## The Involute of a Cubical Parabola

In his remarkable book The Theory of Singularities and its Applications, Vladimir Arnol’d claims that the symmetry group of the icosahedron is secretly lurking in the problem of finding the shortest path from one point in the plane to another while avoiding some obstacles that have smooth boundaries.

Arnol’d nicely expresses the awe mathematicians feel when they discover a phenomenon like this:

Thus the propagation of waves, on a 2-manifold with boundary, is controlled by an icosahedron hidden at an inflection point at the boundary. This icosahedron is hidden, and it is difficult to find it even if its existence is known.

I would like to understand this!

I think the easiest way for me to make progress is to solve this problem posed by Arnol’d:

Puzzle. Prove that the generic involute of a cubical parabola has a cusp of order 5/2 on the straight line tangent to the parabola at the inflection point.

There’s a lot of jargon here! Let me try to demystify it. (I don’t have the energy now to say how the symmetry group of the icosahedron gets into the picture, but it’s connected to the ‘5’ in the cusp of order 5/2.)

A cubical parabola is just a curve like $y = x^3$:

It’s a silly name. I guess $y = x^3$ looked at $y = x^2$ and said “I want to be a parabola too!”

The involute of a curve is what you get by attaching one end of a taut string to that curve and tracing the path of the string’s free end as you wind the string onto that curve. For example:

Here our original curve, in blue, is a catenary: the curve formed by a hanging chain. Its involute is shown in red.

There are a couple of confusing things about this picture if you’re just starting to learn about involutes. First, Sam Derbyshire, who made this picture, cleverly moved the end of the string attached to the catenary at the instant the other end hit the catenary! That allowed him to continue the involute past the moment it hits the catenary. The result is a famous curve called a tractrix.

Second, it seems that the end of the string attached to the catenary is ‘at infinity’, very far up.

But you don’t need to play either of these tricks if you’re trying to draw an involute. Take a point $p$ on a curve $C.$ Take a string of length $\ell,$ nail down one end at $p,$ and wind the string along $C.$ Then the free end of your string traces out a curve $D.$

$D$ is called an involute of $C.$ It consists of all the points you can get to from $p$ by a path of length $\ell$ that doesn’t cross $C.$

So, Arnol’d’s puzzle concerns the involute of the curve $y = x^3.$

He wants you to nail down one end of the string at any ‘generic’ location. So, don’t nail it down at $x = 0, y = 0,$ since that point is different from all the rest. That point is an inflection point, where the curve $y = x^3$ switches from curving down to curving up!

He wants you to wind the string along the curve $y = x^3,$ forming an involute. And he wants you to see what the involute does when it crosses the line $y = 0.$

This is a bit tricky, since the region $y \le x^3$ is not convex. If you nail your string down at $x = -1, y = -1$, your string will have to start out above the curve $y = x^3.$ But when the free end of your string crosses the line $y = 0,$ the story changes. Now your string will need to go below the curve $y = x^3.$

It’s a bit hard to explain this both simply and accurately, but if you imagine drawing the involute with a piece of string, I think you’ll encounter the issue I’m talking about. I hope I understand it correctly!

Anyway, suppose you succeed in drawing the involute. What should you see?

Arnol’d says the involute should have a ‘cusp of order 5/2’ somewhere on the line $y = 0.$

A cusp of order 5/2 is a singularity in an otherwise smooth curve that looks like $y^2 = x^5$ in some coordinates. In a recent post I described various kinds of cusps, and in a comment I mentioned that the cusp of order 5/2 was called a rhamphoid cusp. Strangely, I wrote all that before knowing that Arnol’d places great significance on the cusp of order 5/2 in the involute of a cubical parabola!

Simon Burton drew some nice cusps of order 5/2. The curve $y^2 = x^5$ looks like this:

This is a more typical curve with a cusp of order 5/2:

$(x-4y^2)^2 - (y+ 2x)^5 = 0$

It looks like this:

It’s less symmetrical than the curve $y^2 = x^5.$ Indeed, it looks like a bird’s beak: the word ‘rhamphoid’ means ‘beak-like’.
Arnol’d emphasizes that you should usually expect this sort of shape for a cusp of order 5/2:

It is easy to recognize this curve in experimental data, since after a generic diffeomorphism the curve consists of two branches that have equal curvatures at the common point, and hence are convex from the same side [….]

So, if we draw the involutes of a cubical parabola we should see something like this! And indeed, Marshall Hampton has made a great online program that draws these involutes. Here’s one:

The blue curve is the involute. It looks like it has a cusp of order 5/2 where it hits the line $y = 0.$ It also has a less pointy cusp where it hits the red curve $y = x^3.$ Like the cusp in the tractrix, this should be a cusp of order 3/2, also known as an ordinary cusp.

### Hints

Regarding the easier puzzle I posed above, Arnol’d gives this hint:

HINT. The curvature centers of both branches of the involute, which meet at the point of the inflectional tangent, lie at the inflection point, hence both branches have the same convexity (they are both concave from the side of the inflection point of the boundary).

That’s not what I’d call crystal clear! However, I now understand what he means by the two ‘branches’ of the involute. They come from how you need to change the rules of the game as the free end of your string crosses the line $y = 0.$ Remember, I wrote:

If you nail your string down at $x = -1, y = -1$, your string will have to start out above the curve $y = x^3.$ But when the free end of your string crosses the line $y = 0$, the story changes. Now your string will need to go below the curve $y = x^3.$

When the rules of the game change, he claims there’s a cusp of order 5/2 in the involute.

I also think I finally understand the picture that Arnol’d uses to explain what’s going on:

It shows the curve $y = x^3$ in bold, and three involutes of this curve. One involute is not generic: it goes through the special point $x = 0, y = 0.$ The other two are. They each have a cusp of order 5/2 where they hit the line $y = 0,$ but also a cusp of order 3/2 where they hit the curve $y = x^3.$ We can recognize the cusps of order 5/2, if we look carefully, by the fact that both branches are convex on the same side.

But again, the challenge is to prove that these involutes have cusps of order 5/2 where they hit the line $y = 0.$ A cusp of order 7/2 would also have two branches that are convex on the same side!

Here’s one more hint. Wikipedia says that if we have a curve

$C :\mathbb{R} \to \mathbb{R}^2$

parametrized by arclength, so

$|C^\prime(s)|=1$

for all $s,$ then its involute is the curve

$D :\mathbb{R} \to \mathbb{R}^2$

given by

$D(s) = C(s)- s C^\prime(s)$

Strictly speaking, this must be an involute. And it must somehow handle the funny situations I described, where the involute fails to be smooth. I don’t know it does this.

### 117 Responses to The Involute of a Cubical Parabola

1. But if we refrain from what? Did you mean to post this already? It looks like you weren’t

• John Baez says:

It’s annoyingly easy to hit ‘publish’ when you’re trying to update a draft, and when you do that, the whole world is notified, so people send me puzzled emails if I then go and ‘unpublish’ it. I didn’t want to publish this one today—but okay, it’s done now.

2. jessemckeown says:

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.

• John Baez says:

Thanks for all the erudition!

For those who find this hard to follow, another of Sam Derbyshire’s pictures might help. Just as the tractrix is an involute of the catenary:

so the catenary is the evolute of the tractrix:

You’ll note that the tractrix has a cusp of order 3/2 where it hits the catenary and you have to change your mind a bit about how it’s defined. Similarly, the involutes of the cubical parabola have cusps of order 3/2 when they hit that curve.

Arnol’d is claiming the (generic) involutes of the cubical parabola have cusps of order 5/2 for a subtler reason: that curve has an inflection point. This forces you to change your mind in a subtler way about how an involute is defined.

• Simon Burton says:

I like the connections to wavefronts and pathfinding. Should we think of the tractrix as the ray (path) dual to the tangent lines which are the wavefronts? It looks like it gets refracted around alot and reflects off of the catenary at right angles.

• John Baez says:

I see what you mean, but…

In general, you can think of an involute as a wavefront of some light that moves at constant speed and is not allowed to enter an obstacle.

From this viewpoint, it’s the tractrix that is the wavefront! Each branch of it consists of all points that can be reached by a path of length $\ell$ starting from a particular point $p,$ where the path is not allowed to enter the catenary.

(As mentioned in my post, this example is a bit tricky, because we’re using two different points $p$ for the two different branches of the tractrix. Furthermore, these points are ‘at infinity’—or at least way, way up the catenary.)

3. John Baez says:

This should be helpful: Wikipedia says if we have a curve

$C :\mathbb{R} \to \mathbb{R}^2$

parametrized by arclength, so

$|C^\prime(s)|=1$

for all $s,$ then its involute is the curve

$D :\mathbb{R} \to \mathbb{R}^2$

given by

$D(s) = C(s)- s C^\prime(s)$

I’ll add this to the ‘hints’ in the post.

4. Marshall Hampton says:

Here’s a little Sage cell that draws the example, just made it to make sure I understood some of what you were saying: http://sagecell.sagemath.org/?q=hwedlo

• John Baez says:

Wow, that’s MAGNIFICENT!

Everyone go there and look! He’s taken Arnol’d’s mysterious picture:

and brought it into the 21st century. There’s a slider that lets you see all the involutes of the curve $y = x^3.$ You can see the catastrophe or ‘perestroika’ that occurs when the involute hits the origin, which is an inflection point of $y = x^3.$ But what matters more to me right now are the cusps of order 5/2 that occur in involutes that don’t hit the origin.

Thanks, Marshall.

• John Baez says:

By the way: is there some fairly automatic way to make a nice animated gif showing how the involute changes as the parameter in Marshall’s slider moves? Marshall’s program makes a nice image for each value of that parameter. We’d need a way for Sage to create lots of images as that parameter changes, and bundle them up into a (looped) animated gif.

If someone could do this, I would love to feature it on Visual Insight, and of course credit everyone involved.

• Marshall Hampton says:

Yeah I can do that, gotta teach a class in a bit so it might have to be tonight.

• Marshall Hampton says:

Here is one effort:

Here’s the code in Sage if you want to tweak it (sorry the line breaks will be screwed up):

 var('t') p = (t,t^3) pd = (1,3t^2) sp = sqrt(1+9t^4) invpts = srange(-1,1,.025,include_endpoint=True) def xi(ti,invpt): return ti - numerical_integral(sp,invpt,ti)[0]/sp(t=ti) def yi(ti,invpt): return ti^3 - 3ti^2numerical_integral(sp,invpt,ti)[0]/sp(t=ti) outs = [] for invpt in invpts: p1 = line2d([[xi(i,invpt),yi(i,invpt)] for i in srange(-2,2,.01,include_endpoint=True)], xmin=-1,ymin=-1,xmax=1,ymax=1) p2 = parametric_plot((t,t^3),(-1,1), rgbcolor='red', xmin=-1,ymin=-1,xmax=1,ymax=1) outs.append(p1+p2) animate(outs) 

• John Baez says:

Wow, that’s GREAT! I will use this on Visual Insight. At some point I’ll pester you to electronically ‘sign’ a form giving the American Mathematical Society permission to use this gif—that’s a thing they make me do.

5. Marshall Hampton says:

OK – perhaps I can make a somewhat better one; I cobbled that together pretty fast so I’m sure it could be improved.

• Marshall Hampton says:

All together – probably not useful:

• Simon Burton says:

Oh wow, this is so beautiful. It looks to me like a surface in 3 dimensions, viewed from “above”. It’s as if the involutes unwrap the plane into a more complicated surface in 3 dimensions. And on this surface I suspect the involutes are smooth (they have stationary points at the cusps). This must be what Arnold is calling “the discriminant of the symmetry group of an icosahedron.” I have no idea how this works though.

Here is the figure from Arnold’s book:

• John Baez says:

Simon wrote:

It looks to me like a surface in 3 dimensions, viewed from “above”.

It is. That’s part of what Arnol’d is claiming.

This must be what Arnold is calling “the discriminant of the symmetry group of an icosahedron.”

Right. Perhaps the easiest way to explain it is this—though it’s not ultimately the most beautiful way.

Take the polynomial

$x^5 + ax^2 + bx^2 + c$

For most values of $a,b,c$ the 5 roots of this polynomial are all distinct. For some special values of $a,b,c$ it has ‘repeated roots’: that is, fewer than 5 different roots. Those special values give points $(a,b,c)$ that lie on an interesting surface in 3d space. From a certain view, and perhaps warped a bit, it should look like this:

And this should remind you hugely of the involutes of the cubical parabola!

I wish someone could use a computer to draw the surface I just described. It would be easier if I knew the discriminant of a quintic. This is a function $D(a,b,c,d,e)$ that equals zero when

$x^5 + a x^4 + b x^3 + c x^2 + d x + e$

has repeated roots. Unfortunately $D(a,b,c,d,e)$ is a polynomial with 59 terms—and worse, I don’t know what it is! If we knew it, we could trim it down to handle the special case of polynomials like

$x^5 + ax^4 + bx^2 + c$

In other words, we’d look at $D(a,0,b,0,c).$ A lot of the 59 terms would go away. Then someone good with computers could draw a picture of where the discriminant vanishes.

• Greg Egan says:

The discriminant of:

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

is

$D_x(Q) = c \left(256 a^5 c^2-128 a^4 b^2 c+16 a^3 b^4+2000 a^2 b c^2-900 a b^3 c+108 b^5+3125 c^3\right)$

This in turn has discriminants wrt $a, b, c$ of:

$D_a(D_x(Q)) = 4294967296 c^{13} \left(4 b^5+3125 c^3\right)^2 \left(54 b^5+3125 c^3\right)^3$

$D_b(D_x(Q)) = -256 c^{15} \left(4096 a^5-253125 c\right)^3 \left(16 a^5-3125 c\right)^2$

$D_c(D_x(Q)) = -256 b^{13} \left(4 a^3+25 b\right)^2 \left(4 a^3+27 b\right)^2 \left(128 a^3+675 b\right)^3$

• John Baez says:

Thanks, Greg! What do these higher discriminants let us do, exactly?

Hmm… I’m trying to cajole someone into drawing the surface

$c \left(256 a^5 c^2-128 a^4 b^2 c+16 a^3 b^4+2000 a^2 b c^2-900 a b^3 c+108 b^5+3125 c^3\right) = 0$

(If anyone finds this confusing, they should replace $a,b,c$ by $x,y,z$.) I guess the higher discriminants say something about where various sheets of this surface collide. They do so in a way that’s heavily biased towards the $a,b$ and $c$ axes. But I guess that still could be useful.

• Simon Burton says:

Yes I am attempting to make some plots of this discriminant surface and will post any that look reasonable.

• Greg Egan says:

So far the images I have of the surface are pretty horrible, because (as we discussed in relation to the Capricornoid), capturing cusps accurately can be tricky. A naive plot of the zero set misses the sharp edges of the surface, and unfortunately solving for any one of the variables in terms of the other two also seems to give messy results in Mathematica, due to some kind of numerical artifacts.

BTW, I think Arnol’d omits the plane $c=0$ from the surface he’s drawn, so I’ll concentrate on the remaining factor of the discriminant.

The higher discriminants should let us figure out any genuine, coordinate-independent self-intersections of the surface, because their projections should show up in all three discriminants. So I think the curves:

$\displaystyle{ (a,-\frac{128 a^3}{675},\frac{4096 a^5}{253125}) }$

$\displaystyle{ (a,-\frac{4 a^3}{25},\frac{16 a^5}{3125}) }$

and

$\displaystyle{ (a,0,0) }$

all give sets where the surface intersects itself. All four discriminants (that of the quintic, and the three higher discriminants wrt $a,b,c$) are zero along these curves.

• jessemckeown says:

If we only want to draw the graph of the discriminant, it should be enough to restrict the coefficients in the cubic

$x^3 - p x^2 + q x - r$

such that the product

$(x^2 - 2 u x + u^2) (x^2 - px^2 + qx - r)$

is depressed in the desired way, which is a system of two equations, linear in $p,q,r$; one then has a 2-dimensional parametrization of the discriminant surface. One reasonably-generic singular quintic, then, is

$\displaystyle{ x^5 + (-2u-p) x^4 + (\frac{3u^3}{2} + 2pu^2)x^2 - \frac{u^5}{2-pu^4} }$

• John Baez says:

Jesse: I could only get your last big formula to parse by completely retyping it; I hope I didn’t introduce any errors.

• Greg Egan says:

Here’s an image of the surface of zero discriminant:

The yellow part of the surface is where $c$ is single-valued; the green, blue and orange parts are where $c$ is triple-valued. The straight green line and the curved blue line run along sharp edges of the surface, where it self-intersects and comes to an end (in the real domain); the red line marks a curve where the surface self-intersects and continues.

The 5/2 cusps should lie along the green line, and the 3/2 cusps along the projection into the $(a,b)$ plane of the blue curve, which is a cubic.

• Greg Egan says:

I was hoping that the zero-discriminant surface might be related to the involutes of a cubic in the plane in the simplest possible way … but it’s not.

If you project the blue curve that marks one “fold” in the surface onto the $(a,b)$ plane, you get a cubic: $b=-\frac{128 a^3}{675}$. If you project the red curve that marks the self-intersection of the surface onto the $(a,b)$ plane, you get another cubic: $b=-\frac{4 a^3}{25}$.

So, I was hoping that all the involutes of the first cubic in the plane would self-intersect along the second cubic. But they don’t!

• jessemckeown says:

last term should be two terms: $\frac{u^4}{2} - p u^4$; sorry about that.

• jessemckeown says:

let me know if this doesn’t work. This is a slightly different parametrization; still not quite happy with the results.

• John Baez says:

Marshall drew:

That image may actually be useful. If we look at waves moving in the region $y \ge x^3,$ obeying Huyghen’s principle, their diffraction may be related to that portion of your picture. But I’m a bit confused about the details.

Marshall wrote:

OK – perhaps I can make a somewhat better one; I cobbled that together pretty fast so I’m sure it could be improved.

If you want to improve it, please do. You can post it here. This won’t show up on Visual Insight for a while, since I have a number of posts already lined up.

6. John Baez says:

Simon Burton made a great animation explaining how one particular involute of the curve $y = x^3$ gets its two cusps: the exciting cusp of order 5/2 where the involute hits the $x$ axis, and the less exciting cusp of order 3/2 where the involute hits the curve $y = x^3$:

• Bruce Bartlett says:

Hi Simon – I’m interested to see your code, I’d like to learn how you programmed that in Sage!

• Simon Burton says:

Hi Bruce, I didn’t use sage for this, just plain python and pyx (and some other programs to assemble the gif.) Here is the code: http://pastebin.com/K4kY8ZeS

• Bruce Bartlett says:

Thanks, that’s very useful.

7. Scott Hotton says:

I find it convenient to think of an involute as a roulette
where the rolling curve is a line. It will also be useful to
express the parameterizations of the curves with complex
numbers. The cubical parabola has the parameterization

$z(t) = t + i\, t^3$

The starting position of the rolling line will be the real
axis and we can parameterize it with the arc length of the
cubical parabola so that as the line rolls the speed will be
the same at the contact points of the curves (the no slip
condition).

$s(t) = \int_{0}^{t} | \dot{z}(\tau)| d \tau$

Let $x_0 \in {\bf R}$ be the location of the tracing point
at $t=0$. This is also the location of the involute’s cusp
on the real axis, the “exciting” cusp. In this way the family
of involutes is parameterized by the location of the exciting
cusp. For each $x_0$ the parameterization for the
involute is

$Z(t) = z(t) + ( x_0 - s(t) ) \frac{\dot{z}(t)}{|\dot{z}(t)|}$

From Arnold’s hint the origin is the center of the osculating
circle at the exciting cusp. We can straighten out the
osculating circle with a Mobius transformation that maps
$x_0 e^{i t}$ to the imaginary axis while leaving the real
axis invariant.

$W(t) = \frac{Z(t) - x_0}{Z(t) + x_0}$

The exciting cusp is now at the origin and the curvature
should be $0$ for both branches. The image of the
cubical parabola is a simple closed curve minus the point
$1$. $W(t)$ would be a rational function of
$t$ if it were not for $|\dot{z}(t)|$ and $s(t)$.
Since $t$ only appears with fourth degree under the
radical in $|\dot{z}(t)|$ its Taylor expansion about
$t=0$ only contains terms whose power is a multiple
of $4$.

$|\dot{z}(t)| = 1 + \frac{9}{2} \; t^4 + O(t^8)$

So long as $x_0 \neq 0$ we can algebraically obtain the
first few terms in the Taylor expansion for $s(t)$ and
$W(t)$. I got:

$W(t) = -\frac{6}{5x_0} t^5 + i \, t^2 \left( \frac{3}{2} - \frac{1}{x_0} t \right) + O(t^6)$

This resembles the curve $t^5 + i\, t^2$ but I do not know
if its close enough to help reveal the hidden icosahedron.

8. John Baez says:

Greg drew:

This is great! But I’m having trouble seeing how it’s diffeomorphic to this:

Do you see it?

It should be possible to take a slice of your surface that looks approximately like the blue curve here:

that is, a generic involute of the cubical parabola, containing both a cusp of order 3/2 and a cusp of order 5/2.

I’m having trouble seeing such a slice in your picture, though I see the cusps of order 5/2 very nicely along your green line. You’re making it sound like the cusps of order 3/2 only spring into existence when we project your surface onto the $ab$ plane… while Arnol’d’s picture makes it look like they should already be visible in the 3d picture.

9. John Baez says:

Simon wrote:

This must be what Arnol’d is calling “the discriminant of the symmetry group of an icosahedron.” I have no idea how this works though.

Let me try to explain that. This subject really deserves lots of fancy math jargon, but I’ll try to minimize that.

The symmetry group of the icosahedron, including rotations and reflections, has 120 elements. That’s just enough to move any triangle here to any other triangle:

Certain polynomials in $x,y,z$ are unchanged (or ‘invariant’) when we apply any of the icosahedron symmetries. Apart from constants, the most obvious one has degree 2:

$P(x,y,z) = x^2 + y^2 + z^2$

But there’s another invariant polynomial of degree 6 that we get as follows. The icosahedron has 12 corners, which come in opposite pairs. Choose 6 corners $q_1, \dots, q_6,$ none opposite to each other. Taking the dot product with each of these gives a linear function:

$f_i(x,y,z) = q_i \cdot (x,y,z)$

Multiply all 6 of these linear functions and we get an invariant polynomial!

$Q(x,y,z) = f_1(x,y,z) \cdots f_6(x,y,z)$

There’s another invariant polynomial of degree 10 that we get as follows. The icosahedron has 20 faces, which come in opposite pairs. Choose the midpoints $r_1, \dots, r_{10}$ of 10 faces, none opposite to each other. Taking the dot product with each of these gives a linear function:

$g_i(x,y,z) = r_i \cdot (x,y,z)$

Multiply all 10 of these linear functions and we get an invariant polynomial!

$R(x,y,z) = g_1(x,y,z) \cdots g_{10}(x,y,z)$

It’s not utterly obvious that $Q$ and $R$ are invariant. Because we arbitrarily chose one corner from each opposite pair, and similarly one face from each opposite pair, $Q$ and $R$ could in theory change sign when we apply a symmetry of the icosahedron.

Puzzle. Why don’t they change sign?

Now, it’s a marvelous theorem of Chevalley that every polynomial in $x,y,z$ that’s invariant under the icosahedron symmetries can be expressed as a polynomial in $P,Q,$ and $R.$ Even better, this can be done in a unique way. In other words, $P,Q,R$ don’t obey any polynomial relations like $P^2 Q + P^5 = R.$

To understand the discriminant of the icosahedral symmetry group, we need to think about the function from $\mathbb{R}^3$ to $\mathbb{R}^3$ sending $(x,y,z)$ to

$(P(x,y,z), Q(x,y,z), R(x,y,z))$

This ‘folds up’ 3-dimensional space in a certain fascinating way, which reveals the discriminant.

More later!

• John Baez says:

Here’s the rest of the story.

You can see a bunch of great circles here:

If you imagine this as part of a bigger picture in 3d space, each great circle is the intersection of the sphere with a plane. These planes are called mirrors, because reflecting through any of these planes is a symmetry of the icosahedron.

Even better, every symmetry of the icosahedron can be obtained by a succession of reflections through these mirrors!

The mirrors, taken all together, form a subset of 3-dimensional space. We can map this subset to 3-dimensional space using the function I described above:

$(x,y,z) \mapsto (P(x,y,z), Q(x,y,z), R(x,y,z))$

The result is a new subset of 3-dimensional space, called the discriminant of the icosahedral group.

Arnol’d claims that it looks like this:

Challenge. Can someone make a nice image of it?

Moreover, Arnol’d claims that this subset also looks like the set of points $(a,b,c)$ for which the polynomial

$x^5 + ax^4 + bx^2 + c$

has repeated roots. Greg has drawn that:

So far I’m having trouble seeing why this looks like Arnold’s picture. By “looks like”, I mean that there’s a coordinate transformation (a diffeomorphism) that maps one to the other. This could warp things quite a lot, but it will send cusps of order 5/2 to cusps of order 5/2, etcetera.

Here’s a nice thing about the discriminant. Since the functions $P, Q, R$ are invariant under the icosahedral symmetry group, and all the mirror planes are related to each other by symmetries of the icosahedron, each mirror plane get mapped onto the same set when we apply the function

$(x,y,z) \mapsto (P(x,y,z), Q(x,y,z), R(x,y,z))$

And this set is the discriminant.

Puzzle. How many mirror planes are there?

• Simon Burton says:

I think if Greg could zoom into the origin a bit and also reflect the b axis it would look very similar to the Arnold picture. Also it looks like a part of the orange sheet is missing from left-hand side of that diagram. But it does look like the dark blue curve and the light green curve would correspond to the 5/2 and 3/2 cusps. I see now that when Arnold draws a dashed line he means that that line is hidden behind another surface.

• John Baez says:

Yes, dashed lines are hidden. I don’t see that any orange surface is missing. I could be wrong…

• Simon Burton says:

Ah I got confused, the orange surface is underneath the light blue surface over on the left side. So that’s why you can’t see it.

• Greg Egan says:

I’m afraid the colours here aren’t very systematic. I changed them later, so that the figure would be invariant under sign inversion, but apparently WordPress takes a snapshot of the first version of an image when you link to it, and that’s what’s shown here, not the new file I put on my web site at the same URL.

But the later figure where I lopped off the rectangular corners (and removed the mesh lines) has the new, consistent colouring.

• John Baez says:

Greg wrote:

Apparently WordPress takes a snapshot of the first version of an image when you link to it, and that’s what’s shown here, not the new file I put on my web site at the same URL.

Yes, this rather new policy of there’s causes me endless trouble, because I like to link to pictures on my website, and I sometimes edit those pictures. Even from behind the scenes there seems to be no way to tell a free WordPress blog to take a new snapshot without changing the URL of the image.

I see now you can do it with a plugin, but I believe these plugins are only available for paid blogs.

My problem with using a paid version is not the money but the hassle. David Tanzer has recently proposed that the Azimuth Project improve its blog in a few ways, so we may finally get around to dealing with this.

Anyway, I’m turning your images to clickable links… clicking on them shows the latest version. And I believe that if I post another copy of your image we’ll see the new version. Let’s see:

Nope, dammit! It’s too smart: they’re reusing their stored version. This is fiendish.

Luckily there’s another URL that works for Greg’s pages. Let me try using that:

Hah! This is the new version.

• Greg Egan says:

Taking a diagonal slice that runs directly from the green line to the blue line gives a cross-section where both cusps are more visible, and the result looks more like an involute of a cubic parabola.

Forget everything I said about projections. I was thinking that there was meant to be an isometry with the involute construction, and since the projections of the red and blue curves onto the $(a,b)$ plane are cubics, I thought that would point us in the right direction. But if all that’s expected is a diffeomorphism, projection is probably a complete red herring.

• Greg Egan says:

If I take an icosahedron centred at the origin with unit-length vertices, which are chosen so that they come in 3 sets of 4 that form golden-ratio rectangles in each of the 3 coordinate planes, then the $(x,y)$ plane should be one of the mirror planes.

But the image I get of this plane is rather strange:

$P(x,y,0) = x^2+y^2$

$Q(x,y,0) = \frac{1}{50} x^2 y^2 ((5+\sqrt{5}) x^2+(\sqrt{5}-5) y^2)$

$\displaystyle{ R(x,y,0) = \frac{x^2 y^2(x^2-y^2)^2 ((1165+521 \sqrt{5}) x^2-(7985+3571 \sqrt{5}) y^2)}{14762250} }$

This is $(P(x,y,0), Q(x,y,0), R(x,y,0))$ as defined by the squared magnitude for $P$, the product of dot products with 6 non-opposite vertices for $Q$, and the product of dot products with10 non-opposite face-centres for $R$.

It’s hard to see how this set could be the zero-discriminant set, since the coordinate $x^2+y^2$ is non-negative, but there is no direction in which the zero-discriminant set is similarly constrained.

In what sense are these sets meant to be “the same”? Up to an isometry, up to a linear transformation, or just up to a diffeomorphism?

• John Baez says:

Arnol’d only claims that these things are diffeomorphic. Among other things, he says:

The discriminant of the group $\mathrm{H}_3$ [the icosahedral symmetry group] is shown in Fig. 18:

Its singularities were studied by O. V. Ljashko (1982) with the help of a computer. This surface has two smooth cusped edges, one of order 3/2 and the other of order 5/2. Both are cubically tangent at the origin. Ljashko has also proved that this surface is diffeomorphic to the set of polynomials $x^5 + ax^4 + bx^2 + c$ having a multiple root.

So, that’s something we should be able to see, thanks to your incredible computer skills. I have not been able to find any trace of a paper by O. V. Ljashko.

Maybe I should continue quoting Arnol’d. Next comes the relation to involutes:

The comparison of this discriminant with the patterns of the propagation of the perturbations on a manifold with boundary (studied as early as in the textbook of L’Hopital in the form of the theory of evolutes of plane curves), has led A. B. Givental to the conjecture (later proven by O. P. Scherbak) that this discriminant it locally diffeomorphic to the graph of the multivalued time function in the plane problem on the shortest path, on a manifold with boundary, which is a generic plane curve.

Thus the propagation of the waves, on a 2-manifold with boundary, is controlled by an icosahedron hidden at the inflection point of the boundary. This icosahedron is hidden, and it is difficult to find it even if its existence is known.

Then he discusses the involutes of a cubical parabola. He shows this figure:

Then he says:

Comparing Fig. 20 with Fig. 18, it is easy to guess that the graph of the distance (or time) function, whose level sets are the involutes, is diffeomorphic to the discriminant of the icosahedron symmetry group $\mathrm{H}_3$ (the proof of this fact is not easy at all).

So we are supposed to have 3 diffeomorphic surfaces: one defined using the icosahedron, one defined using the quintic $x^5 + ax^4 + bx^2 + c,$ and one defined using the involutes of the cubical parabola. I have not been able to find proofs in the literature!

• John Baez says:

Actually there is some potentially useful material here:

• O. P. Shcherbak, Wavefronts and reflection groups, Russian Mathematical Surveys 43 (3) (1988), 149–194.

This is not free, but it’s a translation of a paper that’s freely available in its original Russian form. Luckily, I have the translation! It mainly concerns not the case of $\mathrm{H}_3$ but the even more exotic case of $\mathrm{H}_4$: the symmetry group of the 600-cell in 4 dimensions! This case shows up when we consider wavefronts propagating around obstacles in 3 dimension: the 3d analogue of involutes.

• Greg Egan says:

Here’s an image of the “symmetry group discriminant”:

I’ve rescaled some of the coordinates. Since $x$ and $y$ only appear as $x^2$ and $y^2$, all four quadrants are mapped so the same set, so we might as well work with the doubly positive quadrant. If you put polar coordinates on that quadrant, the lines $\theta=0$ and $\theta=\frac{\pi}{2}$ both map to the $P$-axis.

The whole image is a map of a quarter-disk, and the cuspy triangle facing the viewer is the image of a quarter-circle. So it’s easy to see that there are cusps here, but I’m at a loss to see how the surface can be continued through the $P$-axis, as the other surfaces do, rather than coming to an end there. Maybe there’s some fine print that Arnol’d hasn’t mentioned.

• Greg Egan says:

Assuming I haven’t made any mistakes in the algebra, the only thing I can imagine is that this set, which is a many-to-one image of the entire collection of mirror planes, can somehow be “unwrapped” into a version more like Arnol’d’s.

But it’s not obvious to me why such a process would yield, essentially, two copies of the image set glued together in a certain way. Going around the origin in a single mirror plane just sends you back and forth around the image set four times, reversing direction at the P-axis in the image set each time you cross a coordinate axis in the domain. But maybe there’s some elaborate way of traversing the whole complex of mirror planes that can sensibly be interpreted as yielding the desired result.

• John Baez says:

Thanks, that’s a beautiful picture. I bet one of those cusps has order 5/2 (it shows the telltale signs of being rhamphoid, or ‘beak-like’) and the other has order 3/2 (just guessing).

However, the discrepancy with Arnol’d picture makes the story into more of a mystery!

Since Arnol’d is a bigshot, it’s natural to wonder if we’ve made some sort of mistake… or maybe I misunderstood him, or maybe he left out some sort of nuance. Unfortunately he died in 2010, and I don’t know anything written up with more details. I should look around….

Aha, I’ve found two references! The paper by O. P. Shcherbak mentioned above focuses on the $\mathrm{H}_4$ discriminant, but he says these two papers study the $\mathrm{H}_3$ discriminant:

• O.V. Lyashko, The classification of critical points of functions on a manifold with a singular boundary, Funktsional. Anal, i Prilozhen. 17:3 (1983), 28–36. English translation in Functional Anal. Appl. 17:3 (1983), 187–193.

• O.P. Shcherbak, Singularities of a family of evolvents in the neighbourhood of a point of inflection of a curve, and the group $\mathrm{H}_3$ generated by reflections, Funktsional. Anal. i Prilozhen. 17:4 (1983), 70–72. English translation in Functional Anal. Appl. 17:4 (1983), 301–303.

I think I can get ahold of these.

Before I forget it, here’s one idea. The Coxeter group acts not just on $\mathbb{R}^3$ but also $\mathbb{C}^3,$ so there could be different ‘real forms’ of the discriminant. You drew the one where $x,y,z$ are real so $P \ge 0,$ but maybe there’s another one where, say, $x$ and $y$ are real but $x$ is imaginary, so that $P$ can take both positive an negative values.

I’ve never heard people discuss different real forms of discriminants of Coxeter groups, so this is a very tentative idea, but it’s my best attempt so far to get a shape that looks more like what Arnol’d drew.

Now let me find those papers.

• Greg Egan says:

Thanks, John! I agree with the formulas in the paper by Lyashko, up to choices of scale and which coordinates to call $x$ and $y$. After deriving essentially the same map as I did, Lyashko goes on to describe the cubic in three variables whose zero set in $\mathbb{R}^3$ contains the image of the mirror planes, but is larger than that image.

You can get the full zero set in $\mathbb{R}^3$ by including the four possibilities where $x$ and $y$ are either purely real or purely imaginary, and this looks like the complete picture Arnol’d drew:

• John Baez says:

WOW, GREAT!!! This is the picture I’d been dreaming of!

So my idea of letting $y$ be imaginary instead of real was not completely off the mark, but it wasn’t right. I’ll have to think harder about what it means to consider all four possibilities of $x$ and $y$ being real or imaginary. It seems we’re doing a strange mixture of real and complex algebraic geometry here… but I’m probably just not being smart enough.

• Greg Egan says:

I wrote:

Lyashko goes on to describe the cubic in three variables whose zero set in $\mathbb{R}^3$ contains the image of the mirror planes …

I wrote this a bit too hastily: the polynomial in question is cubic in one of its three variables, but its degree is 11.

10. John Baez says:

I want to say a bit more about discriminants of quintics and discriminants of Coxeter groups. We are trying to relate the discriminant of the quintic $x^5 + ax^4 + bx^2 + c$ to the discriminant of the Coxeter group $\mathrm{H}_3$, the symmetry group of the icosahedron. But there’s something simpler that might be related.

First, what’s the discriminant of the Coxeter group $\mathrm{A}_n$? This group is just the symmetric group on $n+1$ letters, $\mathrm{S}_{n+1}.$ This group acts on $\mathbb{R}^{n+1}$ in an obvious way, by permuting the coordinate axes. But it also acts on the $n$-dimensional subspace where the coordinates sum to zero:

$V = \{ r \in \mathbb{R}^{n+1} : \; r_1 + \cdots + r_{n+1} = 0 \}$

and this is how we think of it as a Coxeter group.

Each point in $V$ gives a polynomial like this:

$(x - r_1) \cdots (x - r_{n+1})$

This trick gives all the polynomials of degree $n+1$ whose leading coefficient is 1 and whose roots sum to zero. Two different points of $V$ give the same polynomial iff we can get from one to the other by permuting the coordinate $r_1, \dots, r_{n+1}$.

In other words, we have a map from $V$ to the space $W$ consisting of polynomials of degree $n+1$ whose leading coefficient is 1 and whose roots sum to zero. Two points in $V$ map to the same polynomial in $W$ iff they lie in the same orbit of the Coxeter group.

Generically, $(n+1)!$ different points in $V$ map to each polynomial in $W.$ But the number is smaller for polynomials with repeated roots.

A polynomial has repeated roots iff its discriminant vanishes. The discriminant is very simple, since we’re writing our polynomial in terms of its roots rather than its coefficients! It’s

$\Delta = \prod_{1 \le i < j \le n+1} (r_i - r_j)$

The set on which this vanishes is just the union of all the hyperplanes

$r_i = r_j$

where $1 \le i < j \le n+1.$ These hyperplanes are also the mirror planes for the Coxeter group!

Specializing to $n = 4,$ we see that the discriminant for quintics whose leading term is 1 and whose roots sum to zero vanishes precisely on the mirror planes for the Coxeter group $\mathrm{A}_4 = \mathrm{S}_5.$ Even better, note that these polynomials are precisely those of the form

$x^5 + ax^3 + bx^2 + cx + d$

This is pretty close to what we’re actually interested in: polynomials of the form

$x^5 + ax^4 + bx^2 + c$

We should be able to get those by taking some sort of 3-dimensional slice of the 4-dimensional space $V.$ But I haven’t worked out the details!

By the way, the group $\mathrm{A}_4$ is the symmetry group of a 4-simplex, which looks like this when you project it down to the plane:

The symmetry group of the pentagon is also a Coxeter group, sometimes known as $\mathrm{H}_2.$ So, there’s a relation between this group and $\mathrm{A}_4.$

This is part of a little pattern relating:

the $\mathrm{H}_2$ Coxeter group and the $\mathrm{A}_4$ Coxeter group,
the $\mathrm{H}_3$ Coxeter group and the $\mathrm{D}_6$ Coxeter group,
the $\mathrm{H}_4$ Coxeter group and the $\mathrm{E}_8$ Coxeter group.

For more on that, see “week270”.

But I’m still puzzled about how $\mathrm{H}_3$ is getting related to quintics of the special form $x^5 + ax^4 + bx^2 + c.$

• John wrote:

This is pretty close to what we’re actually interested in: polynomials of the form $x^5 + ax^4 + bx^2 + c$.

Instead of relating the discriminant of $A_4$ to $H_3$, as you’re doing here, perhaps it might be simpler to relate the discriminant of $B_5$ with $H_3$. The orbit space of $B_5$ consists of polynomials of the form

$x^5 + ax^4 + bx^3 + cx^2 + dx + e$,

the point being that there is an $x^4$ term, unlike for $A_4$. So if the discriminant of $H_3$ is somehow obtained from that of $B_5$ by setting the coefficients of $x^3$ and $x$ equal to zero, then we’d be in business.

11. John Baez says:

As I prepare to document all our work on Visual Insight, I have a request for Greg… I hope it’s easy. First, I’m still having trouble understanding this image showing the zero set of the discriminant of $x^5 + ax^4 + bx^2 + c$:

I’d like it to be more obvious that it’s diffeomorphic to this one (assuming that’s actually true):

Simon suggested this:

I think if Greg could zoom into the origin a bit and also reflect the b axis it would look very similar to the Arnol’d picture.

It would also help if the surfaces were made translucent. The “fun” comes from the lines of cusps, some of which are inevitably behind something else.

• John Baez says:

Simon Burton created some nice pictures of slices of the the zero set of the discriminant of $x^5 + ax^4 + bx^2 + c.$
Here is its intersection with the plane $a = -2.2$:

Here is its intersection with the plane $a = +2.2$:

In both of these, the big ticks on the axes are at multiples of 0.5.

This nicely exhibits the relation between this surface and the involutes of the cubical parabola:

If it’s hard to draw this surface in 3d in a way that makes the relation clear, perhaps an animated gif of slices would do the job.

• Greg Egan says:

I’ve tried tinkering with my image of the quintic discriminant in various ways, but none of them look like improvements to me, so you should probably go with some version of Simon’s approach.

• Greg Egan says:

After a bit more tinkering, maybe this is an improvement (this is the quintic discriminant):

• John Baez says:

For some reason I hadn’t seen this until now.

This is just what I hoped for! But it looks so different than the previous images, I can barely believe it’s the same surface.

• Greg Egan says:

Apart from the translucency and the different colouring of the surface (I’m no longer trying to colour-code the different roots for $c$ that sit above each $(a,b)$), I’ve sliced through it at a smaller value for $a$, while expanding the scale along the $a$-axis.

If you follow the surface out to larger $a$ values, as in my previous version, things twist around so that you need to slice obliquely to get a cross-section that looks like the archetypical involute.

12. Let me try and summarize the current status of this thread as I now see it. Please correct me.

No-one has yet solved the original puzzle of Arnol’d : Prove that the generic involute of a cubical parabola has a cusp of order 5/2 on the straight line tangent to the parabola at the inflection point.

We have great graphs, but not yet an analytical proof.

The next claim of Arnold is the following:

Claim A. The discriminant surface $X$ of the icosahedral group is diffeomorphic to

$Y := \{ (a,b,c) : x^5 + ax^4 + bx^2 + c \textrm{ has multiple roots} \}$

Greg Egan has drawn a wondferful picture of $Y$. In order to get a picture that looks like Arnold’s picture of $X$, he needs to add in some “analytically continued” values.

On the other hand, Arnold wasn’t being very precise with regard to real vs complex when he defined what $X$ was.

So we have some good graphical evidence of Claim A, modulo some imprecision.

On the other hand, I have not been able to track down a proof of Claim A in the literature. Perhaps it’s in Lyashko, “Classification of critical points of functions on a manifold with singular boundary”, but if so I can’t find the precise statement.

The final claim of Arnol’d is:

Claim B. The discriminant surface $X$ of the icosahedral group is locally diffeomorphic to the graph $Z$ of the multivalued time function in the plane problem on the shortest path, on a manifold with boundary, which is a generic plane curve.

Here we have some nice pictures of Simon Burton, showing that slices of $Y$ graphically correspond to slices of $Z$. So if we believe Claim A, then this is evidence for Claim B.

Thankfully, we can track down a precise proof of this claim in the literature. It is in Scherbak, “Singularities of families of evolvents in the neighborhood of an inflection point of the curve, and the group $\mathrm{H}_3$ generated by reflections”. Although I don’t understand the proof.

• John Baez says:

Thanks very much for trying to summarize the state of play and point out what remains to be done!

No-one has yet solved the original puzzle of Arnol’d: Prove that the generic involute of a cubical parabola has a cusp of order 5/2 on the straight line tangent to the parabola at the inflection point.

I think that’s right. Scott Votton has an approach that could perhaps be completed with some more thought. Jesse McKeown also has an approach.

I don’t understand either of these approaches as well as I’d like—my main excuse is that I’ve been trying to understand other aspects of this problem. Note that both approaches seem to run into a similar obstacle: the difference between a polynomial that describes a cusp of order 5/2, and a similar-looking polynomial with some higher-order terms. I suspect that these higher-order terms can’t change a cusp of order 5/2 into something else. This is the sort of thing where having a bit more expertise in singularity theory might help a lot.

13. John Baez says:

Bruce wrote:

Claim A. The discriminant surface X of the icosahedral group is diffeomorphic to

$Y := \{ (a,b,c) : x^5 + ax^4 + bx^2 + c \textrm{ has multiple roots} \}$

So we have some good graphical evidence of Claim A, modulo some imprecision.

On the other hand, I have not been able to track down a proof of Claim A in the literature. Perhaps it’s in Lyashko, “Classification of critical points of functions on a manifold with singular boundary”, but if so I can’t find the precise statement.

I don’t think it’s in there. Remember, back in the USSR, Russian mathematicians would hold long seminars where they worked things out in detail. If everyone present was convinced, sometimes they would publish the results with only a sketchy proof. This has often made Western mathematicians unhappy.

I’m still hoping it’s possible to prove Claim A using the known relations between quintics and the icosahedron. Felix Klein wrote a whole book on this, Lectures on the Icosahedron, and luckily there’s a free book which gives a treatment of these ideas that’s a lot easier for modern mathematicians to understand:

• Jerry Shurman, Geometry of the Quintic.

This is hugely fun stuff. Very briefly, the fact that the symmetry group of the icosahedron is almost the Galois group of the general quintic lets you solve the quintic if you can solve the equation $f(w) = z$ where $f$ is a nontrivial rational function on $\mathbb{C}\mathrm{P}^1$ that is invariant under the symmetries of the icosahedron!

The invariant polynomials $P,Q,R$ that Greg Egan is looking at are, I believe, closely related to this business.

Unfortunately I don’t see the significance of the family of quintics

$x^5 + ax^4 + bx^2 + c$

I wrote something here about this issue, but I was led naturally to the so-called depressed quintic, where the quartic term vanishes:

$x^5 + bx^3 + cx^2 + dx + e$

It’s well-known that any polynomial can be transformed by a Tschirnhaus transformation into depressed form, meaning that the next-to-leading coefficient is 0, or in other words, the sum of the roots is zero.

The reason this is important for us is that it means the Galois group of a generic depressed quintic is still that of the generic quintic, namely $S_5$, which is almost the rotational symmetry group of the icosahedron, namely the alternating group $A_5.$ (If we include reflections as symmetries we get not $S_5$ but $A_5 \times \mathbb{Z}/2.$ However, it’s the non-solvable part, the $A_5,$ that’s the really big deal here.)

With more work we can massage any quintic into principal form, where the cubic term also vanishes:

$x^5 + cx^2 + dx + e$

And with even more virtuosic feats of high-school algebra we can do a change of variables to bring any quintic into Bring–Jerrard normal form, where the quadratic term also vanishes:

$x^5 + dx + e$

So, all these special classes of quintics should be closely connected to the alternating group $A_5,$ and thus, I expect, the icosahedron.

But I don’t know what’s so special about quintics like

$x^5 + ax^4 + bx^2 + c$

Maybe your comment holds the key!

• Greg Egan says:

I guess another approach would be to find a diffeomorphism between $\mathbb{R}^3$ and $M^3 \subset \mathbb{R}^5$ that commutes with the actions of $\mathrm{H}_3$, where $M^3$ is the 3-dimensional submanifold of $\mathbb{R}^5$ on which the symmetric polynomials of degree 2 and 4 in the coordinates vanish.

If we could show that, I think it would follow that the orbits on the two spaces were diffeomorphic, and maybe also that the two discriminants — the varieties of irregular orbits of each action — were diffeomorphic.

• Greg Egan says:

I think the suggestion in my comment above is impossible to achieve, at least if the action on $\mathbb{R}^5$ comes from permuting the coordinates according to the permutation the group element induces on the true crosses of the icosahedron, and if $-1$ in $H_3$ acts as $-1$ on $\mathbb{R}^5$. In other words, the elements of $H_3$ with determinant 1 just permute the coordinates on $\mathbb{R}^5$, and if an element $g$ has determinant $-1$, we permute the coordinates by acting with $-g$ and then multiply by $-1$.

In that case, the subset of $M^3$ that is pointwise fixed by the action on $\mathbb{R}^5$ of a reflection in one of the mirror planes will consist of just the origin, which is obviously not diffeomorphic to the whole mirror plane back in $\mathbb{R}^3$.

• John Baez says:

Greg wrote:

I think the suggestion in my comment above is impossible to achieve, at least if the action on $\mathbb{R}^5$ comes from permuting the coordinates according to the permutation the group element induces on the true crosses of the icosahedron, and if $-1$ in $\mathrm{H}_3$ acts as $-1$ on $\mathbb{R}^5$.

Let me try to see if these are reasonable assumptions.

First, we have to be careful because there are 3 different 120-element groups that seem to play a big role in this game, and they’re not isomorphic:

• the symmetric group $\mathrm{S}_5.$ This contains the alternating group $\mathrm{A}_5,$ the rotational symmetry group of the icosahedron, as a normal subgroup, and the quotient is $\mathbb{Z}/2$. But $\mathrm{S}_5$ is not the product of $\mathrm{A}_5$ and $\mathbb{Z}/2$.

• the icosahedron symmetry group, including rotations and reflections, $\mathrm{H}_3.$ This is the product of $\mathrm{A}_5$ and $\mathbb{Z}/2$.

• the binary icosahedral group, meaning the double cover of the rotational symmetry group of the icosahedron. This has $\mathbb{Z}/2$ as a normal subgroup, and the quotient is $\mathrm{A}_5.$

It took years for all this to become second nature to me. I’m happy to see that now it’s on Wikipedia, so people can learn it faster:

• Wikipedia, Icosahedral symmetry: commonly confused groups.

Okay, now to business. You seem to be looking at $\mathbb{R}^5$ with the obvious permutation action of $S_5.$ I think it will be a bit better to look at $\mathbb{C}^5$ with the obvious permutation action of $S_5.$ A point in $\mathbb{C}^5$ describes an ordered 5-tuple of roots of a monic quintic

$(z - r_1) \cdots (z - r_5)$

and $S_5$ acts to permute the labels on the roots. Monic just mean that the coefficient of the highest-order term is 1. Complex polynomials work better than real ones, so I think we want $\mathbb{C}^5.$

Of course the polynomial itself doesn’t know an ordering on its roots. So, the space of monic quintics is $\mathbb{C}^5/\mathrm{S}_5.$ But it can also be seen as $\mathbb{C}^5,$ using the coefficients as coordinates.

On the other hand, $\mathrm{S}_5$ doesn’t act as symmetries of the icosahedron; only $\mathrm{H}_3$ does. What these groups have in common is their subgroup $\mathrm{A}_5.$

We can think of the 5 here as the set of ‘true crosses’ in the icosahedron, meaning things like this:

but we only get the even permutations of these from icosahedron symmetries.

So, I don’t think the assumption you mentioned, about $-1$ in $\mathrm{H}_3$ acting as $-1$ on $\mathbb{R}^5,$ actually holds if we take this group’s most obvious action on $\mathbb{R}^5$ (or $\mathbb{C}^5$.) The element $-1$ in $\mathrm{H}_3$ does not affect a true cross.

• Greg Egan says:

OK, but even if we have $-1$ act as the identity on $\mathbb{C}^5$, an order-5 rotation in $\mathrm{H}_3$ fixes a 1-dimensional subspace in both $\mathbb{R}^3$ and $\mathbb{C}^5$, but the subspace in $\mathbb{C}^5$ will only meet the condition on the symmetric polynomials of degree 2 and 4 being zero at the origin.

For example, the permutation that an order-5 rotation induces on the true crosses might be $(5,1,2,3,4)$, which as a linear operator on the coordinates fixes the subspace $(r,r,r,r,r)$ of $\mathbb{C}^5$, but the symmetric polynomial condition means $r=0$.

• John Baez says:

For some reason it’s taking time for me to absorb this, but thanks. I’ll try to say something more useful tomorrow!

• Bruce Bartlett says:

The invariant polynomials $P,Q,R$ that Greg Egan is looking at are, I believe, closely related to this business.

I would like to understand this. When I skim through Shurman’s Geometry of the Quintic, I haven’t been able to find these functions $P, Q, R$.

In particular, in that book, he starts with a finite rotation group $G$, and then considers it as acting on the Riemann sphere $\mathbb{C}P^1$. The algebra of invariant functions turns out to be freely generated by a single rational function $f$, that is, $(\mathbb{C}P^1)^G = \mathbb{C}(f)$.

But for our purposes, instead of being interested in the action of $G$ on the Riemann sphere, we instead lift $G$ to its double cover $\hat{G} \subset SU(2)$, so that $G$ is thought of as acting on $\mathbb{C}^2$ instead of $\mathbb{C}P^1$. The algebra of invariants is now generated by three functions $P, Q, R$ with a single relation. When $G$ is the icosahedral group, this relation is apparently $P^5 + Q^3 + R^2 = 0$. I don’t know how to relate these two pictures.

• John Baez says:

Bruce wrote:

I don’t know how to relate these two pictures.

I don’t either, but there has to be way. Let me give it a try. It’s probably related to how there’s a map

$f : \mathbb{C}^2 \to \mathbb{C}^3$

sending spinors in $\mathbb{C}^2$ to vectors in $\mathbb{C}^3,$ which actually happen to live in $\mathbb{R}^3.$ This map takes any spinor to the expected value of its angular momentum:

$\psi \mapsto \langle \psi, \sigma_i \psi \rangle$

It’s a quadratic map from the spin-1/2 representation to the spin-1 representation.

The icosahedral group $G$ acts on the spin-1 representation, its double cover $\hat{G}$ acts on the spin-1/2 representation, and the map I just described is ‘equivariant’ in a semi-obvious sense, which involves the double cover $p: \hat{G} \to G.$

By pulling back, I believe we get a map from $G$-invariant polynomial functions on $\mathbb{C}^3$ to $\hat{G}$-invariant polynomial functions on $\mathbb{C}^2.$

We’ve got 3 $G$-invariant polynomial functions on $\mathbb{C}^3,$ namely $P,Q,R.$ These will give 3 $\hat{G}$-invariant polynomial functions on $\mathbb{C}^2,$ but those must obey (at least) one relation.

That’s my guess about how this relation $P^5 + Q^3 + R^2 = 0$ shows up! I.e., it’s not a relation between the original functions $P,Q,R$ on $\mathbb{C}^3,$ but the corresponding functions on $\mathbb{C}^2.$

• John Baez says:

There’s something mildly wrong with my plan here, because the map

$\psi \mapsto \langle \psi, \sigma_i \psi \rangle$

is not quadratic in the complex sense: if you multiply $\psi$ by $a$, you multiply $\langle \psi, \sigma_i \psi \rangle$ by $|a|^2,$ not $a^2$.

Nonetheless there is a linear intertwining operator $\frac{1}{2} \otimes \frac{1}{2} \to 1,$ so there is some quadratic map from the spin-1/2 representation to the spin-1 representation! So I think we need to use that, not what I wrote above.

Another way to put it: the spin-1/2 representation of $\mathrm{SU}(2)$ is isomorphic to its conjugate representation, so the annoying complex conjugate in the inner product is somehow not that big a deal.

• Bruce Bartlett says:

Here is a reference which nicely explains the homeomorphism

$\mathbb{C}^2 / \hat{G} \cong \{ (x,y,z) \in \mathbb{C}^3 : x^2 + y^3 + z^5 = 0\}$

which will probably help in resolving this. The reference is

Kirby and Scharlemann, Eight faces of the Poincare homology 3-sphere, page 128.

They say their proof is a medley of Milnor and Klein.

Something funny happens in the text between page 128 and 129.

• John Baez says:

This looks like a paper worth understanding!

So, these guys write down three rather complicated polynomials $p_1, p_2, p_3$ on $\mathbb{C}^2,$ obeying a relation

$p_1^2 + p_2^3 + p_3^5 = 0$

They are polynomials of degree 30, 20 and 12.

I was hoping to get these from Greg’s icosahedral-invariant polynomials $P, Q, R$ on $\mathbb{C}^3,$ which have degrees 2, 6, and 10. I was actually hoping they were obtained by composing $P, Q, R$ with a quadratic map $\mathbb{C}^2 \to \mathbb{C}^3.$ But that doesn’t work: the degrees don’t work out.

• John Baez says:

Hmm! I’m pretty sure Chevalley’s generators for the icosahedral-invariant polynomials on $\mathbb{C}^3$ have degrees 2, 6 and 10. The 2 comes from $x^2 + y^2 + z^2,$ while the 6 comes from the 12 vertices of the icosahedron and the 10 comes from the 20 faces of the icosahedron, as I explained earlier. But we could also build an invariant polynomial using the 30 edges of the icosahedron using the same trick, and it would have degree 15.

If we take the invariant polynomials of degrees 15, 10 and 6 and compose them with a quadratic map, we’ll get polynomials of degrees 30, 20 and 12. So I bet these are the polynomials Kirby and Scharlemann are working with! These have a chance of obeying the relation

$p_1^2 + p_2^3 + p_3^5 = 0$.

• Greg Egan says:

I’ll try to check JB’s conjecture about the relationships between these various polynomials when I get a chance, but for now I’ll just briefly mention a mental block I had to get past before I could see how to proceed at all!

It seems obvious that there’s an equivariant quadratic map from spin-$\frac{1}{2}$ to spin-1, especially if you take the route where you define spin-1 as the space of symmetric tensors in $\mathbb{C}^2 \otimes \mathbb{C}^2$, with:

$\rho_1(g) v \otimes w = (g v) \otimes (g w)$

for $g \in \mathrm{SU}(2), v, w \in \mathbb{C}^2$.

Then the quadratic map $\phi: \mathbb{C}^2 \to \mathbb{C}^2 \otimes \mathbb{C}^2$ with:

$\phi(v) = v \otimes v$

maps into the subspace of symmetric tensors, and we have:

$\rho_1(g) \phi(v) = (g v) \otimes (g v) = \phi(g v)$

If we choose a suitable orthonormal basis for the symmetric tensors in $\mathbb{C}^2 \otimes \mathbb{C}^2$, we can write $\phi: \mathbb{C}^2 \to \mathbb{C}^3$, with:

$\phi(v) = (v_1^2, \sqrt{2} v_1 v_2, v_2^2)$

But if we look at our simplest invariant polynomial

$\displaystyle{P(x,y,z) = x^2 + y^2 + z^2}$

we have:

$\displaystyle{P(\phi(v)) = (v_1^2 + v_2^2)^2}$

This quantity is not an invariant for SU(2)!

The catch is, we are talking about equivalent representations, but different bases. The version of spin-1 constructed from the symmetrised tensor product is equivalent to the double cover of SO(3) by SU(2), but to switch between the two you still need to use a certain unitary operator, $T$, whose matrix in the bases we’re using is:

$\displaystyle{\left( \begin{array}{ccc} -\frac{1}{\sqrt{2}} & 0 & \frac{1}{\sqrt{2}} \\ -\frac{i}{\sqrt{2}} & 0 & -\frac{i}{\sqrt{2}} \\ 0 & 1 & 0 \end{array} \right)}$

$T$ maps from the spin-1 rep on the tensor product to the spin-1 rep as it acts on $\mathbb{R}^3$. So … what do we get now?

$\displaystyle{T \phi(v) = (\frac{v_2^2-v_1^2}{\sqrt{2}},-\frac{i \left(v_1^2+v_2^2\right)}{\sqrt{2}},\sqrt{2} v_1 v_2)}$

$\displaystyle{P(T \phi(v)) = 0}$

• Greg Egan says:

While I’m looking at equivariant quadratic maps from spin-$\frac{1}{2}$ to spin-1, I might as well include the other nice way of getting spin-1, where you think of the representation space as the space of traceless $2 \times 2$ complex matrices, with the action:

$\rho_1(g) m = g m g^{-1}$

Here, the simplest equivariant quadratic $\phi$ (which agrees with our previous $\phi$ up to a factor) is:

$\displaystyle{\phi(v) = v \otimes (\epsilon v) = \left( \begin{array}{cc} v_1 v_2 & -v_1^2 \\ v_2^2 & -v_1 v_2 \end{array}\right)}$

where:

$\displaystyle{\epsilon = \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array}\right)}$

$\phi(v)$ is traceless, of course, but it also has zero determinant, which corresponds to our previous result:

$\displaystyle{P(\phi(v)) = 0}$

• John Baez says:

Great! I was trying to understand the quadratic map from the spin-1/2 representation to the spin-1 representation a bit better, and hoping to understand why we skip the obvious degree-2 invariant $x^2 + y^2 + z^2$ on the spin-1 rep when we’re using this quadratic map to turn invariants on the spin-1 rep into invariants on the spin-1/2 rep.

Now you’ve done it: the obvious degree-2 invariant on the spin-1 rep becomes zero when we convert it to an invariant on the spin-1/2 rep!

That’s the most satisfying possible explanation. We’re not “skipping” this degree-2 invariant for some obscure reason, it just gives nothing.

• Greg Egan says:

I checked John’s conjecture, and it’s right! That is, we can get three polynomials that obey:

$p_1^2 + p_2^3 + p_3^5 = 0$

by the method he described: composing invariant polynomials of degree 15, 10 and 6 based on the edges, faces and vertices of an icosahedron, with a suitable equivariant quadratic $\phi: \mathbb{C}^2 \to \mathbb{C}^3$:

$\displaystyle{\phi(v) = (\frac{v_2^2-v_1^2}{\sqrt{2}},-\frac{i \left(v_1^2+v_2^2\right)}{\sqrt{2}},\sqrt{2} v_1 v_2)}$

They’re not the ones Kirby and Scharlemann describe, though, not even up to a scale. But there might be some change of basis that will make them the same.

• John Baez says:

Hurrah! There’s something so satisfying about rediscovering these things oneself, not just looking them up. Of course you’re getting most of the satisfaction, not me, since you’re the one actually doing the calculations.

(Unfortunately Mathematica is not advanced enough to feel any of the satisfaction.)

I feel sure this trick must also work for the octahedron, giving polynomials $q_1, q_2, q_3$ with

$q_1^2 + q_2^3 + q_2q_3^3 = 0$

and some variant should work for the tetrahedron, giving polynomials $r_1, r_2, r_3$ with

$r_1^2 + r_2^3 + r_3^4 = 0$

The tetrahedron is different, because opposite to a vertex is not another vertex but the midpoint of a face! So, the trick that applies to the other cases will not give 3 different polynomials here.

• John McKay, A rapid introduction to ADE theory, 1 January 2001.

Now that I look carefully, he even describes a recipe for getting these polynomials. But his recipe is somewhat different than ours, and it may avoid the problem with the tetrahedron.

It’s a very rapid introduction, which however one needs to read and reread for decades to fully understand. McKay figured out how these 3 cases — the tetrahedron, octahedron and icosahedron — are related to $\mathrm{E}_6, \mathrm{E}_6$ and $\mathrm{E}_8.$

So, many of the things we’ve been doing have analogues for the tetrahedron and octahedron, but we’re going straight for the jugular and doing the icosahedral case, which has these additional mysterious relations to wave propagation around boundaries in the plane, and to the equation $x^5 + ax^4 + bx^2 + c = 0.$ And those mysteries remain mysterious to me!

• Greg Egan says:

I looked at the cube/octahedron functions (using our method, not the one you cite by McKay). Here, vertices, edges and faces refer to a cube of side length 2 centred at the origin, and then I tinker with the normalisation later.

$\displaystyle{ P_v(x,y,z) = (-x-y-z) (-x+y-z) (-x-y+z) (-x+y+z) }$
$\displaystyle{ P_e(x,y,z) = (-x-y) (y-x) (-x-z) (z-x) (-y-z) (z-y) }$
$\displaystyle{ P_f(x,y,z) = x y z }$

These have degrees 4, 6 and 3 respectively. When we compose them with the equivariant quadratic, we get polynomials of degree 8, 12 and 6:

$\displaystyle{ p_v(v_1,v_2) = v_1^8+14 v_1^4 v_2^4+v_2^8 }$
$\displaystyle{ p_e(v_1,v_2) = \frac{1}{4} \left(-v_1^{12}+33 v_1^8 v_2^4+33 v_1^4 v_2^8-v_2^{12}\right) }$
$\displaystyle{ p_f(v_1,v_2) = \frac{i v_1 v_2 \left(v_1^4-v_2^4\right)}{\sqrt{2}} }$

If we simply follow the pattern where we give each polynomial an exponent equal to the order of the rotational subgroup that fixes the associated feature of the polyhedron, we can get the same kind of result as with the icosahedron (inserting a suitable choice of normalising factors):

$\displaystyle{ - p_v(v_1,v_2)^3 + 16 p_e(v_1,v_2)^2 + 432 p_f(v_1,v_2)^4 = 0 }$

I don’t see any way to make the degrees add up correctly so that we get something of the form:

$\displaystyle{ q_1^2 + q_2^3 + q_2 q_3^3 = 0 }$

I’m not really sure that McKay is even claiming these relations for his own polynomials; he gives these as equations for “the singularity”, but it’s far beyond my abilities to figure out precisely what he means by that from a single reading.

• John Baez says:

Greg wrote:

I don’t see any way to make the degrees add up correctly so that we get something of the form:

$\displaystyle{ q_1^2 + q_2^3 + q_2 q_3^3 = 0 }$

I’m not really sure that McKay is even claiming these relations for his own polynomials; he gives these as equations for “the singularity”, but it’s far beyond my abilities to figure out precisely what he means by that from a single reading.

I believe he’s claiming his polynomials obey this relation. But mind-reading can be tricky. This is a nice article:

• P. Slowodny, Platonic solids, Kleinian singularities and Lie groups.

On page 7 he write:

Klein obtained the following results. For each finite group $\Gamma \subset \mathrm{SL}(2,\mathbb{C}$ the $\Gamma$-invariant polynomials on $\mathbb{C}^2$ are generated by 3 fundamental invariants $X,Y,Z$ which are subject to a single relation $R(X,Y,Z) = 0.$

He gives a table listing these. Note that every finite subgroup of $\mathrm{SL}(2,\mathbb{C})$ is conjugate to one inside $\mathrm{SU}(2)$, since we can average an inner product over the group action to get an inner product that’s preserved by $\Gamma$. The only options turn out to be our friends the cyclic groups (corresponding via McKay’s magic method to the $\mathrm{A}_n$ Dynkin diagrams), the dihedral groups (corresponding to $\mathrm{D}_n$), and the double covers of the symmetry groups of the tetrahedron, octahedron and icosahedron (corresponding to $\mathrm{E}_6, \mathrm{E}_7, \mathrm{E}_8$).

Sorry, that was a digression. The punchline is that the relation for the octahedron is the one McKay claims:

$X^3 + XY^3 + Z^2$

So maybe you made a calculational mistake, or more likely maybe I was being overoptimistic about how easy it is to find the generators $X,Y,Z.$

I think they must be explained in Slowody’s paper. So far I’m just enjoying his explanation of why these things are called ‘singularities’, and what they look like.

• John Baez says:

Hmm, I can’t find anything in Slowody’s paper that gives a concrete formula for the polynomials $X,Y,Z$!

• Greg Egan says:

The degrees of McKay’s polynomials are the same as the ones we end up with: 8, 12 and 6. These are the count of vertices, edges and faces for a cube (or faces, edges and vertices for an octahedron if you prefer). Our method halves these numbers for the polynomials in three variables, then doubles them again for the polynomials in two variables, but either method ends up with the same degrees.

If we call McKay’s three polynomials $X,Y,Z$, and they obey the relationship:

$\displaystyle{X^3 + X Y^3 +Z^2 = 0}$

then which polynomial should be which? There’s no choice that makes the degrees of all three terms the same, but I guess that’s not necessarily an obstacle; the coefficients of all the individual monomials could still end up being zero, in principle.

Now, the three invariants we get from our method (which do obey a relationship, when suitably normalised, of $p_v^3 + p_e^2 + p_f^4 = 0$), certainly don’t obey one of the form $X^3 + X Y^3 +Z^2 = 0$, for any of the six ways we can assign $X,Y,Z$.

And as far as I can tell, exactly the same is true of McKay’s polynomials, obtained from stereographic projection! That is, they do satisfy the relation $p_v^3 + p_e^2 + p_f^4 = 0$, when suitably normalised, but they don’t obey one of the form $X^3 + X Y^3 +Z^2 = 0$, for any of the six ways we can assign $X,Y,Z$.

I’m happy to believe that something obeys the equation $X^3 + X Y^3 +Z^2 = 0$ … but the invariants that our method yields (and I’m pretty sure that these are, by construction, invariants of the appropriate finite subgroup of SU(2)) certainly don’t obey that equation, and unless I’ve misunderstood McKay’s recipe, neither do his.

• Greg Egan says:

I tried McKay’s construction for the tetrahedron; this is the case where our approach is unable to produce three linearly independent invariants.

And again, what I found was that McKay’s polynomials obey relations, after suitable normalisation, where the polynomial associated with each kind of special point on the polyhedron is raised to a power equal to the order of the rotational subgroup that fixes that point. That is, for the tetrahedron the McKay polynomials obey:

$\displaystyle{ p_v^3 + p_e^2 + p_f^3 = 0 }$

but none of the six ways of assigning $X,Y,Z$ to these polynomials yields a relation of the form:

$\displaystyle{ X^2 + Y^3 + Z^4 = 0 }$

(which is how McKay describes the “singularity”, and – apart from some differences as to which variable gets which exponent – how Slodowy describes the relation satisfied by the generators in this case).

I don’t really understand what’s happening here, but if I’ve misunderstood McKay’s recipe, or made some error in following it, it would be a very strange coincidence that the erroneous results obeyed such simple, systematic relations! So I’m inclined to believe that McKay’s polynomials, which he calls $V, E, F$, are different not only from the $X,Y,Z$ that Slodowy mentions, but different from the $x,y,z$ that McKay mentions on the same page as his construction of $V, E, F$.

I wouldn’t normally appeal to the fact that he should have said something like $V^2 + E^3 + F^4 = 0$ if he meant that, rather than changing notation mid-stream and saying $x^2 + y^3 + z^4 = 0$, because these kinds of shifts happen all the time. But having repeated his construction, and found no way to match up his $x,y,z$ with his $V,E,F$ that makes the equation true, it seems reasonable to conclude that he really wasn’t referring to the same three things, after all.

• Mike Doherty says:

The polynomials Kirby and Scharleman give are (up to numerical factors) on pages 60 and 61 of Klein “Lectures on the Icosahedron…”.

I wonder if the Kirby and Scharleman paper has a page missing between pages 128 and 129 – like Bruce Bartlett I am puzzled by the transition between these two pages.

• Greg Egan says:

In this thesis:

http://math.ucr.edu/home/baez/joris_van_hoboken_platonic.pdf

(which is extremely similar to Slodowy in places, but has a few more details, or maybe just paraphrases Slodowy in a way that I find marginally easier to understand …), there’s a description of the construction of invariants that satisfy the equations everyone talks about. This is in section 4, “Invariant theory of binary polyhedral groups”, starting from p12.

Apparently, to get these particular invariants requires understanding “semi-invariants” and tweaking the McKay construction in some fashion. I haven’t really come to grips with the details, but it does seem clear that in the octahedral and tetrahedral cases, you need to do something slightly different than McKay does on the ADE page.

• Greg Egan says:

I just wanted to mention another reference, “Lectures on representations of finite groups and invariant theory” by Dmitri I. Panyushev, that deals with this subject in a lot more detail.

• John Baez says:

Thanks, Greg! Joris Hoboken’s thesis is very nice, which is why I’d taken the liberty of putting it on my website — it was freely available, but in a way that made me afraid it would disappear someday.

I hadn’t gotten around to looking into it for this issue. I’ll look at it soon.

It’s fascinating that the ‘obvious’ way to get invariants works perfectly for the icosahedron, but apparently fails to give a generating set of invariants for the octahedron and tetrahedron. I’m mainly interested in the icosahedron, but now I’m dying to know what trick is required to handle the other two cases. The tetrahedron has an obvious ‘excuse’ for requiring some trick or other, but not the cube.

I sometimes feel I’m spending my life catching up with Felix Klein. He had a real knack for finding the most beautiful stuff.

While failing to find the formulas for those invariants, I learned something nice from Slowody’s paper. Take the complex surface $S$ we get from the icosahedron:

$p_1^2 + p_2^3 + p_3^5 = 0$

It’s smooth except at the origin, where it has a singularity. There’s a smooth complex surface $\tilde{S}$ that maps down to $S$ in a way that’s one-to-one except at the origin. The points in $\tilde{S}$ that map to the origin form 8 copies of the Riemann sphere.

Draw a dot for each of these spheres. Draw an edge between dots when two of these spheres intersect. You get the $\mathrm{E}_8$ Dynkin diagram!

The same idea works in the other cases, giving $\mathrm{E}_7$ for the octahedron and $\mathrm{E}_6$ for the tetrahedron.

• John Baez says:

Greg wrote:

Apparently, to get these particular invariants requires understanding “semi-invariants” and tweaking the McKay construction in some fashion.

I’m trying to understand this, and it looks like a “semi-invariant” is something that’s invariant “up to a phase”, or more generally up to multiplication by some complex number. So, he’s looking for polynomial functions

$P : \mathbb{C}^2 \to \mathbb{C}$

that obey

$P(g v) = \alpha(g) P(v)$

for all $g$ in our finite group $\Gamma$ and all $v\in \mathbb{C}^2,$ where $\alpha(g)$ is some complex number depending on $g.$ Presumably you can find such semi-invariants and then tweak them a bit to get invariants.

It’s easy to check that $\alpha$ winds up obeying

$\alpha(gh) = \alpha(g) \alpha(h)$

So, you only get semi-invariants with nontrivial $\alpha$ when there are nontrivial homomorphisms

$\alpha : \Gamma \to \mathbb{C}^*$

where $\mathbb{C}^*$ is the multiplicative group of nonzero complex numbers.

Since $\mathbb{C}^*$ is abelian, $\alpha$ must comes from some homomorphism out of the abelianization of $\Gamma.$ That’s the group you get by taking $\Gamma$ and modding out by all commutators $g h g^{-1} h^{-1}.$

The upshot will soon be revealed…

• John Baez says:

… so, there can be a ‘semi-invariant’ polynomial for the finite group $\Gamma$ that’s not an actual invariant only if the abelianization of $\Gamma$ is nontrivial!

For the tetrahedron the relevant group $\Gamma$ is the 24-element binary tetrahedral group, the double cover of the rotational symetries of the tetrahedron. This is also $\mathrm{SL}(2,\mathbb{F}_3),$ and its abelianization is nontrivial: it’s $\mathbb{Z}/3$ according to GroupProps.

For the octahedron the relevant group $\Gamma$ is the 48-element binary octahedral group, the double cover of the rotational symetries of the octahedron. Its abelianization is nontrivial: it’s $\mathbb{Z}/2$ according to GroupProps (if you read between the lines).

For the icosahedron the relevant group $\Gamma$ is the 120-element binary icosahedral group, the double cover of the rotational symetries of the icosahedron. This is also $\mathrm{SL}(2,\mathbb{F}_5),$ and its abelianization is trivial: according to GroupProps.

So, the icosahedral case is ‘better’ than the other two: every semi-invariant is invariant. And the reason is that every element of the binary icosahedral group is a commutator $ghg^{-1}h^{-1}$, unlike the other two cases.

• Greg Egan says:

John wrote:

So, the icosahedral case is ‘better’ than the other two: every semi-invariant is invariant.

Thanks for explaining that!

What also helped me grasp the distinction here was going back and checking what happens if you compute the product of the dot products of ${x,y,z}$ with three linearly independent unit normals to faces of a cube, e.g.:

$P_f(x,y,z) = x y z$

Unlike the case with the faces of an icosahedron, this turns out to be a semi-invariant, only invariant up to a sign. And that’s true even if we restrict to the 24-element group of rotational symmetries of the cube.

It’s worth pointing out that if we were studying the 120-element group of rotations and reflections of an icosahedron, rather than the binary icosahedral group of the same order, there is an obvious semi-invariant: the product of the dot products of ${x,y,z}$ with 15 linearly independent unit vectors that pass through the centres of edges of the icosahedron. This changes sign under an inversion.

• John Baez says:

That cube example is nice.

I had some fun today trying to grok why the abelianization of the binary tetrahedral group is $\mathbb{Z}/3.$ I didn’t completely succeed, but here’s the idea:

Suppose you have a ‘spinor tetrahedron’. This is like an ordinary regular tetrahedron except that its symmetry group is the double cover of the usual rotational symmetry group of the tetrahedron. So, you have to turn it around twice around any axis for it to reach its original state.

For example, if we take our axis to be the line from a vertex to the midpoint of the opposite face, the spinor tetrahedron has not 3 but 6 symmetries that preserve this axis. You can turn it 120° around this axis, and that’s a symmetry, but you have to do this 6 times before you get back where you started.

Similarly for the axis between midpoints of opposite edges: turning the spinor tetrahedron 180° around such an axis is a symmetry, but you have to do this 4 times before you get back where you started.

Now, the thing to grok is this. We can assign to any symmetry of the spinor tetrahedron a ‘twist number’, which is an integer mod 3. The twist number is characterized by two properties:

• if you turn the spinor tetrahedron 120° around any axis from a vertex to the midpoint of the opposite face, and turn it clockwise with the vertex pointing towards you, this symmetry has twist number 1.

• if you do one symmetry $g$ and then another $h,$ the twist number of the composite symmetry $g h$ is the sum of the twist numbers for $g$ and $h.$

It’s also important to grok that while the twist number is well-defined mod 3, it wouldn’t be well-defined mod 6.

(There’s a similar puzzle for the cube, where the twist numbers are well-defined mod 2, but I didn’t think about that.)

• Bruce Bartlett says:

For further reference. I pointed out the article of Kirby and Scharlemann above as a nice reference, observing that something funny happens to the text between page 128 and 129, which was also picked up on by Mike Doherty.

There is a Russian translation of that paper, which fills in the gaps.

I am told that the Russian text at the top of page 150 reads

““The group $I^* < SU(2)$ has the following elements, where $epsilon =$ …"

14. jessemckeown says:

So, a thing that has been niggling me for a little is: the generic quintic (even the generic depressed quintic) has no root at zero, so a rescaling of $x$ allows one to fix the constant term equal to $1$, which means the generic depressed singular quintic mostly lives in a 1-parameter family — but there is a small trouble in that there isn’t a good single parameter that captures these features AND the cusps we’re looking at: so I had to excurse via a Moebius transform to get this picture:

• jessemckeown says:

wordpress is funny. the link is actually preserved in the previous comment, but its contents (an img tag) have been removed…

• Scott Hotton says:

That is interesting. One of the curve’s cusps appears to meet at a point of inflection.

Here is a picture of the cubical parabola and one of its involutes
under the Mobius transformation $(z - x_0)/(z+x0)$.

The red curve is the image of the cubical parabola. The blue curve is the image of the involute. The point $x_0$ is the location of the higher order cusp of the involute before the transformation is applied to the curves. In this case
$x_0 = 5$. The image of the involute of the cubical parabola is not the same as the involute of the image of the cubical parabola.

The image of the cubical parabola and its involute meet at the point 1 in the sense that they meet at the point at infinity and the image of the point at infinity under the Mobius transformation is 1 for any $x_0 \neq 0$.

The image of the cubical parabola’s inflection point is
-1 for any $x_0 \neq 0$. The image of the higher
order cusp is 0 for any $x_0 \neq 0$. The image of the lower order cusp is on the image of the cubical parabola but its not on the cubical parabola’s inflection point. It would be on the cubical parabola’s inflection point in the $x_0 = 0$ case
except the involute’s cusps go away when $x_0 =0$.

I would have posted the picture if I knew how.

• John Baez says:

I fixed Jesse’s picture.

For a long time I thought it was hopeless for commenters to post pictures here: if anyone but me posts a comment containing the usual html for an image:

<img src = “…”>

that portion of the comment is deleted.

Then suddenly I noticed that if someone includes a URL ending in .jpg or .gif, the blog magically displays the picture! I don’t know if that feature is new.

But apparently this doesn’t work for a .png file?

• Scott Horton says:

That is interesting. One of the curve’s cusps appears to meet at a point of inflection.

Here is a picture of the cubical parabola and one of its involutes
under the Mobius transformation $(z - x_0)/(z+x_0)$.

The red curve is the image of the cubical parabola. The blue curve is the image of the involute. The point $x_0$ is the location of the higher order cusp of the involute before the transformation is applied to the curves. In this case
$x_0 = 5$. The image of the involute of the cubical parabola is not the same as the involute of the image of the cubical parabola.

The image of the cubical parabola and its involute meet at the point 1 in the sense that they meet at the point at infinity and the image of the point at infinity under the Mobius transformation is 1 for any $x_0 \neq 0$.

The image of the cubical parabola’s inflection point is -1 for any $x_0 \neq 0$. The image of the higher order cusp is 0 for any $x_0 \neq 0$. The image of the lower order cusp is on the image of the cubical parabola but its not on the cubical parabola’s inflection point. It would be on the cubical parabola’s inflection point in the $x_0 = 0$ case except the involute’s cusps go away when $x_0 =0$.

• jessemckeown says:

Yes, that extra intersection — I’m sure it’s really in my picture, because of how the inversion was necessary to get that very-singular cusp; I’m a tad worried that the resulting cusp picture is actually of a 7/2-cusp, because blowing up to resolve the intersection would stretch the cusp as well.

• John Baez says:

I fixed Jesse’s picture and some other stuff.

For a long time I thought it was hopeless for commenters to post pictures here: if anyone but me posts a comment containing the usual html for an image:

<img src = “…”>

that portion of the comment is deleted.

Then suddenly I noticed that if someone includes a URL ending in .jpg or .gif, the blog magically displays the picture!

I don’t know if that feature is new.

Jesse included a .png, but that works too. Jesse’s mistake was including his .png inside

<a href = “…”>

Simply type the URL, and the picture will appear.

15. Mike Doherty says:

I tried to post last night that the polynomials are on pages 60 and 61 of Klein’s “Lectures on the icosahedron..”, but my post doesn’t seem to have worked.

• John Baez says:

Your comment is visible now; I had to approve it since you were a first-time commenter, and somehow I overlooked it.

Thanks! I obviously need to get ahold of that book; I don’t have it with me now.

16. Bruce Bartlett says:

So, one nice thing that happened recently on this comments thread has been Greg Egan confirming John’s conjecture about the relationship between the invariant polynomials for $\hat{G}$ acting on $\mathbb{C}^2$ and the invariant polynomials for $G$ acting on $\mathbb{C}^3$. Here $G$ is the icosahedral group and $\hat{G} \subset SU(2)$ is its double cover. There seems to remain an issue about whether Mckay’s polynomials are the same as Slowodny’s.

But I’d like to return to the bigger picture, in particular to Claim A of Arnold:

Claim A. The variety of irregular orbits of the action of the icosahedral group $H_3$ on $\mathbb{C}^3$ is isomorphic to the set of polynomials of the form $x^5 + ax^4 + bx^2 + c$ having multiple roots.

There seems to be an even more important claim lurking in the background here, relating the entire orbit space (not just the space of irregular orbits) of reflection groups to spaces of polynomials. The part of the pattern that we know so far seems to go like this:

The orbit space of the action of $A_n$ on $\mathbb{C}^n$ is naturally isomorphic to the space of polynomials of the form $x^{n+1} + a_1 x^{n-1} + \cdots + a_n$.
The orbit space of the action of $B_n$ on $\mathbb{C}^n$ is naturally isomorphic to the space of polynomials of the form $x^n + b_1 x^{n-1} + \cdots + b_n$.
The orbit space of the action of $H_3$ on $\mathbb{C}^3$ is naturally isomorphic to the space of polynomials of the form $x^5 + a x^4 + bx^2 + c$.

Now, the natural map in the first isomorphism is just the roots to coefficients map as explained in TWF261.

I don’t really understand the map in the second isomorphism (Arnold explains it in his book but I don’t understand it).

And, none of us seem to understand at all the map in the third isomorphism.

Perhaps this pattern (of relating orbit spaces of reflection groups to spaces of polynomials) holds for all finite reflection groups. John hinted at that in TWF261. But I have been unable to find references on this. Perhaps someone could point me to a book?

• John Baez says:

I’m glad you’re returning to this “bigger picture”. There’s a lot to explore here, but tonight I just have the energy for one point:

Claim A. The variety of irregular orbits of the action of the icosahedral group $\mathrm{H}_3$ on $\mathbb{C}^3$ is isomorphic to the set of polynomials of the form $x^5 + ax^4 + bx^2 + c$ having multiple roots.

Arnol’d indeed claims this, but Greg showed that this claim is, strictly speaking, false. The reason is that all polynomials $x^5 + ax^4 + bx^2 + c$ with $c= 0$ have zero as a repeated root! Only after we omit the plane $c = 0$ does Greg get a surface that looks diffeomorphic to variety of irregular orbits of the action of the icosahedral group $\mathrm{H}_3$ on $\mathbb{C}^3.$

So, this wrinkle must be taken into account!

• Bruce Bartlett says:

Yes, good point.

Arno’ld attributes the proof of Claim A to Lyashko, and we’re pretty sure he means this paper. Now there’s lots of cool stuff in that paper but I have been unable to find any mention of a quintic polynomial of the form $x^5 + ax^4 + bx^2 + c$. What is going on?

• John Baez says:

When Arnol’d says Lyashko proved X, I don’t think he means “I read a paper in which Lyashko proved something”. I think he means “Lyashko gave a talk in my seminar, in which he proved X.”

Remember, in the good old days before the collapse of the Soviet Union and the exodus of mathematicians, Russian mathematics was seminar-based. Anyone who proved anything about singularity theory would need to explain it to Arnol’d before it was accepted. Publication was often an afterthought in this tradition. That’s why we’re having trouble getting details.

• Bruce Bartlett says:

I think he means “Lyashko gave a talk in my seminar, in which he proved X.”

Scherbak also credits a very similar / identical result to Lyashko in “Singularities of families of evolvents in the neighborhood of an inflection point”, and he actually cites that paper. So perhaps it is in there somewhere.

• Bruce Bartlett says:

I have finally tracked down further details for Arnold’s Claim A and Claim B, in Arnold’s own papers, though I am yet to absorb them. See page 169 of Singularities of systems of rays, where he talks about it using the notion of “reamers”, and page 2696 of Singularities in variational calculus.

• John Baez says:

17. Bruce Bartlett says:

By the way, here’s the solution to the original puzzle of this post (Prove that the generic involute of a cubical parabola has a cusp of order 5/2 on the straight line tangent to the parabola at the inflection point). It was communicated to me by Andre Henriques. The idea is to simply expand everything in a Taylor series up to order $x^5$.

18. Bruce Bartlett says:

Here is an image of the discriminant of the icosahedral group $\mathrm{H}_3$ obtained using Surfer:

(Click to enlarge.)

The algebraic equation is taken straight from Lyashko, a few lines above equation (1).

• John Baez says:

Excellent! I’ve fallen behind you and Greg, because I’m writing a little paper about ideas that came up in the Diamonds and triamonds thread. But at some point I at least want to summarize what you’ve discovered. I have three posts on Visual Insight coming up: the evolute of the cubical parabola on May 1st (which is all written up), the discriminant of the icosahedral group on May 15th (which needs a lot of work. and will profit from your new discoveries), and the discriminant of that quintic on June 1st (ditto). Maybe I should also add one on the swallowtail!

19. Scott Hotton says:

It looks like Etienne Ghys has given the world a holiday present
in the form of a sensational free English tome:

http://perso.ens-lyon.fr/ghys/accueil/

It’s hard for me to describe but I would say that a prominent theme in the book is the combinatorics of singularities. The cubical parabola is discussed very briefly on page 107. Here are five quotes from the book:

“This is a preliminary version. Comments are most welcome …”

“Amazingly, this example of a cross with identified opposite sides has already been considered by Gauss under the name Doppelring. In his remarkable paper “Gauss als geometer”, Stackel relates a conversation between Gauss and Moebius. Gauss observes that the “Doppelring” has a connected boundary. More interestingly, he notes that one can find two disjoint arcs connecting two linked pairs of points on the boundary. I recall that the impossibility of such a configuration in a disc was the crucial point in his proof of the fundamental theorem of algebra.”

“So, Hipparchus was right: there are a(10) = 2 x 103,049 ways of combining 10 assertions, using OR or AND, in the sense just described. … Most mathematicians, including
myself, have a naive idea about Greek mathematics. We believe that it only consists of Geometry, in the spirit of Euclid. The example of the computation by Hipparchus of the tenth Schroeder number may be a hint that the Ancient Greeks had developed a fairly elaborate understanding of combinatorics …”

“We will see that the collection of all singularities, up to homeomorphisms, can be seen as a singular operad and this helps understanding the global picture.

“In his review on the book by Markl, Shinder and Stasheff on operads, John Baez explains one of the motivations for operads.

‘Most homotopy theorists would gladly sell their souls for the ability to compute the homotopy groups of an arbitrary space.'”

• Bruce Bartlett says:

Yes he has given us all a fantastic gift!

• John Baez says:

Hey, this sounds fun. By the way, there’s an interesting operad connected to one of the counting problems described by Hipparchus. I wrote about some of Etienne Ghys’ work on this topic:

• John Baez, The Hipparchus operad, The n-Category Café, 1 April 2013.

(Not an April Fool’s joke!)