Also available at http://math.ucr.edu/home/baez/week229.html
April 13, 2006
This Week's Finds in Mathematical Physics (Week 229)
John Baez
I'm visiting Chicago now. I came just in time for a conference
in honor of Saunders Mac Lane, one of the founders of category
theory, who taught at the University of Chicago for many years
and died last year at the age of 95:
1) Category Theory and its Applications: A Conference in Memory
of Saunders Mac Lane, http://www.math.uchicago.edu/~may/MACLANE/
On Friday there was a memorial service where the friends and family
of Mac Lane spoke about him, and a kind of reminiscence session
where everyone could tell their favorite stories involving him.
Then there were a bunch of math talks, both by people with strong
connections to Mac Lane - Peter Johnstone, Bill Lawvere, Peter Freyd,
Ieke Moerdijk, Peter May and Steve Awodey - and by people
working on higher categories and their applications.
My own connection to Mac Lane is tiny. Everything I do uses his
work, but that's true of many mathematicians: he discovered so much.
Apart from watching him celebrate his 90th birthday at a category
theory conference in Portugal back in 1999, the best moment happened
when I came to Chicago and gave a talk. He invited me up to his office
and we talked a bit. He told me I should write a book explaining
n-categories. I promised I would... I was too shy to say much.
Now I'm trying to write that book, and I just happen to be staying in
Mac Lane's old office, which makes me feel especially obliged to do it.
This office is on the third floor of the Ryerson Physical Laboratory.
It has a very high ceiling, and one wall is lined with two stacks of
metal bookshelves. You'd need a ladder to reach the top! When I spoke
to Mac Lane in his office, they were all full of books. Alas, they're
empty now.
Next time I'll say a bit about Julie Bergner's talk at the Maclane
memorial conference - she spoke about derived categories of quiver
representations and quantum groups. But the conference was so intense
and exhausting that first I need to recover by thinking about something
completely different. So, I'll concentrate on last week's puzzle about
rational tangles.
But first: the astronomy picture of the week!
It'll be more fun after a little background. The northern part of
Mars is very different from the rest. It's much smoother, and the
altitude is much less:
2) Linda M. V. Martel, Ancient floodwaters and seas on Mars,
http://www.psrd.hawaii.edu/July03/MartianSea.html
Why is this?
Many scientists believe the north was an ocean during the Hesperian
Epoch, a period of Martian history that stretches from about 3.5 to
about 1.8 billion years ago. In particular, the beautifully named
"Vastitas Borealis", an enormous plain that covers most of northern
Mars, has textures that may have been formed by an ocean that froze
and then slowly sublimated. (Sublimation is what happens when ice
turns directly into water vapor without actually melting.) Mike Carr
and James Head wrote a paper suggesting that around the end of the
Hesperian, about 30% of the water on Mars evaporated and left the
atmosphere, drifting off into outer space... part of the danger of
life on a planet without much gravity:
3) M. H. Carr and J. W. Head, III, Oceans of Mars: An assessment of
the observational evidence and possible fate, Journal of Geophysical
Research 108 (2003), 5042.
The rest of the water is now frozen at the poles or lurking underground.
And that brings us to our picture. Here's some ice in a crater in
Vastitas Borealis!
4) Water ice in crater at Martian north pole, European Space Agency (ESA),
http://www.esa.int/esaMI/Mars_Express/SEMGKA808BE_3.html
The picture is close to natural color, but the vertical relief is
exaggerated by a factor of 3. The crater is 35 kilometers wide
and 2 kilometers deep. It's incredible how they can get this kind
of picture from satellite photos and lots of clever image processing.
I hope they didn't do *too* much stuff just to make it look pretty.
Next: rational tangles.
In "week228", I asked for help understanding the connection between
rational tangles and the group PSL(2,Z). I got a great reply from
Michael Hutchings, which winds up relating these ideas to the branched
double cover of the sphere by the torus. And, this gives me an excuse
to tell you some stuff I learned from James Dolan about elliptic functions
and a map of the world called "Peirce's quincuncial".
So, let's dive in!
Did you ever try to wrap a sphere around itself twice? Mentally,
I mean? Slit it open, grab it, pull it, stretch it, wrap it around
itself twice, and glue the seams back together?
It's not hard. You just take the Riemann sphere - the complex
numbers together with a point at infinity - and map it to itself
by the function
f(z) = z^2
If you think of the sphere as the surface of the Earth, with zero
at the south pole and infinity as the north pole, this function doubles
the longitude. So, it wraps the sphere around itself twice!
I hope you're visualizing this.
This function is not quite a "double cover", because it's not quite
two-to-one everywhere. Only one point gets mapped to z = 0, namely
itself, and only one point gets mapped to z = infinity, namely itself.
Elsewhere f is two-to-one.
If you walk once around the north pole or south pole, and then apply
the function f to your path, you get a path that goes around these
points TWICE. Summarizing these properties, we call the function
a "branched double cover" of the sphere by itself, with zero and
infinity as branch points.
Now, how about wrapping a torus twice around a sphere?
This too can be done. It turns out there's a nice branched double
cover of the sphere by the torus, which has four branch points.
To visualize this, first take the surface of the Earth and mold it
into a regular octahedron. There will be six corners: the north pole,
the south pole, the east pole, the west pole, the front pole and the
back pole. Now take the octahedron and unfold it like this:
S-----B-----S
| /|\ |
| / | \ |
| / | \ |
| / | \ |
|/ | \|
W-----N-----E
|\ | /|
| \ | / |
| \ | / |
| \ | / |
| \|/ |
S-----F-----S
We get an interesting map of the world, which was invented in 1876
by the American mathematician and philosopher C. S. Peirce while he
was working at the U. S. Coast and Geodetic Survey. This map is
called "Peirce's quincuncial", since when you arrange five dots this
way:
o o
o
o o
it's called a "quincunx". (Somehow this word goes back to the name of
an ancient Roman coin. I don't understand how this pattern is related
to the coin.)
This is how Peirce's quincuncial looks as an actual map:
5) Carlos A. Furuti, Conformal projections,
http://www.progonos.com/furuti/MapProj/Normal/ProjConf/projConf.html
The cool part is that you can tile the plane indefinitely with
this map:
S-----B-----S-----F-----S-----B-----S-----F-----S
| /|\ | /|\ | /|\ | /|\ |
| / | \ | / | \ | / | \ | / | \ |
| / | \ | / | \ | / | \ | / | \ |
| / | \ | / | \ | / | \ | / | \ |
|/ | \|/ | \|/ | \|/ | \|
W-----N-----E-----N-----W-----N-----E-----N-----W
|\ | /|\ | /|\ | /|\ | /|
| \ | / | \ | / | \ | / | \ | / |
| \ | / | \ | / | \ | / | \ | / |
| \ | / | \ | / | \ | / | \ | / |
| \|/ | \|/ | \|/ | \|/ |
S-----F-----S-----B-----S-----F-----S-----B-----S
| /|\ | /|\ | /|\ | /|\ |
| / | \ | / | \ | / | \ | / | \ |
| / | \ | / | \ | / | \ | / | \ |
| / | \ | / | \ | / | \ | / | \ |
|/ | \|/ | \|/ | \|/ | \|
E-----N-----W-----N-----E-----N-----W-----N-----E
This gives a branched cover of the sphere by the plane! It
has branch points at the east, west, front and back poles,
since walking once around a point like that on the above map
corresponds to walking around it twice on the actual Earth.
This is pretty weird, but Peirce cleverly located two of these
branch points in the Pacific Ocean, one in the Atlantic, and
one in the Indian Ocean.
We can be less extravagant and get a branched cover of the
sphere by the torus if we take the smallest parallelogram
whose opposite edges match up:
B
/|\
/ | \
/ | \
/ | \
/ | \
E-----N-----W
/|\ | /|\
/ | \ | / | \
/ | \ | / | \
/ | \ | / | \
/ | \|/ | \
B-----S-----F-----S-----B
\ | /|\ | /
\ | / | \ | /
\ | / | \ | /
\ | / | \ | /
\|/ | \|/
W-----N-----E
\ | /
\ | /
\ | /
\ | /
\|/
B
This would actually be a square if I could draw it right in ASCII.
We can curl this into a torus by gluing together the opposite edges.
There's then an obvious function from this torus to the sphere
sending both points labelled "N" to the north pole, both points
labelled "S" to the south pole, and so on.
This function is mostly two-to-one, but it's one-to-one at the
points labelled E, F, W, and B. After all, there's just *one*
point of each of these sorts in the above picture after we glue
together the opposite edges. There are *two* copies of any other
sort of point.
So, our function is a branched double cover of the sphere by
the torus, which has four branch points. In fact, this function
is quite famous. It's an example of an "elliptic function"!
I explained elliptic functions way back in "week13". Briefly,
what we just did starting with this parallelogram:
B
/|\
/ | \
/ | \
/ | \
/ | \
E-----N-----W
/|\ | /|\
/ | \ | / | \
/ | \ | / | \
/ | \ | / | \
/ | \|/ | \
B-----S-----F-----S-----B
\ | /|\ | /
\ | / | \ | /
\ | / | \ | /
\ | / | \ | /
\|/ | \|/
W-----N-----E
\ | /
\ | /
\ | /
\ | /
\|/
B
actually works for a parallelogram of any shape. The parallelogram
curls up to give a torus, and we get a map from this torus to the
Riemann sphere, called an "elliptic function".
As before, this is a branched double cover with four branch points.
However, where the branch points sit on the sphere depends on the
shape of the parallelogram. By picking the parallelogram carefully,
you can put the branch points wherever you want! Peirce's neat idea
was to put them evenly spaced along the equator - at the east, front,
west and back poles. This is nice and symmetrical.
It's also especially nice to put the branch points at the vertices
of a regular tetrahedron. I'm not sure, but this may give the map
developed by Laurence P. Lee in 1965 - see Furuti's webpage above
for a picture of it.
In fact, these two specially nice locations for branch points
correspond to the two most symmetrical lattices in the plane:
the square one and the hexagonal one. I talked about these in
"week125" - they're really important in the theory of elliptic
functions, and even in string theory.
Anyway: for any parallelogram we can make a map of the Earth that
tiles the plane, with tiles shaped like this parallelogram.
A cool thing about these maps is that they're all "conformal" -
they preserve angles except at the branch points. If you want
to show off, you express this by saying "elliptic functions are
complex analytic".
But now I'm digressing a little. Let's get back on track. What
does all this have to do with rational tangles??
Recall my puzzle from last time. We build rational tangles by
starting with the trivial one, which we call "zero"
| |
| |
| |
| |
and repeatedly doing two operations. The first is a twisting
operation that we call "adding one":
| | | |
| | | |
| | | |
------- -------
| T | |----> | T | = "T + 1"
------- -------
| | \ /
| | /
| | / \
where the box labelled "T" stands for any tangle we've built
so far. The second is a rotation that we call "negative reciprocal":
| | | |
| | | | ____
| | | | / \
------- | ------- |
| T | |----> | | T | | = "-1/T"
------- | ------- |
| | \___/ | |
| | | |
| | | |
Using these tricks we can try to assign a rational number to
any rational tangle. The shocking theorem is that this number
is indeed well-defined, and in fact a complete invariant of
rational tangles.
Every operation built from "adding one" and "negative reciprocal"
looks like this:
az + b
z |-> -------
cz + d
with a,b,c,d integer and ad-bc = 1. The group of these transformations
is called PSL(2,Z). This group acts on the rational numbers together
with a point at infinity (the "rational projective line") by the formula
above. It also acts on rational tangles. The puzzle is to see why these
actions are isomorphic. The proofs I listed in "week228" show it's true;
the problem is to understand what's really going on!
Here's the answer given by Michael Hutching on sci.math.research:
There's a simple topological interpretation of the element of the
rational projective line associated to a rational tangle. I don't know
how to use this to prove the theorem, and I don't know a reference for
it (maybe it is in one of the references you cited). Anyway, regard a
rational tangle as a two-component curve C in the 3-ball B^3 whose four
boundary points are on the 2-sphere S^2. Consider the double branched
cover of B^3 along C. This is a 3-manifold Y whose boundary can be
identified with the 2-torus T^2. (In fact Y is a solid torus.) The
inclusion of T^2 into Y induces a map from H_1(T^2) to H_1(Y), and the
kernel of this map is a one-dimensional subspace of H_1(T^2) = Z^2. If
I am not mistaken, this is the element in question of the rational
projective line.
In other words, we take a 3-dimensional ball and draw a picture
of a rational tangle in it:
.......
. | | .
. ----- .
. | T | .
. ----- .
. | | .
.......
The boundary of this ball is a sphere with 4 points marked. If we
take a branched double cover of the sphere with these as the branch
points, we get a torus T^2. If we take a branched double cover of
the whole ball with everything along the vertical lines as branched
points, we get a solid doughnut Y having T^2 as its boundary.
This gets the torus into the game, and also the branched cover I was
talking about. And this gets the group PSL(2,Z) into the game!
SL(2,Z) is the group of 2x2 matrices with determinant 1. When you
mod out by the matrices +-1, you get PSL(2,Z). But, topologists know
that SL(2,Z) is the "mapping class group" of the torus - the group of
orientation-preserving diffeomorphisms modulo those that can be
smoothly deformed to the identity.
So, something nice is happening.
Even better, the rational first homology group of the torus is Q^2
(pairs of rational numbers), and SL(2,Z) acts in the obvious way,
by matrix multiplication:
a b x ax + by
: |->
c d y cx + dy
It therefore acts on the set of 1-dimensional subspaces of Q^2.
Any such subspace consists of vectors like this:
kx
ky
The subspace is determined by the ratio x/y, which however could be
infinite - so it's just a point in the rational projective line. So,
we get an action of SL(2,Z) on the rational projective line. Indeed
we get an action of PSL(2,Z) since +-1 act trivially. And, you can
easily check that it's the action we've already seen:
a b az + b
: z |-> --------
c d cz + d
In short: "projectivizing" the action of mapping class group of the
torus on its first homology gives the usual action of PSL(2,Z) on the
rational projective line.
What we need next is a natural way to assign to any rational tangle
a 1-dimensional subspace of the homology of the torus. And this is
what Hutchings describes: a rational tangle gives a way of mapping
the torus T^2 into the solid torus Y, and this gives a map on rational
homology
H_1(T^2) -> H_1(Y)
whose kernel is a 1-dimensional subspace of H_1(T^2).
There's more stuff to check....
Personally I've been trying to think of the mapping class group
of the 4-punctured sphere as acting on pictures like this:
.......
. | | .
. ----- .
. | T | .
. ----- .
. | | .
.......
and show that the resulting action on rational tangles factors
through a homomorphism from this mapping class group to PSL(2,Z).
The mapping class group should be generated by the twist
| | | |
| | | |
| | | |
------- -------
| T | |----> | T |
------- -------
| | \ /
| | /
| | / \
and the 90 degree rotation
| | | |
| | | | ____
| | | | / \
------- | ------- |
| T | |----> | | T | |
------- | ------- |
| | \___/ | |
| | | |
| | | |
and our homomorphism should map these to the famous matrices
1 1
T = "shear"
0 1
and
0 -1
S = "90 degree rotation"
1 0
respectively. If this works, and I could figure out the kernel of
this homomorphism and show it acts trivially on rational tangles,
I think I'd be almost done. But, I haven't had time!
By the way, if this works, there's a beautiful little sideshow where
we use as generators of SL(2,Z) not the above matrices but S and
0 -1
ST =
1 1
I explained why these are so great in "week125". S is a symmetry of
the square lattice, while ST is a symmetry of the hexagonal lattice.
The square lattice gives Peirce's quincuncial map, while the hexagonal
one presumably gives Laurence Lee's triangular map!
So, there's some intriguing story about elliptic functions and rational
tangles taking shape before our eyes.... and if I weren't so darn busy,
I'd figure out all the details and write a little paper about it.
Before quitting, there's one more thing I can't resist mentioning.
Any ordered 4-tuple of points (a,b,c,d) in the Riemann sphere gives
a number called its "cross-ratio":
(a-b)(c-d)/(a-d)(c-b)
It's a famous fact that you can find a conformal transformation of
the Riemann sphere mapping one ordered 4-tuple to another if and
only if their cross-ratios are equal!
So, we can play a little trick. Given a lattice we can get a
branched double cover of the Riemann sphere as I sketched earlier.
Then we can use the location of the branch points to calculate a cross
ratio.
But actually, I'm being a bit sloppy here. To compute a cross ratio
from a lattice, we need some extra information to *order* the 4-tuple
of branch points. In other words, if one of the points S is the origin
here:
S-----B-----S-----F-----S-----B-----S-----F-----S
| /|\ | /|\ | /|\ | /|\ |
| / | \ | / | \ | / | \ | / | \ |
| / | \ | / | \ | / | \ | / | \ |
| / | \ | / | \ | / | \ | / | \ |
|/ | \|/ | \|/ | \|/ | \|
W-----N-----E-----N-----W-----N-----E-----N-----W
|\ | /|\ | /|\ | /|\ | /|
| \ | / | \ | / | \ | / | \ | / |
| \ | / | \ | / | \ | / | \ | / |
| \ | / | \ | / | \ | / | \ | / |
| \|/ | \|/ | \|/ | \|/ |
S-----F-----S-----B-----S-----F-----S-----B-----S
| /|\ | /|\ | /|\ | /|\ |
| / | \ | / | \ | / | \ | / | \ |
| / | \ | / | \ | / | \ | / | \ |
| / | \ | / | \ | / | \ | / | \ |
|/ | \|/ | \|/ | \|/ | \|
E-----N-----W-----N-----E-----N-----W-----N-----E
and the lattice is taken just big enough so the pattern repeats,
we need enough information to *label* the points E, F, W and B.
This extra information amounts to "choosing a basis for the
2-torsion subgroup of the plane modulo the lattice". So, the
cross ratio gives a "modular function of level 2".
Hmm, this is getting pretty jargonesque! I don't want to explain the
jargon now, but you can read all about this trick and its consquences
in Lecture 9 here:
6) Igor V. Dolgachev, Lectures on modular forms, Fall 1997/8,
available at http://www.math.lsa.umich.edu/~idolga/modular.pdf
-----------------------------------------------------------------------
Quote of the Week:
Well, what if we consider our lives to be formed of a series of
interlocking practices, including the very important ones of
maintaining a thriving family and community? Then we might learn
from a practice with the pedigree of mathematics - mankind's longest
conversation - about the necessity of certain intellectual and
moral virtues. - David Corfield
-----------------------------------------------------------------------
Addendum: Andrei Sobolevskii points out that the etymology of the word
"quincunx" is explained here:
7) Quincunx, World Wide Words,
http://www.worldwidewords.org/weirdwords/ww-qui2.htm
Very briefly, "quincunx" was a Latin word for "five twelfths", from
"quinque" and "uncia". The latter word is also the root of the word
"ounce". They had a copper coin called the "as" weighing twelve
ounces (!), and the quincunx was apparently not a coin but a symbol for
5/12 of an as - or in other words, 5 ounces of copper.
After reading the above, Peter Dickof clarified and corrected the story:
Love "This Week's Finds" (though I seldom follow it all) and can't
pass up the opportunity to say something.
The as was indeed a unit of currency and also a specific bronze coin.
Early asses (aes = bronzes - hence the AE ligature...) were full Roman
pounds (librae -hence the British pound "L" sign) of 288 scruples with
twelve unciae to the pound. Each uncia was ~27 grams, a modern ounce
near enough. Debasement set in around the time of the first Punic war
and accelerated through the second (Hannibal's), by the end of which
an as, still of bronze, weighed only ~30 grams.
Multiples of the as were minted (misnomer, this was a cast currency):
the decussis (X asses), quincussis (V asses), tressis (III asses),
and the dupondius (two-pounder). The asses were marked with the Roman
numeral I. Common fractions were the semis (half, marked with an S),
triens (third, sometimes called a quatrunx, marked with four "pellets"
or dots), quadrans (quarter, also teruncius, 3 pellets), sextans (sixth,
also biunx, 2 pellets), uncia and semuncia (usually unmarked).
There WERE quincunx coins (5 pellets), and also a dextans (S + 4
pellets), mostly produced by non-Roman Italians. I have appended
photos of three quincunxes produced in Luceria (see Thurlow Vecchi
numbers 274, 281). Note that the "pellets" are sometimes (not
always) arranged in a quincunx. Luceria (modern Lucera) is 2/3
of the way across the boot from Neapolis (Naples).
During the second Punic war, after they captured Syracuse and its
treasure (and killed Archimedes), the Romans introduced the silver
denarius, quinarius and sesterius; they were worth 10, 5, 2 1/2 (IIS)
asses. The denarius is the origin of the "d" for the British shilling,
and was about the size of a dime. Later, circa 141 BC, the value of
silver was re-tariffed so that denarii, quinarii and sestertii became
worth 16, 8, 4 asses; the names did not change but the quinarius and
sestertius became rare.
Julius Caesar doubled the pay of a legionary to 300 sestertii per
installment (stipendium), 3 installments per year.
Later yet, during Imperial Rome, the largest bronze/brass coin minted
(no longer a misnomer) was a sestertius; its weight was less than
30 grams by the time of Claudius and falling, always falling....
Appropriate references are:
Michael H. Crawford, Roman Republican Coinage, Cambridge University
Press, Cambridge, 1974. (See picture of Quincunx on plate XVIII.)
Bradbury K. Thurlow, Italian Cast Coinage, Italian Aes Grave.
Italo G. Vecchi, Italian Aes Rude, Signatum and the Aes Grave of
Sicily. Printed together by Veechi, London, 1979.
Herbert A. Grueber, A Catalogue of the Coins of the Roman Republic
in the British Museum, three volumes, reprinted 1970.
My old pal Squark noted some sloppy language about branched covers.
Here's my reply to what he wrote. I've changed what he wrote a tiny
bit, for cosmetic reasons:
Squark wrote:
>Hello John and everyone!
Hello! Long time no see! How are you doing?
I had written:
>There's a simple topological interpretation of the element of the
>rational projective line associated to a rational tangle. I don't know
>how to use this to prove the theorem, and I don't know a reference for
>it (maybe it is in one of the references you cited). Anyway, regard a
>rational tangle as a two-component curve C in the 3-ball B^3 whose four
>boundary points are on the 2-sphere S^2. Consider the double branched
>cover of B^3 along C.
Squark wrote:
>What is "_the_ double branched cover"? Is there a way to choose a
>canonical one, or is there only one in this case, for some reason?
Good point. I hope there's a specially nice one.
To pick a branched cover of B^3 along C, it's necessary and sufficient
to pick a homomorphism from the fundamental group of B^3 - C to Z/2.
This says whether or not the two sheets switch places as we walk around
C following some loop in B^3 - C.
>In the case of a sphere with 4 points removed it should be easy to
>check.
Yes.
>The fundamental group has 4 generators - a, b, c, d (loops around each
>of the points) and one relation abc = d (since we're on a sphere). Hence,
>it is freely generated by a, b, c (say).
[I changed your presentation slightly here, for my own convenience.]
Right, the fundamental group of the four-punctured sphere is
the free group on 3 generators, F_3. I believe the "specially nice"
homomorphism
f: F_3 -> Z/2
is the one that sends each generator to -1, where I'm thinking
multiplicatively:
Z/2 = {1, -1}
One reason this homomorphism is especially nice is that it also sends
d = abc to -1.
So, if you walk around ANY of the four punctures, the two sheets switch!
This is just what you want for the Riemann surface of an elliptic
integral, as someone else pointed out in another post: there are
four branch points each like the branch point of sqrt(z). It's
also the most symmetrical, beautiful thing one can imagine.
Now let's see if and how this branched cover extends to a branched
cover of the ball B^3 with C (two arcs) removed. The fundamental group
of B^3 - C is the free group on two generators, say X and Y.
The inclusion of the 4-punctured sphere in B^3 - C gives a homomorphism
g: F_3 -> F_2
as follows:
a |-> X
b |-> X^{-1}
c |-> Y
d |-> Y^{-1}
So, to extend our branched cover, we need to write our homomorphism
f: F_3 -> Z/2
as
f = hg
for some homomorphism
h: F_2 -> Z/2
The obvious nice thing to try for h is
X |-> -1
Y |-> -1
It works, and it's unique!
Lee Rudolph added:
Squark wrote:
>What is "_the_ double branched cover"? Is there a way to choose a
>canonical one, or is there only one in this case, for some reason?
John Baez wrote:
>Good point. I hope there's a specially nice one.
In this kind of context, there's always exactly one "double branched
cover" that actually *does* branch doubly over every component of the
proposed branch locus. In particular, in the context of a rational
tangle, of course the pair (B^3,C) is homeomorphic to (B^2,X)xI,
where X is a 2-point set in Int B^2 and the homeomorphism isn't
required to preserve the tangle structure; so the double branched
cover of B^3 branched over C is the product of the double branched
cover of B^2 branched over X with the interval I. Now, because
the branching is *double* at each point of X, and there are *two*
points of X, it follows that the monodromy around the boundary of
B^2 must be trivial, so that we can sew another B^2 to that boundary
and extend the branched double covering over the resulting 2-sphere.
But of course the branched double cover of a 2-sphere over 2 points
is another 2-sphere, the model for the situation being z \mapsto z^2
as a map of the Riemann sphere to itself. Now remove the interior
of the sewed-on second B^2 from the downstairs S^2, and correspondingly
the interiors of its *two* preimage B^2s from the upstairs S^2; you
see that the double cover of B^2 branched over X is an annulus.
(Once you know that, you can see it directly: take an annulus embedded
in R^3 as the cylinder where x^2+y^2=1 and -1\le z\le 1; rotate it
by 180 degrees around the x-axis, and convince yourself that the quotient
space is a 2-disk by considering the fundamental domain consisting
of those points of the annulus with non-negative y-coordinate.)
Then the double cover of B^3 branched over C must be a solid torus.
(Again, now that you know this, you can see it directly: take the
solid torus to be a tubular neighborhood in R^3 of the circle where
x^2+y^2 = 1 and z = 0, and again rotate by 180 degrees around the x-axis
to give yourself the "deck involution".)
Lee Rudolph
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