# Simplicial volume and $0$-strata of
separating filtrations

Hannah Alpert  Auburn University, 221 Parker Hall, Auburn, AL 36849
<EMAIL_ADDRESS>
###### Abstract.

We use Papasoglu’s method of area-minimizing separating sets to give an
alternative proof, and explicit constants, for the following theorem of Guth
and Braun–Sauer: If $M$ is a closed, oriented, $n$-dimensional manifold, with
a Riemannian metric such that every ball of radius $1$ in the universal cover
of $M$ has volume at most $V_{1}$, then the simplicial volume of $M$ is at
most the volume of $M$ times a constant depending on $n$ and $V_{1}$.

##  1\. Introduction

The purpose of this paper is to prove the following theorem.

######  Theorem 1.

Let $M$ be a closed, oriented, $n$-dimensional Riemannian manifold, and let
$\Gamma=\pi_{1}(M)$. Suppose that for all points $\widetilde{p}$ in the
universal cover $\widetilde{M}$ of $M$, we have
$\operatorname{Vol}B(\widetilde{p},1)\leq V_{1}$. Then

$\|M\|_{\Delta}\leq 16^{n}(n!)^{2}\cdot V_{1}\cdot\operatorname{Vol}M,$

where $\|M\|_{\Delta}$ denotes the Gromov simplicial volume of $M$.
Furthermore, if $V_{1}<\frac{1}{n!}$, then the image of the fundamental
homology class of $M$ under the classifying map is zero in
$H_{*}(B\Gamma;\mathbb{Q})$, so $\|M\|_{\Delta}=0$.

Only the constants $16^{n}(n!)^{2}\cdot V_{1}$ and $\frac{1}{n!}$ are new. The
theorem, with non-explicit constants, is proved by Guth in [Gut11] for the
case where $M$ admits a hyperbolic metric; the proof applies to any manifold
with residually finite fundamental group. For the case where $V_{1}$ is close
to zero, Guth adapts the same proof to show in [Gut17] that $M$ has bounded
Urysohn $(n-1)$-width; that is, $M$ admits a map to an $(n-1)$-dimensional
simplicial complex, for which all fibers have diameter at most $2$.
Liokumovich, Lishak, Nabutovsky, and Rotman in [LLNR22] generalize this
Urysohn width theorem to the case where $M$ is not necessarily a manifold, and
Papasoglu in [Pap20] proves the same statement by a shorter method, similar to
the method of minimal hypersurfaces which Guth in [Gut10] adapts from Schoen
and Yau’s papers [SY78, SY79]. Braun and Sauer in [BS21] adapt Guth’s original
proof to generalize it to the case where the fundamental group of $M$ is not
necessarily residually finite. In [Sab22], Sabourau proves a related result,
that if the volume of $M$ is sufficiently small, then in the universal cover
there are balls of all radii greater than $1$ with larger-than-hyperbolic
volume.

Braun and Sauer speculate about whether the method of Papasoglu can be used to
prove their theorems in a shorter way. Here we give an affirmative answer: the
method of Cantor bundles from [BS21] can serve the same role of removing the
assumption that $\pi_{1}(M)$ is residually finite, while the method of
Papasoglu replaces the more complicated method of Guth. Because Papasoglu’s
method is so much simpler, it allows us to give explicit constants in the
theorem statement.

Section 2 gives the properties of simplicial volume that we need to prove
Theorem 1. In Section 3 we prove a version of Theorem 1 with the extra
assumption that every nontrivial loop in $M$ has length greater than $2$, as a
warm-up for proving Theorem 1 in Section 4. Section 5 discusses the extent to
which we can apply an integer simplex straightening method from Campagnolo and
Sauer in [CS19] to recover the theorems about integral foliated simplicial
volume from the Braun–Sauer paper.

_Acknowledgments._ Thank you to Alexey Balitskiy and Caterina Campagnolo for
helpful conversations about the ideas in this paper. The author is supported
in part by the Simons Foundation (Gift Number 965348).

##  2\. Preliminaries on simplicial volume

This section includes some background information on simplicial volume that
links it to the main part of the proof, which is about separating filtrations.
The first subsection includes the definition of simplicial volume and a
theorem bounding the simplicial volume in terms of the number of rainbow
simplices in a vertex-colored cycle. The second subsection includes the
definition of a separating filtration and a lemma bounding the number of
rainbow simplices in terms of the number of points in the $0$-dimensional
stratum of a separating filtration.

###  2.1. Simplex straightening

Let $z=\sum_{i}a_{i}\sigma_{i}$ be a singular $d$-cycle on a space $P$ with
real coefficients $a_{i}\in\mathbb{R}$ and simplices
$\sigma_{i}\colon\thinspace\Delta^{d}\rightarrow P$. The $L^{1}$ norm of $z$,
denoted by $\left\lvert z\right\rvert_{1}$, is $\sum_{i}\left\lvert
a_{i}\right\rvert$, and the simplicial norm of a given homology class is the
infimum of $\left\lvert z\right\rvert_{1}$ over all cycles $z$ representing
the homology class. The simplicial norm of the class $[z]$ is denoted by
$\|[z]\|_{\Delta}$. The simplicial volume of a closed, oriented manifold $M$,
denoted by $\|M\|_{\Delta}$, is the simplicial norm of the fundamental
homology class of $M$. The simplicial norm was introduced by Gromov in
[Gro82].

One foundational property of simplicial norm from [Gro82, Sections 2.3, 3.3]
(or see [Iva87, Theorem 4.1]) is that if $f\colon\thinspace P\rightarrow Q$ is
a continuous map that induces an isomorphism on fundamental group, then the
induced map on homology $f_{*}\colon\thinspace H_{*}(P)\rightarrow H_{*}(Q)$
preserves simplicial norm. Thus, in particular, the simplicial volume of a
manifold $M$ with fundamental group $\Gamma$ is equal to the simplicial norm
of the image of $[M]$ in the homology of the classifying space $B\Gamma$.

Our use of simplicial norm in this paper is based on the following theorem, a
special case of the Amenable Reduction Lemma from [Gro09] (or see [AK16]).
Given a singular cycle $z$ on a space $P$, we define a $\pi_{1}$-killing
vertex coloring of $z$ to be a way to assign colors to the vertices of $z$
such that for each color, if we take the union in $P$ of all edges of $z$ for
which both vertices are that color, then the inclusion of this $1$-complex
into $P$ induces the zero map on $\pi_{1}$. (In particular, if $z$ does not
contain any edges from a vertex to itself, then coloring every vertex a
different color is $\pi_{1}$-killing.) We define a rainbow simplex of such a
coloring to be any simplex in $z$ for which all $d+1$ vertices are different
colors.

######  Theorem 2 ([Gro09]).

Let $z=\sum_{i}a_{i}\sigma_{i}$ be a singular $d$-cycle on $P$, with a
$\pi_{1}$-killing vertex coloring. Then the simplicial norm of the homology
class of $z$ satisfies

$\|[z]\|_{\Delta}\leq\sum_{\mathrm{rainbow}\ \sigma_{i}}\left\lvert
a_{i}\right\rvert.$

###### Proof sketch.

On any space with contractible universal cover, such as the classifying space
$B\Gamma$ where $\Gamma=\pi_{1}(P)$, we can define a notion of simplex
straightening: for each $(d+1)$-tuple of points in the universal cover
$E\Gamma$, the idea is to make a choice of $d$-simplex with those vertices, in
a way that agrees with taking faces, translating by $\Gamma$, and permuting
the vertices (with sign). More formally, instead of literally choosing a
single $d$-simplex, for the permutation invariance we need to choose the
signed average of all of its permutations, and for the $\Gamma$-invariance, in
the case where $\Gamma$ has finite-order elements we need to choose the
average of any $\Gamma$-translations of our simplex that permute its vertices.
Every cycle is homologous to its straightened version, and if a given simplex
lifts to the universal cover in such a way that two of its vertices are the
same, its straightening is zero.

Thus, let $\alpha\colon\thinspace P\rightarrow B\Gamma$ be the classifying
map. The classes $[z]$ and $\alpha_{*}[z]$ have the same simplicial norm. We
homotope $\alpha(z)$ so that all vertices of each color and all edges among
them go to a single point—this is possible because the coloring is
$\pi_{1}$-killing. Then we straighten. The result is homologous to
$\alpha(z)$, and each rainbow simplex contributes the same amount to the
simplicial norm as it did in $[z]$, but each non-rainbow simplex becomes zero.
∎

A special case of the theorem above is when there are no rainbow simplices. In
this case, we can conclude that the simplicial norm is zero, but we can also
say something a bit stronger: the image of the homology class in the
classifying space is torsion.

######  Corollary 3.

Let $P$ be a topological space, let $\Gamma=\pi_{1}(P)$, and let
$\alpha\colon\thinspace P\rightarrow B\Gamma$ be the classifying map. Let $z$
be a singular $d$-cycle on $P$, homologous in $H_{*}(P;\mathbb{R})$ to an
element of $H_{*}(P;\mathbb{Z})$, and suppose that $z$ admits a
$\pi_{1}$-killing vertex coloring that has no rainbow simplices. Then the
class $\alpha_{*}[z]$ is zero in $H_{*}(B\Gamma;\mathbb{Q})$.

###### Proof sketch.

In the proof above, the straightening of the homotoped cycle $\alpha(z)$ is
zero, because there are no rainbow simplices. Thus, $\alpha_{*}[z]$ is zero in
$H_{*}(B\Gamma;\mathbb{R})$. But the change of coefficients from
$H_{*}(B\Gamma;\mathbb{Q})$ to $H_{*}(B\Gamma;\mathbb{R})$ is just the tensor
product with $\mathbb{R}$, which gives an injection, so $\alpha_{*}[z]$ is
zero in $H_{*}(B\Gamma;\mathbb{Q})$ as well. ∎

###  2.2. Triangulating a separating filtration

The next lemma links the idea of counting rainbow simplices with Papasoglu’s
method of area-minimizing separating sets in [Pap20]. Papasoglu’s method is to
find a filtration

$M=Z_{n}\supseteq Z_{n-1}\supseteq\cdots\supseteq Z_{1}\supseteq Z_{0},$

such that each $Z_{i}$ is an $i$-dimensional set, minimizing $i$-dimensional
area subject to the condition that every connected component of
$Z_{i+1}\setminus Z_{i}$ is contained in a ball of radius $R$ in $M$. We
define these sets in terms of Riemannian polyhedra as in [Nab19]. A Riemannian
polyhedron is a finite simplicial complex with a Riemannian metric on every
maximal simplex, such that the metrics agree on common faces. A subpolyhedron
has smoothly embedded faces and carries the induced Riemannian metric.

Our Riemannian polyhedra are pure simplicial complexes, so they have well-
defined volumes. If $P$ is a pure $d$-dimensional Riemannian polyhedron, we
denote its $d$-dimensional volume by $\operatorname{Area}_{d}(P)$. We also
require the smooth part of a subpolyhedron to lie in the smooth part of the
ambient polyhedron. Specifically, if $Z$ is a pure $(d-1)$-dimensional
subpolyhedron of $P$, then we require that for each face of $Z$, if the face
has dimension $k$, its relative interior is embedded in a simplex of $P$ of
dimension at least $k+1$. We say that $Z$ is $R$-separating in $P$ if every
connected component of $P\setminus Z$ is contained in a ball of radius $R$.

We define an $R$-separating filtration of $M$ to consist of Riemannian
polyhedra

$M=Z_{n}\supseteq Z_{n-1}\supseteq\cdots\supseteq Z_{1}\supseteq Z_{0},$

such that each $Z_{i}$ is an $R$-separating subpolyhedron of $Z_{i+1}$. Our
requirement that the smooth part of each level is contained in the smooth part
of the next level implies that every point of $Z_{0}$ has a neighborhood with
a diffeomorphism that sends our filtration to the filtration

$\mathbb{R}^{n}\supseteq\mathbb{R}^{n-1}\supseteq\cdots\supseteq\mathbb{R}^{1}\supseteq\mathbb{R}^{0}.$

Thus, the following lemma shows that if we triangulate consistent with the
filtration, we can produce a $\pi_{1}$-killing coloring with a controlled
number of rainbow simplices.

######  Lemma 4.

Let $M$ be a closed $n$-dimensional Riemannian manifold with an $R$-separating
filtration

$M=Z_{n}\supseteq Z_{n-1}\supseteq\cdots\supseteq Z_{1}\supseteq Z_{0}.$

Suppose that for every ball of radius $R$ in $M$, the map on $\pi_{1}$ induced
by its inclusion into $M$ is the zero map. Then there is a triangulation of
$M$ with a $\pi_{1}$-killing coloring, such that the number of rainbow
simplices is $2^{n}\cdot\\#Z_{0}$.

###### Proof.

We want a triangulation of $M$ such that each $Z_{i}$ is a subcomplex of the
triangulation. To do this, start with $Z_{0}$, which is a finite set of points
and thus is already triangulated. As a Riemannian polyhedron, $Z_{1}$ already
has a triangulation, but we want to choose a refinement that is consistent
with $Z_{0}$. To do this, we start by subdividing each $1$-simplex in $Z_{1}$
that is divided by $Z_{0}$. Continuing, we go up one dimension at a time,
refining the triangulation of each $Z_{i+1}$ so that it is consistent with our
triangulation of $Z_{i}$. Specifically, we look at the simplices of $Z_{i+1}$
that are divided by $Z_{i}$, starting with the $1$-simplices and going up one
dimension at a time. For each $j$-simplex $\Delta^{j}$ divided by $Z_{i}$, for
each component of $\Delta^{j}\setminus Z_{i}$ we triangulate the relative
interior in a way that extends the triangulation we already have on the
boundary. Continuing to subdivide in this way, we obtain a triangulation of
$M$ such that each $Z_{i}$ is a subcomplex.

We color all the points in $M$, such that two points are the same color if and
only if they are in the same level $Z_{i}\setminus Z_{i-1}$ (for
$i=0,\ldots,n$) and they are in the same connected component of
$Z_{i}\setminus Z_{i-1}$. Our initial triangulation of $M$ does not
necessarily have the number of rainbow simplices that we want, but we claim
that its barycentric subdivision does. We observe that in the initial
triangulation, the relative interior of each simplex has only one color, and
that if one simplex is a face of another, the two simplices don’t come from
different components of $Z_{i}\setminus Z_{i-1}$ for the same $i$—either they
are the same color, or they are at different levels of the filtration. Thus,
the corresponding property is true of the barycentric subdivision: if two
vertices are in the same simplex, then either they are the same color, or they
are at different levels. This implies that if a simplex is rainbow, then all
of its vertices are at different levels, and in an $n$-simplex, this means
that among its $n+1$ vertices there must be exactly one at each level
$Z_{0},Z_{1}\setminus Z_{0},\ldots,Z_{n}\setminus Z_{n-1}$.

Thus, in the barycentric subdivision, there are exactly $2^{n}$ rainbow
simplices containing each point of $Z_{0}$: from the point in $Z_{0}$, there
are two directions in $Z_{1}$, and for each, there are two directions in
$Z_{2}$, and so on, and each such chain corresponds to exactly one simplex.
Because the filtration is $R$-separating, each color (and all the edges among
vertices of that color) is contained in a ball of radius $R$, which
contributes nothing to $\pi_{1}(M)$, so the coloring is $\pi_{1}$-killing. ∎

##  3\. Large-systole case

The purpose of this section is to prove the following weaker version of
Theorem 1, in preparation for proving the full version in the next section.

######  Theorem 5.

Let $M$ be a closed, oriented, $n$-dimensional Riemannian manifold, and let
$\Gamma=\pi_{1}(M)$. Suppose that every homotopically nontrivial loop in $M$
has length greater than $2$, and for every $p\in M$ we have
$\operatorname{Vol}B(p,1)\leq V_{1}$. Then

$\|M\|_{\Delta}\leq 16^{n}(n!)^{2}\cdot V_{1}\cdot\operatorname{Vol}M.$

Furthermore, if $V_{1}<\frac{1}{n!}$, then the image of the fundamental
homology class of $M$ under the classifying map is zero in
$H_{*}(B\Gamma;\mathbb{Q})$, so $\|M\|_{\Delta}=0$.

The strategy is to select a $1$-separating filtration that is area-minimizing
at each level, and to bound the number of points in the $0$-dimensional level
$Z_{0}$ in terms of $\operatorname{Vol}M$. Then we can apply the statements
from the previous section to relate the simplicial volume to the number of
points in $Z_{0}$. The next two lemmas show that every point in $Z_{0}$ is in
a ball of fairly large volume.

######  Lemma 6.

Let $P$ be a pure $d$-dimensional Riemannian polyhedron embedded in $M$, and
let $Z$ be an $R$-separating subpolyhedron of $P$. Suppose that
$\operatorname{Area}_{d-1}Z$ is within $\varepsilon$ of the infimal area of
$R$-separating subpolyhedra of $P$. Then for all $p\in M$ and all
$r_{1},r_{2}$ with $0<r_{1}<r_{2}<R$ we have

$\int_{r_{1}}^{r_{2}}\operatorname{Area}_{d-1}(Z\cap B(p,\rho))\
d\rho\leq\operatorname{Area}_{d}(P\cap B(p,r_{2})\setminus
B(p,r_{1}))+2\varepsilon R.$

###### Proof.

Suppose for the sake of contradiction that this is not the case, so we have
$Z,p,r_{1},r_{2}$ violating this inequality. We approximate the distance
function on $M$ by a smooth function that is within a small distance $\delta$
of $M$ and is $(1+\delta)$-Lipschitz. If $\widehat{B}(p,r)$ denotes the ball
of radius $r$ around $p$ computed according to this approximate distance
function, then $B(p,r-\delta)\subseteq\widehat{B}(p,r)\subseteq
B(p,r+\delta)$. We choose $\delta$ small enough that $r_{2}+2\delta<R$ and
$3\delta\operatorname{Area}_{d-1}(Z\cap
B(p,R))+\delta\operatorname{Area}_{d}(P\cap B(p,R))<\varepsilon R$.

For almost all $\rho$, the approximate sphere
$\widehat{S}(p,\rho)=\partial\widehat{B}(p,\rho)$ is transverse to $P$ (more
precisely, $\rho$ is a regular value of the smooth approximate-distance-to-$p$
function on each simplex of $P$) and the function
$\operatorname{Area}_{d}(P\cap\widehat{B}(p,\rho))$ is differentiable at
$\rho$. The coarea inequality says that because the approximate-distance-
to-$p$ function on $P$ is $(1+\delta)$-Lipschitz with fibers
$P\cap\widehat{S}(p,\rho)$, we have

$\operatorname{Area}_{d-1}(P\cap\widehat{S}(p,\rho))\leq(1+\delta)\frac{d}{d\rho}\operatorname{Area}_{d}(P\cap\widehat{B}(p,\rho)).$

On the other hand, replacing $Z\cap\widehat{B}(p,\rho)$ in $Z$ by
$P\cap\widehat{S}(p,\rho)$ gives another $R$-separating subpolyhedron of $P$.
Thus, because $Z$ is area-minimizing up to $\varepsilon$ we have

$\operatorname{Area}_{d-1}(Z\cap\widehat{B}(p,\rho))\leq\operatorname{Area}_{d-1}(P\cap\widehat{S}(p,\rho))+\varepsilon.$

Integrating as $\rho$ varies between $r_{1}+\delta$ and $r_{2}-\delta$, and
observing that $\varepsilon(r_{2}-r_{1}-2\delta)\leq\varepsilon R$, gives

$\int_{r_{1}+\delta}^{r_{2}-\delta}\operatorname{Area}_{d-1}(Z\cap\widehat{B}(p,\rho))\
d\rho\leq(1+\delta)\operatorname{Area}_{d}(P\cap\widehat{B}(p,r_{2}-\delta)\setminus\widehat{B}(p,r_{1}+\delta))+\varepsilon
R,$

and so, plugging in our choice of $\delta$, we have

$\int_{r_{1}}^{r_{2}}\operatorname{Area}_{d-1}(Z\cap B(p,\rho))\
d\rho\leq\operatorname{Area}_{d}(P\cap B(p,r_{2})\setminus
B(p,r_{1}))+2\varepsilon R.$

∎

Applying the lemma above, along with repeated integration, then gives the
following bound on the volumes of balls around points of $Z_{0}$.

######  Lemma 7.

Let $M$ be a closed $n$-dimensional Riemannian manifold. For all
$\varepsilon>0$, there exists an $R$-separating filtration

$M=Z_{n}\supseteq Z_{n-1}\supseteq\cdots\supseteq Z_{1}\supseteq Z_{0},$

such that for all $p\in M$ and all $r_{1},r_{2}$ with $0<r_{1}<r_{2}<R$ we
have

$\operatorname{Area}_{0}(Z_{0}\cap
B(p,r_{1}))\cdot\frac{(r_{2}-r_{1})^{n}}{n!}\leq\operatorname{Vol}B(p,r_{2})+\varepsilon.$

###### Proof.

By induction on $i$ we can prove the following statement: if each $Z_{j}$ is
area-minimizing up to $\varepsilon_{j}$, then for all $0<r_{1}<r_{2}<R$, we
have

$\displaystyle\operatorname{Area}_{0}(Z_{0}\cap
B(p,r_{1}))\cdot\frac{(r_{2}-r_{1})^{i}}{i!}\leq\operatorname{Area}_{i}(Z_{i}\cap
B(p,r_{2}))+$
$\displaystyle+2\varepsilon_{0}R^{i}+2\varepsilon_{1}R^{i-1}+\cdots+2\varepsilon_{i-1}R.$

The base case $i=0$ says $\operatorname{Area}_{0}(Z_{0}\cap
B(p,r_{1}))\leq\operatorname{Area}_{0}(Z_{0}\cap B(p,r_{2}))$, which is true.
The inductive step is obtained by replacing $r_{2}$ by $\rho$ in the inductive
hypothesis, integrating as $\rho$ ranges from $r_{1}$ to $r_{2}$, and applying
Lemma 6 to the right-hand side of the inequality.

For $i=0,\ldots,n-1$ we select
$\displaystyle\varepsilon_{i}=\frac{\varepsilon}{2nR^{n-i}}$, so that

$2\varepsilon_{0}R^{n}+2\varepsilon_{1}R^{n-1}+\cdots+2\varepsilon_{n-1}R=\varepsilon.$

Then plugging $i=n$ into the induction claim gives the desired inequality. ∎

With this lemma we can prove the special case of the main theorem.

###### Proof of Theorem 5.

We apply Lemma 7 with $R=1$, and apply Lemma 4 to get a triangulation of $M$
with $2^{n}$ rainbow simplices for each point in $Z_{0}$. First we prove the
statement about the case $V_{1}<\frac{1}{n!}$. In this case, let $p$ be any
point of $Z_{0}$. Taking $r_{1}\rightarrow 0$ and $r_{2}\rightarrow 1$ in
Lemma 7 gives a contradiction if $\varepsilon$ is sufficiently small. Thus
there are no points in $Z_{0}$, and thus no rainbow simplices. Corollary 3
shows that the image of the fundamental class of $M$ is zero in
$H_{*}(B\Gamma;\mathbb{Q})$.

In the case where $V_{1}\geq\frac{1}{n!}$, we take a maximal collection of
disjoint balls of radius $\frac{1}{4}$, centered at points of $Z_{0}$. The
concentric balls of radius $\frac{1}{2}$ cover all of $Z_{0}$. Let
$B_{1}(\frac{1}{4}),\ldots,B_{k}(\frac{1}{4})$ denote the disjoint balls of
radius $\frac{1}{4}$, and let $B_{1}(\frac{1}{2}),\ldots,B_{k}(\frac{1}{2})$
denote the $Z_{0}$-covering balls of radius $\frac{1}{2}$. We apply Theorem 2,
and take the conclusion of Lemma 7 first with $r_{1}=\frac{1}{2}$ and
$r_{2}\rightarrow 1$, and then with $r_{1}\rightarrow 0$ and
$r_{2}=\frac{1}{4}$, to obtain

$\|M\|_{\Delta}\leq 2^{n}\cdot\\#(Z_{0})\leq
2^{n}\cdot\sum_{i=1}^{k}\\#\left(Z_{0}\cap
B_{i}\left(\frac{1}{2}\right)\right)\leq 2^{n}\cdot\sum_{i=1}^{k}n!\cdot
2^{n}\cdot(V_{1}+\varepsilon)\leq$ $\leq 4^{n}\cdot
n!\cdot(V_{1}+\varepsilon)\cdot\sum_{i=1}^{k}1\leq 4^{n}\cdot
n!\cdot(V_{1}+\varepsilon)\cdot\frac{\operatorname{Vol}M}{\left(\frac{(1/4)^{n}}{n!}-\varepsilon\right)},$

and taking $\varepsilon\rightarrow 0$ gives

$\|M\|_{\Delta}\leq 16^{n}(n!)^{2}\cdot V_{1}\cdot\operatorname{Vol}M.$

∎

##  4\. Main proof

###  4.1. The idea of Cantor bundles

Before getting into the technical setup of the proof of Theorem 1, we begin
with an informal overview of the idea of Cantor bundles from [BS21], which was
developed to extend the proof of Theorem 5 from the case where there may be
short nontrivial loops but $\pi_{1}(M)$ is residually finite, to the case
where $\pi_{1}(M)$ is not residually finite. First we describe what happens
when $\pi_{1}(M)$ is residually finite, and then we describe the Cantor bundle
idea, which is closely analogous.

Let $\Gamma$ be the fundamental group of $M$, and suppose that $\Gamma$ is
residually finite. There are finitely many elements of $\Gamma$ with the
property that their deck transformations on $\widetilde{M}$ move some points
within distance $2$ of themselves. By residual finiteness, there is a finite-
index subgroup of $\Gamma$ that avoids all of these elements (except the
identity), corresponding to a finite-sheeted covering space $\widehat{M}$ of
$M$ without nontrivial loops of length at most $2$. If $\widehat{M}$ has $k$
sheets over $M$, we can think of $\widehat{M}\rightarrow M$ as a locally
trivial bundle with fiber $\\{1,\ldots,k\\}$, and we can also think of this
bundle as the quotient of $\\{1,\ldots,k\\}\times\widetilde{M}$ by the action
of $\Gamma$. Here, the action of $\Gamma$ on $\\{1,\ldots,k\\}$ comes from
identifying $\\{1,\ldots,k\\}$ with the set of cosets
$\Gamma/\pi_{1}(\widehat{M})$.

Instead of looking for a $1$-separating filtration of $M$, we do the same
thing on $\widehat{M}$, or equivalently, we do it equivariantly on
$\\{1,\ldots,k\\}\times\widetilde{M}$. Then, to estimate the simplicial volume
of $M$, instead of triangulating $M$ we triangulate $\widehat{M}$ (or
triangulate $\\{1,\ldots,k\\}\times\widetilde{M}$ equivariantly), multiply
each simplex by $\frac{1}{k}$, and project to $M$ to get a fundamental cycle
for $M$.

To adapt this method to the case where $\Gamma$ is not residually finite,
Braun and Sauer replace the finite set $\\{1,\ldots,k\\}$ by a Cantor set $X$.
The Cantor set admits a free, continuous action of $\Gamma$, as shown by
[HM06]. Thus, our work to find a $1$-separating filtration, and then a
triangulation, is done $\Gamma$-equivariantly on $X\times\widetilde{M}$, or
equivalently is done on the quotient $X\times_{\Gamma}\widetilde{M}$, which is
a locally trivial bundle over $M$ with fiber equal to $X$.

Even though this cover of $M$ or $\widetilde{M}$ now has uncountably many
sheets, over any compact subset of $\widetilde{M}$, we still want only
finitely many different kinds of sheets. To guarantee this property, we
introduce the following definitions. We say that a thick set is a subset of
$X\times\widetilde{M}$ of the form $A\times S$, where $A\subseteq X$ is clopen
and $S\subseteq\widetilde{M}$ is bounded. We say that a thick set has a non-
self-intersecting orbit if it does not intersect any of its other
$\Gamma$-translates. When this is the case, we call the union of the thick set
and all its translates a thick orbit. We verify in the following proposition
that any collection of finitely many thick orbits has this property that there
are only finitely many kinds of sheets over any compact subset.

######  Proposition 8.

Let $\Gamma(A_{1}\times S_{1}),\ldots,\Gamma(A_{r}\times S_{r})$ be any
collection of finitely many thick orbits in $X\times\widetilde{M}$ under the
action of $\Gamma$. Then for any ball $B(\widetilde{p},R)$ in $\widetilde{M}$,
there exists a partition of $X$ into finitely many clopen sets
$X_{1},\ldots,X_{k}$, such that within each set $X_{i}\times
B(\widetilde{p},R)$, every sheet $\\{x\\}\times B(\widetilde{p},R)$ has the
same pattern of thick orbits. That is, for any $i\in\\{1,\ldots,k\\}$, any
$x,y\in X_{i}$, any $\widetilde{q}\in B(\widetilde{p},R)$, and any
$j\in\\{1,\ldots,r\\}$, we have $(x,\widetilde{q})\in\Gamma(A_{j}\times
S_{j})$ if and only if $(y,\widetilde{q})\in\Gamma(A_{j}\times S_{j})$.

###### Proof.

Consider all sets $\gamma A_{j}$, where $\gamma\in\Gamma$ and $A_{j}\times
S_{j}$ is a thick set in our collection such that $\gamma S_{j}$ intersects
$B(\widetilde{p},R)$. There are finitely many such sets, and they generate an
algebra in $X$ under union, intersection, and complement. There are finitely
many minimal sets in this algebra, all of which are clopen, and we choose
$X_{1},\ldots,X_{k}$ to be these sets. Thus, if $x$ and $y$ are in the same
one of these sets, then $x$ and $y$ are in exactly the same sets $\gamma
A_{j}$, so for any $\widetilde{q}\in B(\widetilde{p},R)$, the points
$(x,\widetilde{q})$ and $(y,\widetilde{q})$ are in exactly the same thick
orbits. ∎

In addition to the action of $\Gamma$ on $X$, as in [BS21] we need $X$ to be
equipped with a $\Gamma$-invariant probability measure, $\mu$; [HM06] shows
that a free, continuous action of $\Gamma$ on $X$ can have such a $\mu$. This
means that if $\Gamma(A\times S)$ is a thick orbit, and $S$ is
$d$-dimensional, we can define (with some abuse of notation)

$\operatorname{Area}_{d}(\Gamma(A\times S))=\operatorname{Area}_{d}(A\times
S)=\mu(A)\cdot\operatorname{Area}_{d}(S).$

This allows us to select our $1$-separating filtrations to be (approximately)
area-minimizing at each level $\widetilde{Z}_{i}$, as we did when working
directly with $M$. It also allows us to turn a $1$-separating filtration into
a fundamental cycle for $M$, by triangulating the sheets in a consistent way
and projecting to $M$, with coefficients given by the measures of the relevant
sets in $X$. We record all of the Cantor bundle setup in the following
definition.

######  Definition 9.

Henceforth, $M$ is a closed, oriented, $n$-dimensional Riemannian manifold.
$\Gamma$ is the fundamental group of $M$, and $\widetilde{M}$ is the universal
cover of $M$. We consider a free, continuous action of $\Gamma$ on the Cantor
set $X$, with a $\Gamma$-invariant probability measure $\mu$ on $X$. Then
$\Gamma$ acts on $X\times\widetilde{M}$ by a diagonal action.

###  4.2. Adapting the proof to Cantor bundles

We define a $\Gamma$-equivariant thick polyhedron in $X\times\widetilde{M}$ to
be a union of finitely many thick orbits, such that each thick orbit comes
from thickening a simplex embedded in $\widetilde{M}$. As before, in a given
thick polyhedron, all maximal simplices have the same dimension, and if
$\widetilde{Z}$ is a thick subpolyhedron of a thick polyhedron
$\widetilde{P}$, we require the smooth part of $\widetilde{Z}$ to be inside
the smooth part of $\widetilde{P}$. We say that $\widetilde{Z}$ is
$R$-separating in $\widetilde{P}$ if every connected component of
$\widetilde{P}\setminus\widetilde{Z}$ is contained in a ball of radius $R$ in
some sheet $\\{x\\}\times\widetilde{M}$. We define a $\Gamma$-equivariant
thick $R$-separating filtration of $X\times\widetilde{M}$ to consist of nested
thick polyhedra

$X\times\widetilde{M}=\widetilde{Z}_{n}\supseteq\widetilde{Z}_{n-1}\supseteq\cdots\supseteq\widetilde{Z}_{1}\supseteq\widetilde{Z}_{0}$

such that each $\widetilde{Z}_{i}$ is an $R$-separating thick subpolyhedron of
$\widetilde{Z}_{i+1}$. The analogue of Lemma 4 is the following.

######  Lemma 10.

Consider a $\Gamma$-invariant thick $R$-separating filtration

$X\times\widetilde{M}=\widetilde{Z}_{n}\supseteq\widetilde{Z}_{n-1}\supseteq\cdots\supseteq\widetilde{Z}_{1}\supseteq\widetilde{Z}_{0}.$

There is a fundamental cycle $z=\sum_{i}a_{i}\sigma_{i}$ for $M$ with a
$\pi_{1}$-killing vertex coloring, such that

$\sum_{\mathrm{rainbow}\ \sigma_{i}}\left\lvert
a_{i}\right\rvert=2^{n}\cdot\operatorname{Area}_{0}(\widetilde{Z}_{0}).$

###### Proof.

As in Lemma 4, we refine our triangulations of the sets $\widetilde{Z}_{i}$ so
that they are all consistent, obtaining a structure for $X\times\widetilde{M}$
as a thick simplicial complex, with the various $\widetilde{Z}_{i}$ as
subcomplexes. By Lemma 8 there are finitely many different local arrangements
of simplices to deal with. We color $X\times\widetilde{M}$
$\Gamma$-equivariantly in the following way. First we divide into levels
$\widetilde{Z}_{i}\setminus\widetilde{Z}_{i-1}$ for $i=0,\ldots,n$, and then
we divide each level into connected components. However, each connected
component is contained in one sheet $\\{x\\}\times\widetilde{M}$, and we want
to group the components that differ only vertically. Thus, if two sets
$\\{x\\}\times S$ and $\\{y\\}\times S$ are connected components of some
$\widetilde{Z}_{i}\setminus\widetilde{Z}_{i-1}$, we color them the same color.
If needed, we subdivide our thick triangulation of $X\times\widetilde{M}$ in
the $X$ direction, so that the relative interior of each thick simplex is only
one color.

Taking the barycentric subdivision of this thick triangulation of
$X\times\widetilde{M}$, we obtain $2^{n}$ rainbow simplices for each point of
$\widetilde{Z}_{0}$; however, there are (potentially) uncountably many points
of $\widetilde{Z}_{0}$. Thus, we project to $M$ to obtain a cycle $z$ in the
following way. For each thick orbit $\Gamma(A\times\sigma)$, where $\sigma$ is
an $n$-dimensional simplex of our barycentric subdivision, its contribution to
$z$ is $\mu(A)\cdot\pi(\sigma)$, where
$\pi\colon\thinspace\widetilde{M}\rightarrow M$ is the covering map. The
contribution of each rainbow simplex is equal to the weight of the associated
point in $\widetilde{Z}_{0}$, so taking the sum over all rainbow simplices
gives the desired inequality. ∎

The analogue of Lemma 6 is the following.

######  Lemma 11.

Given $M$, for each $R>0$ there is a constant $m(M,R)$ such that the following
holds. Let $\widetilde{P}$ be a pure $d$-dimensional $\Gamma$-equivariant
thick polyhedron in $X\times\widetilde{M}$, and let $\widetilde{Z}$ be an
$R$-separating subpolyhedron of $P$. Suppose that
$\operatorname{Area}_{d-1}\widetilde{Z}$ is within $\varepsilon$ of the
infimal area of $R$-separating subpolyhedra of $\widetilde{P}$. Then for all
$\widetilde{p}\in\widetilde{M}$, all $r_{1},r_{2}$ with $0<r_{1}<r_{2}<R$, and
all clopen sets $E\subseteq X$ we have

$\displaystyle\int_{r_{1}}^{r_{2}}\operatorname{Area}_{d-1}$
$\displaystyle\left(\widetilde{Z}\cap\left(E\times
B(\widetilde{p},\rho)\right)\right)d\rho\leq$
$\displaystyle\leq\operatorname{Area}_{d}$
$\displaystyle\left(\widetilde{P}\cap\left(E\times\left(B(\widetilde{p},r_{2})\setminus
B(\widetilde{p},r_{1})\right)\right)\right)+2\varepsilon R\cdot m(M,R).$

###### Proof.

The proof closely follows that of Lemma 6, but we might not be able to cut
$\widetilde{Z}\cap(E\times\widehat{B}(\widetilde{p},\rho))$ out of
$\widetilde{Z}$ and replace it by
$\widetilde{P}\cap(E\times\widehat{S}(\widetilde{p},\rho))$, because of worry
about self-intersections. Thus, we need to do the replacement on a smaller
subset of $E$.

There are finitely many elements of $\Gamma$ that move any points of
$\widetilde{M}$ a distance of at most $2R$. As in [BS21], if we put a metric
on $X$, because the action of $\Gamma$ on $X$ is free and $X$ is compact,
there is some minimum distance that the points in $X$ are moved by these
finitely many elements. Thus, we can partition $X$ into clopen sets
$X_{1},\ldots,X_{m}$ that each have diameter less than this minimum distance.
Let $m(M,R)$ be this constant $m$.

For $j=1,\ldots,m$, let $E_{j}=E\cap X_{j}$. Then we can replace
$\widetilde{Z}\cap(E_{j}\times\widehat{B}(\widetilde{p},\rho))$ in
$\widetilde{Z}$ by
$\widetilde{P}\cap(E_{j}\times\widehat{S}(\widetilde{p},\rho))$ to get another
$R$-separating thick subpolyhedron of $\widetilde{P}$, because the sets
$E_{j}\times\widehat{B}(\widetilde{p},\rho)$ have non-self-intersecting
orbits. Because $\widetilde{Z}$ is area-minimizing up to $\varepsilon$, we
have

$\operatorname{Area}_{d-1}(\widetilde{Z}\cap(E_{j}\times\widehat{B}(\widetilde{p},\rho)))\leq\operatorname{Area}_{d-1}(\widetilde{P}\cap(E_{j}\times\widehat{S}(\widetilde{p},\rho)))+\varepsilon.$

Summing over all $j$ gives

$\operatorname{Area}_{d-1}(\widetilde{Z}\cap(E\times\widehat{B}(\widetilde{p},\rho)))\leq\operatorname{Area}_{d-1}(\widetilde{P}\cap(E\times\widehat{S}(\widetilde{p},\rho)))+\varepsilon\cdot
m(M,R).$

The remainder of the proof is the same as that of Lemma 6, carrying the extra
factor of $m(M,R)$ with the $\varepsilon$. ∎

The analogue of Lemma 7 is the following.

######  Lemma 12.

For all $\varepsilon>0$, there exists a $\Gamma$-equivariant thick
$R$-separating filtration

$X\times\widetilde{M}=\widetilde{Z}_{n}\supseteq\widetilde{Z}_{n-1}\supseteq\cdots\supseteq\widetilde{Z}_{1}\supseteq\widetilde{Z}_{0},$

such that for all $\widetilde{p}\in\widetilde{M}$, all $r_{1},r_{2}$ with
$0<r_{1}<r_{2}<R$, and all clopen sets $E\subseteq X$ we have

$\operatorname{Area}_{0}\left(\widetilde{Z}_{0}\cap\left(E\times
B(\widetilde{p},r_{1})\right)\right)\cdot\frac{(r_{2}-r_{1})^{n}}{n!}\leq\operatorname{Area}_{n}\left(E\times
B(\widetilde{p},r_{2})\right)+\varepsilon.$

###### Proof.

The proof follows from Lemma 11 in the same way that the proof of Lemma 7
follows from Lemma 6. The only difference is that we should choose the
constants $\varepsilon_{0},\ldots,\varepsilon_{n-1}$ to be smaller by a factor
of $m(M,R)$. ∎

We are ready to finish the proof of the main theorem.

###### Proof of Theorem 1.

We apply Lemma 12 with $R=1$, and apply Lemma 10 to find a fundamental cycle
for $M$ with total contribution from rainbow simplices equal to
$2^{n}\cdot\operatorname{Area}_{0}\widetilde{Z}_{0}$. To prove the statement
about the case $V_{1}<\frac{1}{n!}$, we observe that if
$A\times\\{\widetilde{p}\\}$ is a thick simplex of $\widetilde{Z}_{0}$, then
the conclusion of Lemma 12 for $E=A$, $r_{1}\rightarrow 0$, and
$r_{2}\rightarrow 1$ gives a contradiction for sufficiently small
$\varepsilon$. Thus, $\widetilde{Z}_{0}$ is empty, and by Corollary 3 the
image of the fundamental class of $M$ is zero in $H_{*}(B\Gamma;\mathbb{Q})$.

In the case where $V_{1}\geq\frac{1}{n!}$, we want to take an equivariant
maximal collection of disjoint balls $\\{x\\}\times
B(\widetilde{p},\frac{1}{4})$, such that the points $(x,\widetilde{p})$ are
all in $\widetilde{Z}_{0}$. To do this, we note that such points
$\widetilde{p}$ are part of finitely many orbits. We consider the orbits one
at a time in order. Given one such $\widetilde{p}$, there is a clopen subset
$A$ of $X$ formed by all $x$ such that $(x,\widetilde{p})$ is in
$\widetilde{Z}_{0}$ but $\\{x\\}\times B(\widetilde{p},\frac{1}{4})$ does not
intersect any of the balls chosen so far.

However, we still need to make $A$ smaller because the translates of $A\times
B(\widetilde{p},\frac{1}{4})$ may intersect. To do this, let
$B=B(\widetilde{p},\frac{1}{4})$, and let $F=\\{\gamma\in\Gamma\ |\ \gamma
B\cap B\neq\emptyset\\}$, a finite set. As in the proof of Lemma 11, we can
take $m=m(M,\frac{1}{4})$ and partition $X$ into clopen sets
$X_{1},\ldots,X_{m}$ such that for all non-identity $\gamma\in F$, each
$X_{j}$ is disjoint from $\gamma X_{j}$. Then we construct sets
$\emptyset=A_{0}\subseteq A_{1}\subseteq\cdots\subseteq A_{m}$ as follows.
Then we let $E=A_{m}$, and $E$ will be our maximal subset of $A$ such that the
translates of $E\times B$ are disjoint. For each $j=1,\ldots,m$ we set $A_{j}$
to be

$A_{j}=A_{j-1}\cup\left((A\cap X_{j})\setminus\bigcup_{\gamma\in F}\gamma
A_{j-1}\right).$

That is, to what we have so far, we add all elements of $A\cap X_{i}$ that do
not cause an intersection with translates of what we have so far.

The resulting set $E\times B$ has a non-self-intersecting orbit, because at
every step $j$, the part of $A\cap X_{j}$ that we add does not create any
self-intersections with itself, nor does it intersect translates of
$A_{j-1}\times B$. It is maximal, in the sense that for every $x\in A\setminus
E$, the set $\\{x\\}\times B$ intersects the orbit of $E\times B$. This is
because $x$ is in some $X_{j}$, and thus would have been added to $E$ at step
$j$ if it did not cause an intersection. And, $E$ is clopen and has nonzero
measure if $A$ does, because if we consider the first set $A\cap X_{j}$ that
has nonzero measure, then the measure of $A_{j}$ is at least the measure of
$A\cap X_{j}$.

Repeating this process finitely many times, once for every orbit in
$\widetilde{Z}_{0}$, we obtain sets $E_{1}\times
B_{1}(\frac{1}{4}),\ldots,E_{k}\times B_{k}(\frac{1}{4})$, such that none of
them or their translates intersect, and they are maximal with this property,
so that replacing their radii $\frac{1}{4}$ by $\frac{1}{2}$ gives sets
$E_{1}\times B_{1}(\frac{1}{2}),\ldots,E_{k}\times B_{k}(\frac{1}{2})$ for
which their orbits cover $\widetilde{Z}_{0}$. We apply Lemma 12 with
$r_{1}=\frac{1}{2}$ and $r_{2}\rightarrow 1$ to give

$\operatorname{Area}_{0}\left(\widetilde{Z}_{0}\cap\left(E_{i}\times
B_{i}\left(\frac{1}{2}\right)\right)\right)\cdot\frac{(\frac{1}{2})^{n}}{n!}\leq\mu(E_{i})\cdot
V_{1}+\varepsilon,$

and then again with $r_{1}\rightarrow 0$ and $r_{2}=\frac{1}{4}$ to give

$\mu(E_{i})\cdot\frac{(\frac{1}{4})^{n}}{n!}\leq\mu(E_{i})\cdot\operatorname{Vol}B_{i}\left(\frac{1}{4}\right)+\varepsilon.$

Then as in the proof of Theorem 5, we sum over all $i$, string the
inequalities together, and take $\varepsilon\rightarrow 0$ to get the desired
conclusion

$\|M\|_{\Delta}\leq 16^{n}(n!)^{2}\cdot V_{1}\cdot\operatorname{Vol}M.$

∎

##  5\. Integral foliated simplicial volume

We have succeeded in proving Theorem 1, which corresponds to Theorems 1.1 and
1.3 of [BS21] and is the main goal of that paper. In the course of their
proof, Braun and Sauer deduce some additional corollaries, which are Theorems
1.4 and 1.5 of their paper. In this section we explore whether our new proof
method also produces these corollaries. The answer is that it does, except
that we need a small hypothesis not included in [BS21]: we assume that the
fundamental group $\Gamma$ does not have any nontrivial elements of finite
order.

These corollaries depend on using coefficients in $\mathbb{Z}$ rather than in
$\mathbb{R}$. Our proof above uses $\mathbb{R}$-coefficients in two ways: once
to weight each thickened simplex by the measure of its Cantor set component,
and once in the simplex straightening step, where each straightened simplex is
chosen to be an average of permuted and translated simplices. For the first
use of $\mathbb{R}$-coefficients, we are in the same situation as Braun and
Sauer. Once we express our proof slightly differently, so that the norm of the
fundamental cycle is estimated in the Cantor bundle rather than downstairs in
$M$ or $B\Gamma$, we can reproduce their proofs of the corollaries. For the
second use of $\mathbb{R}$-coefficients, we need to eliminate the averaging
from the simplex straightening step, which is not a step in the Braun–Sauer
proof. To eliminate the averaging over permutations, we use the work of
Campagnolo and Sauer in [CS19]. I do not know how to make a similar
modification to eliminate the averaging over translations, which is why we
introduce the additional hypothesis not assumed by Braun and Sauer.

The corollaries follow from one main estimate, which we state as Theorem 13
below. The statements of the theorem and corollaries need several definitions
from [BS21]. First is the definition of what can be thought of informally as
the singular homology of a Cantor bundle. Let $P$ be a space with fundamental
group $\Gamma$, so that $\Gamma$ acts on $X\times\widetilde{P}$. As usual,
$C_{d}(\widetilde{P};\mathbb{Z})$ denotes the singular $d$-chains on
$\widetilde{P}$ for each degree $d$. We would like to be able to thicken these
by clopen subsets of the Cantor set $X$. Formally, let $C(X;\mathbb{Z})$ be
the set of functions from $X$ to $\mathbb{Z}$ that are continuous—that is, the
preimage of every integer is a clopen set. Then the homology we are looking
for is that of the chain complex

$C(X;\mathbb{Z})\bigotimes_{\mathbb{Z}[\Gamma]}C_{d}(\widetilde{P};\mathbb{Z}).$

Given a cycle $z=\sum_{i}f_{i}\otimes\sigma_{i}$ in this chain complex, where
none of the $\sigma_{i}$ are $\Gamma$-translates of each other, the analogue
of the $\ell^{1}$ norm of $z$ is

$\left\lvert z\right\rvert_{1}=\sum_{i}\left\lvert\int f_{i}\
d\mu\right\rvert.$

Then the $X$-parametrized integral simplicial norm of a homology class of this
chain complex is the infimum of $\left\lvert z\right\rvert_{1}$ over all
cycles $z$ representing the homology class. The $X$-parametrized integral
simplicial norm of the class $[z]$ is denoted by $\|[z]\|_{\mathbb{Z}}^{X}$.
Given any singular cycle on $P$, we can build a corresponding cycle in the
chain complex we have just constructed, by replacing each simplex $\sigma_{i}$
by $1\otimes\sigma_{i}$. We denote this map on homology by $j_{*}^{P}$.

The theorem below gives an upper bound for the quantity
$\|j_{*}^{B\Gamma}c_{*}[M]\|_{\mathbb{Z}}^{X}$, where $c\colon\thinspace
M\rightarrow B\Gamma$ is the classifying map. This means, then, that we take a
fundamental cycle of $M$, map it to the classifying space $B\Gamma$, find the
corresponding cycle on $X\times E\Gamma$, and construct a homologous cycle
(with integer coefficients) for which the norm is sufficiently small.

######  Theorem 13.

Let $M$ be a closed, oriented, $n$-dimensional Riemannian manifold, such that
for all points $\widetilde{p}$ in the universal cover $\widetilde{M}$ of $M$,
we have $\operatorname{Vol}B(\widetilde{p},1)\leq V_{1}$. Let
$\Gamma=\pi_{1}(M)$, suppose that $\Gamma$ has no nontrivial elements of
finite order, and consider a free, continuous action of $\Gamma$ on the Cantor
set $X$, with a $\Gamma$-invariant probability measure. Then the
$X$-parametrized integral simplicial norm of the image of the fundamental
homology class of $M$ under the classifying map satisfies the inequality

$\|j_{*}^{B\Gamma}c_{*}[M]\|_{\mathbb{Z}}^{X}\leq(n+1)16^{n}(n!)^{3}\cdot
V_{1}\cdot\operatorname{Vol}M.$

First we state the corollaries and summarize how Braun and Sauer deduce them
from this estimate. Then, at the end of this section we show how to modify our
main proof to prove this estimate.

For the first corollary we need the following definitions given in [BS21] and
attributed to [FLPS16, Gro99, Sch05]. Our action of $\Gamma$ on $X$ is an
example of a free probability-measure-preserving action $\alpha$ of $\Gamma$
on a standard probability space, meaning a Polish space where the set of
measurable sets is equal to the Borel $\sigma$-algebra. For such an action,
the norm that we have denoted by $\|j_{*}^{M}[M]\|_{\mathbb{Z}}^{X}$ can also
be denoted by $\left\lvert M\right\rvert^{\alpha}$, and the infimum of all
$\left\lvert M\right\rvert^{\alpha}$ over such actions of $\Gamma=\pi_{1}(M)$
on standard probability spaces is called the integral foliated simplicial
volume of $M$.

The following corollary is the same as Theorem 1.5 of [BS21], because the
assumption that $M$ is aspherical implies our hypothesis about $\Gamma$ not
having finite-order elements.

######  Corollary 14 (Theorem 1.5 of [BS21]).

Let $M$ be a closed, oriented, $n$-dimensional Riemannian manifold, such that
for all points $\widetilde{p}$ in the universal cover $\widetilde{M}$ of $M$,
we have $\operatorname{Vol}B(\widetilde{p},1)\leq V_{1}$, and suppose that
$\widetilde{M}$ is contractible. Then for any free probability-measure-
preserving action $\alpha$ of $\Gamma=\pi_{1}(M)$ on a standard probability
space, we have

$\left\lvert M\right\rvert^{\alpha}\leq(n+1)16^{n}(n!)^{3}\cdot
V_{1}\cdot\operatorname{Vol}M,$

so $(n+1)16^{n}(n!)^{3}\cdot V_{1}\cdot\operatorname{Vol}M$ is an upper bound
on the integral foliated simplicial volume of $M$.

###### Proof.

Because $M$ is aspherical, we can deduce that $\Gamma$ has no non-trivial
elements of finite order. I learned this argument from Caterina Campagnolo.
Suppose to the contrary that $\Gamma$ had a finite cyclic subgroup
$\mathbb{Z}/k$. Then the covering space $\widehat{M}$ of $M$ with fundamental
group $\mathbb{Z}/k$ would be a finite-dimensional classifying space for
$\mathbb{Z}/k$, which is impossible because $\mathbb{Z}/k$ has nontrivial
homology in an unbounded sequence of degrees. Thus, $\Gamma$ has no finite
cyclic subgroup, and we may apply Theorem 13.

The remainder of the proof is identical to the proof in [BS21]. Because
$\widetilde{M}$ is contractible, we can take $B\Gamma$ to be equal to $M$, so
$j_{*}^{B\Gamma}c_{*}[M]=j_{*}^{M}[M]$. This proves the desired bound on
$\left\lvert M\right\rvert^{\alpha}$ in the case where $\alpha$ is our chosen
action of $\Gamma$ on the Cantor set $X$. Then a theorem from [Ele21] shows
that given any other $\alpha$ on any other standard probability space, there
is a suitable embedding of $X$ into that space that allows us to deduce the
conclusion. ∎

The second corollary is the same as Theorem 1.4 of [BS21], except that in part
(1) we add the hypothesis that $\Gamma$ has no nontrivial elements of finite
order. For information about the von Neumann rank we refer to [BS21,
Definition 3.8], and for information about $\ell^{2}$-Betti numbers, Braun and
Sauer cite Lück [Lüc98, Lüc02].

######  Corollary 15 (Theorem 1.4 of [BS21]).

Let $M$ be a closed, oriented, $n$-dimensional Riemannian manifold, such that
for all points $\widetilde{p}$ in the universal cover $\widetilde{M}$ of $M$,
we have $\operatorname{Vol}B(\widetilde{p},1)\leq V_{1}$.

  1. (1)

Let $\Gamma=\pi_{1}(M)$, and suppose that $\Gamma$ has no nontrivial elements
of finite order. Then the von Neumann rank of the image $c_{*}[M]\in
H_{n}(B\Gamma)$ of the fundamental homology class of $M$ under the classifying
map is bounded above by $n(n+1)16^{n}(n!)^{3}\cdot
V_{1}\cdot\operatorname{Vol}M$.

  2. (2)

Suppose that $\widetilde{M}$ is contractible. Then its $\ell^{2}$-Betti
numbers satisfy

$\beta_{i}^{(2)}(M)\leq(n+1)16^{n}(n!)^{3}\cdot V_{1}\cdot\operatorname{Vol}M$

for every $i\in\mathbb{N}$, and the Euler characteristic satisfies

$\left\lvert\chi(M)\right\rvert\leq n(n+1)16^{n}(n!)^{3}\cdot
V_{1}\cdot\operatorname{Vol}M.$

###### Proof.

The proof is the same as in [BS21]. For the statement about von Neumann rank,
Theorem 3.10 of [BS21] states that the von Neumann rank of a class in degree
$d$ is bounded above by $d$ times the corresponding $X$-parametrized integral
simplicial norm. Our fundamental class is in degree $n$, so the bound on its
von Neumann rank is $n$ times the bound from Theorem 13.

For the statement about $\ell^{2}$-Betti numbers, Theorem 3.7 of [BS21],
attributed to [Sch05], states that every $\ell^{2}$-Betti number of a closed,
oriented manifold is bounded above by the $X$-parametrized simplicial norm of
its fundamental class. Using the assumption that $M=B\Gamma$, this is the same
as the quantity bounded by Theorem 13. An aspherical manifold of dimension $n$
can have nonzero $\ell^{2}$-Betti numbers only in degrees $1$ through $n$, and
the alternating sum of $\ell^{2}$-Betti numbers equals the Euler
characteristic, so summing the bound on all $\ell^{2}$-Betti numbers yields
the bound on Euler characteristic. ∎

We turn our attention to proving Theorem 13, which is the estimate that
implies these corollaries. The proof of Theorem 1 needs to be modified only
slightly to produce this estimate. The difference is that we need to prove the
following Lemma 16, which corresponds to Theorem 2, so that we can estimate
the norm of our thickened triangulation in the Cantor bundle without
projecting it down to $M$.

To state the lemma, we need to adapt the definition of $\pi_{1}$-killing
vertex coloring. Let $z=\sum_{i}f_{i}\otimes\sigma_{i}$ be a cycle in the
chain complex
$C(X;\mathbb{Z})\bigotimes_{\mathbb{Z}[\Gamma]}C_{d}(\widetilde{P};\mathbb{Z})$.
Each $f_{i}$ is a $\mathbb{Z}$-linear combination of (possibly infinitely
many) characteristic functions $1_{A}$, where the various sets $A$ are clopen
subsets of $X$. The product of one such $A$ with a vertex of $\sigma_{i}$ is
what we call a thick vertex of $z$. If another simplex $\sigma_{j}$ or one of
its translates shares a boundary face with $\sigma_{i}$, we consider each
thick vertex on that face of $\sigma_{j}$ to be the same thick vertex as its
counterpart in $\sigma_{i}$. A $\pi_{1}$-killing vertex coloring colors the
thick vertices of $z$ such that for each color, if we take all edges of $z$
for which both vertices are that color, and project these to $P$, then this
$1$-complex in $P$ is null-homotopic.

######  Lemma 16.

Let $P$ be a space, let $\Gamma=\pi_{1}(P)$, and suppose that $\Gamma$ has no
nontrivial elements of finite order. Consider a free, continuous action of
$\Gamma$ on the Cantor set $X$, with a $\Gamma$-invariant probability measure
$\mu$. Abusing notation, let $c\colon\thinspace
X\times\widetilde{P}\rightarrow X\times E\Gamma$ be the $\Gamma$-equivariant
map corresponding to the classifying map $c\colon\thinspace P\rightarrow
B\Gamma$. Let $z=\sum_{i}f_{i}\otimes\sigma_{i}$ be a cycle in the chain
complex
$C(X;\mathbb{Z})\bigotimes_{\mathbb{Z}[\Gamma]}C_{d}(\widetilde{P};\mathbb{Z})$,
with a $\pi_{1}$-killing vertex coloring. Then the $X$-parametrized integral
simplicial norm of the image of the homology class of $z$ satisfies

$\|c_{*}[z]\|_{\mathbb{Z}}^{X}\leq(d+1)!\cdot\sum_{\mathrm{rainbow}\
\sigma_{i}}\left\lvert\int f_{i}\ d\mu\right\rvert.$

###### Proof.

The proof follows the proof of Theorem 2: homotope $c(z)$ so that all vertices
of each color and all edges among them go to a single point—or more precisely,
a single $\Gamma$-orbit in $X\times E\Gamma$. Then we straighten, needing to
define the notion of straightening in such a way that each non-rainbow simplex
maps to $0$ and each rainbow simplex maps to a sum of $(d+1)!$ simplices. For
this notion of straightening, we cite Theorem 3.3 of [CS19], which constructs
a chain homotopy equivalence given by replacing each $d$-simplex by a signed
sum of the $(d+1)!$ simplices of its barycentric subdivision, in such a way
that permuting the vertices of the simplex results in multiplying by the sign
of the permutation.

Thus, to define the straightening, we assume that the straightening has
already been constructed for dimensions less than $d$, and consider the
$(d+1)$-tuples of points in $E\Gamma$, with the action of $\Gamma$ on these
tuples. For each orbit, we take one $(d+1)$-tuple in the orbit and select a
$d$-simplex with those vertices, such that the boundary agrees with our lower-
dimensional straightening choices, and take the signed barycentric subdivision
to be our straightening. Translating by $\Gamma$, this defines the
straightening on all $(d+1)$-tuples in the same orbit. Each tuple in the orbit
is obtained by only one translation, because the assumption that $\Gamma$ has
no finite-order elements implies that no $(d+1)$-tuple is mapped to a
permutation of itself by a nontrivial element of $\Gamma$. Thus, the
straightening is well-defined, without the need for averaging and introducing
non-integer coefficients. The straightening of any simplex that has two
vertices at the same point of $E\Gamma$ is $0$, because the transposition that
swaps those two vertices does not affect the straightening, while also
multiplying it by $-1$. ∎

###### Proof of Theorem 13.

The proof follows the proof of Theorem 1. The main change is in the conclusion
of Lemma 10, which becomes the following: there is a cycle
$z=\sum_{i}f_{i}\otimes\sigma_{i}$ in the homology class of $j_{*}^{M}[M]$
with a $\pi_{1}$-killing vertex coloring, such that

$\sum_{\mathrm{rainbow}\ \sigma_{i}}\left\lvert\int f_{i}\
d\mu\right\rvert=2^{n}\cdot\operatorname{Area}_{0}(\widetilde{Z}_{0}).$

No change is needed to prove this statement, except to estimate the norms up
in $X\times\widetilde{M}$ instead of mapping down to $M$.

Applying Lemma 16, we obtain the inequality

$\|c_{*}j_{*}^{M}[M]\|_{\mathbb{Z}}^{X}\leq(n+1)!\cdot
2^{n}\cdot\operatorname{Area}_{0}(\widetilde{Z}_{0}).$

We observe that $c_{*}j_{*}^{M}[M]$ and $j_{*}^{B\Gamma}c_{*}[M]$ mean the
same thing: take the class on $X\times E\Gamma$ corresponding to the
fundamental homology class on $M$. Thus we have

$\|j_{*}^{B\Gamma}c_{*}[M]\|_{\mathbb{Z}}^{X}\leq(n+1)!\cdot
2^{n}\cdot\operatorname{Area}_{0}(\widetilde{Z}_{0}),$

and the remainder of the proof of Theorem 1 proceeds as before, giving

$\|j_{*}^{B\Gamma}c_{*}[M]\|_{\mathbb{Z}}^{X}\leq(n+1)!\cdot
16^{n}(n!)^{2}\cdot V_{1}\cdot\operatorname{Vol}M=(n+1)16^{n}(n!)^{3}\cdot
V_{1}\cdot\operatorname{Vol}M.$

∎

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