Sam Auyeung

Sam Auyeung

The Hopf Degree Theorem and Framed Cobordism

6 minute read

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Let $M,N$ closed topological manifolds of dimension $n$. Then their top integral homology are generated by their respective fundamental classes $[M],[N]$. The degree of a continuous map $f:M \to N$ is defined by what the induced map $f_*:H_n(M,\mathbb{Z}) \to H_n(N,\mathbb{Z})$ does which can only be multiplication by a some integer $k$; $f_*([M])=k[N]$. We’ll denote this as $\deg f = k$.

Theorem: Let $M^n$ be a closed topological $n$ manifold. $f,g:M \to S^n$ are homotopic if and only if $\deg f = \deg g$.

Proof: Let’s consider $S^n$ and attach higher dimensional handles in order to kill off the higher homotopy groups. Thus, we have $S^n \subset K(\mathbb{Z},n)$, the Eilenberg-MacLane space. Now, any map $f:M \to K(\mathbb{Z},n)$ can be deformed to a map $M \to S^n$ because $M$ has, at most, $n$-cells. Hence, the homotopy classes of maps $[M,S^n] = [M,K(\mathbb{Z},n)] = H^n(M,\mathbb{Z}) = \mathbb{Z}$. Now, we just need to show that $[M,S^n] = \mathbb{Z}$ is classified by degree. We can view an element $f \in [M,S^n]$ as telling us what integer to multiply by: $f^*[S^n] = k[M]$ and $k$ is exactly the degree. $\square$

Let’s move into the smooth category and work with smooth manifolds and smooth maps. We can still discuss homotopy and it turns out that continuous maps are always homotopic to smooth ones. For the next part, what I write is largely from Ch. 7 of Milnor’s excellent Topology from a Differential Viewpoint.

The concept of degree still makes sense for smooth maps of course but there’s an alternative definition.

Definition: Let $f:M \to N$ be a smooth map between two smooth manifolds. A regular value $y \in N$ is a points such that for each $x \in f^{-1}(y)$, $df_x:T_xM \to T_yN$ is a surjective linear map.

A very important result is Sard’s Theorem which establishes that the set of regular values for a map $f:M \to N$ is dense in $N$. In fact, since manifolds are locally Euclidean, we can patch together a measure on $N$ and the theorem is that the complement to the set of regular values has measure 0.

When the manifolds have the same dimension, then the expected dimension of the fiber $f^{-1}(y)$ is 0. If we have an assignment of local orientations to the manifolds (the easiest case being a global orientation on orientable manifolds), we can choose bases for any $T_xM$ and $T_yN$. If the fiber is a finite number of points, we can assign a $\pm 1$ to each point $x$ in the fiber, depending on whether $df_x$ preserves the orientation of the basis. The sum of these signs $\sum_{x \in f^{-1}(y)} \text{sgn}(x)$ equals the degree of $f$ as defined above.

Shifting gears for a bit, let’s discuss framed cobordisms which have a rich history for its connection to stable homotopy theory via Pontryagin-Thom.

Definition: Let $X$ be a compact manifold. A framing of a submanifold $Z \subset X$ with codimension n in X is a smooth map $\alpha$ assigning to every $x \in Z$ a basis $\alpha(x)$ for the orthogonal complement of its tangent space $(T_x Z)^\perp \subset T_xX$. The pair $(Z,\alpha)$ is a framed submanifold of $X$.

A smooth map $f: X \to S^n$ with regular value $y\in S^n$ induces a framing on its preimage submanifold $f^{-1}(y)$; the dimension of $X$ does not have to be $n$. For each $x \in f^{-1}(y)$ the derivative $df_x: T_x X \to T_y S^n$ is surjective with kernel $T_x(f^{-1}(y))$. Hence $df_x$ maps the orthogonal complement $T_x(f^{-1}(y))^\perp$ isomorphically onto $T_y S^n$. We can pick an orientation on $S^n$ and a positively oriented basis $\beta$ of $T_y S^n$ which we can pull back via $f$. So we define $(f^{-1}(y),f^*\beta)$ to be the framed manifold called the Pontryagin manifold.

In this context of $X$ being the ambient space, we say that two framed submanifolds $(Z,\alpha)$ and $(Z’,\alpha’)$ of $X$ are framed cobordant if there is a submanifold of $Y \subset X \times [0,1]$ with $\partial Y = (Z\times 0) \sqcup (Z’ \times 1)$ and whose framing agrees with $Z$ and $Z’$. Framed cobordism is an equivalence relation. Note this is more restrictive than the usual definition of framed cobordant which are abstract cobordisms, not requiring an embedding into $X \times [0,1]$. We’ll now present a series of results without proof but it’s fun to try to visualize what’s going on. I think the two lemmas are intuitively plausible.

Lemma: If $y,z$ are two regular values of $f$, then $f^{-1}(y)$ is framed cobordant to $f^{-1}(z)$.

Lemma: Any compact framed submanifold $(Z,\alpha)$ of codimension $n$ in a compact manifold without boundary $X$ occurs as the Pontryagin manifold for some smooth map $f:X \to S^n$. Moreover, any framed cobordism $(Y,\alpha)$ of codimension $n$ in the compact manifold with boundary $X \times [0,1]$ occurs as the preimage manifold for some smooth map $F: X \times [0,1]\to S^n$.

This function $F:X \times [0,1] \to S^n$ might remind you of a homotopy. Indeed, we can use these lemmas to prove a generalization of the Hopf theorem which is:

Theorem: Let $X$ be a smooth, connected, oriented, boundaryless manifold. Two smooth maps $f,g:X \to S^n$ are smoothly homotopic if and only if their Pontryagin manifolds correspond to the same framed cobordism class.

To see why it’s a generalization, we use it to provide another proof of the Hopf theorem in the smooth setting. In the case that $\dim X = n$, the Pontryagin manifold is a finite collection of points and the framing amounts to whether the map preserves the orientation of a basis or reverses it; we get a $+$ or $-$ sign. The definition of degree is simply the sum of these signs. Frame cobordisms would be paths in this context. So pair off points of opposite orientation number in $f^{-1}(y)$ with cobordisms to the empty set until there remain points of a single orientation number, then do the same for $g^{-1} (y)$. The degree is precisely equal to the remaining orientation numbers so we may connect each remaining point in $f^{-1}(y)$ to a remaining point in $g^{-1}(y)$ by a cobordism.

Corollary: If $X$ is as above but nonorientable, then $f,g:X \to S^n$ are homotopic if and only if they have the same degree mod 2.

We can use loops to flip the orientation of a basis when we’re in a nonorientable manifold. Just travel from the point and then return after a global journey to come back, very close to the point. So this allows us to connect any pair of points, not only ones with particular signs.

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