Crinkled Arcs And Brownian Motion

A crinkled arc is a continuous curve that appears as if it is making right-angle turns at every point along its trajectory. Additionally, if you draw a straight line between two recent points and compare that line to a line drawn between any two points in its more distant past, you will find that these two lines are also perpendicular.

On first hearing about the crinkled arc, this feels really weird. It may not be clear that such a curve should exist at all, even if we allow ourselves an infinite-dimensional space to draw it in. However, it turns out that not only do such curves exist, there are ways of constructing them so that their properties begin to feel obvious. In this post, I describe two such ways that crinkled arcs have been constructed, the first by defining it as a path through a space of functions and the second drawing a connection to Brownian motion, the random movement of particles suspended in fluid.

Crinkled Arcs

We begin by giving a definition of what a crinkled arc is.

A crinkled arc is a continuous function \(f\) from the unit interval to a Hilbert space \(X\) such that for any two non-overlapping intervals \([a,b] \subset [0,1]\) and \([c,d] \subset [0,1]\), the chords \(f(b) -f(a)\) and \(f(d)-f(c)\) are orthogonal in the Hilbert space. Their name is derived from this property, as it is observed that “seem to be making a sudden right angle turn at each point”, giving them a crinkled quality.

More formally, if \(X\) is equipped with an inner product \(\langle \cdot, \cdot \rangle\), then we say that \(f: [0,1] \to X\) is a crinkled arc if \(f\) is a continuous function such that for all \(0 \leq a < b \leq c < d \leq 1\), we have \(\langle f(b)-f(a), f(d)-f(c)\rangle = 0\).

This definition is largely paraphrased from Wikipedia, and after this introduction, the article asserts that all crinkled arcs (up to certain normalisation procedures) are of the form

where \({\phi_n}\) are an orthonormal set defining a subspace of \(X\). While this is indeed a crinkled arc (as we will prove), it is not immediately obvious, and in fact that we can construct crinkled arcs in a more intuitive way.

Building an Arc

In A Hilbert Space Problem Book, Halmos presents an exercise and challenging the reader to come up with an example of a curve satisfying the properties of the crinkled arc. What follows is the solution that he presents, and indeed the solution that he claims that everyone who is set this problem finds.

Consider the Hilbert space \(L^2(0,1)\), that is, the space of all square integrable functions over the interval (0,1). For this space, the inner product is defined using the integral of the product of the two functions, so functions \(\psi:[0,1]\to\mathbb{R}\) and \(\varphi:[0,1]\to\mathbb{R}\) are orthogonal if and only if they satisfy \(\int_0^1 \psi(x) \varphi(x) \,dx=0\). Here, we use Greek letters \(\varphi\) and \(\psi\) to denote functions from \((0,1)\) to \(\mathbb{R}\) so as to not confuse them with \(f\), which is a function which maps from the interval \((0,1)\) to real-valued functions, i.e. \(f: [0,1] \to L^2(0,1)\), and for \(t \in [0,1]\), \(f(t):[0,1]\to\mathbb{R}\).

More generally, our Hilbert space’s dot product is defined by this integral \(\langle \psi,\varphi \rangle = \int^1_0 \psi(x)\varphi(x)\,dx\).

Remember that for crinkled arcs, we want \(\langle f(b) - f(a), f(d) - f(c) \rangle = 0\) for all \(0 \leq a < b \leq c < d\). This gives us an idea for how to construct a solution: if the function \(\varphi = f(b)-f(a)\) is zero outside the interval \([a,b]\) and \(\psi = f(d)-f(c)\) is zero outside of the interval \([c,d]\), then there is no value of \(x\) such that \(\varphi(x)\) and \(\psi(x)\) are both non-zero, and therefore their inner product will also be zero.

The easiest way to do this is with indicator functions. Defining \(𝟙_{[0,t]}(x) = \begin{cases}1 \textnormal{ if } x \leq t \\ 0 \textnormal{ if } x > t \end{cases}\).

With a little thought, we can see that defining \(f(t) = 𝟙_{[0,t]}\) gives us what we want: The function given by the difference of two indicator functions with \(t=a\) and \(t=b\) with \(b>a\) is non-zero only for values between \(a\) and \(b\), and the product of \(f(b)-f(a)\) and \(f(d)-f(c)\) is a function with non-zero regions if and only if the intervals \((a,b)\) and \((c,d)\) overlap.

This is demonstrated with the visualisation below. You can set the sliders to see how different values of \(a,b,c\) and \(d\) affect the inner product of the function differences (visualised as the pink area in the bottom plot). When the intervals overlap, there are values at which both differences give a non-zero value at \(x\), contributing to a non-zero inner product. When they don’t overlap, the product is zero everywhere, meaning that the functions are orthogonal.

One last thing we need to check is the continuity of \(f\), which we get from the fact that as for any \(a\) and \(b\), as \(a\rightarrow b\) the norm of the difference of the functions tends to zero. We use the same property of indicator functions we used in computing chord inner products to find:

from which the continuity of \(f\) becomes apparent.

Building A Fourier Basis

We now have a solution, but on the surface it looks very much unlike the solution that we were expecting from earlier. Additionally, while we have a function that satisfies the crinkled arc definition, that definition isn’t very vector space-y; it feels like we’ve been treating \(L^2(0,1)\) more as a set of functions rather than a proper Hilbert space. To fix this, it would be nice if we could define a basis and give the curve some coordinates. So let’s try using the Fourier series to find a basis and to see what the components look like.

Recall that absolutely integrable periodic function with period 2 and only finitely many maxima, minima and discontinuities can be written as a Fourier series as:

for some \(a_0, a_1, \ldots \in \mathbb{R}\) and some \(b_1, b_2, \ldots\in\mathbb{R}\). However, since our functions are in \(L^2(0,1)\) rather than \(L^2(-1, 1)\), meaning we care only about its behaviour over a interval of length 1 rather than 2, we can make life easier for ourselves by imagining that it is extended to be an even function (recall that for an even function, \(f(x)=f(-x)\)). Doing so, we immediately get that all \(b_n\), as coefficients of odd functions, can be shown to be zero.

Given this choice of basis functions, we expect for any \(t\), to be able to write the function \(f(t)\) in terms of these basis functions. In this case our un-normalised basis functions are \(\phi_0(x)=1\) (the constant function) and the set of cosine functions of the form \(\cos(n \pi x)\). The latter need normalising, so that their square integrates to one¹, so our non-constant basis functions are of the form \(\phi_n(x)= \sqrt{2} \cos(n \pi x)\).

With this, we have that \(f (t) = \sum_{n=0}^\infty a_n(t) \phi_n\) for some sequence of \(a_n \in \mathbb{R}\). We find these coefficients using our inner product to find the projection of \(f(t)\) onto \(\phi_n\) for each \(\phi_n\). So for \(n>0\), remembering that \(f(t)\) is the interval indicator function for the interval \([0,t]\):

and for \(n=0\) we have

Remembering the \(\sqrt{2}\) in front of non-constant basis functions, this gives

\(f(t) : x \mapsto t + \sum_{n=1}^\infty \frac{2}{n \pi} \sin (n \pi t) \cos(n \pi x)\).

Reabsorbing the basis functions into \(\phi_n\), we can also write this to as

This solution can be verified using the visualisation below, showing that even with finite approximations of the infinite sum, we get a good approximation of the indicator function.

Building the Standard Solution

In the previous section, we wrote our crinkled arc in terms of a series based on the Fourier expansion of functions. However, when we introduced the crinkled arc a couple of sections ago, we also gave a claimed general solution:

This does not look like our expansion in terms of Fourier basis functions. So what set of basis functions would we need for the standard solution? In this section we are going to use what we know the coefficients look like for the standard solution to guess at a set of basis functions.

Guessing and checking may feel unsatisfying for a couple of reasons: Firstly, we need to worry about whether this basis is even a complete basis for \(L^2(0,1)\) or is otherwise powerful enough to fully describe our functions, and secondly, it feels like we’ve just plucked it from thin air. The first problem can be solved fairly easily ², and while there is a very good reason why this basis is a natural choice, its derivation is non-trivial ³ and is left as a potential topic for a future post.

I wasn’t able to find the form of these basis functions anywhere, though looking at the coefficients, they can be guessed at pretty easily, and we will verify that the guess is correct. Given that the coefficients we are expecting look a lot like the coefficients we just derived for the Fourier series but with \(n\) shifted by \(\frac{1}{2}\), we choose the new basis functions to \(\phi_n\) to have that same shift \(\phi_n(x) = \sqrt{2} \cos \left( (n - \frac{1}{2}) \pi x \right)\). Note that we have to check that \(\sqrt{2}\) remains the correct normalisation constant.

With this change, we can once again calculate the coefficients in the same way to verify our guess:

Since this time there is no \(\phi_0\) coefficient, we get the solution:

This is indeed the form we were looking for. We can once again show that this solution makes sense by plotting, as shown in the following visualisation. It’s interesting to observe that while both series (unsurprisingly) converge to the target function, they also converge faster to each other.

What Does This Have to do With Brownian Motion?

Brownian motion is the random motion of particles suspended in a medium consisting of smaller particles, such as small motes of dust floating in the air or grains of pollen on the surface of water (the latter example being the original case of Brownian motion that Robert Brown observed).

In Brownian motion, the particle’s movement is random and constantly changing, and though over time it will move through the medium, its overall motion is jittery, and its movement over any two non-overlapping intervals is completely independent.

Surprisingly, in describing the crinkled arc, we have also incidentally formulated a representation of Brownian motion, a connection that we will now explore in more detail.

Mathematically, Brownian motion is a stochastic process: a set of real-valued random variables \(\{W_t\}_{t\in[0,1]}\) indexed by real-valued \(t\) and defined by the following four properties:

1. Almost surely, \(W_0 = 0\)
2. \(W_t\) is almost surely continuous in \(t\)
3. For \(0\leq a < b \leq c < d \leq 1\), \(W_b-W_a\) is independent of \(W_d-W_c\)
4. \(W_b-W_a \sim \mathcal{N}(0, b-a)\) for \(0 \leq a \leq b\)

These properties are very similar to the properties we’ve encountered in our construction of the crinkled arc using indicator functions. For our crinkled arc \(f\):

1. \(f(0)\) is the zero function (i.e., \(f(0): x \mapsto 0\) for all \(x\)).
2. \(f\) is continuous.
3. For \(0\leq a < b \leq c < d \leq 1\) \(f(b)-f(a)\) is orthogonal to \(f(d)-f(c)\).
4. \(\lvert \lvert f(b) - f(a) \rvert \rvert = (b-a)\) for \(0 \leq a \leq b\).

These connections are concrete enough that we should expect to be able to construct an isomorphism between the two.

Let’s start off with properties 3 and 4 of Brownian motion. We want that for \(a,b,c,d\) defined in the usual way that \(W_b-W_a \sim \mathcal{N}(0, b-a)\) and \(W_d-W_c \sim \mathcal{N}(0, d-c)\) with the two normal distributions being independent.

We can notice that for our crinkled arc \(f\), the inner product and norm behave in the same way that we would like the covariance and variance to behave. More precisely, we want \(\text{cov}(W_a, W_b)=\langle f(a), f(b)\rangle\) and \(\text{cov} (W_b-W_a, W_d-W_c)\) \(= \langle f(b) -f(a), f(d)-f(c)\rangle\). It sure looks like we want \(W_t\) to live in a space with an inner product structure!

Since \(W_t\) are all real-valued random variables, we can do all the operations on them with regard to adding them together and scaling them to form a vector space, and we can define the inner product to be the covariance (after proving that covariance fits all the necessary criteria for an inner product). But, now we know that \(W_t\)’s live in a Hilbert space, and their trajectory fits all the properties that \(W_t\) has to follow a crinkled arc.

More than that, since we know that \(W_t\) fits the definition for the crinkled arc, there is an orthonormal basis \(\{X_n\}\) such that

Now what are the basis vectors? They’re independent normal distributions with zero mean and unit variance! We get that they’re normally distributed because they are a basis for the space spanned by \(W_t\), which consists solely of zero-mean normally distributed variables (to see that \(W_t\) is normally distributed consider that \(W_t = W_t-W_0\) follows a normal distribution and that \(W_0\) is almost-surely zero), and normal distributions are closed under addition. And by the criterion that the basis form an orthonormal set guarantee unit variance (by normalisation) and independence (by orthogonality).

Though this representation of Brownian motion is well-known in the literature, this is not the usual derivation. Indeed, in proving the generality of the solution for the crinkled arc, Vitale drew the connection the other way, asserting that the crinkled arc is a form of Brownian motion and going from there.

The Fourier cosine basis as a representation of Brownian motion is also well known (see a derivation on YouTube here). Using the same reasoning as for the half-integer cosine series, we get that Brownian motion can also be represented as:

We now show three examples of generating Brownian motion: the first showing it as the limit of a random walk as step size increases, and two using our series approximations to a given number of terms.

Conclusion

This post has described crinkled arcs in three ways: a function describing the curve made by continuously varying a parameter in an interval function, using a set of trigonometric basis functions and as the trajectory of Brownian motion through a Hilbert space of normal distribution random variables.

The original motivation for this post was to clarify for myself the relationship between these three descriptions, giving more space to elucidate the relationship between the crinkled arc and Brownian motion than I was able to find in the literature. I was somewhat surprised to find the definition of the basis functions for the crinkled arc in \(L^2(0,1)\) did not appear in any sources that I found, though would be interested in being pointed to any sources that I missed.

In exploring these connections, I hope that this post can serve as reasonable introduction to the crinkled arc and make its properties seem a little less unintuitive.

Thanks to Andrew Webb for feedback on a draft version of this post.

References

Halmos, P. R. (2012). A Hilbert space problem book (Vol. 19). Springer Science & Business Media. (Problem 4)

Johnson, G. G. (1970, June). A crinkled arc. In Proc. Amer. Math. Soc (Vol. 25, pp. 375-376).

Vitale, Richard A. “Representation of a crinkled arc.” Proceedings of the American Mathematical Society 52.1 (1975): 303-304.

Cite This Post

@misc{wood2025crinkled,
  author = {Wood, Danny},
  title = {Crinkled Arcs and Brownian Motion},
  year = {2025},
  howpublished = {Blog post},
  url = {https://echostatements.github.io/posts/2025/10/crinkled-arcs-and-brownian-motion/}
}