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\begin{document}
\title{{\bf  Homework set 5 - due 10/04/20}\\ {\large Math 5047 \---  Renato Feres} 
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\date{}  
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 This assignment collects a wide range of facts related to formal properties of the Riemannian curvature tensor, spread over several chapters of Lee's text.
Make sure to read and think about all the problems, not only the ones I ask you to turn in.
 
Turn in problems 1(a,b), 2(b) (first Bianchi identity), 4(g),  5(c).
\begin{enumerate}
\item {\bf Ricci identities, Lee's text, Proposition 7.14, page 205.} Let $\nabla$ be a torsion-free connection on $TM$, where the manifold $M$ is not necessarily Riemannian.
We know that $\nabla$ extends to a connection on the general tensor bundle by requiring that the product rule holds for the pairing of vectors and
covectors and for the tensor product. It thus make sense to define, for a general tensor field $\tau$ and $X, Y\in \mathfrak{X}(M)$,
$$R(X,Y)\tau=\nabla_X\nabla_Y\tau -\nabla_Y\nabla_Y\tau -\nabla_{[X,Y]}\tau. $$
Show the following properties of $R(X,Y)$:
\begin{enumerate}
\item For all $f\in C^\infty(M)$, $$R(X,Y)f=0.$$
\item For tensor fields $\tau$ and $\eta$,
$$ R(X,Y)(\tau\otimes \eta)=\left(R(X,Y)\tau \right)\otimes \eta+\tau\otimes R(X,Y)\eta.$$
\item For $\theta\in \Omega^1(M)$ and $Z\in \mathfrak{X}(M)$,
$$ (R(X,Y)\theta)(Z)=-\theta(R(X,Y)Z).$$
\item For a $(k,\ell)$-tensor field $\tau$, covector fields $\theta_1, \dots, \theta_k\in \Omega^1(M)$ and vector fields $X_1, \dots, X_\ell\in \mathfrak{X}(M)$,
\begin{align*}0&=(R(X,Y)\tau)\left(\theta_1, \dots, \theta_k, X_1, \dots, X_\ell\right)\\
&\ \ \ \ \ \ \ \ \ \ \ +\sum_{i=1}^k\tau\left(\theta_1, \dots, R(X,Y)\theta_i, \dots, \theta_\ell\right)+ \sum_{i=1}^\ell\tau\left(\theta_1, \dots, \theta_k, X_1, \dots, R(X,Y)X_i, \dots, X_\ell\right).
\end{align*}
\item If $M$ is Riemannian and $\nabla$ is the Levi-Civita connection we   have
$$ \langle R(X,Y)W,Z\rangle =-\langle   R(X,Y)Z,W\rangle.$$
\end{enumerate}
 Note that (e) is the second property of the next problem (2(b)).




\item {\bf Symmetries of the curvature tensor.}\label{problem 2} (Lee's text, Proposition 7.12, page 202.) Read Proposition 7.12 and the proof of the symmetries of the curvature tensor of a Riemannian manifold $(M, \langle \cdot,\cdot\rangle)$:
\begin{enumerate}
\item $\langle R(W,X)Y,Z\rangle = -\langle R(X,W)Y,Z\rangle$;
\item $\langle R(W,X)Y,Z\rangle = -\langle R(W,X)Z,Y\rangle$;
\item $\langle R(W,X)Y,Z\rangle = \langle R(Y,Z)W,X\rangle$;
\item $\langle R(W,X)Y,Z\rangle +\langle R(X,Y)W,Z\rangle + \langle R(Y,W)X,Z\rangle=0$.
\end{enumerate}
The last item is known as the  {\em first Bianchi identity}.
\begin{enumerate}
\item Use these properties to show:
$$\sum_{\sigma\in S_3} \text{sgn}(\sigma)  R\left(X_{\sigma(1)}, X_{\sigma(2)}\right)X_{\sigma(3)}=0,$$
where $S_3$ is the symmetric group and $\text{sgn}(\sigma)$ is the sign of the permutation $\sigma\in S_3$. 
\item Prove the first Bianchi identity (d).
\item Show that property (c) is a consequence of the others.
\end{enumerate}

\item {\bf The second Bianchi identity.} (Lee's text, Proposition 7.13, page 204.) Read the proof of the {\em second Bianchi identity}, Proposition 7.13. The
identity is more clearly stated as follows: if $\nabla$ is a torsion-free connection on $TM$ (not necessarily metric), then
$$ (\nabla_XR)(Y,Z)+(\nabla_YR)(Z,X)+(\nabla_ZR)(X,Y)=0.$$
Try to give a direct proof of this identity. (This is a bit tedious but more illuminating, I think, than the proof given in Lee's text. That proof uses normal coordinates, which we haven't yet seen.) Recall that
$$(\nabla_XR)(Y,Z)=\nabla_X R(Y,Z)-R(\nabla_XY,Z)-R(Y,\nabla_XZ).$$
(Convince yourself of this last expression. Recall how  $\nabla$, introduced as a covariant derivative on vector fields, gives rise to a covariant derivative on general tensor fields  by imposing the requirement that it  satisfies the product rule for the tensor product and for the pairing of vector and covector.)






\item {\bf Shape operator, second fundamental form, and Gauss curvature.} Let $M$ be an embedded  hypersurface in $\mathbb{R}^{n+1}$. (So $\text{dim}(M)=n$.)
We assume $M$ is oriented. This can be shown to imply the existence of a global unit length vector field $N$ on $M$ such that
$N(p)$ is perpendicular to $T_pM$ at each $p\in M$. The {\em shape operator} of $M$ at $p\in M$ is the linear map $S_p:T_pM\rightarrow T_pM$
defined by $$S_pu=-D_u N$$ for $u\in T_pM$. Here $D$ is the Euclidean connection in $\mathbb{R}^{n+1}$. It should be intuitively clear that
$S_p$, by measuring  changes in the normal vector $N$ along $M$, should somehow encode the bending, or curving, of $M$ in Euclidean space.
This is made more precise here.
\begin{enumerate}
\item Show that $D_uN\in T_pM$ for all $u\in T_pM$. So $S_p$ is indeed a linear map from $T_pM$ to itself.
\item Show that $\langle S_p X, Y\rangle =\langle D_XY,N\rangle$ for any $X,Y\in \mathfrak{X}(M)$. (In particular, the expression on the right on depends on
the values of $X$ and $Y$ at $p$.)
\item Show that $S_p$ is symmetric: for all $u,v\in T_pM$,
$$\langle S_p u, v\rangle_p =\langle u, S_pv\rangle_p.$$
Here the inner product is the ordinary dot product in Euclidean space restricted to $T_pM$. Therefore, $S_p$ is self-adjoint. In particular, the spectral theorem for
self-adjoint (finite dimensional) operators applies, and implies that $T_pM$ has an orthonormal basis of eigenvectors of $S_p$.
\item Let us define the {\em principal directions} of the hypersurface $M$ at $p$ as the unit eigenvectors of $S_p$. The {\em principal curvatures} are
the eigenvalues of $S_p$. Describe the shape operator of the  sphere $$S^n(r)=\left\{x\in \mathbb{R}^{n+1}:|x|=r\right\}$$ of radius $r$
and determine its principal curvatures. 
\item Find the principal directions and principal curvatures at any point on the cylinder of radius $r$ in $\mathbb{R}^3$:
$$ \left\{x=(x_1, x_2, x_3): x_1^2+x_2^2=r^2\right\}.$$
\item Let $M$ be a surface in $\mathbb{R}^3$ with the Riemannian metric induced on $TM$ from the dot product on $\mathbb{R}^3$. Let
$R$ be the curvature tensor on $M$.  Let us suppose that $E_1, E_2$ constitute a local  orthonormal  frame on a neighborhood of $p\in M$
such that, at each $q$ in that neighborhood, $E_1(q), E_2(q)$ are principal directions with principal curvatures $\lambda_1(p), \lambda_2(q)$.
Show that 
$$ \langle R(E_1, E_2)E_2,E_1\rangle = \lambda_1 \lambda_2.$$
The product of the principal curvatures at a point $p$ of a surface in $\mathbb{R}^3$ is the {\em Gauss curvature} of the surface at that point.
Note the remarkable fact that, given the Riemannian metric on $M$ (induced from $\mathbb{R}^3$), the quantity
$\langle R(E_1, E_2)E_2,E_1\rangle$ is obtained intrinsically, without further reference to the ambient space, whereas $\lambda_i$
has to do with the bending of the surface in the Euclidean space. (Note that $\lambda_i$ changes sign if we choose the normal vector field to be $-N$,
but the product $\lambda_1\lambda_2$ is unaffected.) In particular, Gauss's curvature is invariant under local isometry. This 
fact is the statement of Gauss's Theorema Egregium. 
\item Let $e_1, e_2$ be an orthonormal basis for a plane $\sigma$ in $T_pM$ where $M$ is a Riemannian manifold. Show that
$K(\sigma):=\left\langle R_p(e_1, e_2)e_2,e_1 \right\rangle$ does not depend on the choice of orthonormal basis. It is called the {\em sectional curvature}
of $M$ at $p$ associated to the plane $\sigma$. Find the value of $K(\sigma)$ at any point $p\in S^n(r)$ and for any plane $\sigma\subseteq T_pS^n(r)$.
We say that the sphere of radius $r$ has {\em constant sectional curvature}. What is the relation between the sectional curvature of the sphere and
its principal curvatures?
\item The {\em second fundamental form} of a hypersurface $M$ in Euclidean space with unit normal vector field $N$ is the bilinear form on each $T_pM$ defined by
$$\mathbbm{I}_p(u,v)=\langle S_pu,v\rangle N_p.$$
Note that the second fundamental form is symmetric since the shape operator is self-adjoint. More generally, suppose that $M$ is a submanifold  of Euclidean
space of arbitrary codimension. Define $$\mathbbm{I}_p(u,v)=\left(D_uv\right)^\perp.$$
In words, $\mathbbm{I}_p(u,v)$ is the normal component of the Euclidean covariant derivative of any local extension of $v$ in the direction of $u$.
It is not difficult to show that   $\mathbbm{I}_p(u,v)$ is tensorial in $u$ and $v$ and does not depend on the local extension. Check that
$$\left\langle R(u,v)w,z\right\rangle =\left\langle \mathbbm{I}(u,z),\mathbbm{I}(v,w)\right\rangle -\left\langle \mathbbm{I}(u,w),\mathbbm{I}(v,z) \right\rangle.$$
\end{enumerate}



\item {\bf Curvature and differential forms.} (This exercise is related to Problems 4-14 page 113 and  7-5 page 222, of Lee's text.) Let $E_1, \dots, E_n$ be a local orthonormal frame on a Riemannian manifold. Let $\theta_1, \dots, \theta_n$
be the coframe of smooth $1$-forms characterized by the equations $\theta_i(E_j)=\delta_{ij}$. Note that $\theta_i$ is the one form given by $X\mapsto \langle E_i, X\rangle$. If $\nabla$ is the Levi-Civita connection,
define the connection $1$-forms by
$$ \omega_{ij}(X)=\langle E_i, \nabla_{X}E_j\rangle=\theta_i(\nabla_X E_j).$$
\begin{enumerate}
\item Show that $\omega_{ij}=-\omega_{ji}$. Since $\omega^\intercal=-\omega$, we say that $\omega$ is a Lie algebra-valued one-form, for the
Lie algebra $\mathfrak{o}(n)$ of the orthogonal group $O(n)$. 
\item Prove Cartan's first structure equation:
$$d\theta_{i}+\sum_j\omega_{ij}\wedge\theta_j=0. $$
\item Define the curvature $2$-forms $\Omega_{ij}$ as
$ \Omega_{ij}(X,Y)=\left\langle E_i, R(X,Y)E_j\right\rangle,$
where $R$ is the Riemann curvature tensor:
$$ R(X,Y)Z=\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z.$$
Show that $$ \Omega_{ij} = d\omega_{ij} + \sum_k \omega_{ik}\wedge \omega_{kj}.$$
\end{enumerate}




\item {\bf Conformally Euclidean Riemannian manifolds.}   Let $M$ be a non-empty open subset of  $\mathbb{R}^n$
 and let $\eta:M\rightarrow (0,\infty)$ be a smooth positive function. We define a Riemannian metric on $M$
 by
 $$\langle u,v\rangle_p = \frac1{[\eta(p)]^2} u\cdot v $$
 where $u,v \in T_pM$ and $u\cdot v$ denotes the ordinary dot product of vectors in $\mathbb{R}^n$, where  we  identify $T_pM$ with
 $\mathbb{R}^n$. We define an orthonormal frame on $M$ consisting of the vector fields
 $E_i = \eta \frac{\partial}{\partial x_i}$, $i=1, \dots, n$, where the partial derivatives are with respect to the standard coordinates in $\mathbb{R}^n$. 
 Introduce the dual  frame of $1$-forms $\theta_1, \dots, \theta_n$,
 the connection $1$-forms $\omega_{ij}$, and the curvature $2$-forms $\Omega_{ij}$ as in the previous problem. We use the notation
 $\eta_i=\frac{\partial \eta}{\partial x_i}$ and $\eta_{ij}=\frac{\partial^2\eta}{\partial x_i\partial x_j}$.
 \begin{enumerate}
 \item \label{part a} Show that $[E_i,E_j]= \eta_iE_j -  \eta_jE_i$.
 \item Show that 
$d\theta_i=\theta_i\wedge d\log \eta=\sum_j \eta_j \theta_i\wedge \theta_j$.
 \item Show that $\omega_{ij} =\eta_i\theta_j-\eta_j\theta_i.$ (There is more than one way to find the connection forms. One convenient way is to obtain the values of $\omega_{ij}(E_k)$ from the expression  for the Levi-Civita connection
 $$ \langle \nabla_X Y,Z\rangle = \frac12 \left\{ X\langle Y,Z\rangle +Y\langle Z, X\rangle - Z\langle X,Y\rangle - \langle X, [Y,Z]\rangle + \langle Y, [Z,X]\rangle + \langle Z, [X,Y]\rangle \right\}$$
 for any vector fields $X, Y, Z$ on $M$.)
 \item Check that the first structure equation  $d\theta_i +\sum_j \omega_{ij}\wedge \theta_j=0$ holds for this example.
 \item Using the second structure equation $\Omega_{ij} = d\omega_{ij}+\sum_k\omega_{ik}\wedge\omega_{kj}$,
 show that 
  $$\Omega_{ij} =
   \eta \sum_k\left(
   \eta_{jk} \theta_i -\eta_{ik} \theta_j
 \right)\wedge \theta_k
 - \left(\sum_k \eta_k^2\right) \theta_i\wedge \theta_j. 
 $$ 
 \item Suppose $M=\left\{(x_1, \dots, x_n)\in \mathbb{R}^n: x_n>0\right\}$ and that $\eta(x_1, \dots, x_n) =x_n$. Show that
 $\Omega_{ij} = -\theta_i\wedge \theta_j$. 
 \item Using the previous item, show that $M$ (the upper-half space, or {\em real hyperbolic space}) has constant sectional curvature equal to $-1$. 
 \end{enumerate}
 


\item {\bf Ricci and scalar curvatures.} (See Chapter 7, Page 207, Lee's text.)
Let $M$ be a Riemannian manifold and $R$ the Riemann curvature tensor (obtained from the Levi-Civita connection). The {\em Ricci curvature tensor} is defined, at each $p\in M$, as the trace of the linear map $w\in T_pM \mapsto R(w, u)v\in T_pM$:
$$ \text{Ric}_p(u,v):=\text{tr}\left(w\mapsto R(w,u)v\right),$$
for all $u,v\in T_pM$. This defines a smooth tensor field $\text{Ric}$. If $E_1, \dots, E_n$ is a local frame on $M$ (not necessarily orthonormal) and $\theta_1, \dots, \theta_n$ is the dual frame of $1$-forms (so that $\theta_i(E_j)=\delta_{ij}$) then
$$ \text{Ric}(X,Y)=\sum_k\theta_k(R(E_k,X)Y).$$
If the local frame is orthonormal, then
$$ \text{Ric}(X,Y)=\sum_k\left\langle E_k, R(E_k,X)Y\right\rangle.$$
It is clear (given the trace definition) that the tensor defined by the last two expressions does not depend on the choice of local frame. 

The {\em scalar curvature} is a smooth function $S$ on $M$ defined as follows. Let $E_1, \dots, E_n$ be a local orthonormal frame. Then
$$S =  \sum_{k}\text{Ric}\left(E_k,E_k\right). $$
\begin{enumerate}
\item Show that the Ricci curvature tensor is symmetric: $\text{Ric}(X,Y)=\text{Ric}(Y,X)$.
\item Show that the scalar curvature does not depend on the choice of orthonormal frame.
\item Show that the Ricci tensor for the real hyperbolic space (upper-half space of the previous problem) is
$$ \text{Ric}_p(u,v)=-(n-1)\langle u, v\rangle_p$$
where $n$ is the dimension of $M$.
\item Show that the scalar curvature of the real hyperbolic space is
$S=-n(n-1).$
\item You may ignore this item if you don't find it sufficiently interesting. (We won't need it later in the course.) Let $R$ be the Riemann curvature tensor of a Riemannian manifold $M$ and $\nabla$ the Levi-Civita connection. We define the {\em covariant divergence}
of the Ricci tensor at a point $p\in M$  as
$$ \text{div}(\text{Ric})(u) = \sum_{i} (\nabla_{e_i}\text{Ric})(e_i, u),$$
where $e_1, \dots, e_n$ is an orthonormal basis of $T_pM$ and $u\in T_pM$. (The covariant divergence can be defined for more general tensor fields in a similar way.) 
Show that 
$$ \text{div}(\text{Ric})=\frac12 dS.$$
For more information on this equation and its connection with general relativity theory,  see the remarks on 211 of Lee's text.  This formula is stated in Proposition 7.18, page 209.  Take a look at the proof by Penrose tensor diagrams given in my notes on (current) page 346.
\end{enumerate}


\item {\bf Geometry of the orthogonal group.} Let us recall a fact from the homework assignment 1, problem 6. The orthogonal group $O(n)$ is the matrix Lie group
consisting of orthogonal $n\times n$ matrices: $g^\intercal g=I$.  The Lie algebra  $\mathfrak{o}(n)$ may be described in two isomorphic ways: as the space
of left-invariant vector fields on $O(n)$
$$X_A(x)=\sum_{i,j}\left(xA\right)_{ij}\frac{\partial}{\partial x_{ij}} $$
where $A$ is an antisymmetric $n\times n$ matrix: $A^\intercal=-A$. The Lie algebra operation is the Lie bracket of vector fields. This Lie algebra is isomorphic to
the space of antisymmetric $n\times n$ matrices with the commutator bracket: $[A,B]=AB-BA$.  It was proved in problem 6 of homework 1 that these two Lie algebras are isomorphic under the correspondence $A\mapsto X_A$:
$$[X_A,X_B]=X_{[A,B]}. $$ Note that at the identity element $e$ (equal to the identity matrix $I$)
$$X_A(e) =A=\sum_{i,j} A_{ij}\frac{\partial}{\partial x_{ij}}.$$
Also recall that the flow of $X_A$ is given by the exponential map:
$$\Phi_t(g)=ge^{tA}. $$
In particular, if $L_g:h\mapsto gh$ and  $R_g:h\mapsto hg$ denote left- and  right-multiplication in $O(n)$, 
$$ d(L_g)_e X_A(e)=\left.\frac{d}{dt}\right|_{t=0} g e^{tA}=gA=\sum_{i,j}(gA)_{ij}\frac{\partial}{\partial x_{ij}}$$
and 
$$ d(R_g)_e X_A(e)=\left.\frac{d}{dt}\right|_{t=0}  e^{tA}g=Ag=\sum_{i,j}(Ag)_{ij}\frac{\partial}{\partial x_{ij}}.$$
Note that the push-forward $(R_g)_*X_A$ is also a left-invariant vector field since left and right multiplication commute. In fact, convince yourself that
$${R_g}_*X_A = X_{g^{-1}Ag}. $$

Let us define a Riemannian metric on $O(n)$ using the trace as follows: on left-invariant vector fields $X_A, X_B$, 
$$\left\langle X_A,X_B\right\rangle_g:=\frac12\text{tr}\left(A B^\intercal \right). $$ It is not difficult to check that this expression, extended by bilinearity to 
general vector fields, is indeed a Riemannian metric on $O(n)$. 
\begin{enumerate}
\item It is immediate from this definition that the Riemannian metric  is left-invariant. Show  that it is also right-invariant.  We say that this Riemannian metric is {\em bi-invariant}.
\item Show that the Levi-Civita connection $\nabla$ for this metric satisfies the property
$$\nabla_{X_A}X_B=\frac12 X_{[A,B]}. $$
(Use the equation (5.5) in the proof of Theorem 5.10, page 122, of Lee's text.)
\item Show that 
$$R(X_A,X_B)X_C=-\frac14 X_{[[A,B],C]}. $$
\item Show that the sectional curvature $K(\sigma)$ of the plane spanned by $X_A, X_B$ is
$$K=\frac18\frac{\text{tr}\left([A,B][A,B]^\intercal\right)}{Q(A,B)},$$
where $$Q(A,B)=\|X_A\|^2\|X_B\|^2 - \langle X_A,X_B\rangle^2= \frac1{16}\left(\text{tr}\left(AA^\intercal\right)\text{tr}\left(BB^\intercal\right)-2 \left(\text{tr}\left(AB^\intercal\right)\right)^2   \right).$$
Note: If $A,B$ are orthonormal,  we may write (identifying $X_A$ and $A$)
$$K=\frac14\|[A,B]\|^2. $$
 \item The Lie algebra of $O(3)$ is generated by the matrices
 $$L_1=\left(\begin{array}{ccr}0 & 0 & 0 \\0 & 0 & -1 \\0 & 1 & 0\end{array}\right), \ \ L_2=\left(\begin{array}{rcc}0 & 0 & 1 \\0 & 0 & 0 \\-1 & 0 & 0\end{array}\right), \ \ L_3=\left(\begin{array}{crc}0 & -1 & 0 \\1 & 0 & 0 \\0 & 0 & 0\end{array}\right). $$
 It is not difficult to check that these elements form an orthonormal basis of $\mathfrak{o}(3)$ and
 $$[L_1, L_2]=L_3, \ \ [L_2, L_3]=L_1, \ \ [L_3,L_1]=L_2. $$
 Let $u=\sum_i a_iL_i$ and $v=\sum_ib_iL_i$ be an orthonormal basis of a plane $\sigma$ in $T_eO(n)$. Show that
 $$K(\sigma)= \frac14|a\times b|^2=\frac14.$$
 where $|a\times b|$ is the Euclidean norm of the cross-product from Calc. III.
 
 Note: the connected component of $O(3)$, namely, $SO(3)$, is double covered by the sphere $S^3$. (Topologically, $SO(3)$ is homeomorphic to
 projective space $\mathbb{R}P^3$.) So it should not be surprising that the sectional curvature is constant. 
\end{enumerate}

\end{enumerate}


\vspace{.1in}
\begin{figure}[htbp]
\begin{center}
\includegraphics[width=5.5in]{Diagram.pdf}
\caption{\small Proof of Equation    $dS=2\,\text{div(Ric)}.$ If you are interested in figuring out what is going on in this figure, take a look at my notes on tensor calculus.
For a Riemannian metric tensor, the $s_i$ are all $1$. (This proof allows for semi-Riemannian metrics, for which $s_i=\langle e_i,e_i\rangle=\pm1$.)
The very first step in the proof is now show: the trace operation involved in the definition of the Ricci tensor commutes with the divergence operator. (The lasso around $R$ at the very beginning goes through the wire loop defining the trace.)}
\label{Diagram}
\end{center}
\end{figure} 

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