Chapter 11 Exercises (Contd. 2)

Lee Smooth Manifolds

Okay now I’m behind schedule so lemme speed through some of these

Problem 1 (E11-6). Suppose $M$ is a smooth $n$-manifold, $p\in M$, and $y^1,\dots ,y^k$ are smooth real-valued functions defined on a neighborhood of $p$ in $M$. Prove the following statements. (a) If $k=n$ and $\left(d{y}^1{|}_p,\dots ,d{y}^n{|}_p\right)$ is a basis for $T_p^{\ast }M$, then $\left(y^1,\dots ,y^n\right)$ are smooth coordinates for $M$ in some neighborhood of $p$. (b) If $\left(d{y}^1{|}_p,\dots ,d{y}^k{|}_p\right)$ is a linearly independent $k$-tuple of covectors and $k<n$, then there are smooth functions $y^{k+1},\dots ,y^n$ such that $\left(y^1,\dots ,y^n\right)$ are smooth coordinates for $M$ in a neighborhood of $p$. (c) If $\left(d{y}^1{|}_p,\dots ,d{y}^k{|}_p\right)$ span $T_p^{\ast }M$, there are indices $i_1,\dots ,i_n$ such that $\left(y^{i_1},\dots ,y^{i_n}\right)$ are smooth coordinates for $M$ in a neighborhood of $p$.

Proof) (a) Let $(U,(s^i))$ be a smooth chart centered at $p$ on which all $y^i$s are defined. Consider the map $\Phi :U\to {\mathbb{R}}^n$ given by $\Phi (q)=(y^1(q),y^2(q),\dots ,y^n(q))$. In the coordinates of $U$, the differential is given by $d\Phi (q)=(d{y}^i{|}_q{)}^T$ where the row vectors $d{y}^i{|}_q$ are stacked into a matrix. As this is invertible at $p$, the inverse function theorem gives a neighborhood $W$ of $p$ on which $\Phi$ is a diffeomorphism; on $W$ $y^1,\dots ,y^n$ are smooth coordinates, as desired. (b) It suffices to show, for any $\omega \in T_p^{\ast }M$, there exists some smooth real-valued function $y$ defined on some neighborhood of $p$ such that $d{y}_p=\omega$. Take a regular coordinate ball $U$ centered at $p$; let $\omega ={\omega }^{i}d{x}_i$ in these coordinates. Then, simply take $y={\omega }^i{x}_i$ on $U$; clearly, $d{y}_p=\omega$ as desired. (c) We can take a subset of $(d{y}^1{|}_p,\dots ,d{y}^k{|}_p)$ such that the hypotheses of (a) is satisfied; we are done by (a).

Problem 2 (E11-7). In the following problems, $M$ and $N$ are smooth manifolds, $F:M\to N$ is a smooth map, and $\omega \in {\mathfrak{X}}^{\ast }(N)$. Compute $F^{\ast }\omega$ in each case. (a) $M=N={\mathbb{R}}^2$; $F(s,t)=(st,e^t)$; $\omega =xdy-ydx$. (b) $M={\mathbb{R}}^2$ and $N={\mathbb{R}}^3$; $F(\theta ,\phi )=((\cos \phi +2)\cos \theta ,(\cos \phi +2)\sin \theta ,\sin \phi )$; $\omega =z^{2}dx$. (c) $M=\{(s,t)\in {\mathbb{R}}^2:s^2+t^2<1\}$ and $N={\mathbb{R}}^3\setminus \{0\}$; $F(s,t)=(s,t,\sqrt{1-s^2-t^2})$; $\omega =(1-x^2-y^2)dz$.

Solution) (a)

$$F^{\ast }\omega =std(e^t)-e^{t}d(st)=st{e}^{t}dt-e^{t}sdt-e^{t}tds$$

(b)

$$F^{\ast }\omega =(\sin \phi {)}^{2}d((\cos \phi +2)\cos \theta )=-\sin {\phi }^2\sin \theta (2d\theta +\cos \phi d\theta )-\sin {\phi }^3\cos \theta d\phi$$

(c)

$$F^{\ast }\omega =(1-s^2-t^2)d(\sqrt{1-s^2-t^2})=-\sqrt{1-s^2-t^2}(sds+tdt)$$

Problem 3 (E11-8). (a) Suppose $F:M\to N$ is a diffeomorphism, and let $d{F}^{\ast }:T^{\ast }N\to T^{\ast }M$ be the map whose restriction to each cotangent space $T_q^{\ast }N$ is equal to $d{F}_{F^{-1}(q)}^{\ast }$. Show that $d{F}^{\ast }$ is a smooth bundle homomorphism. (b) Let ${\text{Diff}}_1$ be the category whose objects are smooth manifolds, but whose only morphisms are diffeomorphisms; and let $\text{VB}$ be the category whose objects are smooth vector bundles and whose morphisms are smooth bundle homomorphisms. Show that the assignment $M\mapsto T^{\ast }M$, $F\mapsto d{F}^{\ast }$ defines a contravariant functor from ${\text{Diff}}_1$ to $\text{VB}$, called the cotangent functor.

Proof) (a) $d{F}^{\ast }$ clearly covers $F^{-1}$, and as $d{F}_p$ is an isomorphism for any $p$, its dual $d{F}_{F^{-1}(q)}^{\ast }$ is an isomorphism of dual spaces for any $q\in N$. Therefore, $d{F}^{\ast }$ is a smooth bundle homomorphism. It is, in fact, a bundle isomorphism, as it is bijective and its inverse $(d{F}^{\ast }{)}^{-1}$ is also a smooth bundle homomorphism by the same logic. (b) This functor is well-defined as $d{F}^{\ast }$ is a smooth bundle isomorphism, by (a). Identity maps are preserved as ${\text{Id}}_M^{\ast }={\text{Id}}_{T^{\ast }M}$, and composition follows from

$$H=F\circ G\Rightarrow dH=dF\circ dG\Rightarrow d{H}^{\ast }=d{G}^{\ast }\circ d{F}^{\ast }$$

Problem 4 (E11-11). Let $M$ be a smooth manifold, and $C\subseteq M$ be an embedded submanifold. Let $f\in C^{\infty }(M)$, and suppose $p\in C$ is a point at which $f$ attains a local maximum or minimum value among points in $C$. Given a smooth local defining function $\Phi :U\to {\mathbb{R}}^k$ for $C$ on a neighborhood $U$ of $p$ in $M$, show that there are real numbers ${\lambda }_1,\dots ,{\lambda }_k$ (called Lagrange multipliers) such that

$$d{f}_p={\lambda }_{1}d{\Phi }^1{|}_p+\cdots +{\lambda }_{k}d{\Phi }^k{|}_p$$

Proof) As local defining functions are smooth submersions, $d{\Phi }^i{|}_p$ is a linearly independent $k$-tuple of covectors; thus, by 11-6(b) there exists smooth functions $y^{k+1},\dots ,y^n$ defined in some neighborhood of $p$ such that $({\Phi }^1,{\Phi }^2,\dots ,{\Phi }^k,y^{k+1},\dots ,y^n)$ are smooth coordinates in some neighborhood $V$ of $p$; restricting $V$ if necessary, assume it is a regular coordinate ball. Working in these coordinates (and denoting them $(x^i)$), $p$ is a local maximum or minimum on the plane $x^1=x^2=\dots =x^k=0$. Thus, for any $i\ge k+1$, $\partial /\partial x^{i}f=d{f}_p^i=0$, and $d{f}_p$ is in the span of $d{x}^i$ for $i\le k$, which is precisely $d{\Phi }^{.}1{|}_p,\dots ,d{\Phi }^k{|}_p$.

Problem 5 (E11-12). Show that any two points in a connected smooth manifold can be joined by a smooth curve segment.

Proof) Any two points in a connected smooth manifold are joined by a piecewise smooth curve segment $\gamma :[0,1]\to M$. Taking $ϵ>0$ such that $\gamma$ is smooth on $A=[0,ϵ]\cup [1-ϵ,1]$, there is some smooth function $[0,1]\to M$ homotopic to $\gamma$ relative to $A$ by Whitney approximation; this is our desired smooth curve, and we are done.

Problem 6 (E11-13). The length of a smooth curve segment $\gamma :[a,b]\to {\mathbb{R}}^n$ is defined to be the value of the (ordinary) integral

$$\mathrm{L}(\gamma )={\int }_a^b|{\gamma }^{\prime }(t)|dt$$

Show that there is no smooth covector field $\omega \in {\mathfrak{X}}^{\ast }({\mathbb{R}}^n)$ with the property that ${\int }_{\gamma }\omega =\mathrm{L}(\gamma )$ for every smooth curve $\gamma$.

Proof) FTSOC assume such a covector field $\omega$ exists. Clearly, $\omega \ne 0$ so we can assume WLOG $\omega$ is nonzero at the origin. Furthermore, assume WLOG that ${\omega }^1(0)\ne 0$. Consider the smooth curve $\gamma :[-1,1]\to {\mathbb{R}}^n$ given by $\gamma (t)=(t,0,\dots ,0)$ and its time-reverse $\bar{\gamma }$. For any $-1<s\le 1$, let ${\gamma }_s$ be the restriction of $\gamma$ to $[-1,s]$. Then,

$${\int }_{-1}^s|{\gamma }^{\prime }(t)|dt=L({\gamma }_s)={\int }_{-1}^s({\gamma }^{\ast }\omega {)}_{t}dt$$

Both integrands are smooth, so we can differentiate at $s=0$ to obtain $|{\gamma }^{\prime }(0)|=({\gamma }^{\ast }\omega {)}_0(d/dt)=\omega (0)d\gamma (d/dt)=\omega (0)(\partial /\partial x^1)={\omega }^1(0)\Rightarrow {\omega }^1(0)>0$. Doing the same with the time-reversed $\bar{\gamma }$ gives ${\omega }^1(0)<0$, a contradiction.

Problem 7 (E11-14). Consider the following two covector fields on ${\mathbb{R}}^3$:

$$\omega =-\frac{4zdx}{(x^2+1{)}^2}+\frac{2ydy}{y^2+1}+\frac{2xdz}{x^2+1}$$ $$\eta =-\frac{4xzdx}{(x^2+1{)}^2}+\frac{2ydy}{y^2+1}+\frac{2dz}{x^2+1}$$

(a) Set up and evaluate the line integral of each covector field along the straight line segment from $(0,0,0)$ to $(1,1,1)$. (b) Determine whether either of these covector fields is exact. (c) For each one that is exact, find a potential function and use it to recompute the line integral.

Proof) (a) Take the path: $\gamma :[0,1]\to {\mathbb{R}}^3$ given by $\gamma (t)=(t,t,t)$. Then,

$${\gamma }^{\ast }\omega =\left(-\frac{4t}{(t^2+1{)}^2}+\frac{4t}{t^2+1}\right)dt,{\gamma }^{\ast }\eta =\left(-\frac{4t^2}{(t^2+1{)}^2}+\frac{2t}{t^2+1}+\frac{2}{t^2+1}\right)dt$$

and thus ${\int }_0^1{\gamma }^{\ast }\omega =2\ln (2)-1$ and ${\int }_0^1{\gamma }^{\ast }\eta =\ln (2)+1$. (b) As ${\mathbb{R}}^3$ is convex exactness is identical to closedness. $\omega$ is not closed as $\partial {\omega }_x/\partial z\ne \partial {\omega }_z/\partial x$, but $\eta$ is closed as one can quickly verify. Thus, $\eta$ is exact. (c) For any $(a,b,c)$ we take the path from $(0,0,0)$ that follows the $x$, $y$, and $z$ axis, respectively, and evaluate the integral to obtain the potential function $V$ of $\eta$. This can be easily seen to evaluate to $V(a,b,c)=\ln (b^2+1)+2ac/(a^2+1)$. The integral in (a) thus evaluates to $V(1,1,1)-V(0,0,0)=\ln (2)+1$, as expected.

Problem 8 (E11-15). Let $X$ be a smooth vector field on an open subset $U\subseteq {\mathbb{R}}^n$. Given a piecewise smooth curve segment $\gamma :[a,b]\to U$, define the line integral of $X$ over $\gamma$, denoted by ${\int }_{\gamma }X\cdot ds$, as

$${\int }_{\gamma }X\cdot ds={\int }_a^b{X}_{\gamma (t)}\cdot {\gamma }^{\prime }(t)dt$$

where the dot on the right-hand side denotes the Euclidean dot product between tangent vectors at $\gamma (t)$, identified with elements of ${\mathbb{R}}^n$. A conservative vector field is one whose line integral around every piecewise smooth closed curve is zero. (a) Show that $X$ is conservative if and only if there exists a smooth function $f\in C^{\infty }(U)$ such that $X=\mathrm{grad}f$. (b) Suppose $n=3$. Show that if $X$ is conservative, then $\mathrm{curl}X=0$, where

$$\mathrm{curl}X=\left(\frac{\partial X^3}{\partial x^2}-\frac{\partial X^2}{\partial x^3}\right)\frac{\partial }{\partial x^1}+\left(\frac{\partial X^1}{\partial x^3}-\frac{\partial X^3}{\partial x^1}\right)\frac{\partial }{\partial x^2}+\left(\frac{\partial X^2}{\partial x^1}-\frac{\partial X^1}{\partial x^2}\right)\frac{\partial }{\partial x^3}$$

(c) Show that if $U\subseteq {\mathbb{R}}^3$ is star-shaped, then $X$ is conservative on $U$ if and only if $\mathrm{curl}X=0$.

Proof) (a) Let $\omega \in {\mathfrak{X}}^{\ast }U$ be the covector field defined as $\omega =X^{i}d{x}_i$; i.e., the same component functions as $X$ but with vectors replaced with covectors. One can quickly check that

$${\int }_{\gamma }X\cdot ds={\int }_a^b{X}_{\gamma (t)}\cdot {\gamma }^{\prime }(t)dt={\int }_a^b\omega {\gamma }^{\prime }(t)dt={\int }_{\gamma }\omega$$

Thus, $X$ is conservative iff $\omega$ is conservative, which is true iff $\exists V\in C^{\infty }(U)$ such that $\omega =dV$. As, in the coordinates of ${\mathbb{R}}^n$, $\text{grad}V$ has the same component functions as $dV$, $\omega =dV$ iff $X=\text{grad}V$ where $V\in C^{\infty }(U)$. (b) Define $\omega$ as above. If $X$ is conservative then $\omega$ is conservative, which implies $\omega$ is closed. As $X$ and $\omega$ share component functions, clearly $\text{curl}X=0$. (c) If $\text{curl}X=0$, $\omega$ is closed, and as $U$ is star-shaped this implies $\omega$ is conservative. Thus, $X$ is conservative, and we are done.

Problem 9 (E11-16). Let $M$ be a compact manifold of positive dimension. Show that every exact covector field on $M$ vanishes at at least two points in each component of $M$.

Proof) We add the assumption that $M$ is connected and aim to prove every exact covector field vanishes at least twice. Let $\omega =df$ be exact; as $M$ is compact, $f$ is either constant on $M$ or there are points $a$ and $b$ in $M$ where a global maximum and minimum are attained, respectively. If $f$ is constant $\omega$ vanishes at every point, which is at least two as $M$ is of positive dimension. Otherwise $\omega$ vanishes at $a$ and $b$, completing our proof.

Problem 10 (E11-17). Let ${\mathbb{T}}^n={\mathbb{S}}^1\times \cdots \times {\mathbb{S}}^1\subseteq {\mathbb{C}}^n$ denote the $n$-torus. For each $j=1,\dots ,n$, let ${\gamma }_j:[0,1]\to {\mathbb{T}}^n$ be the curve segment

$${\gamma }_j(t)=(1,\dots ,e^{2\pi it},\dots ,1)\text{(with }{e}^{2\pi it}\text{ in the }j\text{th place)}$$

Show that a closed covector field $\omega$ on ${\mathbb{T}}^n$ is exact if and only if ${\int }_{{\gamma }_j}\omega =0$ for $j=1,\dots ,n$.

Proof) It is trivial that if $\omega$ is exact, ${\int }_{{\gamma }_j}\omega =0$. It suffices to show the converse.

We consider the smooth covering map $\mu :{\mathbb{R}}^n\to {\mathbb{T}}^n$ such that $\mu (x^1,x^2,\dots ,x^n)=(e^{2\pi i{x}^1},e^{2\pi i{x}^2},\dots ,e^{2\pi i{x}^n})$. As $\mu$ is a local diffeomorphism, the pullback ${\mu }^{\ast }\omega$ is closed in ${\mathbb{R}}^n$ and therefore exact. Furthermore, let $(p^1,p^2,\dots ,p^n)$ be an arbitrary lattice point, and consider the path $\beta (t)=(p^1,\dots ,p^j+t,\dots ,p^n)$ defined on $[0,1]$ for some arbitrary $1\le j\le n$. One can then observe that ${\gamma }_j=\mu \circ \beta$, and thus

$${\int }_{{\gamma }_j}\omega ={\int }_0^1(d{\gamma }_j^{\ast }{)}_t{\omega }_{\gamma (t)}dt={\int }_0^1(d{\beta }_j^{\ast }\circ d{\mu }^{\ast }{)}_t{\omega }_{\gamma (t)}dt={\int }_{{\beta }_j}{\mu }^{\ast }\omega ={\int }_0^1{\mu }^{\ast }{\omega }^j(\beta (t))dt$$

where ${\mu }^{\ast }{\omega }^j$ is the $j$th component of the covector field. Therefore, the condition that ${\int }_{{\gamma }_j}\omega =0$ for all $j$ is equivalent to the condition that the integral of ${\mu }^{\ast }\omega$ on any path that connects two adjacent lattice points is zero.

Notice that the approach we use above is highly generalizable; any piecewise smooth curve segment $\gamma :[0,1]\to {\mathbb{T}}^n$ can be lifted to another piecewise smooth curve segment $\beta :[0,1]\to {\mathbb{R}}^n$ such that $\gamma =\mu \circ \beta$. Then,

$${\int }_{\gamma }\omega ={\int }_0^1{\gamma }^{\ast }\omega ={\int }_{\beta }{\mu }^{\ast }\omega$$

Consider any piecewise smooth curve segment $\gamma$ with the same start and endpoint and its lift $\beta$ with regard to $\mu$. Then, ${\int }_{\gamma }\omega ={\int }_{\beta }{\mu }^{\ast }\omega$. Recall that ${\mu }^{\ast }\omega$ is exact, and by the nature $\mu$, $\beta$ must begin and start at points $(a^1,a^2,\dots ,a^n)$ and $(a^1+p^1,\dots ,a^n+p^n)$ whose coordinates all differ by some integer $p^i$. Consider the path between these points made of the straight-line path from $(a^1,\dots ,a^n)$ to $(0,\dots ,0)$, the axis-following path from $(0,\dots ,0)$ to $(p^1,\dots ,p^n)$, and the straight-line path from $(p^1,\dots ,p^n)$ to $(a^1+p^1,\dots ,a^n+p^n)$. As the pullback covector field ${\mu }^{\ast }\omega$ is invariant with integral translations, the paths $(a^1,\dots ,a^n)\to (0,\dots ,0)$ and $(p^1,\dots ,p^n)\to (a^1+p^1,\dots ,a^n+p^n)$ cancel out, and as the integral on any path between adjacent lattice points vanishes, the entire expression vanishes. In other words, ${\int }_{\beta }{\mu }^{\ast }\omega =0$, implying ${\int }_{\gamma }\omega =0$. Therefore, $\omega$ is exact, and we are done.