text:
\begin{document}

\title{Macroscopic scalar curvature and local collapsing}

\author[S.~Sabourau]{St\'ephane Sabourau}

\thanks{Partially supported by the ANR project Min-Max (ANR-19-CE40-0014).}

\address{\parbox{\linewidth}{LAMA, Univ Paris Est Creteil, Univ Gustave Eiffel, CNRS, F-94010, Cr\'eteil, France \\
CRM (UMI3457), CNRS, Montr\'eal, QC H3C 3J7, Canada}}

\email{stephane.sabourau@u-pec.fr}

\subjclass[2010]{Primary 53C23; Secondary 53C20, 51K10}

\keywords{Macroscopic scalar curvature, volume growth, Alexandrov/Urysohn width, collapsing, Margulis function, exponential growth.}

\begin{abstract}
Consider a closed Riemannian $n$-manifold~$M$ admitting a negatively curved Riemannian metric.
We show that for every Riemannian metric on~$M$ of sufficiently small volume, there is a point in the universal cover of~$M$ such that the volume of every ball of radius~$r \geq 1$ centered at this point is greater or equal to the volume of the ball of the same radius in the hyperbolic $n$-space.
We also give an interpretation of this result in terms of macroscopic scalar curvature.
This result, which holds more generally in the context of polyhedral length spaces, is related to a question of Guth.
Its proof relies on a generalization of recent progress in metric geometry about the Alexandrov/Urysohn width involving the volume of balls of radius in a certain range with collapsing at different scales.
\end{abstract}

\maketitle

\section{Introduction}

The scalar curvature of a closed Riemannian $n$-manifold~$M$ describes how the volume of infinitesimal balls in~$M$ compares to the volume of infinitesimal balls in the Euclidean $n$-space.
More precisely, the volume expansion of a ball of radius~$r$ centered at $x \in M$ satisfies
\begin{equation} \label{eq:scal}
\vol(B(x,r)) = \omega_n \, r^n \left( 1 - \frac{\scal(M,x)}{6(n+2)} \, r^2 + O(r^3) \right)
\end{equation}
as $r$ goes to zero, where $\scal(M,x)$ is the scalar curvature of~$M$ at~$x$ and $\omega_n$ is the volume of a unit ball in the Euclidean $n$-space; see~\cite[Theorem~3.98]{GHL}.
Understanding the relationship between scalar curvature and the topology of~$M$ is a major problem in Riemannian geometry.

\medskip

In this article, we will be interested in macroscopic versions of the following conjecture attributed to Schoen (which follows from a conjecture of Schoen about the Yamabe invariant of hyperbolic manifolds); see~\cite{schoen} and~\cite{Guth11}.
This conjecture was also stated by Gromov~\cite[3.A]{gro86} with nonsharp constants.

\begin{conjecture}[Schoen] \label{conj}
Let $(M,{\rm hyp})$ be a closed hyperbolic $n$-manifold and let $g$ be another Riemannian metric on~$M$.
If $\scal(g,x) \geq \scal({\rm hyp})$ for every $x \in M$ then
\[
\vol(M,g) \geq \vol(M,{\rm hyp}).
\]

Equivalently, using~\eqref{eq:scal}, if $\vol(M,g) < \vol(M,{\rm hyp})$ then there exists $x_0 \in M$ such that
\[
\vol_g(B(x_0,r)) > \vol_{\rm hyp}(B(r))
\]
for every $r>0$ small enough.
\end{conjecture}

This conjecture is true in dimension~$2$ by the Gauss-Bonnet formula and in dimension~$3$ from Perelman's work; see~\cite[Proposition ~93.10]{KL08}.
In higher dimension, it also holds true for Riemannian metrics close enough to the hyperbolic one, see~\cite[Corollaire~C]{BCG91}, or if one replaces scalar curvature with Ricci curvature, see~\cite{BCG}.

\medskip

Following~\cite{GuthICM}, this leads us to introduce the following notion.
The \emph{macroscopic scalar curvature} of a closed Riemannian $n$-manifold~$M$ at scale~$r$ at~$x \in M$, denoted by $\scal_r(M,x)$, is defined as the unique real~$s$ such that
\[
\vol(B_{\tilde{M}}(\tilde{x},r)) = \vol(B_{\hyp_s^n}(r))
\]
where $\tilde{x}$ is a lift of~$x$ in the universal cover~$\tilde{M}$ of~$M$ and~$\hyp_s^n$ is the simply-connected $n$-dimensional space form with constant curvature~$s$.
It is more conveniently characterized as follows
\[
\scal_r(M,x) \leq s \mbox{ if and only if } \vol(B_{\tilde{M}}(\tilde{x},r)) \geq \vol(B_{\hyp_s^n}(r)).
\]
For example, the macroscopic scalar curvature of a flat torus at any scale is zero.
Note that this property fails if one does not take balls in the universal cover of~$M$, but only in~$M$, in the definition of the macroscopic scalar curvature, as otherwise, it would be positive at a large enough scale.
By~\eqref{eq:scal}, at infinitesimally small scale, we have
\[
\lim_{r \to 0} \scal_r(M,x) = \scal(M,x).
\]

In a different direction, the macroscopic scalar curvature at large enough scale provides information on the exponential growth rate of the volume of balls in the universal cover of~$M$, also known as the volume entropy, a much-studied geometric invariant related to the growth of the fundamental group and the dynamics of the geodesic flow.
This leads us to define
\[
V_{\tilde{M}}(r) = \sup_{\tilde{x} \in \tilde{M}} \vol(B(\tilde{x},r))
\]
as the maximal volume of a ball of radius~$r$ in the universal cover of~$M$.
As explained in~\cite{Guth11} and~\cite{GuthICM}, the celebrated theorem of Besson, Courtois and Gallot~\cite{BCG} on the minimal volume entropy provides a macroscopic version of Schoen's conjecture~\ref{conj} at large enough scales.
Stated in a way suited for comparison (ignoring its rigidity counterpart), this result takes the following form.

\begin{theorem}[Besson-Courtois-Gallot~\cite{BCG}] \label{theo:BCG}
Let $(M,{\rm hyp})$ be a closed hyperbolic $n$-manifold and let $g$ be another Riemannian metric on~$M$.
If $\vol(M,g) < \vol(M,{\rm hyp})$ then there exists~$r_0>0$ (depending on~$g$) such that for every $r \geq r_0$
\[
V_{\tilde{M}}(r) > V_{\hyp^n}(r).
\]

In particular, if $\scal_r(g,x) > \scal_r({\rm hyp})$ for every $r$ large enough and every $x \in M$ then $\vol(M,g) > \vol(M,{\rm hyp})$.
\end{theorem}

A version of this theorem was first established by A.~Katok~\cite{katok} in dimension~$2$ and a nonsharp version was obtained before by Gromov~\cite{gro82} in every dimension.

\medskip

In~\cite{Guth11}, Guth asks for an estimate on~$r_0$ after proving the following nonsharp macroscopic version of Schoen's conjecture.

\begin{theorem}[Guth~\cite{Guth11}] \label{theo:guth}
Let $(M,{\rm hyp})$ be a closed hyperbolic $n$-manifold and let $g$ be another Riemannian metric on~$M$.
Then, for every $r \geq 1$, there exists a constant $\delta_{n,r}>0$ depending only on~$n$ and~$r$, such that if $\vol(M,g) \leq \delta_{n,r} \, \vol(M,{\rm hyp})$ then
\[
V_{\tilde{M}}(r) \geq V_{\hyp^n}(r).
\]

In other words, if $\scal_r(g,x) \geq \scal_r({\rm hyp})$ for every $x \in M$ then $\vol(M,g) \geq \delta_{n,r} \, \vol(M,{\rm hyp})$.
\end{theorem}

Further volume lower bounds have recently been obtained by Alpert and Funano for hypersurfaces in closed manifolds with macroscopic scalar curvature bounded below as a consequence of this result; see~\cite{AF17} and~\cite{A}.

\medskip

Theorem~\ref{theo:guth} gives relatively better volume estimates for unit balls (that is, for $r=1$) than for balls of large radius as the constant~$\delta_{n,r}$ falls off exponentially or faster with~$r$.
In~\cite{Guth11}, Guth suggests that one could try to combine the approaches of~\cite{gro82}, \cite{BCG} and~\cite{Guth11} to obtain a uniform volume estimate with $\delta_{n,r} = \delta_n$ depending only on~$n$.
Such a uniform bound was obtained for surfaces by Karam~\cite{K15}.
In higher dimension, Balacheff and Karam~\cite{BK19} proved a similar result for negatively curved metrics~$g$ using techniques developed in~\cite{gro82}.

\medskip

In this article, we establish the following result in this direction, following a different approach.
See Theorem~\ref{theo:ball}, Corollary~\ref{coro:ball} and Corollary~\ref{coro:final} for more general statements.

\begin{theorem} \label{theo:hyp}
Let $(M,{\rm hyp})$ be a closed hyperbolic $n$-manifold and let $g$ be another Riemannian metric on~$M$.
Then, there exists a constant $\delta'_{n}>0$ depending only on~$n$, such that if \mbox{$\vol(M,g) \leq \delta'_{n}$} then
\[
V_{\tilde{M}}(r) \geq V_{\hyp^n}(r)
\]
for every $r \geq 1$.

In other words, if $\scal_r(g,x) \geq \scal_r({\rm hyp})$ for some $r \geq 1$ and every $x \in M$ then \mbox{$\vol(M,g) \geq \delta_{n}'$.}
\end{theorem}

This result provides a uniform estimate on the radius of balls in the universal cover of~$M$ whose volume is at least the volume of balls of the same radius in~$\hyp^n$.
Furthermore, the balls of radius~$r \geq 1$ of large volume in the universal cover of~$M$ can be assumed to be centered around the same point; see Corollary~\ref{coro:final}.
Actually, we obtain a more general lower bound on the maximal volume of balls that holds for every~$r \geq 0$; see Corollary~\ref{coro:ball}.
However, contrary to Theorem~\ref{theo:guth} and Theorem~\ref{theo:BCG}, the condition of the volume of~$M$ which guarantees the existence of balls of large volume in the universal cover of~$M$ does not involve the volume of the hyperbolic metric in~$M$.
This leads to a stronger condition for manifolds of large hyperbolic volume.
Note that similar versions of Theorem~\ref{theo:hyp} hold for more general spaces, including closed manifolds admitting negatively curved metrics.
See Corollary~\ref{coro:ball} and Corollary~\ref{coro:final}.

\medskip

The overall strategy of the proof of Theorem~\ref{theo:hyp} is inspired by the approach developed in~\cite{S} and extended in~\cite{BS} to prove that the minimal volume entropy of a closed manifold admitting a hyperbolic metric is positive.
(As explained above, this result was first obtained in~\cite{gro82} using bounded cohomology argument, before a sharp version was established in~\cite{BCG} using the so-called barycenter map.)
In~\cite{S}, we show that for every closed Riemannian $n$-manifold admitting a hyperbolic metric, there exist two loops~$\gamma_1$ and~$\gamma_2$ based at the same point, of length $\lesssim \vol(M,g)^{\frac{1}{n}}$, whose homotopy classes generate a subgroup~$\Gamma \leqslant \pi_1(M)$ of positive algebraic entropy.
The proof of this estimate relies on filling techniques developed in~\cite{gro83} to establish systolic inequalities, and more precisely, on a lower bound on the filling radius.
A lower bound on the volume entropy of~$M$ immediately follows from this estimate by standard comparison argument.
Now, in order to derive a lower bound of the volume of balls in the universal cover of~$M$ for any radius, and not simply for asymptotically large radius, we need to show that there is sufficiently volume around the basepoint of the two loops~$\gamma_1$ and~$\gamma_2$ to distribute it in the universal cover of~$M$ under the action of~$\Gamma$.
The approach developed in~\cite{S}, based on filling techniques, does not readily provide any information on where the volume is located and whether there is enough volume around the basepoint of the two loops.
In a different direction, Guth~\cite{Guth11} shows that the volume in the universal cover of a closed aspherical Riemannian manifold is not too diffuse and that a non-negligeable part of it is contained in a ball of given radius, but this result does not say where such a ball is located\footnote{Though we believe we can combine the argument of~\cite{S} with the construction in~\cite{Guth11} to obtain the desired result, we will use different, more elementary, techniques to reach the same conclusion.
These techniques also have the advantage to apply to more general spaces than closed manifolds.}.
Later, building upon on the construction of~\cite{Guth11}, he extended this result into the following theorem.

\begin{theorem}[Guth~\cite{Guth17}] \label{theo:guth2}
Let $M$ be a closed Riemannian $n$-manifold.
If every ball of radius~$R$ in~$M$ has volume at most~$\varepsilon_n R^n$ for a sufficiently small constant~$\varepsilon_n >0$ then the Alexandrov width of~$M$ is at most~$R$.
That is, there exists a continuous map from~$M$ to a simplicial $(n-1)$-complex~$P$ whose fibers have radius at most~$R$.
\end{theorem}

This theorem was generalized by Liokumovich, Lishak, Nabutovsky and Rotman~\cite{LLNR} to metric spaces and Hausdorff content instead of volume.
The proof of this result was greatly simplified by Papasoglu~\cite{P} using the minimal hypersurface approach of Schoen and Yau~\cite{SY79} developed by Guth~\cite{Guth10} in a similar context.
Recently, Nabutovsky~\cite{nab} further extended this approach improving the constant from an exponential bound to a linear one.

\medskip

Now, one can show that the Alexandrov width of a closed Riemannian $n$-manifold~$M$ admitting a hyperbolic metric is bounded from below (up to a multiplicative constant depending only on~$n$) by the minimal length~$L$ such that there exist two loops~$\gamma_1$ and~$\gamma_2$ based at the same point, of length at most~$L$, whose homotopy classes generate a free subgroup of~$\pi_1(M)$; see Section~\ref{sec:margulis}.
This estimate can be interpreted as a bound on Margulis' constant; see Definition~\ref{def:margulis}.
Thus, by Theorem~\ref{theo:guth2}, there exists a ball of radius $\frac{1}{2} L$ in~$M$ of volume at least $c_n L^n$.
As explained above, we would like to show that such a ball~$B$ is centered at the basepoint of the two loops~$\gamma_1$ and~$\gamma_2$ in order to derive an exponential lower bound on the volume of balls in the universal cover~$\tilde{M}$ by taking the translates of~$B$ in~$\tilde{M}$ under the free subgroup induced by the two loops in homotopy.
This does not follow directly from the Alexandrov/Urysohn width estimates of \cite{Guth17}, \cite{LLNR}, \cite{P} or~\cite{nab}.
Instead, we need to extend Theorem~\ref{theo:guth2} by allowing the fibers in the definition of the Alexandrov width to have diameters of various size so that they capture the thick part of~$M$ in an optimal way.

\medskip

This leads us to introduce the following definition.

\begin{definition} \label{def:collapse}
Recall that a compact $n$-polyhedral length space is a finite simplicial $n$-complex endowed with an intrinsic metric (or length structure).
Let $X$ be a compact $n$-polyhedral length space and $\rho:X \to (0,\infty)$ be a function.
The space~$X$ \emph{$\rho$-collapses} if there exists a continuous map $\pi:X \to P$ to a finite simplicial $(n-1)$-complex such that every fiber of~$\pi$ lies in a ball $B(x,\rho(x))$ for some $x \in X$.
\end{definition}

Note that the Alexandrov width of a space~$X$ that $\rho$-collapses is bounded by~$\sup \rho$.
Thus, the bound on the size of the fibers of $\pi:X \to P$ in Definition~\ref{def:collapse} better reflects the shape of~$M$ than the corresponding uniform bound on the size of the fibers of the map in the definition of the Alexandrov width; see Theorem~\ref{theo:guth2}.
The collapsing of~$X$ can also be interpreted in terms of coverings; see Proposition~\ref{prop:mult}.

\medskip

We have the following general theorem which extends Theorem~\ref{theo:guth2} and some of its generalizations in~\cite{LLNR}, \cite{P} or~\cite{nab}.
The point of this theorem is that it allows us to have a better control on where the space~$X$ can collapse or not.

\begin{theorem} \label{theo:0}
Let $X$ be a compact $n$-polyhedral length space and $\rho:X \to (0,\infty)$ be a continuous function.
Suppose that the space~$X$ does not $\rho$-collapse.
Then there exists $x_0 \in X$ such that
\[
\HH_n(B_X(x_0,r)) \geq c_n \, r^n
\]
for every $r \in [0,\rho(x_0)]$, where $c_n$ is an explicit positive constant depending only on~$n$.
\end{theorem}

The proof of this theorem relies on the general approach developed by Papasoglu~\cite{P} and recently further improved by Nabutovsky~\cite{nab}.
As explained above, this approach can be traced back to works of Guth~\cite{Guth10} and, before, of Schoen and Yau~\cite{SY79}.
We will follow the arguments of~\cite{nab}.

\medskip

Now, if~$\rho$ is given by the Margulis function, see Definition~\ref{def:margulis}, and the space~$X$ does not $\rho$-collapse, we can show that, under some condition of the fundamental group of~$X$, there exist two loops of~$X$ based at~$x_0$, of length~$\lesssim \rho(x_0)$, generating a free subgroup in~$\pi_1(X)$ as desired.

\medskip

The article is organized as follows.
In Section~\ref{sec:collapse}, we prove Theorem~\ref{theo:0}.
Then, in Section~\ref{sec:margulis}, we introduce the Margulis function and derive a general version of Theorem~\ref{theo:hyp}.

\section{Collapse and volume localization} \label{sec:collapse}

The goal of this section is to give an interpretation of the collapse of a metric space~$X$ in terms of coverings through the following classical result and to prove Theorem~\ref{theo:0} and related results.
Given a subset~$A \subseteq X$, we denote by~$\bar{A}$ its closure in~$X$.

\begin{proposition} \label{prop:mult}
Let $X$ be a compact $n$-polyhedral length space and $\rho:X \to (0,\infty)$ be a function.
The space~$X$ $\rho$-collapses if and only if there exists a finite covering of~$X$ of multiplicity at most~$n$ by open subsets~$U_i$ whose closures~$\bar{U}_i$ are covered by balls $B(x_i,\rho(x_i))$ with $x_i \in X$.
\end{proposition}

\begin{remark}
The statement of the proposition would still hold by requiring only~$U_i$ to be covered by a ball $B(x_i,\rho(x_i))$.
Though more technical, the current statement simplifies the proofs of forthcoming results, in particular Lemma~\ref{lem:XY}.
\end{remark}

\begin{proof}[Proof of Proposition~\ref{prop:mult}]
Let $\pi:X \to P$ be a continuous map to a finite simplicial $(n-1)$-complex~$P$ such that every fiber of~$\pi$ lies in a ball $B(x,\rho(x))$ for some $x \in X$.
Since $P$ is a finite simplicial $(n-1)$-complex, the finite covering of~$P$ formed of the open stars~${\rm st}(v) \subseteq P$ of the vertices~$v$ of~$P$ has multiplicity at most~$n$.
Subdividing~$P$ if necessary, we can assume that the preimages $U_v = \pi^{-1}({\rm st}(v)) \subseteq X$ of these open stars have their closures~$\bar{U}_v$ lying in the same balls $B(x,\rho(x))$ covering the compact fibers~$\pi^{-1}(v)$.
Furthermore, the subsets~$U_v$ form a finite covering of~$X$ of multiplicity at most~$n$.

Let $\mathcal{U}=\{ U_i \mid i=1,\dots,m \}$ be a finite covering of~$X$ of multiplicity at most~$n$ by open subsets~$U_i$ with $\bar{U}_i \subseteq B(x_i,\rho(x_i))$ for some $x_i \in X$.
Take a partition of unity~$\{ \phi_i \}$ of~$X$ subordinated to~$\{ U_i \}$.
Consider the map $\pi:X \to \Delta^{m-1}$ defined by
\[
\pi(x)=(\phi_1(x),\dots,\phi_m(x))
\]
in the barycentric coordinates of~$\Delta^{m-1}$.
The nerve~$N$ of the covering~$\mathcal{U}$ is a finite simplicial complex with one vertex~$v_i$ for each open set~$U_i$, where $v_{i_0},\dots,v_{i_k}$ span a $k$-simplex of~$N$ if and only if the intersection $\cap_{j=1}^k U_{i_j}$ is nonempty.
By construction, the dimension of the nerve~$N$ is one less than the multiplicity of the covering~$\mathcal{U}$.
That is, $\dim N \leq n-1$.
Now, we identify in a natural way the vertices~$\{ v_i \}$ of~$N$ with the vertices of~$\Delta^{m-1}$.
With this identification, the nerve~$N$ of~$X$ lies in~$\Delta^{m-1}$.
Furthermore, the image of~$\pi$ lies in~$N$ and every fiber of~$\pi:X \to N$ lies in one of the open subsets~$U_i \subseteq B(x_i,\rho(x_i))$.
Hence, the space~$X$ $\rho$-collapses.
\end{proof}

Let us recall Eilenberg's inequality; see~\cite[13.3.1]{BZ}.

\begin{theorem}[Eilenberg's inequality]
Let $f:X \to Y$ be a Lipschitz map between separable metric space (\ie, compact metric spaces).
Then for every $A \subseteq X$ and every $0 \leq m \leq n$, we have
\begin{equation} \label{eq:eilenberg}
\int_Y^{\mkern-20mu *} \HH_{n-m}(A \cap f^{-1}(y)) \, d\HH_m(y) \leq \frac{v_{n-m} v_m}{v_n} \, ({\rm Lip} \, f)^m \, \HH_n(A)
\end{equation}
where $\int^{\mkern-18mu *}$ denotes the upper Lebesgue integral, $v_n$ is the volume of the unit ball in~$\R^n$ and $\HH_n$ is the $n$-dimensional Hausdorff measure.
\end{theorem}

The following result is the equivalent to~\cite[Lemma~2.1]{nab} in our language.

\begin{lemma} \label{lem:n=1}
Let $X$ be a compact $1$-polyhedral length space and $\rho:X \to (0,\infty)$ be a function.
Suppose that for every $x \in X$, there exists a ball $B(x,r_x) \subseteq X$ of radius $r_x \in [0,\rho(x)]$ such that $\HH_1(B(x,r_x)) < r_x$.
Then the space~$X$ $\rho$-collapses.
\end{lemma}

\begin{proof}
By the Eilenberg inequality~\eqref{eq:eilenberg} with $m=n=1$, for every $x \in X$, there exists $\tau_x \in (0,r_x)$ such that the sphere $S(x,\tau_x) \subseteq X$ is empty.
Thus, every connected component of~$X$ lies in a ball $B(x,\rho(x))$ for some $x \in X$.
\end{proof}

The proof of Theorem~\ref{theo:x_0} rests on the following notion, which generalizes the definition given in~\cite[Definition~2.3]{nab}; see also~\cite{P}.

\begin{definition}
Let $X$ be a compact $n$-polyhedral length space and $\rho:X \to (0,\infty)$ be a bounded function.
A compact $(n-1)$-subcomplex~$Y$ of~$X$ \emph{$\rho$-separates}~$X$ if for every connected component~$C$ of~$X \setminus Y$, there exists $x \in X$ (depending on~$C$) such that $\bar{C} \subseteq B_X(x,\rho(x))$.
\end{definition}

The following result is a mere adaptation of~\cite[Lemma~2.5]{nab}.

\begin{lemma} \label{lem:XY}
Let $X$ be a compact $n$-polyhedral length space and $\rho:X \to (0,\infty)$ be a function.
Let $Y$ be a compact $(n-1)$-subpolyhedron of~$X$ which $\rho$-separates~$X$.
Denote by $\rho_Y$ the restriction of~$\rho$ to~$Y$.
Suppose that the space~$Y$ $\rho_Y$-collapses.
Then the space~$X$ $\rho$-collapses.
\end{lemma}

\begin{proof}
We follow the argument presented in~\cite[Lemma~2.5]{nab}.
Recall that the polyhedral space~$Y$ is endowed with the length structure induced from~$X$.
Let $\mathcal{U}=\{U_i\}$ be a finite covering of~$Y$ of multiplicity at most~$n-1$ by open subsets~$U_i$ of~$Y$ with $\bar{U}_i \subseteq B_Y(y_i,\rho(y_i))$ for some $y_i \in Y$.
We can thicken every subset~$U_i$ into an open subset~$V_i$ of~$X$ with $\bar{V}_i \subseteq B_Y(y_i,\rho(y_i))$, without increasing the multiplicity.
(Note that $B_Y(y,r) \subseteq B_X(y,r)$ for every $y \in Y$.)
Adding the connected components of~$X \setminus Y$ to the family~$\{ V_i\}$, we obtain a covering of~$X$ of multiplicity at most~$n$ (one more than the multiplicity of~$\mathcal{U}$) by open subsets of~$X$ whose closures lies in balls $B_X(x,\rho(x))$ with $x \in X$.
Hence, the space~$X$ $\rho$-collapses.
\end{proof}

Denote by $\lambda_n = \frac{v_n}{2 v_{n-1}}$ the coefficient in Eilenberg's inequality with $m=1$; see~\eqref{eq:eilenberg}.
Note that $\lambda_1=1$.

\begin{theorem} \label{theo:x_0}
Let $X$ be a compact $n$-polyhedral length space and $\rho:X \to (0,\infty)$ be a continuous function.
Suppose that the space~$X$ does not $\rho$-collapse.
Then there exists $x_0 \in X$ such that
\[
\HH_n(B_X(x_0,r)) \geq c_n \, r^n
\]
for every $r \in [0,\rho(x_0)]$, where $c_n = \frac{1}{n!} \, \prod_{i=1}^n \lambda_i$.
\end{theorem}

\begin{proof}

Let $\eta>0$ be small enough.
We can $\eta$-approximate the distance function~$d_X(x_0,\cdot)$ to~$x_0$ by a piecewise smooth (or even piecewise linear, after taking sufficiently fine subdivisions) $(1+\eta)$-Lipschitz function whose level sets are $(n-1)$-subpolyhedra.
To avoid burdening the argument with standard approximations, we will assume that the spheres~$S_X(x_0,r)$ are $(n-1)$-subpolyhedra, keeping in mind that the inequalities below, including the Eilenberg inequality, only hold with an extra $1+o(\eta)$ factor with $\eta>0$ arbitrarily small.
Letting $\eta$ go to zero, this will not change the final estimate.

\medskip

Let us show by induction on the dimension~$n$ of~$X$ that for every $\varepsilon \in (0,1)$ and every $\alpha \in (0,1)$, there exists $x_0 \in X$ such that
\begin{equation} \label{eq:lowerX}
\HH_n(B_X(x_0,r)) \geq \frac{1-\varepsilon}{n!} \left( {\textstyle \prod\limits_{i=1}^{n}} \lambda_i \right) r^n
\end{equation}
for every $r \in [\alpha  \, \rho(x_0),\rho(x_0)]$.
The case~$n=1$ with $c_1=1$ is covered by Lemma~\ref{lem:n=1}.

In the general case~$n \geq 2$, fix $\varepsilon \in (0,1)$ and $\alpha \in (0,1)$.
Denote $\varepsilon'=\frac{\varepsilon}{4}$ and $\alpha' = (\frac{\varepsilon}{2})^{\frac{1}{n}} \, \alpha$.
Consider a compact $(n-1)$-subcomplex~$Y \subseteq X$ which $\rho$-separates~$X$ and has minimal volume up to~$\delta$, where $\delta = \frac{\varepsilon'}{(n-1)!} \, \left( \prod_{i=1}^n \lambda_i \right) \, \alpha'^{n-1} \, (\inf \rho)^{n-1} >0$.
By Lemma~\ref{lem:XY}, the space~$Y$ does not $\rho_Y$-collapse.
Thus, by induction (replacing $\varepsilon$ with~$\varepsilon'$ and $\alpha$ with~$\alpha'$), there exists $y_0 \in Y$ such that
\begin{equation} \label{eq:volY}
\HH_{n-1}(B_Y(y_0,s)) \geq \frac{1-\varepsilon'}{(n-1)!}  \left( {\textstyle \prod\limits_{i=1}^{n-1}} \lambda_i \right)  s^{n-1}
\end{equation}
for every $s \in [\alpha'  \, \rho(y_0),\rho(y_0)]$.

Denote $x_0:=y_0$ and let $s \in [\alpha'  \rho(x_0),\rho(x_0))$.
Consider the compact $(n-1)$-subcomplex $Z \subseteq X$ obtained from~$Y$ by replacing $Y \cap B_X(y_0,s)$ with the sphere~$S_X(y_0,s)$.
The resulting space~$Z$ $\rho$-separates~$X$.
Indeed, the closure of the connected component~$B_X(x_0,s)$ of~$X \setminus Z$ is clearly covered by the ball $B_X(x_0,\rho(x_0))$.
Furthermore, all the other connected components of~$X \setminus Z$ are contained in connected components of~$X \setminus Y$ and therefore their closures also lie in balls $B_X(x,\rho(x))$ with $x \in X$.

Now, since $Y$ has minimal volume up to~$\delta$ among all compact $(n-1)$-subpolyhedra that $\rho$-separate~$X$, we derive
\[
\HH_{n-1}(Z) \geq \HH_{n-1}(Y) - \delta.
\]
This implies
\begin{align*}
\HH_{n-1}(S_X(x_0,s)) & \geq \HH_{n-1}(Y \cap B_X(y_0,s)) - \delta \\
 & \geq \HH_{n-1}(B_Y(y_0,s)) - \delta \\
 & \geq \frac{1-\varepsilon'}{(n-1)!}  \left( {\textstyle \prod\limits_{i=1}^{n-1}} \lambda_i \right)  s^{n-1} - \frac{\varepsilon'}{(n-1)!}  \left( {\textstyle \prod\limits_{i=1}^{n-1}} \lambda_i \right)  s^{n-1} \\
 & \geq \frac{1- 2 \varepsilon'}{(n-1)!}  \left( {\textstyle \prod\limits_{i=1}^{n-1}} \lambda_i \right)  s^{n-1}.
\end{align*}
where the second inequality follows from the inclusion $B_Y(y_0,s) \subseteq Y \cap B_X(y_0,s)$ and the third inequality follows from~\eqref{eq:volY} and the bound $s \geq \alpha' (\inf \rho)$.

Thus, by the Eilenberg inequality~\eqref{eq:eilenberg}, we obtain
\[
\HH_n(B_X(x_0,r)) \geq \lambda_n \, \int^{r \mkern-28mu *}_{\alpha'  \rho(x_0)} \HH_{n-1}(S_X(x_0,s)) \, ds \geq \frac{1- 2 \varepsilon'}{n!}  \left( {\textstyle \prod\limits_{i=1}^{n}} \lambda_i \right)  \left( r^n - \alpha'^n \, \rho(x_0)^n \right)
\]
for every $r \in [\alpha  \, \rho(x_0),\rho(x_0)]$.
(Note that $r \geq \alpha' \rho(x_0)$ since $\alpha' \leq \alpha$.)

We want this expression to be greater or equal to~$\frac{1-\varepsilon}{n!} \left( \prod_{i=1}^{n} \lambda_i \right) r^n$ for every $r \in [\alpha  \, \rho(x_0),\rho(x_0)]$.
That is, we want
\[
(\varepsilon - 2 \varepsilon') r^n \geq \alpha'^n \rho(x_0)^n.
\]
Since $r \geq \alpha  \, \rho(x_0)$, it is enough to have
\[
(\varepsilon - 2 \varepsilon') \alpha^n \geq \alpha'^n.
\]
Replacing $\varepsilon'$ and~$\alpha'$ with their expressions in terms of~$\varepsilon$ and~$\alpha$, we observe that this inequality is satisfied, and in fact, is an equality.
Hence,
\begin{equation*}
\HH_n(B_X(x_0,r)) \geq \frac{1-\varepsilon}{n!} \left( {\textstyle \prod\limits_{i=1}^{n}} \lambda_i \right)  r^n
\end{equation*}
for every $r \in [\alpha  \, \rho(x_0),\rho(x_0)]$.

\medskip

Take a decreasing sequence~$(\varepsilon_m)$ converging to zero and fix $\alpha=\varepsilon_m$.
There exists $x_m \in X$ such that the $n$-dimensional Hausdorff measure of~$B_X(x_m,s)$ satisfies a similar lower bound to~\eqref{eq:lowerX} for every $s \in [\varepsilon_m \, \rho(x_m), \rho(x_m)]$ with $\varepsilon=\varepsilon_m$.
By compactness of~$X$, we can assume that $(x_m)$ converges to a point~$x_0 \in X$.
Also, by compactness of~$X$ and continuity of~$\rho$, the function~$\rho$ is bounded.
Let $r \in [0,\rho(x_0)]$.
By the triangle inequality, we have
\[
B_X(x_0,r) \supseteq B_X(x_m,r-|x_0 x_m|).
\]
Thus, for $m$ large enough, we obtain
\[
\HH_n(B_X(x_0,r)) \geq \frac{1-\varepsilon_m}{n!} \left( {\textstyle \prod\limits_{i=1}^{n}} \lambda_i \right) \min\{ r-|x_0 x_m| ,\rho(x_m) \}^n.
\]
Since $|x_0 x_m|$ goes to zero and $\rho$ is continuous, we obtain the desired lower bound for~$\HH_n(B_X(x_0,r))$.
\end{proof}

\begin{remark}
If $X$ is a compact $n$-polyhderon with a piecewise Riemannian metric (\eg, a closed Riemannian $n$-manifold), we can apply the coarea formula, see~\cite[13.4.2]{BZ}, instead of Eilenberg's inequality in the proof of Theorem~\ref{theo:x_0}.
In this case, we can get rid of the product $\prod_{i=1}^n \lambda_i$ in the expression of~$c_n$.
\end{remark}

\begin{remark}
The lower bound~\eqref{eq:lowerX} holds even if $\rho$ is not continuous, provided it is bounded away from zero.
\end{remark}

Let us introduce a couple of definitions: one of topological nature and the other one of geometrical nature.

\begin{definition}
Let $X$ be a connected compact $n$-polyhedral space and $\phi:\pi_1(X) \to G$ be a group homomorphism to a discrete group~$G$.
The space~$X$ is \emph{$\phi$-essential} if the classifying map $\Phi:M \to K(G,1)$ induced by~$\phi$ is not homotopic to a continuous map $X \to P \to K(G,1)$ which factors out through a simplicial $(n-1)$-complex~$P$.
\end{definition}

\begin{definition}
Let $X$ be a connected compact $n$-polyhedral length space and $\phi:\pi_1(X) \to G$ be a nontrivial group homomorphism to a discrete group~$G$.
The \emph{$\phi$-systolic function} of~$X$ is the function $\sys_\phi:X \to (0,\infty)$ defined as
\[
\sys_\phi(X,x) = \inf \{ \length(\gamma) \mid \gamma \mbox{ is a noncontractible loop of } X \mbox{ based at } x \}.
\]
Clearly, the $\phi$-systolic function is continuous and even $2$-Lipschitz.
\end{definition}

We have the following non-collapsing result.

\begin{proposition} \label{prop:sys}
Let $X$ be a $\phi$-essential connected compact $n$-polyhedral length space, where $\phi:\pi_1(X) \to G$ is a nontrivial group homomorphism to a discrete group~$G$.
Then the space~$X$ does not $\frac{1}{2} \sys_\phi$-collapse.
\end{proposition}

\begin{proof}
We follow again an argument presented in~\cite{nab}.
Suppose that $X$ $\frac{1}{2} \sys_\phi$-collapses.
Denote by $\pi:\hat{X} \to X$ the cover corresponding to the subgroup~$\ker \phi \lhd \pi_1(X)$.
By Proposition~\ref{prop:mult}, there exists a finite covering of~$M$ of multiplicity at most~$n$ by open subsets~$U_i$ with $\bar{U}_i \subseteq B(x_i,\frac{1}{2} \sys_\phi(x_i))$ for some $x_i \in X$.
Without loss of generality, we can assume that the open subsets are connected.
Every loop~$\gamma$ in~$\bar{U}_i$ is homotopic to the concatenation of loops of length less than~$\sys_\phi(x_i)$ based at~$x_i$.
(The loops are of the form $\overline{x_ip_s} \cup \overline{p_sp_{s+1}} \cup \overline{p_{s+1}x_i}$, where the points $\{ p_i \}$ are given by a sufficiently fine subdivision of~$\gamma$.)
By definition of the $\phi$-systolic function, each homomorphism $\pi_1(U_i) \to \pi_1(X) \to G$ is trivial, where the first homomorphism is induced by the inclusion $U_i \hookrightarrow X$.
It follows that the preimage~$\pi^{-1}(U_i)$ of each open set~$U_i$ decomposes as the disjoint union of open sets~$\alpha \cdot \hat{U}_i \subseteq \hat{X}$, where $\hat{U}_i$ is a lift of~$U_i$ in~$\hat{X}$ and $\alpha$ runs over $H=\pi_1(X)/\ker \phi$.
Consider the nerve~$N$ of the covering~$\{ U_i \}$ of~$X$ and the nerve~$\hat{N}$ of the covering~$\{ \alpha \cdot \hat{U}_i \}$ of~$\hat{X}$.
We have the following commutative diagram
\begin{center}
\begin{tikzcd}
\hat{X} \arrow{r} \arrow{d}
&\hat{N} \arrow{d} \\
X \arrow{r} &N
\end{tikzcd}
\end{center}
where the horizontal maps correspond to the natural maps from a space to its nerve and the vertical maps are the quotient maps under the natural free actions of the subgroup~$H \leqslant G$.
We can construct an equivariant map $\hat{N} \to \tilde{K}$ to the universal cover~$\tilde{K}$ of~$K=K(G,1)$.
By construction, this map passes to the quotient giving rise to a map $N \to K$ such that the composite $X \to N \to K$ is homotopic to the classifying map induced by~$\phi$.
Since the covering~$\{ U_i \}$ has multiplicity at most~$n$, the nerve of~$N$ is a simplicial complex of dimension at most~$n-1$.
Therefore, the space~$X$ is not $\phi$-essential.
\end{proof}

\section{Margulis' constant and volume of large balls} \label{sec:margulis}

In this section, we introduce the Margulis function and apply the results of the previous section to obtain an exponential lower bound on the maximal volume of balls in the universal cover of a metric with negative curvature, and more generally, of a metric space satisfying similar features.

\medskip

We will need the following definition of the Margulis function.

\begin{definition} \label{def:margulis}
Let $X$ be a compact $n$-polyhedral length space.
Consider a group homomorphism $\phi:\pi_1(X) \to G$ to a discrete group~$G$.
For every $x \in X$ and $\mu >0$, denote by
\[
\Gamma^\mu_{\phi,x} = \langle \phi([\gamma]) \in G \mid \gamma \mbox{ loop of } X \mbox{ based at } x \mbox{ with } \length(\gamma) \leq \mu \rangle
\]
the subgroup of~$G$ generated by the $\phi$-image of the homotopy classes of the loops of~$X$ based at~$x$ of length at most~$\mu$.

The \emph{Margulis function} of~$X$ is the function $\mu_\phi:X \to (0,\infty]$ defined as
\[
\mu_\phi(x) = \sup \{ \mu \mid \Gamma^\mu_{\phi,x} \mbox{ has subexponential growth} \rangle.
\]

Clearly, the Margulis function is continuous and even $2$-Lipschitz.
With this definition, the subgroup~$ \Gamma^\mu_{\phi,x}$ has subexponential growth for every $\mu < \mu_\phi(x)$.
Note that if the image of~$\phi$ has exponential growth, the Margulis function is bounded by $2 \, \diam(X)$.

When $\phi$ is the identity homomorphism, we simply write~$\mu$ for~$\mu_\phi$.
\end{definition}

\begin{remark}
One could replace ``has subexponential growth" with ``is virtually nilpotent" in the definition of the Margulis function since we will be primarily interested in groups satisfying the Tits alternative; see~Definition~\ref{def:tits}.
\end{remark}

We have the following non-collapsing result.

\begin{proposition} \label{prop:simpl}
Every closed $n$-manifold~$M$ with nonzero simplicial volume does not $\frac{1}{2} \mu$-collapse.
\end{proposition}

\begin{proof}
Suppose that $M$ $\frac{1}{2} \mu$-collapses.
By Proposition~\ref{prop:mult}, there exists a finite covering of~$M$ of multiplicity at most~$n$ by open subsets~$U_i$ with $\bar{U}_i \subseteq B(x_i,\frac{1}{2} \mu(x_i))$ for some $x_i \in X$.
Without loss of generality, we can assume that the open subsets~$U_i$ are connected.
The image of~$\pi_1(U_i)$ under the $\pi_1$-homomorphism induced by the inclusion map $U_i \hookrightarrow M$ is generated by the homotopy classes of loops of length at most~$\mu(x_i)$ based at~$x_i$.
By definition of the Margulis function, see Definition~\ref{def:margulis}, this subgroup of~$\pi_1(X)$ has subexponential growth.
Thus, it is amenable.
By Gromov's vanishing simplicial volume theorem, see~\cite{gro82} or~\cite{iva}, it follows that $M$ has zero simplicial volume.
\end{proof}

\begin{remark}
More generally, every closed $n$-manifold with nonzero minimal volume entropy does not $\frac{1}{2} \mu$-collapse; see~\cite[Corollary~2.11]{BS}.
\end{remark}

Let us introduce a quantitative version of the Tits alternative for groups.

\begin{definition} \label{def:tits}
Let $\kappa$ be a positive integer.
A group~$G$ satisfies the \emph{$\kappa$-Tits alternative} if for every symmetric subset~$S$ of~$G$ containing the identity, either $S$ generates a subgroup of subexponential growth, or there exist two elements in~$S^\kappa$ generating a nonabelian free subgroup.
In the latter case, the subgroup generated by~$S$ has exponential growth and its algebraic entropy (or exponential growth rate) with respect to~$S$ is at least~$\frac{\log(3)}{\kappa}$.
\end{definition}

\begin{example} \label{ex}
Our main source of examples of groups satisfying the $\kappa$-Tits alternative is given by fundamental groups of closed Riemanian $n$-manifolds with sectional curvature lying between $-k^2$ and~$-1$ for some $k \geq 1$, where $\kappa=\kappa(n,k)$ depends only on~$n$ and~$k$; see~\cite{DKL19}.
\end{example}

We can now state the main general result of this section.

\begin{theorem} \label{theo:ball}
Let $X$ be a compact $n$-polyhedral length space and $\phi:\pi_1(X) \to G$ be a group homomorphism whose image satisfies the $\kappa$-Tits alternative.
Suppose that $X$ does not $\frac{1}{2} \mu_\phi$-collapse.
Then there exists $\tilde{x}_0 \in \tilde{X}$ such that
\[
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \vol_n(B_{\hyp^n}(C \, r))
\]
for every $r \geq 0$, where $C = C_{n,\kappa} \cdot \min\{ 1,\frac{1}{\sup \mu_\phi} \}$ with $C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$.
\end{theorem}

\begin{proof}

According to Theorem~\ref{theo:x_0}, there exists $x_0 \in X$ such that for every lift $\tilde{x}_0 \in \tilde{X}$ of~$x_0$,
\begin{equation} \label{eq:H>r^n}
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \HH_n(B_X(x_0,r)) \geq c_n \, r^n
\end{equation}
for every $r \in [0, \frac{1}{2} \mu_\phi(x_0)]$, where $c_n = \frac{1}{n!} \, \prod_{i=1}^n \lambda_i$.
For simplicity, denote $\mu_0=\mu_\phi(x_0)$.

\medskip

Let us establish an exponential lower bound for the volume of balls of large enough radius centered at~$\tilde{x}_0$.

\begin{lemma}
For every $r \geq \frac{1}{2} \, \mu_0$, we have
\begin{equation} \label{eq:H>e^r}
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \frac{c_n}{2^n} \, \mu_0^n \left( 3^{\left \lfloor \frac{r-\frac{\mu_0}{2}}{2 \kappa \mu_0} \right \rfloor } -1 \right)
\end{equation}
where $\lfloor \cdot \rfloor$ is the floor function.
\end{lemma}

\begin{proof}
The idea of the proof is to find a nonabelian free subgroup~$F$ of~$\pi_1(X,x_0)$ generated by loops of length at most~$2 \kappa \mu_0$ such that the translates of~$B_{\tilde{X}}(\tilde{x}_0,\frac{\mu_0}{2})$ under the free action of~$F$ on~$\tilde{X}$ are disjoint.

\medskip

Denote by~$S$ the symmetric generating set of~$\Gamma^{2 \mu_0}_{\phi,x_0} \leqslant G$ formed of the elements~$\phi([\gamma])$ where $\gamma$ is a loop of~$X$ based at~$x_0$ of length at most~$2 \mu_0$.
By definition of the Margulis function, see Definition~\ref{def:margulis}, the subgroup $\Gamma^{2 \mu_0}_{\phi,x_0} \leqslant G$ generated by~$S$ has exponential growth.
(One could consider $\Gamma^{\mu_0+\varepsilon}_{\phi,x_0}$ instead of $\Gamma^{2 \mu_0}_{\phi,x_0}$ if one wanted to be more precise.)
Note also that every element of~$S^\kappa$ is the $\phi$-image of the homotopy class of a loop of~$X$ based at~$x_0$ of length at most $2 \kappa \mu_0$.
Thus, since the image of~$\phi$ in~$G$ satisfies the $\kappa$-Tits alternative, there exist two loops~$\gamma_1$ and~$\gamma_2$ based at~$x_0$ of length at most~$2 \kappa \mu_0$ whose homotopy classes~$\alpha_1$ and~$\alpha_2$ generate a nonabelian free subgroup of~$\pi_1(X,x_0)$.

\medskip

Consider a nonabelian free subgroup $F \leqslant \pi_1(X,x_0)$ generated by two (nontrivial) elements $\alpha_1$ and~$\alpha_2$ with $d_{\tilde{X}}(\tilde{x}_0,\alpha_i \cdot \tilde{x}_0) \leq 2 \kappa \mu_0$ such that the orbit $F \cdot \tilde{x}_0$ has a minimal number of points in~$B_{\tilde{X}}(\tilde{x}_0,\mu_0)$.
(The existence of such subgroup is ensured by the previous discussion.)
Suppose that the intersection $F \cdot \tilde{x}_0 \cap B_{\tilde{X}}(\tilde{x}_0,\mu_0)$ has a point~$\alpha \cdot \tilde{x}_0$ different from~$\tilde{x}_0$ with ~$\alpha \in F$.
Denote by~$k$ the smallest positive integer such that $\alpha^k \cdot \tilde{x}_0$ does not lie in $B_{\tilde{X}}(\tilde{x}_0,\mu_0)$.
We have
\begin{align*}
d_{\tilde{X}}(\tilde{x}_0,\alpha^k \cdot \tilde{x}_0) & \leq d_{\tilde{X}}(\tilde{x}_0,\alpha^{k-1} \cdot \tilde{x}_0) + d_{\tilde{X}}(\alpha^{k-1} \cdot \tilde{x}_0,\alpha^k \cdot \tilde{x}_0) \\
 & \leq \mu_0 + d_{\tilde{X}}(\tilde{x}_0,\alpha \cdot \tilde{x}_0) \\
 & \leq 2 \mu_0
\end{align*}
Thus, the displacement of~$\tilde{x}_0$ by~$\alpha^k$ is at most $2 \kappa \mu_0$.
(Note that $\kappa \geq 1$.)
Furthermore, either the subgroup generated by $\alpha_1$ and~$\alpha^k$, or the subgroup generated by $\alpha_2$ and~$\alpha^k$ is nonabelian free.
Denote by $F' \leqslant F$ this nonabelian free subgroup.
Since $k \geq 2$, the point $\alpha \cdot \tilde{x}_0$ does not lie in the orbit of~$F'$.
Therefore, the orbit~$F' \cdot \tilde{x}_0$ of~$F'$ has fewer elements in $B_{\tilde{X}}(\tilde{x}_0,\mu_0)$ than the orbit~$F \cdot \tilde{x}_0$ of~$F$, which contradicts the definition of~$F$.
Thus, the only point of the orbit~$F \cdot \tilde{x}_0$ lying in~$B_{\tilde{X}}(\tilde{x}_0,\mu_0)$ is~$\tilde{x}_0$.
This implies that the translates of~$B_{\tilde{X}}(\tilde{x}_0,\frac{\mu_0}{2})$ under the free action of~$F$ on~$\tilde{X}$ are disjoint.

\medskip

Let $r \geq \frac{1}{2} \, \mu_0$.
The ball $B_{\tilde{X}}(\tilde{x}_0,r)$ contains all the translates of $B_{\tilde{X}}(\tilde{x}_0,\frac{\mu_0}{2})$ under the elements of~$F =\langle \alpha_1, \alpha_2 \rangle$ which can be written as words of length at most
\[
w=\left \lfloor \frac{r-\frac{\mu_0}{2}}{2 \kappa \mu_0} \right \rfloor
\]
with the letters~$\alpha_1$ and~$\alpha_2$.
Since the translates of $B_{\tilde{X}}(\tilde{x}_0,\frac{\mu_0}{2})$ under~$F$ are disjoint, we derive from~\eqref{eq:H>r^n} that
\[
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq c_n \, \frac{\mu_0^n}{2^n} \, (3^w-1)
\]
as desired.
\end{proof}

Let us resume the proof of Theorem~\ref{theo:ball} using the polynomial lower bound~\eqref{eq:H>r^n} and the exponential lower bound~\eqref{eq:H>e^r} on the volume of balls in the universal cover of~$X$.

Recall that
\begin{equation} \label{eq:volH}
\vol_n(B_{\hyp^n}(r)) = \sigma_{n-1} \, \int_0^r \sinh(t)^{n-1} \, dt
\end{equation}
where $\sigma_{n-1}$ is the Euclidean volume of the unit sphere~$S^{n-1}$.
We want to show that
\begin{equation} \label{eq:wanted}
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \vol_n(B_{\hyp^n}(C \, r))
\end{equation}
for every $r \geq 0$, where $C>0$ is a metric-independent constant to determine.
We will consider the following situations:

\begin{enumerate}
\item If $r \geq 5 \kappa \mu_0$ then the estimate~\eqref{eq:H>e^r} yields the following lower bound
\begin{equation} \label{eq:r>5}
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \frac{c_n}{2^n} \, \mu_0^n \left( 3^{\frac{r-\frac{\mu_0}{2}}{4\kappa \mu_0}}-1 \right).
\end{equation}

\item If $r \leq 5 \kappa \mu_0$ then the estimate~\eqref{eq:H>r^n} yields the following lower bound
\begin{equation} \label{eq:r<5}
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \frac{c_n}{2^n (5\kappa)^n} \, r^n.
\end{equation}

\item If $r \geq 1$ then the formula~\eqref{eq:volH} combined with the inequality $\sinh(t) \leq \frac{e^t}{2}$ for $t \geq 0$ yields the following upper bound
\begin{equation} \label{eq:r>1}
\vol_n(B_{\hyp^n}(C r)) \leq \frac{\sigma_{n-1}}{(n-1) 2^{n-1}} \, \left( e^{(n-1) C r} - 1 \right).
\end{equation}

\item If $r \leq 1$ then the formula~\eqref{eq:volH} combined with the inequality $\sinh(t) \geq \sinh(1) \, t$ for $0 \leq t \leq 1$ yields the following upper bound
\begin{equation} \label{eq:r<1}
\vol_n(B_{\hyp^n}(C r)) \leq \frac{\sigma_{n-1}}{n} \, \sinh(1)^{n-1} \, C^n \, r^n.
\end{equation}
\end{enumerate}

We will consider four cases analyzing the different combinations of situations, starting with the more technical cases.

\medskip

\noindent \underline{\it Case 1.}
Suppose that $r \geq 5 \kappa \mu_0$ and $r \geq 1$.
By~\eqref{eq:r>5} and~\eqref{eq:r>1}, the inequality~\eqref{eq:wanted} holds if
\begin{equation} \label{eq:r>5>1}
\frac{c_n}{2^n} \, \mu_0^n \left( 3^{\frac{r-\frac{\mu_0}{2}}{4\kappa \mu_0}}-1 \right) \geq \frac{\sigma_{n-1}}{(n-1) 2^{n-1}} \, \left( e^{(n-1) C r} - 1 \right).
\end{equation}
Comparing the exponential growth rates of the two sides of the inequality, we want $\frac{\log(3)}{4 \kappa \mu_0} \geq (n-1) C$.
That is,
\begin{equation} \label{eq:log3}
C \leq \frac{\log(3)}{4(n-1) \kappa \mu_0}.
\end{equation}
Thus, we need to compare $f(t)=a \, e^{\alpha t} + a'$ and $g(t)= b \, e^{\beta t} +b'$ with $a, b >0$ and $\alpha \geq \beta >0$.
We have $f(t) \geq g(t)$ for every $t \geq t_0$ if $f(t_0) \geq g(t_0)$ and $f'(t_0) \geq g'(t_0)$.
In our case, we need to check the inequality~\eqref{eq:r>5>1} and the inequality
\[
\frac{c_n}{2^n} \, \mu_0^n \, \frac{\log(3)}{4 \kappa \mu_0} \, 3^{\frac{r-\frac{\mu_0}{2}}{4\kappa \mu_0}} \geq \frac{\sigma_{n-1}}{2^{n-1}} \, C \, e^{(n-1) C r}
\]
for $r = \max \{ 5 \kappa \mu_0, 1\}$.

If $5 \kappa \mu_0 \geq 1$, it is enough to have
\[
C \leq \frac{1}{5 (n-1) \kappa \mu_0} \, \log \left( 1+ \frac{(n-1) c_n}{\sigma_{n-1}} \mu_0^n \right)
\]
for the first inequality (since the term between parenthesis in the left-hand side of~\eqref{eq:r>5>1} is bounded from below by~$2$ when $r=5 \kappa \mu_0$) and
\[
\frac{c_n}{2^n} \, \mu_0^n \, \frac{\log(3)}{4 \kappa \mu_0} \, 3^{-\frac{1}{8 \kappa}} \geq \frac{\sigma_{n-1}}{2^{n-1}} \, C
\]
for the second inequality using~\eqref{eq:log3}.
Since $\mu_0 \geq \frac{1}{5 \kappa}$, these conditions are satisfied if $C \leq \frac{C_{n,\kappa}}{\mu_0}$ for some constant $C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$.

If $5 \kappa \mu_0 \leq 1$, it is enough to have
\[
C \leq \frac{1}{n-1} \, \log \left( 1+ \frac{(n-1) c_n}{2 \sigma_{n-1}} \, ( 3^{\frac{1}{8 \kappa \mu_0}} - 1)  \mu_0^n \right)
\]
for the first inequality (since $r - \frac{\mu_0}{2} \geq \frac{r}{2}$ when $r=1$) and
\[
\frac{c_n}{2^n} \, \mu_0^n \, \frac{\log(3)}{4 \kappa \mu_0} \, 3^{\frac{1}{8 \kappa \mu_0}} \geq \frac{\sigma_{n-1}}{2^{n-1}} \, C \, e^{(n-1) C}
\]
for the second inequality.
Since $\mu_0 \leq \frac{1}{5\kappa}$, the expression $\mu_0^{n-1} \, 3^{\frac{1}{8 \kappa \mu_0}}$ is bounded from below by a positive constant depending only on~$n$ and~$\kappa$.
Thus, the previous conditions are satisfied if $C \leq C_{n,\kappa}$ for some constant $C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$.

\medskip

\noindent \underline{\it Case 2.}
Suppose that $r \geq 5 \kappa \mu_0$ and $r \leq 1$.
By~\eqref{eq:r>5} and~\eqref{eq:r<1}, the inequality~\eqref{eq:wanted} holds if
\begin{equation} \label{eq:r>5<1}
\frac{c_n}{2^n} \, \mu_0^n \left( 3^{\frac{r-\frac{\mu_0}{2}}{4\kappa \mu_0}}-1 \right) \geq \frac{\sigma_{n-1}}{n} \, \sinh(1)^{n-1} \, C^n \, r^n.
\end{equation}
Observe that the left-hand side of this inequality divided by~$r^n$ is bounded from below by a positive constant depending only on~$n$ and~$\kappa$ when $5 \kappa \mu_0 \leq r \leq 1$.
Thus, the condition~\eqref{eq:r>5<1} is satisfied if $C \leq C_{n,\kappa}$ for some constant $C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$.

\medskip

\noindent \underline{\it Case 3.}
Suppose that $r \leq 5 \kappa \mu_0$ and $r \geq 1$.
By~\eqref{eq:r<5} and~\eqref{eq:r>1}, the inequality~\eqref{eq:wanted} holds if
\begin{equation} \label{eq:r<5>1}
\frac{c_n}{2^n (5\kappa)^n} \, r^n \geq \frac{\sigma_{n-1}}{(n-1) 2^{n-1}} \, \left( e^{(n-1) C r} - 1 \right).
\end{equation}
Since $1 \leq r \leq 5 \kappa \mu_0$, the left-hand side of this inequality is bounded from below by $\frac{c_n}{2^n (5\kappa)^n}$ and the right-hand side is bounded from above by $\frac{\sigma_{n-1}}{(n-1) 2^{n-1}} \, \left( e^{5(n-1)\kappa \mu_0 \, C } - 1 \right)$.
Thus, the condition~\eqref{eq:r<5>1} is satisfied if $C \leq \frac{C_{n,\kappa}}{\mu_0}$ for some constant $C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$.

\medskip

\noindent \underline{\it Case 4.}
Suppose that $r \leq 5 \kappa \mu_0$ and $r \leq 1$.
By~\eqref{eq:r<5} and~\eqref{eq:r<1}, the inequality~\eqref{eq:wanted} holds if
\begin{equation} \label{eq:r<5<1}
\frac{c_n}{2^n (5\kappa)^n} \, r^n \geq \frac{\sigma_{n-1}}{n} \, \sinh(1)^{n-1} \, C^n \, r^n.
\end{equation}
This condition is satisfied if $C \leq C_{n,\kappa}$ for some constant $C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$.

\medskip

In conclusion, there exists a positive constant~$C_{n,\kappa}$ depending only on~$n$ and~$\kappa$ such that for every $C \leq C_{n,\kappa} \cdot \min\{ 1,\frac{1}{\mu_0} \}$
\begin{equation} \label{eq:end}
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \vol_n(B_{\hyp^n}(C \, r))
\end{equation}
for every $r \geq 0$.
\end{proof}

\forget
\underline{\it Case 1.}
Suppose that $r \leq 1$.
By \eqref{eq:H>r^n} and~\eqref{eq:volH} combined with the inequality $\sinh(t) \leq \sinh(1) \, t$ for every $t \in [0,1]$, it is enough to show that
\[
c_n \, r^n \geq \frac{\sigma_{n-1}}{n} \, \sinh(1)^{n-1} \, C^n \, r^n.
\]
That is, $C \leq \left( \frac{n \, c_n}{\sigma_{n-1} \sinh(1)^{n-1}} \right)^\frac{1}{n}$.

\medskip

\underline{\it Case 2.}
Suppose that $r \geq 1$ and $r \geq \mu_0$.
By \eqref{eq:H>e^r} and~\eqref{eq:volH} combined with the inequality $\sinh(t) \leq \frac{e^t}{2}$ for $t \geq 0$, it is enough to show that
\begin{equation} \label{eq:ee}
\frac{c_n}{2^n} \, \mu_0^n  \left( 3^{\frac{r-\frac{\mu_0}{2}}{2 \kappa \mu_0}} - 1 \right) \geq \frac{\sigma_{n-1}}{(n-1) 2^{n-1}} \, \left( e^{(n-1) C r} - 1 \right)
\end{equation}
for every $r \geq \max\{ 1,\mu_0 \}$.
This requires that $\frac{\log(3)}{2 \kappa \mu_0} \geq (n-1) C$.
That is, $C \leq \frac{\log(3)}{2(n-1) \kappa \mu_0}$.
In this case, the relation~\eqref{eq:ee} only needs to be checked for $r \geq \max\{1,\mu_0\}$.

If $\mu_0 \geq 1$, this leads to the condition
\[
C \leq \min \left\{ \frac{\log(3)}{2(n-1) \kappa \mu_0}, \frac{1}{(n-1) \mu_0} \, \log \left( 1+ \frac{(n-1) c_n}{2 \sigma_{n-1}} \, ( 3^{\frac{1}{4\kappa}} - 1)  \mu_0^n \right) \right\}.
\]
Since $\mu_0 \geq 1$, this condition is satisfied if $C \leq C_{n,\kappa}$ for some constant $C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$.

If $\mu_0 \leq 1$, this leads to the condition
\[
C \leq \min \left\{ \frac{\log(3)}{2(n-1) \kappa \mu_0}, \frac{1}{(n-1)} \, \log \left( 1+ \frac{(n-1) c_n}{2 \sigma_{n-1}} \, ( 3^{\frac{1}{4 \kappa \mu_0}} - 1)  \mu_0^n \right) \right\}.
\]
Since $\mu_0 \leq 1$, this condition is satisfied if $C \leq \frac{C_{n,\kappa}}{\mu_0}$ for some constant $C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$.

\medskip

\underline{\it Case 3.}
Suppose that $r \geq 1$ and $r \leq \mu_0$.
By \eqref{eq:H>r^n} and~\eqref{eq:volH} combined with the inequality $\sinh(t) \leq \frac{e^t}{2}$ for $t \geq 0$, it is enough to show that
\[
c_n \, r^n \geq \frac{\sigma_{n-1}}{(n-1) 2^{n-1}} \, \left( e^{(n-1) C r} - 1 \right)
\]
for $1 \leq r \leq \mu_0$.
This just needs to be checked for $r=\mu_0$, for which we obtain the condition
\[
C \leq \frac{1}{(n-1) \mu_0} \, \log \left( 1+ \frac{(n-1) c_n}{\sigma_{n-1}} \, 2^{n-1}  \mu_0^n \right).
\]
Since $\mu_0 \geq 1$, this condition is satisfied if $C \leq \frac{C_{n}}{\mu_0}$ for some constant $C_{n} >0$ depending only on~$n$.

\medskip

In conclusion, there exists a constant~$C_{n,\kappa} >0$ depending only on~$n$ and~$\kappa$ such that for every $C \leq C_{n,\kappa} \cdot \min\{ 1,\frac{1}{\mu_0} \}$
\begin{equation} \label{eq:end}
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \vol_n(B_{\hyp^n}(C \, r))
\end{equation}
for every $r \geq 0$.
\end{proof}
\forgotten

We immediately derive the following two corollaries.
The first corollary claims that when the volume of~$X$ is small enough, the constant~$C$ in Theorem~\ref{theo:ball} does not depend on the Margulis function, but only on~$n$ and~$\kappa$.

\begin{corollary} \label{coro:ball}
Let $X$ be a compact $n$-polyhedral length space and $\phi:\pi_1(X) \to G$ be a group homomorphism whose image satisfies the $\kappa$-Tits alternative.
Suppose that $X$ does not $\frac{1}{2} \mu_\phi$-collapse.
There exists a constant~$\delta_n>0$ such that if $\HH_n(X) \leq \delta_n$ then there exists $\tilde{x}_0 \in \tilde{X}$ such that
\[
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \vol_n(B_{\hyp^n}(C_{n,\kappa} \, r))
\]
for every $r \geq 0$, where $C_{n,\kappa} >0$ is a constant depending only on~$n$ and~$\kappa$.

In particular, there exists constant~$\delta'_{n,\kappa}>0$ such that if $\HH_n(X) \leq \delta'_{n,\kappa}$ then there exists $\tilde{x}_0 \in \tilde{X}$ such that
\[
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \vol_n(B_{\hyp^n}(r))
\]
for every $r \geq 1$.
\end{corollary}

\begin{proof}
Applying the second inequality of~\eqref{eq:H>r^n} to $r=\frac{1}{2} \mu_\phi(x_0)$, we obtain
\begin{equation} \label{eq:m0}
\mu_\phi(x_0)^n \leq  2^n c_n \, \HH_n(X).
\end{equation}
Thus, if $\HH_n(X) \leq \delta_n := \frac{1}{2^n c_n}$ then $\mu_\phi(x_0) \leq 1$ and we can take $C=C_{n,\kappa}$ in~\eqref{eq:end}, which proves the first part of the corollary.

Recall that the volume of a ball in~$\hyp^n$ grows exponentially with its radius.
Thus, we can choose~$\lambda=\lambda_{n,\kappa} >0$ large enough such that
\[
\vol_n(B_{\hyp^n}(C_{n,\kappa} \, \lambda r)) \geq \lambda^n \, \vol_n(B_{\hyp^n}(r))
\]
for every $r \geq 1$.
Denote by $\lambda X$ the compact $n$-polyhedral length space obtained by multiplying the metric on~$X$ by~$\lambda$.
If $\delta'_{n,\kappa}=\frac{\delta_n}{\lambda^n}$ then $\HH_n(\lambda X) = \lambda^n \, \HH_n(X) \leq \delta_n$.
Therefore, there exists $\tilde{x}_0 \in \widetilde{\lambda X} = \lambda \tilde{X}$ such that
\[
\HH_n(B_{\lambda \tilde{X}}(\tilde{x}_0,t)) \geq \vol_n(B_{\hyp^n}(C_{n,\kappa} \, t))
\]
for every $t \geq 0$.
Hence,
\[
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) = \frac{1}{\lambda^n} \HH_n(B_{\lambda \widetilde{X}}(\tilde{x}_0,\lambda r)) \geq \frac{1}{\lambda^n} \vol_n(B_{\hyp^n}(C_{n,\kappa} \lambda r)) \geq \vol_n(B_{\hyp^n}(r))
\]
for every $r \geq 1$.
\end{proof}

\forget
\begin{corollary}
Let $X$ be a compact $n$-polyhedral length space and $\phi:\pi_1(X) \to G$ be a group homomorphism whose image satisfies the $\kappa$-Tits alternative.
Suppose that $X$ does not $\frac{1}{2} \mu_\phi$-collapse.
Then there exists $\tilde{x}_0 \in \tilde{X}$ such that
\[
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) \geq \vol_n(B_{\hyp^n}(r))
\]
for every $r \geq \HH_n(X)^{\frac{1}{n}}$.
\end{corollary}

\begin{proof}
We argue as in the proof of Corollary~\ref{theo:ball}.

By Theorem~\ref{theo:ball}, we have $r \geq \HH_n(X)^{\frac{1}{n}} \geq a_n \, \sup \mu_\phi$ for $a_n = \frac{1}{2 \sqrt[n]{c_n}}$; see~\eqref{eq:m0}.

In particular, $\min \{1,\frac{1}{\sup \mu_\phi} \} \, r$ is bounded away from zero.
Choose~$\lambda=\lambda_{n,\kappa} >0$ large enough such that
\[
\vol_n(B_{\hyp^n}(C_{n,\kappa} \, {\textstyle \min \{1,\frac{1}{\sup \mu_\phi} \}} \, \lambda r)) \geq \lambda^n \, \vol_n(B_{\hyp^n}(r))
\]
for every $r \geq \HH_n(X)^{\frac{1}{n}}$.
Denote by $\lambda X$ the compact $n$-polyhedral length space obtained by multiplying the metric on~$X$ by~$\lambda$.
By Theorem~\ref{theo:ball}, there exists $\tilde{x}_0 \in \widetilde{\lambda X} = \lambda \tilde{X}$ such that
\[
\HH_n(B_{\lambda \tilde{X}}(\tilde{x}_0,t)) \geq \vol_n(B_{\hyp^n}(C_{n,\kappa} \, {\textstyle \min \{1,\frac{1}{\sup \mu_\phi} \}} \, t))
\]
for every $t \geq 0$.
Hence,
\[
\HH_n(B_{\tilde{X}}(\tilde{x}_0,r)) = \frac{1}{\lambda^n} \HH_n(B_{\lambda \widetilde{X}}(\tilde{x}_0,\lambda r)) \geq \frac{1}{\lambda^n} \vol_n(B_{\hyp^n}(C_{n,\kappa} \, {\textstyle \min \{1,\frac{1}{\sup \mu_\phi} \}} \, \lambda r)) \geq \vol_n(B_{\hyp^n}(r))
\]
for every $r \geq \HH_n(X)^{\frac{1}{n}}$.
\end{proof}
\forgotten

The second corollary is what occurs when we apply the previous result to a closed manifold admitting a metric with negative sectional curvature.

\begin{corollary} \label{coro:final}
Let $M$ be a closed $n$-manifold admitting a metric with sectional curvature lying between $-k^2$ and~$-1$ for some $k \geq 1$.

Then for every Riemannian metric on~$M$ with $\vol(M) \leq \delta'_{n,\kappa}$ for some constant $\delta'_{n,\kappa}>0$ depending only on~$n$, there exists $\tilde{x}_0 \in \tilde{M}$ such that
\begin{equation} \label{eq:C-M}
\vol(B_{\tilde{M}}(\tilde{x}_0,r)) \geq \vol(B_{\hyp^n}(r))
\end{equation}
for every $r \geq 1$.
\end{corollary}

\begin{proof}
As explained in Example~\ref{ex}, the fundamental group of~$M$ satisfies the $\kappa$-Tits alternative for some constant $\kappa=\kappa(n,k)$ depending only on~$n$ and~$k$; see~\cite{DKL19}.
Since $M$ is a closed manifold admitting a metric of negative curvature, its simplicial volume is nonzero; see~\cite{gro82}.
By Proposition~\ref{prop:simpl}, this implies that $M$ does not $\frac{1}{2} \mu$-collapse.
Therefore, the conclusion of Corollary~\ref{coro:ball} holds for~$M$.
\end{proof}


TITLE: Is quantum mechanics truly probabilistic?
QUESTION [3 upvotes]: Probability arises inherently from a lack of information. For example, if I were to take a ball out of a bag with 3 yellow and 2 white balls, I would have a 0.6 probability of getting a yellow and a 0.4 probability of getting a white ball. However, these only apply because I cannot see where each ball is located. However, if I had the information about the forces on each ball as they were thrown into the bag, as well as the angle of my arm as I put it in, I could, without doubt, know for sure which ball I would get.
Hence, is it not possible that the entire probabilistic nature of quantum mechanics arises entirely from our lack of information on phenomenon? I have read up on the Heisenberg uncertainty principle but it seems a bit iffy that physical phenomenon are purely bounded by mathematical theory. Could there be a non-probabilistic explanation for these phenomenon?

REPLY [6 votes]: To answer your question "Hence, is it not possible that the entire probabilistic nature of quantum mechanics arises entirely from our lack of information on phenomenon?" precisely: Yes, it is possible.
Sure, there are different ways of thinking about quantum mechanics and at some point, asking whether the randomness is true or apparent becomes a more philosophical question.
But: There is a formulation of quantum mechanics that is called Bohmian mechanics or de-Broglie-Bohm theory which is entirely deterministic. It describes, in addition to the wave function, particles which have well-defined positions. (Since the theory is non-local, it does not contradict Bell's theorem, of course.)
The probabilities that can be computed from this theory are exactly the same as in Copenhagen quantum mechanics, so it is empirically correct, but the probabilistic nature only comes from our ignorance about the initial values (initial positions of particles), as in classical physics. So to answer all claims that such a thing is impossible in principle, it just is given by this example.


TITLE: Formula of Squaring Sums / Integrals
QUESTION [1 upvotes]: I'm trying to find a proof for the identities (which I use quite often)
$$\left ( \int_{a}^{\infty}f(x)\,dx \right )^2=\int_{a}^{\infty}\int_{a}^{\infty}f(x, y)\,dx\,dy$$
and similarly for the series
$$\left ( \sum_{k=a}^{\infty}a_n \right )^2=\sum_{m=a}^{\infty}\sum_{k=a}^{\infty}a_{m, k}$$
If someone could prove them or give a link, because I can't find them anywhere.. would be nice.
Thanks!

REPLY [2 votes]: For any $a \in \mathbb{R}$, consider the following:
\begin{align}
\left(\int_a^{\infty} f(x)dx \right)^2
&= \left( \int_a^{\infty} f(x) dx \right) \left( \int_a^{\infty} f(y) dy \right)\\
&= \int_a^{\infty} \left( \int_a^{\infty} f(x)dx \right) f(y) dy \\
&= \int_a^{\infty} \int_a^{\infty} f(x)f(y)dxdy.
\end{align}
The same principle holds in the discrete case:
\begin{align}
\left( \sum_{n=a}^{\infty} a_n \right)^2
&= \left( \sum_{m=a}^{\infty} a_m \right) \left( \sum_{n=a}^{\infty} a_n \right)\\
&= \sum_{m=a}^{\infty} \left( \sum_{n=a}^{\infty} a_n \right) a_m \\
& = \sum_{m=a}^{\infty} \sum_{n=a}^{\infty} a_m a_n.
\end{align}


TITLE: What's the parametric equation for the plane through a point (x,y,z) perpendicular to (a,b,c)?
QUESTION [0 upvotes]: Find the parametric vector and Cartesian equations for the following planes:
a. The plane thru point $(2,1,-2)$ perpendicular to vector $(-1,1,2)$.
b. The plane thru the three points $(2,2,-2)$, $(-1,1,2)$ and $(2,3,1)$.
Please help. Studying for an exam and I attempted this problem several times, but my answers do not match the answer doc. Thank you.

REPLY [0 votes]: Fiest be aware that you can define a plan by different means.
One of them is through parametric equations such as
$(1)$ $x=a_1+\lambda u_1 +\mu v_1$, $y=a_2+\lambda u_2 +\mu v_2$ and $y=a_3+\lambda u_3 +\mu v_3$
with $(a_1,a_2,a_3)$ a point of the plane and $(u_1,u_2,u_3)$, $(v_1,v_2,v_3)$ two non colinear vectors.
An other one is through a cartesian equation such as
$(2)$ $ax+by+cz+d=0$
For your question $(a)$, your professor gave you a parametric equation, which is different from the statement of your problem (give the cartesian equation).
In any case, let's find a parametric equation for the problem $(a)$.
Since you already have a point of the plane $(2,1,-2)$, you just need to find two non-colinear vectors of the plane. Each of them should be perpendicular to the normal vector, that is, the dot product should be $0$.
One of the vector could be very easily $(1,1,0)$, whom dot product with $(1,-1,2)$ is clearly $0$.
The second one could be $(0,2,1)$. Since it is not colinear with $(1,1,0)$, this works.
Hence an equation $(2,1,-2)+\lambda (1,1,0)+ \mu (0,2,1)$
This is different from the answer of your professor, but this is OK, there are an infinity of possible parametric equations for the same plan. Bear in mind that a method to find the last vector, instead of finding our one fitting one, could be to find a vector which is normal to both the vector normal to the plane and the one you already found, with the vector cross product. It would be in that case $(1,-1,2) \times (1,1,0)$. It is probably what your professor did.
In the second case, you can take the first point $A$ as your defining point, and the vectors $AB$ and $AC$ as vectors for the parametrization of the plane.
Or you can also (for cartesian equation) find out the three unknowns by resolving this set of equations:
$ax_1+by_1+cz_1+1=0$, $ax_2+by_2+cz_2+1=0$, $ax_3+by_3+cz_3+1=0$


TITLE: Representing dimensions in Dirac delta function results
QUESTION [0 upvotes]: The Dirac delta function can be defined as $$\delta(x)=\frac{1}{2\pi}\int_{-\infty}^\infty e^{itx}dt$$
From this we see that the dirac function has units of $x^{-1}$.

How do we represent the units in cases like the momentum eigenvectors which, when units are included, is represented as
  $$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}$$
  or
  $$\frac{1}{\sqrt{2\pi\hbar}}e^{\frac{\iota px}{\hbar}}\cdot(kg^{\frac{1}{2}}m^{\frac{1}{2}}s^{-\frac{1}{2}})\,?$$
Is there a preferred way to write the units(not constrained to SI units, any other system including natural units too) or do we just leave them out, though it would be dimensionally inconsistent without implied units.

Sources I can find for the momentum eigenvector ignore the units of the delta function without even mentioning.
P.S.
The problem comes when trying to normalize the momentum operator.
Define
$$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$
Over here, $A$ has units of $m^{-\frac{1}{2}}$.
Normalizing it,
$$\int_{-\infty}^\infty\psi^*_{p_1}(x)\psi_{p_2}(x)dx$$
$$=|A|^2\int_{-\infty}^\infty e^{\frac{\iota \left(p_2-p_1\right)x}{\hbar}}dx$$
$$=|A|^22\pi\hbar\delta\left(p_2-p_1\right)$$
Therefore, ignoring consistency of units, and assuming $A$ is positive,
$$A=\frac{1}{\sqrt{2\pi\hbar}}$$
Notice that the units do not match

REPLY [2 votes]: The dimension of the Dirac delta function is the inverse of the dimension of its argument. So if $x$ is a length then $\delta(x)$ has the dimension of inverse length.
In your example, the momentum eigenstate in position representation has the wavefunction
$$
\psi_p(x) = A \mathrm{e}^{i p x / \hbar}
$$
The interpretation of the wavefunction is that $|\psi|^2$ is a probability density, which is dimension of 1/Length. Therefore, $A$ has dimension of $1/\sqrt{\mathrm{Length}}$.
As you say, calculating the scalar product of two momentum eigenfunction with eigenvalues $p_1$ and $p_2$ gives
$$
\int_{-\infty}^\infty \mathrm{d}x \, \psi_{p_1}^*(x) \psi_{p_2}(x)
= |A|^2  \int_{-\infty}^\infty \mathrm{d}x \mathrm{e}^{i(p_2-p_1)x/\hbar}
= |A|^2 2\pi \hbar \delta(p_2 - p_1)
$$
The left-hand side is dimensionless, and therefore so is the right-hand side. $\hbar \delta(p_2-p_1)$ has dimension of Length, so again we get that the dimension of $A$ is $1/\sqrt{\mathrm{Length}}$.

REPLY [0 votes]: I am not entirely sure what you mean with the notation e.g.
$$\frac{1}{\sqrt{2\pi\hbar\cdot(kg^{-1}m^{-1}s)}}e^{\frac{\iota px}{\hbar}}.$$
The momentum eigenstates are $\psi_p(x) = \frac{1}{2\pi\hbar} e^{ipx/\hbar}$, this is true in any unit system. You don't need to multiply with any units. In the SI system, this has units of
$$ \frac{1}{\sqrt{kg\, m^2\, s^{-1}}} . \tag{*}$$

Define
  $$\psi_p(x)=Ae^{\frac{\iota px}{\hbar}}$$
  Over here, $A$ has units of $m^{-1}$.

Note that $\int |\psi_p(x)|^2\, dx \neq 1$ (the state is not normalizable), therefore there's no reason why the dimension of $A$ should be inverse length! What we use instead of normalization is the following:
$$ \langle p' \mid p \rangle = \int \psi_p(x) \psi_{p'}(x)^\ast dx = \delta(p'-p) , $$
here the $\delta$-function has units of $kg^{-1}\, m^{-1}\, s$ which is fully compatible with (*).
Or maybe it helps you to consider
$$ \delta(x') = \langle x' | x=0 \rangle = \int \psi_p(0)^\ast \psi_p(x')\, dp = \frac{1}{2\pi\hbar} \int e^{ipx'/\hbar}\, dp , $$
where the units also match (of course).

REPLY [0 votes]: The units do add up in the example you provide:
You've correctly established that the delta distribution has the dimension of the inverse dimension of its argument. This can be seen in a variety of ways, e.g. by looking at $\int f(x)\delta(x-x_0) dx = f(x_0)$, where the LHS has to have whatever dimensions $f$ has.
With this, we look at the dimensions of $\lvert A \rvert^2 2\pi\hbar \;\delta(p_1-p_2)$:
$$
\left[\lvert A \rvert^2 2\pi\hbar\; \delta(p_1 - p_2) \right] = \textsf{L}^{-1} \left[ \hbar \delta(p_1-p_2) \right] = \textsf{L}^{-1} \frac{\left[ \hbar \right]}{[(p_1-p_2)]} = \textsf{L}^{-1} [x] = \textsf{L}^{-1} \textsf{L}^{1} = 1,
$$where I have used that $\frac{p}{\hbar}$ has to have inverse dimensions of $x$ to make the argument of the exponential dimensionless.

NB: For $\psi_p(x)=A\exp(\frac{ipx}{\hbar})$, we want $\int \lvert \psi \rvert^2 dx$ to represent a probability, which is dimensionless. From this it follows that $A$ has to have dimensions $\textsf{L}^{-\frac12}$.


TITLE: Prove that $|\mathbb{N}\times \mathbb{R}| = |\mathbb{R}|$
QUESTION [1 upvotes]: I'm trying to prove the equality between the cardinalities:
$|\mathbb{N}\times \mathbb{R}| = |\mathbb{R}|$ ($\mathbb{N}\times \mathbb{R}$ is cartesian product from $\mathbb{N}$ to $\mathbb{R}$).
I tried doing that by building injective functions first from $\mathbb{N}\times \mathbb{R}$ to $(0,1)$ and then from $(0,1)$ to $\mathbb{N}\times \mathbb{R}$ so:
$F: \mathbb{N}\times \mathbb{R} \rightarrow (0,1)$ is given by :
$$F(1 , 0 .abcd....) \mapsto 0.abcd..$$
$$F(2 , 0.abcde....) \mapsto 0.abcde..$$
for example
($0.abcd..$ represents a real number in $(0,1)$
I'm not sure if that's right..

REPLY [2 votes]: (I will assume $\Bbb{N}$ starts at $1$). Use the fact that $|\Bbb{R}| = \mathcal{P}(\Bbb{N})$; it suffices to show that $|\mathcal{P}(\Bbb{N})| = |\Bbb{N} \times \mathcal{P}(\Bbb{N})|$.
Clearly,
$$|\Bbb{N} \times \mathcal{P}(\Bbb{N})| \ge |\{1\} \times \mathcal{P}(\Bbb{N})| = |\mathcal{P}(\Bbb{N})|,$$
so it suffices to find a surjection from $\mathcal{P}(\Bbb{N})$ onto $\Bbb{N} \times \mathcal{P}(\Bbb{N})$.
Define such a function as follows. Suppose $S \subseteq \Bbb{N}$. If $S = \emptyset$, map it to $(1, \emptyset)$. Otherwise, $S$ has a minimum element. Map $S$ to $(\min S, (S \setminus \{\min S\}) - \min S)$.
We now show that this map is surjective. If we have a pair $(n, T) \in \Bbb{N} \times \mathcal{P}(\Bbb{N})$, then let $S = \{n\} \cup (T + n)$. Since every $t \in T$ is at least $1$, we have that $n$ is the minimum element of $S$, and clearly the surjection will map $S$ to $T$.
(It's worth pointing out that the map is a bijection too, if you restrict it to $\mathcal{P}(\Bbb{N}) \setminus \{ \emptyset \}$.)

REPLY [1 votes]: To begin with, try solving something simpler: find an injective function $f:\mathbb R\to\mathbb (0,1)$ (this is not trivial, but is commonly known; see, e.g., This MSE thread, or using some manipulation of the $\arctan$ function).
Once you have this, you can `cook up' a new function $F:\mathbb N\times\mathbb R\to \mathbb R$ by setting
$$F(n,x)=f(x)+n\text{ for any }n\in\mathbb N\text{ and }x\in \mathbb R.$$
It should be a simple check that the new function $F$ is injective.


TITLE: Finding 3D vector component form given 3 points
QUESTION [1 upvotes]: I am having a brain fart with a simple methodology here.
Say I have three points in 3 dimensions:
(1,2,0) , (0,0,0) , and (-2,1,0)
The formula to find component form is < t point - i point >
but my question is: What if the initial and terminal points are not specified as above?
How do we know which points to subtract and add from each correctly?

REPLY [2 votes]: The vertices of the triangle are given to be $A(1,2,0)$, $B(0,0,0)$ and $C(-2,1,0)$. The question asks us whether this triangle is acute, obtuse or right.
Recall: A triangle is obtuse if any one angle of the triangle is obuse-angled. It is a right triangle if any one angle is a right angle. Finally, it is an acute triangle if all the three angles are acute. So to do the problem, we need to check whether each of the angles is acute or obtuse or right. I will show the steps for one angle. You can do for the other two angles similarly.
First of all, let $\theta$ be some angle in $(0, \pi)$. Then

$\theta$ is acute $\iff$ $\theta < \frac{\pi}{2}$ $\iff$ $\cos \theta > 0$.
$\theta$ is right $\iff$ $\theta = \frac{\pi}{2}$ $\iff$ $\cos \theta = 0$.
$\theta$ is obtuse $\iff$ $\theta > \frac{\pi}{2}$ $\iff$ $\cos \theta < 0$.

Now, to see if (say) angle $A$ of the triangle $ABC$ is acute/right/obtuse, we need to check whether $\cos \angle BAC$ is positive/zero/negative. But what is $\cos \angle BAC$? It is the angle made by the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. (When you are computing the angle at a particular vertex $v$, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex $v$ as the initial point.) We will first compute these two vectors:
$$
\overrightarrow{AB} = (0,0,0) - (1,2,0) = (-1,-2,0)
$$
$$
\overrightarrow{AC} = (-2,1,0) - (1,2,0) = (-3,-1,0)
$$
Therefore, the angle between these vectors is given by:
$$
\cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|}
= \ldots \tag{1}
$$
Can you take it from here? From the sign of this value, you should be able to decide if angle $A$ is acute/right/obtuse.
Now, do the same procedure for the remaining two angles $B$ and $C$ as well. That should help you solve the problem.
A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.


TITLE: Prove the limit of $\lim_{x \to 0^+}e^{\frac{1}{x}}$ and $\lim_{x \to 0^-}e^{\frac{1}{x}}$
QUESTION [2 upvotes]: I have been given the following task:
Determine whether a limit exists and if so determine it. I need to analyze the following two functions.
$$(1) \lim_{x \to 0^+}e^{\frac{1}{x}}$$
$$(2) \lim_{x \to 0^-}e^{\frac{1}{x}}$$
I know for a fact that a limit for the first function does not exist because for very small positiv values the term $\frac{1}{x}$ approaches infinity.
For the second function a limit does exist. As x approaches very small negative values the term $\frac{1}{x}$ approaches negative infinity. So it follows that $\lim_{x \to 0^-}e^{\frac{1}{x}} = e^{-\infty} = 0.$
However all of these statements are based on intuition and are not formal.
I tried to use the epsilon-delta proof in order to show a limit does or doesn't exist.
For the first function I tried using a contradiction in order to show a limit does not exist. So suppose a limit $L$ does exist and $L > 0$. If $L > 0$ than there exists a $\delta > 0$, so that
$0 < |x - x_0| < \delta \Leftrightarrow 0 < |x - 0| < \delta \Leftrightarrow 0 < x < \delta$ with $|f(x) - L| < \epsilon$. But I don't know how to continue from here. Because normally in an epsilon-delta proof I would now try to find a $\delta$ starting with the term  $|f(x) - L| < \epsilon$ and by using equivalent transformation getting to the term $|x - 0| < \delta$. However I don't know if this is even the right approach in this case.
For the second function I get stuck at the same point. Is it suggestive to even use the epsilon-delta proof for these kind of functions or is there some completely different solution path that is more suitable.

REPLY [1 votes]: So I am going to try to give an answer to $(2)$. As already mentioned above, a limit of $\lim_{x \to 0^-}e^{\frac{1}{x}}$ does exist.
Suppose the limit of $(2)$ is equal to $0$, so $\lim_{x \to 0^-}e^{\frac{1}{x}} = 0.$ I am going to prove this claim by doing an $\epsilon-\delta$ proof.
First of all we need to find a suitable $\delta$. It follows:
$|f(x) - x_0| < \epsilon \\ \Leftrightarrow e^\frac{1}{-x} - 0 < \epsilon \\ \Leftrightarrow e^{-\frac{1}{x}} < \epsilon \\ \Leftrightarrow -\frac{1}{x} < \ln(\epsilon) \\ \Leftrightarrow -1 < \ln(\epsilon)\cdot x \ \ \ (1)\\ \Leftrightarrow -\frac{1}{\ln(\epsilon)} < x \\ \Rightarrow \delta = \frac{1}{\ln(\epsilon)}$
As we found our $\delta$ we can now start with the actual proof.
Let $\epsilon > 0$ be given. Choose $\delta := \frac{1}{\ln(\epsilon)}.$ We need to show from  $|x - x_0| < \delta \Leftrightarrow -x - x_0 < \delta \Leftrightarrow -x - 0 < \delta \Leftrightarrow x > -\delta$ it follows $|f(x) - x_0| < \epsilon.$
$|f(x) - x_0| < \epsilon \\ \Rightarrow e^{-\frac{1}{x}} - 0 = e^{\frac{-1}{x}} \overset{\mathrm{(1)}}{<} e^{\frac{\ln(\epsilon) \cdot x}{x}} = e^{\ln(\epsilon)} = \epsilon \ \ \ \square$
Is the above proof correct?


TITLE: What are the directions generated by the Newton method for $f(Ay)$?
QUESTION [0 upvotes]: Suppose $A \in R^{n \times n}$ and that $A$ is invertible, derive the directions generated by the Newton Method for the following problem
$\min_{y \in R^n} f(Ay)$
I know what it means now but am stuck. I think I have found the gradient but am stuck with the Hessian. Gradient should be $\nabla f(Ay)^T A$ but what about the Hessian?

REPLY [1 votes]: Let $g(y) = f(Ay)$. The Newton direction is $\Delta x = -H_g(y)^{-1} D_g(y)$. By application of the chain rule, $D_g(y) = A^TD_f(Ay)$ and $H_g(y) = A^TH_f(Ay)A$, so
$$\Delta x = -(A^TH_f(Ay)A)^{-1} (A^TD_f(Ay))=-A^{-1}H_f(Ay)^{-1}D_f(Ay).$$


TITLE: Geometric interpretation of determinant
QUESTION [2 upvotes]: I am trying to prove geometrically, without invoking the dot or cross products or orthogonality, that the volume of a parallelepiped formed by vectors $
  \begin{bmatrix}
    a_1 \\
    a_2 \\
    a_3
  \end{bmatrix}$, $\begin{bmatrix}
    b_1 \\
    b_2 \\
    b_3
  \end{bmatrix}$ and $\begin{bmatrix}
    c_1 \\
    c_2 \\
    c_3
  \end{bmatrix}$
is $\det  P = \begin{vmatrix}
    a_1 & b_1 & c_1 \\
    a_2 & b_2 & c_2 \\
    a_3 & b_3 & c_3
  \end{vmatrix}$.
One of the options I see is to argue that any parallelepiped can be created by multiplying the matrix of a cube whose volume is given by
$\det  C= \begin{vmatrix}
    a_1 & 0 & 0 \\
    0 & b_2 & 0 \\
    0 & 0 & c_3
  \end{vmatrix}$ by elementary matrices so $P=CE_1 E_2 ...E_i $. The absolute value of the \det erminants of the transformations which don't change the volume of the cube is one (exchanging columns) or equal to the value by which we have to multiply the volume of the cube in order to get the volume of the parallelepiped (multiplying a column by a constant or adding a multiple of a column to another is equivalent to elongating a vector by that multiple). So the volume of resulting figure is $\det C\det E_1...\det E_i=\det P$.
1.) Is this argument correct?
2.) Is it possible to prove it geometrically? By this I mean similarly to the 2D case here.
3.) Wikipedia says it is possible to find the volume using Cramer's rule on some 2D matrices, how is this done?
Thanks for any advice!

REPLY [3 votes]: Your argument is correct.
With a little effort (but not much), you should be able to modify your argument to make it more geometric.
Here is the the two dimensional version:
If we have a parallelogram spanned by the vectors $\vec{a}$, $\vec{b}$,
then the area of
the parallelogram is the same as the area of the rectangle spanned by $\vec{a}, \vec{b}',$
where $\vec{b}'$ is the component of $\vec{b}$ that is orthogonal to $\vec{a}$,
i.e. $$\vec{b}' = \vec{b} - \dfrac{(\vec{a} \cdot \vec{b})}{(\vec{a} \cdot \vec{a})} \vec{a}.$$
Now the process of going from the matrix $\Bigl( \,  \vec{a} \quad \vec{b} \, \Bigr) $ to the matrix $\Bigl( \, \vec{a} \quad \vec{b}' \, \Bigr)$ involves
right-multiplying by the matrix $\begin{pmatrix} 1 & - (\vec{a} \cdot \vec{b})/
(\vec{a}\cdot\vec{a}) \\ 0 & 1\end{pmatrix}$, whose determinant is $1$.
Thus we see that $\Big( \, \vec{a} \quad \vec{b} \, \Bigr)$
and $\Bigl( \, \vec{a} \quad \vec{b}' \, \Bigr)$ have the same
determinant, and also describe parallelograms with the same area,
the latter being a rectangle.
This reduces you to checking the relationship between area and determinant in the case of a rectangle.  Rotating this rectangle, you can make its edges parallel to the coordinate axes.  Again, a rotation matrix has determinant one, so you are reduced
to checking the relationship between determinants and areas in the case
of a rectangle whose sides are parallel to the coordinate axes.  this
is pretty obvious, and so we are done.

In the three dimensional case, you can argue similarly: you first of all reduce to the case where one face is a rectangle, and you then reduce to the case when the third side is perpendicular to the rectangular face, so that the whole thing
is a cuboid.  These steps involve right multiplying by matrices
which are upper triangular with $1$'s down the diagonal, which thus have det $= 1$.
Now applying a bunch of rotations (again, each has det $= 1$), you can make
your cuboid have sides parallel to the coordinate axes, at which point the
formula is again pretty obvious.

Note that if we allow ourselves one more step --- namely, multiplying by a diagonal matrix (which, geometrically, is a rescaling of each of the coordinate axes) --- then we can start with any (non-degenerate) parallelepiped and convert it to
the standard cuboid with unit length sides sitting at the origin.
In linear algebra terms, this can be restated as the fact that any matrix
with non-zero determinant can be written as the product of a diagonal matrix,
some rotation matrices, and an upper triangular matrix with $1$'s down the diagonal.
Combining the diagonal and upper triangular matrix with $1$'s down the diagonal,
we obtain an upper triangular matrix with non-zero entries down the diagonal.
In other words, any matrix with non-zero determinant can be written as a product
of some rotation matrices with an upper triangular matrix.
This is usually called the $QR$ decomposition in linear algebra textbooks;
in more theoretical treatments it is called the Iwasawa decomposition.
So what I have just given is a geometric description of the $QR$ decomposition.

The difference between what I've described and your argument is that you use
elementary matrices, while I use just upper triangular matrices with $1$'s down
the diagonal.  These arise from the geometric process of projecting one vector
to make it perpendicular to another, which is where the geometric perspective is
coming from.


TITLE: Time And Distance (Train Journey)
QUESTION [0 upvotes]: The average speed of a train in the onward journey is 25% more than that in the return journey. The train halts for one hour on reaching the destination. The total time taken for the complete to and from journey is 17 hours, covering a distance of 800 km. What is the speed of the train in the onward journey?
I know how to solve this question by long method which is very time consuming. Is there any other short way to solve this question?
I solved this question in the following way:
Let x be the distance. total distance is 800 km, so x = 400 km (assuming the train follows the same path in both directions)
Speed outbound V1 = 400/t1,   where t1 is time taken
Speed back V2 = 400/t2, where t2 is time taken
V1 = (1.25)V2
t1 + t2 + 1 = 17
V1 = (1.25)V2
400/t1 = (1.25)(400/t2)
1/t1 = (1.25/t2)
1.25 t1 = t2
t1 + t2 + 1 = 17
t1 + (1.25 t1) + 1 = 17
2.25 t1 = 16
t1 = 7.111 hours
t2 = 8.888 hours
total time = t1 + t2 + 1 = 16.999, check
V1 = 400/7.111 = 56.25 km/hour
V2 = 400/8.888 = 45 km/hour

REPLY [1 votes]: The train halts for one hour at the station, so the total time travelled is 16 hours. Therefore, $v_{avg} = \frac{800}{16}= 50km/h$
You can also find that:  $v_{avg}= \frac{v_1 + v_2}{2}$
Then its easy to find $v_2$:
$v_1 = 1.25v_2$
$100 = 2.25v_2$
$v_2 = 44.4 km/h$
$v_1 = 44.4 * 1.25 = 55.6km/h$


TITLE: Global inverse of $G(x,y)=(\ln (xy),\frac{1}{x^2+y^2})$
QUESTION [0 upvotes]: Let $G(x,y)=(\ln (xy),\frac{1}{x^2+y^2})$, with Dom$(G)=\{(x,y)\in\mathbb{R^2:0<y<x}\}$. I am asked to find the global inverse of $G$.
I have tried to proceed in the following way:
$$u=\ln(xy)=\ln(x)+\ln(y),\quad x=\frac{e^u}{y}$$
$$v=\frac{1}{x^2+y^2}=\frac{1}{\frac{e^{2u}}{y^2}+y^2}=\frac{y^2}{e^{2u}+y^4},\  y=+\sqrt{\frac{1\pm\sqrt{1-4v^2e^{2u}}}{2v}}=\frac{\sqrt{1\pm\sqrt{1-4v^2e^{2u}}}}{\sqrt {2v}}$$
But I'm not sure this is right, as my textbook says correct solution is
$$G^{-1}(u,v)=(\frac{\sqrt{1+2e^uv}+\sqrt{1-2e^uv}}{2\sqrt v}, \frac{\sqrt{1+2e^uv}-\sqrt{1-2e^uv}}{2\sqrt v})$$
Could you help me? Thanks in advance!

REPLY [2 votes]: You seem to be missing the $4$ from the quadratic solution formula!
If the equation is $ax^2+bx+c=0$ the solutions are
$$x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
Applying this to the biquadratic formula $vy^4-y^2+e^{2u}v=0$, which you seems to have done, leads to
$$(y^2)_{1,2}= \frac{1\pm\sqrt{1-{\mathbf  4}v^2u^{2u}}}{2v}$$
which is the same as you got, except for the marked $4$, and the choice of the $\pm$ sign. It turns out that because $y$ is the samller of the solutions ($x$ is the other), you need to use the "-" sign. That is then the same as the solution for $y$ given in the textbook, though it is non-trivial to see that
$$\sqrt{1-\sqrt{1-4v^2u^{2u}}}=\frac{\sqrt{1+2ve^u}-\sqrt{1-2ve^u}}{\sqrt2}$$
Once you know it, you can easily verify it by squaring the right side.
Jane's answer shows you a much easier way to get to the solution, though!


TITLE: Define $f$ on $\Bbb R$ by $f(x)=x^3$ for $x\ge0$ and $f(x)=0$ for $x\lt 0$. Find all $n\in\mathbb N$ such that $f^{(n)}$ exists on all of $\Bbb R$.
QUESTION [0 upvotes]: Define $f$ on $\Bbb R$ by $f(x)=x^3$ for $x\ge0$ and $f(x)=0$ for $x\lt 0$. Find all $n\in\mathbb N$ such that $f^{(n)}$ exists on all of $\Bbb R$.

I am studying for Analysis midterm.
I saw this question and I am not sure what exactly f^(n) means.
Is it talking about a derivative? or something else?
I am not 100% sure. It looks pretty simple.
any tips or suggestions would be great!

REPLY [1 votes]: $f^{(n)}(x)$
means the $n$-th derivative of $f$
at $x$.
For your problem, $f(x)$
is defined on two disjoint domains
as a different function on each domain.
These separate functions
are "nice" in their own domains,
so the only problem can be
where the domains meet,
which is at zero.
So what you need to do
is to see what the derivatives
of each function are at zero
and check if they agree or disagree.
The rest is up to you.
I may have said too much.


TITLE: Why is $u(t)\frac{d }{dt}u(t)=\frac{1}{2}\frac{d}{dt}u(t)^2$?
QUESTION [0 upvotes]: If I have $$u(t)\frac{d }{dt}u(t)$$
how can I write it as
$$
\frac{1}{2}\frac{d}{dt}u(t)^2 \qquad \text{?}
$$
Attempt:
$$
u(t)\frac{d }{dt}u(t)=\frac{d }{dt}\Big (u(t)u(t) \Big )=\frac{d }{dt}u(t)^2 \tag{1}
$$
And just add $1/2$ to compensate for the square:
$$
\frac{1}{2}\frac{d}{dt}u(t)^2
$$
I'm not sure of equation (1), can I really distribute the differential operator?

REPLY [1 votes]: The chain rule states that
$$[f(g(t))]' = f'(g(t))\cdot g'(t).$$
Now set $f(t) = t^2$ whose derivative is $f'(t) = 2t$. Then we have that $f(u(t)) = u(t)^2$. Now apply the chain rule:
$$[u(t)^2]' = [f(u(t))]' = f'(u(t))\cdot u'(t) = 2u(t)\cdot u'(t). $$ Hence by dividing by $2$, we get that
$$\frac12 [u(t)^2]' = u(t)\cdot u'(t), $$
or written with Leibniz notation
$$\frac12\frac{\mathrm d}{\mathrm d t}[u(t)^2] = u(t)\cdot \frac{\mathrm d}{\mathrm d t} u(t).$$


TITLE: Anti-matter black holes radiation
QUESTION [3 upvotes]: It is generally accepted that black hole emits Hawking radiation.
I wonder if this radiation is different in case of black hole created by collapsing anti-matter.
Consider a thought experiment. At the beginning we have antimatter which collapses into black hole. Then black hole evaporates as a result of Hawking radiation.
What is the end result of this radiation: matter, antimatter or mix of both?
If antimatter, then we have a method to distinguish between black hole and anti-black hole - it contradicts to responses to answers to "Anti-Matter Black Holes" that there is no method to find a difference between black hole and anti-black hole.
If matter or mix - then we have strange alchemic transmutation: antimatter (beginning of the experiment) transforms to plain matter.
So maybe anti-black hole emits Hawking radiation consisting of antimatter only.
Another possibility: Hawking radiation consisting only of truly neutral particles (photons, Higgs, etc.). This is strange too because antimatter changed to neutral particles.
What is the end result of anti-black hole evaporation?

REPLY [3 votes]: then we have strange alchemic transmutation: antimatter (beginning of the experiment) transforms to plain matter.

Before we even worry about black holes, note that there is no conservation law that prevents production of antimatter from matter. For example, some nuclei undergo beta decay and emit positrons. This is basically because conservation laws are additive. As a simple example, a photon can do pair production, which is a process in which it disappears and produces an electron and an antielectron. This doesn't violate any conservation laws, because although the electron and antielectron differ in charge and lepton number, the sums of these quantities are zero.
The no-hair theorems tell us that a black hole can be characterized by only three numbers: mass, charge, and angular momentum. That isn't enough information to tell very much at all about the type or types of matter that went into its formation by gravitational collapse -- if it even formed by gravitational collapse at all. It's possible that there are primordial black holes that did not form by gravitational collapse, so that no "matter" went into them.
What is strange from the point of view of particle physics is that black hole evaporation doesn't conserve quantities like baryon number and lepton number. The radiation is predicted to be purely thermal, which means that it can't have an imbalance of, e.g., electrons and antielectrons. Note that in practical terms, we will probably never be able to detect Hawking radiation, and in any case nearly all of it would be photons.


TITLE: Proving $E[X^4]=3σ^4$
QUESTION [22 upvotes]: Given a random variable $X\sim\mathcal N(0,\sigma^2)$, how can we prove that $E[X^4]=3\sigma^4$? I am having trouble even starting with the proof.

REPLY [1 votes]: Here's a solution with no integrals.
Let $Y, Z \sim \mathcal{N}(0, \sigma^2 / 2)$ be independent Gaussian random variables. Then $Y + Z \sim \mathcal{N}(0, \sigma^2)$. We wish to determine the value of $R = \mathbb{E} [(Y + Z)^4]$.
To do so, note that
$$
\begin{align}
R &= \mathbb{E}[(Y + Z)^4] \\
&= \mathbb{E}[Y^4 + 3Y^3Z + 6Y^2Z^2 + 3YZ^3 + Z^4] \\
&= \mathbb{E}[Y^4]
   + 3\mathbb{E}[Y^3]\mathbb{E}[Z]
   + 6\mathbb{E}[Y^2]\mathbb{E}[Z^2]
   + 3\mathbb{E}[Y] \mathbb{E}[Z^3]
   + \mathbb{E}[Z^4] \\
&= R / 4
   + 0
   + 6 \sigma^4 / 4
   + 0
   + R / 4.
\end{align}
$$
Thus $R = R / 2 + 6\sigma^4 / 4$, so $R = 3\sigma^4$.


TITLE: Proving symmetric difference
QUESTION [0 upvotes]: I have been asked to prove that: $(A \cup B)$ \ $(A \cap B) = (A$ \ $B) \cup (B$ \ $A)$
I have got this far in the proof:
$(A \cup B)$ \ $(A \cap B)$
$(x \in (A \cup B)) \land (x \not \in (A \cap B))$ - definition of set difference
$(x \in A \lor x \in B) \land (x \notin A \land  x \notin B)$ - definition of set union and set intersection
I can't see how to go any further, as using the distributive laws may get quite complicated considering there are 2 things either side of the conjunctive operator. Is this the right direction to head or have I started off on the wrong foot?

REPLY [2 votes]: HINT: Go back one step and the next step should be:
$$(x \in A ∨ x \in B) ∧ (x \not \in A \cap B)$$
$$(x \in A ∧ (x \not \in A \cap B)) ∨ (x \in B ∧ (x \not \in A \cap B))$$


TITLE: Finding maxima and minima of $f(x,y)= x^4 + y^4 - 2x^2 - 2y^2 + 4xy$
QUESTION [3 upvotes]: For $f(x,y)=x^4+y^4-2x^2-2y^2+4xy$, I need to find maxima or minima. There are three critical points: $(0,0),(\sqrt2, -\sqrt2),(-\sqrt2,\sqrt2)$
So at $(\sqrt2, -\sqrt2)$, $f$ has minimum value, $-8$ and at $(-\sqrt2,\sqrt2),$ it has same minimum value, $-8$. At $(0,0)$, after inspecting, I get that it has neither maxima or minima. But when we substitute $(0,0),$ we get $0\le f \le8,$ so should not we get $(0,0)$ as a point of maxima?

REPLY [1 votes]: More generally you can just use the Hessian at the critical points. We have
$$
\begin{aligned}
f_{xx}&=12x^2-4,\\
f_{xy}&=4,\\
f_{yy}&=12y^2-4.
\end{aligned}
$$
So at the critical points $P_{1,2}=\pm(\sqrt2,\sqrt2)$ we have
$$H_f(P_{1,2})=\left(\begin{array}{cc}12\cdot2-4&4\\4&12\cdot2-4\end{array}\right).$$
The principal minors $\Delta_1=20$, $\Delta_2=20^2-4^2=384$ are thus both positive. Meaning that the Hessian form is positive definite, and we have local minima.
On the other hand at the remaining critical point $P_3=(0,0)$ the Hessian
$$H_f(P_3)=\left(\begin{array}{cc}12\cdot0-4&4\\4&12\cdot0-4\end{array}\right)$$
is negative semidefinite as the determinant $=\Delta_2=0$.
This is immediately obvious from the quadratic Taylor polynomial at $P_3$ as well, because
$$
T_{2,P_3}(x,y)=-2x^2-2y^2+4xy=-2(x-y)^2.
$$
The nature of this critical point is thus left in the dark by a study of the Hessian alone. Let's do some testing suggested by the form of the quadratic Taylor polynomial. If $x=y=h$, then
$$
f(x,y)=f(h,h)=2h^4-2(h-h)^2=2h^4>0.
$$
On the other hand
$$
f(h,-h)=2h^4-2(h-(-h))^2=-8h^2+2h^4<0
$$
when $h$ is sufficiently close to zero.
This means that $P_3$ is not a local extremal point.
Michael Rozenberg already handled the global minima/maxima.


TITLE: Derivative problem: No $x$ in equation and cubed roots
QUESTION [0 upvotes]: I am having trouble finding this answer to a derivative based problem. If anyone could walk me through this that would be great.
I am aware of all of the basics when it comes to derivatives.
When a problem contains $\dfrac{1}{h}(f(x+h)-f(x))$ I know that secluding the $f(x)$ from this equation helps in finding the derivative $f'(x)$.
However, I have never been faced with a problem like this where there is no $x$ present in the equation for $f(x)$.
Here is the problem:

REPLY [0 votes]: Consider
$f(x) = \sqrt[3]{x^5} $.
Then
$$f'(x) = \lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h} = \lim\limits_{h\to 0}\frac{\sqrt[3]{(x+h)^5}-\sqrt[3]{x^5}}{h} $$
Therefore, by substituting $x = -8$, we get that
$$f'(-8) = \lim\limits_{h\to 0}\frac{\sqrt[3]{(-8+h)^5}-\sqrt[3]{(-8)^5}}{h} $$
That is, the derivative of $f(x)$ evaluated at $x=-8$.


TITLE: Rutherfords experiment gold foil
QUESTION [0 upvotes]: I know there are many similar questions to this but didn't find one specifically like this please refer me to one if there is one already.
One component of his experimental set-up used gold foil, Au.
What change could have happened if he'd used Lithium foil instead what would be some observations he would've seen?

REPLY [1 votes]: Aside of the pedantic issues of isotopes and oxidation, I guess what you're really asking is what would be different if we scattered alpha-particles off a much lighter nucleus.
The important aspect of the Rutherford Scattering Formula is that the scattering rate at any given angle $\propto Z^2$
So reducing the atomic number from 79 to 3 will reduce the rate by a factor of about 700. Since the rate is already quite low at the large angles where Rutherford scattering is notable, the effect would practically disappear.

REPLY [0 votes]: Lithium will be oxidized in a moment, as it is very reactive and unstable on the air. Foil should be very thin, gold is very good to make thin foils.
Also, lithium has 2 stable isotopes. Which one do you mean?

REPLY [0 votes]: There wouldn't have been large changes, the "angle spread" would have been different and the scattering probabilities.
If you want to know how exactly, you need to look up the differential cross section for electron lithium scattering. I can't guide you where to find it but it should certainly available somewhere

REPLY [0 votes]: From the gold foil experiment Rutherford was able to estimate the closest approach of an alpha to the gold nucleus by equating the initial kinetic energy of the alpha to the electric potential energy of the gold nucleus and alpha system at closet approach.
The experimental data was consistent with the theoretical predictions which were based on the assumption the only force between an alpha particle and a nucleus was the Coulomb repulsive force obeying the inverse square law.
The distance of closest approach gave an upper bound to the radius of the nucleus.
Soon after the gold foil experiment, other experiments were performed with other metal being used as targets  and it was found that as the mass of a nucleus decreased scattering angles were more probable than for a target made of gold.
Again the assumption that the only force of interaction was the Coulomb repulsive force was seen to be valid because the experimental data agreed with the theoretic predictions except for large angle deflections.
The scattering angle increases as an alpha particle comes closer to a nucleus.
The departure of the experimental data from the theoretical predictions was thought to be be due to an alpha "touching" a nucleus with the result that there was a second force acting between an alpha and a nucleus.
Noting when experiment and theory did not agree enable an estimate to be made for the radius of a nucleus.
All other things being equal, using Lithium as a target would result in a greater probability of smaller angles of deflection and a greater probability of large angle deflections departing from theoretical predictions, based on the Coulomb force being the only force of interaction between an alpha and a nucleus, because for a given energy of alpha particles an alpha's nearest approach to a lithium nucleus is smaller than that for a gold nucleus.


TITLE: Why is this sum zero?
QUESTION [13 upvotes]: I have been looking at the following sum (for any positive integer $n$)
$$\left(1-\frac{1^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2^2}{n}\right) + \left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3^2}{n}\right) + \ldots $$
Note that the $i$th term in the sum has $i$ factors and  is
$$\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\left(1-\frac{3}{n}\right)\dots \left(1-\frac{i-1}{n}\right)\left(1-\frac{i^2}{n}\right).$$
It seems, amazingly, that the answer is 0. How can one show this?

REPLY [1 votes]: The following holds true for $n\geq 1$
\begin{align*}
\sum_{k=1}^{\infty}\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)=0\tag{1}
\end{align*}

Note, the empty product is set equal to $1$. We start by transforming the product into a somewhat more convenient form .

\begin{align*}
\sum_{k=1}^{\infty}&\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)\\
&=\sum_{k=1}^{n}\left(1-\frac{k^2}{n}\right)\frac{1}{n^{k-1}}\prod_{j=1}^{k-1}(n-j)\tag{2}\\
&=n!\sum_{k=1}^{n}\left(1-\frac{k^2}{n}\right)\frac{1}{n^k}\frac{1}{(n-k)!}\\
&=\frac{n!}{n^n}\sum_{k=0}^{n-1}\left(1-\frac{(n-k)^2}{n}\right)\frac{n^k}{k!}\tag{3}\\
\end{align*}

In (2) we set the upper limit of the sum to $n$ since for values greater than $n$ the product contains a factor zero. In the last step (3) we changed the index summation $k\rightarrow n-k$. In the following we skip the factor $\frac{n!}{n^n}$ and show the sum is equal to zero.

\begin{align*}
\sum_{k=0}&\left(1-\frac{(n-k)^2}{n}\right)\frac{n^k}{k!}\\
&=\frac{1}{n}\sum_{k=0}^{n-1}(n-n^2+2kn-k^2)\frac{n^k}{k!}\\
&=(1-n)\sum_{k=0}^{n-1}\frac{n^k}{k!}+2\sum_{k=1}^{n-1}k\frac{n^k}{k!}
-\frac{1}{n}\sum_{k=2}^{n-1}k(k-1)\frac{n^k}{k!}-\frac{1}{n}\sum_{k=1}^{n-1}k\frac{n^k}{k!}\tag{4}\\
&=(1-n)\sum_{k=0}^{n-1}\frac{n^k}{k!}+2n\sum_{k=0}^{n-2}\frac{n^k}{k!}
-n\sum_{k=0}^{n-3}\frac{n^k}{k!}-\sum_{k=0}^{n-2}\frac{n^k}{k!}\tag{5}\\
&=\frac{n^{n-1}}{(n-1)!}-n\frac{n^{n-1}}{(n-1)!}+n\frac{n^{n-2}}{(n-2)!}\\
&=\frac{n^{n-1}}{(n-1)!}\left(1-n+(n-1)\right)\\
&=0\\
&\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\Box
\end{align*}

Comment:

In (4) we set the lower bounds of the index accordingly and use $k^2=k(k-1)+k$ to be able to cancel out factors of the factorial conveniently.
In (5) we adjust the index accordingly to obtain equal summands for telescoping in the next step.

Added 2015-08-13: A note to the elegant approach of @BarryCipra due to a comment of @hypergeometric.
Let $R$ be a positive integer less than $n$. We obtain from (1)
\begin{align*}
\sum_{k=1}^{\infty}&\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)\\
&=\sum_{k=1}^{R}\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)+
\sum_{k=R+1}^{\infty}\left(1-\frac{k^2}{n}\right)\prod_{j=1}^{k-1}\left(1-\frac{j}{n}\right)\\
&=\prod_{r=1}^R\left(1-\frac{r}{n}\right)\left(\sum_{k=1}^{R}\left(1-\frac{k^2}{n}\right)\prod_{j=k}^R\left(1-\frac{j}{n}\right)^{-1}\right.\\
&\qquad\qquad\qquad\qquad+\left.\sum_{k=R+1}^{\infty}\left(1-\frac{k^2}{n}\right)\prod_{j=R+1}^{k-1}\left(1-\frac{j}{n}\right)\right)\tag{6}\\
&=\prod_{r=1}^R\left(1-\frac{r}{n}\right)\left(R+\sum_{k=R+1}^{\infty}\left(1-\frac{k^2}{n}\right)\prod_{j=R+1}^{k-1}\left(1-\frac{j}{n}\right)\right)\tag{7}
\end{align*}

Comment:

Observe the last line of @BarryCipras representation corresponds to (7) with $R=3$.
Since we know that (7) is valid, we proceed from (6) to (7) by claiming the validity of the identity

\begin{align*}
\sum_{k=1}^{R}\left(1-\frac{k^2}{n}\right)\prod_{j=k}^R\left(1-\frac{j}{n}\right)^{-1}=R\qquad\qquad 1\leq R < n
\end{align*}
This identity is a nice by-product.

When setting $R\geq n$ in (7) we observe, that the leftmost product contains a factor zero. But we can't use this argument in this derivation, since we have to consider in (6) the factors $\left(1-\frac{j}{n}\right)^{-1}$ which are undefined in case $j=n$.