The n-Category Café

Thanks! This formula gives one particular involute of $C(s)$. What’s the general one?

I was surprised by the appearance of a derivative in this formula, because there’s a reverse process to the involute, called the ‘evolute’, and elsewhere Jesse McKeown emphasized an analogy

involute:integral::evolute:derivative

If you don’t know what the evolute of a curve is, another of Sam Derbyshire’s pictures should explain it pretty quick. Just as the tractrix is an involute of the catenary:

so the catenary is the evolute of the tractrix:

A curve has one evolute but a 1-parameter family of involutes, the evolute of the involute is the original curve, etc.

Jesse wrote:

So, to a piecewise convex curve there is an envelope of tangent lines; and to another piecewise convex curve there is an envelope of normal lines; and curve B is an involute of curve A (also pronounced: A is the evolute of B) if the Tangent envelope of A is the Normal envelope of B. Curious fact: The evolute of an algebraic curve is another algebraic curve. Curiouser fact: some algebraic curves are not the evolute of any algebraic curve — an ellipse is a very good bad example. (I also think evolutes are the right way to think about the Four-Vertex Theorem)

This should sound like a similar relationship between integrals (involutes) and derivatives (evolutes), in part because constructing an involute is solving a differential equation, while constructing the evolute is basically dividing by a derivative; but more: because of the string picture described above, (distance to an) involute measures arc length. Corollary: the arclengths between algebraic points on evolutes of algebraic curves are algebraic numbers.

Curiouser and curiouser fact: Apollonius mentioned evolutes at least once, and Archimedes’ Spiral is trying very hard to be a circle involute; but the semicubic parabola, incidentally the evolute of the ordinary parabola, seems to have been the first noticed algebraic curve with piecewise algebraic arclength, around 1659. Rational hypocycloids are further examples (similar to their own evolutes!) but I don’t know who first measured them. Cf Richeson 2013, esp. section 6 (starting on p. 11).

However, none of this has much to do with finite simple Lie algebras or Coxeter groups.

Dan wrote:

Now $D$ is a smooth function. How does that match the singularities shown? Well, $D$ just slows down as you approach those singular points. The image of a smooth curve isn’t necessarily smooth…

That reminds me of a paper we were once trying to write. But this is a very nice point, because it lets us find the cusps in the involute of a curve. If $D$ is an involute of $C$, $D$ can have cusps at the the points $D(s)$ where $D\prime(s) = 0$.

However, this will only be truly helpful if I get a formula for all the involutes of $C$.

… and using it is easy, once remarking that “parametrized by arclength” means $|C'(s)|\equiv 1$ so that the inner product vanishes: $\langle C'(s),C''(s)\rangle = 0$; choosing an orientation, there is a standard unit normal vector $\nu(s)$, with respect to which one defines the curvature of $C$ by $$C''(s) = \kappa(s)\nu(s)$$ and calculates also $$\nu'(s) = - \kappa(s) C'(s)$$ Then calculating everything we might want w.r.t the basis $[C',\nu]$ …

I’m getting a bit worn out for today, but there are some potentially useful (and certainly interesting) clues on the Wikipedia section Cusp singularity: examples. There they mainly teach us how to recognize different kinds of cusps in curves given by an implicit equation

$$f(x,y) = 0$$

not a parametric equation.

It’s also worth looking at the section Cusp singularity: classification in differential geometry, because it’s short and it touches on how these cusps show up in Arnol’d’s ADE classification of singularities.

It mentions that there are some subtleties when working with real plane curves: instead of just type $A_k$, when $k$ is even we get two types $A_k^+$ and $A_k^-$, which have as archetypal examples the curves

$$x^2 \pm y^{k + 1} = 0$$

When $k$ is even these are the same up to a change of coordinates, so we can just say $A_k$.

So, the ‘cusp of order 5/2’ or ‘rhamphoid cusp’ that we seek is a singularity of type $A_4$. This is probably worth remembering when we get deeper into the Dynkin diagram aspects of this story.

So, anyways, as I was trying to say, relative to the tangent/normal basis (and using $\nu' = -\kappa C'$), we have $$C^{(2)} = \kappa \nu$$ $$C^{(3)} = \kappa' \nu - \kappa^2 C'$$ $$C^{(4)} = (\kappa''-\kappa^3)\nu - 3\kappa\kappa' C'$$ $$C^{(5)} = (\kappa^{(3)} - 6\kappa^2\kappa')\nu + (\kappa^4-4\kappa''\kappa - 3 \kappa'^2) C'$$ so that whatever involute has $$D' = - s C''$$ $$D'' = - C'' - s C^{(3)}$$ $$D^{(3)} = -2C^{(3)} - s C^{(4)}$$ $$D^{(4)} = -3C^{(4)} - s C^{(5)}$$ and we are interested in a generic involute where it meets the sinuglar tangent line — w.l.o.g, let’s set $s=1$; the inflection point on the cubic has $\kappa(s=1)=0$ and $\kappa'\neq 0$. So we get $$D' = - \kappa \nu = 0$$ $$D'' = -(\kappa + \kappa')\nu) - \kappa^2 C' = - \kappa'\nu$$ $$D^{(3)} = (-2\kappa' - \kappa'' - \kappa^3)\nu + (-2\kappa^2 + 3\kappa\kappa') C'$$ $${} = -(2\kappa' + \kappa'') \nu$$ $$D^{(4)} = (-3 \kappa'' + 3 \kappa^3 - \kappa^{(3)} - 6\kappa^2\kappa')\nu + (9\kappa\kappa' - \kappa^4 + 4\kappa''\kappa + 3 \kappa'^2) C'$$ $${} = (-3\kappa'' - \kappa^{(3)}) \nu - 3\kappa'^2 C'$$ Near the singular point $D(1)$, where $D'=0$, the curve $D(1+t)$ looks like $$\left(x_1 + \frac{t^4}{8} \kappa'^2 + \dots, y_1 + \frac{t^2}{2} \kappa' - \frac{t^3}{3} (2\kappa' +\kappa'') + \dots\right)$$

So, just for fun, consider the parametric curve $$x=t^4, y = t^2 + t^3$$ One can exactly eliminate $t$ to get $$−x^3−4 x^2 y+x^2−2 y^2 x +y^4 = 0$$ but it might be more helpful to formally solve for $t$ as a power series in $u$ where $y = u^2$.

Thanks, that’s nice!

I did the experiment of posting this blog article both here and on Azimuth. The discussion is mainly occuring there now. In particular, Greg Egan and Simon Burton have created some very nice pictures related to the discriminant of the quintic

$$x^5 + a x^4 + b x^2 + c$$

to check Arnol’ds claim that Lyashko proved that surface is diffeomorphic to this one:

So, I urge everyone interested in this discussion to go there. Later, I will summarize all our results on Visual Insight.

Yes, it is similar but not the same.

The discriminant surface of the quartic is called the swallowtail. It can be described as the set of points $(a,b,c)$ such that

has multiple roots. The two “sharp points” (i.e. cusp points) in a slice of the picture locally look like $y^2 = x^3$.

The discriminant surface of the symmetry group of an icosahedron (John’s “Fig 18” from Arnold’s book) can be described as the set of points $(a,b,c)$ such that

has multiple roots. The two cusp points in a slice of the picture are different, one locally looks like $y^2 = x^3$ and the other “sharper” one like $y^2 = x^5$.