A Can Alexander Duality Help Compute the Homology Groups of R^3 Minus a Tube?

[PsychonautQQ]

Technically speaking, the problem that I'm working on involves taking the complement of a tube around a knot (an embedding of a circle into $R^3$) and calculating the homology group of this space. The approach that I'm using is to use a mayer-vietrois sequence.

So let $K$ be a knot and let $u$ be a tube (or neighborhood) around the knot. We can visualize this as making the knot 'thicker' and thus giving it volume.

I want to calculate the homology group $H_i(R^3 \ u)$. And actually, I was given the homology, I just need to prove it. If $i=0$ then the group equals the infinite cyclic group $Z$, and it also does when $i=1$. However, it equals zero if $i>1$

So yes, I'm trying to understand the homology groups of $H_i(R^3 \ u)$.

In doing so I am using this relative homology group:
$H_i(R^3, cl(R^3 \ u))$ where $cl(R^3 \ u)$ is the closure of $(R^3 \ u)$. I notice that this group is isomorphic to $H_i(u, Bd(u))$,

$H_i(R^3, cl(R^3 \ u))$ $=$ $H_i(u, Bd(u))$
where $=$ is denoting an isomorphism here. I know that these groups are isomorphic by using the excorcism theorem and excorcising $int(R^3 \ u)$ where $int(R^3 \ u)$ is the interior of $R^3 \ u$.

Also, I think it will be worth noting that $R^3 = cl(u) \bigcup (R^3 \ int(u))$.

Also, $(R^3 \ int(u)) = Bd(u)$ = $S^1 \times S^1$

Anyone know much about homology? I'd appreciate some insight!

 

[andrewkirk]

It's several years since I did any homology theory and I'm very rusty, so I'm going out on a limb here. But here goes anyway...

I can see why $H_0(\mathbb R^3-u)\cong \mathbb Z$. It's because any point in $\mathbb R^3$ (a 0-simplex) is a cycle but not a boundary, and generates the entire homology group because subtracting any other point from that point gives the boundary of a 1-simplex. Hence the group is infinite cyclic.

Regarding $H_1$, I presume a generator is any 1-cycle that circumnavigates the tube exactly once, since such a cycle cannot be the boundary of any 2-simplex. 1-cycles that do not circumnavigate the tube at all are boundaries of a 2-simplex, and hence are zero in $H_1$. The tricky bit would be to prove that more complex 1-cycles that may circumnavigate many times, possibly in knotted ways, must differ from the single-circumnavigation simplex by only boundaries of 2-simplices.

The proposal to use the excision theorem (NB, not exorcism theorem, although that sounds like it would be much more exciting) sounds promising.

You haven't said that the tube $u$ is open, but you have implied such an assumption when you take relative homologies, since you assume that $\mathbb R^3 - int(\mathbb R^3 - u) = u$ and $cl(\mathbb R^3-u) - int(\mathbb R^3-u)=Bd(\mathbb R^3-u)$.

If $u$ is open then $H_i(u,Bd(u))$ does not exist, since the second argument is required to be a subspace of the first. We would need to replace it by $H_i(Cl(u),Bd(u))$. Fortunately, that's exactly what we get when we excise $int(\mathbb R^3-u)$ from $\mathbb R^3$.

So we need to compute $H_i(Cl(u),Bd(u))$.

Your observation about the Union is correct. It is also correct that $Md(u)\cong S^1\times S^1$.

My guess is that the Relative Mayer-Vietoris sequence may be a useful tool for the job. The trick is to make a good choice of what $A,B,C,D,X,Y$ should be.

 

[PsychonautQQ]

Thank you. Even with your brilliant insight's, I have failed to proceed with the computations :-(.

 

[mathwonk]

can you use alexander duality? this implies that the homology of the complement of the knot in R^3 depends only on the topology of the circle and not on the manner of its embedding in R^3. Hence you can do the computation for an unknotted circle.

in particular one can see in this case that a generating one cycle would be a loop which, if filled in by a disc in R^3, the disc would intersect the circle once. This corresponds to the fact from duality that a generating one cycle for the complement of the circle, corresponds to a generating one-cocycle for the circle, i.e. something that operates on one - cycles for the circle.

Is this clear? I.e. take a loop in the complement of the knot, representing an elememnt of 1st homology of that complement. If it bounds a disk also in the complement of the knot then it is a trivial one cycle, even homotopically trivial. If noit then the disc it bounds must meet the knot, so by counting the number of intersecvtions of that disc with the knot we get an operator on the generating one cycle for the 1st homology of the knot. In this way each one cycle in the complemnt of the knot gives rise to an operator on one cycles of the knot itself. I.e. there is a map fom one cycles in the knot complement to one cocyles of the knot. This map induces an isomorphism from the first homology of the kmnot complement to the 1st cohomology of the knot.

I admit I just made this up, but it seems so natural that I believe it is correct. And it does agree with the result from the well known alexander duality theorem. I.e. the 1st homology of the knot complement is infinite cyclic.