Riemannian Geometry (Curvature)

The sectional curvature of $M$ at $p$ with respect to $\sigma$

$$
K(p,\sigma)$$

If $M = \mathbb{R}^n$, $K(p, \sigma) = 0$ for all $p$ and all $\sigma$.

Curvature

Definition

The curvature $R$ of a Riemannian manifold $M$ is a correspondence that associates to every pair $X, Y \in \mathcal{X}(M)$ a mapping $R(X, Y):\mathcal{X}(M) \rightarrow \mathcal{X}(M)$ given by

$$
R(X,Y)Z = \nabla_Y \nabla_X Z – \nabla_X \nabla_Y Z + \nabla_{[X,Y]} Z, \;\;\; Z \in \mathcal{X}(M)$$

• Tangent space $T_p M$의 Basis에 대하여 생각해 보면 Since $\left[ \frac{\partial}{\partial x_i}, \frac{\partial}{\partial x_j} \right] = 0$,

  $$
  R\left( \frac{\partial }{\partial x_i}, \frac{\partial }{\partial x_j} \right) \frac{\partial }{\partial x_k} = \left( \nabla_{\frac{\partial }{\partial x_j}} \nabla_{\frac{\partial }{\partial x_i}} – \nabla_{\frac{\partial }{\partial x_i}} \nabla_{\frac{\partial }{\partial x_j}} \right) \frac{\partial}{\partial x_k}$$

Proposition

The curvature $R$ of a Riemannian manifold has the following properties:

• $R$ is bilinear in $\mathcal{X}(M) \times \mathcal{X}(M)$, that is,

  $$
  \begin{align}
  R(fX_1 + gX_2, Y_1) &= fR(X_1, Y_1) + gR(X_2, Y_1) \\
  R(X_1, fY_1 + gY_2) &= fR(X_1, Y_1) + gR(X_1, Y_2) \\
  \end{align}$$

• For any $X,Y \in \mathcal{X}(M)$, the curvature operator $R(X,Y):\mathcal{X}(M) \rightarrow \mathcal{X}(M)$ is linear, that is

  $$
  \begin{align}
  R(X,Y)(Z+W) &= R(X,Y)Z + R(X,Y)W \\
  R(X,Y)fZ &= fR(X,Y)Z
  \end{align}$$

증명은 90P.

Remark

사실 $\nabla_{[X,Y]} Z$ 항의 존재는 Curvature가 위와 같은 Bilinear 특성을 가지도록 추가한 것이다.

Bianchi Identity

$$
R(X,Y)Z + R(Y,Z)X + R(Z,X)Y = 0$$

증명은 91p.

Proposition

$$
\begin{align}
(X, Y, Z, T) + (Y, Z, X, T) + (Z, X, Y, T) &= 0 \\
(X, Y, Z, T) &= – (Y, X, Z, T) \\
(X, Y, Z, T) &= – (Y, X, T, Z) \\
(X, Y, Z, T) = (Z, T, Y, X)
\end{align}$$

Note

• From the Levi-Civita Theorem (p55) for Affine Connection

  $$
  X \langle Y, Z \rangle = \langle \nabla_X Y, Z \rangle + \langle Y, \nabla_X Z \rangle$$

  so that

  $$
  \langle \nabla_Y \nabla_X Z, Z \rangle = Y \langle \nabla_X Z, Z \rangle – \langle \nabla_X Z, \nabla_Y Z \rangle$$

  Proof is (Let $K = \nabla_X Z$)

  $$
  \begin{align}
  Y \langle \nabla_X Z, Z \rangle = Y \langle K, Z \rangle = \langle \nabla_Y K, Z \rangle + \langle K, \nabla_Y Z \rangle \\
  Y \langle K, Z \rangle – \langle K, \nabla_Y Z \rangle = \langle \nabla_Y K, Z \rangle \\
  Y \langle \nabla_X Z, Z \rangle – \langle \nabla_X Z, \nabla_Y Z \rangle = \langle \nabla_Y \nabla_X Z, Z \rangle
  \end{align}$$

• Let $[X,Y] = K$. The proof of

  $$
  \langle \nabla_{[X,Y]} Z, Z \rangle = \frac{1}{2}[X, Y]\langle Z, Z \rangle$$

  From the above equation

  $$
  \begin{align}
  K \langle Z, Z \rangle &= \langle \nabla_K Z, Z \rangle + \langle Z, \nabla_K Z \rangle \\
  K \langle Z, Z \rangle &= 2 \langle \nabla_K Z, Z \rangle
  \end{align}$$

Curvature under Christoffel symbol to Riemann Connection

Set $ \frac{\partial}{\partial x_i} = X_i$

$$
R(X_i, X_j)X_k = \sum_l R_{ijk}^l X_l$$

$R_{ijk}^l$ are the components of the curvature $R$ in $(U,x)$.
If

$$
X= \sum_i u^i X_i, \; Y=\sum_j v^j X_j,\; Z = \sum_k w^k X_k$$

then

$$
R(X,Y)Z = \sum_{i,j,k,l} R_{i.j.k.l}^l X_l$$

• To express $R_{i.j.k.l}^l$ in terms of the coefficients $\Gamma_{ij}^k$

  $$
  \begin{align}
  R(X_i, X_j)X_k &= \nabla_{X_j} \nabla_{X_i} X_k – \nabla_{X_i} \nabla_{X_j} X_k \\
  &= \nabla_{X_j} (\sum_l \Gamma_{ik}^l X_l) – \nabla_{X_i} (\sum_l \Gamma_{jk}^l X_l) \\
  &= \sum_l \frac{\partial}{\partial x_j}\Gamma_{ik}^l X_l + \sum_l \Gamma_{ik}^l \Gamma_{jl}^s X_s – \sum_l \frac{\partial}{\partial x_i} \Gamma_{jk}^l X_l – \sum_l \Gamma_{ik}^l \Gamma_{il}^s X_s
  \end{align}$$

따라서

$$
R_{ijk}^s = \sum_l \Gamma_{ik}^l \Gamma_{jl}^s – \sum_l \Gamma_{jk}^l \Gamma_{il}^s + \frac{\partial}{\partial x_j}\Gamma_{ik}^s – \frac{\partial}{\partial x_i}\Gamma_{jk}^s$$

이에 따라 Curvature Proposition은 다음과 같이 간략화된다.

$$
\begin{align}
R_{ijks} + R_{jkis} + R_{kijs} = 0\\
R_{ijks} = -R_{jiks} \\
R_{ijks} = -R_{ijsk} \\
R_{ijks} = R_{ksij}
\end{align}$$

Sectional Curvature

• Definition $x \wedge y$

  $$
  |x \wedge y| = \sqrt{|x|^2 |y|^2 – \langle x, y \rangle^2}$$

  이것은 그냥 Determinent 이다.

Proposition 3.1

Let $\sigma \subset T_p M$, and let $x, y \in \sigma$ be two linearly independent vectors. Then

$$
K(x,y) = \frac{(x,y,x,y)}{|x \wedge y|^2}$$

does not depend on the choice of the vectors $x,y \in \sigma$.

• $K(x,y)$는 다음에 불변이다.

1. $.K(x,y) = K(y,x)$
2. $.K(x,y) = K(\lambda x, y)$
3. $.K(x,y) = K(x + \lambda y, y)$

Definition (Sectional Curvature)

Given a point $p \in M$ and a two-dimensional subspace $\sigma \subset T_p M$, then real number $K(x,y) = K (\sigma)$ where ${ x, y}$ is any basis of $\sigma$, is called the sectional curvature of $\sigma$ at $p$.

• Scalar 값인 $K (\sigma)$를 통해 $\forall \sigma$, the curvature $R$을 Completely 하게 결정할 수 있다.

Lemma

Let $V$ be a vector space of dimension $\geq 2$ , provided with an inner product $\langle , \rangle$. Let $R:V \times V \times V \rightarrow V$ and $R’:V \times V \times V \rightarrow V$ be tri-linear mappings such that conditions in proposition 2.5 are satisfied by

$$
(x,y,z,t) = \langle R(x,y)z, t \rangle, \;\;\; (x,y,z,t)’ = \langle R'(x,y)z, t \rangle$$

If $x,y$ are two linearly inde;endent vectors, we mat write

$$
K(\sigma) = \frac{(x,y,x,y)}{|x \wedge y |^2}, \;\;\; K'(\sigma) = \frac{(x,y,x,y)’}{|x \wedge y |^2}$$

where $\sigma$ is the bi-dimensional subspace generated by $x$ and $y$. If for all $\sigma \subset V$, $K(\sigma)=K'(\sigma)$, then $R = R’$.

Lemma (p96): Very Important

Let $M$ be a Riemmanian manifold and $p$ a point of $M$.
Define a tri-linear mapping $R’:T_p M \times T_p M \times T_p M \rightarrow T_p M$ by

$$
\langle R'(X,Y,W),Z \rangle = \langle X,W \rangle \langle Y,Z \rangle – \langle Y,W \rangle \langle X,Z \rangle$$

for all $X,Y,W,Z \in T_p M$. Then $M$ has constant sectional curvature equal to $K_0$ if and only if $R = K_0 R’$, where $R$ is the curvature of $M$

proof

Assume that $K(p, \sigma) = K_0$ , $\forall \sigma \subset T_p M$, and set $\langle R'(X,Y,W,Z), Z \rangle = (X,Y,W,Z)’$

$$
\langle R'(X,Y,W), Z \rangle = (X, Y, W, Z)’$$

Lemma 에서 정의된 대로

$$
(X, Y, X, Y)’ = \langle X, X \rangle \langle Y, Y \rangle – \langle X, Y \rangle^2$$

Sectinal curvature의 정의에 의해

$$
K_0 = \frac{(X, Y ,X, Y)}{| X \wedge Y|^2} \Rightarrow (X, Y ,X, Y) = K_0 (\langle X, X \rangle \langle Y, Y \rangle – \langle X, Y \rangle^2) = K_0 R'(X,Y,X,Y)$$

Lemma 3.3 implies that

$$
R(X,Y,W,Z) = K_0 R'(X, Y, W, Z)$$

Corollary 3.5

Let $M$ be a Riemmanian manifold, $p$ a point of $M$ and ${ e_1, \cdots e_n}$$n = \dim M$, an orthonormal basis of $T_p M$.
Define $R_{ijkl} = \langle R(e_i, e_j)e_k, e_l \rangle, i,j,k,l = 1, \cdots n$ Then $K(p, \sigma)=K_0$ for all $\sigma \subset T_p M$, if only if

$$
R_{ijkl} = K_o (\delta_{ik} \delta_{jl} – \delta_{il} \delta_{jk})$$

where

$$
\delta_{ij} =
\begin{cases}
1, &i=j \\
0, &i \neq j
\end{cases}$$

다시말해 $K(p,\sigma)=K_0$, $\forall \sigma \subset T_p M$ if and only if $R_{ijij} = -R_{ijji} = K_0$, $\forall i \neq j$ and $R_{ijkl} = 0$ in other cases.

Ricci Curvature and scalar curvature

• Sectional curvature의 Combination의 한 형태
• Let $x=z_n$ be a unit vector in $T_p M$.
• Take an orthnormal basis $\{z_1, z_2, \cdots z_{n-1} \}$ of the hyperplane in $T_p M$ orthogonal to $x$
• Consider the following averages:
  • Ricci Curvatuere in the direction $x$

    $$
    \text{Ric}_p (x) = \frac{1}{n-1} \sum_{i=1}^{n-1} \langle R(x,z_i)x,z_i \rangle \\$$

  • Scalar curvature at $p$

    $$
    K(p) = \frac{1}{n} \sum_{j=1}^n \text{Ric}_p(z_j) = \frac{1}{n(n-1)} \sum_{ij} \langle R(z_i, z_j) z_i, z_j \rangle$$

Ricci Tensor

• Let $x, y \in T_p M$ and put

$$
Q(x,y) = \text{trace of the mapping} \;\; z \mapsto R(x,z)y$$

• $Q$ is bilinear For orthonormal basis $\{z_1, \cdots z_{n-1},z_n = x \}$ of $T_p M$

$$
\begin{align}
Q(x,y) &= \sum_i \langle R(x,z_i)y, z_i \rangle \\
&= \sum_i \langle R(y,z_i)x, z_i \rangle = Q(y,x)
\end{align}$$

• $Q$ is symmetric and

$$
Q(x,x) = \sum_{i} \langle R(x, z_i)x, z_i \rangle = (n-1)\frac{1}{n-1} \sum_{i} \langle R(x, z_i)x, z_i \rangle = (n-1) \text{Ric}_p(x)$$

• Let a self-adjoint mapping $K$ corresponding the bilinear form $Q$ on $T_p M$

$$
\langle K(x),y \rangle = Q(x.y)$$

• For an orthonormal basis $\{z_1, \cdots z_n \}$

$$
\begin{align}
\text{Trace of K} &= \sum_j \langle K(z_j), z_j \rangle = \sum_j Q(z_j, z_j) \\
&= (n-1)\sum_j \text{Ric}_p(z_j) = n(n-1)K(p)
\end{align}$$

이때, bilinear form$\frac{1}{n-1} Q$를 Ricci Tensor라 한다.

• Let $X_i = \frac{\partial}{\partial x_i}$, $g_{ij} = \langle X_i, X_j \rangle$, and $g^{ij}$ is the inverse matrix of $g_{ij}$ i.e. $\sum_k g_{ik}g^{kj} = \delta_i^j$
• Then the coefficient of the bilinear form $\frac{1}{n-1}Q$ in the basis $\{X_i\}$ are given by

$$
\frac{1}{n-1}R_{ik} = \frac{1}{n-1} \sum_j R_{ijk}^j = \frac{1}{n-1} \sum_j R_{ijks} g^{sj}$$

• The scalar curvature in the coordinate system $(x_i)$ is given by

$$
K = \frac{1}{n(n-1)} \sum_{ik} R_{ik} g^{ik}$$

생각해 보면 Ric Curvature 혹은 Scalar Curvature는 Riemannian Metric 의 Inverse가 필요하다는 것이다.

Lemma 4.1

Let $f:A \subset \mathbb{R} \rightarrow M$ be a parameterized surface. let $(s,t)$ be a usual coordinates of $\mathbb{R}^2$. Let $V = V(s,t)$ be a Vector field alomnng. Then

$$
\frac{D}{dt}\frac{D}{ds}V – \frac{D}{ds}\frac{D}{dt}V = R(\frac{\partial f}{\partial s}\frac{\partial f}{\partial t})V$$

proof

증명 과정은 매우 Simple 하다.

1. Let $V = \sum_i v^i X_i$ where $v^i = v^i(s,t)$, and $X_i = \frac{\partial}{\partial x_i}$
2. Let $f(s,t)=(x_1(s,t), \cdots x_n(s,t))$ so that $\frac{\partial f}{\partial s} = \sum_j \frac{\partial x_j}{\partial s} X_j$ and $\frac{\partial f}{\partial t} = \sum_k \frac{\partial x_k}{\partial t} X_k$

$$
\begin{align}
\frac{D}{\partial s}V &= \frac{D}{\partial s}\sum_i v^i X_i = \sum_i \frac{\partial v^i}{\partial s} X_i + \sum_i v^i \frac{D}{\partial s}X_i \\
\frac{D}{\partial t}\frac{D}{\partial s}V &= \sum_i \frac{\partial^2 v^i}{\partial t \partial s} X_i + \sum_i \frac{\partial v^i}{\partial s} \frac{D}{\partial t} X_i + \sum_i \frac{\partial v^i}{\partial t} \frac{D}{\partial s}X_i + \sum_i v^i \frac{D}{\partial t} \frac{D}{\partial s}X_i
\end{align}$$

그러므로

$$
\frac{D}{dt}\frac{D}{ds}V – \frac{D}{ds}\frac{D}{dt}V = \sum_i v^i \left( \frac{D}{\partial t} \frac{D}{\partial s} X_i – \frac{D}{\partial s} \frac{D}{\partial t} X_i \right)$$

Since

$$
\frac{D}{\partial s}X_i = \nabla_{\frac{\partial f}{\partial s}} X_i = \nabla_{\sum_j \frac{\partial x_j}{\partial s} X_j} X_i = \sum_j \frac{\partial x_j}{\partial s} \nabla_{X_j}X_i$$

그러므로

$$
\begin{align}
\frac{D}{dt}\frac{D}{ds}X_i &= \frac{D }{dt} \left( \sum_j \frac{\partial x_j}{\partial s} \nabla_{X_j}X_i \right) \\
&= \sum_j \frac{\partial^2 x_j}{\partial t \partial s} \nabla_{X_j}X_i + \sum_j \frac{\partial x_j}{\partial s} \nabla_{\sum_k \frac{\partial x_k}{\partial t} X_k}\nabla_{X_j}X_i \\
&= \sum_j \frac{\partial^2 x_j}{\partial t \partial s} \nabla_{X_j}X_i + \sum_{jk} \frac{\partial x_j}{\partial s} \frac{\partial x_k}{\partial t} \nabla_{ X_k}\nabla_{X_j}X_i
\end{align}$$

따라서

$$
(\frac{D}{dt}\frac{D}{ds} – \frac{D}{ds}\frac{D}{dt})X_i = \sum_{jk} \frac{\partial x_j}{\partial s} \frac{\partial x_k}{\partial t} \left( \nabla_{ X_k}\nabla_{X_j}X_i – \nabla_{X_j} \nabla_{ X_k}X_i \right)$$

그러므로, $\sum$ 내부의 각 항이 대응되는 Vector field로 합쳐지므로

$$
(\frac{D}{dt}\frac{D}{ds} – \frac{D}{ds}\frac{D}{dt})V = \sum_{ijk} v^i \frac{\partial x_j}{\partial s} \frac{\partial x_k}{\partial t} R(X_j,X_k)X_i = R(\frac{\partial f}{\partial s}, \frac{\partial f}{\partial t})V$$

Tensor on Riemannian Manifold

Definition : Tensor

A Tensor $T$ of order $r$ on a Riemannian manifold is a multilinear mapping

$$
T : \underset{r \text{ factors}} {\underbrace{\mathcal{X}(M) \times \cdots \times \mathcal{X}(M)}} \rightarrow \mathcal{D}(M)$$

Example : The Curvature Tensor

$$
T : \mathcal{X}(M) \times \mathcal{X}(M) \times \mathcal{X}(M) \times \mathcal{X}(M) \rightarrow \mathcal{D}(M)$$

is defined by

$$
R(X, Y, Z, W) = \langle R(X, Y)Z, W \rangle, \;\;\; X, Y, Z, W \in \mathcal{X}(M)$$

It is a tensor of order 4.
The components in the frame is $\left{ X_i = \frac{\partial}{\partial x_i} \right}$ associated with

$$
R(X_i, X_j, X_k, X_l) = R_{ijkl}$$

Example : Metric Tensor

$G : \mathcal{X}(M) \times \mathcal{X}(M) \rightarrow \mathcal{D}(M)$ is defined by $G(X,Y) = \langle X, y \rangle, \;\; X, Y \in \mathcal{X}(M)$. $G$ is a tensor of order 2. The components in the frame ${ X_i }$ are the coeddicients $g_{ij}$ of the Riemannian metric

Example

The Riemannian Connection $\nabla$ defined by:

$$
\nabla : \mathcal{X}(M) \times \mathcal{X}(M) \times \mathcal{X}(M) \rightarrow \mathcal{D}(M) \\
\nabla(X,Y,Z)= \langle \nabla_X Y,Z \rangle, \;\;\; X, Y, Z \in \mathcal{X}(M)$$

is not a Tensor, because $\nabla$ is not linear with respect to the argument $Y$.

Definition : Covariant Differential $\nabla T$ of T

Let $T$ be a tensor of order $r$. The Covariant Differential $\nabla T$ of T is a tensor of order $(r+1)$ given by

$$
\nabla T (Y_1, \cdots , Y_r, Z) = Z(T(Y_1, \cdots, Y_r)) – T(\nabla_Z Y_1, \cdots, Y_r) – \cdots – T(Y_1, \cdots, Y_{r-1}, \nabla_Z Y_r)$$

For each $Z \in \mathcal{X}(M)$, the covariant derivative $\nabla_Z T$ of $T$ relative to $Z$ is a tensor of order $r$ given by

$$
\nabla_Z T(Y_1, \cdots, Y_r) = \nabla T (Y_1, \cdots, Y_r, Z)$$

Excercise

Let $G$ be a Lie group with a bi-invariant metric $\langle, \rangle$. Let $X, Y, Z \in \mathcal{X}(G)$ be unit left invariant vector field on $G$.

1. $\nabla_X Y = \frac{1}{2}[X,Y]$