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I am reading Burago, Burago & Ivanov's book where they distinguish the notion of a curve and a path in the following way: a path in a topological space $X$ is simply a (continuous) map from a connected subset of a line into $X$, while a curve is an equivalence class of paths under the relation of being a re-parametrization of each other (sort of). EDIT: More precisely, we are talking about the minimal relation of equivalence in which $\gamma_1:I_1\to X$ is equivalent to $\gamma_2:I_2\to X$ whenever there is a continuous non-decreasing $\phi:I_1\to I_2$ such that $\gamma_1=\gamma_2\circ\phi$.

Addition: a little more explicit description of the established equivalence relation is the following lemma: $\gamma_1:I_1\to X$ is equivalent to $\gamma_2:I_2\to X$ if and only if there is a path $\gamma_3:I_3\to X$ and continuous non-decreasing $\phi_i:I_3\to I_i$ such that $\gamma_3=\gamma_i\circ\phi_i$ and if and only if there is a path $\gamma_4:I_4\to X$ and continuous non-decreasing $\psi_i:I_i\to I_4$ such that $\gamma_i=\gamma_4\circ\psi_i$.

Then among other things it is proven that roughly speaking the length (which is a function of paths, but is independent of reparametrization and so can be descended to a function of curves) is lower semi-continuous functional, and the set of all curves of length bounded by a constant is compact. However, this is done in terms of paths and their uniform convergence (paths on closed finite intervals only).

Addition: the length of a path $\gamma:I\to X$ is $\sup\left\{\sum\limits_{k=1}^{n}d\left(\gamma\left(t_{k-1}\right),\gamma\left(t_{k}\right)\right)\left|t_{0}\le t_{1}\le...\le t_{n}\in\ I\right.\right\}$. This quantity does not depend on a parametrization.

Of-course one can simply take the appropriate quotient of say $C([0,1],X)$ with compact-open topology, and make these statements precise in terms of curves.

Hence my question is, what can we say about this quotient? It is not obvious even if it is Hausdorff. If we are living in a metric space, is this quotient metrizable? Note that Hausdorff distance doesn't work since different curves can have the same image.

Also formally we get different quotients depending on the type of the domain of paths ($[a,b],[a,b),(a,b],(a,b)$). Is there any way to put them in one space somehow? It would also make sense to consider closed curves, i.e. having circles as the domains of paths.

References would be highly appreciated.

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  • $\begingroup$ I believe you should indeed make definition of curve more detailed. The way you have it, I believe definition of length is confusing: you can parametrize a subspace homeomorphic to an interval using another interval of arbitrary length. $\endgroup$ – მამუკა ჯიბლაძე Jan 31 '17 at 6:41
  • $\begingroup$ what is wrong with factor-distance: the distance between two curves is the minimal distance between two paths prametrized by them? $\endgroup$ – Fedor Petrov Jan 31 '17 at 7:32
  • $\begingroup$ @მამუკა ჯიბლაძე: done $\endgroup$ – erz Jan 31 '17 at 22:37
  • $\begingroup$ @Fedor Petrov: it is not apparent to me how to show the triangle inequality for these not very explicit classes and besides, this only takes care about one question. $\endgroup$ – erz Jan 31 '17 at 22:40
  • $\begingroup$ I feel (perhaps along the traditional lines) that reparametrizations should be strictly increasing homeomorphisms. The nonincreasing generalization would be a nuisance and they do not give an equivalence (not directly, as you've written). Moreover, and this time more arguably, for the geometric context of curves and their generalizations, I would restrict parametrizations to local homeomorphisms onto the parametrization's range. $\endgroup$ – Włodzimierz Holsztyński Feb 1 '17 at 4:20
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The distance can be defined as inf of $\sup|\gamma_1(t)-\gamma_2(\phi(t))|$ over all continuous increasing reparametrizations $\phi$. This is a distance. (Of course, the length and rectifiability do not depend on reparametrization).

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  • $\begingroup$ what if $\phi$ is not strictly increasing? $\endgroup$ – erz Jan 31 '17 at 8:23
  • $\begingroup$ @erz: strictly or not, does not matter. I take sup over all continuous non-decreasing functions with $\phi(0)=0$ and $\phi(1)=1$. $\endgroup$ – Alexandre Eremenko Jan 31 '17 at 17:48
  • $\begingroup$ I am sorry, but I don't understand. This does not even describe all paths in a curve. Even if it did, I don't see how to show the triangle inequality, since we cannot "reconcile" different parametrizaions. $\endgroup$ – erz Jan 31 '17 at 20:09
  • $\begingroup$ Ok, triangle inequality holds indeed. How to show that this distance is $0$ implies curves being equal? $\endgroup$ – erz Feb 1 '17 at 2:02
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I have been hesitating on whether I should post this long solution, but I was unable to find better. I will just add missing details to the received answers and comments, which all conveyed the same idea on how to put a metric on the space of curves, but did not explain why it is a metric.

First, let me add one more statement from the book that I've quoted. Let $I$ be some fixed closed interval and let $X$ be a metric space. We will call a path $\gamma:I\to X$ unstopping if there is no open subintervals of $I$, where $\gamma$ is a constant (in the book they call such paths never locally constants, but in my opinion it is too mouthful). It is explained in the book that any path is a re-parametrization of an unstopping one, which together with the description quoted in the body of question gives that every curve is of the form $\left\{\gamma\circ\phi\left|\phi\in H\right.\right\}$, where $\gamma:I\to X$ is an unstopping representative and $H$ is the set of all increasing surjection from $I$ onto itself.

Let $G$ be the set of all strictly increasing bijections from $I$ into $I$. It is easy to see that $G$ is dense in $H$ with respect to the uniform topology, and since $I$ is compact, the correspondence $\phi\to\gamma\circ\phi$ is continuous from $H$ into $C\left(I,X\right)$ with respect to the uniform topologies. Thus $\left\{\gamma\circ\phi\left|\phi\in G\right.\right\}$ is dense in $\left\{\gamma\circ\phi\left|\phi\in H\right.\right\}$. Thus we can define the distance between the curves parametrized by unstopping paths $\gamma_{1},\gamma_{2}$ by $\widetilde{d}\left(\widetilde{\gamma_{1}},\widetilde{\gamma_{2}}\right)=\inf\left\{d\left(\gamma_{1},\gamma_{2}\circ\phi\right)\left|\phi\in G\right.\right\}$. Since $G$ is a group it is easy to show the triangle inequality. We are only left with proving non-degeneracy. We will need an additional concept for it.

For a unstopping path $\gamma:I\to X$ define $S_{\gamma}$ to be the collection of all closed subsets $J$ of $I$, which are components of $\gamma^{-1}\left(\gamma\left(J\right)\right)$. It is easy to show the following properties of this family.

Lemma 1

(i) $J\in S_{\gamma}$ if and only if it is a closed connected set, and each of the endpoints of $J$ belongs to $\overline{I\backslash \gamma^{-1}\left(\gamma\left(J\right)\right)}$, unless it is also an endpoint of $I$.

(ii) If $J_{1}, J_{2}\in S_{\gamma}$ and $J_{1}\bigcap J_{2}$, then $J_{1}\bigcap J_{2}\in S_{\gamma}$.

(iii) If $J_{1}, J_{2}\in S_{\gamma}$ and $\gamma\left(J_{1}\right)\subset \gamma\left(J_{2}\right)$, then either $J_{1}\subset J_{2}$, or $J_{1}\bigcap J_{2}=\varnothing$.

(iv) For any $t\in I$ the collection $\left\{int J\left|J\in S_{2},~t\in int J\right.\right\}$ forms a local base of the usual topology of $I$ at $t$. Also $\bigcap\limits_{t\in int J,~J\in S_{\gamma}}\gamma\left(J\right)=\left\{\gamma\left(t\right)\right\}$.

Fix two unstopping paths $\gamma_{1},\gamma_{2}$, such that $\widetilde{d}\left(\widetilde{\gamma_{1}},\widetilde{\gamma_{2}}\right)=0$. Since the uniform distance is at least the Hausdorff distance between the images, we get that that $\gamma_{1}\left(I\right)=\gamma_{2}\left(I\right)$. Without loss of generality, we can assume that $\gamma_{1}\left(I\right)=\gamma_{2}\left(I\right)=X$. Let us refine this last assertion. For $I_{1},I_{2}\subset I$ define $I_{1}\le I_{2}$ if $\sup I_{1}\le \inf I_{2}$, and the same for strict inequalities.

Lemma 2

For any sequence $\left\{I_{n}\right\}_{n=1}^{\infty}$ of closed subintervals of $I$ there is another sequence $\left\{J_{n}\right\}_{n=1}^{\infty}$, such that $\gamma_{1}\left(I_{n}\right)=\gamma_{2}\left(J_{n}\right)$ and whenever $I_{m}\subset I_{n}$ or $I_{m}\le I_{n}$, then $J_{m}\subset J_{n}$ or $J_{m}\le J_{n}$, respectively.

Proof There is a sequence $\left\{\phi_{k}\right\}_{k=1}^{\infty}$ such that $d\left(\gamma_{1},\gamma_{2}\circ\phi_{k}\right)\to 0$. Then the images of $\gamma_{2}\left(\phi_{k}\left(I_{n}\right)\right)$ converge to $\gamma_{1}\left(I_{n}\right)$, $\forall n$ in Hausdorff metric. Note that for each $k$ whenever $I_{m}\subset I_{n}$ or $I_{m}\le I_{n}$, then $\phi_{k}\left(I_{m}\right)\subset \phi_{k}\left(I_{n}\right)$ or $\phi_{k}\left(I_{m}\right)\le \phi_{k}\left(I_{n}\right)$, respectively because $\phi_{k}$ is increasing. Since the hyperspace of a compact $I$ is compact, $\left\{\phi_{k}\left(I_{n}\right)\right\}_{k=1}^{\infty}$ has a converging subsequence for each $n$. Using the Cantor's diagonal argument, we can find a subsequence $\left\{\phi_{l_{k}}\right\}_{k=1}^{\infty}$, and a sequence $\left\{J_{n}\right\}_{n=1}^{\infty}$ of closed subintervals of $I$ such that $\phi_{l_{k}}\left(I_{n}\right)\to J_{n}$, $\forall n$. Since "hypermap" is continuous, we have that $\gamma_{2}\left(J_{n}\right)=\lim\limits_{k\to\infty}\gamma_{2}\left(\phi_{l_{k}}\left(I_{n}\right)\right)=\gamma_{1}\left(I_{n}\right)$. QED

We can modify the previous lemma under the assumption that we are dealing with members of $S_{i}=S_{\gamma_{i}}$.

Lemma 3

For any sequence $\left\{I_{n}\right\}_{n=1}^{\infty}\subset S'_{1}$ there is another sequence $\left\{J_{n}\right\}_{n=1}^{\infty}\subset S'_{2}$, such that $\gamma_{1}\left(I_{n}\right)=\gamma_{2}\left(J_{n}\right)$ and whenever $I_{m}\subset I_{n}$ or $I_{m}< I_{n}$, then $J_{m}\subset J_{n}$ or $J_{m}< J_{n}$, respectively.

Proof We will add some auxiliary sets to our sequence, which will later serve as "boundaries". Assume that $I_{m}< I_{n}$ Then since $\max I_{m}\in\overline{I\backslash \gamma_{1}^{-1}\left(\gamma\left(I_{m}\right)\right)}$ and $\min I_{n}\in\overline{I\backslash \gamma_{1}^{-1}\left(\gamma\left(I_{n}\right)\right)}$, there are $t,s\in\left[\max I_{m},\min I_{n}\right]$, such that $t\le s$, $\gamma_{1}\left(t\right)\not\in\gamma\left(I_{m}\right)$ and $\gamma_{1}\left(s\right)\not\in\gamma\left(I_{n}\right)$. Define $I_{n}^{m}=\left\{s\right\}$. Define $I_{m}^{n}=\left\{t\right\}$ and $I_{n}^{m}=\left\{s\right\}$. Note that $I_{m}\le I_{m}^{n}\le I_{n}^{m}\le I_{n}$.

The original sequence of sets together with all possible $I_{m}^{n}$ is still (at most) countable, so we can apply the previous lemma to it and get the collection $\left\{J'_{n}\right\}_{n=1}^{\infty}\bigcup \left\{J_{m}^{n}\left|I_{m}\bigcap I_{n}=\varnothing\right.\right\}$, which also has additional properties guaranteed by the lemma. Define $J_{n}$ to be the component of $\gamma_{2}^{-1}\left(I_{n}\right)$, which contains $J'_{n}$. Clearly, $J_{n}\in S_{2}$ and $\gamma_{1}\left(I_{n}\right)=\gamma_{2}\left(J_{n}\right)$.

If $I_{m}\subset I_{n}$, then $J'_{m}\subset J'_{n}$, and since $J_{m}, J_{n}\in S_{2}$, $\gamma\left(J_{m}\right)\subset \gamma\left(J_{n}\right)$ and $J_{m}\bigcap J_{m}\supset J'_{m}\ne\varnothing$, by part (iii) of Lemma 1 we conclude that $J_{m}\subset J_{n}$.

If $I_{m}< I_{n}$, then $J'_{m}\le J_{m}^{n}\le J_{n}^{m}\le J'_{n}$. Also $\gamma_{1}\left(I_{m}^{n}\right)=\gamma_{2}\left(J_{m}^{n}\right)$ is a single point, which does not belong to $\gamma_{1}\left(I_{m}\right)=\gamma_{2}\left(J_{m}\right)$. Hence $J_{m}<J_{m}^{n}$, and analogously $J_{n}^{m}<J_{n}$, from which $J_{m}<J_{n}$. QED

From the uniform continuity of $\gamma_{i}$ and previous lemma one can deduce that for any closed $Y\subset X$ which contains more than one point, the number of components $J$ of $\gamma_{1}^{-1}\left(Y\right)$, such that $\gamma_{1}\left(J\right)=Y$ is equal to the analogous number for $\gamma_{2}$.

Let $S'_{i}$ be the subcollection of $S_{i}$ consisting of sets, which are not singletons. Define a map $\phi:S'_{2}\to S'_{1}$ in the following way. Since $\gamma_{2}$ is unstopping, $Y=\gamma_{2}\left(J\right)$ is not a singleton for any $J\in S'_{2}$. Let $J_{1}< J_{2}< ...< J_{n}$ be all the elements of $S_{2}$ such that $\gamma_{2}\left(J_{k}\right)=Y$. Note that $J$ is among $J_{1}, J_{2}, ..., J_{n}$, say $J=J_{k}$. Let $I_{1}< I_{2}< ...< I_{n}$ be all the elements of $S_{1}$ such that $\gamma_{1}\left(I_{k}\right)=Y$. Define $\phi\left(J\right)=I_{k}$.

If $J^{1}, J^{2}\in S'_{2}$ and $J^{1}\subset J^{2}$, then $\phi\left(J^{1}\right)\subset \phi\left(J^{2}\right)$. Indeed if $J^{1}_{1}< J_{2}^{1}< ...< J_{n}^{1}$ are all the elements of $S'_{2}$ such that $\gamma_{2}\left(J^{1}_{k}\right)=\gamma_{2}\left(J^{1}\right)$ and $J^{2}_{1}< J_{2}^{2}< ...< J_{m}^{2}$ are all the elements of $S_{2}$ such that $\gamma_{2}\left(J^{2}_{k}\right)=\gamma_{2}\left(J^{2}\right)$, the statement follows from applying Lemma 3 to the collection of $J_{k}^{j}$.

This last property allows us to extend $\phi$ to the singletons, or with some abuse of notations, to $I$. Namely, for $t\in I$ define $\phi\left(t\right)=\bigcup\limits_{t\in int J,~J\in S_{2}}\phi\left(J\right)$. From the properties of $S_{2}$ and $\phi$, we get that the collection $\left\{\phi\left(J\right)\left|t\in int J,~J\in S_{2}\right.\right\}$ has finite intersection property, and so it has a nonempty intersection. Moreover, since $\gamma_{2}\left(J\right)=\gamma_{1}\left(\phi\left(J\right)\right)$ for any $J\in S'_{2}$, we get that $\gamma_{1}\left(\phi\left(t\right)\right)\subset\bigcup\limits_{t\in int J,~J\in S_{\gamma}}\gamma_{2}\left(J\right)=\left\{\gamma_{2}\left(t\right)\right\}$, and so $\phi\left(t\right)$ is a singleton. Moreover, $\gamma_{1}=\gamma_{2}\circ\phi$. Thus our goal is accomplished once we prove the following lemma.

Lemma 4

$\phi$ is a strictly increasing surjection into $I$.

Proof From the symmetry it is easy to see that $\left\{\phi\left(J\right)\left|t\in int J,~J\in S_{2}\right.\right\}=\left\{J\left|\phi\left(t\right)\in int J,~J\in S_{1}\right.\right\}$, and so $\phi$ is invertible as a map from $I$ into itself; it is also easy to see that the endpoints of $I$ are fixed points of $\phi$.

Finally since the families $\left\{int J, J\in S_{i}\right\}$ form a base of topology of $I$, for each $i=1,2$, it follows that $\phi$ is continuous. Indeed, for any open neighborhood $U$ of $\phi\left(t\right)$ for $t\in I$ there is $J\in S_{1}$ such that $t\in int J$ and $J\subset U$, and then $\phi^{-1}\left(J\right)$ is an open neighborhood of $t$, such that $\phi\left(s\right)\in\phi\left(\phi^{-1}\left(J\right)\right)=J\subset U$, for any $s\in\phi^{-1}\left(J\right)$. QED

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