From: baez@math.removethis.ucr.andthis.edu (John Baez)
Subject: This Week's Finds in Mathematical Physics (Week 283)
Organization: University of California, Riverside
Sender: baez@math.removethis.ucr.andthis.edu (John Baez)
Newsgroups: sci.physics.research,sci.physics,sci.math
Also available at http://math.ucr.edu/home/baez/week283.html
November 10, 2009
This Week's Finds in Mathematical Physics (Week 283)
John Baez
We had a great AMS meeting this weekend at UCR, with far too many
interesting talks going on simultaneously. For example, there were
two sessions on math related to knot theory, one on operator algebras,
one on noncommutative geometry, and one on homotopy theory and higher
algebraic structures! If I could clone myself, I'd have gone to all of
them.
I'd like to discuss some of the talks, and maybe even point you to some
videos. But the videos aren't available yet, so for now I'll just
summarize my own talk on "Who Discovered the Icosahedron", and some
geometry related to that. I'll conclude with a puzzle.
But first - the astronomy pictures of the week!
Galaxies are beautiful things, and there are lots of ways to enjoy
them. Here's the Milky Way in visible light - a detailed panorama
created from over 3000 individual pictures, carefully calibrated to
show large dust clouds:
1) Axel Mellinger, All-sky Milky Way panorama 2.0,
http://home.arcor.de/axel.mellinger/
You can see even more structure in this infrared panorama of the Milky
Way, created by the Spitzer Space Telescope:
2) Astronomy Picture of the Day, GLIMPSE the Milky Way,
http://apod.nasa.gov/apod/ap051216.html
The bright white splotches are star-forming regions. The greenish
wisps are hot interstellar gas. The red clouds are dust and organic
molecules like polycyclic aromatic hydrocarbons (see "week258"). The
darkest patches are regions of cool dust too thick for Spitzer to see
through.
But here's my favorite: the Andromeda Galaxy in viewed in ultraviolet
light:
3) Astronomy Picture of the Day, Ultraviolet Andromeda,
http://apod.nasa.gov/apod/ap090917.html
This was taken by Swift, NASA's ultraviolet satellite telescope.
At this frequency, young hot stars and dense star clusters dominate
the view. It's sort of ghostly looking, no?
Now for my talk on the early history of the icosahedron. This talk on
the early history of the icosahedron continues the tale begun in
"week236" and "week241". Someday it'll get folded into a paper on
special properties of the number 5, and 5-fold symmetry:
4) John Baez, Who discovered the icosahedron?, talk at the Special
Session on History and Philosophy of Mathematics, 2009 Fall Western
Section Meeting of the AMS, November 7, 2009.
Available at http://math.ucr.edu/home/baez/icosahedron/
The dodecahedron and icosahedron are the most exotic of the Platonic
solids, because they have 5-fold rotational symmetry - a possibility
that only exists for regular polytopes in 2, 3 or 4 dimensions. The
dodecahedron and icosahedron have the same symmetry group, because
they are Poincare duals: the vertices of one correspond to faces of
the other. But the icosahedron was probably discovered later. As
Benno Artmann wrote:
The original knowledge of the dodecahedron may have come from
crystals of pyrite, but in contrast the icosahedron is a pure
mathematical creation.... It is the first realization of an
entity that existed before only in abstract thought. (Well,
apart from the statues of gods!)
I'm not sure it's really anything close to the first "realization of
an entity that existed before only in abstract thought". But
it may have been the first "exceptional" object in mathematics -
roughtly speaking, an entity that doesn't fit into any easy pattern,
which is discovered as part of proving a classification theorem!
Other exceptional objects include the simple Lie group E8, and the
finite simple group M12. Intriguingly, many of these exceptional
objects" are related. For example, the icosahedron can be used to
construct both E8 and M12. But the first interesting classification
theorem was the classification of regular polyhedra: convex polyhedra
with equilateral polygons as faces, and the same number of faces
meeting at each vertex. This theorem appears almost at the end of
the last book of Euclid's Elements - Book XIII. It shows that the only
possibilities are the Platonic solids: the tetrahedron, the cube, the
octahedron, the dodecahedron and the icosahedron. And according to
traditional wisdom, the results in this book were proved by Theatetus,
who also discovered the icosahedron!
Indeed, Artmann cites an "an ancient note written in the margins of
the manuscript" of Book XIII, which says:
In this book, the 13th, are constructed the five so-called Platonic
figures which, however, do not belong to Plato, three of the five
being due to the Pythagoreans, namely the cube, the pyramid, and
the dodecahedron, while the octahedron and the icosahedron are due
to Theaetetus.
You may know Theaetetus through Plato's dialog of the same name, where
he's described as a mathematical genius. He's also mentioned in
Plato's dialogue called the Sophist. In the Republic, written around
380 BC, Plato complained that not enough is known about solid geometry:
... and for two reasons: in the first place, no government places
value on it; this leads to a lack of energy in the pursuit of it,
and it is difficult. In the second place, students cannot learn it
unless they have a teacher. But then a teacher can hardly be
found....
Theaetetus seems to have filled the gap: he worked on solid geometry
between 380 and 370 BC, perhaps inspired by Plato's interest in the
subject. He died from battle wounds and dysentery in 369 after Athens
fought a battle with Corinth.
But how certain are we that Theatetus discovered - or at least studied -
the icosahedron? The only hard evidence seems to be this "ancient
note" in the margins of the Elements. But who wrote it, and when?
First of all, if you hope to see an ancient manuscript by Euclid with
a scribbled note in the margin, prepare to be disappointed! All we
have are copies of copies of copies. The oldest remaining fragments
of the Elements date to centuries after Euclid's death: some from a
library in Herculaneum roasted by the eruption of Mount Vesuvius in 79
AD, a couple from the Fayum region near the Nile, and some from a
garbage dump in the Egyptian town of Oxyrhynchus.
There are various lines of copies of Euclid's Elements. Comparing
these to guess the contents of the *original* Elements is a difficult
and fascinating task. Unfortunately, in the fourth century AD, the
Greek mathematician Theon of Alexandria - Hypatia's dad - made a copy
that became extremely popular. So popular, in fact, that for many
centuries European scholars knew no line of copies that hadn't passed
through Theon! And Theon wasn't a faithful copyist: he added extra
propositions, lengthened some proofs, and omitted a few things too.
It seems he wanted to standardize the language and make it easier to
follow. This may have helped people trying to learn geometry -
but certainly not scholars trying to understand Euclid.
In 1808, Francois Peyrard made a marvelous discovery. He found that
the Vatican library had a copy of Euclid's Elements that hadn't
descended through Theon! This copy is now called "P". It dates back
to about 850 AD. I would love to know how Peyrard got his hands on it.
One imagines him rooting around in a dusty basement and opening a
trunk... but it seems that Napoleon somehow took this manuscript from
the Vatican to Paris.
In the 1880s, the great Danish scholar Johan Heiberg used "P" together
with various "Theonine" copies of the Elements to prepare what's still
considered the definitive Greek edition of this book. The
all-important English translation by Thomas Heath is based on this.
As far as I can tell, "P" is the only known non-Theonine copy of
Euclid except for the fragments I mentioned. Heath also used these
fragments to prepare his translation.
This is just a quick overview of a complicated detective story.
As always, the fractal texture of history reveals more complexity
the more closely you look.
Anyway, Heath thinks that Geminus of Rhodes wrote the "ancient note"
in the Elements crediting Theatetus. I'm not sure why Heath thinks
this, but Geminus of Rhodes was a Greek astronomer and mathematician
who worked during the 1st century BC.
In his charming article "The discovery of the regular solids", William
Waterhouse writes:
Once upon a time there was no problem in the history of the regular
solids. According to Proclus, the discoveries of Pythagoras
include "the construction of the cosmic solids," and early
historians could only assume that the subject sprang full-grown
from his head. But a better-developed picture of the growth
of Greek geometry made such an early date seem questionable, and
evidence was uncovered suggesting a different attribution. A
thorough study of the testimony was made by E. Sachs, and her
conclusion is now generally accepted: the attribution to Pythagoras
is a later misunderstanding and/or invention.
The history of the regular solids thus rests almost entirely on a
scholium to Euclid which reads as follows:
"In this book, the 13th, are constructed the 5 figures called
Platonic, which however do not belong to Plato. Three of these
5 figures, the cube, pyramid, and dodecahedron, belong to the
Pythagoreans; while the octahedron and icosahedron belong to
Theaetetus."
Theaetetus lived c. 415-369 B.C., so this version gives a
moderately late date; and it has the considerable advantage of
seeming unlikely. That is, the details in the scholium are not
the sort of history one would naively conjecture, and hence it
is probably not one of the stories invented in late antiquity.
As van der Waerden says, the scholium is now widely accepted
"precisely because [it] directly contradicts the tradition which
used to ascribe to Pythagoras anything that came along."
But probability arguments can cut both ways, and those scholars
who hesitate to accept the scholium do so primarily because it
seems too unlikely. There have been two main sticking places:
first, the earliness of the dodecahedron in comparison with the
icosahedron; and second, the surprising lateness of the octahedron.
The first objection, however, has been fairly well disposed of.
The mineral pyrite (FeS2) crystallizes most often in cubes and
almost-regular dodecahedra; it is quite widespread, being the most
common sulphide, and outstanding crystals are found at a number
of spots in Italy. Moreover it regularly occurs mixed with the
sulphide ores, and underlying the oxidized ores, of copper; these
deposits have been worked since earliest antiquity. Thus natural
dodecahedra were conspicuous, and in fact they did attract
attention: artificial dodecahedra have been found in Italy dating
from before 500 BC. Icosahedral crystals, in contrast, are much
less common. Hence there is no real difficulty in supposing that
early Pythagorean geometers in Italy were familiar with dodecahedra
but had not yet thought of the icosahedron.
Indeed, while I've heard that iron pyrite forms "pseudoicosahedra",
I've never seen one, while the "pyritohedra" resembling regular
dodecahedra are pretty common.
The puzzle of why the octahedron showed up so late seems to have this
answer: it was known earlier, but it was no big deal until the concept
of regular polyhedron was discovered! As Waterhouse says, the discovery
of the octahedron would be like the discovery of the 4rd perfect number.
Only the surrounding conceptual framework makes the discovery
meaningful.
So far, so good. But maybe the Greeks were not the first to discover
the icosahedron! In 2003, the mathematicians Michael Atiyah and Paul
Sutcliffe wrote:
Although they are termed Platonic solids there is convincing
evidence that they were known to the Neolithic people of Scotland
at least a thousand years before Plato, as demonstrated by the stone
models pictured in Fig. 1 which date from this period and are kept
in the Ashmolean Museum in Oxford.
Various people including John McKay and myself spread this story without
examining it very critically. I did read Dorothy Marshall's excellent
paper "Carved stone balls", which catalogues 387 carved stone balls
found in Scotland, dating from the Late Neolithic to Early Bronze Age.
It has pictures showing a wide variety of interesting geometric
patterns carved on them, and maps showing where people have found
balls with various numbers of bumps on them. But it doesn't say
anything about Platonic solids.
In March of 2009, Lieven le Bruyn posted a skeptical investigation of
Atiyah and Sutcliffe's claim. For starters, he looked hard at the
photo in their paper:
... where's the icosahedron? The fourth ball sure looks like one
but only because someone added ribbons, connecting the centers of
the different knobs. If this ribbon-figure is an icosahedron, the
ball itself should be another dodecahedron and the ribbons illustrate
the fact that icosa- and dodecahedron are dual polyhedra. Similarly
for the last ball, if the ribbon-figure is an octahedron, the ball
itself should be another cube, having exactly 6 knobs. Who did adorn
these artifacts with ribbons, thereby multiplying the number of
"found" regular solids by two (the tetrahedron is self-dual)?
Who put on the ribbons? Lieven le Bruyn traced back the photo to
Robert Lawlor's 1982 book Sacred Geometry. In this book, Lawlor wrote:
The five regular polyhedra or Platonic solids were known and
worked with well before Plato's time. Keith Critchlow in his
book Time Stands Still presents convincing evidence that they
were known to the Neolithic peoples of Britain at least 1000
years before Plato. This is founded on the existence of a number
of spherical stones kept in the Ashmolean Museum at Oxford.
Of a size one can carry in the hand, these stones were carved into
the precise geometric spherical versions of the cube, tetrahedron,
octahedron, icosahedron and dodecahedron, as well as some additional
compound and semi-regular solids...
But is this really true? Le Bruyn discovered that the Ashmolean owns
only 5 Scottish stone balls - and their webpage shows a photo of them,
which looks quite different than the photo in Lawlor's book! They
have no ribbons on them. More importantly, they're different shapes!
The Ashmolean lists their 5 balls as having 7, 6, 6, 4 and 14
knobs, respectively - nothing like an icosahedron.
And here is where I did a little research of my own. The library at
UC Riverside has a copy of Keith Critchlow's 1979 book Time Stands
Still. In this book, we see the same photo of stones with ribbons
that appears in Lawlor's book - the photo that Atiyah and Suttcliffe
use. In Critchlow's book, these stones are called "a full set of
Neolithic 'Platonic solids'". He says they were photographed by one
Graham Challifour - but he gives no information as to where they came
from!
And Critchlow explicitly denies that the Ashmolean has an icosahedral
stone! He writes:
... the author has, during the day, handled five of these
remarkable objects in the Ashmolean museum.... I was rapt
in admiration as I turned over these remarkable stone objects
when another was handed to me which I took to be an icosahedron....
On careful scrutiny, after establishing apparent fivefold symmetry
on a number of the axes, a count-up of the projections revealed 14!
So it was not an icosahedron.
It seems the myth of Scottish balls shaped like Platonic solids
gradually grew with each telling. Could there be any truth to it?
Dorothy Marshall records Scottish stone balls with various numbers
of knobs, from 3 to 135 - but just two with 20, one at the National
Museum in Edinburgh, and one at the Kelvingrove Art Gallery and Museum
in Glasgow. Do these look like icosahedra? I'd like to know. But
even if they do, should we credit Scots with "discovering the
icosahedron"? Perhaps not.
So, it seems the ball is in Theaetetus' court.
Here are some references:
The quote from Benno Artmann appeared in a copy of the AMS Bulletin
where the cover illustrates a construction of the icosahedron:
5) Benno Artmann, About the cover: the mathematical conquest of
the third dimension, Bulletin of the AMS, 43 (2006), 231-235.
Also available at
http://www.ams.org/bull/2006-43-02/S0273-0979-06-01111-6/
For more, try this wonderfully entertaining book:
6) Benno Artmann, Euclid - The Creation of Mathematics, Springer,
New York, 2nd ed., 2001. (The material on the icosahedron is not
in the first edition.)
It's not a scholarly tome: instead, it's a fun and intelligent
introduction to Euclid's Elements with lots of interesting digressions.
A great book for anyone interested in math!
I should also get ahold of this someday:
7) Benno Artmann, Antike Darstellungen des Ikosaeders, Mitt. DMV 13
(2005), 45-50.
Heath's translation of and commentary on Euclid's Elements is available
online thanks to the Perseus Project. The scholium crediting Theatetus
for the octahedron and icosahedron is discussed here:
8) Euclid, Elements, trans. Thomas L. Heath, Book XIII,
Historical Note, p. 438. Also available at
http://old.perseus.tufts.edu/cgi-bin/ptext?doc=Perseus%3Atext%3A1999.01.0086&query=head%3D%23566
while the textual history of the Elements is discussed here:
9) Euclid, Elements, trans. Thomas L. Heath, Chapter 5: The Text,
p. 46. Also available at
http://old.perseus.tufts.edu/cgi-bin/ptext?lookup=Euc.+5
Anyone interested in Greek mathematics also needs these books by
Heath, now available cheap from Dover:
10) Thomas L. Heath, A History of Greek Mathematics. Vol. 1: From
Thales to Euclid. Vol. 2: From Aristarchus to Diophantus.
Dover Publications, 1981.
The long quote by Waterhouse comes from here:
11) William C. Waterhouse, The discovery of the regular solids,
Arch. Hist. Exact Sci. 9 (1972-1973), 212-221.
I haven't yet gotten my hold on this "thorough study" mentioned by
Waterhouse - but I will soon:
12) Eva Sachs, Die funf platonischen Koerper, zur Geschichte der
Mathematik und der Elementenlehre Platons und der Pythagoreer,
Berlin, Weidmann, 1917.
I also want to find this discussion of how Peyrard got ahold of the
non-Theonine copy of Euclid's Elements:
13) N. M. Swerlow, The Recovery of the exact sciences of antiquity:
mathematics, astronomy, geography, in Rome Reborn: The Vatican
Library and Renaissance Culture, ed. Grafton, 1993.
Here is Atiyah and Sutcliffe's paper claiming that the Ashmolean
has Scottish stone balls shaped like Platonic solids:
14) Michael Atiyah and Paul Sutcliffe, Polyhedra in physics,
chemistry and geometry, available as math-ph/0303071.
Here is le Bruyn's critical examination of that claim:
15) Lieven le Bruyn, The Scottish solids hoax, March 25, 2009,
http://www.neverendingbooks.org/index.php/the-scottish-solids-hoax.html
Here are the books by Critchlow and Lawlor -speculative books from
the "sacred geometry" tradition:
16) Keith Critchlow, Time Stands Still, Gordon Fraser, London, 1979.
17) Robert Lawlor, Sacred Geometry: Philosophy and Practice,
Thames and Hudson, London, 1982. Available at
http://www.scribd.com/doc/13155707/robert-lawlor-sacred-geometry-philosophy-and-practice-1982
Here's the Ashmolean website:
18) British Archaeology at the Ashmolean Museum, Highlights of the
British collections: stone balls,
http://ashweb2.ashmus.ox.ac.uk/ash/britarch/highlights/stone-balls.html
and here's Dorothy Marshall's paper on stone balls:
19) Dorothy N. Marshall, Carved stone balls, Proc. Soc. Antiq. Scotland,
108 (1976/77), 40-72. Available at
http://www.tarbat-discovery.co.uk/Learning%20Files/Carved%20stone%20balls.pdf
Finally, a bit of math.
In the process of researching my talk, I learned a lot about Euclid's
Elements, where the construction of the icosahedron - supposedly due to
Theaetetus - is described. This construction is Proposition XIII.16,
in the final book of the Elements, which is largely about the Platonic
solids. This book also has some fascinating results about the golden
ratio and polygons with 5-fold symmetry!
The coolest one is Proposition XIII.10. It goes like this.
Take a circle and inscribe a regular pentagon, a regular hexagon, and
a regular decagon. Take the edges of these shapes, and use them as the
sides of a triangle. Then this is a right triangle!
In other words, if
P
is the side of the pentagon,
H
is the side of the hexagon, and
D
is the side of the decagon, then
P^2 = H^2 + D^2
We can prove this using algebra - but Euclid gave a much cooler proof,
which actually find this right triangle hiding inside an icosahedron.
First let's give a completely uninspired algebraic proof.
Start with a unit circle. If we inscribe a regular hexagon in it,
then obviously
H = 1
So we just need to compute P and D. If we think of the unit circle as
living in the complex plane, then the solutions of
z^5 = 1
are the corners of a regular pentagon. So let's solve this equation.
We've got
0 = z^5 - 1 = (z - 1)(z^4 + z^3 + z^2 + z + 1)
so ignoring the dull solution z = 1, we must solve
z^4 + z^3 + z^2 + z + 1 = 0
This says that the center of mass of the pentagon's corners lies right
in the middle of the pentagon.
Now, quartic equations can always be solved using radicals, but it's
a lot of work. Luckily, we can solve this one by repeatedly using
the quadratic equation! And that's why the Greeks could construct
the regular pentagon using a ruler and compass.
The trick is to rewrite our equation like this:
z^2 + z + 1 + z^{-1} + z^{-2} = 0
and then like this:
(z + z^{-1})^2 + (z + z^{-1}) - 1 = 0
Now it's a quadratic equation in a new variable. So while I said this
proof would be uninspired, it did require a tiny glimmer of inspiration.
But that's all! Let's write
z + z^{-1} = x
so our equation becomes
x^2 + x - 1 = 0
Solving this, we get two solutions. The one I like is the golden
ratio:
x = phi = (-1 + sqrt(5))/2 ~ 0.6180339...
Next we need to solve
z + z^{-1} = phi
This is another quadratic equation:
z^2 - phi z + 1 = 0
with two conjugate solutions, one being
z = (phi + sqrt(phi^2 - 4))/2
I've sneakily chosen the solution that's my favorite 5th root of unity:
z = exp(2 pi i / 5) = cos(2pi/5) + i sin(2pi/5)
So, we're getting
cos(2pi/5) = phi/2
A fact we should have learned in high school, but probably never did.
Now we're ready to compute P, the length of the side of a pentagon
inscribed in the unit circle:
P^2 = |1 - z|^2
= (1 - cos(2pi/5))^2 + (sin(2pi/5))^2
= 2 - 2 cos(2pi/5)
= 2 - phi
Next let's compute D, the length of the side of a decagon inscribed
in the unit circle! We can mimic the last stage of the above
calculation, but with an angle half as big:
D^2 = 2 - 2 cos(pi/5)
To go further, we can use a half-angle formula:
cos(pi/5) = sqrt((1 + cos(2pi/5))/2)
= sqrt(1/2 + phi/4)
This gives
D^2 = 2 - sqrt(2 + phi)
But we can simplify this a bit more. As any lover of the golden ratio
should know,
2 + phi = 2.6180339...
is the square of
1 + phi = 1.6180339...
So we really have
D^2 = 1 - phi
Okay. Your eyes have glazed over by now - unless you've secretly been
waiting all along for This Week's Finds to cover high-school algebra
and trigonometry. But we're done. We see that
P^2 = H^2 + D^2
simply says
2 - phi = 1 + (1 - phi)
That wasn't so bad, but imagine discovering it and proving it using
axiomatic geometry back around 300 BC! How did they do it?
For this, let's turn to
20) Ian Mueller, Philosophy of Mathematics and Deductive Structure in
Euclid's Elements, MIT Press, Cambridge Massachusetts, 1981.
This is reputed to be be the most thorough investigation of the
logical structure of Euclid's Elements. And starting on page 257 he
discusses how people could have discovered P^2 = H^2 + D^2 by staring at
an icosahedron!
This should not be too surprising. After all, there are pentagons,
hexagons and decagons visible in the icosahedron. But I was completely
stuck until I cheated and read Mueller's explanation.
If you hold an icosahedron so that one vertex is on top and one is
on bottom, you'll see that its vertices are arranged in 4 horizontal
layers. From top to bottom, these are:
1 vertex on top
5 vertices forming a pentagon: the "upper pentagon"
5 vertices forming a pentagon: the "lower pentagon"
1 vertex on bottom
Pick a vertex from the upper pentagon: call this A. Pick a vertex
as close as possible from the lower pentagon: call this B. A is not
directly above B. Drop a vertical line down from A until it hits the
horizontal plane on which B lies. Call the resulting point C.
If you think about this, or better yet draw it, you'll see that ABC
is a right triangle. And if we apply the Pythagorean theorem to
this triangle we'll get the equation
P^2 = H^2 + D^2
To see this, we only need to check that:
* the length AB equals the edge of a pentagon inscribed in a circle;
* the length AC equals the edge of a hexagon inscribed in a circle;
* the length BC equals the edge of a decagon inscribed in a circle.
Different circles, but of the same radius! What's this radius?
Take all 5 vertices of the "lower pentagon". These lie on a circle,
and this circle has the right radius.
Using this idea, it's easy to see that the length AB equals the edge
of a pentagon inscribed in a circle. It's also easy to see that
BC equals the edge of a decagon inscribed in a circle of the same
radius. The hard part, at least for me, is seeing that AC equals the
edge of a hexagon inscribed in a circle of the same radius... or in
other words, the radius of that circle! (The hexagon seems to be a
red herring.)
To prove this, it would suffice to show the following marvelous fact: the
distance between the upper pentagon and the lower pentagon equals
the radius of circle containing the vertices of the upper pentagon!
Can you prove this?
In Mueller's book, he suggests various ideas the Greeks could
have had about this. Here's one.
Let A be our vertex on the upper pentagon and let B' be the top
vertex. Drop a vertical line from B' until it hits the plane of the
upper pentagon; call the point where it hits C'. Suppose we can prove
that the right triangle AB'C' is congruent to the right triangle ABC.
Then we're done!
Why? Remember, we're trying to show the distance between the upper
pentagon and lower pentagon equals the radius of the circle containing
the vertices of the upper pentagon.
But that's equivalent to showing that AC' is congruent to AC.
To do this, it suffices to show that the right triangles ABC
and AB'C' are congruent! Can you do it?
In the references to Mueller's book, he says the historians
Dijksterhuis (in 1929) and Neuenschwander (in 1975) claimed this
is "intuitively evident". But I don't know if that means
it's easy to prove!
I thank Toby Bartels and Greg Egan for help with this stuff. I also
thank Jim Stasheff for passing on an email from Joe Neisendorfer
pointing out Mellinger's picture of the Milky Way.
-----------------------------------------------------------------------
Quote of the Week:
"Geometry enlightens the intellect and sets one's mind right. All
its proofs are very clear and orderly. It is hardly possible for
errors to enter into geometric reasoning, because it is well arranged
and orderly. Thus the mind that constantly applies itself to geometry
is unlikely to fall into error." - Ibn Khaldun
-----------------------------------------------------------------------
Addendum: Kevin Buzzard explained some of the Galois theory behind why
the pentagon can be constructed with ruler and compass - or in other
words, why the quartic
z^4 + z^3 + z^2 + z + 1 = 0
He wrote:
Now, quartic equations can always be solved using radicals
That's because S_4 is a solvable group, and all Galois groups of
quartics will live in S_4 (and will usually be S_4)...
Luckily, we can solve this one by repeatedly using the quadratic equation!
("this one" being z^4 + z^3 + z^2 + z + 1 = 0 )
...and _that's_ because the Galois group of that *specific*
irreducible polynomial is "only" cyclic of order 4. The splitting
field is Q(zeta_5), which is a cyclotomic field, so has Galois group
(Z/5Z)^*. No Z/3Z factors so no messing around with cube roots, for
example...
So while I said this proof would be uninspired, it did require a
tiny glimmer of inspiration.
With this observation above, I'm trying to convince you that the
proof really _is_ completely uninspired :-) To solve the quartic by
solving two quadratics, you need to locate the degree 2 subfield of
Q(z) (z=zeta_5) and aim towards it (because it's your route to the
solution). This subfield is clearly the real numbers in Q(z), and the
real numbers in Q(z) contains z+(z-bar)=z+z^{-1}. So that's sort of a
completely conceptual explanation of why the trick works and why it's
crucial to introduce z+z^{-1}.
Here zeta_5 is number-theorist's jargon for a "primitive 5th root of
unity", which in turn is number-theorist's jargon for any 5th root of
1 except for 1 itself.
Greg Egan gave a nice modern version of Euclid's original proof of
Prop. XIII.10, which states that if you take take a circle and
inscribe a regular pentagon, a regular hexagon, and a regular decagon,
and make a triangle out of their sides, it's a right triangle!
To understand it, you need to look at this picture:
http://math.ucr.edu/home/baez/pentagon_hexagon_decagon.gif
He writes:
Here's a version of the proof Euclid gave, adapted from
the version JB cited. Rather than proving that various angles here
are identical, I've just written in the (easily established)
numerical values; there's nothing tricky here, so we might as well
take them as given.
Triangle ABF is similar to triangle BFN. So AB/BF = BF/FN = BF/BN,
with the last equality true because the triangles are isosceles
with FN = BN. Thus BF^2 = AB x BN.
Triangle BAK is similar to triangle KAN. So BA/AK = KA/AN. Thus
AK^2 = AB x AN.
Adding our two results, we have: BF^2 + AK^2 = AB x (AN + BN) =
AB^2.
BF is our radius, AK is a decagon side, and AB is a pentagon
side. Well done Euclid.
On Thanksgiving of 2009, I got ahold of Eva Sachs' 1917 book Die Funf
Platonischen Koerper, mentioned above. It's supposed to be the
authoritative tome on the early history of the Platonic solids.
If anyone out there reads German, I'd love a translation of what she
says about the icosahedron and the pentagon-hexagon-decagon identity.
Here are the two most relevant pages:
http://math.ucr.edu/home/baez/icosahedron_sachs_1.jpg
http://math.ucr.edu/home/baez/icosahedron_sachs_2.jpg
http://math.ucr.edu/home/baez/icosahedron_sachs_figure.jpg
I think the first page gets interesting around "Der Satz XIII 10
lautet..." ("Proposition XIII.10 says...") and more interesting after
"Dieser Satz, obwohl planimetrisch, ist durch Betrachtung der ebenen
Geometrie nicht leicht zu finden". ("This proposition, though planar
in character, is not so easy to find through considerations of plane
geometry.")
You'll notice that she focuses our attention on the right triangles ZWQ and
QEP, which are the triangles ABC and AB'C' that I mentioned above:
http://math.ucr.edu/home/baez/icosahedron_with_right_triangles.gif
Proving that these are congruent is the key to the pentagon-decagon-hexagon
identity!
I'm especially curious about the footnote on page 104, and also the remark
further up this page saying "So fand er den Satz XIII 9..." ("This is how
he found Proposition XII.9").
To see Greg Egan's beautiful proof of the pentagon-decagon-hexagon
identity, which meets my challenge above, see "week284".
For more discussion visit the n-Category Cafe:
http://golem.ph.utexas.edu/category/2009/11/this_weeks_finds_in_mathematic_44.html
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