S^3 as a union of two solid tori
arzille
Can anyone explain why S3 can be expressed a a union of two solid
tori (S1 x 2-Disc) with an embedded torus (S1 x S1) as a common
boundary? The book says it follows easily by expressing R3- {solid
torus} as a unions of circle and a straight line and then add a point at
the infinity. How can I use the hint of the book?
Lee Rudolph
I think an easier way to see this is to consider the cartesian
product D of two round 2-disks. On the one hand, D is homeomorphic
to the round 4-disk D^4 (you can take D to be the "unit" bidisk in R^4,
and then easily construct the homeomorphism explicitly), so its
boundary is homeomorphic to the boundary of D^4, namely, S^3.
On the other hand, the given product structure on D gives you
an expression of the boundary of D as the union of two solid
tori intersecting along their common boundary: one solid torus
is the Cartesian product of a round circle with a round 2-disk,
and the other is the Cartesian product of a round 2-disk with
a round circle.
Since you're at UBC, why don't you go talk to Dale Rolfsen?
Lee Rudolph
Chan-Ho Suh
<lrud...@panix.com> wrote:
> arzille <arz...@ubc.edu> writes:
>
> >Dear all,
> > Can anyone explain why S3 can be expressed a a union of two solid
> >tori (S1 x 2-Disc) with an embedded torus (S1 x S1) as a common
> >boundary? The book says it follows easily by expressing R3- {solid
> >torus} as a unions of circle and a straight line and then add a point at
> >the infinity. How can I use the hint of the book?
>
> I think an easier way to see this is to consider the cartesian
> product D of two round 2-disks.
An even easier way is to consider S^3 as the union of two 3-balls.
Picture S^3 as R^3 plus infinity, and put one ball at the origin.
We're going to show this picture of S^3 as the union of these 2 balls
is the same as the union of two solid tori.
Take a solid torus and embed it in a standard way into R^3 plus
infinity. Now fill up the hole in the solid torus with a plug
(homeomorphic to D^2 x I). Voila, it's a 3-ball. The complement of
this 3-ball is clearly a 3-ball also. So the complement of our
original solid torus is a 3-ball with the plug glued on; each end of
the plug glues on the boundary of the 3-ball to give a solid torus (it
can't be a solid Klein bottle).
Thus this *particular* gluing of two solid tori gives S^3. If you pick
different homeomorphisms from the boundary of one solid torus to the
other as gluing maps, you'll in general get something different from
S^3. A lot of somethings. These are called lens spaces. This is
explained in Rolfsen's book.
The same approach shows that S^3 is the union of two genus g
handlebodies for any g. Just plug up each hole in a standardly
embedded genus g handlebody.
As for your hint:
You can get a picture of S^3 - solid torus as foliated by circles in
the way the hint describes by foliating the other torus.
A nice way to see if you get what's going on is to see if you can see a
sweep-out of tori sweeping between two circles, each of which is the
"core" of a solid torus.
[...]
>
> Since you're at UBC, why don't you go talk to Dale Rolfsen?
>
Maybe it's homework from Rolfsen :-)
Lee Rudolph
>In article <cgqcsh$8u0$1...@panix2.panix.com>, Lee Rudolph
><lrud...@panix.com> wrote:
>
>> arzille <arz...@ubc.edu> writes:
>>
>> >Dear all,
>> > Can anyone explain why S3 can be expressed a a union of two solid
>> >tori (S1 x 2-Disc) with an embedded torus (S1 x S1) as a common
>> >boundary? The book says it follows easily by expressing R3- {solid
>> >torus} as a unions of circle and a straight line and then add a point at
>> >the infinity. How can I use the hint of the book?
>>
>> I think an easier way to see this is to consider the cartesian
>> product D of two round 2-disks.
>
>An even easier way is to consider S^3 as the union of two 3-balls.
>Picture
[etc.]
I erred in writing "an easier way to see". I should have written,
and will stand by, "an easier--perhaps the easiest--way to prove".
For, indeed, no power of visualization whatever is required to
prove that S^3 is the union of two solid tori identified along their
common boundary, using the method I described, which is essentially
to "just write down the formulas" that describe the two spaces D^4
and D^2xD^2. On the contrary, there isn't (as far as I know) any
equally simple, equally unaided-by-visualization, "formula" for
writing down higher-genus Heegard splittings of S^3. (It can
be done reasonably simply, reasonably unaided-by-visualization,
using explicit algebraic fibered knots--for instance, a genus-2
splitting of the unit sphere in C^2 is given by the sets of
unit vectors (z,w) in C^2 where the real polynomial function F(z,w)
= Re(z^2+w^3) is non-negative and non-positive; but establishing
the truth of this statement by pure calculation is an order of
magnitude harder than doing the same for genus 1, I think.)
Certainly arzille should develop a power of visualization, and
Chan-Ho's suggestions are good for that; but it's nice to have
more ways to kill a cat than boiling it in butter.
Lee Rudolph
arzille
Paul C. Leopardi
> The book says ... How can I use the hint of the book?
OK, I'll bite. Which book? What page?
I'd rather read the original than your paraphrase, because your paraphrase
does not seem to make any sense.
Chan-Ho Suh
Leopardi <leop...@bigpond.net.au> wrote:
It does make sense. See my reply to Lee Rudolph's reply to arzille.
Lee Rudolph
And actually it makes sense in several ways; not only those in your
reply to my reply to arzille, but also as an approach to the special
case in dimensions 1-and-1 of the general (and just as easily proved
--with a mindless calculation instead of a helpful picture, even!)
result that the join of an m-sphere and an n-sphere is an
(m+n+1)-sphere. (For those playing along at home without their
score-cards, the "join" of topological spaces A and B is, roughly
but correctly, the union of a family of intervals parametrized
by AxB, where the interval I_{a,b} has endpoints a in A and b in
B, and is otherwise disjoint from A, B, and all the other intervals,
and where the union of all these intervals is given the only reasonable
topology. For instance, the join of an interval I and an interval
J is a solid tetrahedron; the locus of midpoints of the join intervals
is a square cross-section of the tetrahedron. Starting from there
and some simple facts about joins and circles, you can fairly quickly
show that the join of two circles C and D is a 3-sphere, and that
the locus of midpoints of the join intervals is a torus which bounds
a solid torus on each side.)
Lee Rudolph
Robin Chapman
Consider S^3 as {(x,y,z,t): x^2 + y^2 + z^2 + t^2 = 1}.
Let A = {(x,y,z,t) in S^3 : x^2 + y^2 <= z^2 + t^2}
and B = {(x,y,z,t) in S^3 : x^2 + y^2 >= z^2 + t^2}.
Convince yourself that A and B are homeomorphic to solid tori
and that A intersect B is homeomorphic to a torus.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"Lacan, Jacques, 79, 91-92; mistakes his penis for a square root, 88-9"
Francis Wheen, _How Mumbo-Jumbo Conquered the World_
Chan-Ho Suh
<lrud...@panix.com> wrote:
> Chan-Ho Suh <s...@math.ucdavis.nospam.edu> writes:
>
> >In article <CztYc.12716$D7....@news-server.bigpond.net.au>, Paul C.
> >Leopardi <leop...@bigpond.net.au> wrote:
> >
> >> I'd rather read the original than your paraphrase, because your paraphrase
> >> does not seem to make any sense.
> >>
> >
> >It does make sense. See my reply to Lee Rudolph's reply to arzille.
>
> And actually it makes sense in several ways; not only those in your
> reply to my reply to arzille, but also as an approach to the special
> case in dimensions 1-and-1 of the general (and just as easily proved
> --with a mindless calculation instead of a helpful picture, even!)
> result that the join of an m-sphere and an n-sphere is an
> (m+n+1)-sphere.
Hmmm...I seem to recall using this explanation a couple years ago when
someone else on sci.math asked why S^3 is the union of two solid tori.
I would say though that I did cover, in essense, this approach also in
my remark about a "sweep-out of tori" between two circles.
I remember RH Bing wrote a whole article about the 3-sphere; that's a
worthwhile read for anyone whose interest has been piqued by all this.
Since our answers are steadily getting more involved, let me not be
amiss, by bringing up the Hopf Fibration.
By foliating each solid torus with circles so that the circles match up
on the torus between them, you can foliate all of S^3 with circles; you
can get the appropriate foliation by foliating the boundary torus with
(1,1) curves and then extending on parallel tori until you hit the core
circle. By identifying each circle to a point, you get a 2-sphere as
the quotient space. This map from S^3 to S^2 is called the Hopf
Fibration.
This is a homotopically nontrivial map and so shows that pi_3(S^2) is
nontrivial.
If you picked your solid tori to be the right size, then these circles
are actually geometric circles (not just in the topological sense).
You can also make an interesting class of metrics on the 3-sphere by
making the arc lengths of all the circles some epsilon. These are the
Berger spheres. They can be used to demonstrate that S^3 collapses
with bounded curvature, with Gromov-Hausdorff limit being a round
2-sphere.