# The Unapologetic Mathematician

## A Family of Nontrivial Homology Classes (part 3)

Sorry this didn’t go up as scheduled last week.

We must now show that the forms we’ve defined are closed, and thus that they indeed will define homology classes. That is, $d\omega=0$. We must also show that they cannot be exact, and thus that the homology classes they define are nontrivial.

We defined $\omega$ to be the pullback $\pi^*\hat{\omega}$ of the form $\hat{\omega}$ on the sphere. Thus we can calculate

$\displaystyle d\omega=d\pi^*\hat{\omega}=\pi^*d\hat{\omega}=\pi^*0=0$

which follows since $\hat{\omega}$ is an $n$-form on the $n$-dimensional sphere, and so no nontrivial $n+1$-form exists on $S^n$ to be equal to $d\hat{\omega}$.

As for exactness: if $\omega=d\eta$ for some $n-1$-form $\eta$, then Stokes’ theorem tells us that

$\displaystyle\int\limits_{S^n}\omega=\int\limits_{S^n}d\eta=\int\limits_{\partial S^n}\eta=\int\limits_\varnothing\eta=0$

Since the sphere has an empty boundary.

But it’s also possible to calculate this integral directly.

\displaystyle\begin{aligned}\int\limits_{S^n}\omega&=\int\limits_{S^n}\pi^*\hat{\omega}\\&=\int\limits_{\pi(S^n)}\hat{\omega}\\&=\int\limits_{S^n}\iota_P(du^0\wedge\dots\wedge du^n)\end{aligned}

Now, if $f:U^n\to S^n$ is any parameterization of the sphere, we set $u^i=f^i(x)$ and calculate

\displaystyle\begin{aligned}\int\limits_{S^n}\omega&=\int\limits_{S^n}\iota_P(du^0\wedge\dots\wedge du^n)\\&=\int\limits_{U^n}f^*\left(\iota_P(du^0\wedge\dots\wedge du^n)\right)\\&=\int\limits_{U^n}\left[du^0\wedge\dots\wedge du^n\right]\left(P(f(x)),f_{*x}\frac{\partial}{\partial x^1},\dots,f_{*x}\frac{\partial}{\partial x^n}\right)\\&=\int\limits_{U^n}\left[du^0\wedge\dots\wedge du^n\right]\left(u^{i_0}\frac{\partial}{\partial u^{i_0}},\frac{\partial u^{i_1}}{\partial x^1}\frac{\partial}{\partial u^{i_1}},\dots,\frac{\partial u^{i_n}}{\partial x^n}\frac{\partial}{\partial u^{i_n}}\right)\end{aligned}

We extend $f$ to depend on a new variable $x_0$ which will vary in some small neighborhood of $1$ by defining $\tilde{f}(x_0,x)=x_0f(x)$, so that we have

$\displaystyle\frac{\partial u^i}{\partial x^0}=u^i$

and so we can continue:

\displaystyle\begin{aligned}\int\limits_{S^n}\omega&=\int\limits_{U^n}\left[du^0\wedge\dots\wedge du^n\right]\left(\frac{\partial u^{i_0}}{\partial x^0}\frac{\partial}{\partial u^{i_0}},\frac{\partial u^{i_1}}{\partial x^1}\frac{\partial}{\partial u^{i_1}},\dots,\frac{\partial u^{i_n}}{\partial x^n}\frac{\partial}{\partial u^{i_n}}\right)\\&=\int\limits_{U^n}\det\left(\frac{\partial u^i}{\partial x^j}\right)\left[du^0\wedge\dots\wedge du^n\right]\left(\frac{\partial}{\partial u^0},\dots,\frac{\partial}{\partial u^n}\right)\\&=\int\limits_{U^n}\det\left(\frac{\partial u^i}{\partial x^j}\right)\bigg\vert_{x^0=1}\,d(x^1,\dots,x^n)\end{aligned}

Now, what is this integral calculating? The first column of the determinant is the vector $f(x)$ which, as the position vector of a point on the sphere, has unit length and points perpendicularly to its surface. The other columns form a basis of the tangent space to the sphere at $f(x)$. At any given point, then, we can change the basis so that the upper-left entry in the matrix is $1$ and all other entries in its row and column are zero. The determinant is thus the determinant of the rest of the matrix, which is the Jacobian determinant of the parameterization, and its integral is thus the volume of the image — the volume of the sphere.

Now, we may not have a formula handy for the $n$-dimensional volume of the $n$-sphere, but it’s certainly not zero! Therefore the closed form $\omega$ cannot have been exact, and so it must define a nontrivial homology class, just as asserted.

December 27, 2011