The Unapologetic Mathematician

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