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 :

It’s a silly name. I guess looked at 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 on a curve Take a string of length nail down one end at and wind the string along Then the free end of your string traces out a curve

is called an **involute** of It consists of all the points you can get to from by a path of length that doesn’t cross

So, Arnol’d’s puzzle concerns the involute of the curve

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

He wants you to wind the string along the curve forming an involute. And he wants you to see what the involute does when it crosses the line

This is a bit tricky, since the region is not convex. If you nail your string down at , your string will have to start out *above* the curve But when the free end of your string crosses the line the story changes. Now your string will need to go *below* the curve

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

A **cusp of order 5/2** is a singularity in an otherwise smooth curve that looks like 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 looks like this:

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

It looks like this:

It’s less symmetrical than the curve 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 It also has a less pointy cusp where it hits the red curve 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 Remember, I wrote:

If you nail your string down at , your string will have to start out

abovethe curve But when the free end of your string crosses the line , the story changes. Now your string will need to gobelowthe curve

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 in bold, and three involutes of this curve. One involute is not generic: it goes through the special point The other two are. They each have a cusp of order 5/2 where they hit the line but also a cusp of order 3/2 where they hit the curve 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 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

parametrized by arclength, so

for all then its involute is the curve

given by

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.

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

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.

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

noticedalgebraic curve with piecewise algebraic arclength, around 1659. Rational hypocycloids are further examples (similarto 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.

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.

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.

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 starting from a particular point 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 for the two different branches of the tractrix. Furthermore, these points are ‘at infinity’—or at least way, way up the catenary.)

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

parametrized by arclength, so

for all then its involute is the curve

given by

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

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

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 You can see the catastrophe or ‘perestroika’ that occurs when the involute hits the origin, which is an inflection point of But what matters more to me right now are the cusps of order 5/2 that occur in involutes that

don’thit the origin.Thanks, Marshall.

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

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 = (1,3

t^2)t^4)sp = sqrt(1+9

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 - 3

ti^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)

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

All together – probably not useful:

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:

Simon wrote:

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

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

Take the polynomial

For most values of the 5 roots of this polynomial are all distinct. For some special values of it has ‘repeated roots’: that is, fewer than 5 different roots. Those special values give points 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 that equals zero when

has repeated roots. Unfortunately 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

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

The discriminant of:

is

This in turn has discriminants wrt of:

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

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

(If anyone finds this confusing, they should replace by .) 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 and axes. But I guess that still could be useful.

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

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 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 threediscriminants. So I think the curves:and

all give sets where the surface intersects itself. All

fourdiscriminants (that of the quintic, and the three higher discriminants wrt ) are zero along these curves.If we only want to draw the graph of the discriminant, it should be enough to restrict the coefficients in the cubic

such that the product

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

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

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

The yellow part of the surface is where is single-valued; the green, blue and orange parts are where 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 plane of the blue curve, which is a cubic.

I was

hopingthat 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 plane, you get a cubic: . If you project the red curve that marks the self-intersection of the surface onto the plane, you get another cubic: .

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!

last term should be two terms: ; sorry about that.

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

Marshall drew:

That image may actually be useful. If we look at waves moving in the region 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:

If you want to improve it, please do. You can post it here. This won’t show up on

Visual Insightfor a while, since I have a number of posts already lined up.Simon Burton made a great animation explaining how one particular involute of the curve gets its two cusps: the exciting cusp of order 5/2 where the involute hits the axis, and the less exciting cusp of order 3/2 where the involute hits the curve :

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

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

Thanks, that’s very useful.

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

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).

Let be the location of the tracing point

at . 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 the parameterization for the

involute is

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

to the imaginary axis while leaving the real

axis invariant.

The exciting cusp is now at the origin and the curvature

should be for both branches. The image of the

cubical parabola is a simple closed curve minus the point

. would be a rational function of

if it were not for and .

Since only appears with fourth degree under the

radical in its Taylor expansion about

only contains terms whose power is a multiple

of .

So long as we can algebraically obtain the

first few terms in the Taylor expansion for and

. I got:

This resembles the curve but I do not know

if its close enough to help reveal the hidden icosahedron.

This is great! I’m thinking it over…

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 plane… while Arnol’d’s picture makes it look like they should already be visible in the 3d picture.

Simon wrote:

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 are unchanged (or ‘invariant’) when we apply any of the icosahedron symmetries. Apart from constants, the most obvious one has degree 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 none opposite to each other. Taking the dot product with each of these gives a linear function:

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

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 of 10 faces, none opposite to each other. Taking the dot product with each of these gives a linear function:

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

It’s not utterly obvious that and are invariant. Because we arbitrarily chose one corner from each opposite pair, and similarly one face from each opposite pair, and 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

everypolynomial in that’s invariant under the icosahedron symmetries can be expressed as a polynomial in and Even better, this can be done in aunique way. In other words, don’t obey any polynomial relations likeTo understand the discriminant of the icosahedral symmetry group, we need to think about the function from to sending to

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

More later!

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,

everysymmetry 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:

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

alsolooks like the set of points for which the polynomialhas 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 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 setwhen we apply the functionAnd this set is the discriminant.

Puzzle.How many mirror planes are there?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.

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

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.

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.

Greg wrote:

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

freeWordPress 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.

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

isometrywith the involute construction, and since the projections of the red and blue curves onto the 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.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 plane should be one of the mirror planes.

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

This is as defined by the squared magnitude for , the product of dot products with 6 non-opposite vertices for , and the product of dot products with10 non-opposite face-centres for .

It’s hard to see how this set could be the zero-discriminant set, since the coordinate 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?

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

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:

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

Then he says:

So we are supposed to have 3 diffeomorphic surfaces: one defined using the icosahedron, one defined using the quintic and one defined using the involutes of the cubical parabola. I have not been able to find proofs in the literature!

Actually there is some potentially useful material here:

• O. P. Shcherbak, Wavefronts and reflection groups,

Russian Mathematical Surveys43(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 but the even more exotic case of : 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.

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

I’ve rescaled some of the coordinates. Since and only appear as and , 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 and both map to the -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 -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.

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.

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 discriminant, but he says these two papers study the 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 inFunctional 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 generated by reflections,

Funktsional. Anal. i Prilozhen.17:4(1983), 70–72. English translation inFunctional 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 but also so there could be different ‘real forms’ of the discriminant. You drew the one where are real so but maybe there’s another one where, say, and are real but is imaginary, so that 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.

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

You can get the full zero set in by including the four possibilities where and are either purely real or purely imaginary, and this looks like the complete picture Arnol’d drew:

This is the picture I’d been dreaming of!WOW, GREAT!!!So my idea of letting 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 and 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.

I wrote:

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

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 to the discriminant of the Coxeter group , the symmetry group of the icosahedron. But there’s something simpler that might be related.

First, what’s the discriminant of the Coxeter group ? This group is just the symmetric group on letters, This group acts on in an obvious way, by permuting the coordinate axes. But it also acts on the -dimensional subspace where the coordinates sum to zero:

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

Each point in gives a polynomial like this:

This trick gives all the polynomials of degree whose leading coefficient is 1 and whose roots sum to zero. Two different points of give the same polynomial iff we can get from one to the other by permuting the coordinate .

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

Generically, different points in map to each polynomial in 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

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

where These hyperplanes are also the mirror planes for the Coxeter group!

Specializing to 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 Even better, note that these polynomials are precisely those of the form

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

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

By the way, the group 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 So, there’s a relation between this group and

This is part of a little pattern relating:

the Coxeter group and the Coxeter group,

the Coxeter group and the Coxeter group,

the Coxeter group and the Coxeter group.

For more on that, see “week270”.

But I’m still puzzled about how is getting related to quintics of the special form

John wrote:

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

,

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

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

Simon suggested this:

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.

Simon Burton created some nice pictures of

slicesof the the zero set of the discriminant ofHere is its intersection with the plane :

Here is its intersection with the plane :

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.

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.

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

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.

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

If you follow the surface out to larger 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.

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 of the icosahedral group is diffeomorphic to

Greg Egan has drawn a wondferful picture of . In order to get a picture that looks like Arnold’s picture of , 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 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 of the icosahedral group is 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.

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

Thankfully, we

cantrack 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 generated by reflections”. Although I don’t understand the proof.Thanks very much for trying to summarize the state of play and point out what remains to be done!

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.

Bruce wrote:

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 where is a nontrivial rational function on that is invariant under the symmetries of the icosahedron!

The invariant polynomials 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

I wrote something here about this issue, but I was led naturally to the so-called

depressedquintic, where the quartic term vanishes: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 , which is almost the rotational symmetry group of the icosahedron, namely the alternating group (If we include reflections as symmetries we get not but However, it’s the non-solvable part, the that’s the really big deal here.)

With more work we can massage any quintic into

principalform, where the cubic term also vanishes: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:Apparently Mathematica will do this change of variables for you!

So, all

thesespecial classes of quintics should be closely connected to the alternating group and thus, I expect, the icosahedron.But I don’t know what’s so special about quintics like

Maybe your comment holds the key!

I guess another approach would be to find a diffeomorphism between and that commutes with the actions of , where is the 3-dimensional submanifold of 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.

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

In that case, the subset of that is pointwise fixed by the action on 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 .

Greg wrote:

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 This contains the alternating group the rotational symmetry group of the icosahedron, as a normal subgroup, and the quotient is . But is not the product of and .

• the icosahedron symmetry group, including rotations and reflections, This

isthe product of and .• the binary icosahedral group, meaning the double cover of the rotational symmetry group of the icosahedron. This has as a normal subgroup, and the quotient is

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 with the obvious permutation action of I think it will be a bit better to look at with the obvious permutation action of A point in describes an ordered 5-tuple of roots of a monic quintic

and acts to permute the labels on the roots.

Monicjust mean that the coefficient of the highest-order term is 1. Complex polynomials work better than real ones, so I think we wantOf course the polynomial itself doesn’t know an ordering on its roots. So, the space of monic quintics is But it can also be seen as using the coefficients as coordinates.

On the other hand, doesn’t act as symmetries of the icosahedron; only does. What these groups have in common is their subgroup

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

evenpermutations of these from icosahedron symmetries.So, I don’t think the assumption you mentioned, about in acting as on actually holds if we take this group’s most obvious action on (or .) The element in does not affect a true cross.

OK, but even if we have act as the identity on , an order-5 rotation in fixes a 1-dimensional subspace in both and , but the subspace in 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 , which as a linear operator on the coordinates fixes the subspace of , but the symmetric polynomial condition means .

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

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 .In particular, in that book, he starts with a finite rotation group , and then considers it as acting on the Riemann sphere . The algebra of invariant functions turns out to be freely generated by a single rational function , that is, .

But for our purposes, instead of being interested in the action of on the Riemann sphere, we instead lift to its double cover , so that is thought of as acting on instead of . The algebra of invariants is now generated by three functions with a single relation. When is the icosahedral group, this relation is apparently . I don’t know how to relate these two pictures.

Bruce wrote:

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

sending spinors in to vectors in which actually happen to live in This map takes any spinor to the expected value of its angular momentum:

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

The icosahedral group acts on the spin-1 representation, its double cover 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

By pulling back, I believe we get a map from -invariant polynomial functions on to -invariant polynomial functions on

We’ve got 3 -invariant polynomial functions on namely These will give 3 -invariant polynomial functions on but those must obey (at least) one relation.

That’s my guess about how this relation shows up! I.e., it’s not a relation between the original functions on but the corresponding functions on

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

is not quadratic in the complex sense: if you multiply by , you multiply by not .

Nonetheless there is a

linearintertwining operator so there issomequadratic 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 is isomorphic to its conjugate representation, so the annoying complex conjugate in the inner product is somehow not that big a deal.

It’s weird that I’ve never thought about this much.

Here is a reference which nicely explains the homeomorphism

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.

This looks like a paper worth understanding!

So, these guys write down three rather complicated polynomials on obeying a relation

They are polynomials of degree 30, 20 and 12.

I was hoping to get these from Greg’s icosahedral-invariant polynomials on which have degrees 2, 6, and 10. I was actually hoping they were obtained by composing with a quadratic map But that doesn’t work: the degrees don’t work out.

Hmm! I’m pretty sure Chevalley’s generators for the icosahedral-invariant polynomials on have degrees 2, 6 and 10. The 2 comes from 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

.

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- to spin-1, especially if you take the route where you

definespin-1 as the space of symmetric tensors in , with:for .

Then the quadratic map with:

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

If we choose a suitable orthonormal basis for the symmetric tensors in , we can write , with:

But if we look at our simplest invariant polynomial

we have:

This quantity is

notan 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, , whose matrix in the bases we’re using is:

maps from the spin-1 rep on the tensor product to the spin-1 rep as it acts on . So … what do we get now?

While I’m looking at equivariant quadratic maps from spin- 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 complex matrices, with the action:

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

where:

is traceless, of course, but it also has zero determinant, which corresponds to our previous result:

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 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 zerowhen 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.

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

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 :

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.

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 with

and some variant should work for the tetrahedron, giving polynomials with

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.

I learned these formulas from this page 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

veryrapid 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 andSo, 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 And those mysteries remain mysterious to me!

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.

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:

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):

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

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.

Greg wrote:

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:

He gives a table listing these. Note that every finite subgroup of is conjugate to one inside , since we can average an inner product over the group action to get an inner product that’s preserved by . The only options turn out to be our friends the cyclic groups (corresponding via McKay’s magic method to the Dynkin diagrams), the dihedral groups (corresponding to ), and the double covers of the symmetry groups of the tetrahedron, octahedron and icosahedron (corresponding to ).

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

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

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.

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

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 , and they obey the relationship:

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

doobey a relationship, when suitably normalised, of ), certainlydon’tobey one of the form , for any of the six ways we can assign .And as far as I can tell, exactly the same is true of McKay’s polynomials, obtained from stereographic projection! That is, they

dosatisfy the relation , when suitably normalised, but theydon’tobey one of the form , for any of the six ways we can assign .I’m happy to believe that

somethingobeys the equation … but the invariants that our method yields (and I’m pretty sure that theseare, 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.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:

but none of the six ways of assigning to these polynomials yields a relation of the form:

(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 , are different not only from the that Slodowy mentions, but different from the that McKay mentions on the same page as his construction of .

I wouldn’t normally appeal to the fact that he

shouldhave said something like if he meant that, rather than changing notation mid-stream and saying , because these kinds of shifts happen all the time. But having repeated his construction, and found no way to match up his with his that makes the equation true, it seems reasonable to conclude that he really wasn’t referring to the same three things, after all.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.

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.

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.

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 setof 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 we get from the icosahedron:

It’s smooth except at the origin, where it has a singularity. There’s a smooth complex surface that maps down to in a way that’s one-to-one except at the origin. The points in 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 Dynkin diagram!

The same idea works in the other cases, giving for the octahedron and for the tetrahedron.

Greg wrote:

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

that obey

for all in our finite group and all where is some complex number depending on Presumably you can find such semi-invariants and then tweak them a bit to get invariants.

It’s easy to check that winds up obeying

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

where is the multiplicative group of nonzero complex numbers.

Since is abelian, must comes from some homomorphism out of the abelianization of That’s the group you get by taking and modding out by all commutators

The upshot will soon be revealed…

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

For the tetrahedron the relevant group is the 24-element binary tetrahedral group, the double cover of the rotational symetries of the tetrahedron. This is also and its abelianization is nontrivial: it’s according to GroupProps.

For the octahedron the relevant group is the 48-element binary octahedral group, the double cover of the rotational symetries of the octahedron. Its abelianization is nontrivial: it’s according to GroupProps (if you read between the lines).

For the icosahedron the relevant group is the 120-element binary icosahedral group, the double cover of the rotational symetries of the icosahedron. This is also 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 , unlike the other two cases.

John wrote:

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 with three linearly independent unit normals to faces of a cube, e.g.:

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 with 15 linearly independent unit vectors that pass through the centres of edges of the icosahedron. This changes sign under an inversion.

That cube example is nice.

I had some fun today trying to grok why the abelianization of the binary tetrahedral group is 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 coverof the usual rotational symmetry group of the tetrahedron. So, you have to turn it aroundtwicearound 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

clockwisewith the vertex pointing towards you, this symmetry has twist number 1.• if you do one symmetry and then another the twist number of the composite symmetry is the sum of the twist numbers for and

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.)

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 has the following elements, where …"

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 allows one to fix the constant term equal to , 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:

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

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 .

The red curve is the image of the cubical parabola. The blue curve is the image of the involute. The point is the location of the higher order cusp of the involute before the transformation is applied to the curves. In this case

. 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 .

The image of the cubical parabola’s inflection point is

-1 for any . The image of the higher

order cusp is 0 for any . 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 case

except the involute’s cusps go away when .

I would have posted the picture if I knew how.

Oh it just replaced the URL with the picture.

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?

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 .

The red curve is the image of the cubical parabola. The blue curve is the image of the involute. The point is the location of the higher order cusp of the involute before the transformation is applied to the curves. In this case

. 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 .

The image of the cubical parabola’s inflection point is -1 for any . The image of the higher order cusp is 0 for any . 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 case except the involute’s cusps go away when .

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.

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.

Ah! Thank you.

And, thank you!

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.

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.

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 acting on and the invariant polynomials for acting on . Here is the icosahedral group and 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 on is isomorphic to the set of polynomials of the form 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 on is naturally isomorphic to the space of polynomials of the form .

The orbit space of the action of on is naturally isomorphic to the space of polynomials of the form .

The orbit space of the action of on is naturally isomorphic to the space of polynomials of the form .

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?

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:

Arnol’d indeed claims this, but Greg showed that this claim is, strictly speaking,

false. The reason is that all polynomials with have zero as a repeated root! Only after we omit the plane does Greg get a surface that looks diffeomorphic to variety of irregular orbits of the action of the icosahedral group onSo, this wrinkle must be taken into account!

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 . What is going on?

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.

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.

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.

Hurrah! I’ll read these.

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 .

Here is an image of the discriminant of the icosahedral group obtained using Surfer:

(Click to enlarge.)

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

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 Insightcoming 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!