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