Cm Math Geometry — Examples

The matrix $R$ has rows $(0,1,0)$, $(-1,0,0)$, $(0,0,1)$, so $R^{\mathsf{T}}R$ has entries: first diagonal $0^2+(-1{)}^2+0^2=1$, second diagonal $1^2+0^2+0^2=1$, third diagonal $1$, and off-diagonal entries $0\cdot 1+(-1)\cdot 0+0\cdot 0=0$, $0\cdot 0+(-1)\cdot 0+0\cdot 1=0$, $1\cdot 0+0\cdot 0+0\cdot 1=0$; hence $R^{\mathsf{T}}R=I_3$, and $\det R=0\cdot (0\cdot 1-0\cdot 0)-1\cdot ((-1)\cdot 1-0\cdot 0)+0=1$, so $R\in SO(3)$.

For the point $p$ with first coordinates $(t,x_1,x_2,x_3)$ the direct substitution $p=o^{\mathrm{\prime }}+t^{\mathrm{\prime }}{u}^{\mathrm{\prime }}+{\sum }_i{x}_i^{\mathrm{\prime }}{e}_i^{\mathrm{\prime }}$ reads $(t,x_1,x_2,x_3)=(t^{\mathrm{\prime }},t^{\mathrm{\prime }}-x_2^{\mathrm{\prime }},1+x_1^{\mathrm{\prime }},x_3^{\mathrm{\prime }})$, whose unique solution is $t^{\mathrm{\prime }}=t$, $x_1^{\mathrm{\prime }}=x_2-1$, $x_2^{\mathrm{\prime }}=t-x_1$, $x_3^{\mathrm{\prime }}=x_3$; comparison with $x^{\mathrm{\prime }}=Rx+vt+a$ gives $Rx+vt+a=(x_2,-x_1,x_3)+(0,t,0)+(-1,0,0)=(x_2-1,t-x_1,x_3)$, so the transformation law of the lemma holds in this chart pair with the displayed $b,v,a,R$.

For the curve $\gamma (s)=(s,s,2s,0)$, with first coordinate representative $x(s)=(s,2s,0)$, the second representative is $y(s)=(2s-1,s-s,0)=(2s-1,0,0)$ by step 2.1, so $y^{\mathrm{\prime }}(s)=(2,0,0)$ and $R{x}^{\mathrm{\prime }}(s)+v=R(1,2,0)+(0,1,0)=(2,-1,0)+(0,1,0)=(2,0,0)$; also $x^{\mathrm{\prime }\mathrm{\prime }}=0$ and $y^{\mathrm{\prime }\mathrm{\prime }}=0$, and $R\cdot 0=0$. This verifies $y^{\mathrm{\prime }}=R{x}^{\mathrm{\prime }}+v$ and $y^{\mathrm{\prime }\mathrm{\prime }}=R{x}^{\mathrm{\prime }\mathrm{\prime }}$ in the worked case.

By [L1], $R^{\mathrm{\prime }}=\nu \left(\begin{array}{ccc}-s_t& -c_t& 0\\ c_t& -s_t& 0\\ 0& 0& 0\end{array}\right)$, and the product $R^{\mathsf{T}}{R}^{\mathrm{\prime }}$ equals $\nu \left(\begin{array}{ccc}{c}_t& s_t& 0\\ -s_t& c_t& 0\\ 0& 0& 1\end{array}\right)\left(\begin{array}{ccc}-s_t& -c_t& 0\\ c_t& -s_t& 0\\ 0& 0& 0\end{array}\right)=\nu \left(\begin{array}{ccc}-c_t{s}_t+s_t{c}_t& -c_t^2-s_t^2& 0\\ s_t^2+c_t^2& s_t{c}_t-c_t{s}_t& 0\\ 0& 0& 0\end{array}\right)=\nu \left(\begin{array}{ccc}0& -1& 0\\ 1& 0& 0\\ 0& 0& 0\end{array}\right)$, using $c_t^2+s_t^2=1$; hence reading off the entries of $\mathrm{\Omega }$ as in clause 2 of [L2] gives $\mathit{\omega }(t)=(0,0,\nu )$ and $\mathrm{\Omega }z=\mathit{\omega }\times z$ for every $z$. Also $R^{\mathsf{T}}R=I_3$ and $\det R=c_t^2+s_t^2=1$, so $R(t)\in SO(3)$; [L1] gives its required $C^2$ regularity.

Take $y(t)=e_1$ for $t\in \mathbb{R}$, so $y^{\mathrm{\prime }}=y^{\mathrm{\prime }\mathrm{\prime }}=0$, $x(t)=R(t)e_1=(c_t,s_t,0)$ and $x^{\mathrm{\prime }\mathrm{\prime }}=-{\nu }^2(c_t,s_t,0)$ by [L1]. Then $R^{\mathsf{T}}{x}^{\mathrm{\prime }\mathrm{\prime }}=-{\nu }^2{R}^{\mathsf{T}}(c_t,s_t,0)=-{\nu }^2{e}_1$; the right-hand side is $\mathit{\omega }\times (\mathit{\omega }\times e_1)=(0,0,\nu )\times (0,\nu ,0)=(-{\nu }^2,0,0)$ by [L3] and step 1.1. The two sides agree.

Take $y(t)=t{e}_1$ for $t\in \mathbb{R}$, so $y^{\mathrm{\prime }}=e_1$, $y^{\mathrm{\prime }\mathrm{\prime }}=0$, $x(t)=t(c_t,s_t,0)$ and $x^{\mathrm{\prime }\mathrm{\prime }}=(-2\nu s_t-t{\nu }^2{c}_t,2\nu c_t-t{\nu }^2{s}_t,0)$ by [L1]. Multiplying by $R^{\mathsf{T}}$ gives $R^{\mathsf{T}}{x}^{\mathrm{\prime }\mathrm{\prime }}=(-t{\nu }^2,2\nu ,0)$; the right-hand side is $2(0,\nu ,0)+t(-{\nu }^2,0,0)=(-t{\nu }^2,2\nu ,0)$ by [L3] and step 1.1, using $2\mathit{\omega }\times e_1=2(0,\nu ,0)$ and $\mathit{\omega }\times (\mathit{\omega }\times t{e}_1)=t(-{\nu }^2,0,0)$. The two sides agree, exhibiting the two rotational terms separately.

The map $F:{\mathbb{R}}^3\to \mathbb{R}$, $F(x)=⟨x,x{⟩}_2$, is smooth because it is a finite sum of coordinate squares and [L2] applies. For $h\in {\mathbb{R}}^3$, $F(x+h)-F(x)-2⟨x,h{⟩}_2=⟨h,h{⟩}_2=\mathrm{\parallel }h{\mathrm{\parallel }}_2^2$, whose ratio to $\mathrm{\parallel }h{\mathrm{\parallel }}_2$ tends to $0$; hence $D{F}_x(h)=2⟨x,h{⟩}_2$. At every $x\in F^{-1}(1)$, the vector $h=x\mathrm{/}2$ has $D{F}_x(h)=1$, so this differential is surjective. The fibre is nonempty since $e_3\in F^{-1}(1)$; [L4] therefore makes $S^2=F^{-1}(1)$ a smooth embedded submanifold of dimension $3-1=2$. Its inclusion is a smooth embedding by [L2], so [L1] applies and gives $G_q(u,v)=m⟨u,v{⟩}_2$ and positive definiteness at every point of the sphere.

The partial derivatives of $\mathrm{\Psi }$ are ${\mathrm{\partial }}_{\theta }\mathrm{\Psi }=(\cos \theta \cos \phi ,\cos \theta \sin \phi ,-\sin \theta )$ and ${\mathrm{\partial }}_{\phi }\mathrm{\Psi }=(-\sin \theta \sin \phi ,\sin \theta \cos \phi ,0)$. For $0<\theta <\pi$ these are linearly independent: a linear combination $\alpha {\mathrm{\partial }}_{\theta }\mathrm{\Psi }+\beta {\mathrm{\partial }}_{\phi }\mathrm{\Psi }=0$ has third component $-\alpha \sin \theta$ and first two components combining as $\cos \theta (\alpha \cos \phi ,\alpha \sin \phi )+\beta (-\sin \theta \sin \phi ,\sin \theta \cos \phi )=0$; the first relation gives $\alpha =0$ because $\sin \theta \mathrm{\ne }0$, and then $\beta =0$ because $\sin \theta \mathrm{\ne }0$. By [L2] this proves that $d\mathrm{\Psi }$ is injective.

The image of $\mathrm{\Psi }$ lies in $S^2$ by [L3]. For $x\in U$, the exclusion of the poles gives $-1<x_3<1$, so [L5] gives a unique $\theta \in (0,\pi )$ with $\cos \theta =x_3$ and $\sin \theta =\sqrt{1-x_3^2}>0$. The pair $(x_1,x_2)\mathrm{/}\sin \theta$ lies on the unit circle and is not $(1,0)$; hence [L5] gives a unique $\phi \in (0,2\pi )$ producing $x$. Conversely an angle in that interval cannot give $(1,0)$ by the same circle bijection. Thus $\mathrm{\Psi }:D\to U$ is bijective. The removed meridian is relatively closed, so $U$ is open in $S^2$. Smoothness follows from [L3] and the embedded-target criterion [L4]. Its differential is an isomorphism between two-dimensional tangent spaces by steps 1.1 and 2.1, so [L4] supplies smooth local inverses everywhere; bijectivity makes these restrictions of its single global inverse. Thus $\mathrm{\Psi }$ is a diffeomorphism onto $U$, and $(U,{\mathrm{\Psi }}^{-1})$ is a smooth chart.

The three inner products of the coordinate vector fields are $⟨{\mathrm{\partial }}_{\theta }\mathrm{\Psi },{\mathrm{\partial }}_{\theta }\mathrm{\Psi }{⟩}_2={\cos }^2\theta {\cos }^2\phi +{\cos }^2\theta {\sin }^2\phi +{\sin }^2\theta =1$, $⟨{\mathrm{\partial }}_{\theta }\mathrm{\Psi },{\mathrm{\partial }}_{\phi }\mathrm{\Psi }{⟩}_2=-\cos \theta \sin \theta \cos \phi \sin \phi +\cos \theta \sin \theta \sin \phi \cos \phi =0$ and $⟨{\mathrm{\partial }}_{\phi }\mathrm{\Psi },{\mathrm{\partial }}_{\phi }\mathrm{\Psi }{⟩}_2={\sin }^2\theta {\sin }^2\phi +{\sin }^2\theta {\cos }^2\phi ={\sin }^2\theta$, using ${\cos }^2+{\sin }^2=1$ for each angle.

Since $\mathrm{\Psi }$ is the parametrization of the inclusion chart, the mass metric in these coordinates is $G_{ij}=m⟨{\mathrm{\partial }}_i\mathrm{\Psi },{\mathrm{\partial }}_j\mathrm{\Psi }{⟩}_2$; by step 3.2 its matrix is $m\left(\begin{array}{cc}1& 0\\ 0& {\sin }^2\theta \end{array}\right)$, so for a coordinate velocity $(\dot{\theta },\dot{\phi })$ the kinetic energy is $T=\frac{1}{2}m({\dot{\theta }}^2+{\sin }^2\theta {\dot{\phi }}^2)$, and the metric is positive definite on the chart domain because $\sin \theta \mathrm{\ne }0$.

Take $S=\{0,e_3\}\subseteq {\mathbb{R}}^3$ and $R=\left(\begin{array}{ccc}-1& 0& 0\\ 0& -1& 0\\ 0& 0& 1\end{array}\right)$. Then $R{e}_3=e_3$ and $R0=0$, while $R\mathrm{\ne }{I}_3$ because $R{e}_1=-e_1$.

The matrix $R$ lies in $SO(3)$: its transpose equals itself, so $R^{\mathsf{T}}R=R^2=\mathrm{diag}(1,1,1)=I_3$, and in the Leibniz sum of [L2] every nonidentity permutation has an index $i$ with $\pi (i)\mathrm{\ne }i$, giving a zero off-diagonal factor. Only the identity permutation contributes, with sign $1$, so $\det R=(-1)\cdot (-1)\cdot 1=1$.

The component functions $F_0(x,y)=2xy$ and $F_1(x,y)=x^2$ are $C^1$, with ${\mathrm{\partial }}_1{F}_0=2x={\mathrm{\partial }}_0{F}_1$; hence $F$ is closed by [L2].

The function $\varphi (x,y)=x^{2}y$ is $C^2$ with ${\mathrm{\partial }}_0\varphi =2xy=F_0$ and ${\mathrm{\partial }}_1\varphi =x^2=F_1$, so $\mathrm{\nabla }\varphi =F$ and $F$ is exact with potential $\varphi$ by [L2].

For $a=0$ the radial formula of [L3] reads ${\varphi }_{\mathrm{r}\mathrm{a}\mathrm{d}}(x,y)={\int }_0^1⟨F(tx,ty),(x,y){⟩}_{2}dt={\int }_0^1(2t^2{x}^{2}y+t^2{x}^{2}y)dt={\int }_0^{1}3x^{2}y{t}^{2}dt$. The integrand is a continuous polynomial on $[0,1]$, hence Riemann integrable by [L4]. By [L4] applied to $G(t):=x^{2}y{t}^3$, whose derivative is $3x^{2}y{t}^2$, the integral equals $G(1)-G(0)=x^{2}y$, so ${\varphi }_{\mathrm{r}\mathrm{a}\mathrm{d}}=\varphi$.

Differentiating the two components of the displayed $x_z$ with respect to $t$ using [L2] gives $x_z^{\mathrm{\prime }}(t)=(-\omega z_1\sin \omega t+z_2\cos \omega t,-{\omega }^2{z}_1\cos \omega t-\omega z_2\sin \omega t)$, whose first component is the second component of $x_z$ and whose second component is $-{\omega }^2$ times the first component of $x_z$; hence $x_z^{\mathrm{\prime }}=F(t,x_z)$ on all of $\mathbb{R}$, and $x_z(0)=(z_1,z_2)=z$ because $\sin 0=0$ and $\cos 0=1$.

The curve of step 1.1 is a solution on all of $\mathbb{R}$; by [L1] and [L3], every solution of the initial value problem with data $(0,z)$ is a restriction of the unique maximal solution $x_z$, whose domain is an open interval. Since a solution with domain $\mathbb{R}$ exists, that maximal domain is all of $\mathbb{R}$; hence $[0,P]\subseteq J_z$ for every $P$ and $D_P={\mathbb{R}}^2$. Substituting $t=P$ in step 1.1 gives the displayed formula for ${\theta }_P(z)$ for every $P>0$.