The Unapologetic Mathematician

Mathematics for the interested outsider

A Family of Nontrivial Homology Classes (part 2)

December 20, 2011 Posted by | Differential Topology, Topology | Leave a comment

A Family of Nontrivial Homology Classes (part 1)

We want to exhibit a family of closed n-forms that aren’t exact, albeit not all on the same space. In fact, there forms will provide models for every possible way a nontrivial homology class can arise.

For each n, we consider the space \mathbb{R}^{n+1}\setminus\{0\} consisting of the normal n+1-dimensional real affine space with the origin removed. Key to our approach will be the fact that we have a “retract” — a subspace \iota:U\hookrightarrow X along with a “retraction mapping” \pi:X\to U such that \pi\circ\iota=1_U. That is, the retraction mapping sends every point in X to some point in U, and the points that were in U to begin with stay exactly where they are. Explicitly in this case, the “punctured” n+1-dimensional space retracts onto the n-dimensional sphere by the mapping p\mapsto\frac{p}{\lvert p\rvert}, which indeed is the identity on the unit sphere \lvert p\rvert=1.

Now, in this space we take the position vector field P(p), which we define by taking the canonical identification \mathcal{I}_p:\mathbb{R}^{n+1}\to\mathcal{T}_p\mathbb{R}^{n+1} and applying it to the vector p itself: P(p)=\mathcal{I}_pp. We also take the canonical volume form du^0\wedge\dots\wedge du^n, and we use the interior product to define the n-form \hat{\omega}=\iota_P(du^0\wedge\dots\wedge du^n).

Geometrically, the volume form measures n+1-dimensional volume near any given point p. Applying the interior product with P is like rewriting the volume form in terms of a different basis so that P(p) is the first vector in the new basis and all the other vectors are perpendicular to that one, then peeling off the first term in the wedge. That is, \hat{\omega} measures n-dimensional volume in the space perpendicular to p — tangent to the sphere of radius \lvert p\rvert at the point p.

Next we restrict this form to S^n, and we pull the result back to all of \mathbb{R}^{n+1} along the retraction mapping \pi:\mathbb{R}^{n+1}\to S^n, ending up with the form \omega. I say that the net effect is that \omega=\frac{1}{\lvert p\rvert^n}\hat{\omega}, but the proof will have to wait. Still, the form \omega is the one we’re looking for.

December 20, 2011 Posted by | Differential Topology, Topology | 3 Comments



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