I need to compute two homology groups:

a) H (S^3\X,Z), where X is a copy of the circle S^1,     
    n
Z is the integers, n >= 0.
 
b) H (X,B), where X is the connected sum of two tori, 
    n
B < X is a circle in one of the tori tangent to the boundary, and n =
0,1,2.  If my description of X and B is unclear, there is a picture of
this on p. 132 of
Hatcher’s Algebraic Topology (circle B, in conjunction with Problem
17(b)).  The relevant chapter is at the link
http://www.math.cornell.edu/~hatcher/AT/ATch2.pdf

[by "<" I mean "is a subset of"]

Any standard result may be used for the two computations, such as
those found in Chapter 2 of Hatcher.  Also, I had some difficulty
pricing this question; if I'm paying too much, keep the difference as
my treat.  If I'm not offering enough, just let me know how long it
took you to answer and I'll tip appropriately to make up for the extra
time.  Thanks in advance.

---

Request for Question Clarification by rbnn-ga on 06 Dec 2002 17:00 PST

For a), I do not know what the \ means in S^3\X . I also do not know
what the notation H_n(Y,Z) means here, unless Z is a subset of S^3\X .

By the way is there any time constraint associated to these two
problems?

---

Request for Question Clarification by mathtalk-ga on 06 Dec 2002 17:17 PST

The link to the PDF with Hatcher's chapter from Algebraic Topology
seems to be broken.  My browser finds the site but shows a blank
screen.  I have Adobe Acrobat Reader 5.1, so I don't think the problem
can be with the version; I see no error messages at all.

-- mathtalk-ga

---

Clarification of Question by dubois-ga on 07 Dec 2002 00:06 PST

To rbnn and mathtalk:

Thank you for your requests for clarification.  S^3\X refers to the
complement of X in S^3.  In other words, X is a subset of S^3, and
S^3\X is all points in S^3 which are not in X.  I am looking for a
computation of the homology group
H_n(S^3\X) in all dimensions (the Z is unnecessary, so please ignore
it).  I apologize if the link is not working; try going to Prof.
Hatcher's homepage located at http://www.math.cornell.edu/~hatcher and
then follow the links to Algebraic Topology.  At that point, you can
choose to download Chapter 2 in either PDF or PS formats.  Lastly, in
terms of time constraints, I would be most grateful if the reply could
be produced by Monday evening or Tuesday morning.  I am actually
trying to prove one more result as well, which I will probably post
here within the next day or so if I cannot come up with a satisfactory
solution on my own (I really should become better versed in AT, but
all my time seems to get sucked away by physics....)

Hi, dubois:

I am interested in solving your problems, but as you know you are on a
short time schedule.  The PDF link did not work for me for some
reason; I determined that Adobe Acrobat Reader (I have version 5.1)
was taking a long time to parse the file, and in the end was unable to
display any part of it (although it seems to "know" it has 88 pages). 
So instead I have downloaded the PostScript version, which I can open
with GSView.

Here is my thinking so far on the first part of your problem; I will
not be offended if another expert is able to solve your problem first.
 I will not lock your problem since your schedule is so short; it may
take a group effort to answer this is so little time.

We can envision S^3 as the one-point compactification of R^3.  If one
excises a circle S^1 from this, then clearly the resulting space is
still (path) connected, but simple loops exist that "entangle" with
the missing S^1 subset and are therefore neither null-homotopic nor
expressible as the boundary of an embedded 2-simplex.

I note that you have asked for the "homology groups" of H_n(S^3\X) and
not the "reduced homology groups".  I'm assuming this is so, and not
an artifact of the notational difficulty of placing a tilde over the
H!  You seem to have done a fine job of finagling the subscripts into
place.

The homology group H_0(K) of a simplicial complex K is always a direct
product of a number of copies of Z corresponding to the number of
(connected) components of K, which as we noted is 1 in this case.  So:

H_0(S^3\X) = Z

One down, infinity minus one to go!  For the sake of discussion, the
reduced zeroth homology group is then trivial:

~
H_0(S^3\X) = {0}

regards, mathtalk-ga

mathtalk computed one homology group, so i think i'll put in a
contribution by computing another infinity-minus-three of them: An
n-dimensional topological manifold without boundary has all homology
groups of order m>n equal to zero.
(This is lemma 6.2 on page 148 of the book "Homology Theory" by James
W. Vick)
Therefore for S^3\X we have H_4 = H_5 = H_6 = ... = 0.

As for H_1, I believe it should be equal to Z, for the reasons
mentioned by mathtalk, although I don't know how to prove this. And
H_2 I would bet is equal to Z, the generator should look like a torus
that enfolds the incised circle. H_3 is a mystery.

Can anybody compute the homology groups for R^3 \ circle ? That would
give us a head start, it might be possible to compute from there using
the Mayer-Vietoris sequence.

Hi, dannidin:

That's a lot of progress! The H_1 case is kind of interesting because
of the relationship to the knot group, the fundamental group of R^3 \
X.  Essentially H_1( R^3 \ X ) is the abelianization of the
fundamental group (homotopy), i.e. mod'ing out the commutator
subgroup.

Now the fundamental groups can vary, depending on how X "sits" inside
R^3, even for "tame" knots.  But evidently from a homology perspective
it should not matter.

regards, mathtalk-ga

Hi again dubois (and mathtalk),

I have a solution for question a. The homology groups of S^3\X are:

H_0 = Z
H_1 = Z
H_ 2 = H_3 = H_4 = ... = 0.

I found this in the same book by Vick that I mentioned in my last
comment. I recently bought this book, wanting to refresh my (very dim)
memory of this beautiful theory and hoping that the financial
investment would give me the necessary incentive... I was reading
through it all weekend to see if I could solve your question, when
this evening I encountered a more general statement as a theorem:
Corollary 1.29 states: If B < S^n is a set homeomorphic to S^k for
0<=k<=n-1, then H(S^n\B) is a free abelian group with two generators,
one in dimension zero and one in dimension n-k-1.

Another surprise: Now that I started writing this comment, I decided
to take a look at Hatcher's book, to see if the proof in Vick's book
uses legitimate tools (it's a bit complicated - uses the
Mayer-Vietoris sequence and some delicate topological arguments); and
I found that this same theorem appears in Hatcher's book: see
Proposition 2B.1(b) on page 169 (there it's phrased in terms of the
reduced homology groups, for the connection see page 110.) The proof
in Hatcher seems to be simpler than the one in Vick. (plus you have
the book so it's simpler for you to read it)

As for question b.: In Vick's book relative homology appears only in
chapter 2. I don't think I'll get there in time to meet your deadline
(just browsing through it does not reveal anything like an immediate
answer). I hope this partial answer still helps you. I would be glad
if you will allow me to post this as an answer. If you want to pay me
for just 1/2 an answer, you may post a new question for the
appropriate sum and explain in the question title that it's for me. If
not - no worries, the knowledge I am gaining from reading this stuff
is worth much more to me ;-)

Cheers,
dannidin

p.s. mathtalk - very interesting comment, I did not realize the
connection to knots. Hatcher says something about it on page 169 -
check it out if you can access the pdf file.

It is very kind of you two to work on the problem without formally
answering it; I sincerely appreciate that.  Very solid progress so
far.  A couple of things.  First, I am indeed looking for the usual
homology groups, rather than the reduced homology groups (although
that tilda would be a tough one...).  Second, in terms of the time
schedule issue: it is true that after Tuesday evening a solution would
mean considerably less to me.  However, I would say that if one of you
is confident that you can solve it by then, just go ahead and do it
even if you think it will take longer than you wish to spend on it for
the asking price.  Then let me know how long it takes you, and I
guarantee I will tip generously for your time (up to doubling the
asking price if necessary) - and same with my other AT question above,
if one of you wishes to solve that one.
   Anyway, thanks again, and please let me know if any other
clarifications are necessary.

I submitted my comment and right after I hit return I saw dannidin's
last comment.  So here's the story.  Dannidin, like mathtalk you went
out of your way to help me and I'll certainly pay you for that.  It
appears that I should have asked these as two separate questions, and
hence, I am going to cancel this questions and pose the two parts as
separate questions.  For the question directed only at you dannidin,
please ignore the very low asking price; I promise you that the tip
will be substantially higher.  However, you'll have to clarify one
thing which I'll explain when I post that question.  Mathtalk, should
you decide to solve one of my other two questions, I will leave you an
additional tip for your helpfulness on this question.  If I am
forgetting anything or one of you has a better suggestion, please let
me know.