Riemannian Metrics Length Distance and Volume

Given: A covariant two-tensor and overlapping smooth coordinate systems.

Riemannian metric and riemannian manifold: A Riemannian metric on a Hausdorff second-countable smooth manifold $M$ is a smooth symmetric covariant two-tensor $g$ such that $g_p(v,v)>0$ for every point $p$ and every nonzero $v\in T_{p}M$. A Riemannian manifold is the pair $(M,g)$. This is a def-smooth-tensor-field giving a def-smooth-bundle-metric on $TM$. Dimension zero is allowed: its zero bilinear form is positive definite because there are no nonzero vectors. The empty manifold has its unique empty metric. Boundaries are allowed where stated, with smoothness understood up to the boundary.

Smoothness of a tensor field is equivalent to smooth coordinate components: A type $(r,s)$ tensor field is smooth if and only if, in every smooth chart, its coordinate component functions are smooth.

Given: A smooth manifold $M$ and countable choice.

Riemannian metric and riemannian manifold: A Riemannian metric on a Hausdorff second-countable smooth manifold $M$ is a smooth symmetric covariant two-tensor $g$ such that $g_p(v,v)>0$ for every point $p$ and every nonzero $v\in T_{p}M$. A Riemannian manifold is the pair $(M,g)$. This is a def-smooth-tensor-field giving a def-smooth-bundle-metric on $TM$. Dimension zero is allowed: its zero bilinear form is positive definite because there are no nonzero vectors. The empty manifold has its unique empty metric. Boundaries are allowed where stated, with smoothness understood up to the boundary.

Every smooth vector bundle admits a smooth bundle metric: Every smooth vector bundle admits a smooth bundle metric.

The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$): The Axiom of Countable Choice, written ${\mathrm{A}\mathrm{C}}_{\omega }$, is the following statement. > For every family $(X_n{)}_{n\in \mathbb{N}}$ of nonempty sets indexed by > $\mathbb{N}$ there is a function $f$ with domain $\mathbb{N}$ such that > $f(n)\in X_n$ for every $n\in \mathbb{N}$. Equivalently, in the vocabulary of def-choice-function: every at most countable family of nonempty sets (def-countable) has a choice function.

Tangent and cotangent bundles extend over a boundary: For a smooth $n$-manifold with boundary, derivations of smooth boundary germs form an $n$-dimensional tangent space at every point, and the usual tangent and cotangent bundles have smooth boundary-chart transition maps.

Smooth partitions of unity exist on manifolds with boundary: Assume ${\mathrm{A}\mathrm{C}}_{\omega }$. Every open cover of a smooth manifold with boundary admits a smooth partition of unity subordinate to it.

Given: A smooth map $F:M\to N$ and a Riemannian metric $h$.

Pullback of a riemannian metric as a tensor: For smooth $F:M\to N$ and a Riemannian metric $h$ on $N$, its pullback tensor is $(F^{\ast }h{)}_p(v,w)=h_{F(p)}(d{F}_{p}v,d{F}_{p}w)$. This is def-pullback-of-a-covariant-tensor-field for the tensor in def-riemannian-metric-and-riemannian-manifold. It is always symmetric and positive semidefinite; the name does not assert positive definiteness. Smoothness and the precise immersion criterion are established next.

Pullback of covariant tensors is smooth and functorial: If $F:M\to N$ is smooth and $T$ is a smooth covariant tensor field on $N$, then $F^{\ast }T$ is a smooth covariant tensor field on $M$. Moreover, $({\mathrm{id}}_M{)}^{\ast }T=T,(G\circ F{)}^{\ast }T=F^{\ast }(G^{\ast }T)$ for every composable smooth map $G$.

Immersions, submersions, and constant-rank maps: Let $F:M^m\to N^n$ be a smooth map. - $F$ is an immersion at $p\in M$ when $d{F}_p$ is injective. - $F$ is a submersion at $p\in M$ when $d{F}_p$ is surjective. - $F$ has constant rank $r$ on $A\subseteq M$ when ${\mathrm{rank}}_{p}F=r$ for every $p\in A$ (def-rank-of-a-smooth-map-at-a-point). The map is an immersion or submersion without qualification when the corresponding pointwise condition holds at every point of $M$.

Given: Isometries of $(M,g)$, and a smooth $F:(P,g_P)\to (Q,g_Q)$ with $F^{\ast }{g}_Q=g_P$ and equal dimensions for the second assertion.

Riemannian isometry and local isometry: An isometry $F:(M,g)\to (N,h)$ is a diffeomorphism with $F^{\ast }h=g$. A local isometry is a smooth local diffeomorphism with $F^{\ast }h=g$. An isometric immersion is a smooth immersion satisfying that same pullback identity. Use def-pullback-riemannian-metric and def-diffeomorphism-and-local-diffeomorphism-of-manifolds. Positivity forces injective differential by prop-pullback-of-a-riemannian-metric-is-riemannian-exactly-for-immersions. In equal dimensions on boundaryless manifolds, the inverse function theorem as applied in the next proposition makes a metric-preserving smooth map a local isometry. At a boundary the definition retains the local-diffeomorphism requirement. An isometric immersion need not have equal source and target dimensions.

Pullback of covariant tensors is smooth and functorial: If $F:M\to N$ is smooth and $T$ is a smooth covariant tensor field on $N$, then $F^{\ast }T$ is a smooth covariant tensor field on $M$. Moreover, $({\mathrm{id}}_M{)}^{\ast }T=T,(G\circ F{)}^{\ast }T=F^{\ast }(G^{\ast }T)$ for every composable smooth map $G$.

The smooth inverse function theorem on manifolds: Let $F:M\to N$ be a smooth map and let $p\in M$. If $d{F}_p:T_{p}M\to T_{F(p)}N$ is an isomorphism, then there are open neighbourhoods $U$ of $p$ and $V$ of $F(p)$ such that $F{\mathrm{\mid }}_U:U\to V$ is a diffeomorphism.

Given: A Riemannian metric with coordinate matrix $G$.

Musical isomorphisms: The musical maps for $g$ are $v^{\mathrm{♭}}=g(v,\cdot )$ and its pointwise inverse ${\alpha }^{\mathrm{♯}}$, characterized by $g({\alpha }^{\mathrm{♯}},v)=\alpha (v)$ for all $v$. For the metric in def-riemannian-metric-and-riemannian-manifold, $v^{\mathrm{♭}}=0$ forces $g(v,v)=0$ and hence $v=0$. Thus $\mathrm{♭}$ is injective between equal-dimensional fibres and bijective; this gives the pointwise inverse. Smooth inverse bundle maps are proved in thm-the-musical-maps-are-smooth-inverse-bundle-isomorphisms. On a zero fibre both are the unique map.

Coordinate criterion for a riemannian metric: A tensor $g={\sum }_{i,j}{g}_{ij}d{x}^i\otimes d{x}^j$ is Riemannian exactly when its coordinate matrix $G=(g_{ij})$ has smooth entries and is symmetric positive definite. Under $J=\mathrm{\partial }x\mathrm{/}\mathrm{\partial }y$ it transforms by $G_y=J^T{G}_{x}J$.

Smoothness of a bundle map is equivalent to smooth local matrices: Let $\mathrm{\Phi }:E\to F$ be a fibrewise linear map over a smooth base map $f:M\to N$. Choose local frames $(e_1,\dots ,e_r)$ for $E$ on $U\subseteq M$ and $(u_1,\dots ,u_s)$ for $F$ on $V\subseteq N$ with $f(U)\subseteq V$. Then $\mathrm{\Phi }$ is smooth on $E{\mathrm{\mid }}_U$ if and only if there are smooth scalar functions $a_{ji}:U\to \mathbb{R}$ such that $\mathrm{\Phi }(e_i(p))={\sum }_{j=1}^s{a}_{ji}(p)u_j(f(p))$ for every $p\in U$.

Given: A Riemannian manifold and a smooth real function $f$.

Riemannian gradient: For a smooth real function $f$, its Riemannian gradient is ${\mathrm{grad}}_{g}f=(df{)}^{\mathrm{♯}}$. prop-exterior-derivative-of-a-function-is-its-differential identifies $df(X)=Xf$. The smooth bundle isomorphism in thm-the-musical-maps-are-smooth-inverse-bundle-isomorphisms therefore makes the gradient a smooth vector field. In coordinates $({\mathrm{grad}}_{g}f{)}^i={\sum }_j{g}^{ij}{\mathrm{\partial }}_{j}f$. Constants, and all functions in dimension zero, have zero gradient.

Given: A Riemannian metric $g$ on a smooth manifold.

The musical maps are smooth inverse bundle isomorphisms: $\mathrm{♭}:TM\to T^{\ast }M$ and $\mathrm{♯}:T^{\ast }M\to TM$ are smooth inverse bundle isomorphisms.

Universal property of the finite-dimensional exterior power: Let $A:V^k\to W$ be an alternating $k$-linear map into a real vector space $W$. Then there is a unique linear map $\tilde{A}:{\bigwedge }^{k}V\to W$ such that $\tilde{A}(v_1\wedge \cdots \wedge v_k)=A(v_1,\dots ,v_k)$ for all $v_1,\dots ,v_k\in V$.

Wedge monomials in a dual basis form a basis: Let $e_1,\dots ,e_n$ be a basis of $V$, with dual basis $e^1,\dots ,e^n$. Then the wedges $e^{i_1}\wedge \cdots \wedge e^{i_k}(1\le i_1<\cdots <i_k\le n)$ form a basis of ${\mathrm{Alt}}^k(V)$.

Universal property of the tensor product for balanced maps into abelian groups: Let $R$ be a unital ring, $M$ a right $R$-module, $N$ a left $R$-module, and $\tau :M\times N⟶M{\otimes }_{R}N,\tau (m,n)=m\otimes n\mathrm{.}$ The map $\tau$ is balanced (def-balanced-and-bilinear-maps). For every abelian group $A$ and every balanced map $b:M\times N\to A$, there is a unique group homomorphism $\bar{b}:M{\otimes }_{R}N⟶A$ such that $\bar{b}(m\otimes n)=b(m,n)$ for all $m,n$. Consequently composition with $\tau$ is a bijection ${\mathrm{Hom}}_{\mathbf{A}\mathbf{b}}(M{\otimes }_{R}N,A)\cong {\mathrm{Bal}}_R(M,N;A)\mathrm{.}$

The elementary tensors of two bases form the product basis of the tensor product: Let $R$ be a commutative ring. If $M$ is free with basis $(e_i{)}_{i\in I}$ and $N$ is free with basis $(f_j{)}_{j\in J}$, then $M{\otimes }_{R}N$ is free with basis $(e_i\otimes f_j{)}_{(i,j)\in I\times J}\mathrm{.}$ Equivalently, the canonical map $R^{(I\times J)}\to M{\otimes }_{R}N$ sending the standard basis vector at $(i,j)$ to $e_i\otimes f_j$ is an isomorphism. This includes an empty basis in either factor.

Given: Two admissible subdivisions of a piecewise $C^1$ curve.

Riemannian speed and length: The Riemannian speed on a $C^1$ piece is $\mathrm{\mid }\dot{\gamma }(t){\mathrm{\mid }}_g=\sqrt{g_{\gamma (t)}(\dot{\gamma }(t),\dot{\gamma }(t))}$. Its length is $L_g(\gamma )={\sum }_j{\int }_{t_{j-1}}^{t_j}\mathrm{\mid }\dot{\gamma }(t){\mathrm{\mid }}_{g}dt$. The curve convention is def-piecewise-c-one-curve-on-a-manifold and the norm is def-pointwise-norm-and-angle-from-a-riemannian-metric. Each integrand is continuous on its closed piece with the one-sided endpoint derivative, hence Riemann integrable and nonnegative. Values chosen at the finitely many corners do not change its integral. For a singleton interval the empty sum is zero; a constant curve also has zero length. Partition independence is established next.

The piecewise-C1 line-integral sums do not depend on the admissible partition: For a piecewise-$C^1$ path $\gamma$, the scalar and vector line-integral sums in def-scalar-and-vector-line-integrals-along-piecewise-c1-paths have the same value for every admissible partition. Thus both line integrals are well-defined.

Given: The maps in the statement, with compact parameter intervals.

Riemannian length is independent of piecewise c one subdivision: Riemannian length is independent of admissible finite subdivision and of corner derivative conventions.

Substitution for a continuous inner map with a Riemann-integrable extension of its interior derivative, without monotonicity or injectivity: Let $c<d$, let $J=[p,q]$ with $p<q$, and let $f:J\to \mathbb{R}$ be continuous. Suppose $\phi :[c,d]\to J$ is continuous on $[c,d]$ and differentiable on $(c,d)$, and that the interior derivative has a Riemann-integrable extension $h:[c,d]\to \mathbb{R}$. Then $(f\circ \phi )h$ is Riemann integrable and ${\int }_c^{d}f(\phi (t))h(t)dt={\int }_{\phi (c)}^{\phi (d)}f(x)dx\mathrm{.}$ The limits on the right are oriented. No injectivity or monotonicity of $\phi$ is required; the identity also covers $\phi (c)>\phi (d)$ and $\phi (c)=\phi (d)$.

Given: Piecewise $C^1$ curves, with matching endpoints for concatenation.

Riemannian length is independent of piecewise c one subdivision: Riemannian length is independent of admissible finite subdivision and of corner derivative conventions.

Line integrals under reversal and concatenation: Let $\gamma$ be a piecewise-$C^1$ path, and let $f$ be a continuous scalar field and $F$ a continuous vector field on a set containing its trace. Then ${\int }_{{\gamma }^{-}}fds={\int }_{\gamma }fds,{\int }_{{\gamma }^{-}}F\cdot d\mathbf{r}=-{\int }_{\gamma }F\cdot d\mathbf{r}\mathrm{.}$ If piecewise-$C^1$ paths $\alpha ,\beta :[0,1]\to {\mathbb{R}}^n$ satisfy $\alpha (1)=\beta (0)$, and $f$ and $F$ are continuous on a set containing both traces, then ${\int }_{\alpha \ast \beta }fds={\int }_{\alpha }fds+{\int }_{\beta }fds,$ ${\int }_{\alpha \ast \beta }F\cdot d\mathbf{r}={\int }_{\alpha }F\cdot d\mathbf{r}+{\int }_{\beta }F\cdot d\mathbf{r}\mathrm{.}$

Given: A compact set $K$ in a single chart.

Coordinate criterion for a riemannian metric: A tensor $g={\sum }_{i,j}{g}_{ij}d{x}^i\otimes d{x}^j$ is Riemannian exactly when its coordinate matrix $G=(g_{ij})$ has smooth entries and is symmetric positive definite. Under $J=\mathrm{\partial }x\mathrm{/}\mathrm{\partial }y$ it transforms by $G_y=J^T{G}_{x}J$.

A product of finitely many compact spaces is compact in the product topology: For every $n\in \mathbb{N}$ (def-natural-numbers) and every family $(X_k{)}_{k<n}$ of compact topological spaces (def-compact-space, def-topological-space), the product ${\prod }_{k<n}{X}_k$ with the product topology (def-product-topology) is compact. In particular a binary product $X\times Y$ of compact spaces is compact, and the empty product, a one-point space, is compact. No choice principle is used beyond lem-finite-choice, which is a theorem of ZF. That is what separates the finite case from the arbitrary one, where the Axiom of Choice is genuinely spent.

A subset of ${\mathbb{R}}^n$ with the product topology is compact exactly when it is closed and bounded, the product topology being the Euclidean metric topology: Let $n\in \mathbb{N}$ with $n\ge 1$, let ${\mathbb{R}}^n$ be the set of functions $n\to \mathbb{R}$ (lem-metrics-on-rn) carrying the product topology of $n$ copies of the usual topology of $\mathbb{R}$ (def-product-topology), and let $d_2$ be the Euclidean metric. Then: 1. The product topology on ${\mathbb{R}}^n$ is the metric topology of $d_2$ (def-metric-topology), so ${\mathbb{R}}^n$ as a product and ${\mathbb{R}}^n$ as a metric space are one topological space, and it is metrizable (def-metrizable-space). 2. A subset $K\subseteq {\mathbb{R}}^n$ is a compact subset for the product topology (def-compact-space) if and only if $K$ is closed in ${\mathbb{R}}^n$ and bounded (def-metric-bounded-diameter). The hypothesis $n\ge 1$ is inherited from lem-metrics-on-rn, which defines ${\mathbb{R}}^n$ and its three metrics only there; for $n=0$ the product is a one-point space and is compact. No choice principle is used: the metric statement it is read off from is proved by bisection (thm-heine-borel-rn).

A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value: Let $(X,d)$ be a nonempty compact metric space (def-metric-compactness, def-metric-space) and let $f:X\to \mathbb{R}$ be continuous (def-metric-continuity), $\mathbb{R}$ carrying its usual metric $d_{\mathbb{R}}(s,t)=\mathrm{\mid }s-t\mathrm{\mid }$ (lem-real-line-is-a-metric-space). Then the image $f[X]$ is bounded above and below (def-bounded-set), and it has a maximum and a minimum (def-max-min): there are points $x_{\max },x_{\min }\in X$ with $f(x_{\min })\le f(x)\le f(x_{\max })\text{for every }x\in X,$ and then $f(x_{\max })=\sup f[X]$ and $f(x_{\min })=\inf f[X]$ (def-complete-ordered-field, def-infimum). Nonemptiness of $X$ is a hypothesis and not an oversight: for $X=\mathrm{\varnothing }$ the image is empty and has neither a supremum nor a maximum. No choice principle is used.

For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide: Let $(X,d)$ be a metric space (def-metric-space) and let ${\mathcal{T}}_d$ be its metric topology (def-metric-topology), so that $(X,{\mathcal{T}}_d)$ is a topological space (def-topological-space) and is metrizable (def-metrizable-space). Then: 1. $(X,d)$ is a compact metric space (def-metric-compactness) if and only if $(X,{\mathcal{T}}_d)$ is a compact topological space (def-compact-space). 2. For every $A\subseteq X$: $A$ is a compact subset of the metric space $(X,d)$ if and only if $A$ is a compact subset of the topological space $(X,{\mathcal{T}}_d)$, the two readings of "compact subset" being the metric subspace $(A,d_A)$ (def-isometry-and-metric-embedding) and the topological subspace $(A,({\mathcal{T}}_d{)}_A)$ (def-subspace-topology-top). Nothing here is a coincidence and nothing is transported. The open-cover condition of def-metric-compactness quantifies over families of subsets open in $(X,d)$, and by def-metric-topology those are exactly the members of ${\mathcal{T}}_d$; so the two conditions are not merely equivalent, they are the same condition written twice. No choice principle is used.

Given: A connected nonempty smooth manifold and points $p,q$.

Piecewise c one curve on a manifold: A piecewise $C^1$ curve in $M$ is a continuous map $\gamma :[a,b]\to M$ with a finite subdivision such that its restriction to each closed piece is $C^1$ in local charts, with one-sided derivatives at piece endpoints. Use the chartwise regularity convention of def-c-r-and-smooth-maps-between-smooth-manifolds and the finite path operations of def-piecewise-c1-path-operations-and-oriented-reparametrizations. Refining a piece into finitely many chart pieces is allowed. No nonzero-velocity hypothesis is imposed: constant segments and pauses are admissible. A singleton parameter interval is interpreted as a constant curve of length zero.

Separation of a topological space, connected and disconnected spaces, clopen sets, and connected subsets: Let $(X,\mathcal{T})$ be a topological space (def-topological-space). - A separation of $X$ is an ordered pair $(U,V)$ of open, nonempty, disjoint subsets of $X$ with $U\cup V=X$. - $X$ is disconnected when a separation of $X$ exists, and connected when none does. - A subset $A\subseteq X$ is a connected subset of $X$ when the space $(A,{\mathcal{T}}_A)$ is connected, ${\mathcal{T}}_A$ being the subspace topology (def-subspace-topology-top). "Disconnected subset" is read the same way. Since $U$ and $V$ are complementary in $X$, each of them is closed as well as open; so a separation is the same thing as a partition of $X$ into two nonempty clopen pieces (def-topological-space). The clopen subsets of $X$ are those that are both open and closed, and $\mathrm{\varnothing }$ and $X$ are always among them. The empty space and the one-point space are connected in this library. Neither admits a separation: a separation requires two nonempty disjoint sets whose union is the whole space, and neither $\mathrm{\varnothing }$ nor a singleton can be written as such a union. So both are connected under the definition above, without any special clause. This is a live convention fork and the competing choice is recorded in rem-connectedness-conventions; nothing on this page depends on which is taken except the reading of the word "connected" applied to those two spaces. Connectedness is a property of a space, not of an ambient pair. The condition above mentions only $(X,\mathcal{T})$. When it is applied to $A\subseteq X$ it is applied to the space $(A,{\mathcal{T}}_A)$, so it does not change if $A$ is regarded as a subspace of some other space inducing the same topology on $A$; in particular a subset of $A$ is connected as a subset of $A$ exactly when it is connected as a subset of $X$, by transitivity of the subspace topology (def-subspace-topology-top). This is why "connected" may be used of a subset with no ambient space named. Spelled out for a subset. $A\subseteq X$ is disconnected exactly when there are open $U,V\subseteq X$ with $A\subseteq U\cup V,U\cap A\mathrm{\ne }\mathrm{\varnothing },V\cap A\mathrm{\ne }\mathrm{\varnothing },U\cap V\cap A=\mathrm{\varnothing },$ because the open sets of $(A,{\mathcal{T}}_A)$ are precisely the traces $U\cap A$. Note the last condition: it asks $U$ and $V$ to be disjoint on $A$, not in $X$. Requiring $U\cap V=\mathrm{\varnothing }$ outright is a strictly stronger demand and is a different notion. The two-point discrete space. Write $\mathbf{2}:=\{0,1\}$ with the discrete topology (def-standard-topologies), in which every subset is open. A separation of $X$ is the same datum as a surjective continuous map $X\to \mathbf{2}$ (def-continuous-map-top): given $(U,V)$, the map sending $U$ to $0$ and $V$ to $1$ is continuous because the preimage of each of the four open subsets of $\mathbf{2}$ is one of $\mathrm{\varnothing }$, $U$, $V$, $X$; given a surjective continuous $\chi :X\to \mathbf{2}$, the pair $({\chi }^{-1}[\{0\}],{\chi }^{-1}[\{1\}])$ is a separation. This reformulation is proved as a theorem on this page and is recorded here only to name $\mathbf{2}$. Separated sets. Two subsets $A_1,A_2\subseteq X$ are separated in $X$ when $\overline{A_1}\cap A_2=\mathrm{\varnothing }\text{and}{A}_1\cap \overline{A_2}=\mathrm{\varnothing },$ closures taken in $X$ (def-interior-closure-boundary-top, thm-closure-characterisation-top). Separated sets are disjoint, since $A_1\subseteq \overline{A_1}$; the converse fails. This is verbatim the condition def-connected-r uses on the real line, transported to an arbitrary space, and the theorem relating it to the definition above is the next lemma on this page. Totally disconnected spaces, and the empty case. The vocabulary for a space all of whose connected subsets are single points is fixed later on this page, together with the components; it is not defined here because it is stated in terms of components.

Given: A connected Riemannian manifold, with its infimum distance.

Riemannian distance on a connected manifold: On a connected Riemannian manifold define $d_g(p,q)=\inf \{L_g(\gamma ):\gamma \text{ is piecewise }{C}^1\text{ from }p\text{ to }q\}$. Lengths are those of def-riemannian-speed-and-length. For each pair $p,q$, lem-any-two-points-in-a-connected-smooth-manifold-can-be-joined-by-a-piecewise-c-one-curve supplies a curve, so the set of lengths is nonempty, contains a finite real number and is bounded below by zero. Applying the least-upper-bound property cor-cauchy-reals-lub-complete to the negatives gives a finite nonnegative infimum. On the empty connected manifold this defines the empty distance function; there are no pairs to evaluate. No minimizing curve is part of this definition.

Length is additive under concatenation and invariant under reversal: Length adds under finite concatenation and is unchanged by reversal.

Local comparison of a riemannian metric with the euclidean metric: For a compact set $K$ contained in one coordinate chart of an $n$-dimensional Riemannian manifold, there are $0<c\le C<\mathrm{\infty }$ such that $c\mathrm{\mid }v{\mathrm{\mid }}^2\le g_x(v,v)\le C\mathrm{\mid }v{\mathrm{\mid }}^2$ for $x\in K$. The dimension-zero assertion is vacuous.

Line-integral estimates by arc length and the supremum of the field: Let $\gamma$ be a piecewise-$C^1$ path of length $L(\gamma )$, let $f$ be a continuous scalar field and $F$ a continuous vector field on its trace, and let $M\ge 0$. 1. If $\mathrm{\mid }f(x)\mathrm{\mid }\le M$ on the trace of $\gamma$, then $\mid {\int }_{\gamma }fds\mid \le ML(\gamma )\mathrm{.}$ 2. If $\parallel F(x){\parallel }_2\le M$ on the trace of $\gamma$, then $\mid {\int }_{\gamma }F\cdot d\mathbf{r}\mid \le ML(\gamma )\mathrm{.}$

Newton–Leibniz needs only continuity on $[a,b]$, differentiability on $(a,b)$, and a Riemann-integrable extension of the interior derivative: Let $a<b$. Suppose $G:[a,b]\to \mathbb{R}$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f:[a,b]\to \mathbb{R}$ is Riemann integrable and $f(x)=G^{\mathrm{\prime }}(x)(a<x<b),$ then ${\int }_a^{b}f=G(b)-G(a)\mathrm{.}$ No derivative of $G$ at either endpoint is assumed, and the two endpoint values assigned to the integrable extension $f$ do not enter the conclusion.

Given: A point $p$ of a connected Riemannian manifold.

Riemannian distance is a metric: $d_g$ is a finite metric on a connected Riemannian manifold.

Local comparison of a riemannian metric with the euclidean metric: For a compact set $K$ contained in one coordinate chart of an $n$-dimensional Riemannian manifold, there are $0<c\le C<\mathrm{\infty }$ such that $c\mathrm{\mid }v{\mathrm{\mid }}^2\le g_x(v,v)\le C\mathrm{\mid }v{\mathrm{\mid }}^2$ for $x\in K$. The dimension-zero assertion is vacuous.