text:
\section{Introduction}
The standard cross-polytope $\crosp^n$ in $\mathbb{R}^n$ is the convex hull of the vectors of the standard basis $(e_i)_1^n$ of $\mathbb{R}^n$ and the centrally symmetric to them vectors
$( - e_i)_1^n$. The cross-polytope is the dual (polar) body for the standard cube $\cube^n = [-1, 1]^n.$ In this paper we study lower and upper bounds on the volume of a projection of $\crosp^n$ onto a $k$-dimensional subspace, that is, we study extremizers of
\begin{equation}
\label{main_pr_original}
F(H_k)  = \vol \left( \crosp^n | H_k\right),
\end{equation}
where $H_k$ is a $k$-dimensional subspace of $\mathbb{R}^n$ and $ \crosp^n | H_k$ denotes the projection of $\crosp^n$ onto $H_k$.
By the duality arguments, this problem can be considered as the dual problem for finding extremizers of
$$
	F_2 (H_k) = \vol \left(\cube^n \cap H_k \right),
$$
whose global extrema are reasonably well-studied
\begin{equation}\label{ineq_main_cube}
\vol \cube^k \leqslant \vol \left(\cube^n \cap H_k \right) \leqslant
\min \left\{\left( \frac{n}{k}\right)^{\frac{k}{2}}, 2^{\frac{n-k}{2}}  \right\} \vol \cube^k,
\end{equation}
where the left-hand side inequality is due to J. Vaaler \cite{vaaler1979geometric} and it is optimal for all $n \geqslant k \geqslant 1;$ and the right-hand side is due to
K. Ball \cite{ball1989volumes} and it is optimal whenever $k | n $ or $2k \geqslant n$
(that is,  the optimal constants are not known only when $2k < n$ and $k \nmid n$).  To get a rather complete survey on cross sections of the cube with the proofs of these inequalities, we refer the reader to the first chapter of Zong's book \cite{zong2006cube}. Very interesting generalization of the Vaaler result was made by M. Meyer and A. Pajor \cite{meyer1988sections} and a new proof of Vaaler's inequality in terms of waists was given  in  \cite{2016arXiv161206926A} recently.

As for the optimal constants, they are mostly unknown.  For example, the following conjecture can be considered as an analogue (or the dual conjecture) for Vaaler's  inequality.
\begin{conjecture}\label{conj_max_vol_proj_crosspolytope}
The volume of the projection of the $n$-dimensional cross-polytope $\crosp^n$ onto a $k$-dimensional subspace is at most the volume of the $k$-dimensional cross-polytope, i.e.
$\vol (\crosp^n | H_k) \leqslant \frac{2^k}{k!}.$ The bound is attained only on coordinate subspaces.
\end{conjecture}

The only known results here are that the conjecture is true in  the hyperplane case $k = n-1$ (see \cite{ball1995mahler}), which is quite simple (since the projections of the surface of the cross-polytope onto a hyperplane covers the projection of the cross-polytope twice almost everywhere) and in the two-dimensional case \cite{Filliman1988}. Also, P. Filliman proved that this conjecture is true for $k=3$ and $n \leqslant 6.$ The  deep generalization of the hyperplane case for the volume of a projection of the $\ell_p^n$ balls was given by A. Naor and F. Barthe \cite{barthe2002hyperplane}. We show that this Conjecture is true in  lower-dimensional cases.

\begin{theorem}\label{th_upper_bound_crossp_lowdimcase}
Conjecture \ref{conj_max_vol_proj_crosspolytope} is true for $k =2,3.$
\end{theorem}

As for Ball's inequalities, the only known optimal $H_k$ are such that the projections of the cube on them are the affine cubes. In \cite{2017arXiv170704442I}, the author proved that   all optimal subspaces  have this property whenever $k \mid n$ in the right-hand side of \reff{ineq_main_cube}.  We think that we have  nice candidates for the optimal subspaces and constants in the Ball's inequalities.
Let $C_\cube(n,k) 2^k$ be the maximum volume of a cross section of  $\cube^n$
by a $k$-dimensional subspace $H_k$ such that $\cube^n \cap H_k$ is an affine cube.
\begin{conjecture}\label{conj_max_vol_sec_cube}
The maximum  volume of a cross section of $\cube^n$ by a
$k$-dimensional subspace $H_k$ is attained on subspaces such that $\cube^n \cap H_k$ is an affine cube, i.e.
$\vol \left(\cube^n \cap H_k\right) \leqslant  C_\cube(n,k) 2^k.$
\end{conjecture}

And we can formulate the dual statements for projections of $\crosp^n.$ Let $C_\crosp (n,k) \frac{2^k}{k!}$ be the minimum volume of a projection of the $\crosp^n$
onto a $k$-dimensional subspace $H_k$ such that $\crosp^n  | H_k$ is an affine cross-polytope.

\begin{conjecture}\label{conj_min_vol_proj_crosspolytope}
The minimum  volume of a projection of  $\crosp^n$  onto a
$k$-dimensional subspace $H_k$ is attained on subspaces such that $\crosp^n | H_k$ is an affine cross-polytope, i.e. $ \vol (\crosp^n | H_k) \geqslant C_\crosp(n,k)\frac{2^k}{k!}.$
\end{conjecture}

It is not hard to prove that
\begin{equation}\label{eq_exact_constants}
C_\cube(n,k) = \frac{1}{C_{\crosp}(n,k)} =
\left(\ceil*{\frac{n}{k}}\right)^{n -k \floor*{\frac{n}{k}}}
\left(\floor*{\frac{n}{k}}\right)^{k - \left(n -k \floor*{\frac{n}{k}} \right)},
\end{equation}
and that these constants are attained  on the same subspaces.
For completeness, we give a proof of \reff{eq_exact_constants} and give a complete description of a $k$-dimensional subspace $H_k$ on which  constants
$C_\cube (n,k)$ and $C_\crosp (n,k)$ are attained  in Lemma \ref{lemma_extremizers}.

We shall notice that K. Ball proved that the constant in Conjecture \ref{conj_max_vol_sec_cube} is optimal whenever $k | n $ or $2k \geqslant n,$
F. Barthe \cite{barthe1998reverse} proved the constant in Conjecture \ref{conj_min_vol_proj_crosspolytope} is optimal whenever $k |n$ and, due to F. Barthe and A. Naor  \cite{barthe2002hyperplane}, it is optimal  in the hyperplane case $k = n-1.$
In \cite[Theorem 1.5]{2017arXiv170704442I}, the author proved that the last two conjectures are true in
case $k|n.$ In the rest,  we can, alas, give only a partial answer to Conjectures \ref{conj_max_vol_sec_cube}, \ref{conj_min_vol_proj_crosspolytope} in the most simple case.
\begin{theorem}\label{th_extr}
	Conjectures \ref{conj_max_vol_sec_cube} and \ref{conj_min_vol_proj_crosspolytope} are
	true for $k=2.$
\end{theorem}

It is well-known that any centrally symmetric polytope is an affine image of the high-dimensional cross-polytope or an affine image of a cross section of the high-dimensional cube. This means that we can reformulate our conjectures in terms of the uniform bounds on the volume of a centrally symmetric polytope in $\mathbb{R}^k.$
We formulate the proper inequalities in Section  \ref{sec_pertubations} after some preliminaries and explanations.

The paper is organized as follows:  After presenting our notation and basic definitions, we  reformulate the problems in terms of the polytopes generated by projections of vectors of an orthonormal basis in
Section \ref{sec_def_pre}. Then we  explain our approach in Section \ref{sec_pertubations}. In Section \ref{sec_simplif}, we simplify the structure of local extremizers of \ref{main_pr_original}. Then, after some geometric constructions, we give a geometrical necessary condition for $H_k$ to be a local maximizer of  \reff{main_pr_original}. Using these results, we prove Theorems \ref{th_extr} and \ref{th_upper_bound_crossp_lowdimcase} in Section \ref{sec_proofs_main_results}. Then, we show that there exist local and non-global maximizers of \ref{main_pr_original}. And the last Section is devoted to a more general setting, we explain how our problems are related to isotropic measures and how one can use the same approach in similar problems.

\section{Defintions and Preliminaries}\label{sec_def_pre}
\label{section_reform}
We use $\crosp^n$ to denote an $n$-dimensional cross-polytope
$\{x |  \sum\limits_1^n \left|x[i]\right| \leqslant 1\}$ in $\mathbb{R}^n.$
Here and throughout the paper   $x[i]$ stands for  $i$-th coordinate of a vector $x.$
As usual, $\{e_i\}_1^n$ is the standard orthonormal basis of $\mathbb{R}^n.$
We use $\langle p,x \rangle$ to denote the \emph{value of a linear functional} $p$ \emph{at a vector}  $x.$

Throughout the paper $H_k$  will be a $k$-dimensional subspace of $\mathbb{R}^n.$
For a convex body $K \subset \mathbb{R}^n$ and a $k$-dimensional subspace $H_k$ of $\mathbb{R}^n$ we denote by
$K \cap H_k$ and $K | H_k$ the section of $K$ by $H_k$ and the orthogonal projection of $K$ onto $H_k,$
respectively.  For  a $k$-dimensional subspace $H_k$ of $\mathbb{R}^n$ and a  convex body  $K \subset H_k$
 we denote by $\vol K$  the $k$-dimensional volume of $K$.
We consider only $n \geqslant k \geqslant 2.$

First of all, it is convenient to identify a projection of the cross-polytope with a convex polytope in $\mathbb{R}^k.$
Let $v_i = P e_i,$ where $P$ is the projection onto $H_k.$  Clearly,
\[
	\crosp^n | H_k =  \mathop{\rm co} \{ \pm v_1, \ldots, \pm v_n\}.
\]
That means that a projection of $\crosp^n$ is determined by the set of vectors
$(v_i)_1^n,$ which are the projections of the orthogonal basis.
To deal with those vectors we introduce several definitions following \cite{2017arXiv170704442I}, \cite{2018arXiv180410055I}.

\begin{definition}
We will say that some vectors $(w_i)_1^n \subset H$ {\it give  a unit decomposition} in  a vector space $H$
if
\begin{equation}
\label{John_cond_3}
\left. \left(\sum\limits_1^n w_i \otimes w_i \right) \right|_H = I_H,
\end{equation}
where $I_H$ is the identity operator in $H$  and $\left. A\right|_H$ is the restriction of an operator  $A$ onto $H.$
\end{definition}

In the following Lemma  we understand $\mathbb{R}^k \subset \mathbb{R}^n$ as the subspace of vectors, whose  last $n-k$ coordinates are zero.
For convenience, we will consider $(w_i)_1^n \subset \mathbb{R}^k \subset \mathbb{R}^n$ to be $k$-dimensional vectors.
\begin{lemma}\label{lemma_equiv_cond}
The following assertions
 are equivalent:
\begin{enumerate}
\item \label{lemma_eq_cond_it1} the vectors  $(w_i)_1^n \subset \mathbb{R}^k$  give a unit decomposition in $\mathbb{R}^k$;
\item \label{lemma_eq_cond_it2} there exists an orthonormal basis $\{ f_1, \ldots, f_n \}$  of  $\mathbb{R}^n$ such that   $w_i$ is the orthogonal projection
of $f_i$ onto $\mathbb{R}^k,$ for any $i \in [n];$
\item \label{lemma_eq_cond_Gram}
$\mathop{\rm Lin}\{w_1, \ldots ,w_n\} = \mathbb{R}^k$ and the Gram matrix $\Gamma$ of vectors $\{w_1, \ldots ,w_n\} \subset \mathbb{R}^k$ is the matrix of a  projection operator from $\mathbb{R}^n$ onto the  linear hull of the rows of the matrix 	$M = (w_1, \ldots, w_n).$
\item \label{lemma_eq_cond_it3}  the $k\times n$ matrix $M = (w_1, \ldots, w_n )$ is a sub-matrix of an orthogonal matrix of order $n$.
\end{enumerate}
\end{lemma}
This Lemma was proven in \cite[Lemma~2.1]{2017arXiv170704442I}.

\begin{definition}
	A  set of $n$ vectors in $\mathbb{R}^k$  which span $\mathbb{R}^k$ is called  an {\it $(n, k)$-frame}.
	If the vectors of an $(n, k)$-frame  $S$ give a unit decomposition, we say that $S$ is an {\it $(n, k)$-uframe}.
	We  use $\Omega(n,k)$ to denote the set of all $(n,k)$-uframes.
\end{definition}

\begin{definition}\label{def_ass_fr}
We will say that the set  $S = \{ w_1, \ldots, w_n\}$ of $n$ vectors of $\mathbb{R}^k$ generates
\begin{enumerate}
\item  a set  $\mathop{\rm co} \{\pm w_1, \cdots, \pm w_n\},$ which we call a projection of the cross-polytope generated by $S$.   We use $ \crosp^n | S$ to denote this set.
\item a matrix $ \sum\limits_{i \in [n]}  w_i \otimes w_i.$ We use $A_S$ to denote this matrix.
\item a  subspace $H_k^{S} \subset \mathbb{R}^n$ which is the linear hull of the rows of the $k \times n$ matrix $(w_1, \ldots, w_n).$
\end{enumerate}
\end{definition}

The simple Lemma \ref{lemma_equiv_cond} from linear algebra allows us to identify vectors $(v_i)_1^n \subset H_k$
with a set of vectors in $\mathbb{R}^k,$ which give a unit decomposition.
To be more precise,
there is an isometric isomorphism between any $k$-dimensional subspace  $H_k \subset \mathbb{R}^n$ and $\mathbb{R}^k.$
Thus, the projections $\crosp^n | H_k$ correspond to $\crosp^n|S,$ where $S$ is the  $(n,k)$-uframe corresponding to this isometry.
Vice versa, given an $(n,k)$-uframe $S= \{w_1, \ldots, w_n\}$ the assertion \reff{lemma_eq_cond_Gram} of Lemma \ref{lemma_equiv_cond}  gives the proper $H_k,$
and this is exactly $H_k^{S}$ defined in Definition \ref{def_ass_fr} (it is  a $k$-dimensional space in case $S$ is an $(n,k)$-frame).
Therefore, the global extrema of \reff{main_pr_original} and the global extrema of
\begin{equation}\label{main_problem_crosp}
 F(S) = \vol (\crosp^n | S), \qquad \mbox{where} \qquad  S \in \Omega(n,k)
\end{equation}
are the same. Moreover,  the local extrema and the local extremizers in both problems coincide. A nice geometric way of endowing a Grassmann manifold with a metric  is to use a largest principal angle between two $k$-dimensional subspaces
(and this is exactly the Hausdorff distance between the unit circles in these subspaces).
Also, we   may consider the natural metric  on the Grassmanian of $k$-dimensional subspaces of $\mathbb{R}^n$
(the Grassmanian is a homogeneous manifold). As shown in  Neretin's paper \cite{neretin2001jordan}, this  metric is the square root of the sum of squares of all principal angles.
Therefore, the two metrics are equivalent.
We can  consider  a set of all $(n,k)$-frames   as a open set of $\mathbb{R}^{nk}$ equipped  with the metric of $\mathbb{R}^{nk}.$
Clearly, $\det B_S$ is  a continuous function of $S$ in this metric. Hence, $B_S S$ is a continuous function on the set of all $(n,k)$-frames, and one can see that a sufficiently  small neighbourhood in $\Omega(n,k)$ of  an $(n,k)$-uframe $S$ is the image of the proper neighbourhood   of $S$ in the set of all $(n,k)$-frames under the mapping $B_S.$
Any element $H_k$ of the Grassmanian can be obtained as a proper $H_k^{S}$ (not in a unique way), and the Hausdorff distance depends smoothly on $S$. Combining this with the previous observation, we get that
a small enough neighbourhood of an $(n,k)$-uframe  $S$ is homeomorphic to the proper neighbourhood of $H_k^{S}$ in the Grassmanian, and the latter covers all  of the  Grassmanian.  But this means that the local extrema are the same  for the both problems.

\section{Perturbations of frames} \label{sec_pertubations}
Our main idea is to transform given $(n,k)$-uframe $S$  to a new one $S'$; and
in case $S$ gives an (local) extremum of \reff{main_problem_crosp} we get the inequality for
$\vol (\crosp^n | S)$ and $\vol (\crosp^n | S')$. Let us explain this approach.

The first observation is that we can transform any $(n,k)$-frame $S = \{v_1, \ldots, v_n\}$ to an $(n,k)$-uframe $S'$ using a linear transformation $L:$ $S' = LS  = \{L v_1, \ldots, L v_n\},$
or, equivalently, any non-degenerate centrally symmetric polytope in $\mathbb{R}^k$ is an affine image of a projection of a high dimension cross-polytope.

For an $(n, k)$-frame $S = \{ v_1, \ldots, w_n\}$, by definition put
$$
    B_S =   A_S^{-\frac{1}{2}} = \left( \sum\limits_{i \in [n]} v_i \otimes v_i \right)^{-\frac{1}{2}}.
$$
The operator $B_S$ is well-defined as the condition $\mathop{\rm Lin} S = \mathbb{R}^k$ implies that $A_S$ is a strictly positive operator.
Clearly,  $B_S$ maps any $(n,k)$-frame $S$ to the $(n,k)$-uframe:
$$
		\sum_{i=1}^{n} B_S v_i \otimes B_S v_i   =
		B_S \left( \sum_{i=1}^{n} v_i  \otimes v_i \right) B_S^T = B_S A_S B_S = I_k.
	$$
This means that our conjectures can be rewritten in the following way.
\begin{conjecture}
	Let $S = \{v_1, \ldots, v_n\}$ be an $(n,k)$-frame and by definition put  $P = \mathop{\rm co} \{\pm v_1, \ldots, \pm v_n\}$ and
	$Q = \cap_1^n \{x \in \mathbb{R}^k \mid |\langle x, v_i \rangle| \leqslant 1\}.$ Then
	\[
	C_{\crosp}(n,k)	\vol \crosp^{k} \leqslant \frac{\vol P}{\det B_S} \leqslant \vol \crosp^{k};
\quad
		\vol \cube^{k} \leqslant {\vol Q} \cdot {\det B_S} \leqslant C_{\cube}(n,k)\vol \cube^{k}.
	\]
\end{conjecture}

Then, in order to obtain properties of extremizers, we will consider a composition of two transformations:
	\begin{equation}\label{eq:framization2}
		S \quad \stackrel{\mathbf{T}}{\longrightarrow} \quad \tilde{S} \quad \stackrel{B_{\tilde{S}}}{\longrightarrow} \quad S',
	\end{equation}
where $\mathbf{T}$ will be chosen in a specific way    and $B_{\tilde{S}}$ just maps
$\tilde{S} = T (S)$ to a new  $(n,k)$-uframe $S',$ for example
\begin{figure}[h]%
\center{
\psfrag{v}[1]{$\hspace{0.3em}v$}
\psfrag{T}[1]{\hspace{2.8em}\raisebox{2.3ex}{$\mathbf{T} : v \to 0$}}
\psfrag{S}[1]{\hspace{3.8em}\raisebox{2.3ex}{$B_{\tilde{S}}$}}
\includegraphics[scale=0.86]{crosspp_Scheme_paper.eps}
}
\caption{Here we map  one vector  to zero.}
\end{figure}

\noindent
Finally, it is easy to write an inequality on the volumes of $\crosp^n | S$ and
$\crosp^n | S':$
\begin{lemma}\label{lemma_n_s_cond}
The maximum (or minimum) in \reff{main_problem_crosp} is attained on an $(n,k)$-uframe $S = \{v_1, \ldots, v_n\}$ iff the following inequality holds for an arbitrary $(n,k)$-frame $\tilde{S}$
\begin{equation}\label{ineq_n_s_cond_crosp}
	\frac{\vol (\crosp^n | \tilde{S})}{\vol (\crosp^n | S)} \leqslant \sqrt{\det A_{\tilde{S}}}
	 \quad ( or \geqslant).
\end{equation}
\end{lemma}
\begin{prf}
	As mentioned above, $B_{\tilde{S}} \tilde{S}$ is an $(n,k)$-uframe, and clearly, $\vol(\crosp^n | B_{\tilde{S}} \tilde{S})= \det B_{\tilde{S}} \vol (\crosp^n | \tilde{S}).$
	The $(n,k)$-uframe  $S$ is a maximizer iff $\vol(\crosp^n | B_{\tilde{S}} \tilde{S}) \leqslant \vol(\crosp^n | S)$ for  an arbitrary $(n,k)$-frame $\tilde{S}.$
	Using these observations  and the definition of $B_S,$ we have
	$$
	\frac{\vol (\crosp^n | \tilde{S})}{\vol (\crosp^n | S)} = \frac{1}{ \det B_{\tilde{S}}} \frac{\vol(\crosp^n | B_{\tilde{S}} \tilde{S})}{\vol (\crosp^n | S)} \leqslant  \frac{1}{ \det B_{\tilde{S}}} = \sqrt{\det A_{\tilde{S}}}.
	$$
\end{prf}

Choosing  a proper simple  transformation $\mathbb{T},$ we may calculate the left-hand side of
\reff{ineq_n_s_cond_crosp} in some geometric terms. We consider several simple transformations:  Scaling one or several vectors, moving one vector to the origin,
mapping one vector to another.
On the other hand, the determinant in the   right-hand  side of
\reff{ineq_n_s_cond_crosp}  can be calculated for the  mentioned above transformations.
In particular, the following first-order approximation of the determinant was obtained by the author in
\cite[Corollary 6.1]{2018arXiv180410055I}.
\begin{lemma}
	 \label{cor_der_det_identity}
For an arbitrary $(n,k)$-uframe $S$
the following identity holds
$$
	\sqrt{\det A_{S'}} =  1 + \sum\limits_{i = 1}^{n} t_i \langle x_i, v_i \rangle + o \left(\sqrt{t_1^2 + \ldots + t_n^2}\right),
$$
where $S'$  is obtained from $S$ by the substitution $v_i \to v_i + t_i x_i,$ $i \in [n]$.
\end{lemma}

Also, since the composition of  projections from $\mathbb{R}^n$ onto $\mathbb{R}^k$ and from $\mathbb{R}^k$
 onto a subspace $H \subset \mathbb{R}^k$ is a projection, we have.
\begin{lemma}\label{lemma_proj_unit_dec}
	Let $H$ be a supspace of $\mathbb{R}^k$ and $P$ be the orthogonal projection onto $H,$
	let $S = \{v_1, \ldots, v_n \}$ be an $(n,k)$-uframe.
	Then   the vectors  $\{Pv_1, \ldots, Pv_n \}$ give a unit decomposition in $H.$
\end{lemma}

\section{Reduction of the problems} \label{sec_simplif}
For a positive integer $n,$  we refer to the set $\{1, 2, \ldots,  n\}$  as $[n].$
The set of all $\ell$-element subsets (or {\it $\ell$-tuple}) of a set $M \subset [n]$ is denoted by ${M \choose \ell}.$

On the other hand, we may consider the matrix  $A_S$ as the matrix of a positive definite operator and choose the most convenient orthonormal basis. For example,
let $M  = I_k  \pm v \otimes v.$ Then
\begin{equation}\label{eq_det_M}
\det M = 1  \pm |v|^2.
\end{equation}
Indeed, it is  sufficient  to consider $M$ in an orthonormal basis of $\mathbb{R}^k$ in which the first basis vector is collinear to $v$. Also, we will often use the identity
\begin{equation}\label{eq_trace_AS}
	\operatorname{tr} A_S = \sum\limits_{1}^n |v_i|^2 = k
\end{equation}
for an $(n,k)$-uframe $S = \{v_1, \ldots, v_n\}.$

By definition put $S \setminus i$  is the set $(v_j)_{j\in[n]}^{j \neq i}$ for an
$(n,k)$-frame $S = \{v_1, \ldots, v_n\},$
 \begin{lemma}
 Let $S'$ be a set of vectors
obtained from an $(n,k)$-uframe $S = \{v_1, \ldots, v_n\}$ by substitution $v_i \to v_i + tx$, where
$t \in  \mathbb{R}, x \in \mathbb{R}^k$ and a fixed index $i$.
If $|v_i| < 1,$ then
\begin{equation}\label{B_S_subst_det}
	\det B_{S'} = \sqrt{ \frac{1+ |B_{S \setminus i} v_i|^2}{1+|B_{S \setminus i} v|^2}}.
\end{equation}
\end{lemma}
\begin{prf}
	Since $|v_i| < 1,$ we have that  $A_{S \setminus i} = I - v_i \otimes v_i$ is a positive definite matrix.
Hence,  $ B_{S \setminus i} > 0.$

	 Applying the affine transformation $B_{S \setminus i}$ to $S'$,  we get a new $(n, k)$-frame
	$\tilde{S} = \{v'_1, \cdots, v'_n\},$ where $v'_j = B_{S \setminus i} v_j, j \in [n] \setminus i,$ and
	$v'_i  =  B_{S \setminus i} v.$
	By the definition of $B_S,$ we have that   $A_{\tilde{S} \setminus i} = I_k$  and $A_{\tilde{S}} = I_k + v'_i \otimes v'_i.$
	Hence, by \reff{eq_det_M}, we get $\det A_{\tilde{S}} = 1 + |v'_i|^2 = 1+ |B_{S \setminus i} v|^2.$

	By the same arguments,
	$$
		\det A_{B_{S \setminus i}S}  = 1+ |B_{S \setminus i} v_i|^2.
	$$

 As  $\det A_S = 1$, we obtain
 $$
 	\det B_{S'} = \frac{1}{\sqrt{\det A_{S'}}}=
 	 \sqrt{\frac{\det A_S}{ \det A_{S'}}} =
 	 \sqrt{\frac{\det A_{B_{S \setminus i} S}}{\det A_{B_{S \setminus i} {S'}}}} =
 	  \sqrt{ \frac{1+ |B_{S \setminus i} v_i|^2}{1+|B_{S \setminus i} v|^2}}.
 $$
 \end{prf}

Using results from the previous section, we can simplify a configuration of extremizers
for \reff{main_problem_crosp}.

The straightforward consequence of the following lemma is that if  $S = \{v_1, \ldots, v_n\}$ is a (local) maximizer for \reff{main_problem_crosp} then the non-zero elements of $\{\pm v_1, \ldots, \pm v_n\}$ are pairwise distinct vertices of $\crosp^n |S.$
\begin{lemma}\label{lemma_simplif_main}
	Let $S = \{v_1, \cdots, v_n\}$ be an $(n, k)$-uframe such that
	$v_i \in \crosp^{n-1} | (S \setminus i).$   Let $\tilde{S}$ be an $(n, k)$-frame obtained from $S$ by substitution $v_i \to 0.$
	By definition put $S' = B_{\tilde{S}} \tilde{S}.$ Then
	$$
		\vol (\crosp^n|S) \leqslant \vol (\crosp^n|S'),
	$$
	and equality holds iff $v_i  = 0.$
\end{lemma}
\begin{prf}
	Indeed, we have that $\crosp^n|S = \crosp^n|\tilde{S}$ and, obviously, $B_{\tilde{S}} \geqslant B_S = I$ (moreover, the equality holds iff $v_i = 0$).  Therefore,
	$$
		\vol (\crosp^n|S')  = \det B_{\tilde{S}} \vol (\crosp^n|\tilde{S}) \geqslant \vol (\crosp^n|\tilde{S})  = \vol (\crosp^n|S).
	$$
\end{prf}
Moreover, we can change $v_i$ in a continuous way $v_i \to (1 - \lambda) v_i,$ where $\lambda \in [0,1],$ while increasing the volume of $\crosp^n | S'.$

Using the same arguments and  \reff{B_S_subst_det}, we get
 \begin{corollary}\label{cor_subst_decrease}
 	Let $\tilde{S}$ be an $(n, k)$-frame obtained from $S$ by substitution  $v_i \to v.$ If $|v_i| < 1,$ $|B_{S \setminus i} v_i| \leqslant |B_{S \setminus i} v|$ and $ \crosp^n | \tilde{S} \subset \crosp^n | S,$ then
for an $(n,k)$-uframe $S\ =  B_{\tilde{S}} \tilde{S}$,  we have
 $$
 	\vol (\crosp^n | S') \leqslant \vol (\crosp^n | S).
 $$
 Equality holds iff  $|B_{S \setminus i} v_i|  = |B_{S \setminus i} v|$ and $ \crosp^n | \tilde{S}= \crosp^n | S.$
 \end{corollary}

Now we are ready to prove some simple properties of extremizers  for \reff{main_problem_crosp}.  Let us start with the minimizers.
\begin{lemma}\label{lemma_about_vertecies}
	If   $\min\limits_{S \in \Omega(n,k)} \vol \left(\crosp^n | S \right)$ is attained on $S = \{v_1, \ldots, v_n\},$
	then
	\begin{enumerate}
	\item all vectors $|v_i|$ are vertices of $\crosp^n | S;$
	\item $|v_j| \leqslant  \sqrt{2} |v_i|,$ for any $i,j \in [n];$
	\item at most $\lfloor \frac{n}{k} \rfloor +1$ vectors of $S$ may be collinear.
	\end{enumerate}
\end{lemma}
\begin{prf}
	1) Suppose vector $v_i$ is not a vertex  of $\crosp^n | S.$
	Then $v_i$ is a convex combination of some other vectors of $S$ and $-S.$
	The same is true for $B_{S \setminus i} v_i$ and $B_{S \setminus i} S.$
	Therefore, by the triangle inequality, we have that $|v_i| < 1$ and there exists $j \in [n]$ such that
	$|B_{S \setminus i} v_i| < |B_{S \setminus i} v_j|.$
	Let $\tilde{S}$ be an $(n, k)$-frame obtained from $S$ by substitution $v_i \to v_j.$
	We have that $|v_i| < 1,$ $|B_{S \setminus i} v_i| < |B_{S \setminus i} v|$ and $ \crosp^n | \tilde{S} \subset \crosp^n | S,$
	Applying Corollary \ref{cor_subst_decrease} for   the
	$(n, k)$-uframe $S$, we get a new $(n, k)$-uframe $S'$ such that
	$\vol \left(\crosp^n | S'\right) < \vol \left(\crosp^n | S \right).$
		This contradicts to the initial choice of $S.$\\

	\noindent
	2)  If $|v_i| \geqslant \frac{1}{\sqrt{2}},$ there is nothing to prove.  Assume $|v_i|  < \frac{1}{\sqrt{2}}$  and
	$|v_j| > \sqrt{2} |v_i|$ for a fixed $j \in [n].$
	Let $\tilde{S}$ be an $(n, k)$-frame obtained from $S$ by substitution $v_i \to v_j.$

	Clearly, $0 < A_{S \setminus i} \leqslant A_S =  I_k$ and, consequently, $B_{S \setminus i} \geqslant I_k.$ Therefore,
	$\left| B_{S \setminus i} v_j\right| \geqslant |v_j|.$
It easy to see that  the affine map  $B_{S \setminus i}$ just scales the space
	in the direction $v_i$ with the coefficient $\frac{1}{\sqrt{1 - |v_i|^2}}.$ That is,
	$B_{S \setminus i} v_i  = \frac{v_i}{\sqrt{1 - |v_i|^2}}.$  One can see that
	$$
		\left|\frac{v_i}{\sqrt{1 - |v_i|^2}} \right| \leqslant \sqrt{2} |v_i|  < |v_j| \leqslant \left|B_{S \setminus i} v_j\right|.
	$$
	Using the same arguments as above,  we get that
	$\vol  \left(\crosp^n |S' \right)  < \vol \left(\crosp^n|{S}\right),$ where $S' = B_{\tilde{S}} {\tilde{S}}$
	This completes the proof of the second assertion of the lemma.
	  \\

	\noindent
	3) Assume the contrary.   Let the vectors $v_1, \cdots, v_d$ be collinear, where $d \geqslant \lfloor\frac{n}{k} \rfloor +2.$
	By the assertion (1), we can assume that $v_1 =\cdots = v_d.$
	Consider the affine transformation $B_{S \setminus 1}.$  We have that $B_{S \setminus 1} v_1 = \cdots =  B_{S \setminus 1} v_d$ and
that the 	vectors $B_{S \setminus 1} v_2, \cdots,  B_{S \setminus 1} v_n$ give a unit decomposition in $\mathbb{R}^k.$ Applying Lemma \ref{lemma_equiv_cond}, we get
	$$
		| B_{S \setminus 1} v_2|^2 + \cdots + | B_{S \setminus 1} v_d|^2  \leqslant 1.
	$$
	Therefore, $|B_{S \setminus 1} v_1| = | B_{S \setminus 1} v_2| \leqslant \sqrt{\frac{1}{d-1}}.$
	Again, since the vectors $ B_{S \setminus 1} v_2, \cdots,  B_{S \setminus 1} v_n$ give a unit decomposition and by Lemma \ref{lemma_equiv_cond}, there exists $j \in [n]$ such that
	$$
	|B_{S \setminus 1} v_j| \geqslant \sqrt{\frac{k}{n -1}} > \sqrt{\frac{k}{n}} > \sqrt{\frac{1}{d-1}}  \geqslant | B_{S \setminus 1} v_1|.
	$$
	Let $\tilde{S}$ be an $(n, k)$-frame obtained from $S$ by substitution $v_1 \to v_j$ and $S' = B_{\tilde{S}} \tilde{S}.$
	Using the same arguments as above,  we get that
	$\vol  \left(\crosp^n | S' \right)  < \vol \left(\crosp^n|{S}\right).$
	This contradicts  the initial choice of $S.$
\end{prf}

And again, in all assertions of this lemma we can transform $S$ to $S'$ in a continuous way while decreasing the volume in a monotonic way.
And now, we are ready to describe all $H_k$ such that $C_{\crosp} (n,k)$ is attained.
\begin{lemma}\label{lemma_extremizers}
The constants $C_{\crosp} (n,k)$ and $C_{\cube} (n,k)$ are given by \reff{eq_exact_constants} and they are attained on the same subspaces.  All such subpspaces are  given by the following rule:
\begin{enumerate}
\item \label{enum_card_1} we partition $[n]$  in $k$ sets such that the cardinalities of any two sets differ in one at most;
\item \label{enum_card_2} let set $\{i_1, \ldots, i_\ell \}$ be one of the sets of partition. Then, choosing arbitrary signs, we write the system of linear equations
$$
	\pm x[i_1] = \ldots = \pm x[i_\ell];
$$
\item \label{enum_card_3} our subspace is the solution of the system of all  equations written for each set of the partition on the step \reff{enum_card_2}.
\end{enumerate}
\end{lemma}
\begin{prf}
By the duality argument, $\crosp^n | H_k$ is an affine cross-polytope iff
$\cube^n \cap H_k$ is an affine cube. Therefore,
	$ \vol \left( \crosp^n | H_k \right) \cdot \vol \left( \cube^n \cap H_k \right) = \vol \crosp^k \cdot \vol \cube^k$ in case $\crosp^n | H_k$ is an affine
	cross-polytope.
	Hence, the first claim is proven.

	Let $H_k$ be such that $C_\crosp (n, k)$ is attained and let
	$\{\pm a_1, \ldots, \pm a_k \}$ be the vertices of $\crosp^n | H_k$.
	By the first assertion of Lemma \ref{lemma_about_vertecies}, we have that the projections $v_i$ of the vectors $e_i$ are the vertices  $\crosp^n | H_k,$ i.e. $v_i$
	coincides with $\pm a_j$ for a proper sign and $j \in [k].$
	Or, equivalently, we partition $[n]$ in $k$ sets  and $H_k$ is the solution of a proper system of linear equations constructed as in
\reff{enum_card_2} and \reff{enum_card_3}, 	except we have not proved that \reff{enum_card_1} holds yet. Let us prove this assertion.
Let $d_i$ vectors of the standard basis project onto a pair $\pm a_i.$
Identifying $H_k$ with $\mathbb{R}^k$
and using Lemma \ref{lemma_equiv_cond}, we conclude that
$a_i$ and $a_j$ are orthogonal whenever $i \neq j.$ Therefore, $|a_i|^2 = \frac{1}{d_i}$ and
\begin{equation}\label{calc_C_crosp_1}
\vol \left( \crosp^n | H_k \right)= \frac{2^k}{k!} \frac{1}{\sqrt{d_1 \cdot \ldots \cdot d_k}}.
\end{equation}
Suppose $d_i \geqslant d_j +2$ for some $i, j \in [k].$ Then
$d_i \cdot d_j  \leqslant (d_i -1)(d_j+1).$ By this and by \reff{calc_C_crosp_1},
we show that \reff{enum_card_1} holds.

It is easy to see that   there are exactly $n - k \floor*{\frac{n}{k}}$ of $d_i$'s equal
$\ceil*{\frac{n}{k}}$ and all others $k - \left(n - k \floor*{\frac{n}{k}}\right)$ are equal to $\floor*{\frac{n}{k}}.$ That is, $C_\crosp (n,k)$ is given by \reff{eq_exact_constants}. This completes the proof.
\end{prf}

The simple Lemma \ref{lemma_simplif_main} allows us to bound efficiently $n$ for a maximizer of \reff{main_problem_crosp}.

\begin{lemma}
	Denote $V (n,k) = \max\limits_{S \in \Omega(n,k)} {\vol \left( \crosp^n |S\right)}$. Then $V (n,k) \leqslant V(k^3, k).$
\end{lemma}
\begin{prf}
Let us show that $V(n, k) \geqslant \frac{2}{k} V (n-1, k-1).$ Indeed, consider
 an $(n-1, k-1)$-uframe $S_1$ such that $\vol \left( \crosp^{n-1} | S_1 \right) = V(n-1, k-1).$ Adding $e_k$ to $S_1,$ we obtain a new  $(n,k)$-uframe, lets call it $S$. Obviously, $\vol \left( \crosp^{n} | S \right) = \frac{2}{k}\vol \left( \crosp^{n-1} | S_1 \right).$ The needed inequality is proven.

 Let $S = \{v_1, \ldots, v_n \}$ be an $(n,k)$-uframe such that
 $\vol \left( \crosp^{n} | S \right) = V(n, k),$ and let $n > k^3.$
By Lemma \ref{lemma_simplif_main}, we conclude that all vectors of $S$ are non-zero vectors (otherwise we can omit them and decrease $n$).
 Since $\sum\limits_1^n |v_i|^2 = k,$ there exists $i$ such that
 $|v_i| \leqslant \sqrt{\frac{k}{n}}.$  By Lemma \ref{lemma_proj_unit_dec},
 the projections of $\{v_1, \ldots, v_n \}$ onto  the hyperplane  $H^i$ perpendicular to $v_i$ give a unit decomposition in $H_i.$ As $v_i$ projects at zero, we may consider  all others projections as an $(n-1, k-1)$-uframe, which we denote $S_1$. On the other hand, we have
\[V (n,k) = \vol \left( \crosp^{n} | S \right) \leqslant 2 |v_i| \vol \left( \crosp^{n-1} | S_1 \right) < \frac{2}{k} V(n-1, k-1).\]
We come to a contradiction.
\end{prf}
\section{Geometric constructions and properties of maximizers}
As mentioned above, to use of Lemma \ref{lemma_n_s_cond}, we need to understand the geometry behind left-hand side of \reff{ineq_n_s_cond_crosp}.
For this purpose we introduce the following definitions.

Let $\{\pm v_1, \ldots , \pm v_n\}$ be a set of pairwise distinct vectors  and the vertex set of a centrally  symmetric polytope $P.$

We use  $\mathcal{F}(v)$ to denote the set of facets of $P$ incident to its vertex $v.$
Then we define a {\it star} of the vertex $v$ of $P$ as follows
$$
N_P(v) =  \bigcup\limits_{F \in \mathcal{F}(v), F \in  \mathcal{F}(-v)} \mathop{\rm co}\{F, 0\}.
$$
By definition put  $Q_P (v) = \overline{P \setminus N_P(v)}$ and
$R_P(v) = \mathop{\rm co} \{ \pm v_i|  v_i \neq \pm v \},$ we call these sets a {\it belt} and a {\it  rest}  of the vertex $v$ in $P,$ respectively.

\begin{figure}[h]%
\center{
\psfrag{v1}[1]{\raisebox{-1ex}{$\hspace{-0.3em}v$}}
\psfrag{v2}[1]{\raisebox{-1ex}{$\hspace{-0.3em}v$}}
\psfrag{v}[1]{\raisebox{-1ex}{$\hspace{-0.3em}v$}}
\psfrag{B}[1]{\hspace{0em}\raisebox{0ex}{$Q_P (v)$}}
\psfrag{S}[1]{\hspace{0em}\raisebox{0ex}{$N_P (v)$}}
\psfrag{R}[1]{\hspace{0em}\raisebox{0ex}{$R_P (v)$}}
\includegraphics[scale=0.46]{crosspp_Def_paper.eps}
}
\vspace{-1.cm}
\caption{The star, the belt and the rest of vertex $v$ for the regular hexagon $P$.}
\end{figure}
\noindent By  symmetry, $N_P(v) = N_P(-v)$ , $Q_P(v) = Q_P(-v)$ and $R_P(v) = R_P(-v).$

Also, for any vertex $v$ of $P$ we have
\begin{equation}\label{volume_P}
\vol{P}  =  \vol Q_P(v)  + \vol N_P(v);
\end{equation}
and
\begin{equation}\label{pos_vol_neighb}
\vol N_P(v) > 0.
\end{equation}

By definition put  $P_\lambda (v) = \mathop{\rm co} \{R(v), \pm\lambda v\}.$
The idea behind this notation is that we will slightly scale one vertex of the projection of  the cross-polytope and will write the first-order necessary condition for such a transformation of a uframe.
\begin{lemma}\label{lemma_vertex_movement}
Let $v$ be a vertex of $P$.
Then $\vol P_\lambda (v)$ is a convex function on $\lambda$
and  there  exists small enough $\varepsilon$ such that
\begin{equation}\label{volume_P_lambda}
\vol{P_\lambda (v)}  =  \vol Q_P(v)  + \lambda \vol N_P(v),
\end{equation}
for $\lambda \in [1, 1 + \varepsilon].$
\end{lemma}
\begin{prf}
By  symmetry, we assume that $\lambda \geqslant 0.$
All affine hyperplanes containing facets of $R_P(v)$ divides a
line $\{t v \; |\; t \in \mathbb{R}\}$ in several intervals.  The function  $\vol P_\lambda (v)$ is linear on each interval. Clearly, we can only increase the slope of the linear functions while increase  $\lambda$. This means that we have proven the convexity of the function.

For small enough positive $\varepsilon$ the interval $(v, (1+ \varepsilon) v]$ belongs to one of the  above mentioned  intervals. It is easy to see that for each facet $F$ incident to the vertex $v$ the volume of $\mathop{\rm co} \{F, \lambda v, 0 \}$  is
$\lambda \vol \mathop{\rm co} \{F,  0 \}$ for $\lambda \geqslant 1.$
By the choice of $\varepsilon$ and by construction,  we see that
\[
\vol{P_\lambda (v)}  =  \vol Q_P(v)  +  \sum\limits_{F \in \mathcal{F}(v), F \in  \mathcal{F}(-v)}\vol (\mathop{\rm co} \{F, \lambda v, 0 \}) =  \vol Q_P(v)  + \lambda \vol N_P(v).
\]
This completes the proof.
\end{prf}

As soon as the function $\vol{P_\lambda (v)}$ is  linear in some open neighborhood of 1,
we can  write a first-order necessary condition for mazimizers and minimizers for
\reff{main_problem_crosp}. Moreover, this is the main observation in our proof of Theorem \ref{th_extr}.
Unfortunately, the function $\vol{P_\lambda (v)}$ is not linear in some open neighborhood of 1 in a general case, for example, see the corresponding polytopes for 3-dimensional cube in Figure \ref{fig_cube}.  One can show that this function is linear in some neighborhood of 1 if and only if all facets incident to $v$ are simplices.
\begin{figure}[h]%
\center{
\includegraphics[scale=0.5]{crosspp_cubepv.eps}
}
\vspace{-1.5cm}
\caption{Scaling the corresponding opposite vertices of the cube  along the green line, we obtain different combinatorial structure and linear coefficients.}
\label{fig_cube}
\end{figure}
\begin{lemma}\label{lemma_nergh_vol_lower_bound}
$$
	\sum\limits_{i \in [n]} \frac{\vol N(v_i)} {\vol P} \geqslant k.$$
The bound is tight iff $P$ is a simplicial polytope.
\end{lemma}
\begin{prf}
Let $F$ be a facet of $P$ with $f$ vertices.
Then the ratio $\frac{\vol(\mathop{\rm co}\{F, 0\})}{\vol P}$ is a summand  of exactly $f$ ratios $\frac{\vol N(v)} {\vol P}.$
Since $f \geqslant k,$ we have that
$$
\sum\limits_{i \in [n]} \frac{\vol N(v_i)} {\vol P} \geqslant k \sum\limits_{F \in \mathcal{F}_P} \frac{\vol(\mathop{\rm co}\{F, 0\})}{\vol P} = k \frac{\vol P }{\vol P} = k,
$$
where $\mathcal{F}_P$ is the set of all facets of $P.$
Obviously, we have equality here only iff $P$ is a simplicial polytope.
\end{prf}
Now we are ready to prove a geometric necessary condition for \reff{main_problem_crosp}.

\begin{theorem}\label{th_ness_cond_max_vol}
Let a  $S = \{v_1, \ldots,  v_n\}$  be  an $(n, k)$-uframe such that a local  maximum of   \reff{main_problem_crosp} is attained on $S$.
Then $\crosp^n |S$  is a simplicial polytope and for every vertex $v$ of $\crosp^n | S$ we have  that  $|v|^2 = \frac{\vol N_{(\crosp^n |S)}(v)}{ \vol \left( \crosp^n |S \right)}.$
\end{theorem}
\begin{prf}
Denote $P  = \crosp^n |S.$
Let $v_i \in S$ be a vertex of $P.$
 Let $\tilde{S}$ be an $(n, k)$-frame obtained from $S$ by substitution $v_i \to  v_i + t v_i.$
 Substituting identities given by Lemma \ref{cor_der_det_identity} and Lemma \ref{lemma_vertex_movement} into inequality \reff{ineq_n_s_cond_crosp}, we get
\[
	1 + t |v_i|^2 + o(t) \geqslant \frac{\vol Q_P(v_i)  +  (1 + t) \vol N_P(v_i)}{\vol P} = 1 + t \frac{\vol N_P(v_i)}{\vol P},
\]
for a sufficiently small positive $t.$  From this, we obtain that $|v_i|^2 \geqslant \frac{\vol N_P(v_i)}{\vol P}$.
By Lemma \ref{lemma_simplif_main}, we have that all non-zero elements of
 $\{\pm v_1, \ldots, \pm v_n\}$ are pairwise distinct vertices of $P.$
Therefore, by \reff{eq_trace_AS}, we have
$$
	k = \sum\limits_{i \in [n] | v_i \neq 0} |v_i|^2 \geqslant \sum\limits_{i \in [n] | v_i \neq 0} \frac{\vol N_P(v_i)}{\vol P}.
$$
And finally, by Lemma \ref{lemma_nergh_vol_lower_bound}, the right-hand side is at least $k$ and it is $k$ if and only if $P$ is a simplicial polytope. Therefore,  all inequalities  are tight.
This completes the proof.
\end{prf}

Theorem \ref{th_ness_cond_max_vol} implies that in case of local maximum we can slightly perturb all vectors of a uframe (for a sufficiently  small perturbation the images of zero vectors of $S$ remain in the interior and don't affect  the volume).
Moreover, we can use all proper transformations described in Section \ref{sec_pertubations}, which, along with Lemma \ref{lemma_n_s_cond}, give us
some geometric restriction for the uframe to give a local maximum.

\section{Proofs of Theorem \ref{th_extr}  and Theorem \ref{th_upper_bound_crossp_lowdimcase}}\label{sec_proofs_main_results}
\noindent {\bf Proof of Theorem \ref{th_extr}.}
It is sufficient to
prove  the lower bound for the volume of  a projection of the
cross-polytope. Since the dual problem for the  cube is a direct consequence of the two-dimensional Mahler inequality \cite{mahler1939minimalproblem}.

Let us show that the assertion of Conjecture \ref{conj_min_vol_proj_crosspolytope} is true for $k=2.$
The main observation is that polygons are simplicial polytopes. Therefore, we can use identity \ref{volume_P_lambda} for the projection.  Let $S = \{v_1, \cdots, v_n\}$ be an $(n,k)$-uframe such that $\vol \left( \crosp^n | S\right)$ is the global minimum of \reff{main_problem_crosp}. Denote $P = \crosp^n | S$ and let $v$ be a vertex of $P.$
Clearly, $v \in \{\pm v_1, \cdots, \pm v_n\},$ and, by Corollary \ref{cor_subst_decrease}, we know that  $\{\pm v_1, \cdots, \pm v_n\}$
are vertices of $P.$ Let exactly $d$ pairs $\pm v_1, \ldots, \pm v_d$ coincide with $\pm v.$
Let $\tilde{S}$ be an $(n,k)$-frame obtained from $S$ by substitution
$v_i  \to (1+t) v_i,$ where $i \in [d]$ and $t \in (-\varepsilon, \varepsilon)$ for a sufficiently small $\varepsilon$.  That is, we do not change a combinatorial type of $P.$
Using identities \reff{eq_det_M}, \reff{volume_P_lambda} and Lemma \ref{lemma_n_s_cond}, we obtain
\[
	1 + t \frac{\vol N_P (v)}{\vol P (v)} \geqslant \sqrt{1 + (2t + t^2)d|v|^2}.
\]
Since $t \in (-\varepsilon, \varepsilon),$ the linear part in $t$ on the both sides
of the inequality coincide, and, as the quadratic part is only in the right-hand side, the coefficient at $t^2$ on the right must be non-positive.
This coefficient is
\[
C=\frac{(d |v|^2)^2 - d |v|^2}{2}.
\]
  Since $S$ is an $(n,k)$-frame and by Lemma \ref{lemma_equiv_cond}, we have that $d|v|^2 \in (0,1].$
Therefore, $C \leqslant 0$ if and only if $d|v|^2 = 1.$
Again, by Lemma \ref{lemma_equiv_cond}, this implies that  $ \langle v_j, v \rangle = 0$ for $j > d,$ and, by \reff{eq_trace_AS}, $P$ has only two pairs of centrally symmetric vertices.
Or, equivalently, $P$ is a rhombus. The optimal constant in this case was found in Lemma \ref{lemma_extremizers}.
\bbox

Actually, we used only that $S$ is a local minimum to prove that $\crosp^n |S$ is a rhombus in Theorem \ref{th_extr}. This means that all local minima are attained on rhombus.  But the maximization problem is more sophisticated in this sense.
\\

\noindent {\bf Proof of Theorem \ref{th_upper_bound_crossp_lowdimcase}}.
The main idea is to use substitution $v \to 0$ for an
appropriately chosen vector $v$ of a maximizer.
Such a transformation is not continuous, and, as will be shown in the next section, we cannot transform an arbitrary $(n,k)$-uframe to obtain the global maximum while increasing the volume of the generated projection of  the cross-polytope in  a  monotonic way.

So, let $S = \{v_1, \cdots, v_n\}$ be an $(n,k)$-uframe such that the global maximum of \reff{main_problem_crosp} is attained on it, and denote $P  =  \crosp^n | S$. By Lemma \ref{lemma_simplif_main},   we assume that $\{\pm v_1, \ldots, \pm v_n\}$ are pairwise distinct vertices of $P.$  Choose a vertex $v$ of $P$ such that $|v| < 1$
(if $|v_i| = 1$ for all $i \in [n]$ then, by Lemma \ref{lemma_equiv_cond}, $n =k$ and vectors of $S$ form an orthonormal basis, and $P$ is the standard cross-polytope).
Let  $t \in (0, 1)$
such that $tv$ is on the boundary of the set $R_P (v)$
(if $t =0$ then $|v| = 1$).
Then, the theorem  follows from the inequality
\begin{equation}\label{ineq_t_v}
	\frac{1}{2} > \frac{x}{1+x} \geqslant t \geqslant \frac{x}{\sqrt{n-1} \sqrt{1 - x^2}},
\end{equation}
where $ x = \sqrt{1 - |v|^2} \in (0, 1).$
Before proving this inequality, let us complete the proof using it.
As a straightforward consequence, we obtain the following chain of equivalent inequalities:
$$
\frac{x}{1+x}  \geqslant \frac{x}{\sqrt{n-1} \sqrt{1 - x^2}} \;\; \Leftrightarrow
\;\; \frac{1-x}{1+x} \geqslant \frac{1}{n-1}  \;\; \Leftrightarrow \;\; 1-\frac{2}{n} \geqslant x.
$$
By identity \reff{eq_trace_AS} , $x$ can be chosen such that $x \geqslant \sqrt{1 - \frac{k}{n}}.$
Substituting this in the last inequality, we obtain
$$
	1 - \frac{4}{n} + \frac{4}{n^2} \geqslant 1 - \frac{k}{n}  \;\;
	\Leftrightarrow \;\; k \geqslant 4 - \frac{4}{n}.
$$
Finally, if $k=2$, $n \geqslant 3$  and if $k = 3$, $n > 4$ we are done.
As mentioned in the Introduction, the  hyperplane case (in particular, $k = 3$ and $n =4$) was proven by K. Ball in \cite{ball1995mahler}.

Let us prove inequality \reff{ineq_t_v}. We assume $v = v_1.$
Let $\tilde{S}$ be an $(n,k)$-frame obtained from $S$ by substitution $v_1 \to 0.$
Then, by \reff{eq_det_M} and by Lemma \ref{lemma_n_s_cond},  we obtain
\begin{equation}\label{th_max_ineq_1}
	\sqrt{1 - |v^2|} \geqslant \frac{\vol \left( \crosp^n | \tilde{S}\right)}{\vol P}.
\end{equation}
But $\crosp^n | \tilde{S}$ is precisely $R_P (v).$
The rest of $v$ in $P$ is the union of the two internally disjoint sets
\begin{enumerate}
\item  the  belt $Q_P(v),$ which, by Theorem \ref{th_extr}, has the volume $(1 -  |v|^2) \vol P;$
\item the intersection  of the rest and the star of the vertex $v;$  this set contains $tv,$ and, therefore, its volume is at least $t \vol N_P(v),$
which is, by Theoren \ref{th_extr},  $t |v|^2 \vol P$.
 \end{enumerate}

 Returning to inequality \reff{th_max_ineq_1}, we obtain
 $$
 	\sqrt{1 - |v|^2} \geqslant (1 - |v|^2) + t |v|^2
 	\;\; \Leftrightarrow \;\; \frac{\sqrt{1 - |v|^2}- (1 - |v|^2)}{|v|^2} \geqslant t
\;\; \Leftrightarrow \;\;  \frac{x}{1+x} \geqslant t.
$$

Let $H$ be the supporting hyperplane to $R_P (v)$ at $tv$ and $\ell$ be its orthogonal complement. Then, by Lemma \ref{lemma_proj_unit_dec}, the projections $\{v'_1, \ldots, v'_n \}$ of vectors of $S$ onto $\ell$ give a unit decomposition in $\ell,$ hence
$\sum\limits_{i \in [n]} |v'_i|^2 =1.$ On the other hand, the projection of $v$ has the length at most $|v|$ and the projections of all others have the length at most $t |v|.$ This means that we have proven  inequality
$
	|v|^2 + (n-1) t^2 |v|^2 \geqslant 1,
$
or, equivalently,
$$
	t \geqslant \frac{x}{\sqrt{n-1} \sqrt{1 - x^2}}.
$$
This completes the proof of inequality \reff{ineq_t_v} and the proof of the theorem.
 \bbox
\section{Local maxima} \label{{sec_local_m}}

\begin{lemma}\label{lemma_local32}
Let $W = \{w_1, w_2, w_3\}$ be an $(3,2)$-uframe such that
$\crosp^n | W$ is the regular hexagon. Then $W$ is  the local maximum of \reff{main_problem_crosp}.
\end{lemma}
\begin{prf}
Let  $\Omega \subset \Omega (3,2)$  be a set of all $(3,2)$-uframes $S= \{v_1, v_2, v_3 \}$ such that
all three vectors $v_1, v_2, v_3$ are on the boundary of $\crosp^n |S.$ It is a closed set.
Hence, there exists the global maximum of \reff{main_problem_crosp} on $\Omega.$
Let it be attained on an $(3,2)$-uframe $S = \{v_1, v_2, v_3\}.$

We begin with excluding the degenerate case when the generated projection of a
cross-polytope is a parallelogram, we assume $v_3 \in [v_1, v_2].$ Then the vector $v_3$ is perpendicular to the segment $[v_1, v_2],$ otherwise we increase the volume and decrease the determinant in \reff{ineq_n_s_cond_crosp}  slightly rotating $v_3.$
Choosing $v_3$ as a direction of the first orthonormal basis of the plane and using the assertion \reff{lemma_eq_cond_it3} of Lemma \ref{lemma_equiv_cond}, we obtain that
\[S = \begin{pmatrix}
1/\sqrt{3} , & 1/\sqrt{3} , & 1/\sqrt{3} \\
\pm 1/\sqrt{2}, &  \mp 1/\sqrt{2}, & 0
\end{pmatrix}   .
\]
Clearly, $\vol \left(\crosp^n |S \right) < \vol \left(\crosp^n |W\right).$

 Now, let $\crosp^n |S$ be a hexagon, see notation in Figure \ref{fig_max}.
 \begin{figure}[h]%
\center{
\psfrag{v1}[1]{\raisebox{-1.6em}{\hspace{-.8em}{$\hspace{0.3em}v_3$}}}
\psfrag{v2}[1]{\raisebox{-1.6em}{\hspace{2.8em}{$\hspace{0.3em}v_2$}}}
\psfrag{v3}[1]{\hspace{-2.8em}{$\hspace{0.3em}v_1$}}
\psfrag{a1}[1]{\raisebox{-1.8em}{$\hspace{3.1em}A_1$}}
\psfrag{a2}[1]{\raisebox{2.9em}{$\hspace{2.3em}A_2$}}
\psfrag{a3}[1]{\raisebox{3.8em}{$\hspace{-1.3em}A_3$}}
\psfrag{a4}[1]{\hspace{-2.8em}{$\hspace{0.3em}A_4$}}
\psfrag{a5}[1]{\raisebox{-2.8em}{$\hspace{0.3em}A_5$}}
\psfrag{a6}[1]{\raisebox{-3.em}{$\hspace{-0.3em}A_6$}}
\psfrag{o}[1]{\hspace{0.8em}\raisebox{1.0ex}{$O$}}
\includegraphics[scale=0.66]{crosspp_Max_paper.eps}
}
\caption{Illustration for Lemma \ref{lemma_local32}.}
\label{fig_max}
\end{figure}

 In this case $S$ is  in the interior of $\Omega,$ and, therefore, we can perturb $S$ slightly and use Lemma \ref{lemma_n_s_cond}:
  \begin{enumerate}
 \item Performing again  a small enough rotation of $v_3,$ we obtain that $v_3 \perp [v_1, v_2].$
 \item Using a  substitution $v_1  \to v_1 + t v_2, v_2  \to v_2 - t v_1$ and inequality \reff{ineq_n_s_cond_crosp}, we get      that the triangles $O A_1 A_2$ and $O A_2 A_3$ have the same area.
 \end{enumerate}
 Therefore, $A_5 A_2$ is a median and an altitude of the triangle $A_1 A_3 A_5,$ hence it is an isosceles triangle. By the symmetry, the triangles $A_1 A_3 A_5$ and
 $A_2 A_4 A_6$ are equilateral and $O$ is their common center.
We conclude that $\crosp^n | S$ is a regular hexagon.
This completes the proof.
\end{prf}
Again, by Lemma \ref{lemma_simplif_main}, we can add zero vectors to $W$ and a new
$(n,2)$-uframe remains a local maximum of \reff{main_problem_crosp}.
Moreover, for  a given local maximum for  $\Omega (n, k)$ we can construct a new local maximum for $\Omega (n+1, k+1).$
\begin{lemma}\label{lemma_local_mx_add_dim}
Let $S \subset \Omega (n ,k)$ give a local maximum of \reff{main_problem_crosp}.
Consider $S_1$ obtained from $S$ as follows
\begin{itemize}
\item considering the standard embedding $\mathbb{R}^k \subset \mathbb{R}^{k+1},$ we extend all vectors of $S$ by adding zero in the last $(k+1)$-st coordinate;
\item add $e_{k+1}.$
\end{itemize}
Then $S_1 \subset \Omega (n+1, k+1)$ and it gives a local maximum of \reff{main_problem_crosp} for $\Omega (n+1, k+1).$
\end{lemma}
\begin{prf}
 By Theorem \ref{th_ness_cond_max_vol}, $\crosp^n | S$ is a simplicial polytope.
 Hence, we can subdivide it into non-overlapping simplices of the type
 $\mathop{\rm co} \{F, 0\},$ where $F$ is a facet of $\crosp^n | S.$ Then $S_1$ is a simplicial polytope as well, and simplices $\mathop{\rm co} \{F ,  0 , \pm e_{k+1}\}$  subdivide $\crosp^{n+1} | S_1.$ Let $S'_1 = \{v_1, \ldots, v_{n+1}\} \subset \Omega (n+1, k+1)$  be a sufficiently  small perturbation of $S_1$ (where $v_{n+1}$ is a perturbation of $e_{k+1}$) and $H^\perp$ be the orthogonal complement to $v_{n+1}$ in $\mathbb{R}^{k+1}.$ Denote by $P_\perp$ the projector onto $H^\perp.$  Then, by linearity and since
 $\crosp^{n+1} |S'_1$ and $\crosp^{n+1} |S_1$ have the same combinatorial structure,
 we have
\begin{gather*}
 	\vol \left(\crosp^{n+1} |S'_1 \right) =
 	\sum  \left( \vol \mathop{\rm co} \{F, 0, v_{n+1}\} + \vol \mathop{\rm co} \{F, 0, -v_{n+1}\}\right)
 	= \\ \sum  \left( \vol \mathop{\rm co} \{P_\perp F, 0, v_{n+1}\} + \vol \mathop{\rm co} \{P_\perp F, 0, -v_{n+1}\}\right) = \vol   \left(\crosp^{n+1} | S_2 \right),
\end{gather*}
where the summations are over all $(k-1)$-dimensional  faces of $\crosp^{n+1} |S'_1$ generated by  the vectors $\{\pm v_1, \ldots, \pm v_{n}\}$ and
$S_2 = \{P_\perp v_1, \ldots, P_\perp v_n, v_{n+1}\}.$

By Lemma \ref{lemma_proj_unit_dec}, $S_3 = \{P_\perp v_1, \ldots, P_\perp v_n\}$ give a unit decomposition in $H^\perp$ and it can be considered as a sufficiently small perturbation of $S.$ Therefore, by the choice of $S$, we have
$$
\vol \left(\crosp^{n+1} |S'_1 \right) = 	\vol   \left(\crosp^{n+1} | S_2 \right) =
\frac{2}{k+1} |v_{n+1}| \vol \left( \crosp^n | S_3 \right) \leqslant
\frac{2}{k+1}  \vol \left( \crosp^n | S \right)  = \vol \left( \crosp^{n+1} |S_1 \right).
$$
That is, $S_1$ gives a local maximum.
\end{prf}

Summarizing,  we have just proved.
\begin{theorem}
For any $n > k \geqslant 2$ there exists a local and not the global maximum of \reff{main_problem_crosp}.
\end{theorem}

Therefore, it is not sufficient to use only a  necessary condition of local maximum (for example, Theorem \ref{th_ness_cond_max_vol}) in order to prove Conjecture \ref{conj_max_vol_proj_crosspolytope}.  Our approach in a lower dimensional case is to map one vector to the origin, and it works for $k =2, 3.$  We don't know whether our approach works in a higher dimensional case. The following questions remain open:
 can we delete one (or even  several) vertex, transform to a new uframe  (using the scheme $S \to \tilde{S} \to S'$) and increase the volume for an arbitrary polytope in $\mathbb{R}^n?$ Is it true that the maximum volume of the projection with $2\ell$ vertices is a decreasing function of $\ell?$

 \section{Isotropic measures}\label{sec_isotropic_measures}
If $\mu$ is a finite Borel measure on $\mathbb{R}^k$, then $\supp(\mu)$ will denote its support.
Recall that a finite
 Borel measure $\mu$ on $\mathbb{R}^k$ is said to be isotropic if
$$
	\int\limits_{\mathbb{R}^k}  v \otimes v \, \,  \mu(dv) =  I_k.
$$

Usually, isotropic measures with support on the unit sphere are considered. Such measures appear naturally in  John's condition for the unique ellipsoid of maximal volume in a convex body.  Let us recall John's well-known result \cite{John}.
\begin{theorem}[F. John]
The Euclidean unit ball $\EE_k$ is the ellipsoid of maximal volume
contained in a convex body $K \in \mathbb{R}^k$ iff $\EE_k \subset K$ and, for some $n \geqslant k$,
there are Euclidean unit vectors $(u_i)_1^n $, on the boundary of $K$, and positive
numbers $(c_i)_1^n$ for which

\begin{enumerate}
\item \label{John_cond_1}  $\sum\limits_1^n  c_i u_i = 0$;
\item \label{John_cond_2} $ \sum\limits_1^n c_i u_i \otimes u_i = I_k,$ the identity on $\mathbb{R}^k.$
\end{enumerate}
\end{theorem}
\noindent
Identity \reff{John_cond_2} is called John's condition.
One can see that John's condition says that the measure
$$
\mu (x)= \sum\limits_{1}^{n} c_i  \, \delta_{u_i} (x)
$$
is an isotropic measure. Using this observation and his version of the Brascamb--Lieb inequality, Keith Ball  \cite{ball1989volumes}  found tight upper bounds on the volume of a cross section of $n$-dimensional cube, and, in \cite{ball1991volume}, he established the celebrated  reverse isoperimetric inequality. Isotropic measures are closely related and studied in the context of the isotropic constant conjecture \cite{Milman1989}, Kannan-Lov{\'a}sz-Simonovits conjecture \cite{Kannan1995}, the thin shell conjecture  and etc.   To get more information about isotropic measures and related problems, we refer the reader  to \cite{giannopoulos1999isotropic}.

We consider  finite Borel  isotropic measures in which all points of the support have the unit mass, i.e., $\mu (x) = \sum\limits_1^n  \delta_{v_i}(x),$ where the vectors
$\{v_1, \ldots, v_n\}$ satisfy
\begin{equation}\label{unit_dec_1}
	\sum\limits_{1}^n v_i \otimes v_i = I_k,
\end{equation}
i.e., they give a unit decomposition in $\mathbb{R}^k.$

Clearly, if unit vectors $(u_i)_1^n$ and positive numbers $(c_i)_1^n$ satisfy John's condition (or, equivalently, $\mu_1 (x)= \sum\limits_{1}^{n} c_i  \, \delta_{u_i} (x)$ is an isotropic measure) then  the vectors $v_i = \sqrt{c_i} u_i,$ for $i \in [n],$
give a unit decomposition in $\mathbb{R}^k$ (or, equivalently, $\mu_2 (x)= \sum\limits_{1}^{n} c_i  \, \delta_{u_i} (x)$ is an isotropic measure). And the inverse statement holds, if vectors
$\{v_1, \ldots, v_n\} $  give a unit decomposition in $\mathbb{R}^k$  then the vectors
$$
u_i = \frac{v_i}{|v_i|} \quad \mbox{for} \quad i \in [n] \quad \mbox{such that}
 \quad |v_i| \neq 0
$$
satisfy John's condition.

In addition, we allow some  vectors of $\{v_1, \ldots, v_n\}$   to coincide.
Considering  this generalization, Lemma \ref{lemma_equiv_cond}  gives a complete geometric description of a property of  a set of vectors $\{v_1, \ldots, v_n\}$  to be  such that $\mu (x) = \sum\limits_1^n  \delta_{v_i}(x)$ is an isotropic measure.
More precisely,  $\mu (x) = \sum\limits_1^n  \delta_{v_i}(x)$ is an isotropic measure if and only if the vectors $\{v_1, \ldots, v_n \}$ are projections of some orthonormal basis of $\mathbb{R}^n$ onto $\mathbb{R}^k.$

We think that there are some  advantages of this approach (to consider the unit masses and the support inside the unit ball):
\begin{itemize}
\item By Lemma \ref{lemma_equiv_cond}, we have a simple geometric description of such measures.
\item By the properties of $B_S,$ we have an explicit expression how to transform a given non-degenerate finite measure to an isotropic measure with  unit masses.
\item By Lemma \ref{lemma_proj_unit_dec}, an orthogonal projections of such a measure is still  an isotropic measure with the unit masses.
\item By Lemma \ref{lemma_n_s_cond}, we have the first order approximation of the corresponding determinants, which allows us to write a necessary condition for different problems similar to \reff{main_pr_original}. For example, the author used the same approach to study of the maximizers of the projections of a cube $I^n = [0,1]^n$ in \cite{2018arXiv180410055I}. We formalize this observation below.
\end{itemize}

Let us explain for what  kind of problems we can use the same approach.   The main observation here is that  we only use  the fact that the volume is a positive homogeneous function: $\vol T (K) = |\det T| \vol K$  in Lemma \ref{lemma_n_s_cond}.
So, let $\Psi $ be a function from the set of all $(n,k)$-frames to $\mathbb{R}^+$  such that
$\Psi(T(S)) = |\det T| \cdot \Psi(S)$ for any $(n,k)$-uframe $S$ and  linear transformation $T.$
Then we can  rewrite the inequality from Lemma \ref{lemma_n_s_cond} in the following way:
\begin{equation}\label{ineq_psi}
\frac{\Psi (\tilde{S})}{ \max\limits_{\Omega (n,k)} \Psi (S)} \leqslant
\sqrt{\det{ A_{\tilde{S}}}} \; .
\end{equation}

Using the same arguments as in Section \ref{sec_def_pre}, we may consider
$\left. \Psi \right|_{\Omega(n,k)}$ to be the function on the Grassmanian  of all $k$-dimensional subspaces of $\mathbb{R}^n.$
For example,  $\left. \Psi \right|_{\Omega(n,k)}$  can be  the volume of the projection of the standard  cross-polytope, a cube $I^n = [0,1]^n,$ the standard simplex $\mathop{\rm co} \{0, e_1, \ldots, e_n\}$ on a $k$-dimensional subspace or, even,
$1 / {\vol \left([-1,1]^n \cap H_k\right)}$ for a $k$-dimensional subspace $H_k$.
We have to notice that F. Filliman used a very similar approach in \cite{Filliman1990}, but his studies did not involve  geometry, which gives us our necessary conditions. However, having a first order approximation for the determinant in the right-hand side of \reff{ineq_psi}, we may get more information about local maximizers for $\Psi.$ For example, by  subdifferential calculus, we have
\begin{corollary}
	If $\Psi$ is a convex function in some neighborhood of its local mazimizer $S \in \Omega (n,k)$
	 then  $\Psi$ is a differentiable function at $S$ and the first derivative can be found from the coincidence  of the linear parts in \reff{ineq_psi}.
\end{corollary}
We use this observation in this paper in a implicit way, and the author used it for the projections of a cube $I^n = [0,1]^n$ in \cite{2018arXiv180410055I}. Actually, the local convexity  of $\Psi$ is a rather typical situation when we consider the volume of the projection of a polytope in $\mathbb{R}^n$ onto a $k$-dimensional subspace, as the latter is piecewise linear on the corresponding Grassmanian by Theorem 1 in \cite{Filliman1990}.

\begin{remark}
As the identity has the largest determinant among all positive-definite operators with the trace  equals $k,$ we may consider  bigger  set of frames than $\Omega (n,k)$ when we study the maxima and the maximizers of $\Psi$. More formally,
we use $\Lambda(n, k)$ to denote the set of all $(n,k)$-frame $S = \{v_1, \ldots, v_n\}$ such that
$\sum\limits_{i=1}^n |v_i|^2 = k.$  Then
\[
	\max\limits_{S \in \Omega(n, k)} \Psi (S) = \max\limits_{S \in \Lambda(n, k)} \Psi (S).
\]
\end{remark}
\nocite{lutwak2004volume}
\bibliographystyle{abbrv}

redpajama_set_name: RedPajamaArXiv


text:
Washington Training Camp
Redskins Release ILB Adam Hayward
Hayward's Redskins career has come to an end.
By Alex Rowsey Jul 22, 2016, 9:04am EDT
Share All sharing options for: Redskins Release ILB Adam Hayward
Photo by Rob Carr/Getty Images
The Washington Redskins have cut veteran Adam Hayward after two seasons troubled often by injuries. He's been in the league for nine years, primarily as a special teams standout.
Signing with the 'Skins a couple years ago, Hayward was expected to be the team's new special teams ace/captain. While STs may have improved marginally, it wasn't due much to Hayward's play as he missed a majority of that first season with a broken tibia. Last year he tore his ACL in preseason and didn't play a regular season down for the Redskins.
Hayward was going to have a tough time making the team anyway after the 'Skins brought in another STs standout ILB in Terrance Garvin and drafted Stephen Daniels.
The roster now stands at 89.
ROSTER MOVE: #Redskins release LB Adam Hayward.
More: https://t.co/jeaL8IFbrE pic.twitter.com/IqFtOeee9l
— Washington Redskins (@Redskins) July 22, 2016

redpajama_set_name: RedPajamaCommonCrawl


text:
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redpajama_set_name: RedPajamaGithub


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text:
Home » News » Urology » Kidney Cancer
New Findings Support Ablation for Small Renal Tumors
Percutaneous ablation offers cancer-related outcomes similar to those of radical nephrectomy but with a lower likelihood of long-term renal insufficiency and complications.
Percutaneous ablation for small renal cell carcinoma (RCC) tumors in well-selected older patients may result in oncologic outcomes similar to those of radical nephrectomy, but with less long-term renal insufficiency and fewer perioperative complications, according to a new study.
In an observational study of 4310 patients aged 66 years or older who received treatment for renal tumors smaller than 4 cm (T1a RCC) from 2006 to 2011, the 5-year RCC-specific survival rate for patients who underwent percutaneous ablation (PA) and radical nephrectomy (RN) was 96% and 95%, respectively, a non-significant difference between treatment arms, Adam D. Talenfeld, MD, MS, of Weill Cornell Medicine/New York-Presbyterian Hospital in New York, and colleagues reported online in Annals of Internal Medicine. The 5-year RCC-specific survival rate was slightly lower for PA compared with partial nephrectomy (PN): 95% vs 98%.
The 5-year overall survival (OS) rates were similar for PA and RN (74% vs 75%) and lower for PA than PN (77% vs 86%).
In adjusted analyses, RCC-specific and all-cause mortality risk did not differ significantly between PA and RN. PA was associated with a non-significant 2-fold increased risk of RCC-specific mortality and a significant 2-fold increased risk of all-cause mortality compared with PN.
The cumulative rates of renal insufficiency 31 to 365 days after PA, PN, and RN were 11%, 9%, and 18%, respectively. Rates of non-urologic complications within 30 days after PA, PN, and RN were 6%, 29%, and 30%, respectively.
The authors concluded that appropriately selected older patients with small renal tumors may expect oncologic outcomes from PA similar to those of RN, with fewer complications and less chronic renal insufficiency. "Although our conclusions about the comparability of oncologic outcomes for PA versus PN are less certain, our findings suggest that patients who receive ablation might have fewer complications," Dr Talenfeld and his colleagues wrote. "This large, population-based, comparative analysis of PA outcomes strengthens the findings of recent institutional studies and raises the level of evidence in support of PA for well-selected older patients with small renal tumors."
For the study, the investigators used the SEER (Surveillance, Epidemiology, and End Results) cancer registry linked to Medicare claims data. The median follow-up for the whole cohort was 52 months for overall survival and 42 months for RCC-specific survival. The median follow-up for the PA, PN, and RN groups was 44, 51, and 55 months, respectively, for OS and 33, 40, and 46 months, respectively, for RCC-specific survival.
Radiofrequency Ablation May Be a Valid Approach for RCC
New Ablative Techniques in Urologic Oncology
Robotic PN May Offer Better Outcomes Than RFA for T1a RCC
In a discussion of study limitations, the investigators pointed out that their use of SEER-Medicare linked files prevented analysis of patients who received treatment after 2011, "possibly reducing generalizability to the newest PA, PN, and RN techniques."
Talenfeld AD, Gennarelli RL, Elkin EB, et al. Percutaneous ablation versus partial and radical nephrectomy for T1a renal cancer. Ann Intern Med. 2018; published online ahead of print.
http://annals.org/aim/article-abstract/2686095/percutaneous-ablation-versus-partial-radical-nephrectomy-t1a-renal-cancer-population
Kidney Cancer Renal Cell Carcinoma
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Sunitinib Alone or After Surgery for mRCC Offers Similar Outcomes
Close more info about New Findings Support Ablation for Small Renal Tumors


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Zuma on the run as sources revealed "Bosasa paid R300,000 a month" to avoid prosecution
NDOLOVU May 16, 2019
Former Bosasa COO Angelo Agrizzi has dropped more bombshells , including claims that former president Jacob Zuma and top NPA…
US NGO sues Malema and EFF for $2 million dollars over assault and deadly threats on Karima Brown
A US-based NGO defending journalists targeted Economic Freedom Fighters (EFF) leader Julius Malema on after he tweeted the personal phone…
See what married Police constable was doing. Photos and video gone viral. (Photos inside)
A married South African police officer's photos having sex with the Nigerian man have been leaked by the Nigerian man…
BREAKING !!! Nkosazana Dlamini-Zuma to become SA's deputy president, replaces Mabuza
According to reports, It is understood that groups in the ANC are pushing for former AU Commission chairperson Nkosazana Dlamini…
Shocking !!! Mugabe goes completely insolvent, auctions properties to settle debts.
Farm tractors, pick-up trucks and combine harvesters from former Zimbabwe president Robert Mugabe's dairy farm will go under the hammer…
"SA will be better under Ramaphosa if he keeps his promises": De Klerk
Former president De Klerk said South Africa will be a better country under ANC President Cyril Ramaphosa after the 2019…
BLF leader cries profusely, 'Whites voted for the FF Plus and our black people didn't vote for us'
BLF Mngxitama is absolutely disappointed that the same black people that he has sacrificed everything for turned their back against…
Couple narrowly escaped death, wife shot in the stomach
NDOLOVU May 7, 2019
A Durban man could have lost "the love of his life" when the couple was shot by unidentified men outside…
Man stabs two children to death.
A Vredenburg family will on Monday identify the bodies of two children who were killed, allegedly by their mother's boyfriend.…
Malema begs white youth to vote EFF on May 8th..
Julius Malema has urged white South Africans to vote for the EFF. He made the comments while speaking at his…


Before completing this activity, you must read chapters 7 & 8 of Freak the Mighty.
How does Kevin "steer" Max when he rides on his shoulders?
What is Kevin comparing Max to by "steering" him?
Where does Max go to escape the gang?
Tony D exchanges his knife for __________ to throw at them.
What does Kevin call the police?
What did the cops call Max? Why?
They called him a hero, because he saved Kevin from the gang.
They called him a coward, because he ran away from the gang.
They called him stupid, because he got mixed up with the gang.
Does Max agree with what the police say about him? Why?
No. He says that Kevin is the hero.
No. He says that they are stupid.
Yes. He thinks he did a wonderful thing.
Yes. He says they are right.
Grim and Gram are __________ and __________ at the police's opinion of Max.
Grim was so amazed that he gave Max __________ to drink.
What does Kevin do every morning?
How many subspecies of ants are there?
Is Kevin a "morning person"? Is Max?
What is a dragon, according to Kevin?
Kevin keeps a __________ in his knapsack.
What does Kevin do when he looks up a word?
How big are "dinosaur brains"?


Those are quite the Football Socks! Good for you, hopefully Aaron Rogers can work with all the new players this year. Still can't get over them losing Jordy Nelson.
I hope Rogers has a better season. My husband is a Bears fan, so I'll hear a lot about it if he doesn't stay healthy.


- 4-port KVM switch with USB 2.0 support and 2.1 surround sound audio.
- One console controls four computers and two additional USB devices.
- 2-port USB 2.0 hub built in.
- Video DynaSync - stores the console monitor's EDID (Extended Display Identification Data) to optimize display resolution.
- Auto Scan Mode for monitoring all computers.
- LED Display for easy status monitoring.
- Easy installation no software required.