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

  1. Mar 21, 2008 #1
    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

    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.

    -----------------------------------------------------------------------
    Previous issues of "This Week's Finds" and other expository articles on
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    http://math.ucr.edu/home/baez/

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    http://math.ucr.edu/home/baez/twfcontents.html

    A simple jumping-off point to the old issues is available at

    http://math.ucr.edu/home/baez/twfshort.html

    If you just want the latest issue, go to

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  2. jcsd
  3. Mar 22, 2008 #2
    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
     
  4. Mar 30, 2008 #3
    "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/
     
  5. Apr 1, 2008 #4
    "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/
     
  6. Apr 2, 2008 #5
    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
     
  7. Apr 27, 2008 #6
    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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