I was interested in putting a complete metric on the space of Jordan curves. Say, just planar Jordan curves contained in $B(\bar{0}, 2) \backslash B(\bar{0}, 1)$ which separates $\bar{0}$ and infinity.

As we can see, the standard Hausdorff metric does not do the job, for example we can have a Cauchy sequence of Jordan curves with thinner and thinner 'fingers', the limit is a circle with a spike, hence not a Jordan curve. (see Figure below)

Hence in some sense, the metric needs to not only 'see' the curve but also what it encloses. I looked a bit into the literature and didn't find anything...Does anyone know such a metric? Any ideas?

An attempt of mine in constructing the metric:

First we can consider the plane as in the Riemann sphere, so our space is merely the set of Jordan curves that separates the unit disk around the North pole and that around the South pole. For each curve we have (up to a Mobius automorphism of the disc) two Riemann maps from the standard unit disc to the North and South hemisphere separated by the curve.

Define the distance between two curves to be the minimum distance between the pair of 'North hemisphere Riemann maps', plus that of the South hemisphere. It's easy to check that this does give a metric. In our 'finger' case above, the $ \{ C_n \} $ is not Cauchy in our metric (because Riemann map will map less and less of the disc inside the finger so there is a subsequence of distance finger-length apart).

I strongly believe this metric is indeed complete, but can't find a good argument.

What can be done is, given a Cauchy sequence of Jordan curves under the above metric, pick a pair of distance minimizing Riemann map for each curve (such thing always exist by a simple compactness argument) then apply Carathéodory, all Riemann maps extends to the homeo on the boundary, so we can take the pointwise limit of the boundary map. Do the same for both North and South hemisphere. All we need to show is this limit is still injective. (I think it should be because if not, either North or South will have a infinitely thin 'neck', so the Riemann maps on that half will not be Cauchy).

I ran into many topological difficulties when trying to make an argument for this (mainly because we know nothing about the limiting curve and it can be really complicated)...Any ideas on how one might go about that would be greatly appreciated. :-P

Figure for my comment on professor Thurston's response:


---A simple case to illustrate that the two Riemann maps does not match up on the equator. i.e. radial lines in the North hemisphere does not 'want' to pass through the thin tube and land on the tip, but radial lines in the South hemisphere does not care. Hence the pre-image of the black arc will be very short in the Northern disc but normal in the Southern disc. However, there of course exists unique homeo $h$ on the boundary to glue them up. I do not know how would that give a homeo from the sphere to itself, though.


2 Answers 2


There is a very good theory of the boundary behavior for Riemann maps, the Caratheodory theory of ends. Riemann maps for an open set extend continuously to the boundary of a disk provided the frontier is locally connected.

I assume your definition is in terms of the Riemann maps that fix the north or south pole and have derivative a positive real number at those points. If so, then Riemann maps depend continuously in the $L^\infty$ metric on Jordan curves as you have described, so this gives a metric.

Is the metric complete? A sequence of the pair (lower hemisphere map, upper hemisphere map) that is Cauchy in the uniform ($L^\infty$) topology converges to a pair of continuous maps that agree on the equator, so the glued maps give a continuous image of the sphere that is a homeomorphism at least in the complement of the equator. If it is not injective, the preimages of points would need to be intervals, otherwise the topology would be destroyed. But that's impossible. If you push forward the measure $ds$ on the equator by a Riemann map, it never has atoms. It is the same as hitting measure for Brownian motion in the image: if you start at the north pole and follow a Brownian path, where does it first arrive at the boundary? This is always a diffuse measure.

So, you're right: it's a complete metric on the set of Jordan curves you described.

Note. Any quasisymmetric map (I won't define it here, but Holder is sufficient) from the circle to the circle arises from a pair of Riemann maps for a Jordan curve, although not necessarily in the annulus you described, and the Jordan curve in this case is known as a quasi-circle. However, there is no known good characterization of which gluing maps from the circle to the circle give Jordan curves in general. Continuity is not sufficient: there are counterexamples using things that are locally (for example) the graph of $\sin(1/x)$ plus the interval $[-1,1]$ on the $y$-axis, where the gluing map for the Riemann maps to the two sides extends continuously across the discontinuity of the graph. Better characterizations of what gluing maps give what topology and geometry are very hard, but of great interest in complex dynamics and some other subjects.

  • $\begingroup$ Dear professor Thurston, It does make things easier to take the unique Riemann map that fixes the pole and has positive real derivative! (For some reason I was taking the inf over all Riemann maps to define the distance, it was stupid) $\endgroup$
    – Conan Wu
    Sep 27, 2010 at 20:17
  • $\begingroup$ However, I'm not sure if I understood your 'limiting map on the sphere': for each Jordan curve in the sequence, we extend the two Riemann maps to a the circle (Catatheodory applies to prime ends but in our case since $J_n$ is already a Jordan curve, this is a homeo). But the two maps would not match on the equator if we put them on the North and South hemisphere. As the picture below suggests: !homeo It is true that there is a unique homeo $h:S^1 \rightarrow S^1$ that identifies the two copies of the equator, do we cone the homeo? $\endgroup$
    – Conan Wu
    Sep 27, 2010 at 20:25
  • $\begingroup$ The problem is I'm not sure what sequence did you say 'converge to a pair of continuous maps that agree on the equator', because Jn by itself does not seem to define a canonical homeomorphism from the sphere to itself. I have only read about quasi-circles in Bowen's paper '[Hausdorff dimension of quasi-circles](springerlink.com/content/fn86113101014518/…)' But that's defined by two isomorphic groups of Mobius transformations from the hemispheres to themselves, I guess you are talking about a different notion of quasi-circles? PS: I'm thrilled to see your reply! $\endgroup$
    – Conan Wu
    Sep 27, 2010 at 21:02
  • $\begingroup$ @Conan Wu, I think I phrased it poorly, or else my thinking was muddy. The images of the circle agree from the two sides, but one should really think of two disks glued together along the boundary by a homeomorphism of the circle that varies with the curve. The homeomorphism uniquely determines the curve, but as I said, not all homeomorphisms define a curve. Really you don't need to worry about quasicircles and quasisymmetric unless for a particular application; they only tell about some Jordan curves. I was just mentioning it in case it's relevant to you. ... cont'd to next comment. $\endgroup$ Sep 27, 2010 at 21:47
  • $\begingroup$ @Conan Wu: If you minimize over all Riemann mappings, I'm worried that the infimum may be very irregular. If you measured the distance between gluing maps, then all smooth curves are distance 0 from a round circle --- by sending n pole, s. pole to two opposite points across the smooth curve. $\endgroup$ Sep 27, 2010 at 21:53

I fear this is possibly a bit too abstract for your purposes, but tells that there is such a complete distance topologically equivalent to the uniform distance, and can be made more concrete (check the construction of the complete metric in Alexandrov theorem on Polish spaces).

The set $\mathcal{J}$ of all Jordan curves $\gamma:S^1\to\mathbb{R^2}$, as a subset of the separable Banach space $X:=C(S^1,\mathbb{R^2})$ is a $G_\delta$ set. Indeed, for all $n\in\mathbb{N}_+$ define $\mathcal{J}_n$ as the set of all $\gamma\in C(S^1,\mathbb{R^2})$ such that $\gamma(s)\ne \gamma(t)$ whenever $|s-t|\geq\frac{1}{n}$: then $\mathcal{J}_n$ is open and $\mathcal{J}=\cap_n \mathcal{J}_n$. By Alexandrov theorem, $\mathcal{J}$ is a Polish space as $X$ itself. $$*$$ [EDIT]. AS you said in your comment below, your aim is to prove that connecting the two boundaries of the annulus $A$ without crossing more than once a generic given Jordan curve winding around the origin, require a path of infinite length. I'd like to add some hints and remarks (here I keep a parametric point of view). Then it seems to me that, not only you don't need to construct explicitly a complete metric for the $G_\delta$ of the Jordan curves, but you actually don't even need a metric at all (i.e. you don't need the Alexandrov thm). You only need a more elementary (and more general) fact: a $G_\delta$ of a Baire space is itself a Baire space. Thus $\mathcal{J}$ is a Baire space.

So you only need to show that for all $k$ the set $L_k$ is a dense $G_\delta$ subset of $\mathcal{J}$, where as said $L_k$ is defined to be the set of all $\gamma\in\mathcal{J}$ with the property that for all $p\in C^0([0,1],A)$ connecting the boundaries of $A$ (say $|p(0)|=1$ and $|p(1)|=2$) and crossing $\gamma$ only once, one has $L(p) > k$.

To show that $L_k$ is dense in $\mathcal{J}$, it is sufficient to approximate in the uniform topology a smooth regular curve $\gamma$. Actually, to make the description simpler I'll show the construction for a (parametrization of the) Y-axis (then the case of $\gamma\in\mathcal{J}$ can also be achieved via some bi-lipschitz transformation). A simple way to make a wiggled curve approximating the $Y$ axis, is by the union of the graphs of the functions $$u_j:[-\epsilon,+\epsilon]\ni x\mapsto\epsilon\sin(x/\epsilon^2)+\epsilon^2j,$$ with $ j \in \mathbb{Z},$ joined on the left and on the right by vertical segments of length $\epsilon^2,$ so as to make a unique curve $\gamma_\epsilon$ close to the $Y$-axis. Clearly, any path $p$, say from $-1$ to $+1$, that cross this curve $\gamma_\epsilon$ only once, has to stay completely between the graphs of $u_j$ and $u_{j+1}$ either for $-\epsilon\leq x\leq 0$ or for $0\leq x\leq \epsilon,$ and therefore has a length at least $O(1/\epsilon)$.

The sets $L_k$ don't seem to be open: though I think there's hope that you can show they are $G_\delta$'s, by means of a decomposition like the one for the $\mathcal{J}_n$ (here key point should be the semicontinuity of the length, and the compactness of curves of finite length un to reparametrization). Hope this helps! $$*$$ PS: a stronger conjecture. A generic $\gamma\in\mathcal{J}$ meets any path of finite length that connects the boundaries of $A$ in an infinite perfect set (maybe it's known. It reminds me of some results about topologic dimension).

  • 1
    $\begingroup$ Yes, this does imply $\mathcal{J}$ completely metrizable~ And so is the space I took (curves separating the two discs) since they form an connected component in Jordan curves inside $B(\bar{0}, 2) \backslash B(\bar{0},1)$ Thank you! I am looking at this because I initially wanted to prove some property being generic amoung Jordan curves (i.e. generically one can not find a finite length arc connecting two given points on different sides of the curve, intereecting the curve only at one point) Thank you very much and I'll try to trace through the Alexandrov construction now...Hope it works~ $\endgroup$
    – Conan Wu
    Sep 26, 2010 at 22:02
  • 1
    $\begingroup$ Oh very nice. In fact, since your result is topological in nature, I think you shouldn't need to construct explicitly a complete metric. For instance, consider, for any $k > 0$, the set $L_k$ of all Jordan curves $\gamma$ in your annulus, such that any curve $p$ connecting $\partial B(\bar{0}, 1)$ and $\partial B(\bar{0}, 2)$ and crossing $\gamma$ exactly once, has lenght $L(p)>k$. This $L_k$ seems to be open (the lenght is LSC wrto uniform convergence) and it is certainly dense. So $\cap_k L_k$ hsould be a dense $G_\delta$ in your set of curves. $\endgroup$ Sep 26, 2010 at 22:51
  • $\begingroup$ Umm...One reason that I did not do it in $C^0(S^1, \mathbb{R}^2)$ is that I'd like to think of Jordan curves as being independent of the parametrization (i.e. just a subset of the space of compact sets, but not under a finer topology than that induced by the Hausdorff metric). Injective maps in $C^0(S^1, \mathbb{R}^2)$ contains a lot more points, but is it 'safe' to inf over all possible parametrizations? Because this puts the Jordan curves inside a weird space. PS: Is it obvious to you that $L_k$ is dense? You are very fast then~ $\endgroup$
    – Conan Wu
    Sep 27, 2010 at 0:20
  • $\begingroup$ About $L_k$ being dense. Well, it seems quite obvious, although putting down a proof in all details may be a bit technical. The point is that the uniform approximation allow leaves a lot of feedom in the local behaviour. So you can approximate in the uniform distance any loop with e.g. a piecewise one that belongs to the set $L_k$ (these locally should look like the first steps of the construction of the Peano curve, in order to ensure that they are in $L_k$). $\endgroup$ Sep 27, 2010 at 7:08
  • $\begingroup$ It does took me some time to figure out exact how to 'wiggle' the curve to so that it will be nowhere 'pieceable' by short arcs though...I guess it's just me then. $\endgroup$
    – Conan Wu
    Sep 27, 2010 at 21:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.