Discriminant of Restricted Quintic

Zero Set of Discriminant of Restricted Quintic – Greg Egan

This image by Greg Egan shows the set of points $(a,b,c)$ for which the quintic $x^5 + ax^4 + bx^2 + c$ has repeated roots. The plane $c = 0$ has been removed.

The fascinating thing about this surface is that it appears to be diffeomorphic to two other surfaces, defined in completely different ways, which we discussed here:

• Involutes of a cubical parabola.

• Discriminant of the icosahedral group.

The icosahedral group is also called $\mathrm{H}_3$ . In his book The Theory of Singularities and its Applications, V. I. Arnol’d writes:

The discriminant of the group $\mathrm{H}_3$ is shown in Fig. 18. Its singularities were studied by O. V. Lyashko (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. Lyashko has also proved that this surface is diffeomorphic to the set of polynomials $x^5 + ax^4 + bx^2 + c$ having a multiple root.

Figure 18 of Arnold’s book is a hand-drawn version of the surface below:

$$\mathrm{H}_3$ Discriminant - Greg Egan$

$\mathrm{H}_3$ Discriminant – Greg Egan

Arnol’d’s claim that the discriminant of $\mathrm{H}_3$ is diffeomorphic to the set of polynomials $x^5 + ax^4 + bx^2 + c$ having a repeated root is not literally true, since all such polynomials with $c = 0$ have a repeated root, and we need to remove this plane to obtain a surface that looks like the discriminant of $\mathrm{H}_3$ . After this correction his claim seems right, but it still deserves proof.

Puzzle. Can you prove the corrected version of Arnol’d’s claim?

References

Arnol’d’s claim appear on page 29 here:

• Vladimir I. Arnol’d, The Theory of Singularities and its Applications, Cambridge U. Press, Cambridge, 1991.

The following papers are also relevant:

• Vladimir I. Arnol’d, Singularities of systems of rays, Uspekhi Mat. Nauk 38:2 (1983), 77-147. English translation in Russian Math. Surveys 38:2 (1983), 77–176.

• O. Y. Lyashko, Classification of critical points of functions on a manifold with singular boundary, Funktsional. Anal. i Prilozhen. 17:3 (1983), 28–36. English translation in Functional Analysis and its Applications 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 Analysis and its Applications 17:4 (1983), 301–303.

• O. P. Shcherbak, Wavefronts and reflection groups, Uspekhi Mat. Nauk 43:3 (1988), 125–160. English translation in Russian Mathematical Surveys 43:3 (1988), 1497–194.

All these sources discuss the discoveries of Arnol’d and his colleagues relating singularities and Coxeter–Dynkin diagrams, starting with the more familiar $\mathrm{ADE}$ cases, then moving on to the non-simply-laced cases, and finally the non-crystallographic cases related to $\mathrm{H}_2$ (the symmetry group of the pentagon), $\mathrm{H}_3$ (the symmetry group of the icosahedron) and $\mathrm{H}_4$ (the symmetry group of the 600-cell).

5 thoughts on “Discriminant of Restricted Quintic”

James Prichard says:

5 June, 2016 at 18:33

Can Klein’s method for solving the quintic be adapted to eliminate $x^2$ & $x^2$ terms instead of $x^4$ & $x^3$ ?
- John Baez says:
  
  5 June, 2016 at 19:48
  
  I don’t know. I have thought about how the special class of quintics discussed here ( $x^5+ax^4+bx^2+c = 0$ ) is related to the special classes that show up when people try to solve the quintic, but I haven’t any good ideas yet! Klein’s ideas are especially tempting, because they involve the icosahedron. But still, I don’t see what’s going on.
Greg Egan says:

2 September, 2016 at 06:36

An explicit diffeomorphism between the three-dimensional space of invariants of the icosahedral symmetry group $(P,Q,R)$ and the coefficients $(a,b,c)$ of the restricted quintics that maps the discriminant into the discriminant is given by:

$(a,b,c) = (P, \alpha Q + \beta P^3, \gamma R + \delta Q P^2 + \epsilon P^5)$

where the constants $\alpha, \beta, \gamma, \delta, \epsilon$ depend on the precise choice of normalisation for the invariants $P, Q, R$ . If we recall that those invariants (defined here) are homogeneous polynomials of degree 2, 6 and 10 in the coordinates $x, y, z$ , this diffeomorphism is the simplest kind of map that is consistent with the fact that $a, b, c$ are homogenous polynomials of degree 1, 3 and 5 in the roots of the quintic. It’s not hard to see that this map has a constant, non-zero Jacobian determinant, which proves that it is a diffeomorphism, and in fact it has a polynomial inverse of the same general form.

The five constants $\alpha, \beta, \gamma, \delta, \epsilon$ can be determined by identifying three curves that appear in both discriminant surfaces, the lines of cusps of type 5/2 and 3/2 and the line of double points, which all take the general parametric form $(a, A_i a^3, B_i b^5)$ or $(P, C_i P^3, D_i P^5)$ , with four of the constants here equal to zero, in those cases where the curve in question is a straight line along a coordinate axis.
- John Baez says:
  
  2 September, 2016 at 07:15
  
  Excellent! So the Puzzle is solved!
- Greg Egan says:
  
  3 September, 2016 at 00:13
  
  Although the particular diffeomorphism depends on the choice of normalisation for the invariants $P,Q,R$ of the icosahedral symmetry group, it’s possible to make the following identification that remains true regardless of that choice.
  
  If $e_1$ and $e_2$ are two orthogonal unit vectors that pass through edge centres of the icosahedron, then the point on the mirror plane:
  
  $m = \sqrt{\frac{1}{2}\left(\left(1+\frac{1}{\sqrt{5}}\right) a + \sqrt{5} \alpha \right)} e_1 + \sqrt{\frac{1}{2}\left(\left(1-\frac{1}{\sqrt{5}}\right) a – \sqrt{5} \alpha \right)} e_2$
  
  when mapped to the invariant space $(P,Q,R)$ and then to the space of quintics $(a,b,c)$ corresponds to the restricted quintic with a coefficient of $a$ for the degree-4 term and a repeated root of $\alpha$ .