremum over |a(y)-a(z)| was taken over such Lipschitz continuous functions, so you do need this.

 

[mma]

The supremum over |a(y)-a(z)| was taken over such Lipschitz continuous functions.

No, the supremum was taken over continuously differentiable functions having sup(||grad a||) ≤ 1. It is shown that sup(||grad a_x||) = 1 (by local geodesic calculation), so we need only be sure that a_x is continuously differentiable. Perhaps that's why we want it to be Lipschitz? Does continuously differentiability property follows from Lipschitz continuity property?

 

[gel]

No, Lipschitz continuous only implies that it is differentiable almost everywhere. That is, except on a set of measure zero. And grad a will be bounded by the Lipschitz constant (=1) wherever it is defined.
I think this is why he says "in this supremum we can use a in C(M) not necessarily smooth; a need only be continuous with grad a (nu-essentially) bounded."

I think that differentiable almost everywhere is enough for this to work. However, I bet you could also smooth out a_x by convolving with smooth functions on a coordinate chart, but the details of that are tricky.

 

[mma]

I'm afraid that I misinterpreted the notation C(M). I automatically identified this with C¹(M), i.e. with the set of continuously differentiable functions. Perhaps because of the presence of grad, that has sense only on differentiable functios. If C(M) stands for the continuous functions then I understand the role of the Lipschitz-condition: this assures the sense of the grad operation at almost all points. Am I right now?

I note, that the text is still misty a little bit for me. He says: "Since we have obtained $$|a(y)-a(x)| \leq \Vert{\mathrm{grad} a\Vert _\infty d(x,y)$$ , we see that a need only be Lipschitz on M --with respect to the distance d-- with Lipschitz constant $$\leq \Vert{\mathrm{grad} a\Vert _\infty$$"
Isn't mismatched the result with the starting condition here?

Oher question: what does the phrase "nu-essentially" mean?

Sorry for my incomprehension, I hope you have enough patience for me.

 

[gel]

grrr, just spent a while on a detailed response and it logged me out!

anyway, I was saying smthing along the lines of

- have not read this paper myself. Just skimmed some of it. C(M) = cts functions would seem like the normal notation though
- "nu-essentially bounded". It must mean that nu = measure on manifold given by metric. then grad defined and bounded on a set S and nu(M-S)=0 (ie, bounded almost-everywhere). This bound is also called the "essential supremum" of |grad(a)| in measure theory.
- Looks like you could prove the result for smooth functions anyway.

 

[gel]

and $\Vert f\Vert_{\infty}$ is the measure theory notation for the minimum real number K>0 such that K>|f| almost everywhere, also known as the essential supremum of |f|. Must be what he means.

 

[mma]

I'm very sorry for you lost your detailed post. Next time don't forget to copy&paste your text somewhere before pressing the preview or submit button. Thank you for repeating it shortly.

 

[mma]

It is shown that sup(||grad a_x||) = 1 (by local geodesic calculation)

I thought originally that this "local geodesic calculation" is trivial. But now I see that I don't really know how is it.

It is clear that taking a short geodesic $$\gamma(s)$$ starting from a(x) and parametrized by its arc length, then

$$d(x,y) = a_x(y) - a_x(x) = \int_0^{d(x,y)}\dot{\gamma}(a_x)|_{\gamma(s)} ds$$

and from this follows $$\dot{\gamma}(a_x) = 1$$.

But this means only that $$g(\mathrm{grad}(a_x), \dot{\gamma}) =1$$, and of course we know that $$\Vert\dot{\gamma}\Vert$$ = 1.

But how follows $$\Vert\mathrm{grad}(a_x)\Vert = 1$$ from this?

 

[mma]

I suspect that $$\mathrm{grad}(a_x) = \dot{\gamma}$$.

Here $$a_x(y) := d(x,y)$$, the distance between x and y (I forgot to mention this in my previous post), and $$\gamma$$ is a geodesic through x, parametrized by its arc length)

So, my question is: how can I prove this?

 

[mma]

It is clear that taking a short geodesic $$\gamma(s)$$ starting from a(x)

Of course strarting from x and not from a(x). Sorry for the mistyping.