Tangent Vectors

Lee Smooth Manifolds

Lee 3-1) Suppose $M$ and $N$ are smooth manifolds with or without boundary, and $F:M\to N$ is a smooth map. Show that $d{F}_p:T_{p}M\to T_{F(p)}N$ is the zero map for each $p\in M$ if and only if $F$ is constant on each component of $M$.

Proof) Suppose that $F$ is constant on each component of $M$. For any $f\in C^{\infty }(N)$ and any $v\in T_{p}M$, $d{F}_p(v)f=v(f\circ F)=0$ as $f\circ F$ is constant in a neighborhood of $p$ (more specifically, the connected component containing $p$). Thus, $d{F}_p(v)$ is the zero derivation, and we are done.

We now prove the converse. We first show if $U\subset M$ is a regular coordinate ball or half-ball, $F$ must be constant on $U$. Assume, FTSOC, that $\exists p,q\in U$ s.t. $F(p)\ne F(q)$. Using bump functions, we can construct some smooth map $f:N\to \mathbb{R}$ such that $f(F(p))=0$ and $f(F(q))=1$. As $U$ is a regular coordinate ball/half-ball, we can also take a smooth function $\gamma :(-1,2)$ such that $\gamma (0)=p$ and $\gamma (1)=q$. Noting that $f\circ F\circ \gamma$ is a map $(-1,2)\to \mathbb{R}$, for any $w\in (-1,2)$,

$$\frac{d(f\circ F\circ \gamma (x))}{dx}(w)={\frac{d}{dx}|}_w(f\circ F\circ \gamma )=d{F}_{\gamma (w)}d{\gamma }_w\left({\frac{d}{dx}|}_w\right)(f)=0_{F(\gamma (w))}(f)=0$$

However, as $f\circ F\circ \gamma (0)=0$ and $f\circ F\circ \gamma (1)=1$, there is a clear contradiction. Furthermore, any two points within the same component of $M$ can be connected via finitely many regular coordinate balls and half-balls, and thus $F$ must be constant on each component of $M$.

Lee 3-5) Let $K=\{(x,y):\max (|x|,|y|)=1\}$. Show that there is no diffeomorphism $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ such that $F({\mathbb{S}}^1)=K$.

Proof) It suffices to show there cannot exist an injective smooth curve $\gamma :J\to {\mathbb{R}}^2$ where $J=(0,1)$ with non-vanishing velocity such that $\gamma (J)=K-\{p\}$, where $p\in K$ is some point. WLOG, assume that $p\ne (-1,1)$. First, as $\gamma$ is smooth, the global differential $d\gamma :TJ\to T{\mathbb{R}}^2$ is also smooth. In particular, define $f:J\to \mathbb{R}$ as $f(j)={\gamma }^{\prime }(j)(x)$ where $x:{\mathbb{R}}^2\to \mathbb{R}$ is the smooth projection map to the first component. $f$ must be smooth by the following argument: the map $f_1:J\to TJ$ such that $f_1(j)={\frac{d}{dt}|}_j$ is smooth, and if we define $f_3:T{\mathbb{R}}^2\to \mathbb{R}$ as $f_3(v)=v(x)$, $f=f_3\circ d\gamma \circ f_1$, and thus it suffices to show $f_3$ is smooth. This is true as $f_3$ has the coordinate representation of a projection onto the third coordinate, which is trivially smooth. More generally, for any smooth manifold $M$ with or without boundary and $h\in C^{\infty }(M)$, the map $H:TM\to \mathbb{R}$ defined by $H(v)=v(h)$ is smooth. WLOG, assume $\gamma (0.5)=(-1,1)$ and that $\gamma (j)$ traverses up the vertical edge when $j<0.5$ and traverses through the horizontal edge when $j>0.5$. $f$ must be strictly nonzero whenever $\gamma$ is on a horizontal edge and zero whenever $\gamma$ is on a vertical edge; by continuity, $f(0.5)=0$. By an analogous argument, ${\gamma }^{\prime }(0.5)(y)=0$, and the velocity of $\gamma$ is forced to vanish at $(-1,1)$. This is a contradiction, completing our proof.

Lee 3-8) Let $M$ be a smooth manifold with or without boundary and $p$ be a point of $M$. Let ${\mathcal{V}}_{p}M$ denote the set of equivalence classes of smooth curves starting at $p$ under the relation ${\gamma }_1\sim {\gamma }_2$ if $(f\circ {\gamma }_1{)}^{\prime }(0)=(f\circ {\gamma }_2{)}^{\prime }(0)$ for every smooth real-valued function $f$ defined in a neighborhood of $p$. Show that the map $\Psi :{\mathcal{V}}_{p}M\to T_{p}M$ defined by $\Psi [\gamma ]={\gamma }^{\prime }(0)$ is well defined and bijective.

Proof) Note that $(f\circ \gamma {)}^{\prime }(0)={\gamma }^{\prime }(0)f$, and thus if ${\gamma }_1\sim {\gamma }_2$, ${\gamma }_1^{\prime }(0)f={\gamma }_2(0{)}^{\prime }f$ for all $f\in C^{\infty }(M)$, implying ${\gamma }_1^{\prime }(0)={\gamma }_2^{\prime }(0)$. Therefore, $\Psi$ is well-defined. For bijectivity, note that any $v\in T_{p}M$ is the velocity of some smooth curve, and thus $v\in \Psi ({\mathcal{V}}_{p}M)$, showing surjectivity. Injectivity is trivial.