# December 31, 2000 {#week163}

If you think numbers start with the number 1, you probably think the
millennium is ending now. I think it ended last year... but either way,
now is a good time to read this book:

1) Georges Ifrah, _The Universal History of Numbers from Prehistory to the Invention of the Computer_, Wiley, New York, 2000.

On the invention of zero:

> Most peoples throughout history failed to discover the rule of
> position, which was discovered in fact only four times in the history
> of the world. (The rule of position is the principle in which a 9,
> let's say, has a different magnitude depending on whether it comes in
> first, second, third... position in a numerical expression.) The
> first discovery of this essential tool of mathematics was made in
> Babylon in the second millennium BCE. It was then rediscovered by the
> Chinese arithmeticians at around the start of the Common Era. In the
> third to fifth centuries CE, Mayan astronomers reinvented it, and in
> the fifth century CE it was rediscovered for the last time, in India.
>
> Obviously, no civilization outside of these four ever felt the need to
> invent zero; but as soon as the rule of position became the basis for
> a numbering system, a zero was needed. All the same, only three of the
> four (the Babylonians, the Mayans, and the Indians) managed to develop
> this final abstraction of number; the Chinese only acquired it through
> Indian influences. However, the Babylonian and Mayan zeroes were not
> conceived of as numbers, and only the Indian zero had roughly the same
> potential as the one we use nowadays. That is because it is indeed the
> Indian zero, transmitted to us through the Arabs together with the
> number-symbols that we call Arabic numerals and which are in reality
> Indian numerals, with their appearance altered somewhat by time, use
> and travel.

Among other things, this book has wonderful charts showing the
development of each numeral. You can see, for example, how the primitive
numeral

                              ____
                              ____
                              ____

slowly evolved to our modern "3". Hmm --- how come this doesn't feel
like progress?

Now, I usually keep my eyes firmly focused on the beauties of nature,
but once in a millennium I feel the need to engage in some politics.
So....

In ["Week 155"](#week155) I talked a lot about polyhedra and their
4-dimensional generalizations, and I referred to Eric Weisstein's
online math encyclopedia since it had lots of nice pictures. Now this
website has been closed down, thanks to a lawsuit by the people at CRC
Press:

2) Frequently asked questions about the MathWorld case, `http://mathworld.wolfram.com/docs/faq.html`

Weisstein published a print version of his encyclopedia with CRC press,
but now they claim to own the rights to the online version as well. So I
urge you all to remember this: *when dealing with publishers, never sign
away the electronic rights on your work unless you're willing to accept
the consequences!*

For example, suppose you write a math or physics paper and put it on the
preprint archive, and then publish it in a journal. They'll probably
send you a little form to sign where you hand over the rights to this
work --- including the electronic rights. If you're like most people,
you'll sign this form without reading it. This means that if they feel
like it, they can now sue you to make you take your paper off the
preprint archive! Journals don't do this yet, but as they continue
becoming obsolete and keep fighting ever more desperately for their
lives, there's no telling what they'll do. Corporations everywhere are
taking an increasingly aggressive line on intellectual property rights ---
as the case of Weisstein shows.

So what can you do? Simple: don't agree to it. When you get this form,
cross out any sentences you refuse to agree to, put your initials by
these deletions, and sign the thing --- indicating that you agree to the
*other* stuff! Keep a copy. If they complain, ask them how much these
electronic rights are worth.

Basically, I think it's time for academics to take more responsibility
about keeping their work easily accessible.

There are lots of things you can do. One of the easiest is to stop
refereeing for ridiculously expensive journals. Journal prices bear
little relation to the quality of service they provide. For example, the
Elsevier-published journal "_Nuclear Physics B_" costs \$12,596 per year
for libraries, or \$6,000 for a personal subscription. The comparable
journal "Advances in Theoretical and Mathematical Physics" costs \$300
for libraries or \$80 for a personal subscription --- and access to the
electronic version is free. So when Nuclear Physics B asks me to referee
manuscripts, I now say "Sorry, I'll wait until your prices go down."

In fact, I no longer referee articles for any journals published by
Elsevier, Kluwer, or Gordon & Breach. If you've looked at their prices,
you'll know why. G&B has even taken legal action against the American
Institute of Physics, the American Physical Society, and the American
Mathematical Society for publishing information about journal prices!

3) Gordon and Breach et al v. AIP and APS, brief of amici curiae of the American Library Association, Association of Research Libraries and the Special Library Association, `http://www.arl.org/scomm/gb/amici.html`

4) AIP/APS prevail in suit by Gordon and Breach, G&B to appeal, `http://www.arl.org/newsltr/194/gb.html`

Of course, the ultimate solution is to support the math and physics
preprint archives, and figure out ways to decouple the refereeing
process from the distribution process.

Okay, enough politics. I was thinking about $4$-dimensional polytopes, and
Eric Weisstein's now-defunct website... but what got me going in the
first place was this:

5) John Stilwell, "The story of the 120-cell", _AMS Notices_ **48** (January 2001), 17--24.

The 120-cell is a marvelous $4$-dimensional shape with 120 regular
dodecahedra as faces. I talked about it in ["Week 155"](#week155),
but this article is full of additional interesting information. For
example, Henri Poincare once conjectured that every compact 3-manifold
with the same homology groups as a 3-sphere must *be* a 3-sphere. He
later proved himself wrong by finding a counterexample: the "Poincare
homology 3-sphere". This is obtained by identifying the opposite faces
of the dodecahedron in the simplest possible way. What I hadn't known
is that the fundamental group of this space is the "binary icosahedral
group", $I$. This is the 120-element subgroup of $\mathrm{SU}(2)$ consisting of all
elements that map to rotational symmetries of the icosahedron under the
two-to-one map from $\mathrm{SU}(2)$ to $\mathrm{SO}(3)$. Now $\mathrm{SU}(2)$ is none other than the
3-sphere... so it follows that $\mathrm{SU}(2)/I$ is the Poincare homology
3-sphere!

When cosmologists study the possility that universe is finite in size,
they usually assume that space is a 3-sphere. In this scenario, barring
sneaky tricks, it's likely that the universe would recollapse before
light could get all the way around the universe. But there's no strong
reason to favor this topology. Some people have checked to see whether
space is a $3$-dimensional torus. In such a universe, light might wrap all
the way around --- so you might see the same bright quasars by looking in
various different directions! People have looked for this effect but not
seen it. This doesn't rule out a torus-shaped universe, but it puts a
limit on how small it could be.

In fact, some physicists have even considered the possibility that space
is a Poincare homology 3-sphere! Can light go all the way around in this
case? I don't know. If so, we might see bright quasars in a pretty
dodecahedral pattern.

Amusingly, Plato hinted at something resembling this in his "Timaeus":

6) Plato, "Timaeus", translated by B. Jowett, in _The Collected Dialogues_, Princeton U. Press, Princeton, 1969 (see line 55c).

This dialog is one the first attempts at doing mathematical physics. In
it, the Socrates character guesses that the four elements earth, air,
water and fire are made of atoms shaped like four of the five Platonic
solids: cubes, octahedra, icosahedra and tetrahedra, respectively. Why?
Well, fire obviously feels hot because of those pointy little tetrahedra
poking you! Water is liquid because of those round little icosahedra
rolling around. Earth is solid because of those little cubes packing
together so neatly. And air... well, ahem... we'll get back to you on
that one.

*But what about the dodecahedron?* On this topic, Plato makes only the
following cryptic remark: "There was yet a fifth combination which God
used in the delineation of the universe with figures of animals."

Huh??? I think this is a feeble attempt to connect the 12 sides of the
dodecahedron to the 12 signs of the zodiac. After all, lots of the signs
of the zodiac are animals. The word "zodiac" comes from the Greek
phrase "zodiakos kuklos", or "circle of carved figures" --- where
"zodiakos" or "carved figure" is really the diminutive of "zoion",
meaning "animal". There may even be a connection between the
dodecahedron and the "quintessence": the fifth element, of which the
heavenly bodies were supposedly made. I know, this is all pretty weird,
but there seems to be some tantalizingly murky connection between the
dodecahedron and the heavens in Greek cosmology.... so it would be cool
if space turned out to be a Poincare homology 3-sphere. But of course,
there's no reason to believe it is.

Okay, enough goofing around. Now let me talk a bit about the exceptional
Jordan algebra and the octonionic projective plane. I'll basically pick
up where I left off in ["Week 162"](#week162) --- but you might want
to reread ["Week 61"](#week61), ["Week 106"](#week106) and
["Week 145"](#week145) to prepare yourself for the weirdness to
come. Also, keep in mind the following three facts about the number 3,
which fit together in a spooky sort of synergy that makes all the magic
happen:

i) An element of $\mathrm{h}_3(\mathbb{O})$ is a $3\times3$ hermitian matrix with octonionic
    entries, and thus consists of 3 octonions and 3 real numbers:
    $$
      \left(
        \begin{array}{ccc}
          a&z^*&y^*
        \\z&b&x
        \\y&x^*&c
        \end{array}
      \right)
      \qquad\mbox{($a,b,c$ in $\mathbb{R}$, $x,y,z$ in $\mathbb{O}$.)}
    $$

ii) The octonions arise naturally from "triality": the relation
    between the three $8$-dimensional irreps of $\mathrm{Spin}(8)$, i.e. the vector
    representation $V_8$, the right-handed spinor representation $S_8^+$, and
    the left-handed spinor representation $S_8^-$.

iii) The associative law $(xy)z = x(yz)$ involves 3 variables.

Let's see how it goes.

First, if we take the 3 octonions in our element of $\mathrm{h}_3(\mathbb{O})$ and identify
them with elements of the three $8$-dimensional irreps of $\mathrm{Spin}(8)$, we get
$$\mathrm{h}_3(\mathbb{O}) = \mathbb{R}^3\oplus V_8\oplus S_8^+\oplus S_8^-.$$
A little calculation then reveals a wonderful fact: while superficially
the Jordan product in $\mathrm{h}_3(\mathbb{O})$ is built using the structure of $\mathbb{O}$ as a
normed division algebra, it can actually be defined using just the
natural map
$$t\colon V_8\times S_8^+\times S_8^-\to\mathbb{R}$$
and the inner products on these 3 spaces. It follows that any element of
$\mathrm{Spin}(8)$ gives an automorphism of $\mathrm{h}_3(\mathbb{O})$. Indeed, $\mathrm{Spin}(8)$ becomes a
subgroup of $\mathrm{Aut}(\mathrm{h}_3(\mathbb{O}))$.

So the exceptional Jordan algebra has a lot to do with geometry in 8
dimensions --- that's not surprising. What's surprising is that it also
has a lot to do with geometry in 9 dimensions! When we restrict the
spinor and vector representations of $\mathrm{Spin}(9)$ to the subgroup $\mathrm{Spin}(8)$,
they split as follows:
$$
  \begin{aligned}
    S_9 &= S_8^+\oplus S_8^-
  \\V_9 &= \mathbb{R}\oplus V_8
  \end{aligned}
$$
This gives an isomorphism
$$\mathrm{h}_3(\mathbb{O}) = \mathbb{R}^2\oplus V_9\oplus S_9$$
and in fact the product in $\mathrm{h}_3(\mathbb{O})$ can be described in terms of natural
maps involving scalars, vectors and spinors in 9 dimensions. It follows
that $\mathrm{Spin}(9)$ is also a subgroup of $\mathrm{Aut}(\mathrm{h}_3(\mathbb{O}))$.

This does not exhaust all the symmetries of $\mathrm{h}_3(\mathbb{O})$, since there are
other automorphisms coming from the permutation group on 3 letters,
which acts on $(a,b,c)$ in $\mathbb{R}^3$ and $(x,y,z)$ in $\mathbb{O}^3$ in an obvious way.
Also, any matrix $g$ in the orthogonal group $\mathrm{O}(3)$ acts by conjugation as
an automorphism of $\mathrm{h}_3(\mathbb{O})$; since the entries of $g$ are real, there is no
problem with nonassociativity here. The group $\mathrm{Spin}(9)$ is 36-dimensional,
but the full automorphism group $\mathrm{h}_3(\mathbb{O})$ is 52-dimensional. In fact, it
is the exceptional Lie group $\mathrm{F}_4$!

However, we can already do something interesting with the automorphisms
we have: we can use them to diagonalize any element of $\mathrm{h}_3(\mathbb{O})$. To see
this, first note that the rotation group, and thus $\mathrm{Spin}(9)$, acts
transitively on the unit sphere in the vector representation $V_9$. This
means we can use an automorphism in our $\mathrm{Spin}(9)$ subgroup to bring any
element of $\mathrm{h}_3(\mathbb{O})$ to the form
$$
  \left(
    \begin{array}{ccc}
      a&z^*&y^*
    \\z&b&x
    \\y&x^*&c
    \end{array}
  \right)
$$
where $x$ is *real*. The next step is to apply an automorphism that makes
$y$ and $z$ real while leaving $x$ alone. To do this, note that the subgroup
of $\mathrm{Spin}(9)$ fixing any nonzero vector in $V_9$ is isomorphic to $\mathrm{Spin}(8)$.
When we restrict the representation $S_9$ to this subgroup it splits as
$S_8^+\oplus S_8^-$, and with some work one can show that $\mathrm{Spin}(8)$ acts on
$S_8^+\oplus S_8^- = \mathbb{O}^2$ in such a way that any element $(y,z)$ in $\mathbb{O}^2$
can be carried to an element with both components real. The final step
is to take our element of $\mathrm{h}_3(\mathbb{O})$ with all real entries and use an
automorphism to diagonalize it. We can do this by conjugating it with a
suitable matrix in $\mathrm{O}(3)$.

To understand the octonionic projective plane, we need to understand
projections in $\mathrm{h}_3(\mathbb{O})$. Here is where our ability to diagonalize
matrices in $\mathrm{h}_3(\mathbb{O})$ via automorphisms comes in handy. Up to
automorphism, every projection in $\mathrm{h}_3(\mathbb{O})$ looks like one of these four
guys:
$$
  \begin{gathered}
    p_0 =
    \left(
      \begin{array}{ccc}
        0&0&0\\0&0&0\\0&0&0
      \end{array}
    \right)
    \qquad\qquad
    p_1 =
    \left(
      \begin{array}{ccc}
        1&0&0\\0&0&0\\0&0&0
      \end{array}
    \right)
  \\p_2 =
    \left(
      \begin{array}{ccc}
        1&0&0\\0&1&0\\0&0&0
      \end{array}
    \right)
    \qquad\qquad
    p_3 =
    \left(
      \begin{array}{ccc}
        1&0&0\\0&1&0\\0&0&1
      \end{array}
    \right)
  \end{gathered}
$$

Now, the trace of a matrix in $\mathrm{h}_3(\mathbb{O})$ is invariant under automorphisms,
because we can define it using only the Jordan algebra structure:
$$\operatorname{tr}(a) = \frac13 \operatorname{tr}(L_a)$$
where $L_a$ is left multiplication by $a$. It follows that the trace of any
projection in $\mathrm{h}_3(\mathbb{O})$ is $0$, $1$, $2$, or $3$.

Remember from ["Week 162"](#week162) that the "dimension" of a
projection $p$ in a formally real Jordan algebra is the largest number $d$
such that there's a chain of projections
$$p_0<p_1<\ldots<p_d=p.$$
In favorable cases, like the exceptional Jordan algebra, the dimension-1
projections become the points of a projective plane, while the
dimension-2 projections become the lines. But what's a practical way to
compute the dimension of a projection? Well, in $\mathrm{h}_3(\mathbb{O})$ the dimension
equals the trace.

Why?

Well, clearly the dimension is less than or equal to the trace, since
$p < q$ implies $\operatorname{tr}(p) < \operatorname{tr}(q)$, and the trace goes up by integer steps. But
on the other hand, the trace is less than or equal to the dimension. To
see this it suffices to consider the four projections shown above, since
both trace and dimension are invariant under automorphisms. Since
$p_0< p_1 < p_2 < p_3$, it is clear that for these projections the
trace is indeed less than or equal to the rank.

So: the points of the octonionic projective plane are the projections
with trace $1$ in $\mathrm{h}_3(\mathbb{O})$, while the lines are projections with trace $2$. A
brutal calculation in Reese Harvey's book:

7) F. Reese Harvey, _Spinors and Calibrations_, Academic Press, Boston, 1990.

reveals that any projection with trace 1 has the form
$$
  \vert\psi\rangle\langle\psi\vert =
  \left(
    \begin{array}{ccc}
      xx^*&xy^*&xz^*
    \\yx^*&yy^*&yz^*
    \\zx^*&zy^*&zz^*
    \end{array}
  \right)
$$
where
$$\vert\psi\rangle = (x,y,z)$$
is a unit vector in $\mathbb{O}^3$ for which $(xy)z = x(yz)$. This is supposed to
remind you of stuff about spinors and the heavenly sphere in
["Week 162"](#week162).

On the other hand, any projection with trace $2$ is of the form $1-p$
where $p$ has trace $1$. This sets up a one-to-one correspondence between
points and lines in the octonionic projective plane. If we use this
correspondence to think of both as trace-$1$ projections, the point $p$ lies
on the line $q$ if and only if $p < 1 - q$. Of course, $p < 1 - q$ iff
$q < 1 - p$. The symmetry of this relation means the octonionic projective
plane is self-dual! This is also true of the real, complex and
quaternionic projective planes. In all cases, the operation that
switches points and lines corresponds in quantum logic to "negation".

Let's use $\mathbb{OP}^2$ to stand for the set of points in the octonionic
projective plane. Given any nonzero element $(x,y,z)$ in $\mathbb{O}^3$ with
$(xy)z = x(yz)$, we can normalize it and then use the above formula to obtain a
point of $\mathbb{OP}^2$, which we call $[(x,y,z)]$. We can make $\mathbb{OP}^2$ into a
smooth manifold by covering it with three coordinate charts: one
containing all points of the form $[(x,y,1)]$, one containing all points
of the form $[(x,1,z)]$, and one containing all points of the form
$[(1,y,z)]$. Checking that this works is a simple calculation. The only
interesting part is to make sure that whenever the associative law might
appear necessary, we can either use the weaker equations
$$
  \begin{aligned}
    (xx)y &= x(xy) 
  \\(xy)x &= x(yx)
  \\(yx)x &= y(xx)
  \end{aligned}
$$
which still hold for the octonions, or else the fact that only triples
with $(xy)z = x(yz)$ give points $[(x,y,z)]$ in $\mathbb{OP}^2$.

Clearly the manifold $\mathbb{OP}^2$ is $16$-dimensional. The lines in $\mathbb{OP}^2$ are
copies of $\mathbb{OP}^1$, and thus 8-spheres. It is also good to work out the
space of lines going through any point. Here we can use self-duality:
since the space of all points lying on any given line is a copy of
$\mathbb{OP}^1$, so is the space of all lines on which a given point lies! So the
space of lines through a point is also an 8-sphere. Everything is very
pretty.

If we give $\mathbb{OP}^1$ the nicest possible metric, its isometry group is $\mathrm{F}_4$:
just the automorphism group of the exceptional Jordan algebra. However,
the group of "collineations" --- i.e., line-preserving transformations -
is a form of the 78-dimensional exceptional Lie group $\mathrm{E}_6$. From stuff
explained last week, the subgroup of collineations that map a point $p$ to
itself and also map the line $1 - p$ to itself is isomorphic to $\mathrm{Spin}(9,1)$.
This gives a nice embedding of $\mathrm{Spin}(9,1)$ in this form of $\mathrm{E}_6$. So the
octonionic projective plane is also related to $10$-dimensional
*spacetime* geometry.

I hope I've got that last part right.... ultimately, this is supposed
to explain why various different theories of physics formulated in 10d
spacetime wind up being related to the exceptional Lie groups! But I'm
afraid that so far, I'm just struggling to understand the basic
geometry.

Happy New Year!