[Tu Differential Geometry] 16. Distance and Volume

This article is one of Manifold, Differential Geometry, Fibre Bundle.

16.1 Distance in a Riemannian Manifold

In a Riemannian manifold we defined the length of a tangent vector $X_p\in T_pM$ to be $$\lVert X_p\rVert =\sqrt{\langle X_p,X_p\rangle}.$$ If $c:[a,b]\rightarrow M$ is a parametrized curve in $M$, its arc length is given by $$l(c)=\int^b_a\lVert c'(t)\rVert dt.$$

We can reparametrize $c(t)$ via a diffeomorphism $t(u):[a_1,b_1]\rightarrow [a,b]$; the reparametrization of $c(t)$ is then $c(t(u)):[a_1,b_1]\rightarrow M$. We say that the reparametrization is orientation-preserving if it preserves the orther of the endpoints: $t(a_1)=a$ and $t(b_1)=b$; it is orientation-reversing if it reverses the order of the endpoints. Since a diffeomorphism $t(u):[a_1,b_1]\rightarrow [a,b]$ is one-to-one, it is either increasing or decreasing. Let $u(t)$ be its inverse. Because $u(t(u))=u$, by the chain rule, $$\frac{dt}{du}\cdot\frac{du}{dt}=1,$$ which shows that $dt/du$ is never zero. So $dt/du$ is either always positive or always negative, depending on whether the reparametrization $t(u)$ preserves or reverse the orientation.

Proposition 16.1. The arc length of a curve $c:[a,b]\rightarrow M$ in a Riemannian manifold $M$ is independent of its parametrization.

Proof. Suppose $c(t(u))$, $a_1\le u\le b_1$ is an orientation-preserving reparametrization. Then $dt/du>0$. By the chain rule, $$\begin{align} \frac{d}{du}c(t(u))=c'(t(u))\frac{dt}{du}.\end{align}$$ It follows that $$\int_{a_1}^{b_1}\left\lVert \frac{d}{du}c(t(u))\right\rVert du =\int_{a_1}^{b_1}\lVert c'(t(u))\rVert \left| \frac{dt}{du}\right| du =\int_{a_1}^{b_1}\lVert c'(t(u))\rVert \frac{dt}{du} du =\int_a^b \lVert c'(t)\rVert dt$$

If $c(t(u))$ is an orientation-reversing parametrization, the calculation is the same except that (i) $|dt/du|=-dt/du$, because $dt/du<0$, and (ii) the limits of integration in the last line are reversed. These two changes cancel out: $$\int_{a_1}^{b_1}\left\lVert \frac{d}{du}c(t(u))\right\rVert du =\int_{a_1}^{b_1}\lVert c'(t(u))\rVert \left| \frac{dt}{du}\right| du =-\int_{a_1}^{b_1}\lVert c'(t(u))\rVert \frac{dt}{du} du =\int_{b_1}^{a_1}\lVert c'(t(u))\rVert \frac{dt}{du} du=\int_a^b \lVert c'(t)\rVert dt$$ So the arc length is still independent of the parametrization.

Given two points $p$ and $q$ on a connected Riemannian manifold, we define the distance between them to be $$d(p,q)=\inf_c l(c)$$ where the infimum is taken over all piecewise smooth curves $c$ from $p$ to $q$ and $l(c)$ is the length of the curve $c$. It is easily verified that the distance function on a Riemannian manifold satisfies the three defining properties of a metric on a metric space.

16.2 Geodesic Completeness

Definition 16.4. A Riemannian manifold is said to be geodesically complete if the domain of every geodesic in it can be extended to $\mathbb{R}$.

Definition 16.5. A geodesic defined on $[a,b]$ is said to be minimal if its length is minimal among all piecewise smooth curves joining its two endpoints.

Theorem 16.6 (Hopf-Rinow theorem). A Riemannian manifold is geodesically complete if and only if it is complete as a metric space in the distance metric d defined above. Moreover, in a geodesically complete Riemannian manifold, any two points may be joined by a minimal geodesic.

16.3 Dual $1$-Forms Under a Change of Frame

(See Chapter 11.)

Many of the constructions in differential geometry are local, in terms of a frame of vector fields and its dual frame of $1$-forms, for example, the connection matrix and the curvature matrix of an affine connection. To see how the construction transforms under a change of frame, it is useful to know how the dual frame of $1$-forms fransforms.

If $e_1,\cdots,e_n$ and $\bar{e}_1,\cdots,\bar{e}_n$ are two $C^\infty$ frames on an open set $U$ in a manifold $M$, then $$\begin{align}\bar{e}_j=\sum_i a^i_je_i\end{align}$$ for some $C^\infty$ function $[a^i_j]:U\rightarrow GL(n,\mathbb{R})$. Let $\theta^1,\cdots,\theta^n$ and $\bar{\theta}^1,\cdots,\bar{\theta}^n$ be the dual frames of $1$-forms, respectively, meaning $$\theta^i(e_j)=\theta^i_j\quad \mbox{and}\quad \bar{\theta}^i(\bar{e}_j)=\delta^i_j,$$ where $\delta^i_j$ is the Kronecker delta. To write $\bar{\theta}^1,\cdots,\bar{\theta}^n$ in terms of $\theta^1,\cdots,\theta^n$, it is best to use matrix notation. In matrix notation we write the frame $e_1,\cdots,e_n$ as a row vector $$e=[e_1\cdots e_n]$$ and the dual frame $\theta^1,\cdots,\theta^n$ as a column vector $$\theta=\begin{bmatrix}\theta^1\\ \vdots\\ \theta^n\end{bmatrix}.$$ The duality is expressed in matrix notation as a matrix product: $$\theta e=\begin{bmatrix}\theta^1\\ \vdots\\ \theta^n\end{bmatrix} \begin{bmatrix} e_1& \cdots & e_n\end{bmatrix}=[\theta^i(e_j)]=[\delta^i_j]=I.$$ Equation (2) translates into $$\bar{e}=[\bar{e}_1\ \cdots\ \bar{e}_n]= [e_1\ \cdots\ e_n] [a^i_j]=eA.$$

Proposition 16.7. Let $e=[e_1\ \cdots\ e_n]$ and $\bar{e}=[\bar{e}_1\ \cdots\ \bar{e}_n]$ be two frames on an open set $U$. If $\bar{e}=eA$ for $A:U\rightarrow GL(n,\mathbb{R})$, then the dual frames $\bar{\theta}$ and $\theta$ are related by $\bar{\theta}=A^{-1}\theta$. What this means that if $\bar{\theta}^i=\sum b^i_j \theta^j$, then $[b^i_j]=A^{-1}$.

Proof. Since $$(A^{-1}\theta)\bar{e}=A^{-1}\theta e A=A^{-1}IA=I,$$ we have $\bar{\theta}=A^{-1}\theta$.

Next we study how $\bar{\theta}^1\wedge \cdots\wedge \bar{\theta}^n$ is related to $\theta^1\wedge \cdots \wedge \theta^n$.

Proposition 16.8 (Wedge product under a change of frame). Let $V$ be a vector space of idmension $n$ and $\theta^1,\cdots,\theta^n$ a basis for the dual space $V^\wedge$. If the $1$-covectors $\bar{\theta}^1,\cdots,\bar{\theta}^n$ and $\theta^1,\cdots,\theta^n$ are related by $\bar{\theta}^i=\sum b^i_j\theta^j$, then with $B=[b^i_j]$, $$\bar{\theta}^1\wedge \cdots\wedge \bar{\theta}^n=(\mbox{det}B)\theta^1\wedge \cdots \wedge \theta^n.$$

16.4 Volume Form

Let $(U,x^1,\cdots,x^n)$ be an arbitrary chart on an oriented Riemannian manifold $M$. Applying the Gram-Schmidt process to the coordinate frame $\partial_1,\cdots,\partial_n$, we can construct an orthonormal fram $e_1,\cdots,e_n$ on $U$. Let $\theta^1,\cdots,\theta^n$ be the dual frame of $1$-form and define $$\omega=\theta^1\wedge \cdots \wedge \theta^n.$$ Then $\omega$ is a nowhere-vanishing $n$-form on $U$.

To see how $\omega$ depends on the choice of an orthonormal frame, let $\bar{e}_1,\cdots,\bar{e}_n$ be another orthonormal frame on $U$. Then there are $C^\infty$ functions $a^i_j:U\rightarrow \mathbb{R}$ such that $$\begin{align}\bar{e}_j=\sum a^i_je_i.\end{align}$$ Since both $e_1,\cdots,e_n$ and $\bar{e}_1,\cdots,\bar{e}_n$ are orthonormal, the change of basis matrix $A=[a^i_j]$ is a function from $U$ to the orthonormal group $O(n)$. Let $\bar{\theta}^1,\cdots,\bar{\theta}^n$ be the dual frame of $\bar{e}_1,\cdots,\bar{e}_n$. By Proposition 16.7, $\bar{\theta}^i=\sum (A^{-1})^i_j\theta^j$. It then follows from Proposition 16.8 that $$\bar{\theta}^1\wedge \cdots\wedge \bar{\theta}^n=(\mbox{det}A^{-1})\theta^1\wedge \cdots \wedge \theta^n.$$ We now assme that $e$ and $\bar{e}$ are both positively oriented. Then the matrix function $A:U\rightarrow SO(n)$ assumes values in the special orthogonal group. Since $A^{-1}\in SO(n)$, $\det (A^{-1})=1$, and so $$\bar{\theta}^1\wedge\cdots\bar{\theta}^n=\theta^1\wedge\cdots\wedge\theta^n.$$ This proves that $\omega=\theta^1\wedge\cdots\wedge\theta^n$ is independent of the choice of the positively oriented orthonormal frame. It is a canonically defined $n$-form on an oriented Riemannian manifold $M$. We call $\omega$ the volumn form of $M$. In case the integral $\int_M\omega$ is finite, we call $\int_M\omega$ the volume of the oriented Riemannian manifold $M$.

The notion of a tangent bundle extends to a manifold $M$ with boundary $\partial M$; hence so does the notion of a Riemannian metric: a Riemannian metric on a manifold $M$ with boundary is a positive-definite symmetric bilinear form on the tangent bundle $TM$. If $M$ is a Riemannian manifold with boundary, then $\partial M$ inherits naturally a Riemannian metric from $M$.

Theorem 16.11 (Volumn Form of Boundary). Let $\mbox{vol}_M$ and $\mbox{vol}_{\partial M}$ be the volume forms on an oriented Riemannian manifold $M$ and on its boundary $\partial M$. Assume that the boundary is given the boundary orientation. If $X$ is the outward unit normal along $\partial M$, then $\mbox{vol}_{\partial M}=\imath_X(\mbox{vol}_M)$, where $\imath_X$ is the interior multiplication with $X$. 

Proof. Let $p$ be a point in the boundary $\partial M$. If $(X,e_2,\cdots,e_n)$ is a positively oriented orthonormal frame field for $M$ in a neighborhood $U$ of $p$, then by the definition of the boundary orientation, $(e_2,\cdots,e_n)$ is a positively oriented orthonormal frame for $\partial M$ in $U\cap \partial M$. Let $\theta^1,\cdots,\theta^n$ be the dual frame to $X$, $e_2,\cdots,e_n$. Clearly, $\theta^2,\cdots,\theta^n$ is dual to $e_2,\cdots,e_n$ on $U\cap \partial M$. So $$\mbox{vol}_M=\theta^1\wedge \cdots\wedge \theta^n\quad \mbox{and}\quad \mbox{vol}_{\partial M} =\theta^2\wedge \cdots\wedge \theta^n.$$ At a point of $\partial M$, $$\imath_X(\mbox{vol}_M)=\imath_X (\theta^1\wedge \cdots\wedge \theta^n)=\sum_j (-1)^{j-1}\theta^j(X)\theta^1\wedge\cdots\wedge \hat{\theta^j}\wedge \cdots\wedge \theta^n=\theta^1(X)\theta^2\wedge\cdots\wedge\theta^n=\theta^2\wedge\cdots\wedge\theta^n=\mbox{vol}_{\partial M}.$$

16.5 The Volume Form in Local Coordinates

On a coordinate chart $(U,x^1,\cdots,x^n)$ of a Riemannian manifold $M$, the volume form $\mbox{vol}_M$, being a nowhere-vanishing top form, is a multiple $fdx^1\wedge\cdots\wedge dx^n$ of $dx^1\wedge\cdots\wedge dx^n$ for some nonvanishing function $f$ on $U$. The next theorem determines $f$ in terms of the Riemannian metric.

Theorem 16.13. Let $(U,x^1,\cdots,x^n)$ be a coordinate chart on a Riemannian manifold $M$, and let $g_{ij}=\langle \partial_i,\partial_j\rangle=\langle \partial/\partial x^i,\partial/\partial x^j\rangle$.  Then the volume form of $M$ on $U$ is given by $$\mbox{vol}_M=\sqrt{\det g}dx^1\wedge \cdots\wedge dx^n.$$ where $g$ is the matrix $[g_{ij}]$.

Proof. Let $e_1,\cdots,e_n$ be an orthonormal frame on $U$, with dual frame of $1$-forms $\theta^1,\cdots,\theta^n$. Then $\partial_j=\sum_i a^i_je_i$ for some matrix-valued function $A=[a^i_j]:U\rightarrow GL(n,\mathbb{R})$. The dual frame to $\partial_1,\cdots,\partial_n$ is $dx^1,\cdots,dx^n$. By Proposition 16.7, $dx^i=\sum_jb^i_j\theta^j$ with $B=[b^i_j]=A^{-1}$. By Proposition 16.8, $$dx^1\wedge \cdots\wedge dx^n=(\det B)\theta^1\wedge \cdots\wedge \theta^n.$$ Hence, $$\mbox{vol}_M=\theta^1\wedge\cdots\wedge\theta^n=(\det B)^{-1}dx^1\wedge \cdots\wedge dx^n=(\det A)dx^1\wedge\cdots\wedge dx^n.$$ To find $\det A$, note that $$g_{ij}=\langle \partial_i,\partial_j\rangle =\langle \sum a^k_ie_k,a^l_je_l\rangle =\sum a^k_ia^l_j\langle e_k,e_l\rangle =\sum a^k_i a^l_j \delta_{kl}=\sum_k a^k_ia^k_j=(A^TA)^i_j.$$ Therefore, $g=[g_{ij}]=A^TA$, so that $$\det g=\det (A^TA)=(\det A)^2.$$ It follows that $$\det A=\sqrt{\det g}$$ and $$\mbox{vol}_M=(\det A)dx^1\wedge\cdots\wedge dx^n=\sqrt{\det g}dx^1\wedge \cdots\wedge dx^n.$$