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\begin{document}
\title{{\bf  Homework set 9  - due 11/15/20}\\ {\large Math 5047 \---  Renato Feres} 
}
\author{}
\date{}  
\maketitle
\thispagestyle{empty}
%\pagestyle{empty}
 

  
Turn in the solutions to problems   6, 7, 8(a,b,c,d), and 9(b).
\begin{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf Reading:}
{\bf The Gauss-Bonnet theorem.} We consider here an oriented $2$-dimensional Riemannian manifold  $M$. Orientation is not essential\----one can do most of the work on the orientation double cover of $M$.
  \begin{definition}[Geodesic curvature]\label{geodesic curvature}
  Let $c:(a,b)\rightarrow M$ be a smooth curve parametrized by arclength. 
 Denote by $N(s)$ the unit vector perpendicular to $T(s)=c'(s)$ such that  $T(s), N(s)$ form a positive
 orthonormal basis of $T_{c(s)}M$ for each $s$. Then
 the number
 $$\kappa_g(s):=\left\langle \frac{\nabla c'}{ds}, N(s)\right\rangle $$
 is called the {\em geodesic curvature} of $c$. 
  \end{definition}
 Note that $c(s)$ is a geodesic if and only if  its geodesic curvature is zero.  
 
 
 \vspace{.1in}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=3in]{GaussBonnetI.eps}
\caption{\small  Definitions for the local version of the Gauss-Bonnet theorem.}
\label{GB1}
\end{center}
\end{figure} 


Consider a region $\mathcal{R}$ homeomorphic to a disc  in the oriented surface $M$. We suppose that
the boundary of $\mathcal{R}$ is  
 a continuous, piecewise smooth simple closed curve $c$ with finitely many smooth curve segments as indicated in Figure \ref{GB1}.
The smooth segments of $c$ are denoted $c_i:[a_i,b_i]\rightarrow M$ and are parametrized by arclength. Also suppose that
$\mathcal{R} $ is contained in an open set $U\subset M$ on which is defined a smooth vector field $e_1$
of unit length.  Let $e_2$ be the unit length vector field orthogonal to $e_1$ such that $e_1, e_2$ form
at each    $p\in U$ a positive basis of $T_pM$.  At each point $c_i(s)$ let $\varphi(s)$ be the angle
that $c_i'(s)$ makes with $e_1(c_i(s))$ as indicated in Figure \ref{GB1}. We call $\varphi(s)$ the {\em turning angle.}
At the corner point connecting $c_{i}$ and $c_{i+1}$, where $c$ is not differentiable, let $\alpha_i$
be the angle between $c_{i}'(b_i)$ and $c'_{i+1}(a_{i+1})$ measured counterclockwise.  Note
that the accumulated {turning angle} that $c'$  makes with $e_1$ as the closed curve is traced around once,
including the jump discontinuities at the corners of $\mathcal{R}$, must be $2\pi$. I will take this
as a fact and omit the proof. In other words, if $\varphi_i(s)$ is the turning angle along $c_i(s)$, then 
\begin{equation}\label{angle}
\sum_i\left[\int_{a_i}^{b_i} \varphi_i'(s)\, ds +\alpha_i\right]=2\pi.
\end{equation} 

 
 \begin{theorem}[Gauss-Bonnet,  local version]\label{gbl} Let $(M, \langle\cdot,\cdot\rangle)$
 be a $2$-dimensional, oriented Riemannian manifold  and let $\mathcal{R}\subset M$ be
 a compact region as indicated in Figure \ref{GB1}. The boundary of $\mathcal{R}$ is
 a piecewise smooth curve with smooth segments $c_i:[a_i,b_i]\rightarrow M$ with turning
  angle $\varphi_i(s)$ having jump discontinuities at the corners given by $\alpha_i$ as indicated in
  the figure. Then 
  $$ \iint_{\mathcal{R}} K\, dA + \int_c\kappa_g(s)\, ds + \sum_i \alpha_i = 2\pi $$
  where $K$ is the Gauss curvature and $k_g$ is the geodesic curvature. 
 \end{theorem}
 \begin{proof}
 
Let us recall a few facts. The Gauss curvature
$K(p)$ is given, 
for any  orthonormal basis  $u, v$ of  $T_pM$,  by
 $$ K(p)=\left\langle R_p(u, v)v, u\right\rangle.$$
 Cartan's structure equations expressed with respect to  a local  orthonormal frame $e_1, e_2$ and
 its dual frame $\theta_1, \theta_2$ give
 $$\left\langle R_p(u, v)v, u\right\rangle=\Omega_{12}(e_1,e_2)=d\omega_{12}(e_1,e_2) $$
 where $\omega_{ij}(u)=\langle e_i,\nabla_{u}e_j\rangle$.
 Thus 
 $$d\omega_{12}= K(p)\, \theta_1\wedge\theta_2 $$
where  $\theta_1\wedge\theta_2$ is the area form of $M$ corresponding to the area measure $dA$. 
At points where $c'(s)$ exists we have (dropping for now the index $i$ to simplify the notation) 
$$c'(s) = \cos(\varphi(s)) e_1(c(s)) + \sin (\varphi(s)) e_2(c(s)). $$
The unit normal vector field $N(s)$ obtained by rotating $c'(s)$ counterclockwise by $\pi/2$ is
$$N(s)= -\sin(\varphi(s)) e_1(c(s))+\cos(\varphi(s))e_2(c(s)). $$
Keeping in mind that $\langle \nabla e_i/ds, e_i\rangle =0$,  we have
$$ \frac{\nabla c'}{ds}= \left(-\varphi'\sin (\varphi) + \sin(\varphi)\left\langle \frac{\nabla e_2}{ds}, e_1\right\rangle\right) e_1+
 \left(\varphi'\cos(\varphi) + \cos(\varphi)\left\langle \frac{\nabla e_1}{ds}, e_2\right\rangle\right) e_2. $$
The normal component of this vector is then
 $$ \kappa_g=\left\langle \frac{\nabla c'}{ds}, N\right\rangle = \varphi' + \left\langle \frac{\nabla e_1}{ds}, e_2\right\rangle= \varphi'-\omega_{12}(c'). $$
 Integrating over each $c_i$ and summing over $i$ (reintroducing the index at this point) gives
 $$\sum_i\left(\int_{a_i}^{b_i} \kappa_g(s)\, ds + \int_{c_i} \omega_{12}\right)    = \sum_i \int_{a_i}^{b_i} \varphi'(s)\, ds = 2\pi - \sum_i \alpha_i.$$
 We now apply Stokes' theorem: $$ \sum_{i} \int_{c_i} \omega_{12}=  \iint_{\mathcal{R}}d\omega_{12}= \iint_{\mathcal{R}} K\, dA.$$ This gives
 $$ \iint_{\mathcal{R}} K\, dA + \int_c \kappa_g(s)\, ds= 2\pi -\sum_i\alpha_i$$
 as we wished to show.
 \end{proof}
 
 \begin{corollary}[Geodesic $n$-gon]
 Let $\mathcal{R}$ be as in the statement of Theorem \ref{gbl} with the additional assumption that
 the smooth segments of the boundary of this region are geodesics. Let as before $\alpha_i$ be
 the turning angles at the corners of $\mathcal{R}$. Then
 $$ \iint_{\mathcal{R}}  K\, dA = 2\pi-\sum_i \alpha_i.$$
 \end{corollary}
 
 Suppose now that $M$ is a compact, oriented $2$-dimensional Riemannian manifold empty boundary.
  
  \vspace{.1in}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=3in]{GaussBonnetII.eps}
\caption{\small Decomposition of $M$ into regions $\mathcal{R}_i$  for the global Gauss-Bonnet theorem. The
edge curves do not need to be geodesics as in this figure. Note that the spherical caps are regions bounded by $4$ sides
with corner angles equal to $\pi$.}
\label{GB1}
\end{center}
\end{figure} 


We may partition $M$ into finitely many nonoverlapping regions $\mathcal{R}_i$, each of the type considered in the local form of 
the Gauss-Bonnet theorem. The regions are assumed to intersect only on parts of their boundary. Each
boundary segment, or edge, is shared by exactly two regions. Let $\alpha_{ij}$ denote the turning angles at the corners 
of the boundary of
  $\mathcal{R}_i$ and $\beta_{ij}$ the corresponding  internal angles, so that  $\alpha_{ij}=\pi-\beta_{ij}$. 
  Let $\#_f$ be the number of regions, $\#_e$ the number of edges and $\#_v$ the number of corner vertices
  of the decomposition. 
  
  \begin{definition}[Euler characteristic]
  The number $\chi(M)=\#_v-\#_e+\#_f$ is known as the {\em Euler characteristic} of the surface. 
  \end{definition}
  
  It is known that the Euler characteristic $\chi(M)$ does not depend on the decomposition $M$ into regions $\mathcal{R}_i$. 
   
   \begin{theorem}[Gauss-Bonnet, global version]
   Let $M$  be a compact, two-dimensional, orientable Riemannian manifold without boundary. Then 
   $$\frac1{2\pi}\iint_{M}K\, dA =  \chi(M) $$
   where $\chi(M)\in \mathbb{Z}$ is the Euler characteristic of $M$. In particular,  $\chi(M)$ does not depend on the decomposition of $M$ into regions $\mathcal{R}_i$ and the integral on the left-hand side does not depend on the Riemannian metric from which $K$ is calculated.
   \end{theorem}
 \begin{proof} We use the decomposition $M=\bigcup_i\mathcal{R}_i$. 
  Note that 
  $\sum_{i,j}\beta_{ij} =2\pi\#_v $
and     $\sum_i\int_{\partial \mathcal{R}_i} \kappa_g(s)\, ds=0.$ The latter  is because 
  each edge is shared by a pair of regions that contribute geodesic curvature integrals with opposite signs.
   Then, using the local form of the Gauss-Bonnet theorem, 
  $$\iint_MK\, dA = \sum_{i} \iint_{\mathcal{R}_i}  K\, dA= \sum_i\left( \iint_{\mathcal{R}_i} K\, dA + \int_{\partial \mathcal{R}_i}\kappa_g(s)\, ds  \right)=\sum_i \left( 2\pi - \sum_j \alpha_{ij}\right)=2\pi \#_f - \sum_{ij}\alpha_{ij}.$$
The sum $\sum_{ij}$ is over all the edges counted with multiplicity two, so the sum has $2\#_e$ terms.
The internal angles $\beta_{ij}$ add up to $2\pi\#_v$ since each vertex accounts for an angle $2\pi$. Therefore,
  $$ \sum_{ij}\alpha_{ij} =\sum_{ij}(\pi-\beta_{ij})=2\pi \#_e - \sum_{ij}\beta_{ij}=2\pi \#_e-2\pi\#_v.$$
  Therefore
  $$\iint_MK\, dA= 2\pi(\#_f-\#_e+\#_v)=2\pi\chi(M). $$
  as we wished to prove.
     \end{proof}

 
 
 
 A  fundamental theorem in topology states that two compact surfaces without boundary (not necessarily  submanifolds of  $\mathbb{R}^3$)
  are homeomorphic to one another if and only  if their Euler characteristics coincide and both surfaces are
  either orientable or non-orientable.  Orientable, compact surfaces without boundary are determined up to
  homeomorphism (equivalently, diffeomorphism) if they have the same Euler characteristics. 
  The model surface with Euler characteristic $\chi(M)=2-2g$
  is homeomorphic to a sphere with $g$ handles attached. 
 
 \vspace{.1in}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=3in]{handles.eps}
\caption{\small A  surface of genus $g=3$ and Euler characteristic $\chi(M)=2-2g= -4$.}
\label{GB1}
\end{center}
\end{figure} 

We will later revisit the proof of the Gauss-Bonnet theorem for surfaces in light of Chern's proof of the general Chern-Gauss-Bonnet theorem.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 

\item {\bf Read Chapter 9, pages 263 to 282, of Lee's text, about the Gauss-Bonnet Theorem for surfaces.}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\item  {\bf Curvature of an regular plane curve.}   Let $c(t)=(x(t),y(t))$  be a regular plane curve, where $t$ is not necessarily arclength parameter. Starting from the definition of the
signed curvature  $\kappa$ given  above (Definition \ref{geodesic curvature}), show that
\begin{enumerate}
\item
$\kappa=\left({\dot{x}\ddot{y}-\dot{y}\ddot{x}}\right)/{\left(\dot{x}^2+\dot{y}^2\right)^{3/2}}.$
Here $\dot{x}=dx/dt$ and $\ddot{x}=d^2x/dt^2$. 
\item Obtain the curvature $\kappa(\theta)$ of an ellipse given in parametric form by 
$ c(\theta)=(a\cos\theta, b\sin\theta).$
 \end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf Shape operator, mean and Gaussian curvatures of surfaces of revolution. }
Let  $(f(u),g(u))$ be a unit-speed  curve (i.e., a curve parametrized by arc length) without self-intersections in the $(y,z)$-plane.
Assume $f(u)>0$, so that $(f(u),g(u))$ can be rotated about the $z$-axis to form  a surface of revolution $M$ in $\mathbb{R}^3$.
A parametrization of the surface of revolution  is
$$\Psi(u,v)= \left(\begin{array}{c}f(u)\cos v \\f(u)\sin v \\g(u)\end{array}\right),\ \  0<v<2\pi. $$
Assume that $\Psi$ is a diffeomorphism onto its image, so that $u,v$ are coordinates on a chart $U$ in $M$.
Then $$e_1=\partial/\partial u, \ \ \ e_2=\partial/\partial v$$ form a basis for the tangent space $T_pM$ for $p\in U$. 
\begin{enumerate}
\item Find the matrix  of the shape operator  of the surface of revolution  with respect to the  basis $e_1, e_2$ at $p$.
\item Compute the mean and Gaussian curvatures  of   $M$ at $p$. (Recall: the Gaussian curvature is the product of the principal curvatures and
the mean curvature is the arithmetic average of the principal curvatures. The principal curvatures are the eigenvalues of the shape operator.)
\end{enumerate}


 
 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\item  {\bf Geometry of tubes  in $\mathbb{R}^3$.} Let $c(s)$ be a smooth closed curve in $\mathbb{R}^2$ parametrized by arc-length, where $\mathbb{R}^2$ is regarded as a plane in $\mathbb{R}^3$. Let $e_1(s):= c'(s)$ (the tangent vector to the curve) and $e_2(s)$ the unit orthogonal vector to 
$e_1(s)$ so that $e_3:=e_1(s)\times e_2(s)$ is the standard basis vector $(0,0,1)$ of $\mathbb{R}^3$.  Define the  parametric surface  $M\subseteq \mathbb{R}^3$ with parametrization
$$\Phi(s,\varphi) = c(s) + r\left[\cos(\varphi) e_2 (s) + \sin(\varphi) e_3\right] $$
for $0\leq \varphi\leq 2\pi$. Note that the image of $\Phi$ is a  tube of radius $r$ with central (plane) curve $c(s)$. 
Let $\kappa(s)$ denote the (signed) curvature of $c$, so that
$$ e'_1(s) = \kappa(s) e_2(s), \ \ e_2'(s) =- \kappa(s) e_1(s).$$ 
We further define the vector fields $E_1, E_2$ on the parametric surface $M$ by
$$E_1(s,\varphi) =\frac{\frac{\partial \Phi}{\partial s}(s,\varphi)}{\| \frac{\partial \Phi}{\partial s}(s,\varphi) \| }, \ \ \ 
E_2(s,\varphi) =\frac{\frac{\partial \Phi}{\partial \varphi}(s,\varphi)}{\| \frac{\partial \Phi}{\partial \varphi}(s,\varphi) \| }, \ \ \
N(s,\varphi) = \cos(\varphi) e_2(s) + \sin(\varphi) e_3. $$
\begin{enumerate}
\item Show that $E_1, E_2$ constitute an orthonormal frame  on  $TM$ (i.e., an orthonormal basis of $T_pM$ at each point
$p=\Phi(s,\varphi)$, and that $N$ is a unit normal vector field to $M$.  
\item Show that the shape operator $S$ of the tube satisfies at each point of $M$:
$$S(E_1(s,\varphi)) = \frac{\kappa(s) \cos(\varphi)}{1-r\kappa(s)\cos(\varphi)} E_1(s,\varphi), \ \ \ S(E_2(s,\varphi))= -\frac1r E_2(s,\varphi).$$
\item What is the Gaussian   curvature of the parametric surface at each point $p=\Phi(s,\varphi)$?
\item Let $\nabla$ denote the Levi-Civita connection on $M$. Obtain the vector fields:
$$ \nabla_{E_1} E_1, \ \ \nabla_{E_1} E_2, \ \ \nabla_{E_2} E_1, \ \ \nabla_{E_2} E_2.$$
\end{enumerate}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \item {\bf Curvature of a $2$-torus.} Show that
 there exists no Riemannian metric  on a two-dimensional torus $T$ such that the Gaussian curvature $K$ is everywhere non-zero.
 

{\bf Solution.}
 By the Gauss-Bonnet theorem, given that the Euler characteristic of the torus $T$ is $0$, we have
 $$\int_{T} K\, \text{vol}_T =0$$
 where $\text{vol}_T$ is the area form on $T$.  But $K$ is a continuous  function on $T$, so it must attain value $0$ somewhere.
\begin{center}
$\diamond $
\end{center}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

\item {\bf Holonomy angle.} Let $M$ be a $2$-dimensional Riemannian manifold and $c:[0,\ell]\rightarrow M$  a smooth simple closed path which is
the boundary of a topological disc $D$.  
We assume that $D$ is contained in the domain $\mathcal{U}$ of a local orthonormal frame $E_1, E_2$. Let $V(s)$, $s\in [0,\ell]$, be a parallel  unit length  vector field
along $c$ and $\Theta(s)$ the angle $V(s)$ in the given frame. More precisely, we have
$$V(s) =  \cos(\Theta(s)) E_1(c(s)) +  \sin(\Theta(s))E_2(c(s)).$$
Let $\omega_{12}$ be the connection form relative to the given orthonormal frame.
\begin{enumerate}
\item  Show that $\Theta'(s)= \omega_{12}(c'(s))$ for each $s\in [0,\ell]$.
\item Show that 
$$ \Theta(\ell) - \Theta(0) = \int_D K\, \text{vol}.$$
Thus by parallel translating a vector along a small loop and measuring the change $\Delta \Theta$ in the angle of the vector from the beginning to the end of the transport ($\Delta$ indicates change; it is not the Laplacian!),
we can measure the curvature of the surface near the loop: $\Delta \Theta\approx \pi r^2 K$, where $r$ is the radius of a very small (approximate) disc.
\end{enumerate}

{\bf Solution.}
\begin{enumerate}
\item Observe that, since $V(s)$ is parallel, $\frac{d}{ds} \langle E_2, V\rangle = \left\langle \frac{DE_2}{ds} , V\right\rangle$ and 
$$
\cos(\Theta(s)) \Theta'(s) = \frac{d}{ds} \langle E_2, V\rangle = \left\langle \frac{DE_2}{ds} , V\right\rangle = \omega_{12}(c'(s)) \langle E_1, V\rangle=
\omega_{12}(c'(s)) \cos(\Theta(s)).
$$
Therefore, $\Theta'(s)= \omega_{12}(c'(s)).$ 
\item By integrating over $[0,\ell]$ and using Stokes' Theorem as in the proof of the Gauss-Bonnet formula for polygons,
$$ \Theta(\ell)-\Theta(0)=\int_0^\ell \omega_{12}(c'(s))\, ds =  \int_{D} d\omega_{12} = \int_D K\, \text{vol}.$$
\end{enumerate} 
\begin{center}
$\diamond $
\end{center}



 
 \item {\bf The Pfaffian.} Let $\mathfrak{so}(2m)$ denote the vector space of $2m\times 2m$ skew-symmetric  real matrices (the Lie algebra of the special orthogonal group $SO(2m)$). We define a map
$$\text{Pf}: \mathfrak{so}(2m)\rightarrow \mathbb{R}$$
as follows. Let $e_1, \dots, e_{2m}$ be a  basis of $\mathbb{R}^{2m}$ and consider for each $A=(a_{ij})\in \mathfrak{so}(2m)$ the alternating tensor
$$ \alpha:=\sum_{i<j} a_{ij} e_i\wedge e_j = \frac12\sum_{i,j=1}^{2m} a_{ij}e_i\wedge e_j.$$
Then the Pfaffian $\text{Pf}(A)$ is the real number such that
$$\alpha^m =\alpha\wedge \cdots \wedge \alpha = m!\,\text{Pf}(A) e_1\wedge e_2\wedge \cdots \wedge e_{2m}.$$
\begin{enumerate}
\item Using the definition, show that the Pfaffian of the  $4\times 4$ matrix 
$$A = \left(\begin{array}{cccc}0 & a_{12} & a_{13} & a_{14} \\-a_{12} & 0 & a_{23} & a_{24} \\-a_{13} & -a_{23} & 0 & a_{34} \\-a_{14} & -a_{24} & -a_{34} & 0\end{array}\right)$$
is given by 
$$ \text{Pf}(A)=a_{12}a_{34}- a_{13}a_{24} + a_{23} a_{14}.$$
Observe that the value of the Pfaffian does not depend on the choice of basis of $\mathbb{R}^{2m}$.
\item Let $J=\left(\begin{array}{cc}\,\,0 & 1 \\ \!\!-1 & 0\end{array}\right)$ and consider the block-diagonal matrix
$$A =\text{diag}\left(a_1J, \dots, a_m J\right). $$ Find the  associated  alternating $2$-vector $\alpha$ and
 show that
$$ \text{Pf}(A) = a_1\cdots a_m.$$
\item If $A\in \mathfrak{so}(2m)$ and $B$ is any $2m\times 2m$ matrix, show that 
$$ \text{Pf}\left(B^\intercal A B \right) = \text{Pf}(A)\, \text{det}(B).$$
It follows that the Pfaffian is an $SO(2m)$-invariant polynomial on $\mathfrak{so}(2m)$. 
\item Show that if $A\in \mathfrak{so}(2m)$, then
$$ \left(\text{Pf}(A)\right)^2 = \text{det}(A).$$
You may take for granted the following fact from matrix algebra (which I encourage  you to try to prove for yourself). There exists an orthogonal matrix
$B\in O(2n)$ such that $$B^\intercal A B =\text{diag}\left(a_1 J, \dots, a_m J\right). $$
\item Convince yourself (no need to write it down) that 
$$ \text{Pf}(A) = \frac1{2^mm!} \sum_{\sigma\in S_{2m}} \text{sign}(\sigma) a_{\sigma(1)\sigma(2)} \cdots a_{\sigma(2m-1)\sigma(2m)}.$$
\end{enumerate}





\item {\bf Differential forms with coefficients in a vector bundle.} Let $M$ be a smooth manifold and $\pi:E\rightarrow M$  a smooth vector bundle. 
We consider the vector bundle of alternating forms with values in $E$:
$\bigwedge^k(T^*M)\otimes E$ and its smooth sections
$$ \Omega^k_M(E):= \Gamma\left(\bigwedge^k(T^*M)\otimes E\right)=\Omega^k(M)\otimes \Gamma(E).$$
Note that the second symbol $\otimes$ is the tensor product of modules.
Elements of $\Omega^k_M(E)$ are, locally, linear combinations of terms of the form $\omega\otimes \xi$ where $\omega$ is an ordinary smooth $k$-form on $M$ and
$\xi$ is a local section of $E$. An  example of an element of $\Omega_M^2(E)$ is given by the curvature tensor $R(\cdot, \cdot)$ of a  connection on 
a vector bundle $F$, where $E=F\otimes F^*\cong \text{End}(F)$. We may say that elements of $\Omega_M(E)=\bigoplus\Omega^k_M(E)$ are differential forms with coefficients in $\Gamma(E)$.

Given a connection $\nabla$ on $E$, we define the {\em  exterior covariant derivative} on $\Omega^*_M(E)$ as the map
$$d^\nabla: \Omega^k_M(E)\rightarrow \Omega_M^{k+1}(E) $$
that satisfies 
\begin{enumerate}
\item $d^\nabla = \nabla$ for $k=0$; thus $\left(d^\nabla \xi\right)(X)=\nabla_X\xi$ for $X\in \mathfrak{X}(M)$.
\item $d^\nabla\left(\omega\otimes \xi\right)= d\omega \otimes \xi + (-1)^k \omega\wedge d^\nabla \xi$.
\end{enumerate}
Show the following:
\begin{enumerate}
\item If $\psi\in \Omega_M^k(E)$ and $X_0, \dots, X_k\in \mathfrak{X}(M)$, then
\begin{align*}
\left(d^\nabla \psi\right)(X_0,\dots, X_k) &= \sum_i ((-1)^i \nabla_{X_i}\left(\psi\left(X_0, \dots, \widehat{X}_i, \dots, X_k\right)\right)\\
& \ \ \ \ \ \ \ \ \ \ +\sum_{i<j}(-1)^{i+j}\psi\left([X_i,X_j], X_0, \dots, \widehat{X}_j, \dots, \widehat{X}_i, \dots, X_k\right).
\end{align*}
\item If $\psi\in \Omega_M^0(E)$ (that is, a smooth section of $E$), then
$$ d^\nabla\circ d^\nabla \xi =  R(\cdot, \cdot)\xi$$
where $R$ is the curvature tensor of $\nabla$. 
\end{enumerate}


 
 \item  {\bf Volume form of a sphere in Cartesian coordinates.}   
Let $S^{n-1}(\rho)$ be the sphere of radius $\rho$ centered 
 at the origin of $\mathbb{R}^n$, where $\rho(x) =\sqrt{x_1^2+\cdots +x_n^2}$
 and $x_1, \dots, x_n$ are the Cartesian coordinates on $\mathbb{R}^n$.
 Orient $S^{n-1}(\rho)$ as the boundary of the solid ball of radius $\rho$.
 \begin{enumerate}
\item Prove that  the volume form on $S^{n-1}(\rho)$ is (the pull-back to the sphere under the inclusion map of) 
 $$ \text{vol}_{S^{n-1}(\rho)} = \frac1\rho\sum_{i=1}^n (-1)^{i-1}x^i \, dx_1\wedge \cdots \wedge \widehat{dx}_i\wedge \cdots \wedge dx_n.$$
 Here the hat $\widehat{dx}_i$ indicates that the term $dx_i$ is absent. 
 
 Note: If $M$ is an oriented codimension-$1$ submanifold of a Riemannian manifold $\tilde{M}$ and $N$ is a unit normal vector to $M$ compatible with the orientation, then the (positive) volume form $\omega$ of $M$ is related to the volume form $\tilde{\omega}$ of $\tilde{M}$  according to 
 $\omega= i_N\tilde{\omega}.$
 Here $i_N$ indicates interior multiplication: 
 $$i_N\tilde{\omega}(\cdot,\dots,\cdot)=\tilde{\omega}(N,\cdot,\dots,\cdot).$$
\item Let $\text{vol}_{\mathbb{R}^n}=dx_1\wedge\cdots \wedge dx_n$ be the standard volume form on $\mathbb{R}^n$.   Show that 
$\text{vol}_{\mathbb{R}^n} =d\rho\wedge \text{vol}_{S^{n-1}(\rho)}$.
\item  Let $Z$ be a vector field on an orientable Riemannian manifold $M$ with volume form $\omega$. The {\em divergence} of $Z$, denoted $\text{div}Z$ or $\text{div}_MZ$,
is the unique function on $M$ such that
$$\mathcal{L}_Z \omega = (\text{div}Z) \omega.  $$
Thus the divergence measures the rate at which the flow of $Z$ changes the volume in $M$ at the linear level. 
Now suppose $Z$ is a vector field in $\mathbb{R}^n$ such that $Z_x\cdot x=0$; that is,  $Z$ is, at each point, tangent to the sphere containing that point.
Show that $\text{div}_{S^{n-1}}Z = \text{div}_{\mathbb{R}^{n}}Z$. In words: the restriction to the sphere of the divergence of $Z$ regarded as a vector field on $\mathbb{R}^n$ is the
divergence of $Z$ regarded as a vector field on the sphere.  (Suggestion: Apply the Lie derivative along $Z$ to $d\rho\wedge \text{vol}_{S^{n-1}}$.)
Also check that the divergence of a vector field $Z=\sum_i f_i\partial/\partial x_i$ in $\mathbb{R}^n$ is
$\text{div}Z=\sum_i \partial f_i/\partial x_i.$
\item A skew-symmetric $n\times n$ real matrix $z$ (an element of the Lie algebra $\mathfrak{so}(n)$) induces a vector field $Z$ on $\mathbb{R}^n$
such that at each $x\in \mathbb{R}^n$,
$$ Z_x = \left.\frac{d}{dt} e^{tz}x\right|_{t=0} = zx = \sum_{i,j}z_{ij} x_j\frac{\partial}{\partial x_i}.$$ 
Show that the restriction of $Z$ to the unit sphere has $0$ divergence.
\item For this final item, recall that if $M$ is an $n$-dimensional manifold and $\omega$ is a smooth $n$-form with compact support (in particular, if $\omega$ is
any $n$-form on a compact $M$) and $F:M\rightarrow M$ is a diffeomorphism then,  for every open set $U\subseteq M$, 
$$\int_{F(U)} \omega =\int_U F^*\omega.  $$
Show that if $Z$ is the vector field on the sphere defined in the previous item and $\Phi_t:{S}^{n-1}\rightarrow S^{n-1}$ is the flow of  $Z$
then, for every open set $U\subseteq S^{n-1}$,  
$$\text{Vol}(\Phi_t(U)):=\int_{\Phi_t(U)}\text{vol}_{S^{n-1}}$$ is constant in $t$.   

\end{enumerate}




\item {\bf Volume form of an $n$-sphere in spherical coordinates and hypersurface area.}  For $n\geq 3$ we define the spherical  coordinates  on $\mathbb{R}^n$ as follows. (See Figure \ref{spherical}.)
Let $r_k$ be the distance of the point $(x_1, \dots, x_k)$ from the origin in $\mathbb{R}^k$:
$$ r_k = \sqrt{x_1^2+\cdots+ x_k^2}.$$
The {\em spherical coordinates} on $\mathbb{R}^2$ are the usual polar coordinates $r=r_2$, $\theta$, $0\leq \theta<2\pi$. 

 \vspace{.1in}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=2.0in]{spherical.eps}
\caption{\small  Spherical coordinates. }
\label{spherical}
\end{center}
\end{figure} 

For $n\geq 3$, if $x=(x_1, \dots, x_n)$, the angle $\varphi_n$ is the angle the vector $x$ makes relative  to the  $x_n$-axis; 
it is determined uniquely by the formula
$$\cos\varphi_n = \frac{x_n}{r_n}, \ \  0\leq \varphi_n<\pi. $$
Project $x$ to $\mathbb{R}^{n-1}$ along  the $x_n$-axis. By induction, the spherical  coordinates $r_{n-1}, \theta, \varphi_3, \dots, \varphi_{n-1}$ of 
the projection $(x_1, \dots, x_{n-1})$ in $\mathbb{R}^{n-1}$ are defined. Then the spherical coordinates of $(x_1, \dots, x_n)$ in $\mathbb{R}^n$ are defined
to be $r_n, \theta, \varphi_3, \dots, \varphi_{n-1}, \varphi_n$. Thus for $k=3, \dots, n$ we have $\cos\varphi_k=x_k/r_k$.
More explicitly,  setting $\rho = r_n$,
\begin{align*}
x_n&= \rho \cos \varphi_n\\
 x_{n-1}&= \rho\sin \varphi_n \cos\varphi_{n-1}\\
 x_{n-2} &= \rho  \sin \varphi_n\sin\varphi_{n-1} \cos \varphi_{n-2}\\
  &\  \  \vdots  \\
  x_3 &=  \rho\sin \varphi_{n} \cdots \sin \varphi_4\cos\varphi_3\\
 x_2 &= \rho\sin \varphi_{n} \cdots \sin \varphi_4\sin\varphi_3\sin\theta\\
 x_1 &=  \rho\sin \varphi_{n} \cdots \sin \varphi_4\sin \varphi_3\cos\theta.
\end{align*}
Note that $r_k = \rho \sin \varphi_n\cdots \sin \varphi_{k+1}$ (or $r_k = \sin{\varphi_{k+1} r_{k+1}}$) for $k<n$. 
Setting $\rho=a$, the above defines a parametrization of $S^{n-1}(a)$, which we may write as $x = F(\theta, \varphi_3, \varphi_4, \dots, \varphi_n)$.
(A {\em parametrization} of a manifold is the inverse of a coordinate map.) Also observe  that, we may identify 
the tangent vectors to the sphere ${\partial}/{\partial \theta}$ and ${\partial}/{\partial \varphi_k}$  with the vectors
${\partial F}/{\partial \theta}$ and ${\partial F}/{\partial \varphi_k}$
 in $\mathbb{R}^n$.

It is not difficult to show that 
$$e_1 =\frac{\partial}{\partial \rho}= \frac{x}{\rho}, \ \ e_2= \frac1{r_2}\frac{\partial}{\partial \theta}, \ \ e_3=\frac1{r_3}\frac{\partial}{\partial\varphi_3}, \ \ \dots,\ \ e_n= \frac{1}{r_n} \frac{\partial}{\partial \varphi_n} $$
constitutes an orthonormal frame along the sphere, where $e_1=\nu$ is a unit normal vector field. 

 \begin{enumerate}
 \item Give the  sphere $S^{n-1}(a)$ the boundary orientation of the closed solid ball of radius $a$. (This is the orientation defined by the form
 $\text{vol}_{S^{n-1}(a)}$.) Show that the volume form on $S^{n-1}(a)$ in spherical  coordinates  is, up to sign,
 $$ \omega = a^{n-1}\left(\sin^{n-2}\varphi_n\right) \left(\sin^{n-3}\varphi_{n-1}\right)\cdots (\sin \varphi_3)\, d\theta \wedge d\varphi_3\wedge \cdots \wedge d\varphi_n.$$
 Note: A smooth $n$-form $\omega$ on an oriented $n$-dimensional Riemannian manifold $M$  is the volume form of $M$ if $\omega_p(u_1, \dots, u_n)=1$ for any positive
 orthonormal  basis $\{u_j\}$ at $p$, for all $p\in M$. 
 \item By integration by parts, show that
$$\int_0^\pi \sin^n \varphi \, d\varphi = \frac{n-1}{n}\int_0^\pi \sin^{n-2}\varphi\, d\varphi. $$
\item Give a numerical expression for $\int_0^\pi \sin^{2k}\varphi\, d\varphi$ and $\int_0^\pi \sin^{2k-1}\varphi\, d\varphi.$
\item  Show that 
$$\frac{\text{Vol}\left(S^{n+1}(1)\right)}{\text{Vol}\left(S^{n-1}(1)\right)} =  \int_0^\pi \sin^n \varphi\, d\varphi  \int_0^\pi \sin^{n-1} \varphi\, d\varphi.$$
Although this wasn't asked in the original version of the problem, it is not difficult to conclude that
$$\frac{\text{Vol}\left(S^{n+1}(1)\right)}{\text{Vol}\left(S^{n-1}(1)\right)} = \frac{2\pi}{n}$$
for $n$ both even and odd. 
\item Show that the volume of the $(2n-1)$-dimensional unit sphere is  $2\pi^n/(n-1)!$. 
 \end{enumerate}



\end{enumerate}
  \end{document}