text:
\section{Preliminary Results.}\label{sec-1}
This section contains various results that are necessary for the proof of the main theorem.  The section is divided into subsections according to the nature of the results presented.

\subsection{Vector fields on surfaces.}
Let $X$ be a normal projective surface defined over an algebraically closed field $k$ of characteristic $p>0$. Let $D$ be a nontrivial vector field on $X$. The next proposition presents a method to find curves on $X$ stabilized by $D$.

\begin{proposition}[Proposition 2.1~\cite{Tz18}]\label{sec1-prop1}
Let $X$ be a normal projective variety defined over an algebraically closed field of characteristic $p>0$. Let $D$ be a non trivial global vector field on $X$ such that either $D^p=0$ or $D^p=D$. Let $\pi \colon X \rightarrow Y$ be the quotient of $X$ by the $\alpha_p$ or $\mu_p$ action on $X$ induced by $D$. Let $L$ be a rank one reflexive sheaf on $Y$ and $M=(\pi^{\ast}L)^{[1]}$. Then $D$ induces a $k$-linear map
\[
D^{\ast} \colon H^0(X,M) \rightarrow H^0(X,M)
\]
with the following properties:
\begin{enumerate}
\item $\mathrm{Ker}(D^{\ast})=H^0(Y,L)$ (considering $H^0(Y,L)$ as a subspace of $H^0(X,M)$ via the map $\pi^{\ast}$).
\item If $D^p=0$ then $D^{\ast}$ is nilpotent and if $D^p=D$ then $D^{\ast}$ is a diagonalizable map whose eigenvalues are in the set $\{0,1,\ldots,p-1\}$.
\item Let $s\in H^0(X,M)$ be an eigenvector of $D^{\ast}$. Then $D(I_{Z(s)})) \subset I_{Z(s)}$. Suppose moreover that $D^{\ast}(s)=\lambda s$, and $\lambda \not= 0$. Then $
(D(I_{Z(s)}))|_V=I_{Z(s)}|_V,$ where $V=X-\pi^{-1}(W)$, $W\subset Y$ is the set of points that $L$ is not locally free.
\end{enumerate}
\end{proposition}

The next proposition gives information about the fixed points of the vector field $D$. It is a generalization of~\cite[Proposition 3.4]{Tz19}.

\begin{proposition}\label{sec1-prop-2}
Let $X$ be a normal projective $\mathbb{Q}$-factorial surface defined over an algebraically closed field of characteristic $p>0$. Let $D$ be a non trivial global vector field on $X$, $A$ a nef and big line bundle on $X$ and $C \in |A|$  a curve such that $D(I_C)\subset I_C$. Let $C=\sum_{i=1}^m n_i C_i$ be the decomposition of $C$ into its irreducible components.
Suppose that one of the following happens:

\begin{enumerate}[label=(\Alph*)]
\item There exists a positive rational number $m$ such that $mA-K_X$ is nef.
\item $K_X$ is a nef and big $\mathbb{Q}$-Cartier divisor.
\end{enumerate}

And in addition,
\begin{enumerate}[label=(\Alph*)]
\item[$(C)$] $A\cdot L<p$, for some nef and big line bundle $L$,
\item[$(D)$] $(m+1)A^2+2A\cdot L<p/d$ (if (A) holds), or $3A^2+K_X\cdot A<p/d$ (if (B) holds) where $d$ is the least common multiple of the indices $d_1,\ldots,d_m$ of $C_1\ldots, C_m$.
\end{enumerate}

Then,
\begin{enumerate}
\item $D$ restrict to a vector field on each $C_i$ such that $L \cdot C_i >0$, $i=1,\ldots, m$.
\item  Let $C_i, C_j$, $i\not= j$, be two distinct components of $C$ both stabilized by $D$. Suppose that at least  one of the following happens
\begin{enumerate}
\item If (A) holds then  $L\cdot C_i\not= 0$, or $L\cdot C_j\not=0$.
 \item If (B) holds, then $K_X\cdot C_i \not= 0$, or $K_X \cdot C_j\not=0$.
\end{enumerate}
Then every point of intersection of $C_i$ and $C_j$  is a fixed point of $D$.

\item Let $C^{\prime}=\sum_{j=1}^sn_j^{\prime}C^{\prime}_j \in |A|$ be another member of $|A|$ which is also stabilized by  $D$. Let $C_i$ be a component of $C$ which is not a component of $C^{\prime}$.  Then, if $A^2<p$ and even without the validity of  (A), (B) or (D), every point of intersection  of $C_i$  with a component $C^{\prime}_j$ of $C^{\prime}$ is a fixed point of $D$.
\end{enumerate}
\end{proposition}

\begin{proof}
The condition $(C)$ implies that $n_i <p$, for any $C_i$ such that $L\cdot C_i>0$, $i=1,\ldots, m$. Therefore from~\cite[Proposition 3.2]{Tz19}, $D(I_{C_i}) \subset I_{C_i}$. Hence $D$ restricts to a vector field on $C_i$ (perhaps the zero one), for every $i$ such that $L\cdot C_i >0$.

 Let $C_i$ and $C_j$ be two distinct irreducible components of $C$ stabilized by $D$, i.e.,   $D(I_{C_i})\subset I_{C_i}$ and $D(I_{C_j})\subset I_{C_j}$. Therefore, $D(I_{C_i}+I_{C_j}) \subset I_{C_i}+I_{C_j}$. Let $R\in C_i \cap C_j$ be a point of intersection, and $U =\mathrm{Spec} A$ be an affine open neighborhood of $R$ in $X$ such that no other point of intersection of $C_i$ and $C_j$ exists in $U$. Let $Q$ be the ideal of $A$ which corresponds to $Q=I_{C_i}|_U+I_{C_j}|_U$. Then $D$ induces a derivation of $A$ and  $D(Q)\subset Q$. Moreover, $r(Q)=\mathbf{m}_A$, where $\mathbf{m}_A$ is the maximal ideal of $A$. I will show that the condition (D) implies that  $\dim_k (A/Q)<p$. Then the rest of the proof is identical as in~\cite[Proposition 3.4]{Tz19}.

\textbf{Claim:}
\begin{gather}\label{eq-claim}
C_i  \cdot C_j <
\begin{cases}
(m+1)A^2+2A\cdot L, & \mathrm{if} \;\; (A) \;\; \mathrm{holds}\\
A^2+3K_X\cdot A, & \mathrm{if} \;\; (B) \;\; \mathrm{holds}
\end{cases}
\end{gather}
for all $i=1,\ldots, m$.

I will prove only the second part. The first is identical and is omitted. Since (B) holds, $K_X$ is nef and big. Then, according to the assumptions, one of $K_X \cdot C_i$ and $K_X \cdot C_j$ is not zero. Without loss of generality we may assume that $K_X \cdot C_i>0$. Then, since $C\in |A|$, and $A$ is nef and big, it follows that
\begin{gather}\label{sec1-eq-5}
A^2\geq A\cdot C_i =n_jC_i \cdot C_j +n_i C_i^2 +\sum_{k\not= i,j}n_k C_k \cdot C_i \geq n_j C_i \cdot C_j +n_i C_i^2.
\end{gather}
Therefore,
\begin{gather}\label{sec1-eq-6}
C_i\cdot C_j <A^2-n_i C_i^2.
\end{gather}
Next I will show that $-C_i^2\leq 2+K_X \cdot C_i$. Let $f \colon X^{\prime}\rightarrow X$ be the minimal resolution of $X$. Let $C_i^{\prime}=f_{\ast}^{-1}C_i$, be the birational transform of $C_i$ in $X^{\prime}$. Then by the adjunction formula for $C_i^{\prime}$ it follows that
\begin{gather}\label{sec1-eq-7}
-(C_i^{\prime})^2=-2p_a(C_i^{\prime})+2+K_{X^{\prime}}\cdot C_i^{\prime}\leq 2+K_{X^{\prime}}\cdot C_i^{\prime}.
\end{gather}
Now there are adjunction formulas
\begin{gather}
f^{\ast}C_i=C^{\prime}_i+ E\\
K_{X^{\prime}}+F=f^{\ast}K_X \nonumber
\end{gather}
Where $E$ and $F$ are effective $f$-exceptional divisors ($F$ is effective because $f$ is the minimal resolution). From these immediately follows that $C_i^2\geq (C_i^{\prime})^2$ and $K_X\cdot C_i \geq K_{X^{\prime}}\cdot C_i^{\prime}$. From these and the equations (\ref{sec1-eq-7}) it follows that
\begin{gather}\label{sec1-eq-8}
-C_i^2\leq 2+K_X\cdot C_i.
\end{gather}
Then the equation (\ref{sec1-eq-6}) becomes
\begin{gather}\label{sec1-eq-88}
C_i \cdot C_j \leq A^2+2n_i+n_i(K_X\cdot C_i).
\end{gather}

Since $K_X$ is nef and $K_X \cdot C_i >0$,  then  $n_i<n_i (K_X \cdot C_i)\leq A\cdot K_X$.  Then the equation (\ref{sec1-eq-88}) gives
\[
C_i\cdot C_j \leq A^2+3A\cdot K_X,
\]
and the claim has been proved.

Let now $d_i$ be the index of $C_i$ in $X$, i.e., $d_iC_i$ is Cartier. Then
\begin{gather}\label{sec1-eq-9}
(d_iC_i) \cdot C_j=\deg (\mathcal{O}_X(d_iC_i)\otimes  \mathcal{O}_{C_j})
\end{gather}
Now restricting the exact sequence
\[
0 \rightarrow \mathcal{O}_X \rightarrow \mathcal{O}_X(d_iC_i) \rightarrow \mathcal{O}_{d_iC_i} (d_iC_i) \rightarrow 0
\]
to $C_j$ we get the exact sequence
\begin{gather}\label{sec1-eq-100}
0 \rightarrow \mathcal{O}_{C_j} \rightarrow \mathcal{O}_{C_j}(d_iC_i) \rightarrow \mathcal{O}_{d_iC_i} (d_iC_i)\otimes \mathcal{O}_{C_j} \rightarrow 0
\end{gather}
But since $C_i$ and $C_j$ are distinct irreducible curves, it follows that
\begin{gather}\label{sec1-eq-10}
\mathcal{O}_{d_iC_i} (d_iC_i)\otimes \mathcal{O}_{C_j}  \cong \mathcal{O}_X(d_iC_i) \otimes (\mathcal{O}_{d_iC_i} \otimes \mathcal{O}_{C_j}) \cong \mathcal{O}_{d_iC_i \cap C_j} = \frac{\mathcal{O}_X}{I_{C_i}^{(d_i)} +I_{C_j}},
\end{gather}
Where $I_{C_i}^{(d_i)}$ is the $d_i$-th symbolic power of $I_{C_i}$. Then by taking Euler characteristics in (\ref{sec1-eq-100}) and using the Riemann-Roch theorem on $C_j$ gives that
\[
(d_iC_i) \cdot C_j=\deg (\mathcal{O}_X(d_iC_i)\otimes  \mathcal{O}_{C_j}) =\dim_k \frac{\mathcal{O}_X}{I_{C_i}^{(d_i)} +I_{C_j}}\geq \dim_k \frac{\mathcal{O}_X}{I_{C_i} +I_{C_j}}
\]
Therefore, $\dim_k(A/Q) < p$, if $(d_iC_i)\cdot C_j <p$, a condition satisfied if (D) holds.

Finally, let $C^{\prime}=\sum_{j=1}^sn_j^{\prime}C^{\prime}_j \in |A|$ be another member of $|A|$ which is an integral curve of $D$. Let $C_i$ be a component of $C$ which is not a component of $C^{\prime}$. Since $A^2<p$, it follows that $C_i \cdot C^{\prime} <p$. Then, since $C^{\prime}$ is Cartier it follows that
\[
\deg (\mathcal{O}_X(C^{\prime}) \otimes \mathcal{O}_{C_i} )<p.
\]
Then, by repeating the previous arguments with $C^{\prime}$ in the place of $d_iC_i$ and $C_i$ in the place of $C_j$, and by considering that, since $C_i$ is not a component of $C^{\prime}$, that
$
 \mathcal{O}_{C^{\prime}}(C^{\prime})\otimes \mathcal{O}_{C_i}\cong \mathcal{O}_{C^{\prime}\cap C_i},
$ we get that
\[
\dim \frac{\mathcal{O}_X}{I_{C_i}+I_{C^{\prime}_j}} \leq \dim \frac{\mathcal{O}_X}{I_{C^{\prime}}+I_{C_i}}=\deg((\mathcal{O}_X(C^{\prime}) \otimes \mathcal{O}_{C_i} )<p,
\]
for any $j=1,\ldots, s$. From this it now follows exactly as in the proof of (2) earlier, that every point of intersection of $C_i$ and $C^{\prime}_j$ is a fixed point of $D$.

This concludes the proof of the proposition.
\end{proof}

The next lemma provides information about the divisorial part of the vector field obtained by blowing up fixed points of a given vector field.
\begin{lemma}\label{sec1-prop-12}
Let $P\in X$ be a smooth point on a surface $X$ defined over an algebraically closed field of characteristic $p>0$. Let $D$ be a nontrivial global vector field on $X$ such that $D^p=D$ and $P$ is a fixed point of $D$. Let $X^{\prime}\stackrel{f}{\rightarrow}X$ be the blow up of $X$ at $P$ and $E$ the $f$-exceptional curve. Suppose that $E$ is contained in the divisorial part of $D^{\prime}$, the lifting of $D$ on $X^{\prime}$. Let $X^{\prime\prime}\stackrel{g}{\rightarrow} X^{\prime}$ be the blow up of $X^{\prime}$ at a point on $E$. Let $D^{\prime\prime}$ be the lifting of $D^{\prime}$ on $X^{\prime\prime}$ and $F$ be the $g$-exceptional curve. Then $F$ is an integral curve of $D^{\prime\prime}$ and is not contained in the divisorial part of $D^{\prime\prime}$.
\end{lemma}

\begin{proof}
By~\cite{RS76} in suitable local coordinates $x, y$ of $X$ at $P$, $D=x\partial/\partial x +\lambda y \partial /\partial y$, $\lambda \in \mathbb{F}_p$. The lemma follows by a straightforward calculation of the blow ups in question and the vector fields $D^{\prime}$ and $D^{\prime\prime}$.
\end{proof}

\subsection{The genus of curves with many components.}

\begin{definition}
Let $C$ be a connected reduced curve. Let $C_i \subset C$, $i=1, \ldots, m$, be connected  reduced curves such that $C=\cup_{i=1}^mC_i$, and $C_i$, $C_j$ do not have any common irreducible components if $i\not= j$. Let $P\in C$ be a point. Then $\varepsilon(P; C_1,\ldots,C_m)$ denotes the number of distinct curves $C_i$ that pass through $P$. In particular, if $C_i$ are the irreducible components of $C$ then we set $\varepsilon(P)=\varepsilon(P; C_1,\ldots,C_m)$.
\end{definition}

\begin{proposition}\label{sec1-prop-5}
Let $C$ be a connected reduced projective curve and suppose that $C=\cup_{i=1}^nC_i$, where $C_i$ is a reduced connected curve for all $i$, and $C_i$, $C_j$ do not have common irreducible components for $i\not= j$.  Then
\[
h^1(\mathcal{O}_C) \geq \sum_{i=1}^n h^1(\mathcal{O}_{C_i})  + \sum_{P}(\varepsilon(P;C_1,\dots, C_n)-1)+1-n,
\]
where $P$ runs over all distinct points of intersection of the curves $C_i$, $i=1,\dots,m$. In particular, if $C_i$ are the irreducible components of $C$, then
\[
h^1(\mathcal{O}_C) \geq \sum_{i=1}^n h^1(\mathcal{O}_{C_i})  + \sum_{P}(\varepsilon(P)-1)+1-n,
\]
\end{proposition}

\begin{proof}
Let $N$ be the cokernel of the natural map $\mathcal{O}_C \rightarrow \oplus_{i-1}^n\mathcal{O}_{C_i}$. Then there exists an exact sequence
\begin{gather}\label{sec1-eq-11}
0 \rightarrow \mathcal{O}_C \rightarrow \oplus_{i-1}^n\mathcal{O}_{C_i}\rightarrow N \rightarrow 0.
\end{gather}

\textbf{Claim:} Let $P\in C$ be a closed point. Then
\begin{gather}\label{sec1-eq-12}
\mathrm{length}(N_P) \geq \varepsilon(P; C_1,\dots, C_n)-1
\end{gather}

Suppose that $P$ belongs to exactly $k$ distinct members of the set  $\{C_1,\ldots,C_n\}$. Let $C_{m_i}$, $i=1,\dots, k$, be these curves. Let $m_P$ be the ideal sheaf of $P$ in $C$. Then from (\ref{sec1-eq-11}) we get the exact exact sequence
\[
\mathcal{O}_C/m_P\rightarrow \oplus_{i=1}^k \mathcal{O}_{C_{m_i}} /m_P \rightarrow N/m_PN \rightarrow 0.
\]
From this it immediately follows that
\[
\mathrm{length}(N_P) \geq \mathrm{length}(N/m_PN) \geq k-1=\varepsilon(P;C_1,\ldots,C_n)-1,
\]
as claimed. The proposition now follows by using this and taking Euler characteristics in (\ref{sec1-eq-11}).
\end{proof}

The previous proposition implies the following result, which will be used  in the proof of the main theorem.

\begin{corollary}\label{sec1-cor-6}
Let $C$ be a reduced connected projective curve such that $C=C_1\cup C_2 \cup C_3 \cup B$, where $C_1, C_2, C_3$ are integral curves and $B$ is a connected reduced curve such  that
\begin{enumerate}
\item There exists a point $P\in C_1 \cap C_2 \cap C_3$,
\item $B\cap C_i \not=\emptyset$, $i=1,2,3$, and that $P\not\in B\cap C_i$, $i=1,2,3$.
\end{enumerate}
Then $h^1(\mathcal{O}_C )\geq 2$.
\end{corollary}

\subsection{Blow up of ample line bundles.}

\begin{proposition}\label{sec1-prop-11}
Let $X$ be a normal projective surface whose singularities are rational double points and $K_X$ is nef and big. Let $A$ be an ample Cartier divisor on $X$ and $f \colon Y \rightarrow X$ be the blow up of a singular point $P\in X$. Then $B=f^{\ast}(2A)+K_Y-E$, is ample on $Y$, where $E=f^{-1}(P)$.
\end{proposition}
\begin{proof}
Since $P\in X$ is a rational double point, $Y$ has also rational double points and $K_Y=f^{\ast}K_X$. By the classification of rational double points, $P\in X$ is one of the following types: $A_n$, $D_n$, $E_6$, $E_7$, $E_8$. Simple calculations of the blow up of a rational double point show the following.

Suppose that $P\in X$ is of type $A_n$. If $n=1$ then $E\cong \mathbb{P}^1$ and $E^2=-2$. If $n\geq 2$, then  $E=E_1+E_2$, $E_1\not= E_2$ and $E_i \cong \mathbb{P}^1$, $i=1,2$. Moreover, $E_i^2=-\frac{n+1}{n+2}$, $i=1,2$, $E_1\cdot E_2 =\frac{1}{n+1}$ and $E^2=-2$. In all other cases, $E=2F$, where $F\cong \mathbb{P}^1$ and $E^2=-2$.

In order to show that $B$ is ample it suffices to show that $B^2>0$ and that $B \cdot C>0$, for every integral curve $C$ in $Y$. Since $K_Y=f^{\ast}K_X$, it follows that
\[
B^2=4A^2+4A\cdot K_X+K_X^2+E^2=4A^2+4A\cdot K_X+K_X^2-2>0.
\]
Suppose that $C$ is $f$-exceptional. Then from the above description of the exceptional set of $f$ it follows easily that $B\cdot C>0$. It remains to consider the case when $C$ is not $f$-exceptional.

\textbf{Claim:} Let $C$ be an integral curve in $Y$ which is not $f$-exceptional. Then
\begin{gather}\label{sec1-eq-00000}
(E\cdot C)^2\leq (-E^2)[K_Y \cdot C+\frac{(A\cdot \tilde{C} )^2}{A^2}+2]=2[K_Y \cdot C+\frac{(A\cdot \tilde{C} )^2}{A^2}+2]
\end{gather}
where $\tilde{C}=f_{\ast}C$.

Indeed. $f^{\ast}A$ is nef and big and hence from the generalized Hodge Index Theorem~\cite[Lemma 3.13]{Tz19} it follows that
\[
(aE+C)^2\leq \frac{(f^{\ast}A\cdot (aE+C))^2}{(f^{\ast}A)^2}=\frac{(A\cdot \tilde{C})^2}{A^2},
\]
for every $a\in \mathbb{R}$. Hence,
\[
E^2 a^2+2(E\cdot C)a +C^2-\frac{(A\cdot \tilde{C})^2}{A^2}\leq 0,
\]
for all $a \in \mathbb{R}$. From this it immediately follows that
\[
(E\cdot C)^2 -E^2[C^2-\frac{(A\cdot C)^2}{A^2}]\leq 0.
\]
Now from the equation (\ref{sec1-eq-7}) it follows that $-C^2\leq 2+K_Y \cdot C$. Therefore the previous inequality gives
\[
(E\cdot C)^2 \leq (-E^2)[-C^2+\frac{(A\cdot C)^2}{A^2}]\leq (-E^2)[K_Y\cdot C+\frac{(A\cdot C)^2}{A^2}+2].
\]
Considering that $E^2=-2$, the claim follows.

Let now
\[
\Lambda=2f^{\ast}A\cdot C+K_Y \cdot C+E\cdot C=2A\cdot \tilde{C} +K_X\cdot \tilde{C}+E\cdot C>0,
\]
since $K_X$ is nef, $A$ ample and $C\not= E$. Then from the inequality (\ref{sec1-eq-00000}) it follows that
\begin{gather*}
\Lambda (B\cdot C)=\Lambda (2f^{\ast}A \cdot C+K_Y\cdot C -E\cdot C)=\Lambda (2A\cdot \tilde{C}+K_X\cdot \tilde{C}-E\cdot C) = \\
(2A \cdot \tilde{C}+K_X\cdot \tilde{C})^2-(E\cdot C)^2\geq
(2A\cdot \tilde{C} +K_X\cdot \tilde{C})^2-(2K_X\cdot \tilde{C}+2\frac{(A\cdot \tilde{C})^2}{A^2}+4)=\\

2(A\cdot\tilde{C})^2(2-\frac{1}{A^2})+(K_X \cdot \tilde{C})(K_X \cdot \tilde{C}+4 (A\cdot \tilde{C})-2)-4
\end{gather*}
Suppose that $K_X \cdot \tilde{C}\not=0$. Then, since $K_X$ is nef and $A$ ample, it follows that $\Lambda (B\cdot C)>0$ and hence $B\cdot C>0$.

Suppose that $K_X \cdot \tilde{C}=0$. Let $g \colon X^{\prime}\rightarrow Y$ be the minimal resolution of $Y$. Then $X^{\prime}$ is also the minimal resolution of $X$. Moreover, $K_{X^{\prime}}=g^{\ast}K_Y$ and $K_{X^{\prime}}= h^{\ast}K_X$, where $h=fg$.

Suppose that $E\cdot C=0$. Then $B\cdot C=2A \cdot \tilde{C}>0$.  Suppose that $E\cdot C >0$. Let $C^{\prime}=g_{\ast}^{-1}C$ be the birational transform of $C$ in $X^{\prime}$. Then $C^{\prime}$ intersects the exceptional set of $h$ over $P\in X$.

Consider next cases with respect to the type of singularity of $P\in X$. I will only do the case when $P\in X$ is of type $A_n$. The rest are similar and are omitted.

Since $P\in X$ is an $A_n$ type singularity and $f$ the blow up of $P$, the dual graph of the exceptional set of $h$ over $P$ is
\[
\underset{E^{\prime}_1}{\bullet}-\underset{F_1}{\circ}-\cdots-\underset{F_{n-2}}{\circ}-\underset{E^{\prime}_2}{\bullet},
\]
where $E_i^{\prime}$ are the birational transforms of $E_i$ in $X^{\prime}$, $i=1,2$. The subgraph with the white bullets is the dual graph of the singularity of $Y$ over $P$, the curves $F_1, \ldots, F_{n-2}$ are the $g$-exceptional curves over $P$. A straightforward calculation shows that
\begin{gather*}
g^{\ast}E_1=E_1^{\prime}+\frac{n-2}{n-1}F_1+\frac{n-3}{n-1}F_2+\cdots + \frac{1}{n-1}F_{n-2},\\\nonumber
g^{\ast}E_2=E_2^{\prime}+\frac{1}{n-1}F_1+\frac{2}{n-1}F_2+\cdots + \frac{n-2}{n-1}F_{n-2}
\end{gather*}
And hence
\begin{gather}\label{sec1-eq-00002}
g^{\ast}(E_1+E_2)=E_1^{\prime}+F_1+\cdots +F_{n-2}+E_2^{\prime}.
\end{gather}
Now since $K_X{\prime}\cdot C^{\prime}=0$, it follows that $(C^{\prime})^2=-2$ and $C^{\prime}\cong \mathbb{P}^1$. Then since $C^{\prime}$ intersects $E$, it follows that $C^{\prime}\cup E_1^{\prime}\cup E_2^{\prime}\cup_{i=1}^{n-2}F_i$ is a connected contractible set of $(-2)$ curves on $X^{\prime}$. Hence their configuration must be of type $A_n$, $D_n$, $E_6$, $E_7$ or $E_8$. In any case, $C^{\prime}$ intersects exactly one of the $h$-exceptional curves with intersection multiplicity 1. Then from the equation (\ref{sec1-eq-00002}) it follows that
\[
C\cdot E=C\cdot (E_1+E_2)=f^{\ast}(E_1+E_2)\cdot C^{\prime}=1.
\]
But then,
\[
B\cdot C=2f^{\ast}A\cdot C +K_{Y}\cdot C-E\cdot C=A\cdot \tilde{C} -1>0,
\]
since $A$ is ample and Cartier on $X$.
\end{proof}

\begin{corollary}\label{sec1-cor-12}
Let $X$ be a canonically polarized surface with rational double points. Suppose that the singular locus of $X$ consists of the points $A^{\ast}_i$ of type $A_{n_i}$, $i=1,\ldots, r$, $D^{\ast}_j$ of type $D_{m_j}$, $j=1,\ldots, s$,
$E^{\ast}_{6,k}$ of type $E_6$, $k=1,\ldots , t$, $E^{\ast}_{7,\nu}$ of type $E_7$, $\nu=1,\ldots, w$ and $E^{\ast}_{8,\mu}$ of type $E_8$, $\mu=1,\ldots, u$. Let $f \colon X^{\prime} \rightarrow X$ be the minimal resolution of $X$ and $E_1, \ldots, E_n$ be the $f$-exceptional curves. Then there exist integers $a_i \geq 0$, $i=1,\ldots, n$ with $\sum_{i=1}^na_i < (2^{m+1}-1)^2K_X^2$ such that $B=(2^{m+1}-1)K_{X^{\prime}}-Z$ is ample on $X^{\prime}$, where $Z=\sum_{i=1}^na_iE_i$ and
\[
m=\sum_{i=1}^r \lceil\frac{n_i}{2}+1\rceil +\sum_{j=1}^sm_j+4t+7w+8u.
\]

\end{corollary}
\begin{proof}
Let $P\in X$ be a singularity of type $A_n$. Then $P\in X$ is resolved after $\lceil\frac{n}{2}+1\rceil$ blow ups. Let $P\in X$ be of type $D_n$. Then $P\in X$ is resolved after $n$ blow ups if $n$ is even and after $n-1$ blow ups if $n$ is odd. An $E_6$ type singularity is resolved after $4$ blow ups, an $E_7$ requires $7$ blow ups and an $E_8$ 8 blow ups. Now by successive applications of Proposition~\ref{sec1-prop-11} starting with $A=K_X$ it follows that there exists an effective divisor $Z=\sum_{i=1}^na_iE_i$ such that $B=(2^{m+1}-1)K_{X^{\prime}}-Z$ is ample.

Since $B$ is ample, $B^2>0$. Then,
\begin{gather}\label{sec1-eq-3333}
0 < B^2=[(2^{m+1}-1)K_{X^{\prime}}-Z]\cdot [(2^{m+1}-1)K_{X^{\prime}}-Z]=\\\nonumber
(2^{m+1}-1)^2K_{X^{\prime}}^2-(2^{m+1}-1)K_{X^{\prime}} \cdot Z-[(2^{m+1}-1)K_{X^{\prime}}-Z]\cdot Z =(2^{m+1}-1)^2K_{X}^2-B\cdot Z,
\end{gather}
since $K_{X^{\prime}}=f^{\ast}K_X$ and hence $K_{X^{\prime}} \cdot Z=0$ and $K_{X^{\prime}}^2=K_X^2$. Hence
\begin{gather*}
B\cdot Z=\sum_{i=1}^n a_i (B\cdot E_i) < (2^{m+1}-1)^2K_X^2,
\end{gather*}
and hence since $B$ is ample, $\sum_{i=1}^n a_i < (2^{m+1}-1)^2K_X^2$. This concludes the proof of the corollary.
\end{proof}

In the case when $X$ has a nontrivial global vector field, a case of interest in this paper, the following holds.

\begin{corollary}\label{sec1-cor-13}
Let $X$ be a canonically polarized surface with rational double points. Suppose that $X$ has a nontrivial global vector field. Let   $f \colon X^{\prime} \rightarrow X$ be its minimal resolution and $E_i$, $i=1,\ldots, n$ the $f$-exceptional curves. Then there exists $a_i \geq 0$, $i=1,\ldots, n$ with $\sum_{i=1}^na_i <(2^m-1)^2K_X^2$ such that  $(2^m-1)K_{X^{\prime}}-Z$ is ample, where  $m=17K_X^2+37$, $Z=\sum_{i=1}^n a_i E_i$.
\end{corollary}
\begin{proof}
From Corollary~\ref{sec1-cor-12} it follows that there exist $a_i\geq 0$ such  $(2^{d+1}-1)K_{X^{\prime}}-Z$ is ample, where $Z=\sum_{i=1}^na_iE_i$ and
\[
d=\sum_{i=1}^r \lceil\frac{n_i}{2}+1\rceil +\sum_{j=1}^sm_j+4t+7w+8u.
\]
Now from~\cite[Corollary 3.5]{Tz19} it follows that $d \leq 12  \chi(\mathcal{O}_X)+11K_X^2$  and from Noether's inequality, $2\chi(\mathcal{O}_X) \leq K_X^2+6$. Hence $d+1\leq 17K_X^2+37$. Since $K_{X^{\prime}}$ is nef and big it follows that $A=(2^m-1)K_{X^{\prime}}-Z$ is ample, where $m=17K_X^2+37$. The corollary now follows from Corollary~\ref{sec1-cor-12}.

\end{proof}

\begin{remarks}
\begin{enumerate}
\item The divisor $Z$ in Corollary~\ref{sec1-cor-12} can be explicitly calculated by using Proposition~\ref{sec1-prop-11} and the resolutions of rational double points. However this would increase the length of the paper without adding anything to the main results. For this reason I opted not to do it and instead give an upper bound in Corollary~\ref{sec1-cor-12} which will be sufficient for the needs of this paper.
\item The minimal resolution of $X$ in Corollary~\ref{sec1-cor-12} is a minimal surface of general type. Corollary~\ref{sec1-cor-12} can be rewritten in the following form: \textit{Let $X$ be a minimal surface of general type. Then $mK_{X}-Z$ is ample on $X$, where $Z$ is an effective divisor supported on the set of $(-2)$ curves on $X$}, where $m$ and $Z$ can be described as in Corrolary~\ref{sec1-cor-12}.
\end{enumerate}
\end{remarks}

\subsection{Contraction of elliptic curves in characteristic $p>0$.}
Let $E \subset X$ be a  elliptic curve in a smooth projective surface defined over an algebraically closed field $k$ of characteristic $p >0$. Suppose  that $E^2<0$. Unlike the case of smooth rational curves with negative self intersection, $E$ cannot always be contracted algebraically.

The next proposition shows that if an elliptic curve is contracted then it is contracted to a Gorenstein singularity.

\begin{proposition}\label{sec1-prop-100}
Let $X$ be a smooth surface and $E\subset X$ an elliptic curve in $X$. Suppose that there exists a projective morphism $f \colon X \rightarrow Y$ with the following properties.
\begin{enumerate}
\item $Y$ is a normal surface.
\item $f(E)$ is a point $P$ in $Y$.
\item $X-E \cong Y-P$.
\end{enumerate}
Then $K_Y$ is Cartier, i.e., $P\in Y$ is a Gorenstein singularity.
\end{proposition}
\begin{proof}
Let $A=\mathcal{O}_X(K_X+E)$. I will show that $L=f_{\ast}A$ is invertible in $Y$.

Suppose that this has been shown. Then there exists an injection $f^{\ast}L \rightarrow A$ and the restriction of this map to $X-E$ is an isomorphism. Therefore $\mathcal{O}_Y(K_Y)|_{Y-P} \cong L|_{Y-P}$. Then, since $Y$ is normal, it follows that $L \cong \mathcal{O}_Y(K_Y)$ and hence $Y$ is Gorenstein as claimed.

It remains to prove that $L=f_{\ast}A$ is invertible.  By the formal functions theorem,
\[
\hat{(f_{\ast}A)}_P \cong \lim_{\leftarrow}H^0(X,A\otimes \mathcal{O}_{nE}),
\]
There are also short exact sequences
\begin{gather}\label{sec1-eq-13}
0 \rightarrow \mathcal{O}_E(-(n-1)E)\rightarrow \mathcal{O}_{nE} \rightarrow \mathcal{O}_{(n-1)E} \rightarrow 0,
\end{gather}
for $n \geq 2$. I will show that for all $n\geq 1$, $A\otimes \mathcal{O}_{nE}\cong \mathcal{O}_{nE}$. For $n=1$, this is simply the adjunction formula for $E$. Assume by induction that
$A\otimes \mathcal{O}_{(n-1)E}\cong \mathcal{O}_{(n-1)E}$. Then tensoring the equation~\ref{sec1-eq-13} with $A$ we get that
\begin{gather}\label{sec1-eq-14}
0 \rightarrow \mathcal{O}_E(-(n-1)E)\rightarrow A\otimes \mathcal{O}_{nE} \rightarrow \mathcal{O}_{(n-1)E} \rightarrow 0,
\end{gather}
This can be seen as an exact sequence of $\mathcal{O}_{nE}$-modules. Such extensions are classified by $\mathrm{Ext}^1_{nE}(\mathcal{O}_{(n-1)E},\mathcal{O}_E(-(n-1)E))$. Now from the equation~\ref{sec1-eq-13} it follows that there exists an exact sequence
\begin{gather*}
0 \rightarrow \mathrm{Hom}(\mathcal{O}_E(-(n-1)E),\mathcal{O}_E(-(n-1)E) \rightarrow \\
\mathrm{Ext}^1_{nE}(\mathcal{O}_{(n-1)E},\mathcal{O}_E(-(n-1)E))
\rightarrow \mathrm{Ext}^1_{nE}(\mathcal{O}_{nE},\mathcal{O}_{E}(-(n-1)E))
\end{gather*}
But
\[
\mathrm{Ext}^1_{nE}(\mathcal{O}_{nE},\mathcal{O}_{E}(-(n-1)E))=H^1(E, \mathcal{O}_{E}(-(n-1)E))=H^0(\mathcal{O}_E((n-1)E)=0,
\]
since $E^2<0$. Therefore,
\[
\mathrm{Ext}^1_{nE}(\mathcal{O}_{(n-1)E},\mathcal{O}_E(-(n-1)E)) \cong k.
\]
Hence there exists exactly two isomorphism classes of extensions of $\mathcal{O}_{(n-1)E}$ by $\mathcal{O}_E(-(n-1)E)$. These are $\mathcal{O}_{nE}$ and $\mathcal{O}_{(n-1)E}\oplus \mathcal{O}_E(-(n-1)E)$.
Considering that $A\otimes \mathcal{O}_E \cong \mathcal{O}_E$, then $A\otimes \mathcal{O}_{nE}$ cannot be the second possibility. Hence $A\otimes \mathcal{O}_{nE}\cong \mathcal{O}_{nE}$, as claimed.

Then by the formal functions theorem,
\[
\hat{(f_{\ast}A)}_P \cong \lim_{\leftarrow}H^0(X,A\otimes \mathcal{O}_{nE})\cong \lim_{\leftarrow}H^0(X, \mathcal{O}_{nE})\cong \hat{\mathcal{O}}_{Y,P}.
\]
But this implies that $f_{\ast}A$ is locally free of rank 1 at $P$. Hence $f_{\ast}$ is an invertible sheaf as claimed.

\end{proof}
The next proposition shows that under certain restrictions an elliptic curve can be contracted.

\begin{proposition}\label{sec1-prop-7}
Let $X$ be a $\mathbb{Q}$-factorial normal projective surface over an algebraically closed field $k$ of characteristic $p>0$. Let $E \subset X$ be a  elliptic curve contained in the smooth part of $X$ such that $E^2<0$. Suppose that either $K_X$ is nef and big or $k=\bar{\mathbb{F}}_p$. Then there exists a projective morphism $f \colon X \rightarrow Y$, such that:
\begin{enumerate}
\item  $f(E)=P$, $P\in Y$ is a closed point.
\item There exists an open set $ V \subset Y$ such that $P\in V$ and $f^{-1}(V)-E \cong V-P$. Moreover, if either $X$ is smooth, or $K_X$ is ample or $k=\bar{\mathbb{F}}_p$, then $V=Y$, i.e., the exceptional set of $f$ is exactly $E$.
\item $K_Y$ is invertible at $P\in Y$.
\item If $k=\bar{\mathbb{F}}_p$, and $X$ is $\mathbb{Q}$-factorial, then $Y$ is also $\mathbb{Q}$-factorial.
\end{enumerate}
\end{proposition}

\begin{proof}
Since $X$ is $\mathbb{Q}$-factorial, there exists a positive integer $m$ such that $mK_X$ is Cartier. Let $A=\mathcal{O}_X(mK_X+mE)$. Then $A$ is an invertible sheaf. Since  $E$ is an elliptic curve contained in the smooth part of $X$, $A\otimes \mathcal{O}_E=\omega^m_E\cong\mathcal{O}_E$ and hence $A\cdot E=0$.

Suppose that $K_X$ is nef and big.  Then,
\[
A^2=m^2(K_X+E)^2=m^2(K_X^2+K_X\cdot E +(K_X+E)\cdot E) =m^2(K_X^2+K_X\cdot E) >0.
\]
Moreover, since $K_X$ is nef, $A\cdot C \geq 0$, for any irreducible curve $C$. Hence $A$ is nef and big.
Therefore by~\cite[Theorem 1.2]{Tan14}, $A$ is semi-ample. Therefore there exists $n>0$ such that $A^{\otimes n}$ is generated by global sections and hence for $n>>0$, $|nA|$ is base point free and defines a birational map $f \colon X \rightarrow Y$, where $Y$ is a normal projective surface. The exceptional set of $f$ are precisely the curves $C$ in $X$ such that $A\cdot C=0$. In particular $E$ is contracted by $f$. This shows part (1) of the proposition.

Let $C\not= E$ be another $f$-exceptional curve. Then, since $K_X$ is nef, $A\cdot C=0$ if and only if $C\cdot E=0$ and $K_X \cdot C=0$. Let $C_i$, $i=1,\ldots, s$ be the $f$-exceptional curves different from $E$. Let $V=Y-f(\cup_{i=1}^sC_i)$. Then $f^{-1}(V)-E \cong V-P$, where $f(E)=P$. Now by applying Proposition~\ref{sec1-prop-100} to the map $f^{-1}(V)\rightarrow V$ it follows that $P\in Y$ is a Gorenstein singularity.

Suppose that  $K_X$ is ample. Then there do not exist curves $C$ on $X$ such that $K_X \cdot C=0$  and hence the exceptional set of $f$ consists of only $E$, hence $V=Y$.

Suppose that $X$ is smooth. Then $X$ is a minimal surface of general type. Then by Corollary~\ref{sec1-cor-12}, there exists an integer $m>0$ and an effective divisor $Z=\sum_{i=1}^na_iF_i$ such that $F_i^2=-2$ for all $i$, and $mK_X-Z$ is ample. After renumbering the curves $F_i$, $i=1,\ldots, n$, there exists an integer $n^{\prime}\leq n$ such that $F_i \cdot E=0$ for all $i \leq n^{\prime}$ and $F_i \cdot E>0$, for all $i>n^{\prime}$. Let $A=m^{\prime}K_X-Z^{\prime}+m^{\prime}E$, where $m^{\prime}=2m^2K_X^2$ and $Z^{\prime}=\sum_{i=1}^{n^{\prime}}a_iF_i$. I will show that $A$ is nef and big and moreover, $A\cdot C>0$ unless $C=E$ in which case $A\cdot E=0$. Indeed. Write
\begin{gather}\label{sec6-eq-1}
A=(m^{\prime}-m)K_X+(mK_X-Z)+(Z-Z^{\prime})+m^{\prime}E.
\end{gather}
Then clearly, since $K_X$ is nef and big, $mK_X-Z$ ample and $Z-Z^{\prime}$ effective, $A\cdot C>0$ for all $C$ such that $C \not\subset Z-Z^{\prime}$ and $C\not= E$. Suppose that $C \subset Z-Z^{\prime}$. Then $Z=F_i$, for some $i>n^{\prime}$ and hence $C\cdot E>0$. Now from the equation (\ref{sec1-eq-3333}) in the proof of Corollary~\ref{sec1-cor-13} it follows that $\sum_{i=1}^na_i < m^2K_X^2$. In particular $a_i < m^2K_X^2$. Then,
\[
A\cdot C>a_iF_i^2+m^{\prime}E \cdot F_i=-2a_i+m^{\prime}E\cdot C\geq -2a_i+2m^2K_X^2>0.
\]
Now let $C=E$. Then since $E$ is elliptic and $E \cdot Z^{\prime}=0$, it follows that
\[
A \otimes \mathcal{O}_E=\mathcal{O}(m^{\prime}K_X+m^{\prime}E)\otimes \mathcal{O}_E=\omega^{m^{\prime}}_E\cong \mathcal{O}_E.
\]
Since $mK_X-Z$ is ample, $d(mK_X-Z)\sim W$, where $W$ is an effective divisor, for $d$ sufficiently large. Hence
\[
dA\sim d(m^{\prime}-m)K_X+W+d(Z-Z^{\prime})+dm^{\prime}E.
\]
Hence, since $A \cdot C>0$  for any curve $C\not= E$, $A\cdot E=0$ and $K_X$ is nef and big, it follows that $A^2>0$. Hence $A$ is nef and big. Therefore by~\cite[Corollary 0.3]{Ke99}, $A$is semiample and hence a multiple $|dA|$ defines a birational map $f \colon X \rightarrow Y$ whose exceptional set is exactly $E$. This shows parts (2) and (3) of the proposition.

Suppose that $k=\bar{\mathbb{F}}_p$, and $K_X$ not necessarily nef and big anymore. Let $H$ be an ample line bundle on $X$ and $m>0$ such that $(H+mE)\cdot E=0$. Then $H+mE$ is nef and big. Therefore by~\cite[Corollary 0.3]{Ke99}, $H+mE$ is semi ample. Hence, as above, some multiple of it defines a birational $X \rightarrow Y$ whose exceptional set is $E$. Moreover, if $X$ is $\mathbb{Q}$-factorial, $Y$ is $\mathbb{Q}$-factorial~\cite[Theorem 4.5]{Tan14}. This shows part (4) of the proposition.

\end{proof}
\begin{remark}
In the complex analytic case, any  elliptic curve in a smooth surface can be contracted to a Gorenstein singularity~\cite{Lau77}. However, Laufer's proof is complex analytic. It is likely that the proof of the previous propositions is known to the experts but since I was not able to find a reference to a proof that works over a general algebraically closed field, I included it in this paper for the convenience of the reader.
\end{remark}

\subsection{Effective base point free theorems in positive characteristic}

The following effective base point freeness result will be used for the proof of the main theorem.

\begin{theorem}[Theorem 4.1, Corollary 5.2~\cite{Wi17}]\label{sec1-prop-8}
Let $X$ be a normal projective surface defined over an algebraically closed field of characteristic $p>0$. Suppose that $X$ has $F$-pure singularities and that its minimal resolution is of general type. Suppose moreover that there exists $m\in \mathbb{Z}$ such that $p$ does not divide $m$ and $mK_X$ is Cartier. Let $A$ be an ample line bundle on $X$. Then
\begin{enumerate}
\item $4mK_X+14mA$ is base point free.
\item $13mK_X+45mA$ is very ample.
\end{enumerate}
\end{theorem}

\begin{remark}
In~\cite{Wi17} the previous theorem is proved with the additional assumption that $X$ is $\mathbb{Q}$-factorial. The condition that $X$ is $\mathbb{Q}$-factorial is used only for the proof of Lemma 2.10~\cite{Wi17} whose statement holds  if the index of $K_X$ is not divisible by $p$ and hence in this case $X$ need not be $\mathbb{Q}$-factorial.
\end{remark}

\subsection{Combinatorial results}
The following simple combinatorial results will be used for the proof of the main theorem.

\begin{lemma}\label{sec1-prop-9}
Let $X$ be a smooth projective surface and $C=\sum_{i=1}^mn_iC_i$ be a divisor on $X$ such that $C_{red}$ is either a chain or a cycle on $X$. Moreover, assume that $C_i$ is a reduced and irreducible curve such that $C_i^2 \leq -2$, for all $i=1,\ldots, m$. Then $C^2\leq 0$.
\end{lemma}

\begin{proof}

Suppose that $C_i^2=-d_i$, $i=1,\ldots, m$. Then by assumption, $d_i \geq 2$ for all $i$.

Suppose that $C_{red}$ is a cycle. The case when $C_{red}$ is a chain is similar and its proof is omitted.  Then
\begin{gather}\label{sec1-eq-15}
-C^2=\sum_{i=1}^kn_i^2d_i-2 \sum_{i=1}^{k-1}n_in_{i+1}-2n_1n_k \geq 2\sum_{i=1}^kn_i^2-2 \sum_{i=1}^{k-1}n_in_{i+1}-2n_1n_k
\end{gather}

Now by using the elementary identity $-2ab=(a-b)^2-a^2-b^2$, the previous inequality becomes
\[
-C^2\geq 2\sum_{i=1}^kn_i^2 +\sum_{i=1}^{k-1}(n_i-n_{i+1})^2 +(n_1-n_k)^2 -2\sum_{i=1}^kn_i^2 \geq 0.
\]
In fact, $C^2<0$ unless $C$ is reduced, and $C_i^2=-2$ for all $i=1,\ldots, k$,

The following completely elementary result will be useful in obtaining bounds for the index of the various surfaces that appear in the proof of the main theorem.

\begin{lemma}\label{sec1-prop-10}
Let $d_i$, $i=1,\ldots, k$ be positive integers such that $\sum_{i=1}^k d_i \leq n$. Let $d$ be the least common multiple of $d_1,\ldots, d_k$. Then
\[
d \leq
\begin{cases}
e^{n/e} , & \mathrm{if}\; n\geq 1\\
\left(\frac{1+\sqrt{1+8n}}{4}\right)^{\frac{-1+\sqrt{1+8n}}{2}}, & \mathrm{if}\;  n\geq  15
\end{cases}
\]
In particular, if $n \geq 15$, then the simplified inequality holds,
\[
d \leq \left(\frac{n}{2}\right)^{\sqrt{\frac{n}{2}}}.
\]

\end{lemma}
\begin{proof}
We can assume that $d_i \not=d_j$, for all $i\not= j$. Suppose then that $1\leq d_1<d_2<\cdots < d_k\leq n$. Then
\[
\frac{k(k+1)}{2}=1+2+\cdots + k < d_1+\cdots +d_k \leq n.
\]
Hence $k(k+1)/2<n$ and therefore $k^2+k<2n$ and hence $k<(-1+\sqrt{1+8n})/2$. Hence by the geometric mean value inequality
\[
d\leq d_1\cdots d_k \leq \left(\frac{d_1+\cdots+ d_k}{k}\right)^k <\left(\frac{n}{k}\right)^k.
\]
The lemma now follows by studying, using elementary calculus methods, the maximum value of the function $f(x)=(n/x)^{x}$ in the interval $[1,(-1+\sqrt{1+8n})/2]$.

In particular, considering that $[1,(-1+\sqrt{1+8n})/2]\subset [1,\sqrt{2n}]$, then by studying the maximum value of $f(x)$ in this interval, we find the second part of the statement of the lemma.
\end{proof}

\end{proof}

config: arxiv


text:
\section{Proofs of Section \ref{sec:optimal_drawing}}
\label{app:proof_proba}

Propositions in Section \ref{subsec:opt_proba} are direct consequences of the minimization of the bounds obtained on the number of measurements in Theorems of Section \ref{sec:main_results}, given the following lemma.

\begin{lemme}
\label{lem:max_over_simplex}
Given $(\gamma_k)_{1\leq k \leq p}$ such that for all $1\leq k \leq p, \gamma_k\geq 0$. Define for $\pi$ a probability distribution on $\{1, \hdots , p \}$, the following function:
$$
K( \pi) := \max_{1\leq k \leq p} \frac{\gamma_k}{\pi_k}.
$$
Then, it holds that
$$
\min_{\pi_k \geq 0 \atop \sum_k \pi_k =1} K(\pi) = \sum_{j=1}^p \gamma_j,
$$
and the unique minimizer $\pi^\star$ of $K$ over discrete probability distributions is for all $1 \leq k \leq p$
$$
\pi_k^\star = \frac{\gamma_k}{\sum_{j=1}^p \gamma_j}.
$$
\end{lemme}

\begin{proof}
The proof is inspired by an intermediary result in \cite{chauffert2013variable}. We give the proof here for completeness. Note that for $\pi^\star$, $K(\pi^\star) = \sum_{j=1}^p \gamma_j$. For a probability distribution $q \neq \pi^\star$, there exists an index $j_0$ such that $q_{j_0} < \pi^\star_{j_0}$ since both sum to 1. Then, $K(q) \geq \gamma_{j_0} / q_{j_0} > \gamma_{j_0} / \pi^\star_{j_0} = \sum_{j=1}^p \gamma_j = K(\pi^\star)$.
\end{proof}

Therefore, the proofs of Propositions \ref{prop_opt_drawing_proba}, \ref{prop_opt_proba_iso}, \ref{prop_opt_drawing_proba_lambda}, \ref{prop_opt_drawing_proba_iso_lambda} easily follow Lemma \ref{lem:max_over_simplex} combined respectively with Theorems \ref{thm:noisy}, \ref{thm:killthetalog}.


TITLE: Connection formulas: Why do we assume asymptotic behavior of the Airy functions?
QUESTION [1 upvotes]: The derivation is obtained from Introduction to Quantum Mechanics by Griffiths
Let assume the turning point occurs at $x=0$, then the WKB solutions right and left to the turning point are:
$$
\psi=\dfrac{C}{\sqrt{|p|}}\exp{\left(-\dfrac{1}{\hbar}\left|\int_0^x pdx\right|\right)} \quad \text{for} \quad x>0,   \tag{8.31a}
$$
$$
\psi=\dfrac{C_1}{\sqrt{p}}\exp{\left(\dfrac{i}{\hbar}\int_0^x pdx\right)}+\dfrac{C_2}{\sqrt{p}}\exp{\left(-\dfrac{i}{\hbar}\int_0^x p dx\right)}\quad \text{for} \quad x<0.\tag{8.31b}
$$
By assuming the potential near $x=0$ to be
$$
V(x) \cong E + V'(0)x \tag{8.32}
$$
and solving the Schrödinger equation, he takes the patching wave function as
$$
\psi_p(x) = aAi(\alpha x) + bBi(\alpha x).\tag{8.37}
$$
Then for $x > 0$, he uses the asymptotic forms of the Airy functions to obtain
$$
\psi_p(x) \cong \dfrac{a}{2\sqrt{\pi}(\alpha x)^{1/4}}\exp(-\dfrac{2}{3}(\alpha x)^{3/2}) + \dfrac{b}{\sqrt{\pi}(\alpha x)^{1/4}}\exp(\dfrac{2}{3}(\alpha x)^{3/2}),\tag{8.40}
$$
and use this to obtain the constant $C$,
My question is, why can we assume $x \gg 0$? Shouldn't this wave function stay in the neighborhood of the origin?

REPLY [2 votes]: Actually, Griffiths does not assume that the position coordinate $x\to \infty$ is unbounded, cf. footnote 8 on p. 288.  To the contrary, he assumes that $x$ does not leave the patching region so that the linear approximation (8.32) to the potential is valid, cf. Fig. 8.9. Instead he assumes that the argument (to the Airy function)
$$ z~\equiv~\alpha x~\gg~1 \tag{8.35}$$
is large, where
$$ \alpha~~\equiv~\left[\frac{2m}{\hbar^2}V^{\prime}(0)\right]^{1/3}. \tag{8.34}$$
Semiclassically, $\alpha\to \infty$ for $\hbar\to 0$. Griffiths makes the assumption $z\gg 1$, since he is only after the leading semiclassical behavior.
For alternative derivations of the WKB connection formulas, see e.g. this related Phys.SE post.


TITLE: Transforming an arbitrary quadrilateral to a unit square
QUESTION [2 upvotes]: In this answer from Pedro Gimeno he proposed the following transformation to map the points of any arbitrary quadrilateral to the unit square

$$\pmatrix{x'\\y'} =
> \pmatrix{u_x&v_x&w_x\\u_y&v_y&w_y}\pmatrix{x\\y\\xy}$$
$$x'=u_xx+v_xy+w_xxy\\ y'=u_yx+v_yy+w_yxy$$
It transforms the unit square in a way controlled by the vectors
  $u=(u_x,u_y), v=(v_x,v_y), w=(w_x,w_y)$ as follows:
Geometric interpretation of u, v, w (sorry, I can't post images)

However, I'd like to do the opposite. I need to integrate on an arbitrary quadrilateral and I'd like to define my integral on the unit square and then map the coordinates of the unit square on the original quadrilateral .
The matrix Pedro proposed, however, is not invertible and I'm not sure a pseudo-inverse matrix is what I'm looking for.
How can I obtain the opposite transformation?

REPLY [0 votes]: Perhaps, you could use perspective projection matrix in homogenous coordinates.

Reversing this operation is just a matter of finding an inverse of this transformation matrix.
This will work for many cases. However, it doesn't map to any quadrilateral. For one, I think, the quadrilateral will still have to be convex.
(Image source: Wikimedia Commons, Author: D.stebani, License: CC BY-SA 4.0)


TITLE: Pareto allocation problem when the utility is decreasing in one good
QUESTION [0 upvotes]: I have to characterize the Pareto optimal allocation of the following problem:
Consider two-agents-two-goods economy. The preferences of the agents are given by the following
utility functions:
$\upsilon_1(x_{1,1},x_{2,1})=x_{1,1}-0.25x^2_{2,1}$,
$\upsilon_2(x_{1,2},x_{2,2})=x_{1,2}+x^{0.5}_{2,2}$
where $x_{j,i}$ denotes the consumption of good j by agent i and agents have the following endowements $e_1=(0,8)$ and $e_2=(16,8)$.
I know that since the second good for the first consumer is a "bad good" in a Pareto optimal allocation $x_{2,1}$ will be equal to zero.
My question is if there is a way to obtain this results mathematically?

REPLY [0 votes]: With positive consumption, the marginal utility of $x_2$ is negative for the first consumer, and positive for the second consumer, $$\frac{\partial u_{1}}{\partial x_2}<0 \text{ and } \frac{\partial u_{2}}{\partial x_2}>0$$. The Pareto optimal allocations $(a,0),(b,16)$, where $a+b=16$.


TITLE: How does Wolfram Alpha come up with the substitution $x = 2\sin u$? Integration/Analysis
QUESTION [6 upvotes]: I have to integrate
$$
\int_0^2 \sqrt{4-x^2} \, dx
$$
I looked at the Wolfram Alpha step by step solution, and first thing it does is it substitutes
$x = 2\sin(u)\text{ and } \,dx = 2\cos(u)\,du$
How does it know to substitute $2\sin(u)$ for $x$?

REPLY [3 votes]: Theóphile's comment is a good answer. I will give another.
The $\sqrt{4-x^2}$ and indeed anything of the form $\sqrt{a^2-b^2}$ --- think $\sqrt{x^2-b^2}$ or $\sqrt{a^2-x^2}$ --- or $x^2+b^2$ can be considered --- via Pythagoras --- as sides of a right-angled triangle.
In this example we have $\sqrt{2^2-x^2}=b\Rightarrow b^2+x^2=2^2$ so that we have a right-angled-triangle with sides $x$, $b=\sqrt{2^2-x^2}$ and hypotenuse $2$.
Draw this triangle. Now choose an angle $\theta$ in the triangle so that you can write
$$\sec/\tan/\sin\theta=\frac{x}{2}.$$
In this case we have $\sin \theta=\frac{x}{2}$.
Now, using the triangle you sketched,
$$\cos\theta=\frac{\sqrt{4-x^2}}{2}\Rightarrow \sqrt{4-x^2}=2\cos\theta.$$
We can also get a handle on $dx$ no problem.
When we eventually antidifferentiate, our answer will be in terms of $\theta$. We can find any of the trig ratios in terms of $x$ using our triangle and $\theta$... well $\theta$ is than angle whose sine is $\frac{x}{2}$ so
$$\theta=\sin^{-1}\left(\frac{x}{2}\right).$$
... not that it arises in this question.


TITLE: The product of two sequences where one is bounded and the other is convergent to zero
QUESTION [0 upvotes]: I'm trying to prove, given only that one sequence $B$ is bounded and the other sequence $A$ is convergent to zero, that their product also converges.
My work probably wouldn't be considered a formal proof but I'm curious if my strategy/thought process is on the right track.
My work:
Since $A$ converges to zero, there exists a point in this sequence where all terms are between $-1$ and $1$.
In this case, whatever behavior $B$ has will be dictated by the terms of $A$.
If $B=0$, then the conclusion is trivial.
Thus, $AB$ converges to zero.

REPLY [1 votes]: Here's a formal one: Let $M>0$ be such that $|y_{n}|\leq M$. Let $\{x_{n}\}$ be such that $x_{n}\rightarrow 0$. Given $\epsilon>0$, for the positive number $\epsilon/M$, there exists some $N$ such that $|x_{n}|<\epsilon/M$ for all $n\geq N$, so $|x_{n}y_{n}|\leq M|x_{n}|<M\cdot\epsilon/M=\epsilon$ for all such $n$, this shows that $x_{n}y_{n}\rightarrow 0$.


TITLE: Let $M$ be a non-trivial $R$-module having no non-trivial submodule; then every non-trivial $R$-homomorphism on $M$ is isomorphism?
QUESTION [1 upvotes]: Let $M$ be an irreducible $R$-module such that $\exists r_0 \in R$, $\exists m_0 \in M$ such that $r_0m_0 \ne 0$. Then how to prove that any $R$-homomorphism $T$ of $M$ into $M$ is either an isomorphism or is trivial ( i.e. $T(m)=0, \forall m \in M$ )

REPLY [0 votes]: Let $T\colon M\to M$ be an $R$-homomorphism such that $T(x)\ne0$ for some $x\in M$.
Then $x\notin\ker T$, which forces $\ker T=\{0\}$, because the $R$-submodules of $M$ are $\{0\}$ and $M$.
Similarly the image of $T$ is an $R$-submodule of $M$ and it is not $\{0\}$, so…