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\subsection*{Group generated by central symmetries}
\label{Group generated by central symmetries}

\begin{pr}
Consider Lobachevsky space of sufficiently large dimension.
Does it admit a discrete cocompact isometric action generated by 
\begin{enumerate}[(a)]
 \item elements of finite order?
 \item central symmetries?
 \item\label{vinberg} rotations around subspaces of codimension $k$?
\end{enumerate}

\end{pr}

Ernest Vinberg \cite{vinberg, vinberg-strong} proved that there are no such actions generated by reflections;
it solves part (\ref{vinberg}) for $k=1$; the remaining questions are completely open.

\subsection*{Milnor's cartography problem}
\label{Milnor's cartography problem}

\begin{pr}
Let $\Omega$ be a round disc of radius $\alpha<\tfrac\pi2$ on the unit sphere $\mathbb{S}^2$
and $\Omega'$ be a convex domain in $\mathbb{S}^2$ with the same area as $\Omega$. Show that there is $(1,\tfrac{\alpha}{\sin\alpha})$-bi-Lipschitz map from $\Omega'$ to the plane.
\end{pr}

The question was asked by John Milnor \cite{milnor-cartography}.
The polar coordinates with the origin at the center of $\Omega$ provide an example of $(1,\tfrac{\alpha}{\sin\alpha})$-bi-Lipschitz map and this constant $\tfrac{\alpha}{\sin\alpha}$ is optimal.

If in the formulation of the problem, one exchanges ``area''  above to ``perimeter'',
then the answer is yes.
It can be proved using the idea of Gary Lawlor \cite{lawlor}.

\subsection*{Convex hull in CAT(0)}
\label{Convex hull in CAT(0)}

Let $X$ be a geodesic metric space.
Recall that a set $C\subset X$ is called \emph{convex} if for any two points $x,y\in C$ every geodesic from $x$ to $y$ lies in $C$. 
The \emph{closed convex hull} of a subset $K\subset X$ is defined as the intersection of all closed convex sets containing $K$.


\begin{pr}
Let $X$ be complete length $\mathop{\rm CAT}(0)$-space and $K\subset X$ be a compact subset.
Is it true that the closed convex hull of $K$ is compact?
\end{pr}


Note that the space is not assumed to be locally compact; otherwise, the statement is evident.
I believe there is a counterexample, maybe even with negatively pinched curvature (in the sense of Alexandrov).

The closed convex hull of a compact set (in particular of a three-point set) might look ugly even in a smooth 3-dimensional manifold; if doubt, check the note of Alexander Lytchak and the author \cite{lytchak-petrunin}.

This problem was mentioned by Mikhael Gromov \cite[6.B$_1\text{(f)}$]{gromov-asymptotic}
and restated by Eva Kopeck\'a and Simeon Reich \cite{kopecka-reich}, 
but the problem seems to be known in mathematical folklore for a long time. 



\subsection*{Two-convexity}
\label{Two-convexity}

Note that an open connected set $\Omega\subset\mathbb R^n$ is convex (in the usual sense) if together with two sides of triangle $\Omega$ contains the whole triangle.
This observation motivates the definition of two-convexity used in the following problem.


\begin{pr}
Let $\Omega$ be a simply-connected open set in $\mathbb R^n$.
Assume that $\Omega$ is two-convex; that is, if 3 faces of arbitrary 3-simplex belong to $\Omega$, then the whole simplex lies in $\Omega$.

Is it true that any component of the intersection of $\Omega$ with any 2-plane is simply-connected?
\end{pr}


If the boundary of $\Omega$ is a smooth hypersurface then the answer is ``yes''.
Indeed, the above property implies that at most one principle curvature of the boundary is negative.
Then the statement follows easily from a Morse-type argument \cite[see Lefschetz theorem in Section~$\tfrac12$ in][]{gromov-SGMC}.
This argument shows that any component of the intersection of $\Omega$ with any affine subspace (not necessarily 2-dimensional) is simply-connected.
By the following example, a general 2-convex open set does not satisfy the latter property.

The answer is ``yes'' if $n=3$; in this case, one can mimic the Morse-type argument;
it also follows from \cite[4.5.2 in][]{AKP-invitation}.

The statement would follow if one could approximate any $\Omega$ by two-convex domains with smooth boundaries.
However, the example below shows that such an approximation does not exist for $n\ge 4$.


\parit{Example.}
We will construct a two-convex simply connected open set $\Omega$ in $\mathbb R^4$ such that intersection $L_{t_0}$ of $\Omega$ with some hyperplane is not simply-connected.
(Note that this is not a counterexample;
it only shows that it is impossible to prove it using smoothing, as indicated in the comments.)

Set 
$$\Pi=\{\,(x,y,z,t)\in\mathbb R^4\mid\,y< x^2\}.$$
Let $\Pi'$ be the image of $\Pi$ under a generic rotation of $\mathbb R^4$,
say $(x,y,z,t)\z\mapsto(z,t,x,y)$.

Note that $\Pi$ is open and two-convex.
Therefore $\Omega=\Pi\cap \Pi'$ is also an open two-convex set.
Note that we can choose coordinates so that $\Omega$ is an epigraph for a function $f\colon\mathbb R^3\to\mathbb R$ like
$$f=\max\{\alpha_1-\beta_1^2,\alpha_2-\beta_2^2 \},$$
where $\alpha_i$ and $\beta_i$ are linear functions.
In particular $\Omega$ is contactable.

Let $L_{t_0}$ be the intersection of $\Omega$ with hyperplane $t=t_0$;
it is a complement of two convex parabolic cylinders in general position. 
If these cylinders have a point of intersection then $\pi_1 L_{t_0}=\mathbb Z$.
 
In particular, the function $f$ cannot be approximated by smooth functions such that their Hessians have at most one negative eigenvalue value at all points.
 
\subsection*{Homeomorphism of torus}

Let $\TT$ denote the 2-dimensional torus.

A homeomorphism $h\:\TT\to \TT$ is called \emph{simple} if it is supported on an embedded open disc;
that is, there is an embedded open disc $\Delta\subset \TT$ such that $h$ is identity outside of $\Delta$.

The \emph{disc norm} of a homeomorphism $h\:\TT\to \TT$ is defined as the least number of simple homeomorphisms $h_1,\dots,h_n$ such that
\[h=h_1\circ\dots\circ h_n.\]

\begin{pr}
Is there a homeomorphism of the 2-dimensional torus with an arbitrary large disc norm?
\end{pr}

Analogous questions for cylinder and sphere are solved, but for M\"obius band is open.
This problem is discussed in by Dmitri Burago, Sergei Ivanov, and Leonid Polterovich \cite{BIP};

\subsection*{Open projection}

Let us denote by $\DD^n$ a closed $n$-dimensional ball.

Recall that a map is called open if the image of any open set is open.

\begin{pr}
Is it possible to construct an embedding $\DD^4\hookrightarrow \mathbb{S}^2\times
\mathbb R^2$
such that the projection  $\DD^4\to \mathbb{S}^2$ is an open map?
\end{pr}


It is easy to construct an embedding $\DD^3\hookrightarrow \mathbb{S}^3$ such that
its composition with Hopf fibration $f_3:\DD^3\to \mathbb{S}^2$ is open.
Therefore there is an open map $f_3\:\DD^3\to\mathbb{S}^2$.

Composing $f_3$ with any open map $\DD^4\to \DD^3$,
one gets an open map $f_4:\DD^4\to \mathbb{S}^2$.

The map $f_3$ is not a projection of embedding  $\DD^3\hookrightarrow \mathbb{S}^2\times\mathbb R$.
(We have $f_3^{-1}(p)=\mathbb{S}^1$ for some $p\in \mathbb{S}^2$ and $\mathbb{S}^1$ cannot be embedded in $\mathbb R$.)  

The question was asked by $\eps$-$\delta$ (an anonymous mathematician)  \cite{epsilon-delta}.

\subsection*{Braid space}

Denote by $\mathbb{B}^n$ the universal covering of $\CC^n$ infinitely branching along the lines $z_i=z_j$ for $i\ne j$.
We suppose that $\CC^n$ is equipped with canonical Euclidean metric, 
and $\mathbb{B}^n$ is equipped with lifted length metric.

In other words, if $\mathbb{W}^n$ denoted the complement of $\CC^n$ to the lines $z_i=z_j$,
then the braid space $\mathbb{B}^n$ is the completion of the universal metric covering of $\mathbb{W}^n$.

\begin{pr}
Is it true that $\mathbb{B}^n$ is $\mathrm{CAT}(0)$ for any $n$? 
\end{pr}

The answer is yes for $n\le 3$.
While the case $n=2$ is trivial.
The case $n=3$ follows from the theorem of Ruth Charney and Michael Davis about branched coverings of 3-sphere \cite{charney-davis};
another proof is given by Dmitri Panov and the author \cite{panov-petrunin-ramification}.

The space $\mathbb{B}^n$ has plenty of convex functions that can be constructed the following way:
given $n$ distinct points in $\CC$ (that is, an element of $\mathbb{W}^n$) consider a circle $z$ in its complement.
Consider the function $\mathbb{B}^n\to \RR$  that returns the greatest lower bound of lengths of closed curves that homotopic of $z$. 
This construction can be used to show that there is no geodesic loops in $\mathbb{B}^n$;
in order to show that $\mathbb{B}^n$ is $\mathrm{CAT}(0)$ one has to show that there are no geodesic diangles.
This construction is inspired by the argument given by Ian Agol \cite{agol-braid} which provides an affirmative answer to a corollary from the conjecture. 

\subsection*{Quotients of Hilbert space}

\begin{pr} Suppose $R$ is a compact simply-connected Riemannian manifold that is isometric to a quotient of the Hilbert space by a group of isometris (or more generally $R$ is the target of a Riemannian submersion from a Hilbert space).
Is it true that $R$ is isometric to a double quotient?
That is, is it true that $R$ is a quotient of compact Lie group $G$ with a bi-invariant metric by a group of isometries?
 
\end{pr}

Any double quotient can appear as a quotient of the Hilbert space by a group of isometries.
The following proof is taken from \cite{lebedeva-petrunin-zolotov}; it was suggested by Alexander Lytchak.
Another proof follows from the construction of Chuu-Lian Terng and Gudlaugur Thorbergsson given in \cite[Section 4]{terng-thorbergsson}.

Denote by $G^n$ the direct product of $n$ copies of $G$.
Consider the map $\phi_n\:G^n\to G/\!\!/H$ defined by
\[\phi_n\:(\alpha_1,\dots,\alpha_n)\mapsto [\alpha_1\cdots\alpha_n]_H,\]
where $[x]_H$ denotes the $H$-orbit of $x$ in $G$.

Note that $\phi_n$ is a quotient map for the action of $H\times G^{n-1}$ on $G^n$ defined by
\[(\beta_0,\dots,\beta_n)\cdot(\alpha_1,\dots,\alpha_n)=(\gamma_1\cdot \alpha_1\cdot\beta_1^{-1},\beta_1\cdot\alpha_2\cdot\beta_2^{-1},\dots,\beta_{n-1}\cdot\alpha_n\cdot\beta_n^{-1}),\]
where $\beta_i\in G$ and $(\beta_0,\beta_n)\in H<G\times G$.

Denote by $\rho_n$ the product metric on $G^n$ rescaled with factor $\sqrt{n}$.
Note that the quotient $(G^n,\rho_n)/(H\times G^{n-1})$ is isometric to $G/\!\!/H\z=(G,\rho_1)/\!\!/H$.
Let $\phi_n\:(G^n,\rho_n)\to G/\!\!/H$ be the corresponding quotient map; 
clearly, $\phi_n$ is a submetry. 

As $n\to\infty$ the curvature of $(G^n,\rho_n)$ converges to zero and its injectivity radius goes to infinity.
Therefore the ultralimit of $(G^n,\rho_n)$ with marked identity element is a Hilbert space $\HH$, and the submetries $\phi_n$ ultraconverge to a submetry $\phi\:\HH\to G/\!\!/H$.


\subsection*{Nested convex surfaces}

\begin{pr}
Describe the Riemannian metrics on $\mathbb{S}^n$ that are isometric to smooth nested convex surfaces;
that is, a complete smooth convex hypersurface in a complete smooth convex hypersurface in ... in a Euclidean space.
\end{pr}

If $n=2$, then by Alexandrov embedding theorem these are all Riemannian metrics with nonnegative curvature.

Direct calculations show that the metric has \emph{nonnegative cosectional curvature};
the latter means that at each point the curvature tensor can be expressed as a linear combination of the curvature tensors of $\mathbb{S}^2\times \RR^{n-2}$ with positive coefficients.
It might happen that any metric with nonnegative cosectional curvature on $\mathbb{S}^n$ is isometric to a nested convex surface.
If one drops completeness from the definition of nested convex surfaces
and assumes that the cosectional curvature is strictly positive,
then we get the answer is yes \cite{petrunin-poly}.

\subsection*{Maximal finite subgroups}

\begin{pr}
Suppose that a group $\Gamma$ acts effectively, isometrically, and totally discontinuously on the Euclidean space $\RR^n$.
Show that $\Gamma$ contains at most $2^n$ maximal finite subgroups up to conjugation.
\end{pr}

Let $F$ be a maximal finite subgroup in $\Gamma$.
Note that the fixed points of $F$ form an affine subset of $\mathbb{E}^n$ and it is mapped to a singular set $S_F$ in the quotient space $X=\mathbb{E}^n/\Gamma$.
Moreover, (1) a conjugation of $F$ does not change the set $S_F$ and, since $F$ is maximal, (2) $S_F$ is a \emph{simple} singular set;
that is, $S_F$ does not have a proper singular subset.
So instead of counting subgroups up to conjugation, we may count simple singular sets in $X$.

It is expected that the maximal number of simple singular sets appears if all of them have dimension 0;
that is, each set $S_F$ is an isolated point in $X$.
If this is indeed true, then the statement would follow.
The latter was shown by  Nina Lebedeva \cite{lebedeva};
she proved that the number of such points indeed can not exceed $2^n$.
This is done by modifying a solution to the following problem of Paul Erdős given by Ludwig Danzer and Branko Gr{\"u}nbaum \cite{danzer-guenbaum}.
\emph{Show that one can choose at most $2^m$ points in $\mathbb{E}^m$ such that any triangle with the vertexes in the chosen points is acute or right.}

\subsection*{Zero-curvature along geodesics}

\begin{pr}
Let $M$ be a surface of genus 2 with a Riemannian metric of nonpositive curvature.
Show that with probability 1, a random geodesic in $M$ runs into the set of negative curvature.
\end{pr}

I learned the problem from 
Dmitri Buragov; it originated from a discussion with Keith Burns, Michael Brin, and Yakov Pesin \cite[see also][]{hertz}.
It seems that \textit{there are no known examples of $M$ with nonperiodic geodesic that runs in the set of zero curvature}.

\subsection*{Ambrose problem}

\begin{pr}
Let $(M,g)$ be a simply-connected complete Riemannian manifold and $p\in M$.
Denote by $h$ the pullback of the metric tensor $g$ to the tangent space $\T_pM$ along the exponential map $\exp_p\:\T_pM\z\to M$;
in general, $h$ might degenerate at some points.
Does $h$ define the manifold $(M,g)$ and the point $p$ up to isometry? 
\end{pr}

This simple-looking question was asked a long time ago by Warren Ambrose \cite{ambrose}.
Affirmative answers were obtained in three cases: (1) in dimension 2, (2) for analytic metrics, and (3) for generic metrics; see \cite{hebla, itoh, ardoy} and the references therein.


\subsection*{Shortcut in a connected set}

\begin{pr}
Let $p$ and $q$ be a pair of points in a compact connected subset $K\subset \RR^n$.
Is it possible to connect $p$ to $q$ by a curve $\gamma$ the length of $\gamma\setminus K$ is arbitrary small?
\end{pr}

Be aware that there are examples of compact connected sets which contain no nontrivial paths; for example, the so-called \emph{pseudoarc}.

In the case $n=2$, an affirmative answer was obtained by Taras Banakh \cite{banakh}.
His argument is based on the following claim

\begin{itemize}
 \item Suppose $K$ is a connected compact set in the plane and $p,q\in K$.
 Then given $\eps>0$ there is a sequence of compact connected subsets $K_0,K_2,\dots K_n$
 with pairs of points $p_i,q_i\in K_i$ for each $i$ such that 
(1) $\diam K_i<\eps$ for each $i$,
(2) $p_0=p$, $q_n=q$, and (3) if $\ell_i$ denotes the line segment $[q_{i-1}q_i]$, then 
 \[\length \ell_1+\dots+\length \ell_n<\eps.\]
\end{itemize}

Assume that the claim is proved.
Let us apply it recursively to each $K_i$ choosing a much smaller value $\eps$; each time we can assume that the number $n$ is taken to be minimal for the given $\eps$.
Consider the closure $L$ of the set of all points $p_i$ and $q_i$ for all iterations; it is a closed subset $L\subset K$.
By adding to $L$ the line segments $\ell_i$ for all the iterations we obtain a curve connecting $p$ to $q$ that spends an arbitrary small length outside of $L$.
Whence the 2-dimensional case follows.
It remains to prove the claim.

Consider a coordinate grid $\Gamma$ on the plane with step $\tfrac\eps2$.
Since the intersection $\Gamma\cap K$ is compact, it can be covered by a finite collection of closed line segments $\{I_1,\dots,I_n\}$ in the grid such that the 
\[\sum_i\length(I_i\setminus K)<\eps.\]

Take a minimal collection of sets $\{K_i\}$ such that each set $K_i$ is a connected component of the intersection of $K$ with a closed square of the grid and the union 
\[\biggl(\bigcup_i I_i\biggr)\cup\biggl(\bigcup_j K_j\biggr)\]
is a connected set containing $p$ and $q$. 
Observe that the collection $\{K_i\}$ is finite and it satisfies the conditions in the claim, assuming that $K_i$ are ordered correctly.
(The choice of point $p_i$ and $q_i$ is straightforward;
the line segments $\ell_j$ can be chosen to lie in the union of $\{I_i\}$.)


\subsection*{Optic fiber}

Suppose that $V$ is a body in $\RR^3$ bounded by two plane figures $F_1$ and $F_2$ (the ends) and a smooth surface $\Sigma$ (a tube).
Suppose that a light ray that is coming into $V$ thru one end and bouncing with perfect reflection from the interior walls of $\Sigma$ will emerge from the other end with probability 1;
moreover, a light ray that emerges from the interior of $V$, after bouncing $\Sigma$ emerges from one of the ends with  probability 1.
In this case, we say that $V$ is an \emph{optic fiber}.

A small tubular neighborhood of a smooth curve $\gamma\:[a,b]$ with disk ends gives an example of optic fiber.
A more general example can be constructed along the same lines using any plane figure bounded by a smooth curve instead of the disk.
To do this one has to choose a  parallel normal frame $e_1,e_2$ along $\gamma$ 
(that is, such that $e_i'(t)\parallel\dot\gamma(t)$ for all $t$) 
and consider the tube $[a,b]\times\mathbb S^1\hookrightarrow\mathbb R^3$ defined as
$$(t,\theta)\mapsto \gamma(t)+x(\theta){\cdot}e_1(t)+y(\theta){\cdot}e_2(t)$$
(The condition that the frame is parallel implies that any normal plane to $\gamma$ cuts tube at right angle.)
This way we get an \emph{optic fiber} with congruent ends.

Are there any constructions of optic fibers different from the one described above? In other words:

\begin{pr}
Is it always possible to slice an optic fiber by planes that cut the walls at right angle?
\end{pr}

In particular, 

\begin{pr}
Is there an optic fiber with noncongruent ends?
\end{pr}

Note that by  Liouville's theorem, it is clear that the ends must have the same area.

If the walls are only assumed to be \emph{piecewise} smooth then one can make an optic fiber with a pair of equidecomposable figures at the ends.
(The construction is the same, but one splits the tube into few on the way and then rearranges them back together.) 

In dimension 2, a line passing through focal points cuts from confocal ellipses an optic fiber. (I learned it from Arseniy Akopyan.)  I do not know smooth 3D examples of that type.

The problem is based on two questions posted on mathoverflow \cite{rourke-optic, petrunin-optic}

\subsection*{Guth's sponge}

\begin{pr}
Show that there is $\eps_n>0$ such that any open set in $\RR^n$ with volume at most $\eps_n$ admits a length-increasing embedding into a unit ball.
\end{pr}

The question was asked by Lary Guth in his thesis \cite{guth-2005}.
It seems that if one removes the injectivity condition, then the answer is ``yes''.
At least it is true in the two-dimensional case; a proof is sketched in \cite{petrunin-guth}.

\subsection*{Involution of 3-sphere}

\begin{pr}
Suppose that a closed geodesic $\gamma$ is the fixed-point set of an isometric involution of $(\mathbb{S}^3,g)$.
Assume that sectional curvature of $g$ is at least $1$.
Is it true that $\length\gamma\le2\cdot\pi$?
\end{pr}

The same question can be asked about polyhedral spaces with curvature at least 1 in the sense of Alexandrov.

If instead of an isometric involution we have an $\mathbb{S}^1$-action, then the answer is yes.
It follows since the quotient space $(\mathbb{S}^3,g)/\mathbb{S}^1$ is a convex disc with sectional curvature at least 1,
and $\gamma$ projects isometrically to its boundary.