Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm reading Madsen and Tornehave's "From Calculus to Cohomology" and tried to solve this interesting problem regarding knots.

Let $\Sigma\subset \mathbb{R}^n$ be homeomorphic to $\mathbb{S}^k$, show that $H^p(\mathbb{R}^n - \Sigma)$ equals $\mathbb{R}$ for $p=0,n-k-1, n-1$ and 0 for all other $p$. Here $1\leq k \leq n-2$.

Now the case $p=0$ is obvious from connectedness and the two other cases are easily solved by applying the fact that

$H^{p+1}(\mathbb{R}^{n+1} - A) \simeq H^p(\mathbb{R}^n - A),~~~~p\geq 1$

and

$H^1(\mathbb{R}^n - A) \simeq H^0(\mathbb{R}^n - A)/\mathbb{R}\cdot 1$

So what is my problem, really?

Now instead let's look at this directly from Mayer-Vietoris. If $\hat{D}^k$ is the open unit disk and $\bar{D}^k$ the closed. Then $\mathbb{R}^n - \mathbb{S}^k = (\mathbb{R}^n - \bar{D}^k)\cup (\hat{D}^k)$ and $(\mathbb{R}^n - \bar{D}^k)\cap (\hat{D}^k) = \emptyset$

Now $H^p(\mathbb{R}^n - \bar{D}^k) \simeq H^p(\mathbb{R}^n - \{ 0 \})$ since $\bar{D}^k$ is contractible. And $H^p(\mathbb{R}^n - \{ 0 \})$ is $\mathbb{R}$ if $p=0,n-1$ and 0 else. Since $\hat{D}^k$ is open star shaped we find it's cohomology to be $\mathbb{R}$ for $p=0$ and 0 for all other $p$.

This yields and exact sequence

$\cdots\rightarrow 0\overset{I^{\*}}\rightarrow H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \overset{J^{\*}}\rightarrow \mathbb{R} \rightarrow 0\cdots$

So due to exactness I find that $\ker(J^*) = \text{Im}(I^*) = 0$' and that $J^*$' is surjective, hence $H^{n-1}(\mathbb{R}^n - \mathbb{S}^k) \simeq \mathbb\{R}$.

But ... If I apply the exact same approach to $p = n-k-1$ my answer would be $0$ for $H^{n-k-1}(\mathbb{R}^n - \mathbb{S}^k)$.

Where does this last approach fail? (Also - my TeX doesn't seem to render properly when published, help?)

share|improve this question

2 Answers 2

up vote 10 down vote accepted

To apply the Mayer-Vietoris sequence, you need subspaces whose interiors cover your space (see e.g. Wikipedia, or Hatcher, p. 149). This is not true in your example, because a k-disk in Rn has empty interior for k<n.

You might also enjoy deriving this result from Alexander duality.


Edit: Carsten makes an excellent point, which is that the hard part is to show that the homology of the complement is independent of the embedding. You did say that this is proved in your book, but I wanted to point out that this is quite difficult and arguably surprising.

1) One embedding that satisfies the conditions of the theorem is the Alexander horned sphere, a "wild" embedding of S2 into S3 (this animation is quite nice too). While it's true that the outer component of the complement has the homology of a point, it is very far from being simply connected -- in fact its fundamental group is not finitely generated. (You can find an explicit description of its fundamental group in Hatcher, p. 170-172.)

2) Every knot is an embedding of S1 into S3. The fundamental group of the complement is a strong knot invariant, and is usually much more complicated than just Z. Since H1 of the knot complement is the abelianization of the knot group, the result you are using implies that all knot groups have infinite cyclic abelianization. This is true (it can be seen nicely from the Wirtinger presentation), but it's not obvious.

3) It is important that the ambient space is a sphere (or equivalently Rn). For a simple example where the theorem breaks down, consider embeddings of S1 into a surface Σg of genus g≥2. Taking g=2 for simplicity, we see that there are three topologically inequivalent ways of embedding a circle into Σ2: A) a tiny loop enclosing a disk; B) a loop encircling the waist of the surface and separating it into two components, each of genus 1; and C) a loop going through one of the handles, which does not separate the surface at all.

The homology groups of Σ2 are H0=Z, H1=Z4, and H2=Z. For both A) and B), the complement of S1 has homology groups H0=Z2 because the curve separates, and H1=Z4. However, for C) we have H0=Z because the complement is connected, and H1=Z3 because we have "interrupted" one of the elements [you can see where it went by looking at the Mayer-Vietoris sequence]. Thus we see that the homology of the complement depends essentially on the embedding into the surface Σg, in contrast with the classical case of embedding a circle into the sphere S2.

share|improve this answer

Your question has been answered by Tom. But I am also not sure if you are aware that the point of the problem was to show that the cohomology of a sphere embedded in euclidean space is independent of the embedding. You seem to think only of the standard embedding. As Tom mentioned, duality is one way to prove this independence.

(This might have been better as a comment, but I am not yet allowed to comment.)

share|improve this answer
    
From Tom's comment I figured that I couldn't use those two sets as a covering, as both of them aren't open! Now I'm not really sure what you are trying to say and my book doesn't say anything about duality (at least not explicitly, anyways). Now the reason is that I only consider $\mathbb{S}^k$ is that it is closed and homeomorphic to $\Sigma$, hence $H^p(\mathbb{R}^n - \mathbb{S}^k)$ = $H^p(\mathbb{R}^n - \Sigma)$. Are there anything I seem not to be aware of? –  Magnus Botnan Nov 24 '09 at 22:56
    
Ok, I do not know the book. Generally, the hard part is to deduce from $\Sigma\approx\mathbb S^k$ that $H^p(\mathbb R^n-\Sigma)\cong H^p(\mathbb R^n-\mathbb S^k)$. It is hard, because the complements are not homotopy equivalent in general. But maybe, this was proven somewhere in the book. –  Carsten Schultz Nov 25 '09 at 22:24
    
Yep, it is proved! (I tried to find the proof in the book through google books but unfortunately those two pages are removed from the preview.) –  Magnus Botnan Nov 26 '09 at 1:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.