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\begin{document}
\title{{\bf  Homework set 10  - due Monday 11/25/24 }\\ {\large Math 5047 \---  Renato Feres} 
}
\author{}
\date{}  
\maketitle
\thispagestyle{empty}
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For the remainder of the course we will use the text {\em Differential Geometry} by Loring W. Tu. The Olin library has the pdf for free download. This assignment is a work in progress. I expect to make changes which, however, should not (greatly) affect the exercises to be turned in.

Turn in problems 1(d),  2, 3(c) 5(b,c).
\begin{enumerate}
\item {\bf Gauss-Bonnet for hypersurfaces.} Let $M$ be an oriented hypersurface in $\mathbb{R}^{2m+1}$. (Thus $M$ is a smooth manifold of  even dimension $n=2m$.) Let $N$ be a unit normal vector field on $M$ compatible with the orientation. Let $Sv:=-D_vN$ be the shape operator of $M$, where
$D$ is the Euclidean connection in $\mathbb{R}^{n+1}$. Recall that $S_p$ is a symmetric operator on $T_pM$ at each $p\in M$. Its eigenvalues
$k_1(p),\dots, k_n(p)$ are the {\em principal curvatures} of $M$ at $p$. The {\em Gauss curvature} of $M$ at $p$ is defined as the product $K(p)=k_1(p) \dots k_n(p)$. Thus $K$ is the determinant of the shape operator.

The purpose of this exercise (mostly reading) is to prove (modulo the statement of the Poincar\'e-Hopft theorem) the Gauss-Bonnet theorem for hypersurfaces of even dimension $n$:
\begin{equation}\label{gaussbonnet}\int_MK\omega =\frac12 \text{Vol}\left(S^n\right) \chi(M).\end{equation}
Here $\omega$ is the volume form on $M$. 

The {\em Gauss map} of the hypersurface is the map $g:M\rightarrow S^n$ defined by $g(p)=N(p)$, where the normal vector $N(p)$ can be   regarded 
as a point in the unit sphere once we translate it to the origin of Euclidean space.  

Finally observe that  the two affine subspaces  $T_pM$ and $T_{g(p)}S^n$
of $\mathbb{R}^{n+1}$ are translates of each other.   Thus the differential $dg_p:T_pM\rightarrow T_{g(p)}S^n$   can be regarded as a linear map on the same vector space and its determinant makes sense. 


\begin{enumerate}
\item Check that the Gauss curvature at $p$ equals the determinant of the differential of $g$. (Note: with the identification of $T_pM$ and
$T_{g(p)}S^n$, the differential  $dg_p$ can be identified with the negative of the shape operator $S_p$. So the determinant of $S_p$ and the determinant
of $dg_p$ differ by $(-1)^n$. But as $n$ is even, the two determinants are the same.)

\item Let $\omega\in \Omega^n(M)$ denote the volume form on $M$ and $\text{deg}(g)$ the degree of the Gauss map. Let us recall here the definition of
degree. Let $g:M\rightarrow S$ be a smooth map between compact, connected oriented manifolds of the same dimension $n$. Let $H^n(M)$ and $H^n(S)$ be
the top degree de Rham  cohomology groups.  By definition $H^n(M)=\Omega^n(M)/d\Omega^{n-1}(M)$. It can be shown (using Stokes's  theorem) that
the map
$$ [\omega]\in H^n(M)\mapsto \int_M\omega\in \mathbb{R}$$
is well-defined. It is also easy to check  that this map is surjective. A deeper result (see, for example, {\em From Calculus to Cohomology: de Rham cohomology and characteristic classes} by Madsen and Tornehave) is that this map is an isomorphism. We then define the degree of $g$ by
the commutative diagram
$$\begin{tikzcd}
H^n(S) \arrow{r}{g^*} \arrow[swap]{d}{\cong} & H^n(M) \arrow{d}{\cong} \\%
\mathbb{R}\arrow{r}{\text{deg}(g)}&\mathbb{R}
\end{tikzcd}
$$
Thus
$$\int_M g^*\omega =\text{deg}(g)\int_S\omega. $$
The degree of $g$ only depends on the homotopy class of $g$. A theorem from differential topology states the following. Choose a regular point $p$ of the
smooth map $g$ and let $q_1, \dots, q_\ell$ be the inverse image of $p$. Let $\text{Ind}(g,q_i)$ be either $1$ or $-1$ depending on whether
$dg_{q_i}$ is orientation preserving or reversing. Then 
$$ \text{deg}(g)=\sum_{q\in g^{-1}(p)}\text{Ind}(g,q).$$
(See Theorem 11.9, page 101, of the above cited book.) In particular, the degree of $g$ is an integer.

If we now take $\omega$ to denote the volume form on $M$, show that
$$\int_MK\omega=\text{deg}(g) \text{Vol}(S^n).$$ 
Note: by the change of variables in integration formula
$$ \int_M K\omega= \int_M \text{det}(dg_p)\omega=\int_M g^*\omega_S,$$
where $\omega_S$ is the volume form on $S^n$.
\item In this and the next item, we show that the degree of the Gauss map $g$ is half the Euler characteristic of $M$.  From this observation we obtain the Gauss-Bonnet formula for hypersurfaces:
$$\int_MK\omega =\frac12 \text{Vol}\left(S^n\right) \chi(M).$$
Let $u$ be a point is $S^n$ such that both $u$ and $-u$ are regular values of  $g$. (This is possible: if $\pi:S^n\rightarrow \mathbb{R}P^n$ is the projection map, pick a regular value of $\pi\circ g$.)  Define the vector field $X\in \mathfrak{X}(M)$ such that $X(p)$ is the orthogonal projection of $-u$ to $T_pM$:
$$X(p):=-u+\left(u\cdot N(p)\right) N(p). $$
A point $p\in M$ is a singular point (zero) of $X$ if and only if  $u=(u\cdot N(p)) N(p)$or,  equivalently, $g(p)=\pm u$. In particular, $u$ is perpendicular to $T_pM$.
Since $u$ and $-u$ are regular values of $g$ and $M$ is compact, $X$ has only finitely many zeros. Observe that, at a zero $p$ of $X$ where $g(p)=\pm u$, 
$$ dX_p w =\pm\left(u\cdot dg_pw\right) u \pm dg_pw$$
for all $w\in T_pM$. But since $g$ and $N$ are really the same (up to translation) we have $dg_pw=D_wN\in T_pM$  so $u\cdot dg_pw=0$. Therefore
$$
dX_p =\begin{cases} +dg_p & \text{ if } g(p)=u\\
-dg_p & \text{ if } g(p)=-u.
\end{cases} $$
In particular, $dX_p$ maps $T_pM$ into itself (at a zero point $p$). It follows that
$$\text{det}\left(dX_p\right)=(\pm 1)^n\text{det}\left(dg_p\right)= \text{det}\left(dg_p\right)$$
since $n$ is even.  In class, we introduced the index of a vector field in dimension $n=2$ in a somewhat informal way. A more precise definition (using the above notion of degree) shows that, if $dX_p$ is an isomorphism (which is the case here since $g(p)$ is a regular value), then $$\text{Ind}(X,p)=\text{det}\left(dX_p\right).$$
We conclude that $\text{Ind}(X,p)=+1$ if $dg_p$ preserves orientation and $-1$ if $dg_p$ reverses orientation.   
The index $\text{Index}(X)$ is the sum of the indices of $X$ at all zeros. Since the zeros of $X$ are located at the inverse images under $g$ of two regular values $\pm u$ we conclude that 
$$ \text{Index}(X)=2\text{deg}(g).$$
We conclude that
$$\int_MK\omega=\frac12\text{Index}(X) \text{Vol}(S^n).$$ 
But according to the Poincar\'e-Hopf theorem (which we discussed briefly in class), the index of $X$ equals the Euler characteristic $\chi(M)$.
\item Verify that Equation (\ref{gaussbonnet}) holds when $M$ is a sphere of radius $R$. 
\end{enumerate}

 

\item {\bf The pfaffian of an even-dimensional  hypersurface.} 
In Homework 9, Problem 8  we defined the pfaffian of a  antisymmetric linear transformation of an even-dimensional vector space. Here we use it to define the
Euler form and compute it for a hypersurface. You may want to refer to the solutions to that problem when answering the exercises enumerated below.

Let $R$ be the curvature tensor of a connection $\nabla$ on an oriented vector bundle $E\rightarrow M$ of even rank $r$. (We will shortly assume that $E=TM$.)
We assume the connection is metric for a given Riemannian metric on $E$.
If $\xi_1, \dots, \xi_r$ is a local, positive orthonormal frame of sections of $E$ and $X, Y\in \mathfrak{X}(M)$, we define the following ordinary $2$-forms on $M$:
$$\Omega_{ij}(X,Y):=\left\langle \xi_i, R(X,Y)\xi_j \right\rangle. $$
We thus obtain a matrix $\Omega$ with entries in $\Omega^2(M)$. This matrix is antisymmetric due to the symmetries of $R$. It thus makes sense
to define a differential form $\text{Pf}(\Omega)\in \Omega^r(M)$.  
 

\begin{enumerate} 
\item Check that $\text{Pf}(\Omega)$ does not depend on the choice of local orthonormal frame. Therefore it defines a global smooth form on $M$.  
 Refer to item (c) of Problem 8 in Homework 9. 
 
 Note:  we obtain directly from the above definition of $\Omega_{ij}$ the following change of frame equation. Let $\xi'_k=\sum_{i}F_{ki}\xi_i$ be a new local, 
 positive orthonormal frame and $\Omega'_{k\ell}$ the curvature forms in this new frame. Then there is a smooth function (locally defined) $F$ taking values in $SO(r)$, where $r$ is the rank of $E$, so that 
 $ \Omega'=F\Omega F^{\intercal}$
 holds. 

\item Let $S$ denote the shape operator (page 235 of Lee's text) of a hypersurface $M\subset \mathbb{R}^{n+1}$. Let $R$ be the curvature tensor
of $M$. Show that
$$\langle R(W,X)Y,Z\rangle =\langle Z, SW\rangle\langle Y,SX\rangle-\langle Y, SW\rangle \langle Z, SX\rangle,  $$
where $W, X, Y,Z\in \mathfrak{X}(M)$. (Use the Gauss Equation, Theorem 8.5, page 230, in Lee's text.)
\item Let $e_1, \dots, e_n$ be a local, positive orthonormal frame  consisting of eigenvectors of $S$ (recall that the shape operator is symmetric)
with eigenvalues (principal curvatures) $k_1, \dots,k_n$. Note that  these $k_i$ are smooth real-valued functions. Let $\theta_1, \dots, \theta_n$ be the dual frame, so that $\theta_i(e_j)=\delta_{ij}$. Show that
$$\Omega_{ij}:=\langle e_i, R(\cdot, \cdot)e_j\rangle=k_ik_j\theta_i\wedge\theta_j. $$
\item Let $m=n/2$. 
Using the notation of Problem $8$, Homework 9 we write
$$\alpha :=\sum_{i<j} \Omega_{ij} e_i\wedge e_j=\frac12\sum_{i,j}\Omega_{ij}e_i\wedge e_j. $$
By the definition of the pfaffian given in that assignment, 
$$\alpha^m =\underbrace{\alpha\wedge\cdots\wedge \alpha}_m=m! \text{Pf}(\Omega) e_1\wedge \cdots\wedge e_{2m}. $$
Show that 
$$ \text{Pf}(\Omega) = K\omega$$
where $K=\prod k_i$ is the Gauss curvature and $\omega$ is the volume form.
\end{enumerate}
Comparing the above expression for the pfaffian with the Gauss-Bonnet formula for hypersurfaces we see that the integrand in the latter formula is, in fact, the pfaffian of curvature. 




\item {\bf Induced connection on a pullback bundle.}\label{first} (See section 22.10, page 210, of Loring Tu's text.) Let $\pi: E\rightarrow M$ be a smooth vector bundle  of rank $r$ over the manifold $M$. Let $f:N\rightarrow M$ be a smooth map and consider the pullback  vector bundle $f^*E$. (This is defined in Section 20.4, page 177.)  Given a connection $\nabla$ on $E$, we wish to define a connection $\overline{\nabla}$ on $f^*E$ that satisfies the following properties:
\begin{enumerate}
\item If  $e$ is a smooth section of $E$  and  $f^*e$ denotes the pullback section $(f^*e)(q):=\left(q,e(f(q))\right)$ then 
$$\overline{\nabla}_u\left(f^*e\right)  = \left(q, \nabla_{d\!f_qu} e\right)$$
for any $u\in T_qN$. (Here $d\!f_qu=(f_*)_qu:T_qN\rightarrow T_{f(q)}M$ is the standard tangent map.)
\item The fundamental properties for a connection. (Definition 10.1, page 72.)
\end{enumerate}

 Show the following:
\begin{enumerate}
\item The above properties uniquely specify a connection on $f^*E$. 
\item  If $e_1, \dots, e_r$ is a local frame defined on an open set $U\subseteq M$ and $\omega_{ij}$ are  the connection $1$-forms for $\nabla$ relative to this frame,
then the pull-back $\overline{\omega}_{ij}:=f^*\omega_{ij}$ are the connection $1$-forms for $\overline{\nabla}$ relative to the local frame $f^*e_1, \dots, f^*e_r$ on
$f^{-1}(U)$. 
\item With the same notation as in the previous item,
let  $\Omega_{ij}$ be the curvature $2$-forms for $\nabla$ relative to the local frame $e_i$. Then 
  $\overline{\Omega}_{ij}=f^*\Omega_{ij}$ are the curvature $2$-form for $\overline{\nabla}$ relative to
the pullback frame. 
\item Let $\overline{R}$ denote the curvature tensor of $\overline{\nabla}$ and $R$ the curvature tensor of $\nabla$. Show that  
$$\overline{R}_q(u,v)\overline{\xi}= \left(q, R_{f(q)}\left(d\!f_qu,d\!f_qv\right)\xi\right)$$
for $u,v\in T_qN$ and $\overline{\xi}=(q,\xi)\in (f^*E)_q$.
\end{enumerate}
Further comments: We may consider the vector bundle $\Lambda^k(T^*M)\otimes E$ over $M$, whose sections are differential $k$-forms with coefficients in $E$.
For such a form $\mu$ we define its pullback under $f$ as
$$ (f^*\mu)_q(v_1, \dots,v_k)= \left(q, \mu_{f(q)}\left(df_qv_1, \dots, df_qv_k\right)\right).$$
If $\mu=\alpha\otimes \xi$ is the tensor product of an ordinary differential form on $M$ and a section of $E$, then the definition just given amounts to
$f^*(\alpha\otimes \xi)=(f^*\alpha)\otimes (f^*\xi)$, where $f^*\alpha$ is the ordinary pullback of differential  forms. If we think of the curvature tensor $R$ as a section
of $\Lambda^2(T^*M)\otimes \text{End}(E)$, then under this definition $\overline{R}=f^*R$ and $\overline{R}_q(v_1,v_2)\overline{\xi}_q = (f^*R)_q(v_1,v_2) (f^*\xi)_q$.
Also note that the definition of $\overline{\nabla}$ implies $\overline{\nabla}(f^*\xi)=f^*(\nabla\xi)$. 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf More on vector bundle-valued differential forms.}\label{two} Let $\pi:E\rightarrow M$ be a smooth vector bundle over the manifold $M$. Recall that differential $k$-forms on $M$ with coefficients in $E$ are defined as sections of the vector bundle $\Lambda^{k}(T^*M)\otimes E$. The space of such sections will be denoted
$\Gamma\left(\Lambda^{k}(T^*M)\otimes E\right)$. In this definition,  $E$  may be replaced with other vector bundles obtained from $E$. We are especially interested in the bundle $\Lambda^*(E^*)$, that is, the exterior algebra bundle  of alternating forms on $E$. This is the vector bundle over $M$ whose fiber over $p\in M$ is 
the exterior algebra $\Lambda^*\left(E_p^*\right)=\bigoplus_k \Lambda^k\left(E_p^*\right)$, where $E_p^*$ is the dual vector space to $E_p$.  
An element $\mu\in \Gamma\left(\Lambda^{k}(T^*M)\otimes \Lambda^\ell (E^*)\right)$  will be called a $(k,\ell)$-differential form.  Note that at any $p\in M$ and
given $v_1, \dots, v_k\in T_pM$, 
$$ \mu_p(v_1, \dots, v_k)\in \Lambda^\ell\left(E^*_p\right).$$
Thus it makes sense to write $ \mu_p(v_1, \dots, v_k)(u_1, \dots, u_\ell)\in \mathbb{R}$ (or $\mathbb{C}$ if $E$ is complex), where the $u_i$ are elements of $E_p$,
and the resulting number is anti-symmetric separately in the $v_i$ and in the $u_j$ arguments.  

For example, if $E$ is Riemannian, with metric $\langle\cdot, \cdot\rangle$ and $\nabla$ is a metric connection on $E$, then
$$\Omega_p(v_1,v_2)(u_1,u_2):=\langle R(v_1,v_2)u_1,u_2\rangle_p$$
defines a section $\Omega\in \Gamma\left(\Lambda^2(T^*M)\otimes \Lambda^2\left(E^*\right)\right)$, where $R$ is the curvature tensor of $\nabla$. 
If $\mu$ is a differential $(r,j)$-form on $M$ and $\nu$ is a differential $(s,k)$-form, we define their wedge product as the $(r+s, j+k)$-form
$\mu\wedge \nu$ as follows:
$$(\mu\wedge\nu)(v_1, \dots, v_{r+s}):=\frac1{r!s!}\sum_{\sigma\in S_{r+s}} \text{sign}(\sigma) \mu(v_{\sigma(1)}, \dots, v_{\sigma(r)})\wedge  \nu(v_{\sigma(r+1)}, \dots, v_{\sigma(r+s)}).$$
Notice the need for the wedge product symbol on the right-hand side of this identity.  This is the wedge product on the exterior algebra on the dual of $E$. 
\begin{enumerate}
\item Show that if $\mu$ is a $(r,i)$-form and $\nu$ is an $(s,j)$-form, then
$$\mu\wedge \nu = (-1)^{rs+ij} \nu\wedge \mu. $$
\item \label{product separable} Show that if $\mu=\alpha\otimes \xi$ and $\nu=\beta\otimes \eta$, then
$\mu\wedge \nu = (\alpha\wedge \beta) \otimes (\xi\wedge \eta).$
\item Suppose $E$ is an oriented Riemannian vector bundle over $M$ of rank $2m$. (The even parity of the rank is not required in this part of the exercise, but will be needed later.)  Let $e_1, \dots, e_{2m}$ be a local positive (relative to the given orientation) orthonormal frame for $E$ over an open set $U\subset M$.
Let $\epsilon_1, \dots, \epsilon_{2m}$ be the dual frame, so that $\epsilon_i=\langle e_i,\cdot\rangle$. Then $\omega=\epsilon_1\wedge \dots\wedge \epsilon_{2m}$ is
the Riemannian volume form on $E$. 
\begin{enumerate} \item Argue that $\omega$ is globally defined.
\item Show that $\omega$ is parallel: $\nabla \omega=0$. 
\end{enumerate}
\item We define the Chern-Euler form $\chi\in \Gamma\left(\Lambda^{2m}(T^*M)\otimes \Lambda^{2m}\left(E^*\right)\right)$ as
$$\chi = \frac{(-1)^m}{m!(2\pi)^m} \Omega^m,$$
where $\Omega$ is the curvature form defined above. (Note: this is a frame-independent version of the curvature matrix  previously defined.) I am writing  the
exterior power  $\Omega\wedge \cdots \wedge \Omega$ as $\Omega^m$.

 Show that
$$\chi = \frac{(-1)^m}{(2\pi)^m} \text{Pf}([\Omega])\otimes \omega. $$
Here $[\Omega]$ is meant to designate the curvature matrix relative to a choice of positive orthonormal frame, whose entries are:
$$\Omega_{ij}(v_1,v_2)=\Omega(v_1,v_2)(e_i,e_j). $$ In other words,
$$\Omega(v_1,v_2) = \sum_{i<j} \Omega_{ij}(v_1,v_2)\epsilon_i\wedge\epsilon_j.  $$
(Suggestion: use the definition of the Pfaffian given in Exercise 1 of Homework set 9.)
\end{enumerate}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \item {\bf More on the exterior covariant derivative.} Different from Homework 9, I'll define here the exterior covariant derivative for a vector bundle $E$ as the map
 $$d^\nabla: \Gamma\left(\Lambda^{k}(T^*M)\otimes E\right)\rightarrow \Gamma\left(\Lambda^{k+1}(T^*M)\otimes E\right)$$
 such that, at a point $p\in M$, 
 \begin{equation}\label{dnabla}\left(d^\nabla \psi\right)\left(v_0, \dots, v_k\right) = \sum_{i}(-1)^i \nabla_{v_i}\psi(V_0, \dots, \hat{V}_i, \dots, V_k)+\sum_{i<j} \psi\left([V_i,V_j], v_0, \dots, \hat{v}_i, \dots, \hat{v}_j, \dots, v_k\right)\end{equation}
for $\psi\in  \Gamma\left(\Lambda^{k}(T^*M)\otimes E\right)$, $v_0, \dots, v_k\in T_pM$, and $V_0, \dots, V_k\in \mathfrak{X}(M)$  such that $V_i(p)=v_i$.  It follows from this definition that
if $\xi$ is a section of $E$, hence an element of $\Gamma\left(\Lambda^{0}(T^*M)\otimes E\right)=\Gamma(E)$, then $d^\nabla\xi =\nabla \xi$; and for separable sections
$\omega\otimes \xi\in  \Gamma\left(\Lambda^{k}(T^*M)\otimes E\right)$, 
\begin{equation}\label{separable}d^\nabla(\omega\otimes\xi) = d\omega \otimes \xi +(-1)^k \omega \wedge d^\nabla\xi.\end{equation}
As above, we are especially interested here in the case where the vector bundle in question is the exterior algebra $\Lambda^*(E^*)$ over a given vector bundle $E$.
Keep in mind that if $\xi, \eta$ are sections of $\Lambda^i\left(E^*\right)$ and $\Lambda^j\left(E^*\right)$, respectively,  and $v\in T_pM$, then
$$ \nabla_v (\xi\wedge\eta) = \left(\nabla_v\xi\right)\wedge \eta + \xi \wedge\left(\nabla_v\eta\right). $$
(The definition in Section 22.5, Proposition 22.7, page 206, in the textbook extends to  tensors over a general vector bundle $E$, not only $TM$. See also the initial remarks in Exercise 6,  of Homework 8.)
Note that, different from expressions involving the exterior derivative, there is no sign in this Leibniz rule for the covariant derivative.
\begin{enumerate}
\item  Show that the definition of $d^\nabla$ given by Equation (\ref{dnabla}) implies the identity (\ref{separable}).
\item If $\mu$ is an $(r,i)$-form and $\nu$ is $(s,j)$-form, show that 
$$d^\nabla(\mu\wedge\nu) =\left(d^\nabla \mu\right)\wedge \nu +(-1)^r \mu \wedge d^\nabla\nu. $$
\item If $f:N\rightarrow M$ is a smooth map and $E\rightarrow M$ is a smooth vector bundle over $M$, let $\overline{\nabla}$ be the pullback connection on $f^*E$. Show
that $f^* \circ d^\nabla=d^{\overline{\nabla}}\circ f^*$.
\item Let $\Omega$ be the curvature $(2,2)$-form defined in the previous exercise for a given metric  connection $\nabla$ on the Riemannian vector bundle $E$. Show that 
$d^\nabla \Omega=0$.
This is an expression of the Second Bianchi Identity. Also show that $d^\nabla R=0$. The latter identity does not depend on the connection being metric and on $E$ being Riemannian.
\item Recalling from the previous exercise the relation between $\Omega^m$ and the Pfaffian, show that if $E$ has rank $2m$, then $\text{Pf}([\Omega])$ is a closed
$2m$-form (of the standard kind) on $M$. 
\end{enumerate}



\item {\bf The canonical section and canonical form on $\pi^*E$.} Using the bundle map $\pi:E\rightarrow M$ we can pull back $E$ under $\pi$. Then $\pi^*E$ is a vector bundle over the manifold $E$. 
By the definition of pullback bundles,  elements of
$\pi^*E$ have the form $(e,e')\in E\times E$ such that $\pi(e) = \pi(e')$. 
Another vector bundle we can define over the manifold $E$ is the {\em vertical bundle} $V$.   The vector fiber at each $e\in E$ is the kernel of $d\pi_e$:
$V_e =\{\xi\in T_eE: d\pi_e\xi=0\}. $  Note that $V_e$ is linearly isomorphic to $E_{\pi(e)}$, with the isomorphism given by the map
$$\mathcal{I}_e: e'\in  E_{\pi(e)}\mapsto \left.\frac{d}{dt}\right|_{t=0}\left(e+te'\right)\in  V_e. $$

Given a connection $\nabla$ on $E$, we can define the {\em horizontal} subbundle of $TE$, denoted $H$. The fiber $H_e$ is defined as the kernel of a map $\mathcal{K}_e:T_eE\rightarrow E_{\pi(e)}$ which we define as follows. An element $\xi\in T_eE$ can be represented by a curve $e(t)\in E$, so that
$e(0)=e$ and $e'(0)=\xi$. Then $e(t)$ is a section of $E$ over the curve $\gamma(t)=\pi(e(t))\in M$. Therefore it makes sense to define
$$\mathcal{K}_e\xi =\left.\frac{\nabla e}{dt}\right|_{t=0}\in E_{\pi(e)}. $$
Note that $\mathcal{K}_e\circ \mathcal{I}_e$ is the identity map on $E_{\pi(e)}$. Moreover, the rank of $H$ is equal to the dimension of $M$ and $d\pi_e:H_e\rightarrow T_{\pi(e)}M$ and $\mathcal{K}_e:V_e\rightarrow E_{\pi(e)}$ are linear isomorphisms for each $e$, as can be easily checked. 
Also observe that $\mathcal{I}_e\circ\mathcal{K}_e$ is the identity on $V_e$.
From these definitions we obtain a splitting
of $TE$ as a direct sum of subbundles: $TE=V\oplus H$.  

The pullback bundle  $\pi^*E$ admits a canonical section $\mathcal{s}\in \Gamma(\pi^*E)$ such that $\mathcal{s}(e)=(e,e)$.  Given the isomorphism of $\pi^*E$ and $V$ that you will establish below, $\mathcal{s}$ may be regarded as a section of $V$, in which case it is  given by
  $\mathcal{s}(e)=\mathcal{I}_ee$. 
  Further define the $(0,1)$-form $\eta$ on $\pi^*E$  (which we may regard as a section of the dual vertical bundle $V^*$ over $E$):
$$\eta_e\left(\mathcal{I}_ee'\right)=\left\langle \mathcal{s}(e), \mathcal{I}_ee'\right\rangle_e = \left\langle e,e'\right\rangle_{\pi(e)}.$$ We are using here the pullback metric to $\pi^*E$,  
still denote by $\langle\cdot,\cdot\rangle$.  I note that $\overline{\nabla}$ is a metric connection for this metric.

Finally, we define the {\em connection} $(1,1)$-form $\theta=d^{\overline{\nabla}}\eta \in \Gamma\left(T^*E \otimes V^*\right)$. 
  
\begin{enumerate}
\item Show that $\pi^*E$ and $V$ are isomorphic vector bundles over $E$. 
\item If $\overline{\nabla}$ is the pullback connection on the vector bundle $\pi^*E\cong V\rightarrow E$  and $\xi\in T_eE$, show that 
$ \overline{\nabla}_\xi \mathcal{s} = \mathcal{I}_e  \mathcal{K}_e \xi.$ 
\item Let $\xi\in T_eE$ and $e'\in E_{\pi(e)}$. Show that $\theta_e(\xi)\left(\mathcal{I}_ee'\right)=\left\langle \mathcal{K}_e\xi, e'\right\rangle_{\pi(e)}$.
\item Show that
$d^{\overline{\nabla}}\theta_e\left(\xi_1,\xi_2\right)\left(u\right)=\left(\pi^*\Omega\right)_e\left(\xi_1, \xi_2\right)(\mathcal{s}(e),u)$ for $\xi_1, \xi_2\in T_eE$ and $u\in V_{e}$.
\item Let $E_1\subseteq E$ denote the subbundle of $E$ whose fiber at $p\in M$ consists of the vectors of unit length in $E_p$. We call it the {\em sphere bundle} of $E$. (It is a fiber bundle over $M$ but, naturally,  not a vector bundle.) If $\xi$ is tangent to $E_1$ at $e\in E_1$, show that
$\theta_e(\xi)(\mathcal{s}(e))=0.$
\item Consider the $(2m-1,2m)$-form $\Pi_j:=\eta\wedge\theta^{2j-1}\wedge \pi^*\Omega^{m-j}$.
Let $\xi_1, \dots, \xi_{2m-1}$ be tangent vectors to $(E_1)_p$ at $e\in E_1$ (thus the $\xi_i$ are vertical vectors; that is, $\xi_i\in V_e$) and let $X_0, X_1, \dots, X_{2m-1}$ form a positive orthonormal basis of $(\pi^*E)_e$ such that $X_0=\mathcal{s}(e)$. Show that
$$\Pi_j(\xi_1, \dots, \xi_{2m-1})(X_0, \dots, X_{2m-1})=0$$
for $j<m$ and 
$$\Pi_m(\xi_1, \dots, \xi_{2m-1})(X_0, \dots, X_{2m-1})=(2m-1)!\omega_e\left(\xi_1, \dots, \xi_{2m-1}\right)$$
where $\omega$ is the volume form on the fiber of $E_1$ at $p\in M$. 
\end{enumerate}

\item Keeping with the notation of the previous exercise, define $\Pi:= \sum_{j=1}^m c_j\Pi_j$ where
$$c_j=(-1)^{m+j+1}\frac{2^{m-j+1}(j-1)!}{(m-j)!(2j-1)!\pi^m}. $$ Then $\Pi\in \Gamma\left(\Lambda^{2m-1}(T^*E)\otimes\Lambda^{2m}(\pi^*E^*)\right)$.
Show that $d^{\overline{\nabla}}\Pi = \pi^*\chi$ on $E_1$.  

Note that  $\Pi$ and $\pi^*\chi$ are forms on $E$ (or $E_1$) with coefficients in the vector bundle
$\Lambda^{2m}(E^*)$. Since the latter bundle is trivial for an oriented $E$, with the volume form $\omega$ serving as a global section, we can
write $\Pi$ and $\pi^*\chi$ as a product of ordinary forms on the base manifold $E_1$ tensor $\omega$. Using the same notations $\Pi$ and $\pi^*\chi$
for these ordinary forms, then $\pi^*\chi=d\Pi$, where $d$ is now the ordinary exterior derivative. Another point to observe is that
the integral of $\Pi$ over $(E_1)_p$ is $-1$. This is due to the previous item in this exercise, and the choice of the constant $c_m$. 



 
 \item  {\bf The Hopf bundle and Chern number.}  
In what follows, elements of $\mathbb{C}P^1$ will be written as $[z]=[z_1:z_2]$ (so-called {\em homogeneous coordinates}) to indicate the equivalence class represented by $z\in \mathbb{C}^2\setminus \{(0,0)\}$
under the action of the multiplicative group of non-zero complex numbers. (Or, equivalently, $z\in S^3$ under the action of $U(1)$.)


The Hopf bundle is the bundle of unit length vectors   of a complex line bundle over $\mathbb{C}P^1$ called the {\em tautological bundle}, which
is  a rank $1$ (complex) subbundle $L$ of the trivial bundle $\text{pr}_1:\mathbb{C}P^1\times \mathbb{C}^2\rightarrow \mathbb{C}P^1$ defined by the property
  that the fiber  $L_{[z]}$ of $L$ at $[z]$  the one-dimensional complex subspace in $\mathbb{C}^2$ spanned by $z$ itself (thus the name {\em tautological}). 
I'll used the same letter $\pi$ to denote the base point map $\pi:L\rightarrow \mathbb{C}P^1$.  By definition, $$L=\{([z],\lambda z): [z]\in \mathbb{C}P^1, \lambda\in \mathbb{C}\}=\{(\ell, z)\in \mathbb{C}P^1\times\mathbb{C}^2: z\in \ell\}, \ \ \pi(\ell, z)=\ell.$$

The trivial vector bundle $\mathbb{C}P^1\times \mathbb{C}^2$ can be endowed with  a  Hermitian metric by giving $\mathbb{C}^2$ the standard (complex valued) inner product $$\langle (z_1,z_2),(w_1,w_2)\rangle= \overline{z}_1w_1+\overline{z}_2w_2.$$
(Note: as a matter of taste, I deviate from  the textbook convention by placing the complex conjugate bar over the first vector argument of the inner product.)
The Hopf bundle is then the bundle of unit length vectors of $L$. (It is a {\em principal bundle} over $\mathbb{C}P^1$ with structure group $U(1)$, according to the definitions at the beginning of Chapter 6 of the textbook, although we don't need this fact for this assignment.) We thus have $S^3\subseteq L\subseteq \mathbb{C}P^1\times \mathbb{C}^2$, and these are all
bundles over projective space.


Our  goal   is to compute the {\em Chern number} of $L$, which is the integral over $S^2\cong \mathbb{C}P^1$ of the Chern class $c_1(L)$.

The line bundle $L$ with the  Hermitian metric $\langle\cdot, \cdot\rangle$ can be given a metric connection as follows. For each $[z]\in \mathbb{C}P^1$, we have
the orthogonal direct sum decomposition $\mathbb{C}^2 = L_{[z]}\oplus L_{[z]}^\perp$ where $L_{[z]}$ is the fiber of $L$ over $[z]$ and  $$L_{[z]}^\perp=\left\{u\in \mathbb{C}^2:\langle z, u\rangle=0\right\}.$$
 Let $\Pi_{[z]}:\mathbb{C}^2\rightarrow L_{[z]}$ be the resulting  projection map. We now  define a connection $\nabla$ on smooth sections of $L$ as follows:
 given $\xi\in \Gamma(L)$ and $u\in T_{[z]}\mathbb{C}P^1$, 
 \begin{equation}\label{connection} \nabla_u\xi = \Pi_{[z]}D_u\xi\end{equation}
 where $D$ is the ordinary derivative of a $\mathbb{C}^2$-valued (equivalently, $\mathbb{R}^4$-valued) function on $\mathbb{C}P^1$:
 $$ D_u(f_1+if_2, f_3+if_4)= \left(uf_1+iuf_2, uf_3+uf_4\right).$$
 
 We will need the connection and curvature forms of $\nabla$ for convenient choices of local trivializations of $L$.  But first note that
 if $\xi:U\rightarrow S^3\subseteq L$ is a section of $L$ over $U\subseteq \mathbb{C}P^1$, then the corresponding connection form is
 $\omega_p(u)=\langle \xi(p), \nabla_u\xi\rangle=\langle \xi(p), D_u\xi\rangle$   for  $u\in T_p\mathbb{C}P^1$. This being an actual $1$-form on $U$, the associated curvature form is 
 simply $\Omega = d\omega +\omega\wedge \omega=d\omega$. The effect of taking sections of $L$ of unit length (hence in $S^3$) is that
 $\text{Re}\left( \langle \xi, D_u\xi\rangle\right)=0$.    In fact, 
 $$0=u\langle \xi, \xi\rangle = \left\langle \nabla_u\xi, \xi\right\rangle+\left\langle \xi, \nabla_u\xi\right\rangle = \left\langle \xi, \nabla_u\xi\right\rangle +\overline{\left\langle \xi, \nabla_u\xi\right\rangle }=2\text{Re}\left(\left\langle \xi, \nabla_u\xi\right\rangle \right). $$ 
 Therefore
 \begin{equation}\label{connection form}\omega_p(u) = 2i \text{Im}\left(\langle \xi(p), D_u\xi\rangle \right). \end{equation}
  
 
 Observe that if $\eta$ is another section also  defined on $U$, then $\eta = f\xi$ where $f$ is a complex-valued function on $U$ of unit modulus: $|f(p)|=1$. 
 Then $\omega_\eta =\omega_\xi +  f^{-1}df$ and $d\omega_\eta=d\omega_\xi$. This means that the curvature form $\Omega$ will define a global closed form 
 on $\mathbb{C}P^1$.  The Chern class is then  (according to the definition on page 235 of the textbook in the special case of rank $1$) $c_1(L)=\frac{i}{2\pi}\Omega$ and
 the Chern number, our ultimate goal here, is the value of  $\int_{\mathbb{C}P^1} c_1(L)$. 
 
 

We now choose trivializing neighborhoods and sections of the Hopf bundle.  
Let  $\{U_-, U_+\}$ be the  open cover of $\mathbb{C}P^1$ given by 
 $$U_-:=\left\{[z_1:z_2]\in \mathbb{C}P^1: z_2\neq 0\right\}, \ \ U_+ := \left\{[z_1:z_2]\in \mathbb{C}P^1: z_1\neq 0\right\}.$$ We define coordinates $\varphi_\pm :U_{\pm}\rightarrow \mathbb{C}$
 by $\varphi_-([z_1:z_2])=z_-:=z_1/z_2$ and $\varphi_+([z_1:z_2])=z_+:=z_2/z_1$. (These give us a smooth (in fact, holomorphic) atlas on $\mathbb{C}P^1$.)
 On $U_\pm$ we define the section $\xi_\pm$ as follows:
 \begin{equation}\label{sections}
 \xi_-([z_1:z_2]):= \frac1{\sqrt{1+|z_-|^2}} \left(\begin{array}{c}z_- \\1\end{array}\right), 
 \ \ \xi_+([z_1:z_2]):= \frac1{\sqrt{1+|z_+|^2}} \left(\begin{array}{c} 1 \\ z_+\end{array}\right)
 \end{equation}
 
Associated to these sections we have connection forms $\omega_-, \omega_+$.  You will show in one of the exercises below that on $U_-\cap U_+$
\begin{equation}\label{f}
\omega_+ = \omega_ + + \overline{f} df
\end{equation}
where $f([z_1:z_2]) = z_+/|z_+|$.  
\begin{center}
$\diamond $
\end{center}
 Before moving forward, let us try to get a better sense of what parts of $S^2$ are covered by $U_\pm$. 
 Let us define the map $F:\mathbb{C}P^1\rightarrow S^2$ such that $F([z_1:z_2])=(x_1, x_2,x_3)$ where
  $$x_1 = 2 \text{Re}\left(z_1\overline{z}_2\right), \ \ x_2 = 2\text{Im}\left(z_1\overline{z}_2\right), \ \ x_3= |z_1|^2-|z_2|^2. $$
 We are assuming here that $|z_1|^2+|z_2|^2=1$. It is not difficult to see from this expression that $F$ is injective. To see that it is surjective, observe that 
 $$ F\left(
 \left[
 \cos\left(\frac{\varphi}{2}\right)e^{\frac{\psi+\theta}{2}i}:
 \sin\left(\frac{\varphi}{2}\right)e^{\frac{\psi-\theta}{2}i}
 \right]
 \right)=\left(\sin\varphi\cos\theta, \sin\varphi \sin\theta, \cos\varphi\right).$$
 On the right-hand side we have an arbitrary point on   $S^2$  expressed in spherical coordinates by allowing $\theta \in [0,2\pi)$ and $\varphi\in [0,\pi]$.
 Changing $\psi$ does not change the point $[z_1:z_2]$ ($\psi$ parametrizes the $U(1)$ fiber of the Hopf bundle). From this description 
 we see that $[1:0]$ is mapped to the North Pole  $N=(0,0,1)^\intercal$ and  $[0:1]$ to the South Pole $S=(0,0,-1)^\intercal$, while
 $\left[\frac1{\sqrt{2}}: \frac{e^{i\theta }}{\sqrt{2}}\right]$ is mapped to the equator ($x_3=0$). Also note that the function $f$ in (\ref{f})
 maps  $\left[\frac1{\sqrt{2}}: \frac{e^{i\theta }}{\sqrt{2}}\right]$ to $e^{i\theta}$. We have that $(U_+, \varphi_+)$ parametrizes a region of $S^2$ that contains the northern hemisphere $S^2_+$ while $(U_-,\varphi_-)$ parametrizes a region containing the southern hemisphere $S^2_-$. As $\theta$ increases in $\left[\frac1{\sqrt{2}}: \frac{e^{i\theta }}{\sqrt{2}}\right]$ we traverse the equation in the positive direction if we regard the equation as the boundary of the northern hemisphere, and in the negative direction relative to the boundary orientation of the southern hemisphere.   
 
 
 Also observe that $f\left(\left[\frac1{\sqrt{2}}: \frac{e^{i\theta }}{\sqrt{2}}\right]\right)= e^{i\theta}$ and that 
 the pull-back (or restriction) of $\overline{f} df$ to the equator  is 
 $$e^{-i\theta} d\left(e^{i\theta}\right) = e^{-i\theta}e^{i\theta} i d\theta = id\theta. $$ 
 Finally observe that 
 $$\int_{\mathbb{C}P^1}\Omega = \int_{S^2_+}d\omega_+ +\int_{S^2_-} d\omega_- = \int_{\partial S^2_+}\omega_+ + \int_{\partial S^2_-}\omega_-= \int_{\partial S^+}(\omega_+-\omega_-)=\int_0^{2\pi} id\theta = i2\pi.  $$
 We conclude that
 $$\int_{\mathbb{C}P^1}c_1(L) =  \int_{\mathbb{C}P^1}\frac{i\Omega}{2\pi} =  \frac{i}{2\pi} 2\pi i = -1.$$
 Therefore the Chern number of the tautological line bundle over $\mathbb{C}P^1$ is $-1$. 
 \begin{center}
$\diamond $
\end{center} 
 
 
 
 
  
 
 \begin{enumerate}
 %\item Show that $\nabla$ (defined above by (\ref{connection})) is indeed a connection on the line bundle $L$ over $\mathbb{C}P^1$, and that it is a metric connection for the given Hermitian metric. 
 \item Let $L^{\otimes n}$ be the line bundle over $\mathbb{C}P^1$ given by the tensor product $L\otimes \dots \otimes L$ ($n$ times). Show that the
 Chern number of $L^{\otimes n}$  is $-n$. (As a lemma, show that if $L_1, L_2$ are two line bundles then the first Chern class of  their tensor product  is $c_1(L_1\otimes L_2)=c_1(L_1)+c_1(L_2)$.)
 \item Let $L^*$ denote the dual bundle to the tautological line bundle $L$. Show that $c_1(L^*)=-c_1(L)$. In particular, $L^*$ has   Chern number $1$. (Note: $L\otimes L^*$ is the bundle of endomorphisms of $L$. This line bundle has a nowhere vanishing  section given by the identity map on each fiber of $L$. )
 \end{enumerate}
 Knowing that the trivial line bundle has Chern number $0$, we conclude from the above that every integer $n\in \mathbb{Z}$ is the Chern number of some complex line bundle over $\mathbb{C}P^1$ (and line bundles with different Chern numbers are not isomorphic).
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
 \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} 

