# 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

## A Family of Nontrivial Homology Classes (part 2)

We continue investigating the differential forms we defined last time. Recall that we started with the position vector field $P(p)=\mathcal{I}_pp$ and use the interior product to produce the $n$-form $\hat{\omega}=\iota_P(du^0\wedge\dots\wedge du^n)$ on the punctured $n+1$-dimensional space $\mathbb{R}^{n+1}\setminus\{0\}$. We restrict this form to the $n$-dimensional sphere $S^n$ and then pull back along the retraction mapping $\pi:p\mapsto\frac{p}{\lvert p\rvert}$ to get the form $\omega=\pi^*\hat\omega$.

I’ve asserted that $\omega=\frac{1}{\lvert p\rvert^n}\hat{\omega}$, and now we will prove it; let $\{v_1,\dots,v_n\}$ be $n$ tangent vectors at $p$ and calculate

\displaystyle\begin{aligned}{}[\omega(p)]&(v_1,\dots,v_n)\\&=[[\pi^*\hat{\omega}](p)](v_1,\dots,v_n)\\&=[\hat{\omega}(\pi(p))](\pi_{*p}v_1,\dots,\pi_{*p}v_n)\\&=\left[\hat{\omega}\left(\frac{p}{\lvert p\rvert}\right)\right]\left(\frac{1}{\lvert p\rvert}\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}v_1,\dots,\frac{1}{\lvert p\rvert}\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}v_n\right)\\&=\frac{1}{\lvert p\rvert^n}\left[\hat{\omega}\left(\frac{p}{\lvert p\rvert}\right)\right]\left(\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}v_1,\dots,\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}v_n\right)\\&=\frac{1}{\lvert p\rvert^n}\left[[du^0\wedge\dots\wedge du^n]\left(\frac{p}{\lvert p\rvert}\right)\right]\left(P\left(\frac{p}{\lvert p\rvert}\right),\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}v_1,\dots,\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}v_n\right)\\&=\frac{1}{\lvert p\rvert^n}\left[[du^0\wedge\dots\wedge du^n]\left(\frac{p}{\lvert p\rvert}\right)\right]\left(\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}P(p),\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}v_1,\dots,\mathcal{I}_{\frac{p}{\lvert p\rvert}}\mathcal{I}_p^{-1}v_n\right)\\&=\frac{1}{\lvert p\rvert^n}[du^0\wedge\dots\wedge du^n]\left(\mathcal{I}_p^{-1}P(p),\mathcal{I}_p^{-1}v_1,\dots,\mathcal{I}_p^{-1}v_n\right)\\&=\frac{1}{\lvert p\rvert^n}\left[[du^0\wedge\dots\wedge du^n](p)\right](P(p),v_1,\dots,v_n)\\&=\left[\left[\frac{1}{\lvert p\rvert^n}\hat{\omega}\right](p)\right](v_1,\dots,v_n)\end{aligned}

as asserted. Along the way we’ve used two things that might not be immediately apparent. First: the derivative $\pi_*$ works by transferring a vector from $\mathcal{T}_p\mathbb{R}^{n+1}$ to $\mathcal{T}_{\frac{p}{\lvert p\rvert}}\mathbb{R}^{n+1}$ and scaling down by a factor of $\frac{1}{\lvert p\rvert}$, which is a consequence of the linear action of $\pi_*$ and the usual canonical identifications. Second: the volume form on $\mathcal{T}_p\mathbb{R}^{n+1}$ can be transferred to essentially the same form on $\mathbb{R}^{n+1}$ itself by using the canonical identification $\mathcal{I}_p$.

December 20, 2011

## 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

## Simply-Connected Spaces and Cohomology

We’ve seen that if a manifold $M$ is simply-connected then the first degree of cubic singular homology is trivial. I say that the same is true of the first degree of de Rham cohomology.

Indeed, say that $M$ is simply-connected, so that any closed curve $c$ can be written as the boundary $c=\partial S$ of some surface $S$. Then we take any closed $1$-form $\omega$ with $d\omega=0$. Stokes’ theorem tells us that

$\displaystyle\int\limits_c\omega=\int\limits_{\partial S}\omega=\int\limits_Sd\omega=\int\limits_S0=0$

for any closed curve $c$. But this means that every closed $1$-form $\omega$ is path-independent, and path-independent $1$-forms are exact. And so we conclude that $H^1(M)=0$, as asserted.

It will (eventually) turn out that the fact that both $H_1(M)$ and $H^1(M)$ vanish together is not a coincidence, but is in fact an example of a much deeper correspondence between homology and cohomology — between topology and analysis.

December 17, 2011

## Path-Independent 1-Forms are Exact

Today we prove the assertion from last time: if $\omega$ is a $1$-form on a manifold such that for every closed curve $c$ we have

$\displaystyle\int_c\omega=0$

then $\omega=df$ for some function $f$. As we saw last time, the condition on $\omega$ is equivalent to the assertion that the line integral of $\omega$ over any curve $c$ depends only on the endpoints $c(0)$ and $c(1)$, and not on the details of the path $c$ at all.

So, let’s define a function. In every connected component of $M$, pick some base-point $p$. As an aside, what we really want are the arc components of $M$, but since $M$ is pretty topologically sweet the two concepts are the same. Anyway, if $x$ is in the same component as the selected base-point $p$, we pick some curve $c$ from $p$ to $x$ and define

$\displaystyle f(x)=\int\limits_c\omega$

Remember here that the choice of $c$ doesn’t matter at all, since we’re assuming that $\omega$ is path-independent, so this gives a well-defined function given the choice of $p$.

Incidentally, what would happen if we picked a different base-point $q$? Then we could pick a path $c'$ from $q$ to $p$ and then always choose a path $d$ from $q$ to $x$ by composing $c':q\to p$ and $c:p\to x$. Doing so, we find

$\displaystyle g(x)=\int\limits_d\omega=\int\limits_{c'}\omega+\int\limits_c\omega=K+f(x)$

So the only difference the choice of a base-point makes is an additive constant over the whole connected component in question, which will make no difference once we take their differentials.

Anyway, we need to verify that $df=\omega$. And we will do this by choosing a vector field $X$ and checking that $X(f)=df(X)=\omega(X)$. So, given a point $x$ we may as well choose $x$ itself as the base-point. We know that we can choose an integral curve $c$ of $X$ through $x$, and we also know that

$\displaystyle[X(f)](c(0))=\frac{d}{dt}f(c(t))\bigg\vert_{t=0}$

for an integral curve. For any $t$, we can get a curve $c_t$ from $x=c(0)$ to $c(t)$ by defining $c_t(s)=c(st)$. And so we calculate (in full, gory detail):

\displaystyle\begin{aligned}{}[X(f)](x)&=[X(f)](c(0))\\&=\frac{d}{dt}f(c(t))\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_{c_t}\omega\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_{[0,1]}c_t^*\omega\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[[c_t^*\omega](s)]\left(\frac{d}{ds}\bigg\vert_s\right)\,ds\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[\omega(c_t(s))]\left({c_t}_{*s}\frac{d}{ds}\bigg\vert_s\right)\,ds\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[\omega(c(st))]\left(tc_{*(st)}\frac{d}{dt}\bigg\vert_{st}\right)\,ds\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[\omega(c(st))]\left(c_{*(st)}\frac{d}{dt}\bigg\vert_{st}\right)t\,ds\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^1[\omega(c(st))]\left(c_{*(st)}\frac{d}{dt}\bigg\vert_{st}\right)\,d(st)\bigg\vert_{t=0}\\&=\frac{d}{dt}\int\limits_0^t[\omega(c(u))]\left(c_{*u}\frac{d}{dt}\bigg\vert_u\right)\,du\bigg\vert_{t=0}\\&=[\omega(c(t))]\left(c_{*t}\frac{d}{dt}\bigg\vert_t\right)\bigg\vert_{t=0}\\&=[\omega(c(0))]\left(c_{*0}\frac{d}{dt}\bigg\vert_0\right)\\&=[\omega(x)](X_x)\end{aligned}

So, having verified that at any point $x$ we have $X(f)=df(X)=\omega(X)$, we conclude that $\omega=df$ for the given function $f$, and is thus exact.

December 15, 2011

## Simply-Connected Spaces

We say that a space is “simply-connected” if any closed curve $c$ with $c(0)=c(1)=p$ is homotopic to a constant curve that stays at the single point $p$. Intuitively, this means that any loop in the space can be “pulled tight” without getting caught up on any “holes”.

It turns out that this is equivalent to saying that every closed curve is the boundary of some parameterized square. Indeed, consider the following diagram I’ve drawn with the help of Geogebra:

This is a picture of the homotopy cylinder. The domain of a curve is the interval $[0,1]$, so the domain of the homotopy cylinder is the square $[0,1]$. I’ve labeled the sides to describe what the homotopy does to them: the lower edge $(x,0)$ follows the curve $c$; the upper edge $(x,1)$ is the constant point $p$; the two sides are also constant at $p$, meaning that we’re holding the curve’s ends fixed as we perform the homotopy. And so the homotopy is exactly a continuous (or smooth) map from the square into our space, and the boundary of the parameterized square is exactly the curve $c$. The converse — that any parameterized square can be homotoped to look like this — shouldn’t be hard to see.

So what does this mean for homology? Well, for cubic singular homology it means that $H_1(M)$ is exact if $M$ is simply-connected. Indeed, if $C$ is a closed $1$-chain, then it must be made up of a formal sum of curves. Any curve which isn’t already closed must have a start and an end, and the end must be the start of another curve, or else the boundary points of $C$ wouldn’t cancel off. We can break $C$ up — possibly non-uniquely — into a collection of closed curves, each of which is the boundary of some parameterized square, by the above argument. Thus $C$ is itself the boundary of this collection of squares; since all closed $1$-chains are exact, the first homology vanishes.

December 14, 2011

## The “Hairy Ball Theorem”

We can use the concept of degree to prove the (in)famous “hairy ball theorem”. That is, any smooth vector field defined on the whole sphere $S^n$ for $n$ even must vanish at some point. The name comes from thinking of a “hair” growing out of each point on the sphere, and trying to “comb” them all to lay flat — tangent — against the sphere. If this can be done, then the direction each hair points defines a unit vector at its point, and combing smoothly means we have a smooth vector field. The assertion is that such a “combing” is impossible.

First, I say that if $n$ is even then the antipodal map sending a point $p\in S^n$ to $-p$ is orientation-reversing. Indeed, we first extend to the larger space $\mathbb{R}^{n+1}$; the antipodal map $-1:\mathbb{R}^{n+1}\to\mathbb{R}^{n+1}$ is clearly orientation-reversing, as we can see by taking its Jacobian. This matrix has $n+1$ eigenvalues, all equal to $-1$, so the determinant is $(-1)^{n+1}=-1$.

If $p$ is the position vector of a point in $S^n$, then we consider the canonical identification $\mathcal{I}_p$ of $\mathbb{R}^{n+1}$ with $\mathcal{T}_p\mathbb{R}^{n+1}$ and calculate

$\displaystyle(-1)_*\mathcal{I}_pu=\mathcal{I}_{-p}(-u)=-\mathcal{I}_{-p}u$

So if $\{\mathcal{I}_pu_i\}$ is a positively-oriented basis of $\mathcal{T}_p\mathbb{R}^{n+1}$, then $\{-\mathcal{I}_{-p}u_i\}$ is a negatively-oriented basis of $\mathcal{T}_{-p}\mathbb{R}^{n+1}$.

Now, the position vector field $p\mapsto\mathcal{I}_pp$ is $-1$-related to itself; if $\mathcal{I}_pp$ is used as the first vector in a positively-oriented basis of $\mathcal{T}_p$, then $-\mathcal{I}_{-p}p=\mathcal{I}_{-p}(-p)$ is the first vector in the antipodal basis, which we know is negatively-oriented. But this means that the positively-oriented basis of $\mathcal{T}_pS^n$ must be flipped to a negatively-oriented basis of $\mathcal{T}_{-p}S^n$.

So what does this mean? The identity map has degree $1$, while we can calculate that the antipodal map has degree $-1$. Since these are different, the two maps must not act identically on homology, and therefore cannot be homotopic.

But now let’s assume that $X$ is an everywhere-nonzero vector field which, without loss of generality, we may assume to have constant length $1$ — use the inner product we get by taking $\mathcal{T}_pS^n\subseteq\mathcal{T}_p\mathbb{R}^{n+1}$ and divide by the original length to normalize. For each point $p$ we can define the great circle

$\displaystyle c_p(t)=(\cos(\pi t))p+(\sin(\pi t))\mathcal{I}_p^{-1}X(p)$

This is a curve that lies on the sphere for every $p$. Indeed, we can check that

\displaystyle\begin{aligned}\langle c_p(t),c_p(t)\rangle=&\langle(\cos(\pi t))p+(\sin(\pi t))\mathcal{I}_p^{-1}X(p),(\cos(\pi t))p+(\sin(\pi t))\mathcal{I}_p^{-1}X(p)\rangle\\=&\langle(\cos(\pi t))p,(\cos(\pi t))p\rangle\\&+\langle(\cos(\pi t))p,(\sin(\pi t))\mathcal{I}_p^{-1}X(p)\rangle\\&+\langle(\sin(\pi t))\mathcal{I}_p^{-1}X(p),(\cos(\pi t))p\rangle\\&+\langle(\sin(\pi t))\mathcal{I}_p^{-1}X(p),(\sin(\pi t))\mathcal{I}_p^{-1}X(p)\rangle\\=&(\cos(\pi t))^2\langle p,p\rangle\\&+\cos(\pi t)\sin(\pi t)\langle p,\mathcal{I}_p^{-1}X(p)\rangle\\&+\sin(\pi t)\cos(\pi t)\langle\mathcal{I}_p^{-1}X(p),p\rangle\\&+(\sin(\pi t))^2\langle\mathcal{I}_p^{-1}X(p),\mathcal{I}_p^{-1}X(p)\rangle\\=&(\cos(\pi t))^2+(\sin(\pi t))^2=1\end{aligned}

So if we define $H:S^n\times[0,1]\to S^n$ by $H(p,t)=c_p(t)$, then we have a homotopy from the identity map on $S^n$ to the antipodal map, which is exactly what we just showed could not exist. Thus we conclude that no such non-vanishing vector field can exist on an even-dimensional sphere.

December 13, 2011

## Calculating the Degree of a Proper Map

As I asserted yesterday, there is a simple formula for the degree of a proper map $f:M\to N$. Pick any regular value $q\in N$ of $f$, some of which must exist, since the critical values have measure zero in $N$. For each $p\in f^{-1}(q)$ we define $\mathrm{sgn}_pf$ to be $+1$ or $-1$ as $f$ is orientation-preserving or orientation-reversing. Then I say that

$\displaystyle\deg f=\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf$

The key is the inverse function theorem: the Jacobian of $f$ must have maximal rank at $p$, so there’s some $U\subseteq M$ around $p$ on which $f:U\to f(U)\subseteq N$ is a diffeomorphism. This is true for each $p\in f^{-1}(q)$, so we get an open $U_p$ around each $p$. Since $f$ is proper, there are only finitely many such $p$, and they’re all separated from each other, so we can shrink each $U_p$ so that they’re all disjoint. Then we can set $V\subseteq N$ to be the intersection of all the $f(U_p)$, and we can further shrink each $U_p$ until its image is exactly $V$.

That is, we’ve started with a regular value $q\in N$ and its preimage $f^{-1}(q)$, consisting of a bunch of points $p\in M$; we’ve used the fact that $f$ is proper to widen it to an open set $V$ and its preimage, consisting of a bunch of disjoint open sets $U_p\subseteq M$. Since $f:U_p\to V$ is a diffeomorphism for each $p\in f^{-1}(q)$, $f$ is constantly orientation-preserving or orientation-reversing over all of $U_p$, though of course each $U_p$ can be different.

Now if we pick some $\omega\in\Omega_c^n(N)$ supported in $V$, then $f^*\omega$ is supported in the disjoint union of the $U_p$. So we can calculate

\displaystyle\begin{aligned}\int\limits_Mf^*\omega&=\int\limits_{\biguplus_{p\in f^{-1}(q)}U_p}f^*\omega\\&=\sum\limits_{p\in f^{-1}(q)}\int\limits_{U_p}f^*\omega\\&=\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf\int\limits_{f(U_p)}\omega\\&=\left(\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf\right)\int\limits_V\omega\\&=\deg f\int\limits_V\omega\end{aligned}

Since the coefficient $\deg f$ is independent of the choice of $\omega$, this proves the assertion.

December 10, 2011

## The Degree of a Map

This will be a nice, simple definition. As we saw last time, the top-degree compactly-supported de Rham cohomology is easy to calculate for a connected, oriented manifold $M$. Specifically, we know that $H_c^n(M)\cong\mathbb{R}$.

So, let’s say that $f:M\to N$ is a smooth map from one connected, oriented $n$-manifold to another. The pullback $f^*$ induces a map on cohomology: $f^*:H_c^n(N)\to H_c^n(M)$. But since each of these cohomology spaces is (isomorphic to) $\mathbb{R}$, $f^*$ must act on them by multiplication by some number. That is, if $\omega\in\Omega_c^n(N)$ is a closed $n$-form, and if the integral of $\omega$ over $N$ is $I$, then $f^*\omega\in\Omega_c^n(M)$ is also a closed $n$-form, and the integral of $f^*\omega$ over $M$ is $d_fI$, where $d_f$ is some real number that depends only on the function $f$ and not on the form $\omega$. That is, the same $d_f$ works for all $\omega$.

We call this number the “degree” of $f$, and write $\deg f$. It turns out that when $f$ is a proper map — one for which the preimage of any compact set is compact, which is always the case when $M$ is itself compact — that we have a good way of calculating it. If $q\in N$ is a regular value of $f$, some of which which will always exist, then the preimage of $q$ consists of a finite collection of points $p\in M$. For each one we calculate the “signum” of $f$ at $p$ — written $\mathrm{sgn}_pf$ — to be $+1$ if $f_{*p}$ is orientation-preserving and $-1$ if $f_{*p}$ is orientation-reversing. I say that

$\displaystyle\deg f=\sum\limits_{p\in f^{-1}(q)}\mathrm{sgn}_pf$

While we will not prove this until next time, there is one immediate consequence: if $f$ is not onto, then any point in $N\setminus f(M)$ has an empty preimage, and so we find that $\deg f=0$. In a sense, the degree of $f$ is counting how many times the image of $M$ covers that of $N$, considering orientation.

December 9, 2011

## Nonvanishing Compactly-Supported de Rham Cohomology

Last time we saw that compactly-supported de Rham cohomology is nonvanishing in the top degree for $\mathbb{R}^n$. I say that this is true for any oriented, connected $n$-manifold $M$. Specifically, if $\omega\in\Omega_c^n(M)$, then the integral of $\omega$ over $M$ is zero if and only if $\omega=d\eta$ for some $\eta\in\Omega_c^{n-1}(M)$. That the second statement implies the first should be obvious.

To go the other way takes more work, but it’s really nothing much new. Firstly, if $\omega$ is supported in some connected, parameterizable open subset $U\subseteq M$ then we can pull back along any parameterization and use the result from last time.

Next, we again shift from our original assertion to an equivalent one: $\omega_1$ and $\omega_2$ have the same integral over $M$, if and only if their difference is exact. And again the only question is about proving the “only if” part. A partition of unity argument tells us that we only really need to consider the case where $\omega_i$ is supported in a connected, parameterizable open set $U_i\subseteq M$; if the integrals are zero we’re already done by using our previous step above, so we assume both integrals are equal to $c\neq0$. Dividing by $c$ we may assume that each integral is $1$.

Now, if $p_0\in M$ is any base-point then we can get from it to any other point $p\in M$ by a sequence of connected, parameterizable open subsets $W_i$. The proof is basically the same as for the similar assertion about getting from one point to anther by rectangles from last time. We pick some such sequence taking us from $U_1$ to $U_2$, and just like last time we pick a sequence of forms $\alpha_i$ supported in $W_{i-1}\cap W_i$. Again, the differences between $\omega_1$ and $\alpha_1$, between $\alpha_i$ and $\alpha_{i+1}$, and between $\alpha_N$ and $\omega_2$ are all exact, and so their sum — the difference between $\omega_1$ and $\omega_2$ is exact as well.

And so we conclude that the map $\Omega_c^n(M)\to\mathbb{R}$ given by integration is onto, and its kernel is the image of $\Omega_c^{n-1}(M)$ under the exterior derivative. Thus, $H_c^n(M)\cong\mathbb{R}$, just as for $\mathbb{R}^n$.

December 8, 2011