32
$\begingroup$

Reading Serre's letter to Gray , I wonder if now modern expositions of the themes in Klein's book exist. Do you know any?

Improve this question
$\endgroup$
2

9 Answers 9

Reset to default
37
$\begingroup$

"Geometry of the Quintic" is available for free at my website.

Jerry Shurman

Improve this answer
$\endgroup$
1
17
$\begingroup$

I got interested in this subject last year (2011) and just got round to writing up some notes which I hope may be of use. I also have a python script hosted here, which implements Klein's icosahedral solution of the quintic, as well as a brief summary of what it does here.

The geometry is easy to summarise: using a radical transformation, a quintic can be put in the form $y^5 + 5\alpha y^2 + 5\beta y + \gamma = 0$. The vector of ordered roots of such a quintic lies on the quadric surface $\sum y_i = \sum y_i^2 = 0$ in $\mathbb{P}^4$ and the reduced Galois group $A_5$ acts on the two families of lines in this doubly-ruled surface by permuting coordinates. The $A_5$ actions on these families, parameterized by $\mathbb{P}^1$, are equivalent to the action of the group of rotations of an icosahedron on its circumsphere and the quintic thus defines a point in the quotients — the icosahedral invariants of a quintic. Inverting either of these quotients (e.g., using the hypergeometric functions given below) is sufficient to allow us solve quintic (in rational functions).

Here's how it looks for a quintic in the simpler form: $$ y^5 + 5y + \gamma = 0 $$ (In fact any quintic can be put in this form using only radical transformations.)

Given such a quintic, set: $$ \nabla = \sqrt{\gamma^4 + 256}\\ Z = \frac{1}{2\cdot 1728}[2\cdot 1728 + 207\gamma^4 + \gamma^8 - \gamma^2 (81 + \gamma^4)\nabla]\\ z = \frac{{}_2F_1(\frac{31}{60}, \frac{11}{60}; \frac{6}{5}; Z^{-1})} {(1728Z)^{1/5}{}_2F_1(\frac{19}{60}, -\frac{1}{60}; \frac{4}{5}; Z^{-1})} $$ and: $$ f(z) = z(z^{10} + 11z^5 - 1)\\ H(z) = -(z^{20} + 1) + 228(z^{15} - z^5) - 494z^{10}\\ T(z) = (z^{30} + 1) + 522(z^{25} - z^5) - 10005(z^{20} + z^{10})\\ B(z) = -1 - z - 7(z^2 - z^3 + z^5 + z^6) + z^7 - z^8\\ D(z) = -1 + 2z + 5z^2 + 5z^4 - 2z^5 - z^6 $$ Then: $$ y = -\gamma\cdot\frac{f(z)}{H(z)/B(z)} - \frac{7\gamma^2 + 9\nabla}{2\gamma(\gamma^4 + 648)} \cdot\frac{D(z)T(z)}{f(z)^2H(z)/B(z)} $$ is a root.

Replacing $z$ with $e^{2\pi\nu i/5}z$ for $\nu=1, 2, 3, 4$ provides all the other roots.

Even in this rather gross explicit form, the link with regular solids is visible:

  • The roots of $f, H, T$ are, respectively, the locations of the projection of the vertices, face centres, and edge midpoints of a regular icosahedron onto its circumsphere (once this circumsphere has been identified with the extended complex plane by stereographic projection).

  • The roots of the last two polynomials, $B, D$ are, respectively, the locations of the vertices and face centres of a regular cube inscribed in the icosahedron.

Improve this answer
$\endgroup$
2
  • $\begingroup$ The quintic in your answer is $y^5+5y+1$, not $y^5+y+1$? Did you put the 5 in order to simplify the formulas? $\endgroup$
    – coudy
    Commented Sep 4 at 15:14
  • $\begingroup$ @coudy The factor of 5 is there "for historical reasons". In other words, Klein put it in and when I wrote these notes I decided to make it easy for people to compare my formulae with his so I kept it. As to why Klein included it, I'm afraid I don't know (I may once have known ¯\_(ツ)_/¯). $\endgroup$
    – Oliver Nash
    Commented Sep 6 at 13:17
12
$\begingroup$

I covered Klein's "Lectures on the Icosahedron" in a modern way in my doctoral thesis:

Elliptic Curves and Icosahedral Galois Representations, Stanford University (1999) http://www.math.purdue.edu/~egoins/notes/thesis.pdf

A much shorter and more direct exposition is my publication in IMRN:

Icosahedral $\mathbb Q$-Curve Extensions, Mathematical Research Letters 10, 205–217 (2003) http://intlpress.com/site/pub/files/_fulltext/journals/mrl/2003/0010/0002/MRL-2003-0010-0002-00019947.pdf

Improve this answer
$\endgroup$
12
$\begingroup$

There is a (german) new edition of Klein's "Vorlesungen über das Ikosaeder ..." by Peter Slodowy (1993) with a large (about 80 pages) section of comments and remarks about new developments.

Improve this answer
$\endgroup$
11
$\begingroup$

Chapter 5 of McKean and Moll's "Elliptic Curves" explores the circle of ideas around Ikosaeder.I'm not sure if you'd consider this sufficiently "modern" - it's certainly a contemporary book but it doesn't use, say, scheme-theoretic language.

Improve this answer
$\endgroup$
3
  • 3
    $\begingroup$ It looks like our identical answers crossed paths! You beat me by one minute, so I'll delete my answer. $\endgroup$
    – Andy Putman
    Commented Dec 21, 2009 at 19:18
  • $\begingroup$ lol, you're right - I caught it just before you deleted. Very gallant of you! and hey, in math, every minute counts. :-) $\endgroup$
    – Alon Amit
    Commented Dec 21, 2009 at 19:58
  • 2
    $\begingroup$ As an aside to everyone else -- McKean and Moll's book is really beautiful! I read large chunks of it as an undergraduate, and I still go back to it periodically. $\endgroup$
    – Andy Putman
    Commented Dec 21, 2009 at 20:12
11
$\begingroup$

In Glimpses of algebra and geometry by Gabor Toth, chapter 25 is devoted to Klein's main result.

Improve this answer
$\endgroup$
5
$\begingroup$

There is the outrageously expensive "Geometry of the quintic" by Jerry Shurman, which discusses both Klein and Doyle-McMullen approaches (and then some more).

Improve this answer
$\endgroup$
4
  • $\begingroup$ "expensive", yes. $\endgroup$
    – lhf
    Commented Dec 21, 2009 at 22:51
  • $\begingroup$ since it's softcover with horrible print quality - yes it is. $\endgroup$
    – David Lehavi
    Commented Dec 22, 2009 at 5:07
  • 4
    $\begingroup$ See Jerry Shurman's answer for a link to a free copy ;) $\endgroup$
    – Dr Shello
    Commented Jan 12, 2011 at 8:33
  • 1
    $\begingroup$ Re: Dr. Shello - yes, Shurman remarked a few months ago on MO he made the text publicly available. This answer predates his release. $\endgroup$
    – David Lehavi
    Commented Jan 12, 2011 at 9:03
3
$\begingroup$

In my PH Thesis work http://systembit.es/schwarz.htm

In my papier I have asociatted a Riemann Surface to each Schwarz function triangle. After that ,I´ve got genus and geometric density of spherical tesselation expresions in a new method. Then I see Their Poincare Groups ( of each above Riemann Surfaces) as index normal finite subgroup of Г(2) (Thanks to Modular Function Lambda). Then I calculate signature of these fuchisian Groups. Finally I see there are only nine ( of above Riemann Surfaces) more Dihedrical cases.

I think my idea is a new interpretation of Schwarz triangles , different one to the Famous Schwarz Classification based on 14 Schwarz triangles +Dihedrical cases.

Alfonso García [email protected]

Improve this answer
$\endgroup$
2
$\begingroup$

You could also take a look at Section 1.6 of Finite Mobius Groups, Immersion of Spheres, and Moduli, by Gabor Toth.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .