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\begin{document}
\title{{\bf  Homework set 8 - due 11/01/20}\\ {\large Math 5047 \---  Renato Feres} 
}
\author{}
\date{}  
\maketitle
\thispagestyle{empty}
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Turn in problems   1, 2, 8, 11(b). In problems 8 and 11(b) (and all others), you may freely quote from statements made in  other exercises from this assignment.
\begin{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf The gradient, divergence and Laplacian.} Let $(M, \langle\cdot, \cdot\rangle)$ be an oriented Riemannian manifold with volume form $\omega$ and $f:M\rightarrow \mathbb{R}$ a smooth
function. The {\em gradient} of $f$ is the vector field denoted $q\mapsto \text{grad}_qf$ such that,
for all $v\in T_qM$,
$$\langle \text{grad}_qf, v\rangle = df_p(v). $$
Here $df_pv:= vf$ is the directional derivative of $f$ along $v$. The {\em divergence} of a vector field $X$ on $M$ is the function $q\mapsto \text{div}_qX$ such
that $$ \mathcal{L}_X\omega = (\text{div}X) \omega$$
where $\mathcal{L}_X$ is the Lie derivative along $X$. The {\em Laplacian} of a smooth function $f$ is 
$$\Delta f = \text{div}(\text{grad}f).$$
 Let $E_1, \dots, E_n$ be an orthonormal frame on the open set $\mathcal{U}\subseteq M$ and $\nabla$ the Levi-Civita connection on $M$. Recall 
 that the connection $1$-forms relative to the frame $\{E_i\}$ are defined as $\omega_{ij}(v)=\langle E_i, \nabla_v E_j\rangle$. 
\begin{enumerate}
\item Show that $\text{grad}f=\sum_{i=1}^n (E_if) E_i$ for any smooth function $f$ on $\mathcal{U}$.
\item Show that $\text{div} X= \sum_{i=1}^n E_i\langle E_i, X\rangle +\sum_{i,j=1}^n \langle E_j,X\rangle \omega_{ij}(E_i)$ for any smooth vector field $X$ on $\mathcal{U}$.
\item Show that $\Delta f = \sum_{i=1}^n E_iE_if + \sum_{i,j=1}^n \omega_{ij}(E_i)E_jf$ for any smooth function $f$ on $\mathcal{U}$. 
\item Show that if $f, g$ are smooth functions on $M$,
$$\Delta(fg) = f\Delta g + g\Delta f + 2\langle \text{grad} f, \text{grad} g\rangle.$$
\item We say that the orthonormal frame $\{E_i\}$ is {\em geodesic} at $q\in \mathcal{U}$ if $(\nabla_{E_i}E_j)_q=0$ for all $i, j$. Convince yourself that
a geodesic frame at $q$ exists on some neighborhood of $q$, but you don't have to write it down. (The key idea is this: Let $\mathcal{U}$ be  the image under $\text{Exp}_q$ of a small ball in $T_qM$ centered at the origin. Let $e_1, \dots, e_n$ be an orthonormal basis of $T_qM$. Define $E_i$ at $p\in \mathcal{U}$ as the parallel transport of $e_i$
along the unique geodesic from $q$ to $p$ contained in $\mathcal{U}$. 
)
 Conclude that, relative to a geodesic orthonormal frame $\{E_i\}$ at $q$ on a sufficiently small neighborhood  $\mathcal{U}$ of $q$, the divergence of $X$ and the Laplacian of $f$ at $q$ (but not necessarily at other points of $\mathcal{U}$) are
$$\text{div}_qX= \sum_{i=1}^n E_i(\langle E_i,X\rangle)(q), \ \ \ (\Delta f)(q) = \sum_{i=1}^n E_i(E_if)(q). $$
\end{enumerate}
 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf Harmonic functions are constant on a compact manifold.}  For this problem, you may need to review Stokes' theorem on a manifold with boundary:
$$\int_{\partial M} \nu = \int_M d\nu$$
where $\nu$ is a compactly supported $(n-1)$-form on the $n$-dimensional manifold $M$. 
\begin{enumerate}
\item Show that if $X$ is a (compactly supported) smooth vector field on  the oriented Riemannian manifold $M$, then 
$$ \int_M (\text{div} X) \text{vol}_M = \int_{\partial M}  \langle N,X\rangle \, \text{vol}_{\partial M} $$
where
$\text{vol}_M$ is the volume form of $M$ and 
 $\text{vol}_{\partial M}$ is the induced volume form on $\partial M$. 
This is the {\em divergence theorem}. In words: the integral of the divergence of the vector field $X$ is the flux of that vector field across the boundary.
\item Using the divergence theorem, show that if $f$ is a smooth function on a compact Riemannian manifold $M$ without boundary, then
$$ \int_M (\Delta f) \text{vol}_M = 0.$$
(This is the statement that the volume form is harmonic in the weak sense.)
\item Suppose $f$ is a harmonic function on the compact, connected Riemannian manifold $M$ without boundary; that is, $\Delta f=0$. Show that $f$ must be constant. 
(Suggestion: Start with $0=\int_M \Delta \left(f^2\right)\text{vol}_M$ and express the integral in terms of the gradient of $f$ using an identity from 
the previous problem.)
\end{enumerate}



%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf Geodesics in lens spaces.} Identify $\mathbb{R}^4$ with $\mathbb{C}^2$ by letting  $(x_1, x_2, x_3, x_4)$ correspond to $(x_1+ix_2, x_3+ix_4)$. Let 
$$ S^3=\left\{(z_1, z_2)\in \mathbb{C}^2:|z_1|^2+|z_2|^2=1\right\},$$
and let $h:S^3\rightarrow  S^3$ be given by
$$h(z_1,z_2)=\left(e^{\frac{2\pi i}q} z_1, e^{\frac{2\pi i r}{q}}z_2\right), \ \ (z_1, z_2)\in S^3,$$
where $q$ and $r$ are relatively prime integers, $q>2$.
\begin{enumerate}
\item Show that $G=\left\{\text{id}, h, \cdots, h^{q-1}\right\}$ is a group of isometries of the sphere $S^3$, with the usual metric, which operates  in  a totally discontinuous manner. The manifold
$S^3/G$ is called a {\em lens space}.
\item Consider $S^3/G$ with the metric induced by the projection $p:S^3\rightarrow S^3/G$. Show that all the geodesics of $S^3/G$ are closed but can have different lengths.
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%



 %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf Exterior derivative expressed in terms of a connection.} If  $\omega$ is a differential $k$-form on a smooth manifold $M$ equipped with
a torsion-free (i.e., symmetric)  connection $\nabla$, show that 
$$ d\omega(X_0, \dots, X_k) =\sum_{i=0}^k (-1)^i \left(\nabla_{X_i}\omega\right)\left(X_0, \dots, \widehat{X}_i, \dots, X_k\right),$$
where $X_0, X_1, \dots, X_k\in \mathfrak{X}(M)$ and the hat symbol $\widehat{X}$ indicates that the corresponding vector field is dropped. 
 
 Note: A useful identity involving the exterior product which, unfortunately, is not in the appendix of Lee's text, is the following:
\begin{align*}(d\omega)\left(X_0, \dots, X_k\right)&= \sum_{i=0}^k (-1)^i X_i\left(\omega\left(X_0, \dots, \widehat{X}_i, \dots, X_k\right)\right)\\
 & \ \ \ \ \ \ \ \ \ \ +\sum_{i<j} (-1)^{i+j} \omega\left([X_i,X_j], X_0, \dots, \widehat{X}_j, \dots, \widehat{X}_i,\dots, X_k\right).
 \end{align*}
 You may take it for granted.  
 
 


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item {\bf Riemannian metric on the tangent bundle of a Riemannian manifold.} It is possible to introduce a Riemannian metric on the tangent bundle
 $N=TM$ of a Riemannian manifold  $M$ in the following manner.  
 Let $(p,v)\in  N$ and $V, W$ be tangent vectors in $N$ at $(p,v)$. Choose
 curves 
 $\alpha:t\mapsto (p(t),v(t))$ and $\beta:s\mapsto (q(s),w(s))$ in $N$ 
 with $p(0)=q(0)=p$, $v(0)=w(0)=v$, and $V=\alpha'(0)$, $W=\beta'(0)$. Define an inner product on $N$ by 
\begin{equation}\label{inner product} \left\langle V,W\right\rangle_{(p,v)}:=\left\langle d\pi_{(p,v)} V,d\pi_{(p,v)} W\right\rangle_p+\left\langle \frac{Dv}{dt}(0),\frac{Dw}{ds}(0)\right\rangle_p,\end{equation}
 where $d\pi_{(p,v)}$  is the differential of the base point projection $\pi:N\rightarrow M$.  It is not difficult to show that this inner product is well-defined (i.e., it is independent of the choices of $\alpha$ and $\beta$) and that it  indeed defines a Riemannian metric on $N$. This is often called the {\em Sasaki} metric on $TM$.
 
 A vector $V$ at $(p,v)\in N$ that is orthogonal (in the metric (\ref{inner product}))  to the fiber $\pi^{-1}(p)\cong T_pM$ is called a {\em horizontal vector}. A curve
$t\mapsto (p(t),v(t))$ in $N$ is {\em horizontal} if its tangent  vector  is horizontal  for all $t$.
Here are some facts we know (from other homework assignments or from class):
\begin{itemize}
\item   The  curve $t\mapsto (p(t),v(t))$ is horizontal  if and only if the vector field $v(t)$ is  parallel along $p(t)\in M$. 
\item We define the {\em geodesic vector field}  $Z$ as  the horizontal vector field on $N$ such that 
$d\pi_{(p,v)}Z(p,v)=v$ for all $(p,v)\in N$. The flow line $t\mapsto \Phi_t(p,v)$ of the flow $\Phi_t$ of  $Z$ projects under $\pi$ to the geodesic on $M$
with initial conditions $(p,v)$.
\end{itemize}

Problems:
   \begin{enumerate}
   \item Let $V$ be a vertical vector in $T_{(p,v)}N$. Recall that  we may regard $V$ as a vector in $T_pM$ due to the canonical isomorphism between $T_pM$ and the vertical subspace of $T_{(p,v)}N$:
   $$V\in T_pM\mapsto \left.\frac{d}{dt}(v+tV)\right|_{t=0}\in T_{(p,v)}N. $$
     Show that $J(t):=d(\pi\circ\Phi_t)_{(p,v)}V$ is a Jacobi field  with initial conditions $J(0)=0$
   and $\frac{DJ}{dt}(0)=V$.
     \item Let $V$ be a horizontal vector in $T_{(p,v)}N$ and $w:=d\pi_{(p,v)}V$.   Show that $J(t):=d(\pi\circ\Phi_t)_{(p,v)}V$ is a Jacobi field  with initial conditions $J(0)=w$
   and $\frac{DJ}{dt}(0)=0$.
   \end{enumerate}



\item     {\bf Riemannian submersions}.
Read about Riemannian products (Chapter 2, page 20) and Riemannian submersions (Chapter 2, page 21) in Lee's text.
 Let $\overline{M}$ and $M$ be manifolds of dimension $n+k$ and $n$, respectively.  A differentiable mapping $f:\overline{M}\rightarrow M$ is called {\em submersion} if $f$ is surjective, and
for all $\overline{p}\in\overline{M}$, the differential $df_{\overline{p}}:T_{\overline{p}}\overline{M}\rightarrow T_{f(\overline{p})}M$ has rank $n$.  In this case, for all $p\in M$, the {\em fiber} $f^{-1}(p)=F_p$
is a submanifold of $\overline{M}$ (by the implicit function theorem) and a tangent vector of $\overline{M}$, tangent to some $F_p$, $p\in M$, is called a {\em vertical vector} of the submersion. If, in addition, $\overline{M}$ and
$M$ have Riemannian metrics, the submersion $f$ is said to be {\em Riemannian} if  $df_p:T_p\overline{M}\rightarrow T_{f(p)}M$ preserves lengths of vectors orthogonal to $F_p$, for all $p\in \overline{M}$. 
\begin{enumerate}
\item If $M_1\times M_2$ is the Riemannian product,  then the natural projections $\pi_i:M_1\times M_2\rightarrow M_i$, $i=1,2$, are Riemannian submersions.
\item If the tangent bundle $N=TM$ is given the Riemannian  metric as in the previous exercise, then the projection $\pi:TM\rightarrow M$ is a Riemannian submersion. 
\end{enumerate}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%


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


\item   {\bf Connection of a Riemannian submersion.} Let $f:\overline{M}\rightarrow M$ be a Riemannian submersion. A vector $\overline{x}\in T_{\overline{p}}\overline{M}$ is {\em horizontal} if it is orthogonal  to the fiber. The tangent space $T_{\overline{p}}\overline{M}$ then admits a decomposition 
$$T_{\overline{p}}\overline{M} = \left(T_{\overline{p}}\overline{M} \right)^h\oplus \left(T_{\overline{p}}\overline{M} \right)^v,$$ where $\left(T_{\overline{p}}\overline{M} \right)^h$ and $\left(T_{\overline{p}}\overline{M} \right)^v$ denote the subspaces of horizontal and vertical vectors, respectively.  If $X\in \mathfrak{X}(M)$, the {\em horizontal lift} $\overline{X}$ of $X$ is the horizontal field defined by $df_{\overline{p}}\left(\overline{X}(\overline{p})\right) = X(f(p))$. You may take for granted (but think about it!) that $\overline{X}$ is a smooth vector field.
\begin{enumerate}
\item Show that $\overline{X}$ is a smooth vector field.
\item Let $\nabla$ and $\overline{\nabla}$ be the Riemannian connections of $M$ and $\overline{M}$, respectively. Show that
$$ \overline{\nabla}_{\overline{X}}\overline{Y}=\overline{\nabla_XY} +\frac12 \left[\overline{X},\overline{Y}\right]^v,\ \ \ X, Y\in \mathfrak{X}(M),$$
where $Z^v$ is the vertical component of $Z$. 
\item $\left[\overline{X},\overline{Y}\right]^v\left(\overline{p}\right)$ depends  only on $\overline{X}\left(\overline{p}\right)$ and $\overline{Y}\left(\overline{p}\right)$.
\end{enumerate} 


Hint for (b):  Let $X,Y,Z\in \mathfrak{X}(M)$. Let  $T\in \mathfrak{X}\left(\overline{M}\right)$ be a vertical field. Observe that
$$ \left\langle\overline{X},T\right\rangle= \left\langle\overline{Y},T\right\rangle= \left\langle\overline{Z},T\right\rangle=0, \ \ \overline{X}\left\langle \overline{Y},\overline{Z}\right\rangle=X\langle Y,Z\rangle.$$
Also $$df\left[\overline{X},T\right]=0, \ \ \ [X,Y]=\left[df \overline{X},df \overline{Y}\right]=df\left[ \overline{X},\overline{Y}\right]$$
and 
$$T\left\langle  \overline{X}, \overline{Y}\right\rangle =0.$$
From this we can conclude that 
$$\left\langle \left[\overline{X},\overline{Y}\right],\overline{Z}\right\rangle =\langle [X,Y],Z\rangle, \ \  \ \  \left\langle \left[\overline{X},T\right],\overline{Y}\right\rangle =0.$$
We can now use the expression for the Levi-Civita connection in Theorem 5.10 to obtain
$$\left\langle\overline{\nabla}_{\overline{X}}\overline{Y},\overline{Z} \right\rangle= \langle \nabla_XY,Z\rangle,  \ \ \ 
2\left\langle\overline{\nabla}_{\overline{X}}\overline{Y},T \right\rangle= \left\langle T, \left[\overline{X},\overline{Y}\right]\right\rangle.$$
This implies the desired identities.
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

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

\item  {\bf Curvature of a Riemannian submersion.}  Let $f:\overline{M}\rightarrow M$ be a Riemannian submersion. Let $X, Y, X, W\in \mathfrak{X}(M)$
and $\overline{X}, \overline{Y}, \overline{Z}, \overline{W}$ be their horizontal lifts. Let $R$ and $\overline{R}$ be the curvature tensors of $M$ and $\overline{M}$, respectively. Prove that 
\begin{enumerate}
\item $$\left\langle R(X,Y)Z,W\right\rangle\circ f=
\left\langle\overline{R}(\overline{X},\overline{Y})\overline{Z},\overline{W}\right\rangle  -\frac14\left\langle \left[\overline{X},\overline{Z}\right]^v, \left[\overline{Y},\overline{W}\right]^v \right\rangle
+\frac14\left\langle \left[\overline{Y},\overline{Z}\right]^v, \left[\overline{X},\overline{W}\right]^v \right\rangle -\frac12\left\langle \left[\overline{Z},\overline{W}\right]^v, \left[\overline{X},\overline{Y}\right]^v \right\rangle.$$
\item If $\sigma$ is the plane generated by the orthonormal vectors $X,Y\in \mathfrak{X}(M)$ and
$\overline{\sigma}$ is the plane generated by $\overline{X}$, $\overline{Y}$, then 
$$K(\sigma)=\overline{K}(\sigma) +\frac34 \left\| \left[\overline{X}, \overline{Y}\right]^v\right\|^2 \geq \overline{K}(\overline{\sigma}). $$
\end{enumerate}

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



\item  {\bf The Hopf bundle and $\mathbb{C}P^1$.}  (This is only reading.)
The complex projective space of complex dimension $1$
 is defined as the set of complex lines (one-dimensional complex vector subspaces) in $\mathbb{C}^2$. Formally, it is defined as the 
 quotient of $\mathbb{C}^2\setminus \{0\}$ (the $2$-dimensional complex vector space minus the origin) under the equivalence relation that
identifies any  two non-zero vectors that are collinear.   Equivalently, let $S^3$ denote the $3$-dimensional unit sphere, regarded as the submanifold of $\mathbb{C}^2$
consisting of pairs $(z_1, z_2)$ such that $|z_1|^2+|z_2|^2=1$, and define on it the action by the group $U(1)$ of unit modulus complex numbers:
$$\left(e^{i\theta}, (z_1,z_2)\right)\mapsto \left(e^{i\theta}z_1, e^{i\theta}z_2\right).$$
Then $\mathbb{C}P^1$ is the quotient of $S^3$ under this action. Notice that the orbits are circles, so we should expect the quotient to be a $2$-dimensional (real) manifold.  In fact, $\mathbb{C}P^1$ has the structure of a smooth manifold diffeomorphic to $S^2$. (It is also a $1$-dimensional complex manifold.) 
The quotient map defines a  {\em circle bundle} $\pi:S^3\rightarrow S^2$ called the {\em Hopf bundle}. 


There are several ways to show that $\mathbb{C}P^1$ is diffeomorphic to $S^2$. I find the following, inspired by a view from quantum theory,
especially interesting. Let $M$ denote the space of self-adjoint $2\times 2$ complex matrices of trace $0$. Such matrices can
be written as
$${\sigma}\cdot x := x_1\sigma_1 +x_2\sigma_2 + x_3\sigma_3 = \left(\begin{array}{cc}x_3 & x_1+ix_2 \\x_1-ix_2 & -x_3\end{array}\right)$$
where 
$$ \sigma_1 :=\left(\begin{array}{cc}0 & 1 \\1 & 0\end{array}\right) , \ \ \sigma_2:=\left(\begin{array}{cc}0 &\!\! -i \\i & 0\end{array}\right),\ \ \sigma_3:= \left(\begin{array}{cc}1 & 0 \\0 & \!\!-1\end{array}\right)$$
are the so called {\em Pauli matrices}. The $x_j$ are real.
We are interested in $x\in S^2$.
Note that $$(\sigma\cdot x)^2=|x|^2 \left(\begin{array}{cc}1 & 0 \\0 & 1\end{array}\right)=I$$ where $|x|^2=x_1^2+x_2^2+x_3^2=1$ and $I$ is the identity matrix.
The matrices $\sigma\cdot x$ for $x\in S^2$ have eigenvalues $\pm 1$. In fact, consider the matrices $P_{\pm}(x) =\frac12(I\pm\sigma\cdot x)$.
They are the orthogonal projections to the $1$-dimensional subspaces of $\mathbb{C}^2$ spanned by the eigenvalues $\pm 1$.  This is a consequence of the
easy to verify matrix identities (where $P^*$ indicates the transpose-conjugate, or matrix adjoint):
$$P_{\pm}(x)^*= P_{\pm}(x), \ \ P_{\pm}(x)^2=P_{\pm}(x), \ \ P_-(x)P_+(x)=P_+(x)P_-(x)=0, \ \ P_-(x)+P_+(x)=I, \ \  \sigma\cdot x P_{\pm}(x)=\pm P_{\pm}(x).$$
We have in this way  used $S^2$ to parametrize all the (both) self-adjoint and unitary $2\times 2$ complex matrices having distinct eigenvalues.
Let $u(x)$ be a  unit length eigenvector in $\mathbb{C}^2$ associated to the eigenvalue $1$ of $P_+(x)$. Thus $u(x)\in S^3$ and  $P_+(x)u(x)=u(x)$.  Observe that
$P_+(x)$ is the rank-$1$ orthogonal projection matrix  given by   
$$v\mapsto P_+(x)v = \langle u(x),v\rangle u(x)$$
for $v\in \mathbb{C}^2$.
Here
$\langle \cdot, \cdot\rangle$ is the Hermitian inner product in $\mathbb{C}^2$ given by:
$ \langle (z_1,z_2),(w_1,w_2)\rangle= \overline{z}_1w_1+\overline{z}_2w_2.$


We may write $P_+(x)=u(x)\otimes u(x)^*$, where $u(x)^*$ is the dual vector to $u(x)$. (In quantum theory notation, using Dirac's bra-kets, $u(x)\otimes u(x)^*=|x\rangle \langle x|$.)



The outcome of the above discussion is this: if we identify each element of projective space $\mathbb{C}P^1$ with   the orthogonal projection operator $P_+(x)$ in the Hilbert space $\mathbb{C}^2$ onto the eigenspace of $\sigma\cdot x$ for eigenvalue $1$, we obtain a map $S^2\rightarrow  \mathbb{C}P^1$
which it is not difficult to show is a diffeomorphism. (In this way, the complex projective space can be regarded as a submanifold of a  three-dimensional (over $\mathbb{R}$) space of matrices.)  Note that this map is well-defined since 
a different choice $e^{i\theta}u(x)$ of  unit length eigenvector will give the same projection map $u(x)\otimes u(x)^*$. 
On the other hand, each projection can be written as the image of a point in $S^3$ under the map
$$\pi:S^3=\{z=(z_1,z_2)\in \mathbb{C}^2:|z_1|^2+|z_2|^2=1\}\rightarrow \mathbb{C}P^1\cong S^2$$ such that $\pi(z)=z\otimes z^*$. Observe that $\pi\left(e^{\lambda i}z\right)=\pi(z)$ since
$$\pi\left(e^{\lambda i} z\right)= \left(e^{\lambda i}z\right)\otimes  \left(e^{\lambda i}z\right)^* =e^{\lambda i}e^{-\lambda i} z\otimes z^* = z\otimes z^*=\pi(z).  $$
 So we have, as expected, that $\mathbb{C}P^1$ is a quotient of $S^3$ under the diagonal action of the circle group $U(1)$ on $S^3$: $$e^{\lambda i}(z_1,z_2)=\left(e^{\lambda i}z_1,e^{\lambda i}z_2\right).$$
 The identification between $S^2$ and $\mathbb{C}P^1$ is accomplished by the bijection $x\mapsto \sigma\cdot x$. 
The resulting map $\pi:S^3\rightarrow S^2$ is, again, the Hopf bundle.
\begin{center}
$\diamond $
\end{center}


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

\item  {\bf The complex projective space $\mathbb{C}P^n$.}   On $\mathbb{C}^{n+1}\setminus \{0\}= \left\{Z=(z_0, \dots, z_n)\neq 0: z_j = x_j + i y_j\right\}$
define the equivalence relation on $\mathbb{C}^{n+1}\setminus \{0\}$:
$ Z=(z_0, \dots, z_n)\sim W =(w_0, \dots, w_n)$ if $z_j=\lambda w_j$ for all $j$ where $\lambda\in \mathbb{C}\setminus \{0\}$.
The equivalence class represented by $Z$\----the complex line through the origin containing $Z$\----will be denoted $[Z]$. The
quotient $$\mathbb{C}P^n:=\mathbb{C}^{n+1}\setminus \{0\}/\sim$$
is called the {\em complex projective space} of complex dimension $n$.
\begin{enumerate}
\item Show that $\mathbb{C}P^n$ has a differentiable structure of a manifold of real dimension $2n$. 
Note: The proof  is in Example C.19, page 413, of Lee's text.
\item Let $\langle Z,W\rangle =\overline{z}_0w_0+\cdots+\overline{z}_nw_n$ be the Hermitian product on $\mathbb{C}^{n+1}$,
where the bar denotes complex conjugation. Identify $\mathbb{C}^{n+1}\approx \mathbb{R}^{2n+2}$ by writing $z_j=x_j+iy_j=(x_j,y_j)$. Show that
$$S^{2n+1}=\left\{N\in \mathbb{C}^{n+1}\approx \mathbb{R}^{2n+2}: \langle N,N\rangle=1\right\} $$
is the unit sphere in $\mathbb{R}^{2n+2}$.
\item Show that   $\sim$ induces on $S^{2n+1}$ the following equivalence relation:
$Z\sim W$ if $W=e^{i\theta}Z$. Establish that there exists a differentiable map (the Hopf bundle) $f:S^{2n+1}\rightarrow \mathbb{C}P^n$,
such that 
$$f^{-1}([Z])=\left\{e^{i\theta}N\in S^{2n+1}: N\in [Z]\cap S^{2n+1}, 0\leq \theta\leq 2\pi\right\}=[Z]\cap S^{2n+1}. $$
\item Show that $f$ is a submersion. (This is already in Example C.19, so nothing more  to do here.)
\end{enumerate}


 
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\item   {\bf Curvature of the complex projective space.} (This is similar to Problem 8-13, page 258 of Lee's text.) Define a Riemannian metric on $\mathbb{C}^{n+1}\setminus \{0\}$ in the following way: If $Z\in \mathbb{C}^{n+1}\setminus \{0\}$ and $V, W\in T_Z(\mathbb{C}^{n+1}\setminus \{0\})$,
$$ \langle V,W\rangle_Z =\frac{\text{Real}(\langle V,W\rangle_0)}{\langle Z,Z\rangle_0}$$
where $\langle V,W\rangle_0 :=\overline{v}_0w_0+\cdots+\overline{v}_nw_n$.
Observe that the metric $\langle\cdot, \cdot\rangle$ restricted to $S^{2n+1}\subseteq \mathbb{C}^{n+1}\setminus \{0\}$ coincides with the metric induced from $\mathbb{R}^{2n+2}$.
\begin{enumerate}
\item  Show that, for  all $0\leq \theta\leq 2\pi$, $e^{i\theta}:S^{2n+1}\rightarrow S^{2n+1}$ is an isometry, and that, therefore, it is possible to define a  Riemannian metric on $P^n(\mathbb{C})$ in such a way that the submersion
$f$ defined in the previous problem  is Riemannian.
\item Show that, in this metric,  the sectional curvature of $P^n(\mathbb{C})$  is given by 
$$K(\sigma)=1+3\cos^2\varphi, $$
where $\sigma$ is generated by the orthonormal  pair $X, Y$, $\cos\varphi= \left\langle  \overline{X},i\overline{Y}\right\rangle$, and $\overline{X}, \overline{Y}$ are the horizontal  lifts
of $X$ and $Y$, respectively. In particular, $1\leq K(\sigma)\leq 4$. 
\end{enumerate}


Hint for part (b): Let $Z$ be the position vector describing  $S^{2n+1}$. Since
$$\left(\frac{d}{d\theta} e^{i\theta} Z\right)_{\theta=0} =iZ$$ 
$ iZ\in T_Z S^{2n+1}$  and it is vertical.  Let $\overline{\nabla}$ be the Levi-Civita connection of  $\mathbb{R}^{2n+2}\approx \mathbb{C}^{n+1}$ and
$X,Y\in \mathfrak{X}(\mathbb{C}P^n)$. Take $\alpha:(-\epsilon, \epsilon)\rightarrow S^{2n+1}$ with $\alpha(0)=Z$, $\alpha'(0)=\overline{X}$. Then
$$
\left(\overline{\nabla}_{\overline{X}} i Z\right)_Z=\left.\frac{d}{dt} i Z\circ \alpha(t)\right|_{t=0}
=\left. i\alpha(t) \right|_{t=0}=i\alpha'(0)=i\overline{X}.
$$
Therefore
$$\left\langle \left[\overline{X},\overline{Y}\right], iZ\right\rangle=\left\langle \overline{\nabla}_{\overline{X}}\overline{Y}-\overline{\nabla}_{\overline{Y}}\overline{X}, iZ\right\rangle = -\left\langle i \overline{X},\overline{Y}\right\rangle +\left\langle i\overline{Y},\overline{X}\right\rangle=2\cos\varphi.$$
Now use the above facts about the sectional curvature of a Riemannian submersion.
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 \end{enumerate}




 \end{document}
 
 