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

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3The link should go to page 550 of the book, presumably. – Mariano Suárez-Álvarez Dec 21 '09 at 14:42
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Related MO question: Can the unsolvability of quintics be seen in the geometry of the icosahedron?. – Joseph O'Rourke Oct 11 '19 at 12:18
9 Answers
"Geometry of the Quintic" is available for free at my website.
Jerry Shurman

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14To save other people a few clicks, Jerry's webpage is http://people.reed.edu/~jerry/ – James Cranch Jun 10 '14 at 10:59
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.

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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
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.
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.

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3It looks like our identical answers crossed paths! You beat me by one minute, so I'll delete my answer. – Andy Putman Dec 21 '09 at 19:18
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lol, you're right - I caught it just before you deleted. Very gallant of you! and hey, in math, every minute counts. :-) – Alon Amit Dec 21 '09 at 19:58
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2As 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. – Andy Putman Dec 21 '09 at 20:12
In Glimpses of algebra and geometry by Gabor Toth, chapter 25 is devoted to Klein's main result.

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

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1Re: Dr. Shello - yes, Shurman remarked a few months ago on MO he made the text publicly available. This answer predates his release. – David Lehavi Jan 12 '11 at 09:03
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 alfonso@systembit.es

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

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