I am trying to prove Euler's spiral is an isometric embedding of $\mathbb{R}$ into $\mathbb{R}^2$ with bounded image. Here is the definition of the spiral:

$(*) \, \,\al(t)= (\int_0^{t} \cos(s^2)ds,\int_0^{t} \sin(s^2)ds)$

$\al$ is clearly a smooth isometric immersion.


(1) Why is $\al$ injective? (There is a picture, but I want a proof...)

Edit: Christian Blatter (see answer below) gave a proof based on Kneser's Nesting Theorem. The theorem also implies the boundedness of $\operatorname{Image}(\al)$. I still wonder whether there is a "direct" proof based only on the formula $(*)$.

Perhaps one can prove that the distance of $\al(t)$ from the (suitable) limit point decreases with $|t|$? (A naive attempt to do so didn't work).

(2) Why is $\al$ an embedding? (i.e homeomrphism onto its image)

Intuitively, there are "no troubles" from a topological point of view, since the path approaches its accumulation points only "from one direction" (Compare for the "figure-eight" in contrast).

(3) Why is the image bounded? That is, how to show the integrals are bounded?

Edit: This follows from Dirichlet's test for improper integrals. (See levap's answer)

Any reference (or self-contained proof of course) would be welcome. I have tried googling for some time but in vain.

  • $\begingroup$ I am not sure I understand what you are asking. The derivative $\dot \alpha(t)= (\cos(t^2),\sin(t^2))$ lies on the unit circle, so $\alpha$ is an isometric immersion. I do not know how to prove it's an embedding. (This was mentioned in this comment: math.stackexchange.com/questions/1790879/… ) $\endgroup$ May 20, 2016 at 17:08

2 Answers 2


From $$\alpha(t)=\int_0^t e^{is^2}\>ds$$ it follows that $\dot \alpha(t)=e^{it^2}$ has absolute value $1$, hence $\alpha$ is a locally isometric immersion. From ${\rm arg}\bigl(\dot\alpha(t)\bigr)=t^2$ it then follows that the curvature $$\kappa(t)={d\over dt}{\rm arg}\bigl(\dot\alpha(t)\bigr)=2t$$ is strictly monotonically increasing for $-\infty<t<\infty$. Kneser's Nesting Theorem then implies (i) that all points $\alpha(t)$ with $t>1$ are contained in the oscillating circle at $\alpha(1)$, and (ii) that the map $\alpha(\cdot)$ is globally injective.

  • $\begingroup$ Thanks! It's a nice theorem and I wasn't familiar with it. However, I am not sure I understand why it implies that all points $\alpha(t)$ with $t>1$ are contained in the oscillating circle at $\alpha(1)$. Would you please elaborate on this deduction? $\endgroup$ May 21, 2016 at 13:58
  • $\begingroup$ OK, after a deeper look, I now see that this is a part of the proof of the theorem, even though it's not stated explicitly in the "title". Thanks again. $\endgroup$ May 21, 2016 at 14:16

Regarding your third question, set $x(t) = \int_0^t \cos(s^2) \, ds$. The function $x$ is continuous so to show that $x$ is bounded it is enough to show that $\lim_{x \to \pm \infty} x(t) = \pm \int_0^{\infty} \cos(s^2) \, ds$ exists. We have

$$ \int_0^{\infty} \cos(s^2) \, ds = \int_0^1 \cos(s^2) \, ds + \int_1^{\infty} \cos(s^2) \, ds = \int_0^1 \cos(s^2) \, ds + \int_1^{\infty} \frac{\cos(u)}{2\sqrt{u}} \, du$$

and the latter integral converges using Dirichlet's test for improper integrals. Similarly for $y(t)$.

Now, I'm not sure how to easily answer your first and second question. However, if your purpose is to construct more or less explicitly an isometric embedding of $\mathbb{R}$ into $\mathbb{R}^2$ with a bounded image, I can suggest the following way which I find more direct:

Consider the map $\pi \colon \mathbb{R} \times \mathbb{R}_{+} \rightarrow \mathbb{R}^2$ given by $\pi(\theta, r) = (r \cos \theta, r \sin \theta)$. This map is a local diffeomorphism and in particular an immersion. Now, choose any bounded monotone increasing function $r \colon (-\infty, \infty) \rightarrow \mathbb{R}_{+}$ and consider the curve $\gamma(t) = \pi(t, r(t))$. The curve $\gamma$ is injective (since $r$ is injective), an immersion (since $\pi$ is a local diffeomorphism) and an embedding (since $$ \gamma((t_0 - \delta, t_0 + \delta)) = \gamma(\mathbb{R}) \cap \pi((t_0 - \delta, t_0 + \delta) \times (r(t_0 - \delta), r(t_0 + \delta))) $$ and the right hand side is open as $\pi$ is open). If $|r(t)| \leq M$ then the image of $\gamma$ lies inside the ball of radius $M$ inside $\mathbb{R}^2$. Now take a unit-length reparametrization of $\gamma$ and you obtain an isometric embedding.

For example, if $r(t) = \arctan(t) + \frac{\pi}{2}$ then the curve $\gamma$ looks like a two-sided spiral that spirals toward $(0,0)$ as $t \to -\infty$ and toward a limit circle of radius $\pi$ centered around $(0,0)$ as $t \to \infty$ (drew using wolfram alpha):

enter image description here

  • $\begingroup$ Nice! By the way, the reason I was interested in this, is that using this fact (existence of an isometric embedding of $\mathbb{R}$ in $\mathbb{R}^2$ with bounded image), one can prove that $\mathbb{R}^n$ with the standard metric can be isometrically embedded in a sphere or a flat torus (of suitable dimensions). Thus by the Nash embedding theorem, spheres and tori are "universal", just like Euclidean spaces. If you are interested in this corollary, you can see some details here: math.stackexchange.com/a/1793136/104576 $\endgroup$ May 21, 2016 at 8:04

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