Degree and orientation

 Preliminary exam prep

Topology

 Recall that a topological manifold is a Hausdroff space in which every point has a neighborhood homeomorphic to $\R^n$ for some $n$. An orientation on $M$ is a choice of basis of $\R^n$ in each neighborhood such that every path in $M$ keeps the same orientation in each neighborhood. Every manifold $M\owns x$ appears in a long exact sequence (via relative homology) with three terms
$$H_n(M-\{x\}) \tov{f} H_n(M) \tov{g} H_n(M,M-\{x\}).$$
The first term is 0, because removing a point from an $n$-dimensional space leaves only its $(n-1)$-skeleton, which is at most $(n-1)$-dimensional. For $U$ a neighborhood of $x$ in $M$, the last term (via excision) is
$$H_n(M-U^c, M-\{x\}-U^c) = H_n(U, U-\{x\})\cong H_n(\R^n, \R^n-\{x\}) \cong H_n(\R^n, S^{n-1}),$$
which in turn fits into a long exact sequence whose interesting part is
$$H_n(\R^n)\to H_n(\R^n,S^{n-1}) \to H_{n-1}(S^{n-1})\to H_{n-1}(\R^n),$$
and since the first and last terms are zero, $H_n(M,M-\{x\})=\Z$. Since $f$ is zero, $g$ into $\Z$ must be injective, meaning that $H_n(M)=\Z$ or 0.

Theorem: Let $M$ be a connected compact (without boundary) $n$-manifold. Then

1. if $M$ is orientable, $g$ is an isomorphism for all $x\in M$, and
2. if $M$ is not orientable, $g=0$.

Definition: Let $f:M\to N$ be a map of connected, oriented $n$-manifolds. Since $H_n(M)=H_n(N)$ is infinite cyclic, the induced homomorphism $f_*:H_n(M)\to H_n(N)$ must be of the form $x\mapsto dx$. The number $d$ is called the degree of $f$.

In the special case when we are computing the degree for a map $f:S^n\to S^n$, by excision we get
$$\deg(f) = \sum_{x_i\in f^{-1}(y)} \deg\left(H_n(U_i,U_i-x_i)\tov{f_*} H_n(V,V-y)\right),$$ for any $y\in Y$, some neighborhood $V$ of $y$, and preimages $U_i$ of $V$. This is called the local degree of $f$.

Geometry

Let $M$ be a smooth $n$-manifold. Recall $\Omega_M^r$ is the space of $r$-forms on $M$ and $d^r:\Omega^r_M\to \Omega^{r+1}_M$ is the differential map. Also recall the de Rham cohomology groups $H^r(M) = \text{ker}(d^r)/\text{im}(d^{r-1})$.

Definition: An $n$-manifold $M$ is orientable if it has a nowhere-zero $n$-form $\omega\in \Omega^n_M$. A choice of $\omega$ is called an orientation of $M$.

We also have a map $H^n(M)\to \R$, given by $\alpha\mapsto \int_M\alpha$, where the integral is normalized by the volume of $M$, so that integrating 1 across $M$ gives back 1. It is immediate that this doesn't make sense when $M$ is not compact, but when $M$ is compact and orientable, we get that $H^n(M)\neq0$. Indeed, if $\eta\in \Omega^{n-1}_M$ with $d\eta =\omega$, by Stokes' theorem we have
$$\int_M\omega = \int_Md\eta = \int_{\dy M}\eta = \int_\emptyset\eta = 0,$$
as $M$ has no boundary (since it is a manifold). But $\omega$ is nowhere-zero, meaning the first expression on the left cannot be zero. Hence $\omega$ is not exact and is a non-trivial element of $H^n(M)$.

Theorem: Let $M$ be a smooth, compact, orientable manifold of dimension $n$. Then $H^n(M)$ is one-dimensional.

Proof: The above discussion demonstrates that $\dim(H^n(M))\>1$. We can get an upper bound on the dimension by noting that the space of $n$-forms on $M$, given by $\Omega^n_M = \bigwedge^n(TM)^*$, has elements described by $dx_{i_1}\wedge \cdots \wedge dx_{i_n}$, with $\{i_1,\dots,i_n\}\subset \{1,\dots,n\}$. By rearranging the order of the $dx_{i_j}$, every element looks like $\alpha dx_1\wedge \cdots\wedge dx_n$ for some real number $\alpha$. Hence $\dim(\Omega^n_M) \leqslant 1$, so $\dim(H^n(M))$ is either 0 or 1. Therefore $\dim(H^n(M))=1$. $\square$

Definition: Let $f:M\to N$ be a map of smooth, compact, oriented manifolds of dimension $n$. Since $H^n(M)$ and $H^n(N)$ are 1-dimensional, the induced map $f^*:H^n(N)\to H^n(M)$ must be of the form $x\mapsto dx$. The number $d$ is called the degree of $f$. Equivalently, for any $\omega\in \Omega^n_N$,
$$\int_M f^*\omega = d\int_N\omega$$

References: Hatcher (Algebraic Topology, Chapters 2, 3.3), Lee (Introduction to Smooth Manifolds, Chapter 17)