Riemannian Metrics

There is one more crucial ingredient which we need to introduce for dealing with manifolds: lengths and angles. Given a smooth manifold, since we know what it means for a curve in the manifold to be smooth, and we have a well-defined notion of the tangent vector to a curve, all we need in order to have a notion of distance on the manifold is a way of defining the speed of a curve — that is, the length of its tangent vector.

Metric Tensors

Metric Tensor & Semi-Riemannian Manifold

A (Semi-Riemannian) metric tensor $g$ on $M$ is a $(2,0)$-tensor, i.e. a 2-covariant tensor, which at each $p\in M$, takes a pair of tangent vectors $u,v\in T_p(M)$ and returns a real number $g_p(u,v)$ such that

• $g_p$ is symmetric, that is $g_p(u,v)=g_p(v,u)$.
• $g_p$ is nondegenerate, that is for every $u\ne 0$, there is a $v$ such that $g_p(u,v)\ne 0$.

A semi-Riemannian manifold is a smooth manifold furnished with a metric tensor.

Riemannian Metric & Riemannian Manifold

A Riemannian metric $g$ on a smooth manifold $M$ is a nonnegative metric tensor, that is $g_p(u,v)\ge 0$ for all $u\in T_p(M)$ and $g_p(u,u)=0$ if and only if $u=0$. A pair $(M,g)$ is called a Riemannian manifold.

e.g.

• The standard inner product on Euclidean space is a trivial example of a Riemannian metric.
• Consider $S^2\subset {\mathbb{R}}^3$ with standard spherical coordinates $(\theta ,\varphi )$ with $\theta \in [0,\pi ]$ and $\varphi \in [0,2\pi ]$, the metric tensor from ${\mathbb{R}}^3$ is given by $g=\text{dd}{\theta }^2+{\sin }^2\theta \text{dd}{\varphi }^2.$

Metric Signature

The signature of a metric tensor $g$ is the pair of integers $(p,q)$ (or sometimes the integer difference $p-q$) that records, respectively, the number of positive and negative eigenvalues of the symmetric matrix representation of the metric at a point. Sometimes it may also be denoted as an explicit list such as $(+,-,-,-)$ or $(-,+,+,+)$.

e.g. A Riemannian metric has signature $(n,0)$.

Killing Vector Field

A killing vector field on a semi-Riemannian manifold is a vector field $X$ for which the Lie derivative of the metric tensor $g$ vanishes: ${\text{L}}_{X}g=0$.

Remannian Metric on Manifold

Induced Metric (Pullback Metric)

Let $(M,g_M)$ be a Riemannian manifold and $N$ be a smooth manifold. Given a smooth map $\iota :N\to M$, the induced metric (often referred to as the pullback metric), denoted as $g_N$ or ${\iota }^{\ast }{g}_M$, is defined on $N$ by the following equation: $g_N(X,Y):=g_M(d{\iota }_x(X),d{\iota }_x(Y))$for any point $x\in N$ and tangent vectors $X,Y\in T_{x}N$. Here, $d{\iota }_x:T_{x}N\to T_{\iota (x)}M$ represents the differential (or pushforward) of the map $\iota$ at point $x$.

Proposition

Let $(M,g_M)$ be a Riemannian manifold, $N$ be a smooth manifold, and $\iota :N\to M$ be a smooth map. If $\iota$ is an immersion, then the induced metric $g_N:={\iota }^{\ast }{g}_M$ is a valid Riemannian metric on $N$.

Proof

To qualify as a Riemannian metric, $g_N$ must be a symmetric, bilinear, and positive-definite covariant 2-tensor field on $N$.

• Symmetry and Bilinearity: These properties are naturally inherited from the metric $g_M$ due to the linearity of the differential $d{\iota }_x$.
• Positive-Definiteness: This is where the immersion condition is strictly required. By the definition of an immersion, the differential $d{\iota }_x:T_{x}N\to T_{\iota (x)}M$ is injective (it has a trivial kernel) for all $x\in N$. To prove positive-definiteness, suppose $g_N(X,X)=0$ for some vector $X\in T_{x}N$. By our definition: $g_M(d{\iota }_x(X),d{\iota }_x(X))=0$ Because $g_M$ is a valid Riemannian metric on $M$, it is strictly positive-definite. Therefore, the only way the inner product of a vector with itself can be zero is if the vector itself is the zero vector: $d{\iota }_x(X)=0$ Since $d{\iota }_x$ is injective, the only vector that maps to $0$ is the zero vector. Thus: $X=0$ This confirms that $g_N(X,X)>0$ for all non-zero vectors $X$, proving that $g_N$ is positive-definite and therefore a true Riemannian metric.

Space $\text{Riem}(M)$

Existence of Riemannian Metrics

Let $M$ be a smooth (paracompact) manifold. Then there exists a Riemannian metric on $M$. Equivalently, the space of Riemannian metrics on $M$, denoted as $\text{Riem}(M)$, is not empty.

Proof

Let $\{U_{\alpha }\}$ be an open coordinate cover of $M$, with corresponding coordinate charts ${\varphi }_{\alpha }:U_{\alpha }\to {\mathbb{R}}^n$. Because $M$ is paracompact, we can choose a smooth partition of unity (PoU), denoted as $\{{\rho }_{\alpha }\}$, that is subordinated to this open cover. By definition, this satisfies:

1. ${\rho }_{\alpha }:M\to [0,1]$ is a smooth function.
2. The support of ${\rho }_{\alpha }$ is contained within $U_{\alpha }$, i.e., $\text{supp}({\rho }_{\alpha })\subseteq U_{\alpha }$.
3. ${\sum }_{\alpha }{\rho }_{\alpha }(x)=1$ for all $x\in M$. For each local chart, we can pull back the standard Euclidean metric $g_{\text{euc}}$ from ${\mathbb{R}}^n$ using the differential of the coordinate map $D{\varphi }_{\alpha }:T_{x}M\to T_{{\varphi }_{\alpha }(x)}{\mathbb{R}}^n\cong {\mathbb{R}}^n$. We construct a global metric $g$ on $M$ by taking a convex combination of these local pullback metrics, weighted by the partition of unity: $g_x(X,Y):={\sum }_{\alpha }{\rho }_{\alpha }(x)g_{\text{euc}}(D{\varphi }_{\alpha }(X),D{\varphi }_{\alpha }(Y))$ where $X,Y\in T_{x}M$. To verify that $g$ is a valid Riemannian metric, its symmetry and bilinearity follow naturally from $g_{\text{euc}}$. We must prove that $g$ is positive-definite. Since $g_{\text{euc}}$ is positive-definite and ${\rho }_{\alpha }(x)\ge 0$, it is clear that for any tangent vector $X$: $g_x(X,X)={\sum }_{\alpha }{\rho }_{\alpha }(x)g_{\text{euc}}(D{\varphi }_{\alpha }(X),D{\varphi }_{\alpha }(X))\ge 0$ Now, suppose $g_x(X,X)=0$. Because ${\sum }_{\alpha }{\rho }_{\alpha }(x)=1$, there must exist at least one index $\alpha$ such that ${\rho }_{\alpha }(x)>0$. For the sum to be zero, the term corresponding to this $\alpha$ must be exactly zero: $g_{\text{euc}}(D{\varphi }_{\alpha }(X),D{\varphi }_{\alpha }(X))=0$ Since the Euclidean metric $g_{\text{euc}}$ is strictly positive-definite, this requires that: $D{\varphi }_{\alpha }(X)=0$. Because ${\varphi }_{\alpha }$ is a diffeomorphism onto its image, its differential $D{\varphi }_{\alpha }$ is a linear isomorphism (specifically, it is injective). The only vector in the kernel of an isomorphism is the zero vector, which implies: $X=0$. Thus, $g_x(X,X)=0⟺X=0$. A convex combination of local metrics using a partition of unity preserves the positive-definite property. Therefore, $g$ is a globally defined Riemannian metric, proving that $\text{Riem}(M)\ne \varnothing$.

• Conformal Rescaling: The space of Riemannian metrics is infinite. If you have a valid metric $g\in \text{Riem}(M)$, then for any smooth function $f\in C^{\infty }(M)$, the rescaled metric $e^{2f}g$ is also a valid Riemannian metric (known as a conformal transformation).
• Local Coordinate Representation: Locally, within a coordinate patch $U_{\alpha }$ of dimension $\dim M=n$, using the coordinate tangent frame $\{{\partial }_{x_1^{\alpha }},\dots ,{\partial }_{x_n^{\alpha }}\}$, the metric can be expressed as a tensor field: $g{|}_{U_{\alpha }}={\sum }_{i,j}{g}_{ij}^{\alpha }d{x}_i^{\alpha }\otimes d{x}_j^{\alpha }$where the individual components of the metric tensor are given by: $g_{ij}^{\alpha }=g{|}_{U_{\alpha }}({\partial }_{x_i^{\alpha }},{\partial }_{x_j^{\alpha }})$

Length and Distance

Length of a Curve

Let $\gamma :[a,b]\to M$ be a piecewise smooth curve in a Riemannian manifold $(M,g)$. The length $L[\gamma ]$ of $\gamma$ is defined by $L[\gamma ]={\int }_a^b{g}_{\gamma (t)}({\gamma }^{\prime }(t),{\gamma }^{\prime }(t){)}^{\frac{1}{2}}\text{dd}t$ The arc-length of $\gamma$ is independent of the parametrization of $\gamma$.

Distance

Given two points $p$ and $q$ in a Riemannian manifold $(M,g)$, the distance $d(p,q)$ between $p$ and $q$ is defined by $d(p,q)=\inf \left\{L[\gamma ]|\gamma :[a,b]\to M\text{ piecewise smooth satisfying }\gamma (a)=p,\gamma (b)=q\right\}$

Proposition

If $M$ is a Riemannian manifold with metric $g$, then $M$ is a metric space with the distance function $d$ defined above. The metric topology agrees with the manifold topology.

Action

The action of a piecewise smooth curve $\gamma :[a,b]\to X$ is $\mathcal{A}(\gamma ):={\int }_a^b{\left|\frac{d\gamma }{dt}\right|}^{2}dt.$