Loose ends of smooth manifolds

 Preliminary exam prep

Here we round up some theorems that have escaped previous roundings-up. Let $X,Y$ be smooth manifolds and $f:X\to Y$ a smooth map.

Theorem: (Inverse function theorem) If $df_p$ is invertible for some $p\in M$, then there exist $U\owns p$ and $V\owns f(p)$ connected such that $f|_U:U\to V$ is a diffeomorphism.

Corollary: (Stack of records theorem) If $\dim(X)=\dim(Y)$, then every regular value $y\in Y$ has a neighborhood $V\owns y$ such that $f^{-1}(Y)=U_1\sqcup \cdots \sqcup U_k$, where $f|_{U_i}:U_i\to V$ is a diffeomorphism.

Proof: Since $y\in Y$ is a regular value, $df_x$ is surjective for all $x\in f^{-1}(y)$. Since $\dim(X)=\dim(Y)$ and $df_x$ is linear, $df_x$ is an isomorphism, hence invertible. By the inverse function theorem, there exist $U\owns x$ and $V\owns y$ connected such that $f|_U:U\to V$ is a diffeomorphism. Before we actually apply this, we need to show that $f^{-1}(y)$ is a finite set.

First we note that by the preimage theorem, since $y$ is a regular value, $f^{-1}(y)$ is a submanifold of $X$ of dimension $\dim(X)-\dim(Y)=0$. Next, if $f^{-1}(y) = \{x_i\}$ were infinite, since $X$ is compact, there would be some limit point $p\in X$ of $\{x_i\}$. But then by continuity,
\[
y = \lim_{i\to \infty}\left[f(x_i)\right] = f\left(\lim_{i\to\infty}\left[x_i\right]\right) = f(p),\]so $p\in f^{-1}(y)$. But then either $p$ cannot be separated from other elements of $f^{-1}(y)$, meaning $f^{-1}(y)$ is not a manifold, or the sequence $\{x_i\}$ is finite in length. Hence $f^{-1}(y) = \{x_1,\dots,x_k\}$. Let $U_i\owns x_i$ and $V_i\owns y$ be the sets asserted to exist by the inverse function theorem (the $U_i$ may be assumed to be disjoint without loss of generality). Let $V = \bigcap_{i=1}^k V_i$ and $U_i' = f^{-1}(V)\cap U_i$, for which we still have $f|_{U_i'}:U_i'\to V$ a diffeomorphism. $\square$

Theorem: (Classification of manifolds) Up to diffeomorphism,

• the only 0-dimensional manifolds are collections of points,
• the only 1-dimensional manifolds are $S^1$ and $\R$, and
• the only 2-dimensional compact manifolds are $S^2\# (T^2)^{\#n}$ or $S^2\#(\R\P^2)^{\#n}$, for any $n\geqslant 0$.

Compact 2-manifolds are homeomorphic iff they are both (non)-orientable and have the same Euler characteristic. Note that
\[
\chi\left(S^2\# (T^2)^{\#n}\right) = 2-2n,
\hspace{1cm}
\chi\left(S^2\#(\R\P^2)^{\#n}\right) = 2-n.
\]
These surfaces are called orientable (on the left) and non-orientable (on the right) surfaces of genus $n$.

Theorem: (Stokes' theorem) For $X$ oriented and $\omega\in \Omega^{n-1}_X$, $\int_X d\omega = \int_{\dy X} \omega$.

Proposition: The tangent bundle $TX$ is always orientable.

Proof: Let $U,V\subset X$ with $\vp:U\to \R^n$ and $\psi:V\to \R^n$ trivializing maps, and $\psi\circ \vp^{-1}:\R^n\to \R^n$ the transition function. To show that $TX$ is always orientable, we need to show the Jacobian of the induced transition function (determinant of the derivative) on $TX$ is always non-negative. On $TU$ and $TV$, we have trivializing maps $(\varphi,d\varphi)$ and $(\psi,d\psi)$, giving a transition function
\[
(\psi\circ \varphi^{-1}, d\psi \circ d\varphi^{-1}) = (\psi\circ \varphi^{-1}, d(\psi \circ \varphi^{-1})). 
\]
The Jacobian of this is
\[
\det(d(\psi\circ \varphi^{-1}, d(\psi \circ \varphi^{-1}))) = \det(d(\psi\circ \varphi^{-1}), d(\psi \circ \varphi^{-1}))) = \det(d(\psi\circ \varphi^{-1}))\cdot \det(d(\psi \circ \varphi^{-1})) \>0,
\]
and since $d(\psi\circ \varphi^{-1})\neq 0$ (as $\psi \circ \varphi^{-1}$ is a diffeomorphism, its derivative is an isomorphism), the result is always positive. $\square$

References: Lee (Introduction to smooth manifolds, Chapter 4), Guillemin and Pollack (Differential topology, Chapter 1

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)