The Groove of Math

or in other words, the Devil is in the Details

Random Exercise #4: The nonexistent 5-Manifold in Bredon’s book

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Dear all, sorry for not having posted anything so far but my master thesis (stable classification of certain families of four manifolds) and PhD interviews have been keeping me very busy 🙂

This is a quick exercise which I was asked to solve by a friend of mine and since I found it funny I decided to spell some more details about it in order to show the importance of the Bockstein exact sequence. This could be a typical exam exercise in “Topology 2” here in Bonn, so dear fellows students, read and comment if anything is unclear!

The following is exercise 7, page 366 from Bredon’s Topology and Geometry:

Show that there can be no 5-manifold M with

screen-shot-2017-02-13-at-18-07-17

Clearly one has to reason by absurd, so let us assume that there exists indeed such manifold M. The absurd will be given by the Bockstein homomorphism.

First of all, a quick application of UCT gives us the following:

screen-shot-2017-02-13-at-18-09-33

Let us denote with a \in H^2(M; Z/3) a generator of Z/3. Let b \in H^3(M;Z/3) a generator of Z/3.

Claim 1 We can choose b=\beta(a), where \beta is the Bockstein homomorphism associated to the s.e.s.

0 \to Z/3 \to Z/9 \to Z/3 \to 0

Proof: It’s enough to prove that \beta is an isomorphism. Let us have a look at the relevant piece of the exact sequences for

0 \to Z \to Z \to Z/3 \to 0  and 0 \to Z/3 \to Z/9 \to Z/3 \to 0

screen-shot-2017-02-13-at-18-14-51

exactness gives us that \overline{\beta} is an isomorphism and naturality w.r.t. coefficients gives us that \beta is an isomorphism as claimed.

Claim 2 a\cup b \neq 0 \in H^5(M;Z/3).

Proof: Start by noticing that H^2(M;Z/3)=\hom (H_2(M;Z/3);Z/3), since we are working with field coefficients. Therefore we can consider (i.e. it exists) an element a_* \in H_2(M;Z/3) as the dual of a. By Poincaré Duality we have that for k=1 or 2 (but not 0!!), k\cdot a_* = b\cap [M]. Therefore the pairing:

Screen Shot 2017-02-13 at 18.32.17.png

proves that a \cup b \neq 0 \in H^5(M;Z/3).

In order to finish the exercise, let us compute \beta(a^2), recall that by assumption H^4(M;Z/3)=0:

0= \beta(0) = \beta(a^2)= 2a\beta(a)= 2\cdot k \neq 0

where we used the fact that the Bockstein for our s.e.s. is a derivation (you can find it, as an exercise on Bredon’s book. You can prove it by hand via diagram chasing of the definition).

This is a clear absurd, which concludes our proof.

Note that the exercise can be generalized to 5-dim Poincaré complexes X with the same homology groups but H_2(X;Z)=Z/p, with p any odd prime. Think about it, so you can see where we used what, which is always important to know.

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Author: RicPed

Young mathematician :)

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