Quail: Cl. mnfld. story

I did this around 1/5/09.  Here M is always an n-manifold. Here orientation always refers with respect to some ring R.  Some generalitiesH_n(M|A):= H_n(M,M - A) is shorthand for local homology.  For B \subset A \subset X there’s an inclusion (X, X - A) \to (X, X - B) giving natural maps H_n(M|A) \to H_n(M|B).

For me one of the big motivations for this story is the following result.

If M is closed and connected then M is compact. This is an easy corollary of 

Orientation result:

a) M orientable implies H_n(M) \to H_n(M|x) is iso.

b) M non orientable implies H_n(M) \to H_n(M|x) is injective with image \{r | 2r = 0\}

c) H_i(M) = 0 for i > n.

pf of compact: take a generator g \in H_n(M) and define an orientation via x \mapsto g_x where g_x \in H_n(M|x). Now g_x \ne 0 only for x in the closure of the image of some cycle \Delta^n \to X representing g.

If g \in H_i(M) misses a point x then it is contained in the complement M - x, hence represents zero in H_n(M|x).

By the result g_x is always nonzero; on the other hand the prev. comment says it can only be nonzero on the image of cycle which has compact image, hence g \colon \Delta^n \to M is surjective and M is compact.

The orientation result is a special case of

Orientation Lemma:

A) A \subset M with A compact, and \sigma \colon M \to M_R then \exists \alpha \in H_n(M|A) s.t. \sigma(x) = \alpha_x for x \in A.

B) H_i(M|A) = 0 for i> n.

proof: Step 1, Reduce to the case M = \mathbb{R}^n.  

For A \subset M arb. right A = A_1 \cup ...\cup A_m where A_i = A \cap U_i and \mathbb{R}^n \cong U_i \subset M. (sim. to cov. spaces, van kampfen, schemes).  Now use an induction argument.

The case n = 1 is covered by the reduction hypothesis.  If it holds for A_1 \cup ... \cup A_{m-1}, then adding A_m, have that it holds for A_1 \cap A_{m}\bigcup ... \bigcup A_{m-1} \cap A_m .  

The induction argument is finished because the result for A,B, A\cap B implies it for A\cup B via the Mayer Veitoris sequence

0 \to H_n(A \cup B) \to \begin{smallmatrix} H_n(A) \\ \oplus \\ H_n(B) \end{smallmatrix} \to H_n(A\cap B) \to ...

B) now follows easily via the l.e.s by things being squeezed by zeros. As for A), basically get classes \alpha_A, \alpha_B, \alpha_{A \cap B} and the l.e.s shows they glue.  As for uniqueness the idea is \alpha_x = o for all x implies \alpha = 0, this is proved via it holds for A,B, etc. and the first map in l.e.s is injective.

Step 2, M = \mathbb{R}^n, and reduce to the case A is convex.  In fact if it holds for convex things it holds for a countable union of convex things via a similar argument as step 1.   

Assuming it holds for convex things, the idea is to treat an arb. \alpha \in H_n(M|A) as if its coming from slightly bigger K which is a union of convex things.  

\alpha \colon \Delta^n \to M with \delta \alpha \in M - A, then the boundary is compact and it has positive distance from A and so each point x \in A has a closed ball nbd not intersection \partial \alpha.  The union of these balls is K.  The key idea, is by construction the relative cycle \alpha \colon \Delta^n \to M = \mathbb{R}^n and it lifts \alpha \in H_n(M|A).

B) can now be proved by taking \alpha to be generator of H_i(M|K), it lifts to zero so it must have been zero to begin with.  As for A), uniqueness uses \alpha_x = 0 for all x \in A means its zero for all x \in K, this is proved via a commutative triangle having legs

H_n(M|K) \to H_n(M|B) \to H_n(M|x)

H_n(M|K) \to H_n(M|x)

where B is any ball B \subset K.  For existence just take the answer for any ball B \supset K.

Step 3, M = \mathbb{R}^n and A convex.

Here the result is evident via H_n(M|A) \cong H_n(M|pt).  B) That the higher homology is zero is clear.  As for A), uniqueness is clear for similar argument as before.  Existence is just taking a generator of H_n(M|pt) \cong H_{n-1}(S^{n-1}) = \mathbb{Z}.




About this entry