text:
\section{Introduction}

Let $d, n \geq 2$ be such that $n \geq d$. Let also $N_{d,n} = \binom{n+d}{d}$ denote the number of monomials of degree $d$ in $n+1$ variables. Ordering monomials lexicographically, degree $d$ hypersurfaces in $\mathbb{P}^n$ which are defined over $\mathbb{Q}$ are parametrized by
\begin{equation*}
\mathbb{V}_{d,n} = \mathbb{P}^{N_{d,n}-1}(\mathbb{Q}).
\end{equation*}
From now on, we thus view the above hypersurfaces as elements of $\mathbb{V}_{d,n}$. We note that the assumption $n \geq d$ implies that a generic element of $\mathbb{V}_{d,n}$ is a smooth Fano hypersurface.

We now introduce an exponential height $H : \mathbb{P}^n(\mathbb{Q}) \to \mathbb{R}_{> 0}$. In order to do so, for any
$N \geq 1$ we let $\mathbb{Z}_{\mathrm{prim}}^N$ be the set of $(c_1, \dots, c_N) \in \mathbb{Z}^N$ such that
$\gcd(c_1, \dots, c_N) = 1$, and we also let $|| \cdot ||$ denote the Euclidean norm in $\mathbb{R}^N$. For
$x \in \mathbb{P}^n(\mathbb{Q})$, we choose $\mathbf{x} = (x_0, \dots, x_n) \in \mathbb{Z}_{\mathrm{prim}}^{n+1}$ such that
$x = (x_0 : \dots : x_n)$ and we set
\begin{equation}
\label{Definition H}
H(x) = ||\mathbf{x}||^{n+1-d}.
\end{equation}
Note that if $V \in \mathbb{V}_{d,n}$ then $H$ is an anticanonical height on $V$.

The goal of this article is to investigate the smallest height of a rational point on a hypersurface $V$ as $V$ runs over the space
$\mathbb{V}_{d,n}$. Therefore, for $V \in \mathbb{V}_{d,n}$, we define
\begin{equation*}
\mathfrak{M}(V) =
\begin{cases}
\displaystyle{\min_{x \in V(\mathbb{Q})}} H(x), & \textrm{if } V(\mathbb{Q}) \neq \emptyset, \\
\infty, & \textrm{if } V(\mathbb{Q}) = \emptyset.
\end{cases}
\end{equation*}

We note here that it would be more natural to consider elements of $\mathbb{V}_{d,n}$ up to isomorphism, but this seems to be a very hard task. We thus content ourselves with the point of view which was adopted for instance by Poonen and Voloch \cite{MR2029869} and by Bhargava \cite{BhargavaCubics}.

We now introduce an ordering of the space $\mathbb{V}_{d,n}$ by defining the height of $V \in \mathbb{V}_{d,n}$ as $||\mathbf{a}_V||$, where $\mathbf{a}_V \in \mathbb{Z}_{\mathrm{prim}}^{N_{d,n}}$ is any of the two primitive coefficient vectors associated to $V$. Finally, for $A \geq 1$, we let
\begin{equation*}
\mathbb{V}_{d,n}(A) = \left\{ V \in \mathbb{V}_{d,n} : ||\mathbf{a}_V|| \leq A \right\}.
\end{equation*}

It is well known that establishing upper bounds for $\mathfrak{M}(V)$ in terms of $||\mathbf{a}_V||$ is a hard problem as this would give a positive answer to Hilbert's tenth problem for the field of rational numbers. In the case $d=2$, Cassels \cite{MR0069217} proved that for any $V \in \mathbb{V}_{2,n}$ such that $V(\mathbb{Q}) \neq \emptyset$, we have
\begin{equation}
\label{Cassels}
\mathfrak{M}(V) \ll ||\mathbf{a}_V||^{n(n-1)/2},
\end{equation}
and Kneser \cite{MR0081306} constucted an example proving that the exponent $n(n-1)/2$ is optimal. It is also worth noting that Browning and Dietmann \cite[Corollary $1$]{MR2396125} have showed that for generic quadratic forms this exponent can be slightly improved if
$n \geq 4$.

In the case $d > 2$, the results are unfortunately sparse. The interested reader is invited to refer to Masser's survey \cite{MR1975456}. We also note that more recently, in the case $d=3$ and $n \geq 16$, Browning, Dietmann and Elliott \cite[Theorem $2$]{MR2885594} have proved that for $V \in \mathbb{V}_{3,n}$, we have $\mathfrak{M}(V) \ll ||\mathbf{a}_V||^{360000}$.

This situation being clearly unsatisfactory, Elsenhans and Jahnel have undertaken a number of theoretical and numerical investigations for the families of diagonal quartic threefolds \cite{MR2333743}, general cubic surfaces \cite{MR2641182} and diagonal cubic surfaces
\cite{MR2651158, MR2676747}. Their results have led them to suggest that, if $V$ belongs to any of these three families of hypersurfaces and if $V(\mathbb{Q}) \neq \emptyset$, then we might have the upper bound
\begin{equation}
\label{Expectation}
\mathfrak{M}(V) \ll \frac1{\tau(V)^{1+\varepsilon}},
\end{equation}
for any fixed $\varepsilon > 0$ and where $\tau(V)$ denotes the Tamagawa number appearing in the definition of the constant introduced by Peyre (see \cite{MR1340296}) in the context of Manin's conjecture (see \cite{MR974910}). Moreover they have showed
\cite[Theorem $2$.$2$]{MR2333743} that the expectation \eqref{Expectation} with $\varepsilon = 0$ does not hold in general. Note finally that the quantities $1/\tau(V)$ and $||\mathbf{a}_V||$ are expected to have comparable size for most $V \in \mathbb{V}_{d,n}$ (see
\cite[Theorem $3$.$4$.$3$]{MR2333743} and \cite[Theorem, Section $1$.$5$]{MR2651158}).

The following theorem is our main result and shows that the orders of magnitude of $\mathfrak{M}(V)$ and $||\mathbf{a}_V||$ are indeed closely related.

\begin{theorem}
\label{Theorem 1}
Let $d \geq 2$ and $n \geq d$ with $(d,n) \neq (2,2)$. Let also $\psi : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ be such that
$\psi(u) = o(u)$ as $u \to \infty$. Then we have
\begin{equation*}
\lim_{A \to \infty} \frac{\# \left\{ V \in \mathbb{V}_{d,n}(A) : \mathfrak{M}(V) > \psi(||\mathbf{a}_V||) \right\}}{\# \mathbb{V}_{d,n}(A)} = 1.
\end{equation*}
\end{theorem}

Theorem~\ref{Theorem 1} is pertinent as we expect that a positive proportion of hypersurfaces in $\mathbb{V}_{d,n}$ admit a rational point for any $d \geq 2$ and $n \geq d$ with $(d,n) \neq (2,2)$. More precisely, a conjecture of Poonen and Voloch \cite[Conjecture $2$.$2$]{MR2029869} implies in particular that
\begin{equation*}
\liminf_{A \to \infty} \frac{\# \left\{ V \in \mathbb{V}_{d,n}(A) : V(\mathbb{Q}) \neq \emptyset \right\}}{\# \mathbb{V}_{d,n}(A)} > 0.
\end{equation*}
We note that this lower bound is proved in forthcoming work of Browning, the author and Sawin. Moreover, it transpires from this investigation that if $(d,n) \neq (3,3)$ then Theorem~\ref{Theorem 1} is optimal in the sense that for any $\eta >0$, we have
\begin{equation*}
\limsup_{A \to \infty}
\frac{\# \left\{ V \in \mathbb{V}_{d,n}(A) : \mathfrak{M}(V) > \eta ||\mathbf{a}_V|| \right\}}{\# \mathbb{V}_{d,n}(A)} < 1.
\end{equation*}

The only case which is not covered by Theorem~\ref{Theorem 1} is the case $(d,n)=(2,2)$ of conics. The situation is fundamentally different here as we have
\begin{equation*}
\frac1{(\log A)^{1/2}} \ll \frac{\# \left\{ V \in \mathbb{V}_{2,2}(A) : V(\mathbb{Q}) \neq \emptyset \right\}}{\# \mathbb{V}_{2,2}(A)} \ll \frac1{(\log A)^{1/2}}.
\end{equation*}
The upper bound and the lower bound have respectively been obtained by Serre \cite{MR1075658} and Hooley \cite {MR2300456}.

As already mentioned, Kneser showed that the exponent $n(n-1)/2$ in the upper bound \eqref{Cassels} could not be improved. The following theorem implies in particular that in the case $n=2$, the exponent $1$ is actually optimal for $100 \%$ of all conics admitting a rational point.

\begin{theorem}
\label{Theorem 1'}
Let $\psi : \mathbb{R}_{>0} \to \mathbb{R}_{>0}$ be such that $\psi(u)=o\left(u/(\log u)^{1/2}\right)$ as $u \to \infty$. Then we have
\begin{equation*}
\lim_{A \to \infty} \frac{\# \left\{ V \in \mathbb{V}_{2,2}(A) : \psi(||\mathbf{a}_V||) < \mathfrak{M}(V) \leq 3 ||\mathbf{a}_V|| \right\}}
{\# \left\{ V \in \mathbb{V}_{2,2}(A) : V(\mathbb{Q}) \neq \emptyset \right\}} = 1.
\end{equation*}
\end{theorem}

We remark that the analogous theorem for the family of diagonal conics can be established by appealing to a result of the author
\cite[Lemma $4$]{MR3456025} and the work of Hooley \cite{MR1199934}.

We finally state another result which is the key tool in the proofs of Theorems~\ref{Theorem 1} and~\ref{Theorem 1'} and which is definitely of independent interest. For $V \in \mathbb{V}_{d,n}$ and for any subset $U \subset V$, we define the counting function
\begin{equation}
\label{Definition N(B)}
N_U(B) = \# \left\{ x \in U(\mathbb{Q}) : H(x) \leq B \right\}.
\end{equation}
If $V \in \mathbb{V}_{d,n}$ is a smooth hypersurface such that $V(\mathbb{Q})$ is Zariski dense in $V$, a corrected version of Manin's conjecture \cite{MR974910} states that there should exist a thin subset $T \subset V$ such that
\begin{equation}
\label{Conjecture Manin}
N_{V \smallsetminus T}(B) = c_V B (\log B)^{\rho - 1} (1 + o(1)),
\end{equation}
where $c_V > 0$ is Peyre's constant (see \cite{MR1340296}) and $\rho$ denotes the rank of the Picard group of $V$.

The following theorem can naturally be viewed as an average version of Manin's conjecture over the space $\mathbb{V}_{d,n}$. We note that here and throughout Sections~\ref{Section Manin} and~\ref{Section smallest height} all the implied constants depend at most on $d$ and
$n$.

\begin{theorem}
\label{Theorem 2}
Let $d \geq 2$ and $n \geq d$. There exists a constant $C_{d,n} >0$ such that for $A \geq B^{1/(n+1-d)}$, we have
\begin{equation*}
\frac1{\# \mathbb{V}_{d,n}(A)} \sum_{ V \in \mathbb{V}_{d,n}(A)} N_V(B) =
C_{d,n} \frac{B}{A} \left( 1 + O \left( \frac{B^{1/(n+1-d)}}{A} + \frac{\log B}{B^{1/(n+1-d)}} \right) \right).
\end{equation*}
\end{theorem}

Since the rank of the Picard group of a generic element of $\mathbb{V}_{d,n}$ is equal to $1$, we see that Theorem~\ref{Theorem 2} agrees with the expectation \eqref{Conjecture Manin}. In addition, we note that the constant $C_{d,n}$ is explicit and its expression is given at the end of Section~\ref{Section Manin}. Also, we remark that the factor $\log B$ in the error term is only necessary if $n = d$.

The analog of Theorem~\ref{Theorem 2} in the case $d \geq 3$ and $n = d-1$ of intermediate type hypersurfaces is worth of attention. If we replace the exponent $n+1-d$ by $1$ in the definition \eqref{Definition H} of the height function $H$ then our methods show that there exists a constant $c_d >0$ such that for $A \geq B$, we have
\begin{equation*}
\frac1{\# \mathbb{V}_{d,d-1}(A)} \sum_{ V \in \mathbb{V}_{d,d-1}(A)} N_V(B)
= c_d \frac{\log B}{A} \left( 1 + O \left( \frac{B}{A} + \frac1{\log B} \right) \right).
\end{equation*}
It is interesting to note that in the case $d=4$, this estimate is in agreement with the unpublished observation of van Luijk that the number of rational points of height bounded by $B$ on a K$3$ surface should, in favourable circumstances, asymptotically behave like
$(\log B)^{\rho}$.

Let us give a quick sketch of the proof of Theorem~\ref{Theorem 2}. The first step consists in translating the statement into lattice point counting problems. Then, for each lattice to be considered, an astute combinatorial argument enables us to explicitly construct a maximal family of linearly independent vectors. Moreover, we crucially observe that we can control the norms of these vectors. Finally, we check that feeding this observation into classical geometry of numbers results allows us to complete the proof.

The topics addressed in this article have not been much studied. However, Br\"{u}dern and Dietmann have proved
\cite[Theorem $1$.$4$]{MR3177289} the analog of Theorem~\ref{Theorem 1} for families of diagonal hypersurfaces under the assumption $n \geq 2d-1$, while Theorem~\ref{Theorem 1} holds in the entire Fano range $n \geq d$.

We conclude this introduction by describing what we can achieve for families of Fano complete intersections. We let $r > 1$ and
$d_1, \dots, d_r \geq 2$ and we assume that $n \geq d_1 + \cdots + d_r$. Intersections of $r$ hypersurfaces in $\mathbb{P}^n$ of respective degrees $d_1, \dots, d_r$ and which are defined over $\mathbb{Q}$ are parametrized by
\begin{equation*}
\mathbb{V}_{\mathbf{d},n} = \mathbb{P}^{N_{d_1,n}-1}(\mathbb{Q}) \times \cdots \times \mathbb{P}^{N_{d_r,n}-1}(\mathbb{Q}),
\end{equation*}
where $\mathbf{d} = (d_1, \dots, d_r)$. Using the Segre embedding it is thus natural to define the height of
$V = (V_1, \dots, V_r) \in \mathbb{V}_{\mathbf{d},n}$ as $||\mathbf{a}_{V_1}|| \cdots ||\mathbf{a}_{V_r}||$. Unfortunately, this definition seems to create insurmountable difficulties as one faces situations in which $V$ lies in cuspidal regions of the space
$\mathbb{V}_{\mathbf{d},n}$. However, our techniques show that the analogs of Theorems~\ref{Theorem 1} and~\ref{Theorem 2} hold if
$n \geq r-1+d_1 + \cdots + d_r$ provided that we define the height of $V$ as $\max \{ ||\mathbf{a}_{V_i}|| : i \in \{1, \dots, r \} \}$.

\subsection*{Acknowledgements}

It is a pleasure for the author to thank Philipp Habegger, Philippe Michel and Ramon Moreira Nunes for interesting conversations. In addition, the author is extremely grateful to Tim Browning for his very careful reading of an earlier version of the manuscript, and for making a number of comments which greatly improved the presentation and the content of this article. This work was initiated while the author was working as an Instructor at the \'{E}cole Polytechnique F\'{e}d\'{e}rale de Lausanne. The financial support and the wonderful working conditions that the author enjoyed during the four years he worked at this institution are gratefully acknowledged. The research of the author is integrally funded by the Swiss National Science Foundation through the SNSF Professorship number $170565$ awarded to the project
\textit{Height of rational points on algebraic varieties}. Both the financial support of the SNSF and the perfect working conditions provided by the University of Basel are gratefully acknowledged.

\section{Geometry of numbers}

\subsection{Lattice point counting estimates}

We start by setting some notation and recalling several classical definitions. For $N \geq 1$, we let $\langle \cdot, \cdot \rangle$ be the Euclidean inner product in $\mathbb{R}^N$ and we recall that $|| \cdot ||$ denotes the Euclidean norm in $\mathbb{R}^N$. Also, for
$u > 0$, we set
\begin{equation*}
\mathcal{B}_N(u) = \left\{ \mathbf{y} \in \mathbb{R}^{N} : ||\mathbf{y}|| \leq u \right\},
\end{equation*}
and we let $V_N$ denote the volume of the unit ball $\mathcal{B}_N(1)$ in $\mathbb{R}^N$.

A lattice $\Lambda \subset \mathbb{R}^N$ is a discrete subgroup of $\mathbb{R}^N$. If $\Lambda$ is a lattice then the dimension of the subspace $\operatorname{Span}_{\mathbb{R}} (\Lambda)$ is called the rank of $\Lambda$. Also, if $\Lambda$ is a lattice of rank $R$ then its determinant
$\det (\Lambda)$ is defined as the $R$-dimensional volume of the fundamental parallelepiped spanned by any basis of $\Lambda$. Therefore, if $(\mathbf{b}_1, \dots, \mathbf{b}_R)$ is a basis of $\Lambda$ and $\mathbf{B}$ is the $N \times R$ matrix whose columns are the vectors $\mathbf{b}_1, \dots, \mathbf{b}_R$ then
\begin{equation}
\label{Definition det}
\det (\Lambda) = \sqrt{ \det (\mathbf{B}^T \mathbf{B})}.
\end{equation}

In addition, if a lattice $\Lambda \subset \mathbb{R}^N$ is a subset of $\mathbb{Z}^N$ then $\Lambda$ is said to be integral. Moreover, we say that an integral lattice $\Lambda$ is primitive if it is equal to the maximal integral lattice in $\operatorname{Span}_{\mathbb{Q}} (\Lambda)$, that is if
$\Lambda = \operatorname{Span}_{\mathbb{Q}} (\Lambda) \cap \mathbb{Z}^N$.

Given a rank $R$ lattice $\Lambda \subset \mathbb{R}^N$, we let $\lambda_1(\Lambda), \dots, \lambda_R(\Lambda)$ denote its successive minima with respect to the unit ball $\mathcal{B}_N(1)$. We recall that they are defined for $i \in \{1, \dots, R\}$ by
\begin{equation}
\label{Definition successive}
\lambda_i(\Lambda) = \inf \left\{ u \in \mathbb{R}_{> 0} : \dim ( \operatorname{Span}_{\mathbb{R}} ( \Lambda \cap \mathcal{B}_N(u) ) ) \geq i \right\}.
\end{equation}
We clearly have $\lambda_1(\Lambda) \leq \dots \leq \lambda_R(\Lambda)$ and the celebrated second Theorem of Minkowski (see for instance \cite[Chapter VIII, Theorem V]{MR1434478}) states in particular that
\begin{equation}
\label{Estimate Minkowski}
\det (\Lambda) \leq \lambda_1(\Lambda) \cdots \lambda_R(\Lambda) \ll \det (\Lambda),
\end{equation}
where the implied constant depends at most on $R$.

Finally, for any lattice $\Lambda \subset \mathbb{R}^N$ and any bounded region $\mathcal{R} \subset \mathbb{R}^N$, we set
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{R}) = \# (\Lambda \cap \mathcal{R}),
\end{equation*}
and if $\Lambda$ is integral, we also set
\begin{equation}
\label{Definition N star}
\mathcal{N}^{\ast}(\Lambda; \mathcal{R}) = \# \left( \Lambda \cap \mathbb{Z}_{\mathrm{prim}}^N \cap \mathcal{R} \right).
\end{equation}

We use the convention that empty products and empty summations are respectively equal to $1$ and $0$. The three following lemmas build on a classical result of Schmidt \cite[Lemma~$2$]{MR0224562} which provides a handy estimate for the number of lattice points lying in a Euclidean ball.

\begin{lemma}
\label{Lemma 1}
Let $N \geq 1$ and $R \in \{1, \dots, N\}$. Let $\Lambda \subset \mathbb{R}^N$ be a lattice of rank $R$. Let also $Y>0$ be such that the ball $\mathcal{B}_N(Y)$ contains $R$ linearly independent vectors of the lattice $\Lambda$. For $T \geq Y$, we have
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) = V_R \frac{T^R}{\det (\Lambda)} \left( 1 + O \left( \frac{Y}{T} \right) \right),
\end{equation*}
where the implied constant depends at most on $R$.
\end{lemma}

\begin{proof}
The result of Schmidt \cite[Lemma $2$]{MR0224562} states that
\begin{equation}
\label{Schmidt}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) = V_R \frac{T^R}{\det (\Lambda)} +
O \left( \sum_{i=1}^R \frac{T^{R-i}}{\lambda_1(\Lambda) \cdots \lambda_{R-i}(\Lambda)} \right).
\end{equation}
Using Minkowski's estimate \eqref{Estimate Minkowski} we thus deduce
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) = V_R \frac{T^R}{\det (\Lambda)}
\left( 1 + O \left( \sum_{i=1}^R \frac{\lambda_{R-i+1}(\Lambda) \cdots \lambda_{R}(\Lambda)}{T^i} \right) \right).
\end{equation*}
Recalling the definition \eqref{Definition successive} of the successive minima, we see that the assumption that the ball $\mathcal{B}_N(Y)$ contains $R$ linearly independent vectors of the lattice $\Lambda$ means exactly that we have $\lambda_i(\Lambda) \leq Y$ for any
$i \in \{1, \dots, R \}$. Therefore, we obtain
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) = V_R \frac{T^R}{\det (\Lambda)} \left( 1 + O \left( \sum_{i=1}^R \frac{Y^i}{T^i} \right) \right),
\end{equation*}
and the assumption $T \geq Y$ allows us to complete the proof.
\end{proof}

The following result is concerned with integral lattices. We note that in the present work we only use it in the case $R_0=R-2$ but this general version may be useful in other contexts.

\begin{lemma}
\label{Lemma 2}
Let $N \geq 1$ and $R \in \{1, \dots, N\}$. Let $\Lambda \subset \mathbb{R}^N$ be an integral lattice of rank $R$. Let also $Y \geq 1$ be such that the ball $\mathcal{B}_N(Y)$ contains $R$ linearly independent vectors of the lattice $\Lambda$. For any
$R_0 \in \{0, \dots, R-1 \}$ and $T \leq Y$, we have
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) \ll T^{R-R_0-1} \left( \frac{T Y^{R_0}}{\det (\Lambda)} + 1 \right),
\end{equation*}
where the implied constant depends at most on $R$.
\end{lemma}

\begin{proof}
We start by noting that the estimate \eqref{Schmidt} implies in particular that
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) \ll
\frac{T^R}{\det (\Lambda)} + \sum_{i=1}^R \frac{T^{R-i}}{\lambda_1(\Lambda) \cdots \lambda_{R-i}(\Lambda)}.
\end{equation*}
Using Minkowski's estimate \eqref{Estimate Minkowski} we see that for any $R_0 \in \{0, \dots, R-1 \}$, we have
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) \ll
\frac{T^R}{\det (\Lambda)} \left( 1 + \sum_{i=1}^{R_0} \frac{\lambda_{R-i+1}(\Lambda) \cdots \lambda_R(\Lambda)}{T^i} \right) + \sum_{i=R_0+1}^R \frac{T^{R-i}}{\lambda_1(\Lambda) \cdots \lambda_{R-i}(\Lambda)}.
\end{equation*}
By assumption, the lattice $\Lambda$ is integral so we have $\lambda_i(\Lambda) \geq 1$ for any $i \in \{1, \dots, R \}$. In addition, as already noted in the proof of Lemma~\ref{Lemma 1}, the assumption that the ball $\mathcal{B}_N(Y)$ contains $R$ linearly independent vectors of the lattice $\Lambda$ amounts to saying that $\lambda_i(\Lambda) \leq Y$ for any $i \in \{1, \dots, R \}$. As a result, we deduce
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) \ll \frac{T^R}{\det (\Lambda)} \left( 1 + \sum_{i=1}^{R_0} \frac{Y^i}{T^i} \right) +
\sum_{i=R_0+1}^R T^{R-i}.
\end{equation*}
Since $T \leq Y$, this finally gives
\begin{equation*}
\mathcal{N}(\Lambda; \mathcal{B}_N(T)) \ll \frac{T^{R-R_0} Y^{R_0}}{\det (\Lambda)} + T^{R-R_0-1},
\end{equation*}
which completes the proof.
\end{proof}

The next result applies to primitive lattices and will be particularly useful in the proof of Theorem~\ref{Theorem 2}.

\begin{lemma}
\label{Lemma 3}
Let $N \geq 3$ and $R \in \{3, \dots, N\}$. Let $\Lambda \subset \mathbb{R}^N$ be an integral and primitive lattice of rank $R$. Let also $Y \geq 1$ be such that the ball $\mathcal{B}_N(Y)$ contains $R$ linearly independent vectors of the lattice $\Lambda$. For $T \geq Y$, we have
\begin{equation*}
\mathcal{N}^{\ast}(\Lambda; \mathcal{B}_N(T)) =
\frac{V_R}{\zeta(R)} \cdot \frac{T^R}{\det (\Lambda)} \left( 1 + O \left( \frac{Y}{T} \right) \right) + O (T \log Y),
\end{equation*}
where the implied constants depend at most on $R$.
\end{lemma}

\begin{proof}
A M\"{o}bius inversion gives
\begin{equation*}
\mathcal{N}^{\ast}(\Lambda; \mathcal{B}_N(T)) = \sum_{\ell \leq T}
\mu(\ell) \left( \mathcal{N} \left( \left( \frac1{\ell} \cdot \Lambda \right) \cap \mathbb{Z}^N; \mathcal{B}_N \left( \frac{T}{\ell} \right) \right) - 1 \right),
\end{equation*}
where the term $-1$ accounts for the zero vector. Since $\Lambda$ is a primitive lattice, for any integer $ \ell \geq 1$ we have
\begin{equation*}
\left( \frac1{\ell} \cdot \Lambda \right) \cap \mathbb{Z}^N = \Lambda.
\end{equation*}
Splitting the summation over $\ell$ into two different summations depending on whether $\ell \leq T/Y$ or $T/Y < \ell \leq T$, we thus obtain
\begin{equation}
\label{Separate}
\begin{split}
\mathcal{N}^{\ast}(\Lambda; \mathcal{B}_N(T)) = & \
\sum_{\ell \leq T/Y} \mu(\ell) \mathcal{N} \left(\Lambda; \mathcal{B}_N \left(\frac{T}{\ell} \right) \right) \\
& + \sum_{T/Y < \ell \leq T} \mu(\ell) \mathcal{N} \left(\Lambda; \mathcal{B}_N \left(\frac{T}{\ell} \right) \right) + O(T).
\end{split}
\end{equation}

First, it follows from Lemma~\ref{Lemma 1} that
\begin{equation*}
\sum_{\ell \leq T/Y} \mu(\ell) \mathcal{N} \left(\Lambda; \mathcal{B}_N \left(\frac{T}{\ell} \right) \right) =
\sum_{\ell \leq T/Y} \mu(\ell) V_R \frac{T^R}{\ell^R \det (\Lambda)} \left( 1 + O \left( \frac{\ell Y}{T} \right) \right).
\end{equation*}
Since $R \geq 3$, we deduce
\begin{equation}
\label{Equation 1}
\sum_{\ell \leq T/Y} \mu(\ell) \mathcal{N} \left(\Lambda; \mathcal{B}_N \left(\frac{T}{\ell} \right) \right) =
\frac{V_R}{\zeta(R)} \cdot \frac{T^R}{\det (\Lambda)} \left( 1 + O \left( \frac{Y}{T} \right) \right).
\end{equation}
Next, an application of Lemma~\ref{Lemma 2} with $R_0=R-2$ yields
\begin{equation*}
\sum_{T/Y < \ell \leq T} \mu(\ell) \mathcal{N} \left(\Lambda; \mathcal{B}_N \left(\frac{T}{\ell} \right) \right) \ll
\sum_{T/Y < \ell \leq T} \left( \frac{T^2 Y^{R-2}}{\ell^2 \det (\Lambda)} + \frac{T}{\ell} \right).
\end{equation*}
Therefore, we get
\begin{equation*}
\sum_{T/Y < \ell \leq T} \mu(\ell) \mathcal{N} \left( \Lambda; \mathcal{B}_N \left(\frac{T}{\ell} \right) \right) \ll
\frac{T Y^{R-1}}{\det (\Lambda)} + T \log Y,
\end{equation*}
and since $T \geq Y$, this gives
\begin{equation}
\label{Equation 2}
\sum_{T/Y < \ell \leq T} \mu(\ell) \mathcal{N} \left( \Lambda; \mathcal{B}_N \left(\frac{T}{\ell} \right) \right) \ll
\frac{T^{R-1} Y}{\det (\Lambda)} + T \log Y.
\end{equation}
Combining the estimates \eqref{Separate} and \eqref{Equation 1} and the upper bound \eqref{Equation 2} allows us to complete the proof.
\end{proof}

\subsection{The determinant of orthogonal lattices}

If $\Lambda \subset \mathbb{R}^N$ is an integral lattice we define the lattice $\Lambda^{\perp}$ orthogonal to $\Lambda$ by
\begin{equation*}
\Lambda^{\perp} = \left\{ \mathbf{y} \in \mathbb{Z}^N :
\forall \mathbf{z} \in \Lambda \ \langle \mathbf{z}, \mathbf{y} \rangle = 0 \right\}.
\end{equation*}
In other words, $\Lambda^{\perp}$ is the set of all vectors with integral coordinates in the orthogonal complement of
$\operatorname{Span}_{\mathbb{Q}}(\Lambda)$. The lattice $\Lambda^{\perp}$ is thus primitive by definition. Moreover, if $\Lambda$ is an integral lattice of rank $R$ then $\Lambda^{\perp}$ has rank $N-R$. Indeed, if $(\mathbf{b}_1, \dots, \mathbf{b}_{N-R})$ is a $\mathbb{Q}$-basis of the orthogonal complement of $\operatorname{Span}_{\mathbb{Q}}(\Lambda)$ such that $\mathbf{b}_1, \dots, \mathbf{b}_{N-R} \in \mathbb{Z}^N$ then $\Lambda^{\perp}$ contains $\mathbb{Z} \mathbf{b}_1 \oplus \dots \oplus \mathbb{Z} \mathbf{b}_{N-R}$. Finally, we remark that if $\Lambda$ is a primitive lattice (see for instance \cite[Corollary of Lemma~$1$]{MR0224562}) then
\begin{equation}
\label{Equality det orthogonal}
\det (\Lambda^{\perp}) = \det (\Lambda).
\end{equation}

For $\mathbf{c} \in \mathbb{Z}^N$, it is convenient to set $\Lambda_{\mathbf{c}} = (\mathbb{Z} \mathbf{c})^{\perp}$, that is
\begin{equation*}
\Lambda_{\mathbf{c}} = \left\{ \mathbf{y} \in \mathbb{Z}^N : \langle \mathbf{c}, \mathbf{y} \rangle = 0 \right\}.
\end{equation*}
If $\Lambda$ is an integral lattice with a specified basis then the following lemma provides us with a formula for the determinant of the lattice
$\Lambda^{\perp}$ in terms of the given basis of $\Lambda$. We note that in this work we only use the case where $\Lambda$ has rank $1$ but this general result will be useful in future works.

\begin{lemma}
\label{Lemma 4}
Let $N \geq 1$ and $k \in \{1, \dots, N-1\}$. Let also $\mathbf{c}_1, \dots, \mathbf{c}_k \in \mathbb{Z}^N$ be linearly independent vectors. Then $\Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k}$ is a primitive lattice of rank $N - k$. Moreover, we have
\begin{equation*}
\det ( \Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k} ) =
\frac{\det (\mathbb{Z} \mathbf{c}_1 \oplus \dots \oplus \mathbb{Z} \mathbf{c}_k)}
{\mathcal{G}(\mathbf{c}_1, \dots, \mathbf{c}_k)},
\end{equation*}
where $\mathcal{G}(\mathbf{c}_1, \dots, \mathbf{c}_k)$ denotes the greatest common divisor of the $k \times k$ minors of the
$N \times k$ matrix whose columns are the vectors $\mathbf{c}_1, \dots, \mathbf{c}_k$.
\end{lemma}

\begin{proof}
By assumption, the lattice $\mathbb{Z} \mathbf{c}_1 \oplus \dots \oplus \mathbb{Z} \mathbf{c}_k$ has rank $k$, so the obvious equality
\begin{equation*}
( \mathbb{Z} \mathbf{c}_1 \oplus \dots \oplus \mathbb{Z} \mathbf{c}_k)^{\perp} =
\Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k}
\end{equation*}
shows that $\Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k}$ is a primitive lattice of rank $N-k$.

Therefore, the equality \eqref{Equality det orthogonal} gives
\begin{equation}
\label{Equality determinants}
\det ((\Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k})^{\perp}) =
\det ( \Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k}).
\end{equation}
Let $(\mathbf{d}_1, \dots, \mathbf{d}_k)$ be a basis of the lattice
$(\Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k})^{\perp}$. Let also $\mathbf{C}$ and
$\mathbf{D}$ be the $N \times k$ matrices whose columns are respectively the vectors $\mathbf{c}_1, \dots, \mathbf{c}_k$ and
$\mathbf{d}_1, \dots, \mathbf{d}_k$. Since the lattice $\mathbb{Z} \mathbf{c}_1 \oplus \dots \oplus \mathbb{Z} \mathbf{c}_k$ is a sublattice of $(\Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k})^{\perp}$, there exists a $k \times k$ matrix
$\mathbf{M}$ with integral entries such that $\mathbf{C} = \mathbf{D} \mathbf{M}$. Recalling the definition \eqref{Definition det} of the determinant of a lattice, we see that
\begin{align*}
\det (\mathbb{Z} \mathbf{c}_1 \oplus \dots \oplus \mathbb{Z} \mathbf{c}_k) & = \sqrt{\det (\mathbf{C}^T\mathbf{C})} \\
& = |\det(\mathbf{M})| \cdot \sqrt{\det (\mathbf{D}^T\mathbf{D})} \\
& = |\det(\mathbf{M})| \cdot \det ((\Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k})^{\perp}).
\end{align*}
As a result, the equality \eqref{Equality determinants} yields
\begin{equation}
\label{Equality M}
\det ( \Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k} ) =
\frac{\det (\mathbb{Z} \mathbf{c}_1 \oplus \dots \oplus \mathbb{Z} \mathbf{c}_k)}{|\det(\mathbf{M})|}.
\end{equation}

Now, for any $N \times k$ matrix $\mathbf{F}$ and any $(i_1, \dots, i_{N-k}) \in \{1, \dots, N \}^{N-k}$ satisfying
$i_1 < \dots < i_{N-k}$, we let $[\mathbf{F}]_{(i_1, \dots, i_{N-k})}$ be the $k \times k$ matrix formed by removing from $\mathbf{F}$ its rows of indices $i_1, \dots, i_{N-k}$. Since we have
\begin{equation*}
[\mathbf{C}]_{(i_1, \dots, i_{N-k})} = [\mathbf{D}]_{(i_1, \dots, i_{N-k})} \cdot \mathbf{M},
\end{equation*}
we deduce that
\begin{equation}
\label{Equality minors}
\det ( [\mathbf{C}]_{(i_1, \dots, i_{N-k})} ) = \det ([\mathbf{D}]_{(i_1, \dots, i_{N-k})}) \cdot \det (\mathbf{M}).
\end{equation}
But the lattice $(\Lambda_{\mathbf{c}_1} \cap \dots \cap \Lambda_{\mathbf{c}_k})^{\perp}$ is primitive so
\cite[Chapter I, Corollary~$3$]{MR1434478} implies that its basis $(\mathbf{d}_1, \dots, \mathbf{d}_k)$ can be extended into a basis of
$\mathbb{Z}^N$. Therefore, it follows from \cite[Chapter I, Lemma $2$]{MR1434478} that
$\mathcal{G}(\mathbf{d}_1, \dots, \mathbf{d}_k) = 1$ and the equality \eqref{Equality minors} gives
\begin{equation*}
\mathcal{G}(\mathbf{c}_1, \dots, \mathbf{c}_k) = |\det (\mathbf{M})|.
\end{equation*}
Recalling the equality \eqref{Equality M}, we see that this completes the proof.
\end{proof}

\subsection{The key combinatorial argument}

For $d \geq 1$ and $n \geq 0$, we introduce the Veronese embedding $\nu_{d,n} : \mathbb{R}^{n+1} \to \mathbb{R}^{N_{d,n}}$ defined by listing all the monomials of degree $d$ in $n+1$ variables using the lexicographical ordering. The following lemma is the key tool in the proof of Theorem~\ref{Theorem 2}.

\begin{lemma}
\label{Lemma 5}
Let $d \geq 1$ and $n \geq 0$. Let also $\mathbf{x} \in \mathbb{Z}_{\mathrm{prim}}^{n+1}$. Then the ball
$\mathcal{B}_{N_{d,n}}(||\mathbf{x}||)$ contains $N_{d,n}-1$ linearly independent vectors of the lattice
$\Lambda_{\nu_{d,n}(\mathbf{x})}$.
\end{lemma}

\begin{proof}
We proceed by induction on $n$. If $n=0$, the result is clear. Assume now that for some $n \geq 1$ the result holds for the integer $n-1$, and let $(x_0, \dots, x_n)$ denote the coordinates of the vector $\mathbf{x}$.

We first treat the case where $x_0 = 0$. We observe that the first $N_{d,n}-N_{d,n-1}$ coordinates of the vector $\nu_{d,n}(\mathbf{x})$ are equal to $0$. Therefore, the first $N_{d,n}-N_{d,n-1}$ vectors of the canonical basis of $\mathbb{R}^{N_{d,n}}$ belong to the lattice
$\Lambda_{\nu_{d,n}(\mathbf{x})}$. Moreover, they also belong to the ball $\mathcal{B}_{N_{d,n}}(||\mathbf{x}||)$ since
$\mathbf{x} \neq \boldsymbol{0}$. Let $E$ be the subspace of $\mathbb{R}^{N_{d,n}}$ defined by the vanishing of the first
$N_{d,n}-N_{d,n-1}$ coordinates. The induction hypothesis applied to the lattice $\Lambda_{\nu_{d,n}(\mathbf{x})} \cap E$ provides us with $N_{d,n-1} -1$ linearly independent vectors of this lattice in the ball $\mathcal{B}_{N_{d,n}}(||\mathbf{x}||)$. We thus obtain a family of $N_{d,n}-1$ linearly independent vectors of the lattice $\Lambda_{\nu_{d,n}(\mathbf{x})}$ belonging to the ball
$\mathcal{B}_{N_{d,n}}(||\mathbf{x}||)$, as desired.

We now deal with the case where $x_0 \neq 0$. For any monomial $P(T_0, \dots, T_n)$ different from $T_0^d$, we let
$i \in \{1, \dots, n\}$ be the least integer such that $T_i \mid P(T_0, \dots, T_n)$ and we define
\begin{equation*}
Q(T_0, \dots, T_n) = \frac{P(T_0, \dots, T_n)}{T_i} T_0.
\end{equation*}
It is crucial to note that $Q$ occurs before $P$ in the lexicographical ordering. We then let $\mathbf{v}(P) \in \mathbb{Z}^{N_{d,n}}$ be the vector with all coordinates equal to $0$, except the $Q$-coordinate equal to $-x_i$ and the $P$-coordinate equal to $x_0$. By construction, we have
\begin{equation*}
- x_i Q(\mathbf{x}) + x_0 P(\mathbf{x}) = 0,
\end{equation*}
which implies that $\mathbf{v}(P) \in \Lambda_{\nu_{d,n}(\mathbf{x})}$. Moreover, we clearly have $||\mathbf{v}(P)|| \leq ||\mathbf{x}||$ and thus $\mathbf{v}(P) \in \mathcal{B}_{N_{d,n}}(||\mathbf{x}||)$. Since we have associated a vector $\mathbf{v}(P)$ to each monomial $P$ except $T_0^d$, we see that we have constructed a family of $N_{d,n}-1$ vectors of the lattice $\Lambda_{\nu_{d,n}(\mathbf{x})}$ belonging to the ball $\mathcal{B}_{N_{d,n}}(||\mathbf{x}||)$. In addition, since $x_0 \neq 0$, the matrix of this family of vectors is in echelon form and so they are linearly independent. This completes the proof.
\end{proof}

\section{An average version of Manin's conjecture}

\label{Section Manin}

We now have all the tools we need to establish Theorem~\ref{Theorem 2}.

\begin{proof}[Proof of Theorem~\ref{Theorem 2}]
Recall the definition \eqref{Definition N(B)} of the counting function $N_V(B)$. We let
\begin{equation*}
S_{d,n}(A,B) = \sum_{ V \in \mathbb{V}_{d,n}(A)} N_V(B),
\end{equation*}
and we note that
\begin{equation*}
S_{d,n}(A,B) = \sum_{\substack{x \in \mathbb{P}^n(\mathbb{Q}) \\ H(x) \leq B}}
\# \left\{ V \in \mathbb{V}_{d,n}(A) : x \in V(\mathbb{Q}) \right\}.
\end{equation*}
It is convenient to set
\begin{equation*}
\Xi_{d,n}(B) = \left\{ \mathbf{x} \in \mathbb{Z}_{\mathrm{prim}}^{n+1} : ||\mathbf{x}|| \leq B^{1/(n+1-d)} \right\}.
\end{equation*}
Recalling the definition \eqref{Definition N star} of the quantity
$\mathcal{N}^{\ast}(\Lambda_{\nu_{d,n}(\mathbf{x})}; \mathcal{B}_{N_{d,n}}(A))$, we remark that
\begin{equation*}
S_{d,n}(A,B) =
\frac1{4} \sum_{\mathbf{x} \in \Xi_{d,n}(B)} \mathcal{N}^{\ast}(\Lambda_{\nu_{d,n}(\mathbf{x})}; \mathcal{B}_{N_{d,n}}(A)).
\end{equation*}

We have $||\mathbf{x}|| \leq B^{1/(n+1-d)}$, so Lemma~\ref{Lemma 5} shows that the ball
$\mathcal{B}_{N_{d,n}} \left(B^{1/(n+1-d)}\right)$ contains $N_{d,n}-1$ linearly independent vectors of the lattice
$\Lambda_{\nu_{d,n}(\mathbf{x})}$. In addition, $\Lambda_{\nu_{d,n}(\mathbf{x})}$ is a primitive lattice and by assumption
$A \geq B^{1/(n+1-d)}$. We can thus apply Lemma~\ref{Lemma 3} to deduce that
\begin{align*}
\mathcal{N}^{\ast}(\Lambda_{\nu_{d,n}(\mathbf{x})}; \mathcal{B}_{N_{d,n}}(A)) = & \ \frac{V_{N_{d,n}-1}}{\zeta(N_{d,n}-1)} \cdot
\frac{A^{N_{d,n}-1}}{\det (\Lambda_{\nu_{d,n}(\mathbf{x})})} \left( 1 + O \left( \frac{B^{1/(n+1-d)}}{A} \right) \right) \\
& + O (A \log B).
\end{align*}
Moreover, using the fact that $\mathbf{x} \in \mathbb{Z}_{\mathrm{prim}}^{n+1}$ we see that Lemma~\ref{Lemma 4} gives
\begin{equation*}
\det (\Lambda_{\nu_{d,n}(\mathbf{x})}) = ||\nu_{d,n}(\mathbf{x})||,
\end{equation*}
and it follows that
\begin{equation*}
\det (\Lambda_{\nu_{d,n}(\mathbf{x})}) \ll B^{d/(n+1-d)}.
\end{equation*}
Hence, our assumption $A \geq B^{1/(n+1-d)}$ implies in particular that
\begin{equation*}
\frac{A^{N_{d,n}-1}}{\det (\Lambda_{\nu_{d,n}(\mathbf{x})})} \cdot \frac{B^{1/(n+1-d)}}{A} \gg A \log B.
\end{equation*}
As a result, we obtain
\begin{equation*}
S_{d,n}(A,B) = \frac{V_{N_{d,n}-1}}{4\zeta(N_{d,n}-1)} A^{N_{d,n}-1} T_{d,n}(B)
\left( 1 + O \left( \frac{B^{1/(n+1-d)}}{A} \right) \right),
\end{equation*}
where we have set
\begin{equation*}
T_{d,n}(B) = \sum_{\mathbf{x} \in \Xi_{d,n}(B)} \frac1{||\nu_{d,n}(\mathbf{x})||}.
\end{equation*}
Furthermore, an elementary application of Lemma~\ref{Lemma 3} yields
\begin{equation*}
\# \mathbb{V}_{d,n}(A) = \frac{V_{N_{d,n}}}{2 \zeta(N_{d,n})} A^{N_{d,n}} \left(1 + O \left(\frac1{A} \right) \right).
\end{equation*}
Therefore, we see that
\begin{equation}
\label{Estimate S}
S_{d,n}(A,B) = \gamma_{d,n} \frac{\# \mathbb{V}_{d,n}(A)}{A} T_{d,n}(B)
\left( 1 + O \left( \frac{B^{1/(n+1-d)}}{A} \right) \right),
\end{equation}
where
\begin{equation*}
\gamma_{d,n} = \frac1{2} \cdot \frac{V_{N_{d,n} - 1}}{V_{N_{d,n}}} \cdot \frac{\zeta(N_{d,n})}{\zeta(N_{d,n}-1)}.
\end{equation*}

Our next task is to establish an asymptotic formula for the quantity $T_{d,n}(B)$. First, a M\"{o}bius inversion yields
\begin{equation*}
T_{d,n}(B) = \sum_{\ell \leq B^{1/(n+1-d)}} \frac{\mu(\ell)}{\ell^d}
\sum_{\substack{\mathbf{z} \in \mathbb{Z}^{n+1} \smallsetminus \{ \boldsymbol{0}\} \\ ||\mathbf{z}|| \leq B^{1/(n+1-d)}/\ell}} \frac1{||\nu_{d,n}(\mathbf{z})||}.
\end{equation*}
We now check that for $Z \geq 1$, we have
\begin{equation}
\label{Goal T(B)}
\sum_{\substack{\mathbf{z} \in \mathbb{Z}^{n+1} \smallsetminus \{ \boldsymbol{0}\} \\ ||\mathbf{z}|| \leq Z}}
\frac1{||\nu_{d,n}(\mathbf{z})||} = \left( \int_{\mathcal{B}_{n+1}(1)} \frac{\mathrm{d} \mathbf{t}}{||\nu_{d,n}(\mathbf{t})||} \right) Z^{n+1-d}
\left( 1 + O \left( \frac{\log Z}{Z} \right) \right),
\end{equation}
where the factor $\log Z$ in the error term is only necessary if $n = d$. Indeed, an application of partial summation gives
\begin{equation*}
\sum_{\substack{\mathbf{z} \in \mathbb{Z}^{n+1} \smallsetminus \{ \boldsymbol{0}\} \\ ||\mathbf{z}|| \leq Z}}
\frac1{||\nu_{d,n}(\mathbf{z})||} = \int_{0}^{\infty} \# \left\{ \mathbf{z} \in \mathbb{Z}^{n+1} \smallsetminus \{ \boldsymbol{0} \} :
\begin{array}{l l}
||\mathbf{z}|| \leq Z \\
||\nu_{d,n}(\mathbf{z})|| \leq u
\end{array}
\right\} \frac{\mathrm{d} u}{u^2}.
\end{equation*}
Furthermore, for $u \geq 1$ it follows from the work of Davenport \cite{MR43821} that
\begin{align*}
\# \left\{ \mathbf{z} \in \mathbb{Z}^{n+1} \smallsetminus \{ \boldsymbol{0} \} :
\begin{array}{l l}
||\mathbf{z}|| \leq Z \\
||\nu_{d,n}(\mathbf{z})|| \leq u
\end{array}
\right\} = & \ \operatorname{vol} \left( \left\{ \mathbf{v} \in \mathbb{R}^{n+1} :
\begin{array}{l l}
||\mathbf{v}|| \leq Z \\
||\nu_{d,n}(\mathbf{v})|| \leq u
\end{array}
\right\} \right) \\
& + O \left( \min \left\{Z^n, u^{n/d} \right\} \right).
\end{align*}
In addition, if $u < 1$ we clearly have
\begin{align*}
\# \left\{ \mathbf{z} \in \mathbb{Z}^{n+1} \smallsetminus \{ \boldsymbol{0} \} :
\begin{array}{l l}
||\mathbf{z}|| \leq Z \\
||\nu_{d,n}(\mathbf{z})|| \leq u
\end{array}
\right\} = & \ \operatorname{vol} \left( \left\{ \mathbf{v} \in \mathbb{R}^{n+1} :
\begin{array}{l l}
||\mathbf{v}|| \leq Z \\
||\nu_{d,n}(\mathbf{v})|| \leq u
\end{array}
\right\} \right) \\
& + O \left( u^{(n+1)/d} \right).
\end{align*}
Therefore, we get
\begin{equation*}
\sum_{\substack{\mathbf{z} \in \mathbb{Z}^{n+1} \smallsetminus \{ \boldsymbol{0}\} \\ ||\mathbf{z}|| \leq Z}}
\frac1{||\nu_{d,n}(\mathbf{z})||} = \int_{\mathcal{B}_{n+1}(Z)} \left( \int_{||\nu_{d,n}(\mathbf{v})||}^{\infty} \frac{\mathrm{d} u}{u^2} \right)
\mathrm{d} \mathbf{v} + O \left( \mathcal{E}_{d,n}(Z) \right),
\end{equation*}
where
\begin{equation*}
\mathcal{E}_{d,n}(Z) = \int_{0}^{1} \frac{\mathrm{d} u}{u^{2-(n+1)/d}} + \int_{1}^{\infty} \min \left\{ Z^n, u^{n/d} \right\} \frac{\mathrm{d} u}{u^2}.
\end{equation*}
Moreover, an elementary calculation provides
\begin{equation*}
\mathcal{E}_{d,n}(Z) \ll
\begin{cases}
\log Z, & \textrm{if } n=d, \\
Z^{n-d}, & \textrm{if } n > d.
\end{cases}
\end{equation*}
We thus derive in particular
\begin{equation*}
\sum_{\substack{\mathbf{z} \in \mathbb{Z}^{n+1} \smallsetminus \{ \boldsymbol{0}\} \\ ||\mathbf{z}|| \leq Z}}
\frac1{||\nu_{d,n}(\mathbf{z})||} = \int_{\mathcal{B}_{n+1}(Z)} \frac{\mathrm{d} \mathbf{v}}{||\nu_{d,n}(\mathbf{v})||}
+ O \left( Z^{n-d} \log Z \right).
\end{equation*}
The equality \eqref{Goal T(B)} follows since the change of variables $\mathbf{v} = Z \mathbf{t}$ gives
\begin{equation*}
\int_{\mathcal{B}_{n+1}(Z)} \frac{\mathrm{d} \mathbf{v}}{||\nu_{d,n}(\mathbf{v})||} = \left( \int_{\mathcal{B}_{n+1}(1)} \frac{\mathrm{d} \mathbf{t}}{||\nu_{d,n}(\mathbf{t})||} \right) Z^{n+1-d}.
\end{equation*}
As a result, we deduce
\begin{equation*}
T_{d,n}(B) = \left(\int_{\mathcal{B}_{n+1}(1)} \frac{\mathrm{d} \mathbf{t}}{||\nu_{d,n}(\mathbf{t})||} \right) B
\sum_{\ell \leq B^{1/(n+1-d)}} \frac{\mu(\ell)}{\ell^{n+1}} \left( 1 + O \left( \frac{\ell \log B}{B^{1/(n+1-d)}} \right) \right),
\end{equation*}
which eventually gives
\begin{equation*}
T_{d,n}(B) = \frac1{\zeta(n+1)} \left(\int_{\mathcal{B}_{n+1}(1)} \frac{\mathrm{d} \mathbf{t}}{||\nu_{d,n}(\mathbf{t})||} \right) B
\left( 1 + O \left( \frac{\log B}{B^{1/(n+1-d)}} \right) \right).
\end{equation*}

Recalling the estimate \eqref{Estimate S}, we see that we have proved that
\begin{equation*}
S_{d,n}(A,B) = C_{d,n} \# \mathbb{V}_{d,n}(A) \frac{B}{A}
\left( 1 + O \left( \frac{B^{1/(n+1-d)}}{A} + \frac{\log B}{B^{1/(n+1-d)}} \right) \right),
\end{equation*}
where
\begin{equation*}
C_{d,n} = \frac1{2 \zeta(n+1)} \cdot \frac{V_{N_{d,n} - 1}}{V_{N_{d,n}}} \cdot \frac{\zeta(N_{d,n})}{\zeta(N_{d,n}-1)} \cdot
\int_{\mathcal{B}_{n+1}(1)} \frac{\mathrm{d} \mathbf{t}}{||\nu_{d,n}(\mathbf{t})||},
\end{equation*}
which completes the proof of Theorem~\ref{Theorem 2}.
\end{proof}

\section{The smallest height of a rational point}

\label{Section smallest height}

In this section we show that Theorems~\ref{Theorem 1} and~\ref{Theorem 1'} follow from Theorem~\ref{Theorem 2}. We start with the proof of Theorem~\ref{Theorem 1}.

\begin{proof}[Proof of Theorem~\ref{Theorem 1}]
Our aim is to prove that
\begin{equation}
\label{Goal Theorem 1}
\lim_{A \to \infty} \frac{\# \left\{ V \in \mathbb{V}_{d,n}(A) : \mathfrak{M}(V) \leq \psi(||\mathbf{a}_V||) \right\}}
{\# \mathbb{V}_{d,n}(A)} = 0.
\end{equation}
Let $\eta \in (0,1)$. By assumption, if $A$ is sufficiently large then for any $u \geq A^{1/2}$, the inequality
$\psi(u) \leq \eta u$ holds. Since we have
\begin{align*}
\# \left\{ V \in \mathbb{V}_{d,n}(A) : \mathfrak{M}(V) \leq \psi(||\mathbf{a}_V||) \right\} = & \
\# \left\{ V \in \mathbb{V}_{d,n}(A) :
\begin{array}{l l}
||\mathbf{a}_V|| > A^{1/2} \\
\mathfrak{M}(V) \leq \psi(||\mathbf{a}_V||)
\end{array} \right\} \\
& + O\left(A^{N_{d,n}/2}\right),
\end{align*}
we deduce
\begin{equation*}
\# \left\{ V \in \mathbb{V}_{d,n}(A) : \mathfrak{M}(V) \leq \psi(||\mathbf{a}_V||) \right\} \ll
\# \left\{ V \in \mathbb{V}_{d,n}(A) : \mathfrak{M}(V) \leq \eta A \right\} + A^{N_{d,n}/2}.
\end{equation*}
Moreover, we clearly have
\begin{equation*}
\# \left\{ V \in \mathbb{V}_{d,n}(A) : \mathfrak{M}(V) \leq \eta A \right\} \leq
\sum_{ V \in \mathbb{V}_{d,n}(A)} N_V(\eta A).
\end{equation*}
Since $A \geq (\eta A)^{1/(n+1-d)}$ we can apply Theorem~\ref{Theorem 2}. We deduce in particular that
\begin{equation*}
\sum_{ V \in \mathbb{V}_{d,n}(A)} N_V(\eta A) \ll \eta \cdot \# \mathbb{V}_{d,n}(A).
\end{equation*}
As a result, we eventually obtain
\begin{equation*}
\limsup_{A \to \infty} \frac{\# \left\{ V \in \mathbb{V}_{d,n}(A) : \mathfrak{M}(V) \leq \psi(||\mathbf{a}_V||) \right\}}{\# \mathbb{V}_{d,n}(A)} \ll \eta.
\end{equation*}
This upper bound holds for any $\eta \in (0,1)$ and the equality \eqref{Goal Theorem 1} thus follows, which completes the proof of
Theorem~\ref{Theorem 1}.
\end{proof}

We now furnish the proof of Theorem~\ref{Theorem 1'}.

\begin{proof}[Proof of Theorem~\ref{Theorem 1'}]
It follows from the work of Davenport \cite{MR86105} that for any $V \in \mathbb{V}_{2,2}(A)$, we have
$\mathfrak{M}(V) \leq 3 ||\mathbf{a}_V||$ if and only if $V(\mathbb{Q}) \neq \emptyset$. Therefore, we observe that our aim is to prove that
\begin{equation}
\label{Goal Theorem 1'}
\lim_{A \to \infty} \frac{\# \left\{ V \in \mathbb{V}_{2,2}(A) : \mathfrak{M}(V) \leq \psi(||\mathbf{a}_V||) \right\}}
{\# \left\{ V \in \mathbb{V}_{2,2}(A) : V(\mathbb{Q}) \neq \emptyset \right\}} = 0.
\end{equation}
Let $\eta \in (0,1)$. Proceeding as in the proof of Theorem~\ref{Theorem 1}, we obtain
\begin{equation}
\label{Upper bound 1}
\# \left\{ V \in \mathbb{V}_{2,2}(A) : \mathfrak{M}(V) \leq \psi(||\mathbf{a}_V||) \right\} \ll
\sum_{ V \in \mathbb{V}_{2,2}(A)} N_V \left(\eta \frac{A}{(\log A)^{1/2}} \right) + A^3.
\end{equation}
Moreover, we have $A \geq \eta A/(\log A)^{1/2}$ so we can apply Theorem~\ref{Theorem 2}. We deduce in particular that
\begin{equation}
\label{Upper bound 2}
\sum_{ V \in \mathbb{V}_{2,2}(A)} N_V \left(\eta \frac{A}{(\log A)^{1/2}} \right) \ll
\eta \frac{\# \mathbb{V}_{2,2}(A)}{(\log A)^{1/2}}.
\end{equation}

In addition, the result of Hooley \cite[Theorem]{MR2300456} can be rewritten as
\begin{equation*}
\sum_{\substack{V \in \mathbb{V}_{2,2}(A) \\ V(\mathbb{Q}) \neq \emptyset}} \left\lfloor \frac{A}{||\mathbf{a}_V||} \right\rfloor \gg
\frac{\# \mathbb{V}_{2,2}(A)}{(\log A)^{1/2}}.
\end{equation*}
For $\varepsilon \in (1/A^{1/2},1)$, we split the summation over $V$ depending on whether $||\mathbf{a}_V|| \leq \varepsilon A$ or
$||\mathbf{a}_V|| > \varepsilon A$. We get
\begin{equation*}
\frac{\# \mathbb{V}_{2,2}(A)}{(\log A)^{1/2}} \ll A
\sum_{\substack{V \in \mathbb{V}_{2,2}(\varepsilon A) \\ V(\mathbb{Q}) \neq \emptyset}} \frac1{||\mathbf{a}_V||} + \frac1{\varepsilon} \cdot \# \left\{ V \in \mathbb{V}_{2,2}(A) \smallsetminus \mathbb{V}_{2,2}(\varepsilon A) : V(\mathbb{Q}) \neq \emptyset \right\}.
\end{equation*}
Furthermore, an application of partial summation yields
\begin{align*}
\sum_{\substack{V \in \mathbb{V}_{2,2}(\varepsilon A) \\ V(\mathbb{Q}) \neq \emptyset}} \frac1{||\mathbf{a}_V||} = & \
\frac1{\varepsilon A} \cdot \# \left\{ V \in \mathbb{V}_{2,2}(\varepsilon A) : V(\mathbb{Q}) \neq \emptyset \right\} \\
& + \int_1^{\varepsilon A} \frac{\# \left\{ V \in \mathbb{V}_{2,2}(t) : V(\mathbb{Q}) \neq \emptyset \right\}}{t^2} \mathrm{d} t.
\end{align*}
We thus obtain
\begin{equation*}
\frac{\# \mathbb{V}_{2,2}(A)}{(\log A)^{1/2}} \ll
\frac1{\varepsilon} \cdot \# \left\{ V \in \mathbb{V}_{2,2}(A) : V(\mathbb{Q}) \neq \emptyset \right\}
+ A \int_1^{\varepsilon A} \frac{\# \left\{ V \in \mathbb{V}_{2,2}(t) : V(\mathbb{Q}) \neq \emptyset \right\}}{t^2} \mathrm{d} t.
\end{equation*}
Moreover, it follows from the upper bound of Serre \cite[Exemple $4$]{MR1075658} that
\begin{equation*}
A \int_1^{\varepsilon A} \frac{\# \left\{ V \in \mathbb{V}_{2,2}(t) : V(\mathbb{Q}) \neq \emptyset \right\}}{t^2} \mathrm{d} t \ll
\varepsilon^5 \frac{\# \mathbb{V}_{2,2}(A)}{(\log A)^{1/2}}.
\end{equation*}
Hence, we get
\begin{equation*}
\frac{\# \mathbb{V}_{2,2}(A)}{(\log A)^{1/2}} \ll
\frac1{\varepsilon} \cdot \# \left\{ V \in \mathbb{V}_{2,2}(A) : V(\mathbb{Q}) \neq \emptyset \right\}
+ \varepsilon^5 \frac{\# \mathbb{V}_{2,2}(A)}{(\log A)^{1/2}}.
\end{equation*}
As a result, if $A$ is sufficiently large then by choosing $\varepsilon$ small enough independently of $A$, we derive
\begin{equation*}
\# \left\{ V \in \mathbb{V}_{2,2}(A) : V(\mathbb{Q}) \neq \emptyset \right\} \gg \frac{\# \mathbb{V}_{2,2}(A)}{(\log A)^{1/2}}.
\end{equation*}

Recalling the upper bounds \eqref{Upper bound 1} and \eqref{Upper bound 2}, we eventually deduce
\begin{equation*}
\limsup_{A \to \infty} \frac{\# \left\{ V \in \mathbb{V}_{2,2}(A) : \mathfrak{M}(V) \leq \psi(||\mathbf{a}_V||) \right\}}
{\# \left\{ V \in \mathbb{V}_{2,2}(A) : V(\mathbb{Q}) \neq \emptyset \right\}} \ll \eta.
\end{equation*}
This upper bound is valid for any $\eta \in (0,1)$ so the equality \eqref{Goal Theorem 1'} follows, which completes the proof of
Theorem~\ref{Theorem 1'}.
\end{proof}

\bibliographystyle{amsalpha}

redpajama_set_name: RedPajamaArXiv


Does your cat like to squeeze into the fruit bowl when no one's watching? Do they twitch all through their body during sleep? And what's their take on cardboard boxes? We could keep going with the oddball questions, but chances are, you've either asked them yourself already – or you've got some similarly confusing cat behaviors happening at home. While most feline antics are perfectly normal, sometimes curious behaviors signal problems on the horizon.
Clicking – When cats see prey that they want to chase and/or sink their teeth into, but can't, this clicking, chattering noise allows them to vent their frustration. Since their hunting skills are so deeply rooted, when they have no way to channel these instincts, this endearing noise is made.
Kneading – When cats "make biscuits" they are reliving their first few weeks of life. Nursing was made possible by pressing on their mother's abdomen, and the reward was sweet, life-sustaining milk. When grown cats do this, they are self-soothing and tapping into those satisfying early moments.
Head Butting – Quite possibly one of the most affectionate cat behaviors, head butting or "bunting" is a way to say hello to a favorite person. Cats will not do this with just anyone.
Table Manners – Cats enjoy high places because the view is incomparable. Whether it's the counter or table top, cats tend to find the highest horizontal surface and claim it as their own. Depending on your point of view, this can be cute or irritating. What's worse is that cats also enjoy batting things (like glasses full of water) off the surface, seemingly just to watch it fall, break, and splatter everywhere. Why do they do this? Because they can! They might also be vying for your attention. Please be watchful of your cat at great heights; they may not always land on their feet.
"Presents" – Many cat owners can attest to this quirky feline trait. Cats are known for bringing their owners treats of the wild, that is, rodents, lizards, birds, etc. for their snacking pleasure. Whether this is because cats think we cannot hunt, or they do it as a token of affection, it's one of the most unsettling cat behaviors.
Some cat behaviors can actually foretell dangerous conditions. For example, it might be amusing to find a cat gravitating to water, but if they're consistently hanging around the sink, water fountain, hose, or toilet, it's important to rule out diabetes or kidney disease.
Likewise, a cat that yowls regularly throughout the night might just be burning off some energy, or they might actually be in pain. Hacking up hairballs is a normal feline action, but they shouldn't be ignored if they persist.
Cats age faster than we do, and they mask their symptoms. Daily observation of cat behaviors, coupled with routine preventive care exams, reinforce a proactive approach to health and wellness.
If our team at Oakhurst Veterinary Hospital can assist you with any questions or concerns, we hope you'll let us know.


la dolce vita: italy rome mission.
there are two missions in italy, the milan mission covers the north, and the rome mission in the south. sorella wightman is serving in the italy rome mission.
michael grant waddoups (64) and anna kay nield waddoups have two children and are from the west jordan utah westland stake. brother waddoups serves as a primary teacher and ward missionary and is a former counselor in a stake presidency, high councilor, bishop, temple preparation and marriard relations teacher and missionary in the italy mission. he is formerly the president of the utah state senate. born in idaho falls, idaho, to grant england waddoups and mary keith blattner johnson.
sister waddoups serves as a primary teacher and is a former temple ordinance worker, stake yound women president, ward primary president, ward young women president and relief society president. she was born in afton, wyoming to willard earl and lorna hurd neild.


text:
Hitler, Stalin and I: An Oral History
The oral history of a renowned Czech writer, whose optimism and faith in people survived grueling experiences under authoritarian regimes.
Heda Margolius Kov ly (1919-2010) was a renowned Czech writer and translator born to Jewish parents. Her bestselling memoir, Under a Cruel Star: A Life in Prague, 1941-1968 has been translated into more than a dozen languages. Her crime novel Innocence; or, Murder on Steep Street--based on her own experiences living under Stalinist oppression--was named an NPR Best Book in 2015.
In the tradition of Studs Terkel, Hitler, Stalin and I is based on interviews between Kov ly and award-winning filmmaker Helena Trest kov . In it, Kov ly recounts her family history in Czechoslovakia, starving in the deprivations of Lodz Ghetto, how she miraculously left Auschwitz, fled from a death march, failed to find sanctuary amongst former friends in Prague as a concentration camp escapee, and participated in the liberation of Prague. Later under Communist rule, she suffered extreme social isolation as a pariah after her first husband Rudolf Margolius was unjustly accused in the infamous Sl nsky Trial and executed for treason. Remarkably, Kov ly, exiled in the United States after the Warsaw Pact invasion in 1968, only had love for her country and continued to believe in its people. She returned to Prague in 1996.
Heda had an enormous talent for expressing herself. She spoke with precision and was descriptive and witty in places. I admired her attitude and composure, even after she had such extremely difficult experiences. Nazism and Communism afflicted Heda's life directly with maximum intensity. Nevertheless, she remained an optimist.
Helena Trest kov has made over fifty documentary films. Hitler, Stalin and I has garnered several awards in the Czech Republic and Japan.
Author: Heda Margolius Kovály
Publisher: Doppelhouse Press
Review Citation(s):
Shelf Awareness 03/13/2018
Heda Margolius Kovály (1919-2010), a Czech writer and translator, was born in Prague to Jewish parents. Heda spent the years of the Second World War in a ghetto, Auschwitz and other concentration camps, escaped from a death march, and took part in the Prague uprising against the Nazis in May 1945. After the war Heda worked at various Prague publishing houses as a graphic designer. In 1952, her first husband, Rudolf Margolius (1913-1952), was convicted in the Stalinist Slánský Trial. The 1968 Soviet occupation of Czechoslovakia forced Heda into exile in the United States. She translated over two dozen books and her celebrated memoir, Under a Cruel Star, was first published in 1973 and has since been translated into many languages. Her crime novel Innocence appeared in Czech in 1985 and in English in 2015. Heda returned to Prague in 1996 where she died.
Helena Trestíková is a documentary film director born in Prague and studied at the Prague Film and Television School of the Academy of Performing Arts. Since 1974, Helena has made over fifty documentary films mostly on the themes of long term human relationships and was awarded number of prizes including the European Film Academy 2008 - Prix Arte. The film Hitler, Stalin and I, based on her interview with Heda, was first shown on Czech television in 2001 and subsequently received the Festival Award Special Commendation at the 2002 Japan Film Festival; the ELSA award for the best Czech TV documentary film by the Czech Film and Television Academy in 2002, and the Gold Kingfisher award for the best documentary film at the Festival of Czech Films, Plzen in 2003.
Ivan Margolius is an architect, translator and author of memoirs, books and articles on art, architecture, engineering, design and automobile history. Ivan, son of Rudolf and Heda, was born in Prague, where he studied architecture and in London following his arrival to the United Kingdom in 1966. He practiced architecture at Foster and Partners, Skidmore, Owings and Merrill, Yorke Rosenberg Mardall and collaborated on books with Norman Foster, Richard Rogers and Jan Kaplický.

redpajama_set_name: RedPajamaCommonCrawl


Our friends at Savarin are Hudson Valley based creatives who wanted to pursue the simple idea that the American flag can be presented as a work of art. They set out to recreate the flag with a meticulous eye for the finer details in a way that it could exist in any home and compliment any interior. They also extended their manufacturing to small pennants representing the community and towns that make up the Hudson Valley as well as custom designs.
All of their materials are made and sourced in the U.S. including Woolrich located in Woolrich, PA. Established in 1830, Woolrich is the longest continually operating woolen mill in the U.S. Savarin's leather work is sourced from Horween Leather Co. in Chicago.(est 1905) and the copper rivets are made by C.S. Osborne (est. 1826) in Harrison, NJ. For his t-shirts the organic grown cotton hails from Shamrock, TX. from a manufacturer called S.O.S. Texas from Shamrock, TX. U.S. Organic grown and made cotton t-shirts.


You're prepared to buy a home. You know what you are able to afford. You have figured out exactly what your ideal mortgage transaction is going to be. Now you have to choose between working with a realtor plus looking at pre-owned homes or even working with a builder upon new homes. A new house is advantageous simply because it can just that - new. There are some reasons it's better.
Many people avoid look at new homes as there is a stigma out there exactly where people believe they are more costly than pre-owned homes. In most cases, this is just the opposite. Brand new home prices are established by builders. They are not attached with the homes - these people simply set the price based on market. Pre-owned home costs are set by proprietors. They are attached to the houses - and many believe their own homes are worth a lot more than they are because it can "theirs. "
A variety of homebuilders have price incentives to truly get you into the door. When you buy new homes, the contractor may be the one paying a few or all of the closing expenses - which can save you lots of money.
In case you shop pre-owned homes, you might be limited on location. You might find a home in a community that you want, but you may not like the house where the house resides. The particular pickier you get (gated local community, cul-de-sac, etc . ) the particular harder it is to find what you need. With new homes, you select a community that a builder will be working in. You can then select your house site from what is accessible.
Along with pre-owned homes, you are restricted because the previous owners possess selected everything for you. If you need to change something, you are going to need to pay out of pocket to change this. With new homes, you can make all of the selections. A few of the things you will get to pick out are the carpets, the cabinets as well as the countertops. Some builders may also allow you to choose paint colour, floor plans and much more.
Regardless of whether you have issues in the very first few years or not, there are likely to be things that have taken location within the house that you don't learn about. With new homes, you might be the original owner so you learn about the entire history of the home once you move in.
New homes are usually better than pre-owned homes to get a variety of reasons, as you can see. Before you decide to ignore homebuilders because you believe they are out of your price range, reconsider.


text:
158 W. Food Corp. company details
C/o Kwan Young Kim
Kwan Young Kim
Fortlee
158 W. Food Corp. is a firm based at 158 W 44Th Street, New York. It was registered as a domestic business corporation in New York county, under id number 2409705 on 08/17/1999, 22 years ago.
The business is, at present, controlled by a manager - Kwan Young Kim from Fortlee.
This company is licensed by the DCA as a tobacco retail dealer and is located at 158, W 44Th St, New York, Ny (40° 45.406' N 73° 59.08' W). You can contact the company by calling (212)944-1234.
158 W. Food Corp.
W 44th St
We gathered reports about 8 Department of Consumer Affairs inspections. A DCA inspection (industry: grocery-retail) was finished on 11/15/2019 - the outcome was "pass". There was also an inspection on 11/15/2019 (industry: grocery-retail) that ended with "a violation issued" result. Finally, there was an inspection on 02/07/2018 (industry: grocery-retail) that ended with "a violation issued" result.
11/15/2019 Manhattan 158 W 44th St
NY 10036 158 W. Food Corp. Grocery-Retail - 808 Pass
Tai Chi Chuan Center, Inc.
Foreign Not-for-profit Corporation
Dis Research Ltd.
Silk Realty Group LLC
Coa 99 Hudson, LLC
Search Office Space (sos) North America, Inc.


text:
#ifndef __JUCE_HEADER_BDEAC9DB91241E30__
#define __JUCE_HEADER_BDEAC9DB91241E30__

//[Headers]     -- You can add your own extra header files here --
#include "JuceHeader.h"
#include <functional>
//[/Headers]

//==============================================================================
/**
                                                                    //[Comments]
    An auto-generated component, created by the Introjucer.

    Describe your class and how it works here!
                                                                    //[/Comments]
*/
class HomeComponent  : public Component,
                       public ButtonListener,
                       public ComboBoxListener
{
public:
    //==============================================================================
    HomeComponent (std::function <void()> listButtonCallback, std::function <void(int)> comboBoxCallback);
    ~HomeComponent();

    //==============================================================================
    //[UserMethods]     -- You can add your own custom methods in this section.
    typedef std::function <void()> ListButtonCallbackFunction;
    //[/UserMethods]

    void paint (Graphics& g);
    void resized();
    void buttonClicked (Button* buttonThatWasClicked);
    void comboBoxChanged (ComboBox* comboBoxThatHasChanged);

private:
    //[UserVariables]   -- You can add your own custom variables in this section.
    std::function <void()> listButtonCallback;
    std::function <void(int)> comboBoxCallback;
    //[/UserVariables]

    //==============================================================================
    ScopedPointer<TextButton> listButton;
    ScopedPointer<ComboBox> comboBox;

    //==============================================================================
    JUCE_DECLARE_NON_COPYABLE_WITH_LEAK_DETECTOR (HomeComponent)
};

//[EndFile] You can add extra defines here...
//[/EndFile]

#endif   // __JUCE_HEADER_BDEAC9DB91241E30__

redpajama_set_name: RedPajamaGithub


text:
Russia is testing its disconnection from the global internet
by Nicholas Flynn
Cybersecurity exercises were conducted, according to the Russian newspaper Kommersantwhich evokes the scenario of a defensive initiative decided by Moscow or that of a measure imposed from outside.
Correspondent in Moscow
In the context of the war in Ukraine, and therefore of increased penalties but also the multiplication of cyberattacks, Moscow is preparing for its possible disconnection from the global internet, whether it is a defensive initiative decided by the Russian authorities or a measure imposed from outside.
An isolation of the Russian Internet from the rest of the world was considered by the Kremlin last March, after the outbreak of the "special military operationin Ukraine, according to the newspaper Kommersant. It also reveals that several cybersecurity exercises have been carried out this year with the participation of banks, telecom operators and the largest internet companies to test the ability of the Russian network to operate autonomously.
Traffic redirection to Asia
Exercises justified by the geopolitical situation, analyzes the Kommersant. Quoted by the newspaper, Deputy Prime Minister Dmitry Chernychenko says the number of cyberattacks has increased by 80% this year. "If last year, the financial sector was the main target, this year it is the public sector that is targeted", emphasizes Dmitry Chernychenko.
Read alsoUkraine-Russia: behind the scenes of the cyber war
Faced with these threats, an Internet disconnection plan was considered in the spring by the Russian operator Rostelecom, indicates the Kommersant which states, quoting an expert, that "If Russia is cut off from European exchange points, it will redirect Internet traffic to Asia". Several times evoked by Vladimir Putin, the project of an Internet "sovereignhas been overshadowed in recent months by restrictions on access to certain sites (Twitter and Instagram, in particular). A digital isolation of Russia, insofar as it is feasible, would also risk accentuating state control over personal data and individual freedoms, some experts and human rights defenders fear.
SEE ALSO – War in Ukraine: the potential loss of the Starlink internet network worries the front
Bypassed Constraints
From outside Russia, several Internet traffic access providers (such as LINX) have already disconnected their services from Rostelecom or MegaFon. Conversely, "US cloud provider Clouflare has refused to work in the Russian Federation and the State Department has urged not to disconnect Russia from the internet", Write the Kommersant. For the major state actors, and in particular the intelligence services, the possibility of obtaining information remains preferable to the absence of data…
As early as last March, the Russian authorities took measures to protect state sites: all websites, in particular those of public services, were ordered to switch to Russian domain names, to abandon hosting abroad and strengthen their password policy. But six months later, these constraints are proving to be largely circumvented, in particular by regional state sites which continue to use codes downloaded from abroad, notes the Kommersant. With as explanation, continues the newspaper, the fact in particular that the fines envisaged remain minimal, around 5000 rubles (83 euros approximately at the current price).
Read also"I am a sponge who receives information": how to tell the war in Ukraine well?
Furthermore, the disconnection of the Russian segment of the Internet from the World Wide Web would have consequences for the connectivity of the entire global network, explain the experts, such as Ivan Begtin, who underlines, for example, that "Traffic from Kazakhstan and parts of Asian countries passes through the Russian Federation". Even if closed models of the Internet exist, in Iran or in China, this specialist judges a "shutdown(closure) of Russia unlikely, leaning more towards a "internal hardening and access restrictions".
SEE ALSO – US concerned about 'destabilizing' Africa by China and Russia
We want to thank the author of this post for this amazing content
Global Internet of Things (IoT) Testing Market Overview,…
Great disconnection from the Internet against the backdrop…
How to plan the disconnection of the WiFi of your Internet…
Can Russia Sabotage the Global Internet by Attacking…
Russia tightens its control over the Internet and social…
Russia loses part of its interconnection with the Internet
Categories Internet Tags disconnection, Global, internet, Russia, Testing
Orange: this fiber offer + 80 GB mobile plan is a great Christmas deal
15 hot Christmas photo spots in Paris 2022 to enjoy the magical illuminations


text:
Home » Featured articles » Pentatonix Manila 2020 » Pentatonix world tour Manila 2020 » PTXManila2020 » Philippines will always be Pentatonix's home
Philippines will always be Pentatonix's home
By: Karen Meets World
In: Featured articles Pentatonix Manila 2020 Pentatonix world tour Manila 2020 PTXManila2020
Philippines will always be Pentatonix's home Karen Meets World 12:44 PM
Philippines will always be Pentatonix's home.
Pentatonix, a five-member a capella singing group composed of Kevin Olusola (beatboxer), Scott Hoying (baritone lead), Mitch Grassi (counter-tenor), Matt Sallee (vocal base lead), and Kirstie Maldonado (mezzo-soprano lead) was astonished at the energy and singing voice Filipino Pentaholics (fandom name) showed them.
Scott even shared "I'll tell you some places that we go on tour make us feel so welcome and I have to say that the Philippines makes us more welcome than anywhere in the world. You know the energy and everything was so amazing. It was such an unforgettable night and we feel right at home here".
The show started with Pentatonix performing "Daft Funk" medley and was followed by "High Hopes". Pentaholics also swayed their hips in tune to "Havana" and "Señorita".
Rich harmonies filled our ears as we heard the music "Evolution of Pentatonix" throughout the years. But not only that... the evolution also gave the audience treats showcasing Ariana Grande's songs such as "Thank You Next", "One Last Time", and "Break Free" to name a few.
One of the highlights of the night was Kevin's cello beatboxing solo performance.
Pentaholics smoothly blend in with Pentatonix as they sang "I Want It That Way" along with them.
The rules are simple if you know the song just sing along... #PTXManila2020 pic.twitter.com/7An6NCv6D9
— Manila Concert Junkies (@mnlcncertjunkie) February 8, 2020
Another highlight occurred when Pentaholics lit up a beautiful white ocean on the songs "Imagine" and "Hallelujah".
Fans gave @PTXofficial a beautiful white ocean during the songs Imagine and Hallelujah! #PTXTheWorldTourMNL #PTXManila2020 pic.twitter.com/3k3cF24bhc
It was truly a night to remember. No doubt that Manila tour was their favorite show they ever done.
Pentatonix Live in Manila 2020 was brought to us by Wilbros Live.
Photos by Caren Gudiaga for Manila Concert Junkies
Tags: Featured articles Pentatonix Manila 2020 Pentatonix world tour Manila 2020 PTXManila2020


Awesome Cycling Jersey - Kathy's original artwork highlights beautiful flowers and mountains in a flattering style that will turn heads and create fun conversations wherever you ride. No words- just beautiful art. Get your unique colorful jersey today and go on a long wonderful bike ride and enjoy!


text:
The Non-Chopsticks User's Guide to Eating with Chopsticks By MELISSA SALVA
Of course you know how to eat with chopsticks. Just use them like giant tweezers and that's all there is to it. If you think wielding them takes up effort that you feel is better utilized by actual eating, you are right. Why suffer? Whether you are dining in China, Korea, Japan, or Vietnam where meals are always served with chopsticks, here's how you can show your host or the restaurant that you'd rather eat with a spear.
Stick chopsticks into your bowl of rice. The meal started ten minutes ago, but you were only able to get the first morsel of food in your mouth after you dropped it thrice. Before then, the round carrot slice had been sliding around in the oyster sauce, escaping your chopsticks' grasp. What better way to rest your chopsticks than upright in your bowl of rice? It's easier to resume your grip and, if you're lucky, a small clump of rice may be stuck at the end, ready to put in your mouth. Your hosts may stare momentarily, wince, or try not to make eye contact with you for a while. That's only because the first thing they learn about chopsticks as children is to never, ever do this—in Chinese and Japanese rituals, this is how they feed the dead, while in Vietnamese rituals, it looks like how joss sticks are placed when praying to dead ancestors. Avoid eating with the elderly. They won't appreciate the visual.
Use a chopstick to spear a piece of food. Using chopsticks the proper way, like tongs, you are reduced to a diet of steamed Japonica and noodles. Really, there are so many things you can eat with ease if you stabbed them with your chopstick. The fat cube of fried tofu. Tempura dipped in the sauce. The quail egg in birthday noodles! While you're at it, improve your grasp by holding the chopstick in your fist instead of with just your fingers. But if you're trying to maintain the appearance of good breeding or would like to uphold the reputation of your parents, it's best not to do this during dates and engagement dinners, where your parents will be unfairly judged based on the silly criterion of how you handle chopsticks.
Show your knowledge of other uses for chopsticks. Use them as drumsticks to provide entertainment at the table while waiting for the food to arrive. Mimic a walrus. Pretend you're a vampire with extra-long teeth. Poke your little brother. You've seen this before—stick one up your nose and, depending on who you're with, hear genuine or nervous laughter.
Tapping bowls with chopsticks is associated with how beggars attract attention, but once you start rapping like Snoop Dogg, people at the table will be begging you for more, right?
Use them when gesticulating—it really does help get your point across. Since they are an extension of your hand, you can move bowls and other things around the table and point at objects (and people, too!) with them.
You can even use them as fashion accessories. Anyway, it's only in Japan where they really notice the difference between chopsticks and the kanzashi that geisha wear in their hair. Everywhere else, a stick is a stick—like, who knows the difference between a turtle and a tortoise? Chopsticks are useful for many things, and as far as you're concerned, the last thing they're useful for is eating.
Take up your chopsticks to signal the start of the meal. You're pretty good at picking up social cues. For instance, at a company dinner, you immediately noticed that everyone started eating when chopsticks were taken up by the Senior Director. At a family meal, elders may shuffle to their seats at a glacial pace or chat everybody up before anyone can begin eating. You thought you just heard your seatmate's stomach growl. Meanwhile, the roasted duck is getting cold. Do everyone a favor. But unless you're the oldest or most senior person in the room, be alert when taking up chopsticks especially in Hong Kong and Japan. The elders might stab you with theirs.
Use your own chopsticks to serve yourself and others. Meals served family-style usually have serving chopsticks or spoons for every dish that is meant for sharing. If they don't, use your own. You've eaten with them, but it's not like you've bitten them, gnawed on the tips, sucked sticky food and sauces from them, licked or washed them clean in your bowl of soup. Yet.
If you don't like a piece of food, put it back where you got it. You couldn't tell beef spare ribs from pork, which you're avoiding, but it was a good thing your host told you—you were just about to take a bite. Put that pork spare rib in XO sauce back in the serving dish for someone who wants it. Someone calling dibs on it before you could put it back? Pass it straight to his chopsticks. The Japanese shudder at this practice because this is how they transfer the remains of the deceased during funeral rites. But you're in a Chinese restaurant, and it's pretty obvious that what you got there is a spare rib, not a rib.
Got a huge helping of catfish that tastes weird to you? Dump it in a used rice bowl. The Koreans are fussy about this sort of thing, but if you're not going to eat it, might as well put it somewhere to make space in your plate. It's not like you're secretly tossing it to the potted plant nearby or feeding it to the family pet under the table. Now that would be rude.
By doing any of these things, actually by doing even just the first one, you will confirm that a) you're an uneducated foreigner who is best regarded with equal parts amusement and pity, b) you're not really focused on the sensibilities of your host ("Sitting down for meals means eating, not socializing!"), or c) hunger really does make one uncivilized. Not very good impressions with which to leave your hosts, but hey, at least you are sure to have a spoon and fork in your place setting next time. If there's a next time.
Written by Melissa Salva
Revenge of the Tinola
Manila magic: A night at Gugu Room
The Devil's Invention


> Do you think other flaky test could be impacted by the same issue?
which may be the same issue or the timeout `50ms` is too short to get updated data.
leader to learner ,which may be a salient cause for flaky test.


Quote: On May 22, 2014 Hope Hudson and her classmates gathered together at the Hope Center to watch Hope ride her new horse Diablo. Hope's classmates decided to raise money to purchase Hope her own horse. With the help of family and friends and our generous community they raised over $3,000. After much research, the center's Certified Riding Instructor Krystal Evans, found Diablo in Oklahoma. Diablo had the temperament and skills needed for Hope and was the breed Hope has always dreamed of. In addition to purchasing "Diablo" the funds were used for necessary tack including a side saddle. The class gathering included lunch, games, and Hope showing her riding skills with her newest friend Diablo.
Like many little girls, Hope has always had a passion for horses. When asked why she loves horses she stated, "I love horses because they are majestic and beautiful animals with free spirits. They are calming to watch and you can feel their emotions when you pet them." She collects Breyer horses, draws pictures of horses, and watches horse races on TV and dreams of one day being a jockey. Hope is a beautiful girl inside and out. Her smile and laughter are contagious. You would never know she has had a few challenges in her life. She was born blind in her right eye and at 10 months old she had heart surgery for a heart defect. In 2011 she was diagnosed with Hajdu-Cheney Syndrome, a rare connective tissue disorder. The connective tissue is a strong fibrous tissue that supports and joins other body tissues and parts. The ailment can lead to abnormal development of bones and joints, as well as a decrease in bone mass and changes in the skull and jawbone. Her prognosis all depends on her body; there is no cure for HCS so doctors continue to treat her symptoms.
In 2012, when given a chance for a wish to come true thanks to the Make-A-Wish Foundation, the Kentucky Derby was a natural destination for Hope. While there, Hope and her family befriended horse trainer Doug O'Neill and his team whose colt "I'll Have Another" ended up winning the Derby. Anyone who has watched a horse race knows that sometimes, against all odds, out comes a winner!
Hope has gone through numerous surgeries over the years for her symptoms. In 2009 she had a corrective lens implanted. Other surgeries were performed to relieve the pain from her symptoms not knowing if they would guarantee success. She is often left wheel-chair bound. She has spells when she loses feeling in her limbs for hours at a time. Although she may feel pain, she never shows it. In 2013 her classmates decided to help Hope with another dream of hers. They decided to get Hope her very own horse and started "A Horse for Hope" campaign.
As her name implies…. "Hope" is what she gives us and we are eternally grateful.
What lead to the formation of this group?
Hope Hudson, who is now 14 years old, suffers from the rare connective tissue disorder Hajdu-Cheney syndrome, which causes neck pain, headaches, slurred speech and vision problems. She has had numerous surgeries which often left her wheel-chair bound. Her inspiration, determination, love and passion for horses, led us to the decision to name the organization, "The Hope Therapeutic Horsemanship Center" in her honor.


text:
Mecca Bingo in St Helens launch special zone for carers
Mecca Bingo club in St Helens is launching Mecca Full House on December 3
Mecca Bingo clubs across the country including St Helens will be launching Mecca Full House on December 3 - a dedicated recreational zone exclusively for carers every Tuesday throughout the month.
Working alongside Carers Trust, the Mecca Full House leisure zone will be available to all carers on Tuesday afternoons throughout December, giving them the opportunity to socialise with others within the caring community.
There are an estimated 6.5 million unpaid carers in the UK. Many are caring round the clock for a relative with a serious illness or disability, making it difficult for the carer to find time for leisure and rest, or even a short respite from the pressures of caring.
Carers Trust is a national charity that provides information, help and support for carers of all ages and backgrounds, many of whom experience isolation and loneliness as a result of their caring role.
This Christmas, Mecca Full House will provide an opportunity for the bingo community to embrace those who need support and company at a time when they need it the most.
Christmas time can leave carers feeling pressurised with many stating they find the Christmas period the most isolating and stressful of the year.
For most of us at Christmas, we are surrounded by loved ones and attending festive parties and functions, but for many carers it can be a difficult month with limited spare leisure time to enjoy the festive fun.
Mecca Full House will allow carers to take some much-needed time out of their hectic schedules and day-to-day responsibilities to socialise with others.
Mecca Full House will be available every Tuesday in December and will offer a space for both the carer and the person they care for to attend, enjoy a free game of bingo if they wish and fully embrace the Christmas spirit.
Paul Harrison, General Manager at Mecca Bingo St Helens, said: "Christmas is always a magical time of year here in the club and we are proud to be launching our Mecca Full House zone alongside the Carers Trust.
"Our recreational area will provide unpaid carers in the area, who do not have many opportunities to enjoy the festivities, a safe space to relax in. There's always something special about our local community coming together in the club to kick start the festive season."
Svetlana Kirov, Director of Fund-raising and Communications at Careers Trust, said: "Many of the carers we work with tell us that Christmas can reinforce their sense of loneliness and isolation.
"They find it hard to join in festive fun because of the need to be constantly looking after a relative who cannot take care of themselves.
Residents invited to festive multi-faith ceremony in St Helens
Grab a festive bargain at Willowbrook Hospice Christmas Fayre
"So, we are thrilled that Mecca Bingo is launching its Mecca Full House initiative to make sure that unpaid carers, right across the country, have a place to go where they can enjoy a few hours away from caring and have some festive fun."
To attend a Mecca Club, you do need to be aged 18 years or over and new customers must bring photo ID.


text:
Kenia nahm an den Olympischen Winterspielen 2006 im italienischen Turin mit einem Athleten teil.

Teilnehmer nach Sportarten

Ski nordisch
 Philip Boit
 Männer, 15 km klassisch: 53:32,4 min. (→ 91.)

Weblinks

Kenia
Winterspiele 2006
Wintersport (Kenia)

redpajama_set_name: RedPajamaWikipedia


text:
Two Relais Desserts meetings are organized every year
The first is a work seminar and takes place in the fall. It gathers Relais Desserts members and their guests, partners of the association exclusively.
The other focuses on communications and happens in the spring. It is open to the public and presents the Haute Pastry world; it is the Relais Desserts International Meetings.
The autumn seminar
The autumn seminar is an opportunity for all members of the association to share their knowledge and know-how, to discover new products, new techniques and to meet the association's partners.
This trade seminar also allows the whole group to meet and nurture the association's spirit based on its values of sharing and exchange.
It takes place at the Ecole Ferrières, in Ferrières en Brie (77).
Relais Desserts International meetings
Each spring, an association member organizes the Relais Desserts International Meetings in his city. During several days, he welcomes all the members and their spouses and proposes an agenda, which focuses on company life and the discovery of regional pastries.
Thanks to an exceptional evening open to the public, these meetings are the opportunity to introduce the Haute Pastry world as well as the latest Association Pastry chefs' creations.
This gathering also allows the whole group to keep up with training through an intense work program that goes from customer relations—often handled by spouses—to marketing 4.0, including tomorrow's commerce and new trends in gastronomy.
It is also an opportunity to preserve the Association's values of exchange and sharing.
In April 2020, the International Meetings should take place in Padoue (Italy), at M. Luigi Biasetto's. This seminar has been cancelled due to the coronavirus sanitary crisis.
Pastry chefs of Relais Desserts from all over the world met in the capital of gastronomy, in Lyon, at Richard Sève, Sébastien Bouillet and Jean-Paul Pignol to exchange their know-how and their profession, but also to allow Lyon fans to taste their creations on the theme of "market pastry". The occasion to celebrate, once again, the Haute Pâtisserie.
Just like every spring, the Relais Desserts pastry chefs met in full to celebrate their association during four ultra-gourmet days. The bests of the sweet went to the other country of chocolate: Belgium. It was an opportunity to discover Jean-Philippe Darcis's universe and boutiques.
Marc Ducobu in Waterloo and Bernard Proot of the del Rey house in Antwerp also opened their doors to their colleagues. The Wittamer house signed the centrepiece of the beautiful "Chefs' afternoon snacks" at the Spa Casino.
Vincent Guerlais welcomed the seminar in Nantes. On the agenda: visit of the Guerlais House boutiques and the association members' latest creations. But it was also the opportunity to work together on client relations, tomorrow's trade, the new trends in gastronomy…
This year, celebrating the Association 35th Anniversary, François Payard in New York entertained his Relais Desserts counterparts and chose to organize with them a charity cocktail for City Harvest, an organization dear to him as they already work together for Macaroon's Day. And so, $72,000 was collected for City Harvest.
The Relais Desserts members also discovered their host's universe and followed a custom-made schedule. They enjoyed these few days and visited the newly open Valrhona School in Brooklyn.
Damien and Lucien Moutarlier had the pleasure of welcoming the Association for three days in Montreux (Switzerland). The Pastry chefs discovered backstage three of their four candy stores in Montreux, in Chexbres and in Lausanne and were able to exchange ideas during the scheduled work sessions. From visits to surprise activities, the Moutarliers proudly presented what makes the Montreux region's wealth.
They also had the opportunity to discover another Relais Desserts member's universe: Sébastien Brocard, whose business is on the French side of the Geneva Lake.
Patricia and Luc Guillet, Relais Desserts members since 1991, were honored to welcome their peers, passionate and masters of haute-pastry.
They met for three days of exchange and sharing in order to reinvent their profession and make the high level of haute-pastry go even higher.
On the agenda, work sessions, lectures, but also moments of conviviality and brotherhood during the chefs' picnic and the evening gala.
Patrick and Xavier Berger hosted the Relais Desserts pastry chefs in Tarbes (French Basque country).
During the course of two days, the members visited Xavier and Patrick's boutiques and laboratories, they also discovered the towns of Tarbes and Pau, met with some Bigorre ham makers as well as visited some wine cellars in Madiran.
On the third day, the Relais Desserts met Basque country's famous conviviality at Lionel and Isabelle Raux's in Bayonne.
The Association members also met in:
Mulhouse in 2012
Nice in 2011
Madrid and Alicante in 2010
Vichy in 2009
Venice in 2008
Fontainebleau in 2007


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