Here is yet another, completely Riemannian application. This statement has
been proven by Perelman, a sketch of its proof is included in an appendix to
[Petrunin 2003]. The proof is based on the following observation: if $\Omega$
is an open subset of a Riemannian manifold and
$F\colon\Omega\to\mathbb{R}^{\ell+1}$ is a tight map with strictly concave
coordinate functions, then its level sets $F^{-1}(x)$ inherit the lower
curvature bound.

  1. $\diamond$

_Continuity of the integral of scalar curvature._ Given a compact Riemannian
manifold $M$, let us define $\mathcal{F}(M)=\int_{M}\operatorname{Sc}$. Then
$\mathcal{F}$ is continuous on the space of Riemannian $m$-dimensional
manifolds with uniform lower curvature and upper diameter bounds.535353In fact
$\mathcal{F}$ is also bounded on the set of Riemannian $m$-dimensional
manifolds with uniform lower curvature, this is proved in [Petrunin 2007] by a
similar method.

##  9 Please deform an Alexandrov’s space.

In this section we discuss a number of related open problems. They seem to be
very hard, but I think it is worth to write them down just to indicate the
border between known and unknown things.

The main problem in Alexandrov’s geometry is to find a way to vary
Alexandrov’s space, or simply to find a nearby Alexandrov’s space to a given
Alexandrov’s space. Lack of such variation procedure makes it impossible to
use Alexandrov’s geometry in the way it was designed to be used:

For example, assume you want to solve the Hopf conjecture545454i.e. you want
to find out if $S^{2}\times S^{2}$ carries a metric with positive sectional
curvature.. Assume it is wrong, then there is a volume maximizing Alexandrov’s
metrics $d$ on $S^{2}\times S^{2}$ with curvature $\geqslant 1$555555There is
no reason to believe that this metric $d$ is Riemannian, but from Gromov’s
compactness theorem such Alexandrov’s metric should exist.. Provided we have a
procedure to vary $d$ while keeping its curvature $\geqslant 1$, we could find
some special properties of $d$ and in ideal situation show that $d$ does not
exist.

Unfortunately, at the moment, except for boring rescaling, there is no
variation procedure available. The following conjecture (if true) would give
such a procedure. Although it will not be sufficient to solve the Hopf
conjecture, it will give some nontrivial information about the critical
Alexandrov’s metric.

######  9.1.1.

Conjecture. The boundary of an Alexandrov’s space equipped with induced
intrinsic metric is an Alexandrov’s space with the same lower curvature bound.

This also can be reformulated as:

9.1.1${}^{\prime}\mskip-3.0mu\mskip-3.0mu\mskip-3.0mu$. Conjecture. Let $A$ be
an Alexandrov space without boundary. Then a convex hypersurface in $A$
equipped with induced intrinsic metric is an Alexandrov’s space with the same
lower curvature bound.

This conjecture, if true, would give a variation procedure. For example if $A$
is a non-negatively curved Alexandrov’s space and $f\colon A\to\mathbb{R}$ is
concave (so $A$ is necessarily open) then for any $t$ the graph

$A_{t}=\\{(x,t{\cdot}f(x))\in A\times\mathbb{R}\\}$

with induced intrinsic metric would be an Alexandrov’s space. Clearly
$A_{t}\buildrel\mathrm{GH}\over{\longrightarrow}A$ as $t\to 0$. An analogous
construction exists for semiconcave functions on closed manifolds, but one has
to take a _parabolic cone_ 565656see footnote 24 on page 24 instead of the
product.

It seems to be hopeless to attack this problem with purely synthetic methods.
In fact, so far, even for a convex hypersurface in a Riemannian manifold,
there is only one proof available (see [Buyalo]) which uses smoothing and the
Gauss formula575757In fact in this paper the curvature bound is not optimal,
but the statement follows from nearly the same idea; see [AKP].. There is one
beautiful synthetic proof (see [Milka 1979]) for a convex surface in the
Euclidian space, but this proof heavily relies on Euclidean structure and it
seems impossible to generalize it even to the Riemannian case.

There is a chance of attacking this problem by proving a type of the Gauss
formula for Alexandrov’s spaces. One has to start with defining a curvature
tensor of Alexandrov’s spaces (it should be a measure-valued tensor field),
then prove that the constructed tensor is really responsible for the geometry
of the space. Such things were already done in the two-dimensional case and
for spaces with bilaterly bounded curvature, see [Reshetnyak] and [Nikolaev]
respectively. So far the best results in this direction are given in
[Perelman-DC], see also section 8.2 for more details. This approach, if works,
would give something really new in the area.

Almost everything that is known so far about the intrinsic metric of a
boundary is also known for the intrinsic metric of a general extremal subset.
In [Perelman–Petrunin 1993], it was conjectured that an analog of conjecture
9.1.1 is true for any _primitive extremal subset_ , but it turned out to be
wrong; a simple example was constructed in [Petrunin 1997]. All such examples
appear when codimension of extremal subset is $\geqslant 3$. So it still might
be true that

######  9.1.2.

Conjecture. Let $A\in\text{{\nnn Alex}}(\kappa)$, $E\subset A$ be a primitive
extremal subset and $\operatorname{codim}E=2$ then $E$ equipped with induced
intrinsic metric belongs to $\text{{\nnn Alex}}(\kappa)$

The following question is closely related to conjecture 9.1.1.

######  9.1.3.

Question. Assume $A_{n}\buildrel\mathrm{GH}\over{\longrightarrow}A$,
$A_{n}\in\text{{\nnn Alex}}^{m}(\kappa)$, $\operatorname{dim}A=m$ (i.e. it is
not a collapse).

Let $f$ be a $\lambda$-concave function of an Alexandrov’s space $A$. Is it
always possible to find a sequence of $\lambda$-concave functions $f_{n}\colon
A_{n}\to\mathbb{R}$ which converges to $f\colon A\to\mathbb{R}$?

Here is an equivalent formulation:

9.1.3.′ Question. Assume $A_{n}\buildrel\mathrm{GH}\over{\longrightarrow}A$,
$A_{n}\in\text{{\nnn Alex}}^{m}(\kappa)$, $\operatorname{dim}A=m$ (i.e. it is
not a collapse) and $\partial A=\varnothing$.

Let $S\subset A$ be a convex hypersurface. Is it always possible to find a
sequence of convex hypersurfaces $S_{n}\subset A_{n}$ which converges to $S$?

If true, this would give a proof of conjecture 9.1.1 for the case of a
_smoothable Alexandrov’s space_ (see page 7.3.1).

In most of (possible) applications, Alexandrov’s spaces appear as limits of
Riemannian manifolds of the same dimension. Therefore, even in this reduced
generality, a positive answer would mean enough.

The question of whether an Alexandrov space is smoothable is also far from
being solved. From Perelamn’s stability theorem, if an Alexandrov’s space has
topological singularities then it is not smoothable. Moreover, from [Kapovitch
2002] one has that any space of directions of a smoothable Alexandrov’s space
is homeomorphic to the sphere. Except for the 2-dimensional case, it is only
known that any polyhedral metric of non-negative curvature on a 3-manifold is
smoothable (see [Matveev–Shevchishin]). There is yet no procedure of smoothing
an Alexandrov’s space even in a neighborhood of a regular point.

Maybe a more interesting question is whether smoothing is unique up to a
diffeomorphism. If the answer is positive it would imply in particular that
any Riemannian manifold with curvature $\geqslant 1$ and
$\operatorname{diam}>\tfrac{\pi}{2}$ is diffeomorphic(!) to the standard
sphere, see [Grove–Wilhelm] for details. Again, from Perelman’s stability
theorem ([Perelman 1991]), it follows that any two smoothings must be
homeomorphic. In fact it seems likely that any two smoothings are PL-
homeomorphic; see [Kapovitch 2007, question 1.3] and discussion right before
it. It seems that today there is no technique which might approach the general
uniqueness problem (so maybe one should try to construct a counterexample).

One may also ask similar questions in the collapsing case. In [PWZ] there were
constructed Alexandrov’s spaces with curvature $\geqslant 1$ which can not be
presented as a limit of an (even collapsing) sequence of Riemannian manifolds
with curvature $\geqslant\kappa>\tfrac{1}{4}$. In [Kapovitch 2005] there were
found some lower bounds for codimension of collapse with arbitrary lower
curvature bound to some special Alexandrov’s spaces, see section 7.3 for more
discussion. It is expected that the same spaces (for example, the spherical
suspension over the Cayley plane) can not be approximated by sequence of
Riemannian manifolds of any fixed dimension and any fixed lower curvature
bound, but so far this question remains open.

##  Appendix A Existence of quasigeodesics

This appendix is devoted to the proof of property 4 on page 4, i.e.

######  A.0.1.

Existence theorem. Let $A\in\text{{\nnn Alex}}^{m}$, then for any point $x\in
A$, and any direction $\xi\in\Sigma_{x}$ there is a quasigeodesic
$\gamma\colon\mathbb{R}\to A$ such that $\gamma(0)=x$ and $\gamma^{+}(0)=\xi$.

Moreover if $E\subset A$ is an extremal subset and $x\in E$,
$\xi\in\Sigma_{x}E$ then $\gamma$ can be chosen to lie completely in $E$.

The proof is quite long; it was obtained by Perelman around 1992; here we
present a simplified proof similar to [Perelman–Petrunin QG] which is based on
the gradient flow technique. We include a complete proof here, since otherwise
it would never be published.

Quasigeodesics will be constructed in three big steps.

A.1

Monotonic curves $\longrightarrow$ convex curves.

A.2

Convex curves $\longrightarrow$ pre-quasigeodesics.

A.3

Pre-quasigeodesics $\longrightarrow$ quasigeodesics.

In each step, we construct a better type of curves from a given type of curves
by an extending-and-chopping procedure and then passing to a limit. The last
part is most complicated.

The second part of the theorem is proved in the subsection A.4.

###  A.0 Step 0: Monotonic curves

As a starting point we use radial curves, which do exist for any initial data
(see section 3), and by lemma 3.1.2 are monotonic in the sense of the
following definition:

######  A.0.1.

Definition. A curve $\alpha(t)$ in an Alexandrov’s space $A$ is called
_monotonic_ with respect to a parameter value $t_{0}$ if for any
$\lambda$-concave function $f$, $\lambda\geqslant 0$, we have that function

$t\mapsto\frac{f\circ\alpha(t+t_{0})-f\circ\alpha(t_{0})-\tfrac{\lambda}{2}{\cdot}t^{2}}{t}$

is non-increasing for $t>0$.

Here is a construction which gives a new monotonic curve out of two. It will
be used in the next section to construct _convex curves_.

######  A.0.2.

Extention. Let $A\in\text{{\nnn Alex}}$, $\alpha_{1}[a,\infty)\to A$ and
$\alpha_{2}\colon[b,\infty)\to A$ be two monotonic curves with respect to $a$
and $b$ respectively.

Assume

$a\leqslant b,\ \ \alpha_{1}(b)=\alpha_{2}(b)\ \ \text{and}\ \
\alpha^{+}_{1}(b)=\alpha^{+}_{2}(b).$

Then its joint

$\beta\colon[a,\infty)\to A,\ \
\beta(t)=\left[\begin{matrix}\alpha_{1}(t)&\text{if}&t<b\\\
\alpha_{2}(t)&\text{if}&t\geqslant b\end{matrix}\right.$

is monotonic with respect to $a$ and $b$.

Proof. It is enough to show that

$t\mapsto\frac{f\circ\alpha_{2}(t+a)-f\circ\alpha_{1}(a)-\tfrac{\lambda}{2}{\cdot}t^{2}}{t}$

is non-increasing for $t\geqslant b-a$. By simple algebra, it follows from the
following two facts:

  1. $\diamond$

$\alpha_{2}$ is monotonic and therefore

$t\mapsto\frac{f\circ\alpha_{2}(t+b)-f\circ\alpha_{2}(b)-\tfrac{\lambda}{2}{\cdot}t^{2}}{t}$

is non-increasing for $t>0$.

  2. $\diamond$

From monotonicity of $\alpha_{1}$,

$\displaystyle(f\circ\alpha_{2})^{+}(b)$
$\displaystyle=d_{\alpha_{1}(b)}f(\alpha_{1}^{+}(b))=$
$\displaystyle=(f\circ\alpha_{1})^{+}(b)\leqslant$
$\displaystyle\leqslant\frac{f\circ\alpha_{1}(b)+f\circ\alpha_{1}(a)-\tfrac{\lambda}{2}(b-a)^{2}}{b-a}.$

∎

###  A.1 Step 1: Convex curves.

In this step we construct _convex curves_ with arbitrary initial data.

######  A.1.1.

Definition. A curve $\beta\colon[0,\infty)\to A$ is called _convex_ if for any
$\lambda$-concave function $f$, $\lambda\geqslant 0$, we have that function

$t\mapsto f\circ\beta(t)-\tfrac{\lambda}{2}{\cdot}t^{2}$

is concave.

Properties of convex curves. Convex curves have the following properties; the
proofs are either trivial or the same as for quasigeodesics:

  1. 1.

A curve is convex if and only if it is monotonic with respect to any value of
parameter.

  2. 2.

Convex curves are $1$-Lipschitz.

  3. 3.

Convex curves have uniquely defined right and left tangent vectors.

  4. 4.

A limit of convex curves is convex and the natural parameter converges to the
natural parmeter of the limit curves (the proof the last statement is based on
the same idea as theorem 7.3.3).

The next is a construction similar to A.0.2 which gives a new convex curve out
of two. It will be used in the next section to construct _pre-quasigeodesics_.

######  A.1.2.

Extention. Let $A\in\text{{\nnn Alex}}$, $\beta_{1}\colon[a,\infty)\to A$ and
$\beta_{2}\colon[b,\infty)\to A$ be two convex curves. Assume

$a\leqslant b,\ \ \beta_{1}(b)=\beta_{2}(b)\ \ \text{and}\ \
\beta^{+}_{1}(b)=\beta^{+}_{2}(b)$

then its joint

$\gamma\colon[a,\infty)\to A,\ \
\gamma(t)=\left[\begin{matrix}\beta_{1}(t)&\text{if}&t\leqslant b\\\
\beta_{2}(t)&\text{if}&t\geqslant b\end{matrix}\right.$

is a convex curve.

Proof. Follows immidetely from A.0.2 and property 1 above.

######  A.1.3.

Existence. Let $A\in\text{{\nnn Alex}}$, $x\in A$ and $\xi\in\Sigma_{x}$. Then
there is a convex curve $\beta_{\xi}\colon[0,\infty)\to A$ such that
$\beta_{\xi}(0)=x$ and $\beta_{\xi}^{+}(0)=\xi$.

Proof. For $v\in T_{x}A$, consider the radial curve

$\alpha_{v}(t)=\operatorname{gexp}_{x}(tv)$

According to lemma 3.1.2 if $|v|=1$ then $\alpha_{v}$ is $1$-Lipschitz and
monotonic. Moreover, straightforward calculations show that the same is true
for $|v|\leqslant 1$.

Fix $\varepsilon>0$. Given a direction $\xi\in\Sigma_{x}$, let us consider the
following recursively defined sequence of radial curves $\alpha_{v_{n}}(t)$
such that $v_{0}=\xi$ and $v_{n}=\alpha^{+}_{v_{n-1}}(\varepsilon)$. Then
consider their joint

$\beta_{\xi,\varepsilon}(t)=\alpha_{v_{\lfloor
t/\varepsilon\rfloor}}(t-\varepsilon\lfloor t/\varepsilon\rfloor).$

Applying an extension procedure A.0.2 we get that
$\beta_{\xi,\varepsilon}\colon[0,\infty)\to A$ is monotonic with respect to
any $t=n{\cdot}\varepsilon$.

By property 1 on page 1, passing to a partial limit
$\beta_{\xi,\varepsilon}\to\beta_{\xi}$ as $\varepsilon\to 0$ we get a convex
curve $\beta_{\xi}\colon[0,\infty)\to A$.

It only remains to show that $\beta_{\xi}^{+}(0)=\xi$.

Since $\beta_{\xi}$ is convex, its right tangent vector is well defined and
$|\beta^{+}_{\xi}(0)|\leqslant\nobreak 1$585858see properties 3 and 2, page 3.
On the other hand, since $\beta_{\xi,\varepsilon}$ are monotonic with respect
to $0$, for any semiconcave function $f$ we have

$\displaystyle d_{x}f(\beta^{+}_{\xi}(0))$
$\displaystyle=(f\circ\beta_{\xi})^{+}(0)\leqslant$
$\displaystyle\leqslant\lim_{\varepsilon_{i}\to
0}(f\circ\beta_{\xi,\varepsilon})^{+}(0)=$ $\displaystyle=d_{x}f(\xi).$

Substituting in this inequality $f=\operatorname{dist}_{y}$ with
$\measuredangle(\uparrow_{x}^{y},\xi)<\varepsilon$, we get

$\langle\beta^{+}_{\xi}(0),\uparrow_{x}^{y}\rangle>1-\varepsilon$

for any $\varepsilon>0$. Together with $|\beta^{+}_{\xi}(0)|\leqslant 1$
(property 2 on page 2), it implies that

$\beta^{+}(0)=\xi.$

∎

###  A.2 Step 2: Pre-quasigeodesics

In this step we construct a _pre-quasigeodesic_ with arbitrary initial data.

######  A.2.1.

Definition. A convex curve $\gamma\colon[a,b)\rightarrow A$ is called a pre-
quasigeodesic if for any $s\in[a,b)$ such that ${|\gamma^{+}(s)|}>0$, the
curve $\gamma^{s}$ defined by

$\gamma^{s}(t)=\gamma\left(s+\frac{t}{|\gamma^{+}(s)|}\right)$

is convex for $t\geqslant 0$, and if ${|\gamma^{+}(s)|}=0$ then
$\gamma(t)=\gamma(s)$ for all $t\geqslant s$.

Let us first define entropy of pre-quasigeodesic, which measures “how far” a
given pre-quasigeodesic is from being a quasigeodesic.

######  A.2.2.

Definition. Let $\gamma$ be a pre-quasigeodesic in an Alexandrov’s space.

The _entropy_ of $\gamma$, $\mu_{\gamma}$ is the measure on the set of
parameters defined by

$\mu_{\gamma}((a,b))=\ln|\gamma^{+}(a)|-\ln|\gamma^{-}(b)|.$

Here are its main properties:

  1. 1.

The entropy of a pre-quasigeodesic $\gamma$ is zero if and only if $\gamma$ is
a quasigedesic.

  2. 2.

For a converging sequence of pre-quasigeodesics $\gamma_{n}\to\gamma$, the
entropy of the limit is a weak limit of entropies,
$\mu_{\gamma_{n}}\rightharpoonup\mu_{\gamma}$.

It follows from property 4 on page 4.

The next statement is similar to A.0.2 and A.1.2; it makes a new pre-
quasigeodesic out of two. It will be used in the next section to construct
_quasigeodesics_.

######  A.2.3.

Extention. Let $A\in\text{{\nnn Alex}}$, $\gamma_{1}\colon[a,\infty)\to A$ and
$\gamma_{2}\colon[b,\infty)\to A$ be two pre-quasigeodesics. Assume

$a\leqslant b,\ \ \gamma_{1}(b)=\gamma_{2}(b),\ \ \gamma^{-}_{1}(b)\ \
\text{is polar to}\ \ \gamma^{+}_{2}(b)\ \ \text{and}\ \
|\gamma^{+}_{2}(b)|\leqslant|\gamma^{-}_{1}(b)|$

then its joint

$\gamma\colon[a,\infty)\to A,\ \
\gamma(t)=\left[\begin{matrix}\gamma_{1}(t)&\text{if}&t\leqslant b\\\
\gamma_{2}(t)&\text{if}&t\geqslant b\end{matrix}\right.$

is a pre-quasigeodesic. Moreover, its entropy is defined by

$\mu_{\gamma}|_{(a,b)}=\mu_{\gamma_{1}},\ \
\mu_{\gamma}|_{(b,c)}=\mu_{\gamma_{2}}\ \ \text{and}\ \
\mu_{\gamma}(\\{b\\})=\ln|\gamma^{+}(b)|-\ln|\gamma^{-}(b)|.$

Proof. The same as for A.0.2. ∎

######  A.2.4.

Existence. Let $A\in\text{{\nnn Alex}}$, $x\in A$ and $\xi\in\Sigma_{x}$. Then
there is a pre-quasigeodesic $\gamma\colon[0,\infty)\to A$ such that
$\gamma(0)=x$ and $\gamma^{+}(0)=\xi$.

Proof. Let us choose for each point $x\in A$ and each direction
$\xi\in\Sigma_{x}$ a convex curve $\beta_{\xi}\colon[0,\infty)\to A$ such that
$\beta_{\xi}(0)=x$, $\beta_{\xi}^{+}(0)=\xi$. If $v=r\xi$, then set

$\beta_{v}(t)=\beta_{\xi}(rt).$

Clearly $\beta_{v}$ is convex if $0\leqslant r\leqslant 1$.

Let us construct a convex curve
$\gamma_{\varepsilon}\colon[0,\infty)\rightarrow M$ such that there is a
representation of $[0,\infty)$ as a countable union of disjoint half-open
intervals $[a_{i},\bar{a}_{i})$, such that
$|\bar{a}_{i}-a_{i}|\leqslant\varepsilon$ and for any
$t\in[a_{i},\bar{a}_{i})$ we have

$|\gamma_{\varepsilon}^{+}(a_{i})|\geqslant|\gamma_{\varepsilon}^{+}(t)|\geqslant(1-\varepsilon)\cdot|\gamma_{\varepsilon}^{+}(a_{i})|.$
$None$

Moreover, for each $i$, the curve
$\gamma_{\varepsilon}^{a_{i}}\colon[0,\infty)\to A$,

$\gamma_{\varepsilon}^{a_{i}}(t)=\gamma_{\varepsilon}\left({a_{i}}+\frac{t}{|\gamma_{\varepsilon}^{+}({a_{i}})|}\right)$

is also convex.

Assume we already can construct $\gamma_{\varepsilon}$ in the interval
$[0,t_{\max})$, and cannot do it any further. Since $\gamma_{\varepsilon}$ is
1-Lipschitz, we can extend it continuously to $[0,t_{\max}]$. Use lemma 1.3.9
to construct a vector $v^{*}$ polar to $\gamma^{-}_{\varepsilon}(t_{\max})$
with $|v^{*}|\leqslant|\gamma^{-}_{\varepsilon}(t_{\max})|$. Consider the
joint of $\gamma_{\varepsilon}$ with a short half-open segment of $\beta_{v}$,
a longer curve with the desired property. This is a contradiction.

Let $\gamma$ be a partial limit of $\gamma_{\varepsilon}$ as $\varepsilon\to
0$. From property 4 on page 4, we get that for almost all $t$ we have
$|\gamma^{+}(t)|=\lim|\gamma_{\varepsilon_{n}}^{+}(t)|$. Combining this with
inequality $(*)$ shows that for any $a\geqslant 0$

$\gamma^{a}(t)=\gamma\left({a}+\frac{t}{|\gamma^{+}({a})|}\right)$

is convex. ∎

###  A.3 Step 3: Quasigeodesics

We will construct quasigeodesics in an $m$-dimensional Alexandrov’s space,
assuming we already have such a construction in all dimensions $<m$. This
construction is much easier for the case of an Alexandrov’s space with only
$\delta$-strained points; in this case we construct a sequence of special pre-
quasigeodesics only by extending/chopping procedures (see below) and then pass
to the limit. In a general Alexandrov’s space we argue by contradiction, we
assume that $\Omega$ is a maximal open set such that for any initial data one
can construct an $\Omega$-quasigeodesic (i.e. a pre-quasigeodesic with zero
entropy on $\Omega$, see A.2.2), and arrive at a contradiction with the
assumption $\Omega\not=A$.

The following extention and chopping procedures are essential in the
construction:

######  A.3.1.

Extention procedure. Given a pre-quasigeodesic $\gamma\colon[0,t_{\max})\to A$
we can extend it as a pre-quasigeodesic $\gamma\colon[0,\infty)\to A$ so that

$\mu_{\gamma}(\\{t_{\max}\\})=0.$

Proof. Let us set $\gamma(t_{\max})$ to be the limit of $\gamma(t)$ as $t\to
t_{\max}$ (it exists since pre-quasigeodesics are Lipschitz).

From Milka’s lemma A.3.2, we can construct a vector $\gamma^{+}(t_{\max})$
which is polar to $\gamma^{-}(t_{\max})$ and such that
$|\gamma^{+}(t_{\max})|=|\gamma^{-}(t_{\max})|$. Then extend $\gamma$ by a
pre-quasigeodesic in the direction $\gamma^{+}(t_{\max})$. By A.2.3, we get

$\mu_{\gamma}\\{t_{\max}\\}=\ln|\gamma^{+}(t_{\max})|-\ln|\gamma^{-}(t_{\max})|=0.$

∎

######  A.3.2.

Milka’s lemma (existence of the polar direction). For any unit vector
$\xi\in\Sigma_{p}$ there is a polar unit vector $\xi^{*}$, i.e.
$\xi^{*}\in\Sigma_{p}$ such that

$\langle\xi,v\rangle+\langle\xi^{*},v\rangle\geqslant 0$

for any $v\in T_{p}$.

The proof is taken from [Milka 1968]. That is the only instance where we use
existence of quasigeodesics in lower dimensional spaces.

Proof. Since $\Sigma_{p}$ is an Alexandrov’s $(m-1)$-space with curvature
$\geqslant 1$, given $\xi\in\Sigma_{p}$ we can construct a quasigeodesic in
$\Sigma_{p}$ of length $\pi$, starting at $\xi$; the comparison inequality
(theorem 5(5iv)) implies that the second endpoint $\xi^{*}$ of this
quasigeodesic satisfies

$|\xi\,\eta|_{\Sigma_{q}}+|\eta\,\xi^{*}|_{\Sigma_{q}}=\measuredangle(\xi,\eta)+\measuredangle(\eta,\xi^{*})\leq\pi\
\ \text{for all}\ \ \eta\in\Sigma_{p},$

which is equivalent to the statement that $\xi$ and $\xi^{*}$ are polar in
$T_{p}$. ∎

######  A.3.3.

Chopping procedure. Given a pre-quasigeodesic $\gamma\colon[0,\infty)\to A$,
for any $t\geqslant 0$ and $\varepsilon>0$ there is $\bar{t}>t$ such that

$\mu_{\gamma}\left((t,\bar{t})\right)<\varepsilon[\vartheta+\bar{t}-t],\ \
\bar{t}-t<\varepsilon,\ \ \vartheta<\varepsilon,$

where

$\vartheta=\vartheta(t,\bar{t})=\measuredangle\left(\gamma^{+}(t),\uparrow_{\gamma(t)}^{\gamma(\bar{t})}\right).$

$\gamma(t)$$\gamma(\bar{t})$$\vartheta$$\gamma$

Proof. For all sufficiently small $\tau>0$ we have

$\vartheta(t,t+\tau)<\varepsilon$

and from convexity of $\gamma^{t}$ it follows that

$\mu\left((t,t+\tau/3)\right)<C{\cdot}\vartheta^{2}(t,t+\tau).$

The following exercise completes the proof. ∎

######  A.3.4.

Exercise. Let the functions $h,g\colon\mathbb{R}_{+}\to\mathbb{R}_{+}$ be such
that for any sufficiently small $s$,

$h(s/3)\leqslant g^{2}(s),\ s\leqslant g(s)\ \hbox{and}\ \lim_{s\to 0}g(s)=0.$

Show that for any $\varepsilon>0$ there is $s>0$ such that

$h(s)<10{\cdot}g^{2}(s)\ \hbox{and}\ g(s)\leqslant\varepsilon.$

Construction in the $\mathbf{\delta}$-strained case. From the extension
procedure, it is sufficient to construct a quasigeodesic $\gamma\colon[0,T)\to
A$ with any given initial data $\gamma^{+}(0)=\xi\in\Sigma_{p}$ for some
positive $T=T(p)$.

The plan: Given $\varepsilon>0$, we first construct a pre-quasigeodesic

$\gamma_{\varepsilon}\colon[0,T)\to A,\ \ \ \gamma_{\varepsilon}^{+}(0)=\xi$

such that one can present $[0,T)$ as a countable union of disjoint half-open
intervals $[a_{i},\bar{a}_{i})$ with the following property ($\vartheta$ is
defined in the chopping procedure A.3.3):

$\mu\left([a_{i},\bar{a}_{i})\right)<\varepsilon{\cdot}\vartheta(a_{i},\bar{a}_{i}),\
\ \bar{a}_{i}-a_{i}<\varepsilon,\ \ \vartheta(a_{i},\bar{a}_{i})<\varepsilon.$
$None$

Then we show that the entropies $\mu_{\gamma_{\varepsilon}}([0,T))\to 0$ as
$\varepsilon\to 0$ and passing to a partial limit of $\gamma_{\varepsilon}$ as
$\varepsilon\to 0$ we get a quasigeodesic.

Existence of $\gamma_{\varepsilon}$: Assume that we already can construct
$\gamma_{\varepsilon}$ on an interval $[0,t_{\max})$, $t_{\max}<T$ and cannot
construct it any further, then applying the extension procedure A.3.1 for
$\gamma_{\varepsilon}\colon[0,t_{\max})\to A$ and then chopping it (A.3.3)
starting from $t_{\max}$, we get a longer curve with the desired property;
that is a contradiction.

Vanishing entropy: From $(\star)$ we have that

$\mu_{\gamma_{\varepsilon}}([0,T))<\varepsilon\cdot\left[T+\sum_{i}\vartheta(a_{i},\bar{a}_{i})\right].$

Therefore, to show that $\mu_{\gamma_{\varepsilon}}([0,T))\to 0$, it only
remains to show that $\sum_{i}\vartheta(a_{i},\bar{a}_{i})$ is bounded above
by a constant independent of $\varepsilon$.

That will be the only instance, where we apply that $p$ is $\delta$-strained
for a small enough $\delta$.

It is easy to see that there is $\varepsilon=\varepsilon(\delta)\to 0$ as
$\delta\to 0$ and $T=T(p)>0$ such that there is a finite collection of points
$\\{q_{k}\\}$ which satisfy the following property: for any $x\in B_{T}(p)$
and $\xi\in\Sigma_{x}$ there is $q_{k}$ such that
$\measuredangle(\xi,\uparrow_{x}^{q_{k}})<\varepsilon$. Moreover, we can
assume $\operatorname{dist}_{q_{k}}$ is $\lambda$-concave in $B_{T}(p)$ for
some $\lambda>0$.

Note that for any convex curve $\gamma\colon[0,T)\to B_{T}(p)\subset A$, the
measures $\chi_{k}$ on $[0,T)$, defined by

$\chi_{k}((a,b))=(\operatorname{dist}_{q_{k}}\circ\gamma)^{-}(b)-(\operatorname{dist}_{q_{k}}\circ\gamma)^{+}(a)+\lambda{\cdot}(b-a),$

are positive and their total mass is bounded by $\lambda T+2$ (this follows
from the fact that $\operatorname{dist}_{q_{k}}$ is $\lambda$-concave and
1-Lipschitz).

Let $x\in B_{T}(p)$, and $\delta$ be small enough. Then for any two directions
$\xi,\nu\in\nobreak\Sigma_{x}$ there is $q_{k}$ which satisfies the following
property:

$\tfrac{1}{10}{\cdot}\measuredangle_{x}(\xi,\nu)\leqslant
d_{x}\operatorname{dist}_{q_{k}}(\xi)-d_{x}\operatorname{dist}_{q_{k}}(\nu)\ \
\ \text{and}\ \ \ d_{x}\operatorname{dist}_{q_{k}}(\nu)\geqslant 0.$ $None$

Substituting in this inequality

$\xi=\gamma^{+}(a_{i})/|\gamma^{+}(a_{i})|,\ \ \
\nu=\uparrow_{\gamma(a_{i})}^{\gamma(\bar{a}_{i})},$

and applying lemma A.3.5, we get

$\vartheta(a_{i},\bar{a}_{i})=\measuredangle(\xi,\nu)\leqslant
10{\cdot}\sum_{n}\chi_{k}([a_{i},\bar{a}_{i})).$

Therefore

$\sum_{i}\vartheta(a_{i},\bar{a}_{i})\leqslant 10{\cdot}N{\cdot}(\lambda
T+2),$

where $N$ is the number of points in the collection $\\{q_{k}\\}$. ∎

######  A.3.5.

Lemma. Let $A\in\text{{\nnn Alex}}$, $\gamma\colon[0,t]\to A$ be a convex
curve $|\gamma^{+}(0)|=1$ and $f$ be a $\lambda$-concave function,
$\lambda\geqslant 0$. Set $p=\gamma(0)$, $q=\gamma(t)$, $\xi=(\gamma)^{+}(0)$
and $\nu=\uparrow_{p}^{q}$. Then

$d_{p}f(\xi)-d_{p}f(\nu)\leqslant(f\circ\gamma)^{+}(0)-(f\circ\gamma)^{-}(t)+\lambda{\cdot}t,$

provided that $d_{p}f(\nu)\geqslant 0$.

$p$$q$$\xi$$\nu$$\gamma$

Proof. Clearly,

$f(q)\leq
f(p)+d_{p}f(\nu){\cdot}|pq|+\tfrac{\lambda}{2}{\cdot}|pq|^{2}\leqslant
f(p)+d_{p}f(\nu)t+\tfrac{\lambda}{2}{\cdot}t^{2}.$

On the other hand,

$f(p)\leqslant
f(q)-(f\circ\gamma)^{-}(t){\cdot}t+\tfrac{\lambda}{2}{\cdot}t^{2}.$

Clearly, $d_{p}f(\xi)=(f\circ\gamma)^{+}(0)$, whence the result. ∎

What to do now? We have just finished the proof for the case, where all points
of $A$ are $\delta$-strained. From this proof it follows that if we denote by
$\Omega_{\delta}$ the subset of all $\delta$-strained points of $A$ (which is
an open everywhere dense set, see [BGP, 5.9]), then for any initial data one
can construct a pre-quasigeodesic $\gamma$ such that
$\mu_{\gamma}(\gamma^{-1}(\Omega_{\delta}))=0$. Assume $A$ has no boundary;
set $\mathfrak{C}=A\backslash\Omega_{\delta}$. In this case it seems unlikely
that we hit $\mathfrak{C}$ by shooting a pre-quasigeodesic in a generic
direction. If we could prove that it almost never happens, then we obtain
existence of quasigeodesics in all directions as the limits of quasigeodesics
in generic directions (see property 6 on page 6) and passing to doubling in
case $\partial A\not=\varnothing$. Unfortunately, we do not have any tools so
far to prove such a thing595959It might be possible if we would have an analog
of the Liouvile theorem for “pre-quasigeodesic flow”. Instead we generalize
inequality $(*)$.

######  A.3.6.

The $\mathbf{(*)}$ inequality. Let $A\in\text{{\nnn Alex}}^{m}(\kappa)$ and
$\mathfrak{C}\subset A$ be a closed subset. Let $p\in\mathfrak{C}$ be a point
with $\delta$-maximal $\operatorname{vol}_{m-1}\Sigma_{p}$, i.e.

$\operatorname{vol}_{m-1}\Sigma_{p}+\delta>\inf_{x\in\mathfrak{C}}\\{\operatorname{vol}_{m-1}\Sigma_{p}\\}.$

Then, if $\delta$ is small enough, there is a finite set of points
$\\{q_{i}\\}$ and $\varepsilon>0$, such that for any
$x\in\mathfrak{C}\cap\bar{B}_{\varepsilon}(p)$ and any pair of directions
$\xi\in\Sigma_{x}\mathfrak{C}$606060$\Sigma_{x}\mathfrak{C}$ is defined on
page 29. and $\nu\in\Sigma_{x}$ we can choose $q_{i}$ so that

$\tfrac{1}{10}{\cdot}\measuredangle_{x}(\xi,\nu)\leqslant
d_{x}\operatorname{dist}_{q_{i}}(\xi)-d_{x}\operatorname{dist}_{q_{i}}(\nu)\ \
\ \text{and}\ \ \ d_{x}\operatorname{dist}_{q_{k}}(\nu)\geqslant 0.$

Proof. We can choose $\varepsilon>0$ so small that for any
$x\in\bar{B}_{\varepsilon}(p)$, $\Sigma_{x}$ is almost bigger than
$\Sigma_{p}$.616161i.e. for small $\delta>0$ there is a map
$f\colon\Sigma_{p}\to\Sigma_{x}$ such that $|f(x)f(y)|>|xy|-\delta$. Since
$\operatorname{vol}_{m-1}\Sigma_{p}$ is almost maximal we get that for any
$x\in\mathfrak{C}\cap\bar{B}_{\varepsilon}(p)$, $\Sigma_{x}$ is almost
isometric to $\Sigma_{p}$. In particular, if one takes a set $\\{q_{i}\\}$ so
that directions $\uparrow_{p}^{q_{i}}$ form a sufficiently dense set and
$\measuredangle q_{i}pq_{j}\approx\tilde{\measuredangle}_{\kappa}q_{i}pq_{j}$,
then directions $\uparrow_{x}^{q_{i}}$ will form a sufficiently dense set in
$\Sigma_{x}$ for all $x\in\mathfrak{C}\cap\bar{B}_{\varepsilon}(p)$.

Note that for any $x\in\mathfrak{C}\cap\bar{B}_{\varepsilon}(p)$ and
$\xi\in\Sigma_{x}\mathfrak{C}$, there is an almost isometry
$\Sigma_{x}\to\Sigma(\Sigma_{\xi}\Sigma_{x})$ such that $\xi$ goes to north
pole of the spherical suspension
$\Sigma(\Sigma_{\xi}\Sigma_{x})=\Sigma_{\xi}T_{x}$. 626262Otherwise, taking a
point $y\in\mathfrak{C}$, close to $x$ in direction $\xi$ we would get that
$\operatorname{vol}_{m-1}\Sigma_{y}$ is essentially bigger than
$\operatorname{vol}_{m-1}\Sigma_{x}$, which is impossible since both are
almost equal to $\operatorname{vol}_{m-1}\Sigma_{p}$.

Using these two properties, we can find $q_{i}$ so that
$\uparrow^{\nu}_{\xi}\approx\uparrow_{\xi}^{\uparrow_{x}^{q_{i}}}$ in
$\Sigma_{\nu}(\Sigma_{x}A)$ and
$\measuredangle(\xi,\uparrow_{x}^{q_{i}})>\tfrac{\pi}{2}$, hence the statement
follows. ∎

Now we are ready to finish construction in the general case. Let us define a
subtype of pre-quasigeodesics:

######  A.3.7.

Definition. Let $A\in\text{{\nnn Alex}}$ and $\Omega\subset A$ be an open
subset. A pre-quasigeodesic $\gamma\colon[0,T)\to A$ is called
$\Omega$-quasigeodesic if its entropy vanishes on $\Omega$, i.e.

$\mu_{\gamma}(\gamma^{-1}(\Omega))=0$

From property 2 on page 2, it follows that the limit of
$\Omega$-quasigeodesics is a $\Omega$-quasigeodesic. Moreover, if for any
initial data we can construct an $\Omega$-quasigeodesic and an
$\Omega^{\prime}$-quasigeodesic, then it is possible to construct an
$\Omega\cup\Omega^{\prime}$-quasigeodesic for any initial data; for
$\Upsilon\Subset\Omega\cup\Omega^{\prime}$, $\Upsilon$-quasigeodesic can be
constructed by joining together pieces of $\Omega$ and
$\Omega^{\prime}$-quasigeodesics and $\Omega\cup\Omega^{\prime}$-quasigeodesic
can be constructed as a limit of $\Upsilon_{n}$-quasigeodesics as
$\Upsilon_{n}\to\Omega\cup\Omega^{\prime}$.

Let us denote by $\Omega$ the maximal open set such that for any initial data
one can construct an $\Omega$-quasigeodesic. We have to show then that
$\Omega=A$.

Let $\mathfrak{C}=A\backslash\Omega$, and let $p\in\mathfrak{C}$ be the point
with almost maximal $\operatorname{vol}_{m-1}\Sigma_{p}$. We will arrive to a
contradiction by constructing a $B_{\varepsilon}(p)\cup\Omega$-quasigeodesic
for any initial data.

Choose a finite set of points $q_{i}$ as in A.3.6. Given $\varepsilon>0$, it
is enough to construct an $\Omega$-quasigeodesic
$\gamma_{\varepsilon}\colon[0,T)\to A$, for some fixed $T>0$ with the given
initial data $x\in\bar{B}_{\varepsilon}(p)$, $\xi\in\Sigma_{x}$, such that the
entropies $\mu_{\gamma_{\varepsilon}}((0,T))\to 0$ as $\varepsilon\to 0$.

The $\Omega$-quasigeodesic $\gamma_{\varepsilon}$ which we are going to
construct will have the following property: one can present $[0,T)$ as a
countable union of disjoint half-open intervals $[a_{i},\bar{a}_{i})$ such
that

$\text{if}\ \ \
\frac{\gamma^{+}(a_{i})}{|\gamma^{+}(a_{i})|}\in\Sigma_{\gamma(a_{i})}\mathfrak{C}\
\ \ \text{then}\ \ \
\mu_{\gamma}([a_{i},\bar{a}_{i}))\leqslant\varepsilon{\cdot}\vartheta(a_{i},\bar{a}_{i})$

and

$\text{if}\ \ \
\frac{\gamma^{+}(a_{i})}{|\gamma^{+}(a_{i})|}\not\in\Sigma_{\gamma(a_{i})}\mathfrak{C}\
\ \ \text{then}\ \ \ \mu_{\gamma}([a_{i},\bar{a}_{i}))=0$

Existence of $\gamma_{\varepsilon}$ is being proved the same way as in the
$\delta$-strained case, with the use of one additional observation: if

$\frac{\gamma^{+}(t_{\max})}{|\gamma^{+}(t_{\max})|}\not\in\Sigma_{\gamma(a_{i})}\mathfrak{C}$

then any $\Omega$-quasigeodesic in this direction has zero entropy for a short
time.

Then, just as in the $\delta$-strained case, applying inequality A.3.6 we get
that $\mu_{\gamma_{\varepsilon}}(0,T)\to 0$ as $\varepsilon\to 0$. Therefore,
passing to a partial limit $\gamma_{\varepsilon}\to\gamma$ gives a
$B_{\varepsilon}(p)\cup\Omega$-quasigeodesic $\gamma\colon[0,T)\to A$ for any
initial data in $B_{\varepsilon}(p)$. ∎

###  A.4 Quasigeodesics in extremal subsets.

The second part of theorem A.0.1 follows from the above construction, but we
have to modify Milka’s lemma A.3.2:

######  A.4.1.

Extremal Milka’s lemma. Let $E\subset T_{p}$ be an extremal subset of a
tangent cone then for any vector $v\in E$ there is a polar vector $v^{*}\in E$
such that $|v|=|v^{*}|$.

Proof. Set $X=E\cap\Sigma_{p}$. If $\Sigma_{\xi}X\not=\varnothing$ then the
proof is the same as for the standard Milka’s lemma; it is enough to choose a
direction in $\Sigma_{\xi}X$ and shoot a quasigedesic $\gamma$ of length $\pi$
in this direction such that $\gamma\subset X$ ($\gamma$ exists from the
induction hypothesis).

If $X=\\{\xi\\}$ then from the extremality of $E$ we have
$B_{\pi/2}(\xi)=\Sigma_{p}$. Therefore $\xi$ is polar to itself.

Otherwise, if $\Sigma_{\xi}X=\varnothing$ and $X$ contains at least two
points, choose $\xi^{*}$ to be closest point in $X\backslash\xi$ from $\xi$.
Since $X\subset\Sigma_{p}$ is extremal we have that for any
$\eta\in\Sigma_{p}$
$\measuredangle_{\Sigma_{p}}\eta\xi^{*}\xi\leqslant\tfrac{\pi}{2}$ and since
$\Sigma_{\xi}X=\varnothing$ we have
$\measuredangle_{\Sigma_{p}}\eta\xi\xi^{*}\leqslant\tfrac{\pi}{2}$. Therefore,
from triangle comparison we have

$|\xi\eta|_{\Sigma_{p}}+|\eta\xi^{*}|_{\Sigma_{p}}=\measuredangle(\xi,\eta)+\measuredangle(\eta,\xi^{*})\leqslant\pi$

∎

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