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\begin{document}
\title{{\bf  Homework set 3 - due 09/20/24}\\ {\large Math 5047} 
}
\author{}
\date{}  
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Turn in problems 3, 4, 5, 6.
\begin{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf Read all of Chapter 4 of Lee's text.}
\item {\bf Read Chapter 5 of Lee's text up to page 26 (end of section `Connections on Abstract Riemannian Manifolds').}
\item {\bf Parallel transport.}  Let $M$ be a Riemannian manifold. Consider the mapping
$$P = P_{c,t_0,t}:T_{c(t_0)}M\rightarrow T_{c(t)}M $$
defined by: $P_{c,t_0,t}(v)$, $v\in T_{c(t_0)}M$, is the vector obtained by parallel transporting the vector $v$ along the curve $c$. Show that $P$ is an isometry
and that, if $M$ is oriented, $P$ preserves the orientation.


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf Recovering the connection from parallel transport.} Let $X$ and $Y$ be differentiable vector fields on a Riemannian manifold $M$. Let $p\in M$ and let $c:I\rightarrow M$
be an integral curve of $X$ through $p$, i.e.
$c(t_0)=p$ and $\frac{dc}{dt}=X(c(t))$. Prove that the Riemannian connection of $M$ satisfies
$$(\nabla_XY)(p)=\left.\frac{d}{dt}(P_{c,t_0,t})^{-1}(Y(c(t)))\right|_{t=t_0}, $$
where $P_{c,t_0,t}:T_{c(t_0)}M\rightarrow T_{c(t)}M$ is the parallel transport along $c$, from $t_0$ to $t$. (This shows how the connection can be recovered from the concept
of parallelism.)


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item  {\bf Levi-Civita connection of a submanifold.} Let $f:M^n\rightarrow \overline{M}^{n+k}$ be an immersion of a differentiable manifold $M$ into a Riemannian manifold $\overline{M}$.
Assume that $M$ has the Riemannian metric induced by $f$. That is, the Riemannian inner product  on $T_pM$ is the restriction to $T_pM$ of
the Riemannian inner product on $T_p\overline{M}$.    Let $p\in M$ and let $U\subseteq M$ be a neighborhood of $p$ such that $f(U)\subseteq \overline{M}$ is a submanifold of $\overline{M}$. Further, suppose that $X, Y$ are differentiable vector fields on $f(U)$ which extend to differentiable vector fields $\overline{X}, \overline{Y}$ on an open  set of $\overline{M}$.  Define $$\left(\nabla_XY\right)(p)= \text{ tangential component of } \left(\overline{\nabla}_{\overline{X}}\overline{Y}\right)(p),$$
where $\overline{\nabla}$ is the Levi-Civita connection of $\overline{M}$. Prove that $\nabla$ is the Levi-Civita connection of $M$.   


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf  Covariant derivative of vector field over a constant curve.}  Let $M$ be a Riemannian  manifold and let $p$ be a point of $M$. 
Consider a constant curve $f:I\rightarrow M$ given by $f(t)=p$, for all $t\in I$. Let $V$  be  a vector field along $f$ (that is,
$V$ is a differentiable mapping of $I$ into $T_pM$). Show that $\frac{DV}{dt}=\frac{dV}{dt}$, that is to say, the covariant derivative  coincides with the usual
derivative of $V:I\rightarrow T_pM.$



\item {\bf Horizontal and vertical subbundles of $TTM$.} This extended discussion contains a very useful characterization of connections on a vector bundle $N$.  We will be a bit sketchy. See, for example, {\em Differential Geometric Structures} by  Walter P. Poor for details omitted here. 

We begin by considering    $N=TM$, the tangent bundle of a smooth manifold $M$,  and $\pi :N\rightarrow M$ the
base-point projection. It will be convenient to indicate a point  $(p,v)\in N$ simply as $v$, with $\pi(v)=p$. At each $v\in N$ define the {\em vertical} subspace
$\mathcal{V}_v\subseteq T_vN$ to be the kernel of $d\pi_v:T_vN\rightarrow T_pM$.  The disjoint union  $\mathcal{V}=\coprod_{v\in TM}\mathcal{V}_v$
with the natural base-point projection $\mathcal{V}\rightarrow N$, which maps $\xi\in \mathcal{V}_v$ to $v$, is easily shown to be a smooth vector subbundle of $TN$. Notice that $\mathcal{V}_v$ is the tangent space at $v$ of the fiber $\pi^{-1}(p)$, where $p=\pi(v)$. 

In general, there is no canonical way to select a complementary subbundle to $\mathcal{V}$, that is, a vector subbundle $\mathcal{H}$ of $TN$ such that
$TN=\mathcal{V}\oplus \mathcal{H}$ (direct sum of vector bundles). A choice of such  {\em horizontal} subbundle is (essentially) a choice of connection on $TM$, as you will show. 

\begin{enumerate}
\item Show that each vertical subspace $\mathcal{V}_v\subseteq T_vN$ is canonically isomorphic to $T_pM$, $p=\pi(v)$.  Specifically,  check that
the map   
$$\mathcal{I}_v:w\in T_pM\mapsto \left.\frac{d}{dt}\right|_{t=0} (v+tw)\in \mathcal{V}_v$$
is an isomorphism of vector spaces. 
\item Let $\nabla$ be an affine connection on $TM$.  For each $v\in N$, $p=\pi(v)$, and $w\in T_pM$, let $t\mapsto c(t)$ be a smooth curve in $M$ representing $w$ (i.e., such that $c(0)=p$, $c'(0)=w$).  Let $V(t)$ be the parallel transport of $v$ along $c(t)$ in $M$. Then $t\mapsto V(t)$
is a smooth curve in $N$ such that $V(0)=v$ and $V'(0)\in T_vN$. Check that   
$$ d\pi_v V'(0) = c'(0)=w.$$ 
It can be shown in local coordinates, and by appealing to the smooth dependence on initial conditions of solutions of the linear system of ordinary differential equations from which parallel transport is obtained, that
 $V'(0)$ does not depend on the choice of representative curve $c(t)$ of $w$. (You don't need to prove this, but convince yourself that it is indeed true.)
 Now define $\mathcal{H}_v$ to be the vector subspace of $T_vN$  defined as the image of the linear map
 $$w\in T_pM\mapsto V'(0)\in T_vN. $$
 It can also be shown that this  is indeed a linear map, hence $\mathcal{H}_v$ is indeed a subspace of $T_vN$, called 
  the {\em horizontal} subspace. The inverse $$K_v:\mathcal{H}_v\rightarrow T_pM$$ is a linear isomorphism called the {\em connection map}.
 
 \item Check that $T_vN=\mathcal{V}_v\oplus \mathcal{H}_v$, the direct sum of the vertical and horizontal subspaces. (Since 
 both $\mathcal{V}_v$ and $\mathcal{H}_v$ have dimension $\text{dim}(M)$, which is half of the dimension of $N$,  it suffices to check that 
  $\mathcal{V}_v\cap \mathcal{H}_v=\{0\}.$)
  \item \label{homogeneous} For each $a\in \mathbb{R}$, let $\mu_a:N\rightarrow N$ be defined by $\mu_a(v)=av$. Show that 
  $d(\mu_a)_v:T_vN\rightarrow T_{av}N$ maps $\mathcal{H}_v$ to $\mathcal{H}_{av}$. We say that the horizontal subbundle is {\em homogeneous}.  (The vertical subbundle is also homogeneous.)
\item (This remark won't be needed later, but you may find it interesting if you saw modules in an algebra course.) A short exact sequence of modules over a fixed ring
$$ 0\rightarrow A\overset{a}{\rightarrow} B\overset{b}{\rightarrow} C\rightarrow 0$$
is called {\em split exact} if it is isomorphic to the exact sequence:
$$ 0\rightarrow A\overset{i}{\rightarrow} A\oplus C\overset{p}{\rightarrow} C\rightarrow 0 $$
where $i$ is inclusion and $p$ is projection.  What we have obtained above is that an affine connection on $TM$ gives rise to 
a splitting of the short exact sequence of modules over $C^\infty(M)$:
$$\mathfrak{X}(M) \overset{\mathcal{I}}{\rightarrow} \Gamma(TN)\overset{\pi_*}{\rightarrow} \mathfrak{X}(M),$$
so that $\Gamma(TN)\cong \Gamma(\mathcal{V})\oplus \Gamma(\mathcal{H})$, a direct sum of modules over $C^\infty(M)$. 
\item Let $\mathcal{H}$ be any complementary vector subbundle of  $\mathcal{V}$ in $TN$ that satisfies the homogeneity property of the above item (\ref{homogeneous}). Let $$K_v:T_vN=\mathcal{V}_v\oplus \mathcal{H}_v\rightarrow \mathcal{V}_v\cong T_pM$$ be the projection to the  vertical subspace,
which is canonically isomorphic to  $T_pM$. Show that
$$\nabla_v X:=K_vdX_pv,$$
for all $X\in \mathfrak{X}(M)$,
defines a connection on $TM$. Check that parallel translation with respect to this connection has the following description:
Let $t\mapsto c(t)$ be a smooth curve on $M$ and $v\in T_{c(0)}M$. Then $c$ has a unique lift    $\overline{c}(t)$, a curve in $N$ 
such that $\pi\circ \overline{c}(t)=c(t)$, $\overline{c}(0)=v$, and $\overline{c}'(t)\in \mathcal{H}_{\overline{c}(t)}$. Note that
$\overline{c}(t)$ can be viewed as a vector field along $c(t)$ in $M$. 

\item Convince yourself that all that we have done above works just as well if we replace $TM$ with a general  vector bundle $\pi:N\rightarrow M$.
That is, a connection on $N$ is equivalent to a splitting of $TN$ into a direct sum $\mathcal{V}\oplus \mathcal{H}$ of vertical and horizontal  bundles, where
 for each $e\in N$, $\mathcal{V}_e$ is canonically isomorphic to the fiber $N_p$, $p=\pi(e)$, and $\mathcal{H}_e$ is homogeneous and  isomorphic to $T_pM$. Parallel transport in $M$ amounts to the horizontal lifting of curves in $M$.

\end{enumerate}


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item  {\bf Parallel transport on $S^2$.} Let $S^2\subseteq \mathbb{R}^3$ be the unit sphere, $c$ an arbitrary parallel of latitude on $S^2$ and $v_0$ a tangent vector to $S^2$ at a point of $c$. Convince yourself that the following claim holds. Consider the cone tangent to $S^2$ that intersects $S^2$ at $c$.
Then the parallel transport of $v_0$ along $c$ is the same whether it is performed on the sphere or on the cone. Furthermore, parallel translation on the cone is obtained by cutting it along a ray and unrolling it flat on $\mathbb{R}^2$. Then parallel translation on the cone  is then simply vector space translation in  $\mathbb{R}^2$.

 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 
 \item {\bf Review  of exterior calculus (or calculus of differential forms).}  The following are all definitions. 
\begin{enumerate}
\item  {\bf Interior product}. Let $X$ be a vector field on the smooth manifold $M$. Given a $k$-form $\theta$ on $M$, we define the $(k-1)$-form
$i_X\theta$ to be $0$ if $k=0$ and
$$ (i_X\theta)(Y_1, \dots , Y_{k-1}) = \theta(X, Y_1, \dots, Y_{k-1})$$
 if $k\geq 1$, where $Y_1, \dots, Y_{k-1}$ are smooth vector fields. 
 \item {\bf Lie derivative of differential forms.}  Let $X$ be a smooth vector field on a smooth manifold $M$. Recall that to $X$ one can associate its (local) flow $\Phi_t$ such that
 for any given $p\in M$, $\Phi_t(p)$ is the integral curve of $X$ with initial condition $\Phi_0(p)=0$, defined for some open interval in $t$ containing $0$.
 In other words, $\gamma(t)=\Phi_t(p)$ satisfies the initial value problem
 $$\gamma'(t) = X_{\gamma(t)},\ \ \gamma(0)=p. $$
 From the general theory of differential equations we know that there is a unique solution (over a maximal interval for $t$)
 and that $\Phi_t$ defines a local flow of diffeomorphisms: $\Phi_{t+s} = \Phi_t\circ \Phi_s$ whenever the composition makes sense. In addition, $\Phi_0$ is the identity diffeomorphism. (Under certain conditions,
 e.g., when the manifold is compact, we know that $\Phi_t$ is a diffeomorphism of $M$ for all $t\in \mathbb{R}$. In such cases we say that $\Phi_t$ defines a flow on $M$.) If $\theta$ is a differential form on $M$, it makes sense (for $t$ sufficiently close to $0$) to define the pullback $\Phi_t^*\theta$.
 The {\em Lie derivative} of  $\theta$ with respect to $X$ at $p\in M$ is defined by
 $$ (\mathcal{L}_X\theta)_p :=\lim_{t\rightarrow 0}\frac{(\Phi_t^*\theta)_p- \theta_p}{t}=\left.\frac{d}{dt}(\Phi_t^*\theta)_p\right|_{t=0}.$$
A similar definition applies to vector fields if we define $\Phi_t^*Y:=(\Phi_{-t})_*Y$, where $Y$ is a smooth vector field on $M$.  (With a little thought, it
 is not difficult to figure out how to define the Lie derivative for a general tensor field. We may return to this later.)
\end{enumerate}

 

\item {\bf Interior multiplication as a signed derivation.}  Show that $i_X$ satisfies the following signed product rule:
 $$i_X(\theta \wedge\eta) = i_X\theta\wedge \eta + (-1)^k \theta \wedge i_X\eta $$
 where $\theta$ is a $k$-form and $\eta$ is an arbitrary differential form. (This is a pointwise operation; there are no derivatives actually involved. Note:
 You can find the proof of this fact in  John Lee's ``Introduction to Smooth Manifolds,''  second edition, Lemma 14.13, page 358.
 You can get  a pdf of this text from the Olin library. )
 
   
 
 
 \item {\bf Properties of the Lie derivative}
  \begin{enumerate}
 \item   Show that if $f$ is a smooth function on $M$ (a $0$-form) then $\mathcal{L}_Xf=Xf$.
 \item   Show that $\mathcal{L}_X$ commutes with the exterior derivative: $\mathcal{L}_Xd=d\mathcal{L}_X$.
 \item  Show that if $\theta_1, \theta_2$ are arbitrary smooth forms,  $$\mathcal{L}_X(\theta_1\wedge \theta_2) = \mathcal{L}_X \theta_1\wedge \theta_2+
\theta_1\wedge  \mathcal{L}_X\theta_2$$ 
 \item   Show that
 $$ \mathcal{L}_XY=[X,Y].$$
 I suggest the following approach: first recall that if $Y$ is a vector field on a manifold $M$ and $F:M\rightarrow N$ is a smooth map into another manifold $N$,
 then the derivative map  $F_{*,p}:T_pM\rightarrow T_{F(p)}N$ satisfies
 $$(F_{*,p} Y_p)f = Y_{p}(f\circ F)$$
 where $f$ is any smooth function on $N$. (Convince yourself of this basic fact!) Let now $F$ be a diffeomorphism. 
 Then the push-forward of $Y$ under $F$ is defined as the vector field $F_*Y$ on $N$ given at any $q=F(p)\in N$ by
 $$ (F_*Y)_{q} = F_{*,p}Y_p. $$
 Now suppose that $f$ and $Y$ are vector fields on $N$.
 It follows from the above remarks and the definition of pull-back of functions ($F^*f=f\circ F$) that
 $$(Yf)(F(p)) = (F^{-1}_*Y)_p (F^*f)(p). $$ 
 Now let $F=\Phi_t$ (the local flow of $X$) and take derivatives in $t$ at $t=0$.
 
  

 \item \label{part e} If $\theta$ is a differential $k$-form and $X, Y_1, \dots, Y_k$ are smooth vector fields,
 $$X\theta(Y_1, \dots, Y_k) = (\mathcal{L}_X\theta)(Y_1, \dots, Y_k) + \theta(\mathcal{L}_XY_1, Y_2, \dots, Y_k)+\cdots +\theta(Y_1, \dots, Y_{k-1}, \mathcal{L}_XY_k). $$
 As a suggestion, you may begin by observing that if $F:M\rightarrow N$ is a diffeomorphism between smooth manifolds, $\theta$ a $k$-form on $N$
 and  $Y_1, \dots, Y_k$ vector fields on $N$, then 
 $$(\theta(Y_1, \dots, Y_k))\circ F = (F^*\theta)\left(\left(F^{-1}\right)_*Y_1, \dots, \left(F^{-1}\right)_*Y_k\right). $$
 Now let $N=M$, set $F=\Phi_t$ the local flow of $X$, and take the derivative in $t$ on both sides of the equation.
 \item   Show that for any smooth form $\theta$ and smooth vector field $X$,
 $$\mathcal{L}_X\theta = (di_X+i_Xd)\theta.$$
 Here are some potentially useful remarks: 
 From the general properties of the exterior derivative (See the beginning of A.5, page 299 of Tu's text), if $\theta$ is a $k$-form and $\omega$ is
 a general smooth form, then $$d(\theta\wedge \omega)=d\theta\wedge \omega+(-1)^k \theta\wedge d\omega.$$
 Using the previous problem concerning the interior product, show that the operation $\mathcal{M}_X= di_X+i_Xd$ satisfies
 $$\mathcal{M}_X(\theta\wedge \omega)  = \mathcal{M}_X\theta \wedge \omega + \theta\wedge \mathcal{M}_X\omega. $$
 We need to show that $\mathcal{M}_X=\mathcal{L}_X$. From the previous remark, it suffices to show this equality of operations on smooth functions and
 $1$-forms of the type $f dg$. Now, both $\mathcal{L}_X$ and $\mathcal{M}_X$ commute with $d$ so, in fact, it suffices to check that the operations are 
 the same on functions.  
 
 

 \item  Show that if $\theta$ is a differential $1$-form and $X_1, X_2$ are smooth vector fields, then
 $$d\theta(X_1, X_2)= X_1\theta(X_2) - X_2\theta(X_1)- \theta([X_1, X_2]). $$
  Suggestion: express this identity in terms of the exterior derivative, the Lie derivative, and the interior product.
  
 \item  Show that if $\theta$ is a smooth $2$-form and $X_1, X_2, X_3$ are smooth vector fields, then
 $$d\theta(X_1, X_2, X_3) = X_1\theta(X_2, X_3) - X_2\theta(X_1, X_3) +X_3 \theta(X_1, X_2)- \theta([X_1,X_2],X_3)+ \theta([X_1, X_3],X_2) -\theta([X_2, X_3],X_1).  $$
 Suggestion: follow the same pattern of proof as the previous problem.
\end{enumerate}

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

 
 \end{enumerate}
 
 
 \end{document}
 
 
 
  
 \vspace{.1in}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=1.5in]{skrew.eps}
\caption{\small  A screw motion in $\mathbb{R}^3$ consists of a rotation by $\omega t$ about an axis $l$
followed by translation parallel to $l$ by $h\omega t/2\pi$, where $h$ is the {\em pitch} of the screw.}
\label{screw}
\end{center}
\end{figure} 

