Vector Bundles (Contd.)

Lee Smooth Manifolds Theorem 1 (Lee E10.15): Let $V$ be a finite-dimensional real vector space, and let $G_k(V)$ be the Grassmannian of $k$-dimensional subspaces in $V$. Let $T\subset G_k(V)\times V$ be defined by

$$T=\{(S,v)\in G_k(V)\times V:v\in S\}$$

Show that $T$ is a smooth rank-$k$ subbundle of the product bundle $G_k(V)\times V\to G_k(V)$, called a tautological vector bundle over $G_k(V)$.

Proof) Take an arbitrary $n-k$ dimensional vector space $S$ and let $U_S$ be the subset of $G_k(V)$ composed of $k$-dimensional spaces that trivially intersect $S$. Taking an arbitrary element $V$ of $U_S$we can construct smooth coordinates on $U_S$ by taking an arbitrary basis $\{v_i\}$ of $V$ and associating the $k$-dimensional subspace spanned by $\{(v_i,s_i)\}$ where $s_i\in S$ with $(s_1,s_2,\dots ,s_k)$. With these coordinates, $U_S$ is diffeomorphic to the matrix space ${\mathbb{R}}_{k\times (n-k)}$, and one can create a smooth frame of $T$ by letting the subspace $U\in U_S$ point to $(v_1,s_1)$ in the first element of the frame, $(v_2,s_2)$ in the second frame, and so on. As this is represented in coordinates as a projection taking the $i$th column of the matrix, each of these sections are smooth. Thus, $T$ is a valid rank-$k$ subbundle of $G_k(V)\times V$.

Theorem 2 (Lee E10.16): Show that the tautological vector bundle over $G_1({\mathbb{R}}^2)$ is smoothy isomorphic to the Möbius bundle.

Proof) The first step is to convert the base manifold $G_1({\mathbb{R}}^2)$ into ${\mathbb{S}}^1$; as we have a fairly complete theory of smooth bundle isomorphisms over a fixed base, the remaining portion of the problem is simple. Note that the one-dimensional Grassmannian $G_1({\mathbb{R}}^n)$ is diffeomorphic to ${\mathbb{RP}}^{n-1}$.

Let $E$ be a smooth bundle over ${\mathbb{S}}^1$ such that $E=({\mathbb{S}}^1\setminus (1,0))\times \mathbb{R}{\cup }_{\sim }({\mathbb{S}}^1\setminus (-1,0))\times \mathbb{R}$ where $(p,u)\sim (q,v)$ iff $p=q$ and $u=p_{y}v$ where $p_y=1$ if $p$ is above the $x$ axis and $p_y=-1$ otherwise (i.e. a smooth bundle with two smooth trivializations with a sign flip transition). Define a diffeomorphism $\Phi :G_1({\mathbb{R}}^2)\to {\mathbb{S}}^1$ such that $[e^{ix}]\in G_1({\mathbb{R}}^2)$ where $0<x\le \pi$ maps to $e^{2ix}\in \mathbb{S}$. Create a smooth trivialization on $G_1({\mathbb{R}}^2)\setminus (y=0)$ by mapping mapping $([\hat{u}],\hat{u}t)$ where $\hat{u}$ is a unit vector above the $x$ axis into $(\Phi ([\hat{u}]),t)$, and let $\phi$ be the smooth identity diffeomorphism with the trivialization over $({\mathbb{S}}^1\setminus (1,0))$. Now create a smooth trivialization on $G_1({\mathbb{R}}^2)\setminus (x=0)$ by mapping $([\hat{u}],\hat{u}t)$ where $\hat{u}$ is a unit vector right of the $y$ axis into $(\Phi ([\hat{u}]),t)$, and let $\phi$ be the smooth identity diffeomorphism with the trivialization over $({\mathbb{S}}^1\setminus (-1,0))$. One can manually check these definitions agree and define a smooth bundle isomorphism. $E$ has the same transition function network over these two trivializations as the Möbius bundle, and thus the two are isomorphic. Therefore, the tautological bundle over $G_1({\mathbb{R}}^2)$ is smoothly isomorphic to the Möbius bundle.

We now pivot to subbundle and submanifolds. Recall that if $E$ is a smooth vector bundle over $M$ and $S$ is an immersed submanifold of $M$, there is a corresponding smooth structure on $E{|}_S$. In the embedded case, $E{|}_S$ is an embedded submanifold of $E$ and thus has a unique smooth structure; however, this is not the case if $S$ is merely immersed. We must then invoke the smooth manifold chart lemma with explicitly denoted smooth trivializations descending from those of $E\to M$.

Theorem 3 (Lee E10.14): Suppose $M$ is a smooth manifold with or without boundary, and $S\subseteq M$is an immersed submanifold with or without boundary. Identifying $T_{p}S$ as a subspace of $T_{p}M$ for each $p\in S$ in the usual way, show that $TS$ is a smooth subbundle of $TM{|}_S$.

Proof) This theorem becomes quite simple once you go over the hurdle of understanding what $TM{|}_S$ and $TS$ actually means. Given any open $U\subset S$ such that there is a slice chart $V\subset M$ of $U$ (note it is not necessary that $V\cap S=U$), $TM{|}_S$ is defined such that the restriction of the smooth trivialization on $V$ is a smooth trivialization on $U$. Represent the trivialization on $V$ (using slice coordinates) as $\Phi :(p,v)\mapsto (p,s^1,s^2,\dots ,s^m)$ where $m=\text{dim}M$; defining smooth sections ${\sigma }_i$ on $U$ ($1\le i\le \text{dim}S$) that map $u\in U$ to $(u,0,\dots ,0,1,0,\dots ,0)$ such that $s^j={\delta }_i^j$ gives smooth sections that span $TS$ at every point; thus, $TS$ is a smooth subbundle of $TM{|}_S$.

Theorem 4 (Lee E10.17): Suppose $M\subseteq {\mathbb{R}}^n$ is an immersed submanifold. Prove that the ambient tangent bundle $T{\mathbb{R}}^n{|}_M$ is isomorphic to the Whitney sum $TM\oplus NM$, where $NM\to M$ is the normal bundle.

Proof) Define $\Phi :TM\oplus NM\to T{\mathbb{R}}^n{|}_M$ such that $(p,v,w)\mapsto (p,v+w)$ where $v\in T{M}_p$ and $w\in N{M}_p$. $\Phi$ clearly restricts to a linear isomorphism on every fiber; it suffices to show $\Phi$ is a diffeomorphism. Fix any $p\in M$ and let $U\subset M$ be a neighborhood of $p$ such that it has a slice chart $V\subset {\mathbb{R}}^n$. As in Theorem 3, there exists smooth sections ${\sigma }_1,\dots ,{\sigma }_m$ (where $m=\text{dim}M$) on $U$ that span $TM$ at every point, and, as $U$ is closed in $V$, there is some neighborhood $W\subset V$ of $p$ in ${\mathbb{R}}^n$ such that there exists a local frame $\{\widetilde{{\sigma }_i}{\}}_{i=1}^n$ on $W$ that agrees with ${\sigma }_i$ on $U\cap W$ for $1\le i\le m$. Using Gram-Schmidt, from this we obtain an orthogonal frame $\{{\sigma }_i^{\prime }\}$ where ${\sigma }_1^{\prime },\dots ,{\sigma }_m^{\prime }$ spans $TM$ and ${\sigma }_{m+1}^{\prime },\dots ,{\sigma }_n^{\prime }$ spans $NM$ for any point on $U\cap W$. Using these sections to define local trivializations (and thus smooth charts) for $T{\mathbb{R}}^n{|}_M$, $TM$, and $NM$ we find that $\Phi$ is the identity diffeomorphism on these coordinates.