Metric geometry on manifolds:
two lectures

Let $M$ be a smooth connected manifold. A metric tensor on $M$ is a choice of positive definite quadratic forms $g_{p}$ on each tangent space $\mathrm{T}_{p}M$ that depends continuously on the point; that is, in any local coordinates of $M$ the components of $g$ are continuous functions.

A Riemannian manifold $(M,g)$ is a smooth manifold $M$ with a choice of metric tensor $g$ on it.

The $g$-length of a Lipschitz curve $\gamma\colon[a,b]\to M$ is defined by

$$\mathop{\rm length}\nolimits_{g}\gamma=\int_{a}^{b}\sqrt{g(\gamma^{\prime}(t),\gamma^{\prime}(t))}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}dt.$$

The $g$-length induces a metric metric on $M$; it is defined as the greatest lower bound to lengths of Lipschitz curves connecting two given points; the distance between a pair of points $x,y\in M$ will be denoted by

$$|{x}-\nobreak{y}|_{g}\quad\text{or}\quad\mathrm{dist}_{x}(y)_{g}.$$

The corresponding metric space $\mathcal{M}$ will be called Riemannian space.

Moreover, there is a continuous Riemannian metric $g$ on $\mathbb{R}^{2}$ such that the corresponding Riemannian space admits an isometry to the Euclidean palne but the induced map $\iota\colon\mathbb{R}^{2}\to\mathbb{R}^{2}$ is not differentiable at some point.

The exercise above shows that in general the smooth structure is not uniquely defined on Riemannian space. Therefore in general case one has to distinguish between Riemannian manifold and the corresponding Riemannian space altho there is almost no difference.¹¹1In fact a straightforward smoothing procedure shows that isometry between Riemannian spaces can be approximated by diffeomorphisms between underlying manifolds; in particular these manifolds are diffeomorphic. Also, if the metric tensor is smooth, then it is not hard to show that Riemannian space remembers everything about the Riemannian manifold, in particular the smooth structure; it is a part of the so-called Myers–Steenrod theorem [16].

The following observation states the key property of Riemannian spaces; it will be used to extend results from Euclidean space to Riemannian spaces.

Since the metric tensor is continuous, the restriction of $s\circ L$ to a small neighborhood of $x$ is $e^{\mp\varepsilon}$-bilipschitz. ∎

Let $(M,g)$ be an $n$-dimensional Riemannian manifold. If a Borel set $R\subset M$ is covered by one chart $\iota\colon U\to M$, then its volume (briefly, $\operatorname{\rm vol}\nolimits R$ or $\operatorname{\rm vol}\nolimits_{n}R$) is defined by

$$\operatorname{\rm vol}\nolimits R\mathrel{:=}\int_{\iota^{-1}(R)}\sqrt{\det{g}}.$$

In the general case we can subdivide $R$ into a countable collection of regions $R_{1},R_{2}\dots$ such that each region $R_{i}$ is covered by one chart $\iota_{i}\colon U_{i}\to M$ and define

$$\operatorname{\rm vol}\nolimits R\mathrel{:=}\operatorname{\rm vol}\nolimits R_{1}+\operatorname{\rm vol}\nolimits R_{2}+\dots$$

The chain rule for multiple integrals implies that the right-hand side does not depend on the choice of subdivision and the choice of charts.

Similarly, we define integral along $(M,g)$. Any Borel function $u\colon M\to\nobreak\mathbb{R}$, can be presented as a sum $u_{1}+u_{2}+\cdots$ such that the support of each function $u_{i}$ can be covered by one chart $\iota_{i}\colon U_{i}\to M$ and set

$$\int_{p\in\mathcal{M}}u(p)\mathrel{:=}\sum_{i}\left[\int_{x\in U_{i}}u_{i}\circ s(x){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sqrt{\det{g}}\right].$$

In particular

$$\operatorname{\rm vol}\nolimits R=\int_{p\in R}1.$$

Let $\mathcal{X}$ be a metric space and $R\subset\mathcal{X}$. The $\alpha$-dimensional Hausdorff measure of $R$ is defined by

$$\mathrm{haus}_{\alpha}R\mathrel{:=}\lim_{\varepsilon\to 0}\,\inf\left\{\,{\sum_{n\in\mathbb{N}}(\operatorname{\rm diam}\nolimits A_{n})^{\alpha}}\,:\,{\begin{aligned} &\operatorname{\rm diam}\nolimits A_{n}<\varepsilon\ \text{for}\\ &\text{for each}\ n,\text{all}\ A_{n}\\ &\text{are closed, and}\\ &\bigcup_{n\in\mathbb{N}}A_{n}\supset R.\end{aligned}}\,\right\}.$$

For properties of Hausdorff measure we refer to the classical book of Herbert Federer [12]; in particular, $\mathrm{haus}_{\alpha}$ is indeed a measure and $\mathrm{haus}_{\alpha}$-measurable sets include all Borel sets.

The following observation follows from 1A and Rademacher’s theorem:

According to Haar’s theorem, a measure on $n$-dimensional Euclidean space that is invariant with respect to parallel translations is proportional to volume. Observe that

• $\diamond$

  A ball in $n$-dimensional Euclidean space of diameter $1$ has unit Hausdorff measure.

• $\diamond$

  A unit cube in $n$-dimensional Euclidean space has unit volume.

Therefore, for any Borel region $R\subset\mathbb{E}^{n}$, we have

$$\operatorname{\rm vol}\nolimits_{n}R=\tfrac{\omega_{n}}{2^{n}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathrm{haus}_{n}R,$$

➊

where $\omega_{n}$ denotes the volume of a unit ball in the $n$-dimensional Euclidean space.

Applying ➊ ‣ 1B together with 1B, we get that the inequalities

$$\operatorname{\rm vol}\nolimits_{n}R\lessgtr e^{\pm 2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}n{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\varepsilon}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\tfrac{\omega_{n}}{2^{n}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathrm{haus}_{n}R$$

hold for any $\varepsilon>0$. Since $\varepsilon>0$ is arbitrary, we get that ➊ ‣ 1B holds in $n$-dimensional Riemannian spaces. More precisely:

Since the Hausdorff measure is defined in pure metric terms, the proposition gives another way to prove that the volume does not depend on the choice of chars and subdivision of $R$.

The identity in this proposition will be used to define volume of any dimension. Namely, given an integer $k\geqslant 0$, the $k$-volume is defined by

$$\operatorname{\rm vol}\nolimits_{k}\mathrel{:=}\tfrac{\omega_{k}}{2^{k}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathrm{haus}_{k}.$$

By 1B, if $A$ is a subset of $k$-dimensional submanifold $\mathcal{N}\subset\mathcal{M}$, then the two definitions of $\operatorname{\rm vol}\nolimits_{k}A$ agree; but the latter definition works for a wider class of sets.

• (a)

  $f$ is injective; that is, if $f(x)=f(y)$, then $x=y$.

• (b)

  For any $c<1$, the map $f$ is locally $[c,1]$-bilipschitz; that is, for any point in $\mathcal{M}$ there is a neighborhood $\Omega$ and $\varepsilon>0$ such that the inequality

  $$c\leqslant\frac{|f(x)-f(y)|_{\mathcal{N}}}{|x-y|_{\mathcal{M}}}\leqslant 1$$

  holds for any pair of distinct points $x,y\in\Omega$.

Suppose that $f\colon\mathcal{M}\to\mathcal{N}$ is a Lipschitz map between $n$-dimensional Riemannian spaces $\mathcal{M}$ and $\mathcal{N}$. Then by Rademacher’s theorem the differential $d_{p}f\colon\mathrm{T}_{p}\mathcal{M}\to\mathrm{T}_{f(p)}\mathcal{N}$ is defined at almost all $p\in\mathcal{M}$; that is, the differential defined at all points $p\in\mathcal{M}$ with exception of a subset with vanishing volume.

The differential is a linear map; it defines the Jacobian matrix $\mathrm{Jac}_{p}f$ in orthonormal frames of $\mathrm{T}_{p}$ and $\mathrm{T}_{f(p)}\mathcal{N}$. The determinant of $\mathrm{Jac}_{p}f$ will be denoted by $\mathrm{jac}_{p}$. Note that the absolute value $|\mathrm{jac}_{p}|$ does not depend on the choice of the orthonormal frames.

The identity in the following proposition is called area formula.

Note that Jacobian of a $e^{\mp\varepsilon}$-bilipschitz map between $n$-dimensional Riemannian manifolds (if defined) has determinant in the range $e^{\mp n{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\varepsilon}$. Applying 1B and the area formula in $\mathbb{E}^{n}$, we get the following approximate version of the needed identity for any $u\geqslant 0$:

$$\int_{p\in\mathcal{M}}u(p){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}|\mathrm{jac}_{p}f|\lessgtr e^{\pm 3{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}n{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\varepsilon}\int_{q\in\mathcal{N}}\sum_{p\in f^{-1}(q)}u(p).$$

Since $\varepsilon>0$ is arbitrary, we get that the area formula holds if $u\geqslant 0$. Finally, since both sides of the area formula are linear in $u$, it holds for any $u$. ∎

The following inequality is called area inequality:

In particular, if $|\mathrm{jac}_{p}f|\leqslant 1$ almost everywhere in $A$, then

$$\operatorname{\rm vol}\nolimits A\geqslant\operatorname{\rm vol}\nolimits[f(A)].$$

Suppose that $f\colon\mathcal{M}\to\mathbb{R}$ is a Lipschitz function defined on an $n$-dimensional Riemannian space $\mathcal{M}$. Then by Rademacher’s theorem, the differential $d_{p}f\colon\mathrm{T}_{p}\mathcal{M}\to\mathbb{R}$ and the gradient $\nabla_{p}f\in\mathrm{T}_{p}\mathcal{M}$ are defined at almost all $p\in\mathcal{M}$.

The identity in the following proposition is a partial case of the so-called coarea formula. (The general coarea formula deals with the maps to the spaces of arbitrary dimension, not necessary $1$.)

The following corollary is a partial case of the so-called coarea inequality;

Suppose that $f$ is 1-Lipschitz. Then for any Borel subset $A\subset M$ we have

$$\operatorname{\rm vol}\nolimits_{n}A\geqslant\int_{x\in\mathbb{R}}\operatorname{\rm vol}\nolimits_{n-1}[A\cap L_{x}]{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}dx.$$

➊

The right-hand side in ➊ ‣ 1C is called coarea of the restriction $f|_{A}$.

To prove the corollary, choose $u$ to be the characteristic function of $A$ and apply the coarea formula. ∎

A closed connected region in a Riemannian manifold bounded by hypersurface will be called Riemannian manifold with boundary. We always assume that the hypersurface can be realized locally as a graph of Lipschitz function in a suitable chart. In this case one can define $g$-length, $g$-distance, and $g$-volume the same way as we did for usual Riemannian manifolds.

This is a partial case of the theorem proved by Abram Besicovitch [6].

Denote by $A$, $A^{\prime}$, and $B$, $B^{\prime}$ the opposite faces of the square $\square$. Consider two functions

	$\displaystyle f_{A}(x)$	$\displaystyle\mathrel{:=}\min\{\,\mathrm{dist}_{A}(x)_{g},1\,\},$
	$\displaystyle f_{B}(x)$	$\displaystyle\mathrel{:=}\min\{\,\mathrm{dist}_{B}(x)_{g},1\,\}.$

Let $\bm{f}\colon\square\to\square$ be the map with coordinate functions $f_{A}$ and $f_{B}$; that is, $\bm{f}(x)\mathrel{:=}(f_{A}(x),f_{B}(x))$.

Indeed,

$$x\in A\quad\Longrightarrow\quad\mathrm{dist}_{A}(x)_{g}=0\quad\Longrightarrow\quad f_{A}(x)=0\quad\Longrightarrow\quad\bm{f}(x)\in A.$$

Similarly, if $x\in B$, then $\bm{f}(x)\in B$. Further,

$$x\in A^{\prime}\quad\Longrightarrow\quad\mathrm{dist}_{A}(x)_{g}\geqslant 1\quad\Longrightarrow\quad f_{A}(x)=1\quad\Longrightarrow\quad\bm{f}(x)\in A^{\prime}.$$

Similarly, if $x\in B^{\prime}$, then $\bm{f}(x)\in B^{\prime}$.

By 1D, it follows

$$\bm{f}_{t}(x)=t{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}x+(1-t){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\bm{f}(x)$$

defines a homotopy of maps of the pair of spaces $(\square,\partial\square)$ from $\bm{f}$ to the identity map; that is, $(t,x)\mapsto\bm{f}_{t}(x)$ is a continuous map and if $x\in\partial\square$, then $\bm{f}_{t}(x)\in\partial\square$ for any $t\in[0,1]$.

It follows that $\deg\bm{f}=1$; that is, $\bm{f}$ sends the fundamental class of $(\square,\partial\square)$ to itself.²²2Here and further, we assume that homologies are taken with the coefficients in $\mathbb{Z}_{2}$, but you are welcome to play with other coefficients. In particular $\bm{f}$ is onto.

Suppose that Jacobian matrix $\mathrm{Jac}_{p}\bm{f}$ of $\bm{f}$ is defined at $p\in\square$. Choose an orthonormal frame in $\mathrm{T}_{p}$ with respect to $g$ and the standard frame in the target $\square$. Observe that the differentials $d_{p}f_{A}$ and $d_{p}f_{B}$ written in these frames are the rows of $\mathrm{Jac}_{p}\bm{f}$. Evidently $|d_{p}f_{A}|\leqslant 1$ and $|d_{p}f_{B}|\leqslant 1$. Since the determinant of a matrix is the volume of the parallelepiped spanned on its rows, we get

$$|\mathrm{jac}_{p}\bm{f}|\leqslant|d_{p}f_{A}|{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}|d_{p}f_{B}|\leqslant 1.$$

Since $\bm{f}\colon\square\to\square$ is a Lipschitz onto map, the area inequality (1C) implies that

$$\operatorname{\rm vol}\nolimits(\square,g)\geqslant\operatorname{\rm vol}\nolimits\square=1.$$

∎

If the $g$-distances between the opposite sides are $d_{1},\dots,d_{n}$, then following the same lines one get that $\operatorname{\rm vol}\nolimits(\square,g)\geqslant d_{1}\cdots d_{n}$. Also note that in the proof we use topology of the $n$-cube only once, to show that the map $f$ has degree one. Taking all this into account we get the following generalization of 1D:

• (a)

  Use Besicovitch inequality to show that

  $$\operatorname{\rm area}\nolimits(\mathbb{S}^{1}\times[0,1],g)\geqslant\tfrac{1}{2}.$$

• (b)

  Modify the proof of Besicovitch inequality using coarea inequality (1C) to prove the optimal bound

  $$\operatorname{\rm area}\nolimits(\mathbb{S}^{1}\times[0,1],g)\geqslant 1.$$

• (a)

  Generalize 1D to noncontinuous metric tensor $g$ described the following way: there are two Riemannian metric tensors $g_{1}$ and $g_{2}$ on $M$ and a subset $V\subset M$ bounded by a Lipschitz hypersurface $\Sigma$ such that $g=g_{1}$ at the points in $V$ and $g=g_{2}$ otherwise.

• (b)

  Use part 1D to prove the following: Let $V$ be a compact set in the $n$-dimensional Euclidean space $\mathbb{E}^{n}$ bounded by a Lipschitz hypersurface $\Sigma$. Suppose $g$ is a Riemannian metric on $V$ such that

  $$|{p}-\nobreak{q}|_{g}\geqslant|{p}-\nobreak{q}|_{\mathbb{E}^{n}}$$

  for any two points $p,q\in\Sigma$. Show that

  $$\operatorname{\rm vol}\nolimits(V,g)\geqslant\operatorname{\rm vol}\nolimits(V)_{\mathbb{E}^{n}}.$$

Show that

$$\operatorname{\rm area}\nolimits(\mathbb{S}^{2},g)\geqslant\tfrac{1}{1000}.$$

Try to show that

$$\operatorname{\rm area}\nolimits(\mathbb{S}^{2},g)\geqslant\tfrac{1}{2},\quad\operatorname{\rm area}\nolimits(\mathbb{S}^{2},g)\geqslant 1,\quad\text{or}\quad\operatorname{\rm area}\nolimits(\mathbb{S}^{2},g)\geqslant\tfrac{4}{\pi}$$

Show that the inequality might be strict.

Let $\mathcal{M}$ be a compact Riemannian space. The systole of $\mathcal{M}$ (briefly $\mathop{\rm sys}\nolimits\mathcal{M}$) is defined to be the least length of a noncontractible closed curve in $\mathcal{M}$.

Let $\Lambda$ be a class of closed $n$-dimensional Riemannian spaces. We say that a systolic inequality holds for $\Lambda$ if there is a constant $c$ such that

$$\mathop{\rm sys}\nolimits\mathcal{M}\leqslant c{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sqrt[n]{\operatorname{\rm vol}\nolimits\mathcal{M}}$$

for any $\mathcal{M}\in\Lambda$.

In the following lecture we will show that systolic inequality holds for many manifolds, in particular for torus of arbitrary dimension.

The following proposition follows immediately from the definitions of Hausdorff measure (Section 1B).

The following exercise provides a weak analog of the Besicovitch inequality that works for arbitrary metrics.

Recall that in $n$-dimensional Riemannian spaces we have

$$\tfrac{\omega_{n}}{2^{n}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\mathrm{haus}_{n}=\operatorname{\rm vol}\nolimits_{n}.$$

Note that $\tfrac{\omega_{n}}{2^{n}}<1$ if $n\geqslant 2$. Therefore, the conclusion in 1F is weaker than in 1D (the assumptions are weaker as well).

One can redefine systolic inequality on $n$-dimensional manifolds using the Hausdorff measure $\mathrm{haus}_{n}$ instead of the volume. It is straightforward to prove analogs of the exercises 1E–1E with this definition.

The optimal constants in the systolic inequality are known only in the following three cases:

• $\diamond$

  For real projective plane $\mathbb{R}\mathrm{P}^{2}$ the constant is $\sqrt{\pi/2}$ — the equality holds for a quotient of a round sphere by isometric involution. The statement was proved by Pao Ming Pu [20].

• $\diamond$

  For torus $\mathbb{T}^{2}$ the constant is $\sqrt{2}/\sqrt[4]{3}$ — the equality holds for a flat torus obtained from a regular hexagon by identifying opposite sides; this is the so-called Loewner’s torus inequality.

• $\diamond$

  For the Klein bottle $\mathbb{R}\mathrm{P}^{2}\#\mathbb{R}\mathrm{P}^{2}$ the constant is $\sqrt{\pi}/2^{3/4}$ — the equality holds for a certain nonsmooth metric. The statement was proved by Christophe Bavard [2].

The proofs of these results use the so-called uniformization theorem available in the 2-dimensional case only. These proofs are beautiful, but they are too far from metric geometry. A good survey on the subject is written by Christopher Croke and Mikhail Katz [11].

An analog of Exercise 1D with Hausdorff measure instead of volume does not hold for general metrics on a manifold. In fact there is a nondecreasing sequence of metric tensors $g_{n}$ on $M$, such that (1) $\operatorname{\rm vol}\nolimits(M,g_{n})<1$ for any $n$ and (2) $\mathrm{dist}_{g_{n}}$ converges to a metric on $M$ with arbitrary large Hausdorff measure of any given dimension; such examples were constructed by Dmitri Burago, Sergei Ivanov, and David Shoenthal [7].

A collection of functions $\{\psi_{i}\}$ that meets the conditions in 2A is called a partition of unity subordinate to the covering $\{V_{i}\}$.

Set

$$\psi_{k}(x)=\frac{\varphi_{k}(x)}{\Phi(x)}.$$

Observe that by construction the functions $\psi_{i}$ meet the conditions in the proposition. ∎

Let $\{V_{1},\dots,V_{k}\}$ be a finite open cover of a compact metric space $\mathcal{X}$. Consider an abstract simplicial complex $\mathcal{N}$, with one vertex $v_{i}$ for each set $V_{i}$ such that a simplex with vertices $v_{i_{1}},\dots,v_{i_{m}}$ is included in $\mathcal{N}$ if the intersection $V_{i_{1}}\cap\dots\cap V_{i_{m}}$ is nonempty.

The obtained simplicial complex $\mathcal{N}$ is called the nerve of the covering $\{V_{i}\}$. Evidently $\mathcal{N}$ is a finite simplicial complex — it is a subcomplex of a simplex with the vertices $\{v_{1},\dots,v_{k}\}$.

Note that the nerve $\mathcal{N}$ has dimension at most $n$ if and only if the covering $\{V_{1},\dots,V_{k}\}$ has multiplicity at most $n+1$; that is, any point $x\in\mathcal{X}$ belongs to at most $n+1$ sets of the covering.

Suppose $\{\psi_{i}\}$ is a partition of unity subordinate to the covering $\{V_{1},\dots,V_{k}\}$. Choose a point $x\in{\mathcal{X}}$. Note that the set

$$\{v_{i_{1}},\dots,v_{i_{n}}\}=\left\{\,{v_{i}}\,:\,{\psi_{i}(x)>0}\,\right\}$$

form vertices of a simplex in $\mathcal{N}$. Therefore

$$\bm{\psi}\colon x\mapsto\psi_{1}(x){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}v_{1}+\psi_{2}(x){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}v_{2}+\dots+\psi_{k}(x){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}v_{n}.$$

describes a Lipschitz map from ${\mathcal{X}}$ to the nerve $\mathcal{N}$ of $\{V_{i}\}$. In other words, $\bm{\psi}$ maps a point $x$ to the point in $\mathcal{N}$ with barycentric coordinates $(\psi_{1}(x),\dots,\psi_{k}(x))$.

Recall that the star of a vertex $v_{i}$ (briefly $\mathop{\rm Star}\nolimits_{v_{i}}$) is defined as the union of the interiors of all simplicies that have $v_{i}$ as a vertex. Recall that $\psi_{i}(x)>0$ implies $x\in V_{i}$. Therefore we get the following:

Then there is a Lipschitz map $\bm{\psi}\colon\mathcal{X}\to\mathcal{N}$ such that $\bm{\psi}(V_{i})\subset\nobreak\mathop{\rm Star}\nolimits_{v_{i}}$ for every $i$.

Suppose $A$ is a subset of a metric space $\mathcal{X}$. The radius of $A$ (briefly $\mathop{\rm rad}\nolimits A$) is defined as the least upper bound on the values $R>0$ such that $\mathrm{B}{}(x,R)\supset A$ for some $x\in\mathcal{X}$.

• $\diamond$

  Observe that

  $$\operatorname{\rm width}\nolimits_{0}\mathcal{X}\geqslant\operatorname{\rm width}\nolimits_{1}\mathcal{X}\geqslant\operatorname{\rm width}\nolimits_{2}\mathcal{X}\geqslant\dots$$

  for any compact metric space $\mathcal{X}$. Moreover, if $\mathcal{X}$ is connected, then

  $$\operatorname{\rm width}\nolimits_{0}\mathcal{X}=\mathop{\rm rad}\nolimits\mathcal{X}.$$

• $\diamond$

  Usually width is defined using diameter instead of radius, but the results differ at most twice. Namely, if $r$ is the $n$-th radius-width and $d$ — the $n$-th diameter-width, then $r\leqslant d\leqslant 2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}r$.

• $\diamond$

  Note that Lebesgue covering dimension of $\mathcal{X}$ can be defined as the least number $n$ such that $\operatorname{\rm width}\nolimits_{n}\mathcal{X}=0$.

• $\diamond$

  Another closely related notion is the so-called macroscopic dimension on scale $R$; it is defined as the least number $n$ such that $\operatorname{\rm width}\nolimits_{n}\mathcal{X}<R$.

What about quasiconverse? That is, suppose a simply connected compact metric space $\mathcal{X}$ has macroscopic dimension at most 1 on scale $R$, is it true that any closed curve $\gamma$ in $\mathcal{X}$ can be contracted in its $100{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}R$-neighborhood?

The following exercise gives a good reason for the choice of term width; it also can be used as an alternative definition.

A Riemannian polyhedron is defined as a finite simplicial complex with a metric tensor on each simplex such that the restriction of the metric tensor to a subsimplex coincides with the metric on the subsimplex.

The dimension of a Riemannian polyhedron is defined as the largest dimension in its triangulation. For Riemannian polyhedrons one can define length of curves and volume the same way as for Riemannian manifolds.

The obtained metric space will be called Riemannian polyhedron as well. A triangulation of Riemannian polyhedron will always be assumed to have the above property on the metric tensor.

Further we will apply the notion of width only to compact Riemannian polyhedrons. If $\mathcal{P}$ is an $n$-dimensional Riemannian polyhedron, then we suppose that

$$\operatorname{\rm width}\nolimits\mathcal{P}\mathrel{:=}\operatorname{\rm width}\nolimits_{n-1}\mathcal{P}.$$

Suppose that $\mathcal{P}$ is an $n$-dimensional Riemannian polyhedron; in this case we will use short cut $\operatorname{\rm vol}\nolimits$ for $\operatorname{\rm vol}\nolimits_{n}$. Let us define volume profile of $\mathcal{P}$ as a function returning largest volume of $r$-ball in $\mathcal{P}$; that is, the volume profile of $\mathcal{P}$ is a function $\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}\colon\mathbb{R}_{+}\to\mathbb{R}_{+}$ defined by

$$\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}(r)\mathrel{:=}\sup\left\{\,{\operatorname{\rm vol}\nolimits\mathrm{B}{}(p,r)}\,:\,{p\in\mathcal{P}}\,\right\}.$$

Note that $r\mapsto\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}(r)$ is nondecreasing and

$$\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}(r)\leqslant\operatorname{\rm vol}\nolimits\mathcal{P}$$

for any $r$. Moreover, if $\mathcal{P}$ is connected, then the equality $\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}(r)=\operatorname{\rm vol}\nolimits\mathcal{P}$ holds for $r\geqslant\mathop{\rm rad}\nolimits\mathcal{P}$.

Note that if $\mathcal{P}$ is a connected 1-dimensional Riemannian polyhedron, then

$$\operatorname{\rm width}\nolimits\mathcal{P}=\operatorname{\rm width}\nolimits_{0}\mathcal{P}=\mathop{\rm rad}\nolimits\mathcal{P}.$$

Since $\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}(R)\leqslant\operatorname{\rm vol}\nolimits\mathcal{P}$ for any $R>0$, we get the following:

The proof of 2E will be given at the very end of this section, after discussing separating polyhedrons.

Let us start three technical statements. The first statement can be obtained by modifying a smoothing procedure for functions defined on Euclidean space.

A function $f$ defined on a Riemannian polyhedron $\mathcal{P}$ is called piecewise smooth if there is a triangulation of $\mathcal{P}$ such that restriction of $f$ to every simplex is smooth.

The following statement can be proved by applying the classical Sard’s theorem to each simplex of a Riemannian polyhedron.

The following statement can be proved by applying the coarea inequality (1C) to the restriction of $f$ to each simplex of the polyhedron and summing up the results.

The proof reminds the proof of the following statement about minimal surfaces: if a point $p$ lies on an compact area-minimizing surface $\Sigma$ and $\partial\Sigma\cap\mathrm{B}{}(p,r)=\varnothing$, then

$$\operatorname{\rm area}\nolimits(\Sigma\cap\mathrm{B}{}(p,r))\leqslant\tfrac{1}{2}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\operatorname{\rm area}\nolimits\mathbb{S}^{2}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}r^{2}.$$

By Sard’s theorem (2E), for almost all values $c\in\nobreak(r_{0}+\nobreak\delta,r_{1}-\delta)$, the level set

$$\tilde{S}_{c}(p)=\left\{\,{x\in\mathcal{P}}\,:\,{\widetilde{\mathrm{dist}}_{p}(x)=c}\,\right\}$$

is a Riemannian polyhedron of dimension at most $n-1$. Since $\delta$ is small, the coarea inequality (2E) implies that $c$ can be chosen so that in addition the following inequality holds:

	$\displaystyle\operatorname{\rm vol}\nolimits_{n-1}\tilde{S}_{c}(p)$	$\displaystyle\leqslant\tfrac{1}{r_{1}-r_{0}-2{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\delta}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\operatorname{\rm vol}\nolimits_{n}[\mathrm{B}{}(p,r_{1})]<$
		$\displaystyle<\tfrac{1}{r_{1}-r_{0}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}(r_{1})+\tfrac{\varepsilon}{2}.$

Suppose $\mathcal{Q}$ is an $R$-separating subpolyhedron in $\mathcal{P}$ with almost minimal volume; say its volume is at most $\tfrac{\varepsilon}{2}$-far from the greatest lower bound. Note that cutting from $\mathcal{Q}$ everything inside $\tilde{S}_{c}(p)$ and adding $\tilde{S}_{c}(p)$ produces a $R$-separating subpolyhedron, say $\mathcal{Q}^{\prime}$.¹¹1If $\operatorname{\rm dim}\nolimits\tilde{S}_{c}(p)<n-1$, then it might happen that $\operatorname{\rm dim}\nolimits\mathcal{Q}^{\prime}<n-1$; so, by the definition, $\mathcal{Q}^{\prime}$ is not separating. It can be fixed by adding a tiny $(n-1)$-dimensional piece to $\mathcal{Q}^{\prime}$.

Since $\mathcal{Q}$ has almost minimal volume, we have

$$\operatorname{\rm vol}\nolimits_{n-1}[\mathcal{Q}\cap\mathrm{B}{}(p,r_{0})_{\mathcal{P}}]-\tfrac{\varepsilon}{2}\leqslant\operatorname{\rm vol}\nolimits_{n-1}S_{c}(p).$$

Therefore

$$\operatorname{\rm vol}\nolimits_{n-1}[\mathcal{Q}\cap\mathrm{B}{}(p,r_{0})_{\mathcal{P}}]\leqslant\tfrac{1}{r_{1}-r_{0}}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}(r_{1})+\varepsilon.$$

➊

Recall that $\mathcal{Q}$ is equipped with the induced length metric; therefore $|{p}-\nobreak{q}|_{\mathcal{Q}}\geqslant|{p}-\nobreak{q}|_{\mathcal{P}}$ for any $p,q\in\mathcal{Q}$; in particular,

$$\mathrm{B}{}(p,r_{0})_{\mathcal{Q}}\subset\mathcal{Q}\cap\mathrm{B}{}(p,r_{0})_{\mathcal{P}}$$

for any $p\in\mathcal{Q}$ and $r_{0}\geqslant 0$. Hence, ➊ ‣ 2 implies the lemma. ∎

Note that $\{V_{1},\dots,V_{k}\}$ can be converted into an open covering of a small neighbourhood of $\mathcal{Q}$ in $\mathcal{P}$ without increasing the multiplicity. This can be done by setting

$$V_{i}^{\prime}=\bigcup_{x\in V_{i}}\mathrm{B}{}(x,r_{x}),$$

where $r_{x}\mathrel{:=}\tfrac{1}{10}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\inf\left\{\,{|{x}-\nobreak{y}|}\,:\,{y\in\mathcal{Q}\backslash V_{i}}\,\right\}$.

By adding to $\{V_{i}^{\prime}\}$ all the components of $\mathcal{P}\backslash\mathcal{Q}$, we increase the multiplicity by at most 1 and obtain a covering of $\mathcal{P}$. The statement follows since $\operatorname{\rm dim}\nolimits\mathcal{P}=\operatorname{\rm dim}\nolimits\mathcal{Q}+\nobreak 1$. ∎

Suppose that the $(n-1)$-dimensional case is proved. Consider an $n$-dimensional Riemannian polyhedron $\mathcal{P}$ and suppose

$$n{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sqrt[n]{\operatorname{\rm VolPro}\nolimits\mathcal{P}(R)}<R$$

for some $R>0$. Let $\mathcal{Q}$ be an $R$-separating subpolyhedron in $\mathcal{P}$ provided by 2 for a small $\varepsilon>0$.

Applying 2 for $r=\tfrac{n-1}{n}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}R$ and $R$, we have that

	$\displaystyle\operatorname{\rm VolPro}\nolimits_{\mathcal{Q}}(r)$	$\displaystyle<\frac{1}{R-r}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\operatorname{\rm VolPro}\nolimits_{\mathcal{P}}(R)+\varepsilon<$
		$\displaystyle<\frac{n}{R}{\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\left(\frac{R}{n}\right)^{n}=$
		$\displaystyle=\left(\frac{r}{n-1}\right)^{n-1};$

that is, $(n-1){\hskip 0.5pt\cdot\nobreak\hskip 0.5pt}\sqrt[n-1]{\operatorname{\rm VolPro}\nolimits\mathcal{Q}(r)}<r$. Since $\operatorname{\rm dim}\nolimits\mathcal{Q}=n-1$, by the induction hypothesis, we get that

$$\operatorname{\rm width}\nolimits\mathcal{Q}\leqslant r<R.$$

It remains to apply 2. ∎