This Week's Finds in Mathematical Physics (Week 261)

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March 19, 2008
This Week's Finds in Mathematical Physics (Week 261)
John Baez

Sorry for the long pause! I've been busy writing. For example:
a gentle introduction to category theory, focusing on its role
as a "Rosetta Stone" that helps us translate between four languages:

1) John Baez and Mike Stay, Physics, topology, logic and
computation: a Rosetta Stone, to appear in New Structures in
Physics, ed. Bob Coecke. Available at
http://math.ucr.edu/home/baez/rosetta.pdf

The idea is to take this chart and make it really precise:

PHYSICS TOPOLOGY LOGIC COMPUTATION
Hilbert space manifold proposition data type
operator cobordism proof program

In each case we have a kind of "thing" and a kind of "process"
going between things. But it turns out we can make the analogies
much sharper and more detailed than that.

The hard work has already been done by many researchers.
People working on topological quantum field theory have seen
how cobordisms - spacetimes going from one slice of space to
another - are analogous to operators between Hilbert spaces.
The "Curry-Howard correspondence" makes the analogy between
proofs and programs precise. Girard's work on "linear logic"
sets up an analogy between operators and proofs. And so on....

We're just trying to present these analogies in an easy-to-read
form, all in one place. I hope that pondering them will help us
break down some walls separating disciplines. In more optimistic
moments, I even thnk they represent the first steps toward a
general theory of systems and processes! Then I remember that
scientists are trained to distrust such grand visions, and for
good reasons. Time will tell.

Okay, now for some math. This Week will be an ode to the number 3.

But first - the nebula of the week!

2) Hubble finds an hourglass nebula around a dying star,
http://hubblesite.org/newscenter/archive/releases/nebula/planetary/1996/07/

It looks like the eye of Sauron in Tolkien's Lord of the Rings trilogy.
It's not. It's a planetary nebula 8000 light years away, called
MyCn 18 - or, more romantically, the Engraved Hourglass Nebula.

The colors look unreal. They are.

H-alpha light is shown as as green, but it's actually red. This is
the light hydrogen emits when its one electron jumps from its n = 3
state to its n = 2 state.

N II light is shown as red, and it actually is. This is light
from singly ionized nitrogen.

O III light is shown as blue, but it's actually green. This is
light from doubly ionized oxygen.

Furthermore, the colors have been adjusted so that regions where
H-alpha and O II overlap are orange.

Okay, so the colors are fake. But how did this weird nebula form?
You can see a clue if you pay attention: the bright white dwarf
star isn't located exactly at the center! It's a bit to the left!
This paper, written by the folks who took the photograph, argues
that it has an unseen companion:

3) Raghvendra Sahai et al, The Etched Hourglass Nebula MyCn 18.
I: Hubble space telescope observations, The Astronomical Journal
118 (1999), 468-476. Also available at
http://www.iop.org/EJ/article/1538-3881/118/1/468/990080.text.html

This paper tackles the difficult problem of modelling the nebula:

4) Raghvendra Sahai et al, The Etched Hourglass Nebula MyCn 18.
II: A spatio-kinematic model, The Astronomical Journal
110 (2000), 315-322. Also available at
http://www.iop.org/EJ/article/1538-3881/119/1/315/990248.text.html

It doesn't seem that the white dwarf alone could have produced all
the glowing gas we see here. A red giant companion could help.
But, there are lots of mysteries.

That shouldn't be surprising. Even the simplest things can be quite
rich in complexity. I want to illustrate this with a little ode to the
number 3. I'll start off slow, and ramp up to a discussion of how all
these mathematical entities are locked in a tight embrace:

the trefoil knot
cubic polynomials
the group of permutations of 3 things
the three-strand braid group
modular forms and cusp forms

As I kind of intermezzo, I'll talk about how to solve the cubic
equation. We all learn about quadratic equations in school:
they're the bread and butter of algebra, right after linear
equations. Cubics are trickier, but studying them can give you
a lifetime's worth of fun.

Let's start with the trefoil knot. This is the simplest of knots.
You can even draw it on the surface of a doughnut! Just take a pen
and draw a curve that winds around your doughnut three time in one
direction as it winds twice in the other direction:

5) Center for the popularisation of mathematics, Torus knots,
http://www.popmath.org.uk/sculpmath/pagesm/torus2.html

Mathematically, the surface of a doughnut is called a "torus".
We can describe a point on the torus by two angles running
from 0 to 2pi - the "latitude" and "longitude". But another name
for such an angle is a "point on the unit circle". If we think
of the unit circle in the complex plane, this gives us a nice
equation for the trefoil:

u^2 = v^3

Here u and v are complex numbers with absolute value 1. The equation
says that as u moves around the unit circle, v moves around 2/3 as
fast. So, the set of solutions is a curve on the torus that winds
around thrice in one direction while it winds around twice in the
other direction - a trefoil knot!

We can also drop the restriction that u and v have absolute
value 1. Then the equation u^2 = v^3 is famous for other reasons -
it's related to cubic equations!

As you've probably heard, there's a formula for solving cubic equations,
sort of like the quadratic formula, but bigger and badder. It goes
back to some Italians in the 1500s who liked to challenge each other
with equations and make bets on who could solve them: Scipione del
Ferro, Niccolo Tartaglia and Gerolamo Cardono.

Imagine we're trying to solve a cubic equation. We can always divide
by the coefficient of the cubic term, so it's enough to consider
equations like this:

z^3 + Az^2 + Bz + C = 0

If we could solve this and find the roots a, b, and c, we could
write it as:

(z - a)(z - b)(z - c) = 0

But this means

A = -(a + b + c)
B = ab + bc + ca
C = -abc

Note that A, B, and C don't change when we permute a, b, and c.
So, they're called "symmetric polynomials" in the variables a, b,
and c.

You see this directly, but there's also a better explanation:
the coefficients of a polynomial depend on its roots, but they
don't change when we permute the roots.

I can't resist mentioning a cool fact, which is deeply related to
the trefoil: *every* symmetric polynomial of a, b, and c can be
written as a polynomial in A, B, and C - and in a unique way!

In fact, this sort of thing works not just for cubics, but for
polynomials of any degree. Take a general polynomial of degree n
and write the coefficients as functions of the roots. Then these
functions are symmetric polynomials, and *every* symmetric polynomial
in n variables can be written as a polynomial of these - and in a
unique way.

But, back to our cubic. Note that -A/3 is the average of the three
roots. So, if we slide z over like this:

x = z + A/3

we get a new cubic equation for which the average of the three
roots is zero. This new cubic equation will be of this form:

x^3 + Bx + C = 0

for some new numbers B and C. In other words, the "A" in this new
cubic is zero, since we translated the roots to make their average
zero.

So, to solve cubic equations, it's enough to solve cubics like
x^3 + Bx + C = 0. This is a great simplification. When you first
see it, it's really exciting. But then you realize you have no idea
what to do next! This must be why it's called a "depressed cubic".

In fact, Scipione del Ferro figured out how to solve the "depressed
cubic" shortly after 1500. So, you might think he could solve any
cubic. But, NEGATIVE NUMBERS HADN'T BEEN INVENTED YET. This
prevented him from reducing any cubic to a depressed one!

It's sort of hilarious that Ferro was solving cubic equations before
negative numbers were worked out. It should serve as a lesson: we
mathematicians often work on fancy stuff before understanding the
basics. Often that's why math seemss hard! But often it's impossible
to discover the basics except by working on fancy stuff and getting
stuck.

Here's one trick for solving the depressed cubic x^3 + Bx + C = 0.
Write

x = y - B/(3y)

Plugging this in the cubic, you'll get a quadratic equation in y^3,
which you can solve. From this you can figure out y, and then x.

Alas, I have no idea what this trick means. Does anyone know? Ferro
and Tartaglia used a more long-winded method that seems just as sneaky.
Later Lagrange solved the cubic yet another way. I like his way
because it contains strong hints of Galois theory.

You can see all these methods here:

6) Cubic function, Wikipedia,
http://en.wikipedia.org/wiki/Cubic_equation

So, I won't say more about solving the cubic now. Instead, I want to
explain the "discriminant". This is a trick for telling when two
roots of our cubic are equal. It turns out to be related to the
trefoil knot.

For a quadratic equation ax^2 + bx + c = 0, the two roots are equal
precisely when b^2 - 4ac = 0. That's why b^2 - 4ac is called the
"discriminant" of the quadratic. The same idea works for other
equations; let's see how it goes for the cubic.

Suppose we were smart enough to find the roots of our cubic

x^3 + Bx + C = 0

and write it as

(x - a)(x - b)(x - c) = 0

Then two roots are equal precisely when

(a - b)(b - c)(c - a) = 0

The left side isn't a symmetric polynomial in a, b, and c; it changes
sign whenever we switch two of these variables. But if we square it,
we get a symmetric polynomial that does the same job:

D = (a - b)^2 (b - c)^2 (c - a)^2

This is the discriminant of the cubic! By what I said about symmetric
polynomials, it has to be a polynomial in B and C (since A = 0). If
you sweat a while, you'll see

D = -4B^3 - 27C^2

So, here's the grand picture: we've got a 2-dimensional space of
cubics with coordinates B and C. Sitting inside this 2d space is a
curve consisting of "degenerate" cubics - cubics with two roots the
same. This curve is called the "discriminant locus", since it's where
the discriminant vanishes:

4B^3 + 27C^2 = 0

If we only consider the case where B and C are real, the discriminant
locus looks like this:

|C
o |
o |
o |
-----------o-------------
o | B
o |
o |
|

It's smooth except at the origin, where it has a sharp point called
a "cusp".

Now here's where the trefoil knot comes in. The equation for the
discriminant locus:

4B^3 + 27C^2 = 0

should remind you of the equation for the trefoil:

u^2 = v^3

Indeed, after a linear change of variables they're the same!
But, for the trefoil we need u and v to be *complex* numbers.
We took them to be unit complex numbers, in fact.

So, the story is this: we've got a 2-dimensional *complex* space
of complex cubics. Sitting inside it is a *complex* curve, the
discriminant locus. In our new variables, it's this:

u^2 = v^3

If we intersect this discriminant locus with the torus

|u| = |v| = 1

we get a trefoil knot. But that's not all!

Normal folks think of knots as living in ordinary 3d space, but
topologists often think of them as living in a 3-sphere: a sphere
in 4d space. That's good for us. We can take this 4d space to be
our 2d complex space of complex cubics! We can pick out spheres in
this space by equations like this:

|u|^2 + |v|^3 = c (c > 0)

These are not round 3-spheres, thanks to that annoying third power.
But, they're topologically 3-spheres. If we take any one of them
and intersect it with our discriminant locus, we get a trefoil knot!
This is clear when c = 2, since then we have

|u|^2 + |v|^3 = 2

and

u^2 = v^3

which together imply

|u| = |v| = 1

But if you think about it, we also get a trefoil knot for any other
c > 0. This trefoil shrinks as c -> 0, and at c = 0 it reduces to a
single point, which is also the cusp here:

|u
| o
| o
| o
-----------o-------------
| o v
| o
| o
|

We don't see trefoil knots in this picture because it's just a
real 2d slice of the complex 2d picture. But, they're lurking in
the background!

Now let me say how the group of permutations of three things
gets into the game. We've already seen the three things: they're
the roots a, b, and c of our depressed cubic! So, they're three
points on the complex plane that add to zero. Being a physicist
at heart, I sometimes imagine them as three equal-mass planets,
whose center of mass is at the origin.

The space of possible positions of these planets is a 2d complex
vector space, since we can use any two of their positions as
coordinates and define the third using the relation

a + b + c = 0

So, there are three coordinate systems we can use: the (a,b)
system, the (b,c) system and the (c,a) system. We can draw all
three coordinate systems at once like this:

b
\ /
\ /
\ /
\ /
--------o--------a
/ \
/ \
/ \
/ \
c

The group of permutations of 3 things acts on this picture
by permuting the three axes. Beware: I've only drawn a 2-dimensional
*real* vector space here, just a slice of the full 2d complex space.

Now suppose we take this 2d complex space and mod out by the
permutation symmetries. What do we get? It turns out we get *another*
2d complex vector space! In this new space, the three coordinate axes
shown above become just one thing... but this thing is a curve, like
this:

o
o
o
o
o
o
o


Look familiar? Sure! It's just the discriminant locus we've
seen before.

Why does it work this way? The explanation is sitting before us.
We've got two 2d complex vector spaces: the space of possible
*ordered triples of roots* of a depressed cubic, and the space of
possible *coefficients*. There's a map from the first space to
the second, since the coefficients are functions of the roots:

B = ab + bc + ca
C = -abc

These functions are symmetric polynomials: they don't change when
we permute a, b, and c. And, it follows from what I said earlier
that we can get *any* symmetric polynomial as a function of these -
under the assumption that a+b+c = 0, that is.

So, the map where we mod out by permutation symmetries of the roots
is exactly the map from roots to coefficients.

The lines in this picture are places where two roots are equal:

c=a
\ /
\ /
\ /
\ /
--------o-------- b=c
/ \
/ \
/ \
/ \
a=b

So, when we apply the map from roots to coefficients, these
lines get mapped to the discriminant locus:

|
o |
o |
o |
-----------o-------------
o |
o |
o |
|

You should now feel happy and quit reading... unless you
know a bit of topology. If you *do* know a little topology,
here's a nice spinoff of what we've done. Though I didn't say
it using so much jargon, we've already seen that space of
nondegenerate depressed cubics is C^2 minus a cone on the
trefoil knot. So, the fundamental group of this space is the
same as the fundamental group of S^3 minus a trefoil knot.
This is a famous group: it has three generators x,y,z, and three
relations saying that:

x conjugated by y is z
y conjugated by z is x
z conjugated by x is y

On the other hand, we've seen this space is the space of triples
of distinct points in the plane, centered at the origin, mod
permutations. The condition "centered at the origin" doesn't
affect the fundamental group. So, this fundamental group is
another famous group: the "braid group on 3 strands". This
has two generators:

\ / |
/ | X
/ \ |

and

| \ /
| / Y
| / \

and one relation, called the "Yang-Baxter equation" or "third
Reidemeister move":

\ / | | \ /
/ | | /
/ \ | | / \
| \ / \ / |
| / = / | XYX = YXY
| / \ / \ |
\ / | | \ /
/ | | /
/ \ | | / \

So: the 3-strand braid group is *isomorphic* to the fundamental
group of the complement of the trefoil! You may enjoy checking
this algebraically, using generators and relations, and then
figuring out how this algebraic proof relates to the geometrical
proof.

I find all this stuff very pretty...

... but what's really *magnificent* is that most of it generalizes
to any Dynkin diagram, or even any Coxeter diagram! (See "week62"
for those.)

Yes, we've secretly been studying the Coxeter diagram A_2, whose
"Coxeter group" is the group of permutations of 3 things, and whose
"Weyl chambers" look like this:


\ /
\ /
\ /
\ /
--------o--------
/ \
/ \
/ \
/ \

Let me just sketch how we can generalize this to A_{n-1}. Here
the Coxeter group is the group of permutations of n things, which
I'll call n!.

Let X be the space of n-tuples of complex numbers summing to 0.
X is a complex vector space of dimension n-1. We can think of
any point in X as the ordered n-tuple of roots of some depressed
polynomial of degree n. Here "depressed" means that the leading
coefficient is 1 and the sum of the roots is zero. This condition
makes polynomials sad.

The permutation group n! acts on X in an obvious way. The
quotient X/n! is isomorphic (as a variety) to another complex
vector space of dimension n-1: namely, the space of depressed
polynomials of degree n. The quotient map

X -> X/n!

is just the map from roots to coefficients!

Sitting inside X is the set D consisting of n-tuples of roots where
two or more roots are equal. D is the union of a bunch of
hyperplanes, as we saw in our example:


\ /
\ /
\ /
\ /
--------o--------
/ \
/ \
/ \
/ \

Sitting inside X/n! is the "discriminant locus" D/n!, consisting of
*degenerate* depressed polynomials of degree n - that is, those with
two or more roots equal. This is a variety that's smooth except for
some sort of "cusp" at the origin:

o
o
o
o
o
o
o



The fundamental group of the complement of the discriminant locus
is the braid group on n strands. The reason is that this group
describes homotopy classes of ways that n points in the plane can
move around and come back to where they were (but possibly permuted).
These points are the roots of our polynomial.

On the other hand, the discriminant locus is topologically the cone
on some higher-dimensional knot sitting inside the unit sphere in
C^{n-1}. So, the fundamental group of the complement of this knot
is the braid group on n strands.

This relation between higher-dimensional knots and singularities
was investigated by Milnor, not just for the A_n series of Coxeter
diagrams but more generally:

7) John W. Milnor, Singular Points of Complex Hypersurfaces,
Princeton U. Press, 1969.

The other Coxeter diagrams give generalizations of braid groups
called Artin-Brieskorn groups. Algebraically you get them by taking
the usual presentations of the Coxeter groups and dropping the
relations saying the generators (reflections) square to 1.

If you like braid groups and Dynkin diagrams, Artin-Brieskorn groups
are irresistible! For a fun modern account, try:

8) Daniel Allcock, Braid pictures for Artin groups, available as
arXiv:math.GT/9907194.

But I'm digressing! I must return and finish my ode to the number 3.
I need to say how modular forms get into the game!

I'll pick up the pace a bit now - if you're tired, quit here.

Any cubic polynomial P(x) gives something called an "elliptic curve".
This consists of all the complex solutions of

y^2 = P(x)

together with the point (infinity, infinity), which we include to
make things nicer.

Clearly this elliptic curve has two points (x,y) for each value of x
*except* for x = infinity and the roots of P(x), where it just has one.
So, it's a "branched double cover" of the Riemann sphere, with branch
points at the roots of our cubic and the point at infinity.

In fact, this elliptic curve has the topology of a torus, at least
when all the roots of our cubic are different. If you have trouble
imagining a torus that's a branched double cover of a sphere, ponder
this:

7) Carlos Furuti, Peirce's quincuncial map,
http://www.progonos.com/furuti/MapProj/Normal/ProjConf/projConf.html

This square map of the Earth is an unwrapped torus; each point of the
Earth shows up lots of times. If we wrap it up just right, we get a
branched double cover of the sphere! Can you spot the branch points?
For a lot more explanation, read "week229".

Now, way back in "week13", I turned this story around. I started with
a torus formed as the quotient of the complex plane by a lattice -
and showed how to get an elliptic curve out of it. I wrote the
equation for this elliptic curve in "Weierstrass form":

y^2 = 4x^3 - g2 x - g3

By a simple change of variables, this is equivalent to a depressed
cubic:

y^2 = x^3 + Bx + C

So, we can think of g2 and g3 as coordinates on our 2d space of
depressed cubics! They're just rescaled versions of our coordinate
functions B and C.

What's the big deal? Well, g2 and g3 are famous examples of
"modular forms" - whatever those are. In fact, it's a famous fact
that every modular form is a polynomial in g2 and g3.

I defined modular forms back in "week142", where I summarized the
Taniyama-Shimura-Weil theorem: the big theorem about modular forms
that implies Fermat's Last Theorem. So, you can reread the definition
there if you're curious. But if you've never seen it before, it's
a bit intimidating. A modular form of weight w is a function on the
space of lattices that transforms in a bizarre certain way, satisfying
a certain growth condition... blah blah blah.

It's important stuff, and incredibly cool once you get a feel for it.
But suppose we're trying to explain modular forms more simply. Then
we can avoid a lot of technicalities if we just say a modular form is
a polynomial on the space of depressed cubics! In other words, a
polynomial in our friends B and C.

Then we can make some definitions. The "weight" of the modular form

B^i C^j

is 4i+6j. Okay, I admit this sounds arbitrary and weird without
a lot more explanation. But better: a "cusp form" is a modular
form that vanishes on the discriminant locus. Then we can see
every cusp form is the product of the discriminant 4B^3 + 27C^2
and some other modular form... and we can use this to work out
lots of basic stuff about modular forms.

So, I hope you now see how tightly entwined all these ideas are:

the trefoil knot
cubic polynomials
the group of permutations of 3 things
the three-strand braid group
modular forms and cusp forms

At this point I should give credit where credit is due. As usual,
I've been talking to Jim Dolan, and many of these ideas come from
him. But also, you can think of this Week as an expansion of the
remarks by Joe Christy and Swiatowslaw Gal in the Addenda to
"week233".

Finally, I should say that my low-budget approach to modular forms
mostly just handles so-called "level 0" modular forms - the basic
kind, defined using the group

Gamma = PSL(2,Z)

More exciting are modular forms that transform nicely only for a
*subgroup* of Gamma. Jim and I are just beginning to understand
these. But the modular forms for Gamma(2) fit nicely into today's
ode! Here Gamma(2) is the subgroup of Gamma consisting of matrices
congruent to the identity matrix mod 2. What does this have to do
with my ode to the number 3? Well,

Gamma/Gamma(2) = PSL(2,F_2)

and this is isomorphic to the group of permutations of 3 things!

So, as a final flourish, I claim that:

Modular forms for Gamma(2) are polynomials on the space X
consisting of roots of depressed cubics:

X = {(a,b,c): a,b,c complex with a + b + c = 0}

Modular forms for Gamma are polynomials on the space X/3!
consisting of coefficients of depressed cubics:

X/3! = {(B,C): B,C complex}

The obvious quotient map X -> X/3! sends roots to coeffficients:

(a,b,c) |-> (B,C) = (ab + bc + ca, abc)

and this induces the inclusion of modular forms for Gamma into
modular forms for Gamma(2):

B |-> ab + bc + ca
C |-> abc

I hope this is all true!

Modular forms for Gamma(2) are particularly nice. A good example
is the *cross-ratio*, much beloved in complex analysis. If you want
to learn more about this stuff, try:

8) Igor V. Dolgachev, Lectures on modular forms, Fall 1997/8,
available at http://www.math.lsa.umich.edu/~idolga/modular.pdf

especially chapter 9 for level 2 modular forms. Also:

9) Henry McKean and Victor Moll, Elliptic Curves: Function Theory,
Geometry, Arithmetic, Cambridge U. Press, 1999.

especially chapter 4.

-----------------------------------------------------------------------
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Answers and Replies

  • #2
Uncle Al
John Baez wrote:
>
> Also available as http://math.ucr.edu/home/baez/week261.html
>
> March 19, 2008
> This Week's Finds in Mathematical Physics (Week 261)
> John Baez

[snip erudition]

> Let's start with the trefoil knot. This is the simplest of knots.
> You can even draw it on the surface of a doughnut! Just take a pen
> and draw a curve that winds around your doughnut three time in one
> direction as it winds twice in the other direction:
>
> 5) Center for the popularisation of mathematics, Torus knots,
> http://www.popmath.org.uk/sculpmath/pagesm/torus2.html

[snip]

A trefoil knot is chiral. One would be negligent to ignore cases
where enantiomorphic forms beget diastereotopic results. There's lots
of fun to be had in putting the wrong shoe on a foot.

> You can see all these methods here:
>
> 6) Cubic function, Wikipedia,
> http://en.wikipedia.org/wiki/Cubic_equation

[snip]

http://www.1728.com/cubic.htm
http://www.akiti.ca/Quad3Deg.html

Genius isn't knowing everything, geuius is knowing where to find it
(then not crediting your sources).

> and one relation, called the "Yang-Baxter equation" or "third
> Reidemeister move":
>
> \ / | | \ /
> / | | /
> / \ | | / \
> | \ / \ / |
> | / = / | XYX = YXY
> | / \ / \ |
> \ / | | \ /
> / | | /
> / \ | | / \
>
> So: the 3-strand braid group is *isomorphic* to the fundamental
> group of the complement of the trefoil! You may enjoy checking
> this algebraically, using generators and relations, and then
> figuring out how this algebraic proof relates to the geometrical
> proof.


The above pair is not chiral - they are superposable by a 180-degree
rotation in the plane. If a trefoil knot is chiral (it is) and those
figures are not (they aren't), something is missing, yes?

Enantiomorphic chiral mass distributions cannot be distinguished in
Newton (Green's function), General Relativity (Equivalence Principle),
and string theory (BRST invariance). Enantiomorphic chiral mass
distributions can be distinguished in teleparallel gravitation. One
of those two sets is subject to empirical falsification as a parity
Eotvos experiment (in quartz) or a parity calorimetry experiment (in
benzil). Both materials are *achiral* as formula units but
crystallize in enantiomorphic space groups P3(1)21 (right-handed screw
axes) and P3(2)21 (left-handed screw axes).

Just because it shouldn't work doesn't mean it won't (Yang and Lee).

--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
 
  • #3
David Park
"John Baez" <baez@math.removethis.ucr.andthis.edu> wrote in message
news:frsuj1$age$1@glue.ucr.edu...
>
> Here's one trick for solving the depressed cubic x^3 + Bx + C = 0.
> Write
>
> x = y - B/(3y)
>
> Plugging this in the cubic, you'll get a quadratic equation in y^3,
> which you can solve. From this you can figure out y, and then x.
>
> Alas, I have no idea what this trick means. Does anyone know? Ferro
> and Tartaglia used a more long-winded method that seems just as sneaky.
> Later Lagrange solved the cubic yet another way. I like his way
> because it contains strong hints of Galois theory.
>


I think there is a way to better understand this trick - but perhaps not
to suggest it in the first place.

Recapping the first steps we first eliminated the z^2 term in the cubic
equation by a shift along the real axis.

ztoxrule = z -> x+s and solve for s to eliminate the quadratic term.

For the second substitution I would start with a more general form,
which will lead to a more symmetrical expression:

xtoyrule = x -> a y + b/y

When we make this substitution it turns out that we can solve for a and
b such that the coefficients of the y and 1/y terms are zero and the
coefficients of the y^3 and 1/y^3 terms are equal. Part of the solution
is a = b. (I'm not certain if this is related to the polynomials being
symmetric.) The substitution rule then becomes:

xtoyrule = x -> a(y + 1/y)

The substitution is invariant to whether we substitute y or the
reciprocal of y and this is part of understanding the trick.

Again, when we substitute with this rule and do some manipulation we can
obtain a quadratic equation of the form:

w + 1/w = constant

where w = y^3. The quadratic equation will have two roots and it is
clear that the roots must be reciprocals of each other. (A similar
argument must hold with the less symmetrical substitution but it is not
as transparent.) Therefore we only have to consider one of the solutions
because the backtracking substitutions are invariant to the use of
reciprocals. What we have to be able to do is solve a quadratic
equation, and calculate the three roots of w = y^3. (I don't know if the
last was easy or difficult in 1500 CE) That gives us the three
solutions.

To illustrate with a specific case, consider the polynomial 1 - 2 z + 3
z^2 + z^3. The three solutions, obtained with Mathematica as a check
are:

{-3.62737, 0.313683- 0.421053 I, 0.313683+ 0.421053 I}

Using the above procedure the specific substitution rules are:

ztoxrule = z -> -1 + x
xtoyrule = x -> Sqrt[5/3] (1/y + y)
ytowrule = {{y -> w^(1/3)}, {y -> -(-1)^(1/3) w^(1/3)}, {y -> (-1)^(2/3)
w^(1/3)}}

The quadratic equation that is solved becomes:

1/w + w == -3 Sqrt[3/5]

with the solutions: wsol = {{w -> 1/10 (-3 Sqrt[15] - Sqrt[35])}, {w ->
1/10 (-3 Sqrt[15] + Sqrt[35])}}

Doing the reverse substitutions (using a Mathematica statement) using
only the first solution gives the same roots as Mathematica.

z
% /. ztoxrule;
% /. xtoyrule;
% /. ytowrule;
% /. First[wsol] // Simplify;
% // N // Chop
z
{0.313683+ 0.421053 I, 0.313683- 0.421053 I, -3.62737}

--
David Park
djmpark@comcast.net
http://home.comcast.net/~djmpark/ [Broken]
 
Last edited by a moderator:
  • #4
David Park
"John Baez" <baez@math.removethis.ucr.andthis.edu> wrote in message
news:frsuj1$age$1@glue.ucr.edu...
>
> Here's one trick for solving the depressed cubic x^3 + Bx + C = 0.
> Write
>
> x = y - B/(3y)
>
> Plugging this in the cubic, you'll get a quadratic equation in y^3,
> which you can solve. From this you can figure out y, and then x.
>
> Alas, I have no idea what this trick means. Does anyone know? Ferro
> and Tartaglia used a more long-winded method that seems just as sneaky.
> Later Lagrange solved the cubic yet another way. I like his way
> because it contains strong hints of Galois theory.
>



As a addition comment on the meaning of the trick I would like to add to my
previous post:

The transformation x -> a(y + 1/y) changes an odd order polynomial x to an
even order polynomial in y. Then if we can get rid of the middle terms we
will end with a quadratic form.

--
David Park
djmpark@comcast.net
http://home.comcast.net/~djmpark/ [Broken]
 
Last edited by a moderator:
  • #5
Gerard Westendorp
David Park wrote:

> "John Baez" <baez@math.removethis.ucr.andthis.edu> wrote in message
> news:frsuj1$age$1@glue.ucr.edu...
>> Here's one trick for solving the depressed cubic x^3 + Bx + C = 0.
>> Write
>>
>> x = y - B/(3y)
>>
>> Plugging this in the cubic, you'll get a quadratic equation in y^3,
>> which you can solve. From this you can figure out y, and then x.
>>
>> Alas, I have no idea what this trick means. Does anyone know? Ferro
>> and Tartaglia used a more long-winded method that seems just as sneaky.
>> Later Lagrange solved the cubic yet another way. I like his way
>> because it contains strong hints of Galois theory.
>>

>
> I think there is a way to better understand this trick - but perhaps not
> to suggest it in the first place.
>
> Recapping the first steps we first eliminated the z^2 term in the cubic
> equation by a shift along the real axis.
>
> ztoxrule = z -> x+s and solve for s to eliminate the quadratic term.
>
> For the second substitution I would start with a more general form,
> which will lead to a more symmetrical expression:
>
> xtoyrule = x -> a y + b/y


It struck me that both substitutions are *Mobius transformations*.
Now that makes sense:

Mobius transformations are the only substitutions that are both
single-valued (as opposed to for example the square root), and whose
inverses are also single valued. This is needed, because we do not want
substitutions to generate a larger set of solutions than the original
equation.
Since they form a group under composition, we can put all substitutions
into a single Mobius transformation:

x = (az + b)/(cz + d)

If you do the math, multiplying out the denominators, you get a cubic in
which each coefficient depends on {a,b,c,d). Choose these such that the
cubic simplifies to

x^3 = 1

Then, after solving this, inverse-Mobius transform back to z.
Call the Mobius transform that did this (M).

This also makes sense geometrically:

If you look at a picture of a quadrtatic, for example
http://www.xs4all.nl/~westy31/Geometry/Geometry.html#Riemann_surfaces
you see there is a symmetry in it: it is symmetrical under 180 degree
rotations about a certain point (z0). The symmetry operation can be
expresed as a 2 element group of Mobius transformations:

{x=z, x=-(z-z0)}

For a Cubic, the symmetry does not immediately seem obvious, because the
3 roots of a cubic need not lie on an equilateral triangle in the
complex plane. However, the *do* lie on a Mobius-transformed equilateral
triangle! The Mobius-transform in question is again (M).
M can also be used to build the symmetry group of the Cubic, by
conjugating 120 degree rotations

{ x=z , x = M [z*exp(i/3 2pi) M^-1] , x = M [z*exp(2i/3 2pi)] M^-1}

You can tile the complex plane with a conformal hexagon that is
invariant under this group. Each hexagon is a copy of a torus, and the
torus is the Riemann surface of the cubic.
Since the cubic is an elliptic curve, here must also be some modular
form associated with it, but I do not know how to do that.

The method I described above should work also for quartics, because we
have 4 parameters (a,b,c,d) to satisfy four equations, to put the
quartic into a form

x^4 = D

I havn't actually tried this. I am also not sure that a quartic has a
symmetry group that is a conjugate of 90 degree rotations. I think it
was more like the 'Klein Vierergruppe'. Need to look at that. (Probably,
the equations are not independent)

But for equations of order higher than or equal to 5, the method fails.

I had better start making some pictures of quintics and quartics, to see
what their symmetry looks like.

Gerard
 
  • #6
Gerard Westendorp
Gerard Westendorp wrote:
[..]

>> xtoyrule = x -> a y + b/y

>
> It struck me that both substitutions are *Mobius transformations*.


To be fair, I should mention that this is actually *not* a Mobius
transformation. (David Park pointed this out in an email)
Also, finding a Mobius transformation that transforms a cubic to
x^3 = 1
is in practice about as difficult as solving the cubic.

Still, the symmetry interpretation of this is cool.

[..]

> { x=z , x = M [z*exp(i/3 2pi) M^-1] , x = M [z*exp(2i/3 2pi)] M^-1}
>
> You can tile the complex plane with a conformal hexagon that is
> invariant under this group. Each hexagon is a copy of a torus, and the
> torus is the Riemann surface of the cubic.
> Since the cubic is an elliptic curve, here must also be some modular
> form associated with it, but I do not know how to do that.


This is also not correct:

An elliptic curve has a squared y:

y^2 = x^3 + ax + b

So it is not the same as a cubic. The symmetry group of a cubic is
isomorphic to a 120 degree rotation, the Riemann surface is not a torus
but the Riemann sphere.

I'll be away for a week. Maybe when I get back, I will understand how
the square in the elliptic curve gives rise to the torus.


Gerard
 

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