text:
\begin{document}

\title{Weyl and Zariski chambers on K3 surfaces}
\author{Thomas Bauer and Michael Funke}
\date{June 2, 2010}
\maketitle
\thispagestyle{empty}

\begin{abstract}
   The big cone of every K3 surface admits two natural chamber
   decompositions: the decomposition into Zariski chambers, and
   the decomposition into simple Weyl chambers. In the present
   paper we compare these two decompositions and we study their
   mutual relationship: First, we give a numerical criterion for
   the two decompositions to coincide. Secondly, we study the
   mutual inclusions of Zariski and simple Weyl chambers.
   Finally, we establish the fact that -- even though the
   decompositions themselves may differ -- the number of Zariski
   chambers always equals the number of simple Weyl chambers.
\end{abstract}

\section*{Introduction}

   The purpose of this note is to study two natural chamber
   decompositions of the big cone on K3 surfaces: On the one hand
   one has the decomposition into Zariski chambers, which is
   important when considering base loci and volumes of line
   bundles (see \cite{BKS}), and on the other hand one has the
   well-known decomposition into simple Weyl chambers, which is
   given by the hyperplanes that are dual to the negative curves.
   We compare these two decompositions and we study their
   geometry -- by determining the mutual inclusions of chambers
   and by asking for the number of chambers into which the big
   cone is divided.

   For a more detailed description, consider a smooth projective
   variety $X$ of dimension $n$
   and a divisor $D$ on $X$. One defines the
   \textit{volume} of $D$ as the real number
   $$
      {\rm vol}_X(D)\stackrel{\rm def}{=}\limsup_k{\frac{h^0(X,kD)}{k^n/n!}}\ , \
   $$
   and one calls $D$ \textit{big} if its volume is positive. The
   big cone $\BigCone(X)$ is then the cone in the N\'eron-Severi
   vector space $\NS(X)\tensor\R$ that is generated by the big
   divisors. These have recently attracted a lot of attention, as
   it turned out that many geometric, cohomological, and
   numerical aspects of the picture that one has classically for
   ample line bundles extend to big divisors (see
   \cite{nak1,PAG,elmnp1,elmnp}). The main result of \cite{BKS}
   states that on surfaces the big cone admits a locally finite decomposition
   into rational locally polyhedral subcones such that
   \begin{itemize}\compact
   \item
      on each subcone the volume
      function is given by a single polynomial of degree two, and
   \item
      in the interior of each of the subcones the stable base
      loci are constant.
   \end{itemize}
   Both facts are explained by the variation of the Zariski
   decomposition of big divisors: In the interior of the subcones
   the support of the negative part of the Zariski decomposition
   is constant. These subcones have therefore been called
   \emph{Zariski chambers} in~\cite{BKS}.

   Suppose now that $X$ is a K3 surface. In that case,
   an interesting second point of view is provided by the
   \textit{Weyl chamber decomposition}. We consider here in
   particular \textit{simple} Weyl chambers,
   which are defined by the intersection
   behaviour of big divisors with $(-2)$-curves
   (see Sect.~\ref{sect-chambers} for details).

   Given that there are two natural decompositions of the big
   cone of a K3 surface, it is an obvious task to compare them.
   In \cite[Sect.~3.2]{BKS} the erroneous claim was made that the
   two decompositions always coincide. They may however differ,
   depending on the geometry of the surface, and our first result
   gives the precise condition when this is the case:

\begin{introtheorem}\label{introthm-equality-criterion}
   Let $X$ be a K3 surface. Then the following conditions are
   equivalent.
   \begin{itemize}\compact
   \item[(i)]
      The interiors of the Zariski chambers coincide with the
      simple Weyl chambers.
   \item[(ii)]
      There is no pair of
      $(-2)$-curves $C_1,C_2\subset X$ such that
      $$
         C_1\cdot C_2= 1 \,.
      $$
   \end{itemize}
   There exist K3 surfaces such that (i) and (ii) hold, and
   there exist K3 surfaces where these conditions do not hold.
\end{introtheorem}

   This theorem is a consequence of the following more general
   result about the geometry of Zariski and Weyl chambers. To
   state it, we will use the following notation: When $\mathcal
   S=\set{\liste C1r}$ is a set of $(-2)$-curves whose
   intersection matrix is negative definite, then $Z_{\mathcal
   S}$ will denote the Zariski chamber consisting of the big divisors
   whose negative part is supported on $\longunion C1r$,
   and $W_{\mathcal S}$ will
   denote the simple Weyl chamber consisting of the divisors that
   intersect $\liste C1r$ negatively and all other $(-2)$-curves
   positively (see Sect.~\ref{sect-chambers} for the details).
   Furthermore, we will write $\interior Z_S$ to denote the
   interior of the chamber $Z_S$.

\begin{introtheorem}\label{introthm-inclusions}
   Let $X$ be a K3 surface, and let $\mathcal S$ be a set of
   $(-2)$-curves on $X$ whose intersection matrix is negative
   definite.
   \begin{itemize}\compact
   \item[(i)]
      We have $W_{\mathcal S}\subset Z_{\mathcal S}$ if and only
      if the following condition holds: If $C'$ is a curve with
      $C'\notin\mathcal S$ such that the intersection matrix of
      the set $\mathcal S\cup\set{C'}$ is
      negative definite, then $C'\cdot C=0$ for all $C\in\mathcal
      S$.
   \item[(ii)]
      We have $\interior Z_{\mathcal S}\subset W_{\mathcal S}$ if and only
      if $C_1\cdot C_2\ne 1$ for all curves $C_1,C_2\subset S$.
   \end{itemize}
\end{introtheorem}

   We will study in Sect.~\ref{sect-example} in detail examples
   of K3 surfaces where various cases of inclusions and
   equalities of chambers occur.

   Given the fact that the decompositions may well be different,
   it is nice and somewhat surprising that on any K3 surface
   there are precisely as many Zariski chambers as there are Weyl
   chambers:

\begin{introtheorem}
   Let $X$ be a K3 surface. Then there is a canonical bijection
   between the set of Zariski chambers in $\BigCone(X)$ and the
   set of simple Weyl chambers in $\BigCone(X)$.
\end{introtheorem}

\section{Chamber decompositions}\label{sect-chambers}

\paragraph{Zariski chambers.}
   We recall very briefly the Zariski chamber decomposition from~\cite{BKS}.
   Let $X$ be a smooth projective surface over $\mathbb C$. To any big and nef
   $\R$-divisor $P$, one associates the \textit{Zariski
   chamber} $\Sigma_P$, which by definition
   consists of all divisors in
   $\BigCone(X)$ such
   that the irreducible curves
   in the negative part of
   the Zariski decomposition of $D$ are precisely the curves $C$
   with
   $P\cdot C=0$.
   One has by \cite[Lemma~1.6]{BKS}:

\begin{quote}\it
   For two big and nef divisors
   $P$ and $P'$, the Zariski
   chambers $\Sigma_P$ and $\Sigma_{P'}$ are
   either equal or disjoint.
   The Zariski chambers yield a decomposition of the
   big cone.
\end{quote}
   If $H$ is an ample divisor, then the interior of the
   chamber $\Sigma_H$ is the
   ample cone, its closure is the nef cone.
   (Note that $\Sigma_H$ itself need not be open or closed.)
   By way of abbreviation,
   we will refer to this chamber as the \emph{nef chamber} in the
   sequel.
   We will make frequent use of the following basic
   observation.

\begin{lemma}\label{lemma-neg-def}
   The set of Zariski chambers on a smooth projective surface
   $X$ that are different from the
   nef chamber is in bijective correspondence with the set of
   reduced divisors on $X$ whose intersection matrix is negative
   definite.
\end{lemma}

\begin{proof}
   The statement is Proposition~1.1 from \cite{BFN}. For the
   benefit of the reader, we briefly give the argument.
   For a chamber $\Sigma_P$ we consider the irreducible
   curves $\liste C1r$ with $P\cdot C_i=0$. Then the divisor
   $C_1+\dots+C_r$ has negative definite intersection matrix
   thanks to the index theorem.
   Conversely, given a reduced divisor
   $C_1+\dots+C_r$ with negative definite intersection matrix,
   we consider
   the divisor
   $$
      D \eqdef H+k(C_1+\dots+C_r) \,,
   $$
   where $H$ is a fixed ample divisor and $k$ a positive integer.
   This divisor is big, and for $k\gg0$ the negative part of
   its Zariski decomposition will have
   $C_1\cup\dots\cup C_r$ as its support.
   (The latter fact can for instance
   be seen from the computation of the
   Zariski decomposition according to \cite{Bau}.)
\end{proof}

   It will be useful to introduce two abbreviations: If $D$ is a
   big divisor on a surface, then $\Null(D)$ will denote the set
   of irreducible curves $C$ such that $D\cdot C=0$. If
   $D=P_D+N_D$
   is the Zariski decomposition of $D$, with nef part $P_D$ and
   negative part $N_D$, then $\Neg(D)$ will
   denote the set of components of $N_D$.
   In this notation, the Zariski chamber associated with a big
   and nef divisor $P$ is
   by definition
   $$
      \Sigma_P=\sset{D\in\BigCone(X)\with\Neg(D)=\Null(P_D)} \,.
   $$

\paragraph{Weyl chambers.}
   Let now $X$ be a K3 surface. Apart from the Zariski chamber
   decomposition, there is a second natural decomposition of the
   big cone: the decomposition into Weyl chambers. Consider the
   set $\mathcal R(X)$ of irreducible $(-2)$-curves (smooth
   rational curves of self-intersection $-2$) on $X$, also referred to as
   the set of \textit{simple roots} on $X$. Via the intersection
   product, each of the curves $C\in\mathcal R(X)$ defines a
   hyperplane $C^\perp=\set{D\with D\cdot C=0}$ in $\NS_\R(X)$.
   The connected components of the complement
   $$
      \BigCone(X)\setminus\bigcup_{C\in\mathcal R(X)}C^\perp
   $$
   yield a decomposition of (a dense open subset of)
   the big cone into subcones, the
   \textit{simple Weyl chambers}. The term \textit{simple}
   relates here to the fact that classically one considers not
   only simple roots, i.e, irreducible $(-2)$-curves, but
   \textit{roots}, i.e., $(-2)$-classes; taking chambers with
   respect to $(-2)$-classes leads in general to a finer
   decomposition, whose sets are called \textit{Weyl chambers}.
   Also note that classically one considers instead of the big
   cone $\BigCone(X)$ the \textit{positive cone} $C^+(X)$, i.e.,
   the cone of divisors $D$ with $D^2>0$ and $D\cdot H>0$ for
   some ample $H$. This cone, which is a subcone of the big cone,
   enjoys the advantage of being invariant under the Weyl group.
   In our situation, however, when comparing with the Zariski
   chamber decomposition, it is more natural to work in the big
   cone.

\paragraph{Comparing decompositions.}

   Our first result compares the two decompositions of
   $\BigCone(X)$.
   Proposition 3.9 of \cite{BKS} asserts that the Zariski chamber
   decomposition coincides on K3 surfaces with the Weyl chamber
   decomposition, but its proof is erroneous.
   Instead, the two decompositions may differ, and
   the following result gives the precise condition when this
   happens.

\begin{theorem}\label{thm-equality-criterion}
   Let $X$ be a K3 surface. Then the following conditions are
   equivalent.
   \begin{itemize}\compact
   \item[(i)]
      The interiors of the Zariski chambers coincide with the
      simple Weyl chambers.
   \item[(ii)]
      There is no pair of
      $(-2)$-curves $C_1,C_2\subset X$ such that
      $$
         C_1\cdot C_2= 1 \,.
      $$
   \end{itemize}
   There exist K3 surfaces such that (i) and (ii) hold, and
   there exist K3 surfaces where these conditions do not hold.
\end{theorem}

   The equivalence of (i) and (ii) is a consequence of
   Propositions \ref{prop-weil-in-zariski} and
   \ref{prop-zariski-in-weil} below, which
   give criteria for
   mutual inclusions of Zariski and Weyl chambers.
   We thought,
   however, that it might be useful to provide a quick
   direct argument
   for this basic result right away.

\begin{proof}
   Suppose first that condition (ii) does not hold, i.e., that
   there is a pair of $(-2)$-curves $C_1,C_2$
   with $C_1\cdot C_2=1$. Choose an ample divisor $H$ and
   consider the big divisor
   $$
      D=H+a_1C_1+a_2C_2
   $$
   for positive rational numbers $a_i$. Its Zariski decomposition
   is of the form
   $$
      D=(H+x_1C_1+x_2C_2)+(b_1C_1+b_2C_2) \,,
   $$
   where $x_i\le a_i$ and $b_i\le a_i$ are non-negative rational numbers
   satisfying
   $x_i+b_i=a_i$ and
   \begin{equation}\label{determine-Z}
      (H+x_1C_1+x_2C_2)\cdot C_i=0
   \end{equation}
   for $i=1,2$.
   We claim that we can choose the numbers $a_1,a_2$ in such a way that
   $$
      \Neg(D)=\set{C_1,C_2}
      \quad\mbox{and}\quad
      D\cdot C_1>0,\ D\cdot C_2<0 \,.
      \eqno(*)
   $$
   Granting $(*)$ for a moment, we consider the big divisor
   $$
      D' = H+k(C_1+C_2)
   $$
   for $k>0$.
   Thanks to the equality $C_1\cdot C_2=1$ we
   have $D'\cdot C_1<0$ and $D'\cdot C_2<0$ for $k\gg 0$, and hence
   $\Neg(D')=\set{C_1,C_2}$ for $k\gg 0$.
   So we see that $D$ and $D'$ lie
   in the same Zariski chamber, whereas they lie
   in different simple Weyl chambers.

   Turning to the proof of $(*)$, note first
   that the $x_i$ are independent of the $a_i$,
   since they are uniquely determined by the system of linear equations
   \eqnref{determine-Z}, whose coefficient matrix is negative
   definite.
   Keeping this in mind, we see from the equations
   \be
      &&D\cdot C_1=N\cdot C_1=(b_1C_1+b_2C_2)C_1=-2b_1+b_2 \\
      &&D\cdot C_2=N\cdot C_2=(b_1C_1+b_2C_2)C_2=b_1-2b_2
   \ee
   that $(*)$ will be fulfilled if we take
   $$
      a_1=x_1+1 \quad\mbox{and}\quad a_2=x_2+3 \,.
   $$

   Suppose now that condition (ii) holds.
   Let $D$ be a big divisor that does not lie on the boundary of
   any Zariski chamber.
   It is enough to show that if
   $C$ is any $(-2)$-curve, then
   we have $D\cdot C<0$ if $C\in\Neg(D)$, and $D\cdot C>0$
   otherwise.
   Let $D=P+N$ be the Zariski decomposition.
   Then $N=\sum b_iC_i$ with positive rational numbers $b_i$ and
   $(-2)$-curves $C_i$. Thanks to negative definiteness and
   condition (ii), we have $C_i\cdot C_j=0$ for $i\ne j$.
   Therefore if $C=C_1$, say, then
   $$
      D\cdot C=N\cdot C=b_1 C_1^2=-2b_1<0 \,.
   $$
   If $C\not\in\Neg(D)$, then clearly $D\cdot C\ge 0$.
   But in fact we have $D\cdot C>0$, since otherwise
   $C\in\Null(P)$, which by \cite[Prop.~1.7]{BKS}
   would imply that $D$ lies on the boundary
   of a Zariski chamber.

   Finally, we wish to show that there are cases where (i) and
   (ii) hold and cases where these conditions do not hold.
   First,
   there are smooth quartic surfaces in $\P^3$ that contain a
   pair of intersecting lines (see Sect.~\ref{sect-example}),
   and hence
   (ii) does not
   hold on such surfaces.
   Secondly, in order to get examples where (ii) does hold, one could
   take
   K3 surfaces of Picard number one -- there
   are no $(-2)$-curves at all on such surfaces, so that (ii) is trivially
   satisfied.
   More substantially, there are examples of K3
   surfaces of Picard number three containing three
   $(-2)$-curves, where no two of them intersect with
   intersection number one
   (see
   Proposition~\ref{prop-K3-coinciding}).
\end{proof}

   We just saw that the Zariski chamber decomposition may differ
   from the decomposition into simple Weyl chambers.
   By contrast, our next result shows
   that, somewhat
   surprisingly, the \textit{number}
   of Zariski chambers always equals the number of simple Weyl
   chambers.

\begin{theorem}\label{thm-bijection}
   Let $X$ be a K3 surface. Then there is a canonical
   bijection between the
   set of Zariski chambers in $\BigCone(X)$ and the set of simple
   Weyl chambers in $\BigCone(X)$.
\end{theorem}

   Note that the number of chambers may well be infinite, as the
   number of $(-2)$-curves may be infinite
   (cf.~\cite[Remark~7.2]{Kov}).

\begin{proof}
   As before,
   denote by $\mathcal R(X)$ the set of \textit{simple roots}, i.e.,
   the set of $(-2)$-curves on $X$. The set of simple Weyl
   chambers on $X$
   is in bijective correspondence with the set
   $$
      \mathcal W(X)=\sset{\tab[c]{subsets $\mathcal S\subset\mathcal R(X)$ such that
         there is a divisor $D\in\BigCone(X)$ \\ with $D\cdot C<0$ if
         $C\in\mathcal S$
         and $D\cdot C>0$ if $C\in\mathcal R(X)\setminus\mathcal
         S$}}\,,
   $$
   whereas, thanks to
   Lemma~\ref{lemma-neg-def}, the set of Zariski chambers on $X$ is in bijective
   correspondence with the set
   \begin{equation}\label{ZX}
      \mathcal Z(X) =\sset{\tab[c]{finite subsets $\mathcal S\subset\mathcal R(X)$ whose
      intersection matrix \\ is negative definite}}\cup\emptyset
      \,.
   \end{equation}
   (In both cases, the empty set $\emptyset$ corresponds to the
   nef chamber.)
   We will show that $\mathcal W(X)=\mathcal Z(X)$.
   To see this, consider first
   a non-empty set $\mathcal S=\set{\liste C1r}\in\mathcal Z(X)$, and fix an
   ample divisor $H$.
   Then for any non-negative rational numbers $a_i$, the divisor
   $$
      D=H+\sum_i a_iC_i
   $$
   is big.
   We claim that we can choose the $a_i$ in such a way that
   \begin{equation}\label{arrange-a-i}
      D\cdot C_i<0 \qquad\mbox{for all $i$.}
   \end{equation}
   Since clearly $D\cdot C>0$ for all curves $C$ different from
   the $C_i$,
   it follows then that the set $\mathcal S$ is contained
   in $\mathcal W(X)$. To prove \eqnref{arrange-a-i}, consider the
   following system of linear equations for the $a_i$,
   $$
      D\cdot C_j=H\cdot C_j+\sum_{i=1}^r a_iC_i\cdot C_j = -1 \qquad\mbox{for
      $j=1,\dots,r$.}
   $$
   Its coefficient matrix is negative definite and has
   non-negative entries outside of the diagonal. An elementary
   result (see Lemma~\ref{lemma-inverse-matrix})
   implies then that all entries of
   its inverse matrix are $\le 0$. Therefore the solutions
   $a_i$
   are non-negative, and we are done.

   Conversely, consider a non-empty
   set $\mathcal S\in\mathcal W(X)$. The existence of a
   big divisor $D$ with $D\cdot C<0$ for $C\in\mathcal S$ implies
   that $\mathcal S$ is a finite set $\set{\liste
   C1r}$ and that the intersection matrix of $\liste C1r$ is negative
   definite (because the negative part of the Zariski
   decomposition of $D$ must contain the curves $C_i$).
   Therefore $\mathcal S\in\mathcal Z(X)$.
\end{proof}

\section{Inclusions of chambers}\label{sect-inclusions}

   We now give a more detailed description of the mutual
   inclusions of Weyl and Zariski chambers. We continue to use
   the notation $\mathcal Z(X)$ from \eqnref{ZX} for the set that
   consists of all
   sets of $(-2)$-curves whose intersection
   matrix is negative definite. For $\mathcal
   S\in\mathcal Z(X)$, we will write $Z_{\mathcal S}$ for the
   Zariski chamber supported by $\mathcal S$, and $W_{\mathcal
   S}$ for the simple Weyl chamber defined by $\mathcal S$. (In other
   words, $Z_{\mathcal S}$ consists of the big divisors whose
   negative part has support $\bigcup\mathcal S$, and
   $W_{\mathcal S}$ consists of the big divisors that have
   negative intersection with the curves in $\mathcal S$ and
   positive intersection with all other $(-2)$-curves.)

   The following two propositions yield
   Theorem~\ref{introthm-inclusions} from the introduction. As
   mentioned before, this result implies in particular the
   equivalence assertion in Theorem~\ref{introthm-equality-criterion}.

\begin{proposition}\label{prop-weil-in-zariski}
   Let $X$ be a K3 surface, and let $\mathcal S\in\mathcal Z(X)$.
   We have
   $$
      W_{\mathcal S}\subset Z_{\mathcal S}
   $$
   if and only if the following condition holds: If $C'$ is a
   curve with $C'\notin\mathcal S$ such that the intersection
   matrix of the set $\mathcal S\cup\set{C'}$ is negative
   definite, then $C'\cdot C=0$ for all $C\in\mathcal S$.
\end{proposition}

\begin{proof}
   Suppose that the stated condition holds. We need to show that
   one has $\Neg(D)=S$ for every divisor $D\in W_S$. We may
   certainly assume $\mathcal S\neq\emptyset$. Let then $\mathcal
   S =\set{\liste C1r}$ and $D\in W_{\mathcal S}$, so that
   the inequality
   $D\cdot C_i < 0$ holds for $i=1,\dots,r$. In the Zariski
   decomposition $D=P_D+N_D$ we have $P_D\cdot C_i\geq 0$, and
   hence $N_D\cdot C_i<0$ for all $i$. Therefore we obtain in any
   event the inclusion $\mathcal S\subset\Neg(D)$.

   To see the converse inclusion $\Neg(D)\subset\mathcal S$,
   assume by way of contradiction that there is a curve
   $C'\in\Neg(D)$ with $C'\notin\mathcal S$. From our condition
   we get $C_i\cdot C'=0$ for all $i=1,\dots,r$. Let now
   $\bigcup_{j=1}^{s}C'_j$ be the connected component in
   $\Neg(D)$ containing $C'$. Then none of the curves $C'_j$ can
   be contained in $S$, and therefore
   $$
      0\le D\cdot C'_j = N_D\cdot C'_j = \sum_{i=1}^{s}a'_iC'_i\cdot C'_j
   $$
   for $j=1,\dots,s$. Then Lemma~\ref{lemma-inverse-matrix} implies
   that all $a'_i$ must be $\le 0$. But this is a contradiction
   with the fact that $N_D$ is effective. Thus $\Neg(D)=\mathcal
   S$, which means that $D\in Z_{\mathcal S}$.

   To proof the other direction we show: If there is a curve
   $C'\notin\mathcal S$ such that $\mathcal S \cup \set{C'}$ has
   negative definite intersection matrix and if there is a curve
   $C\in\mathcal S$ such that $C\cdot C'=1$, then there is a
   divisor $D$ with $D\in W_{\mathcal S}$ but $D\notin
   Z_{\mathcal{S}}$. Assume that $C'$ is such a curve. Then there
   is by Lemma \ref{lemma-neg-def} a big divisor $B$ with
   $\Neg(B)=\mathcal S \cup \set{C'}$. We can write its Zariski
   decomposition as
   $$
      B=P_B+N_B=P_B + b'C' +\sum_{i=1}^{r}b_i C_i \,,
   $$
   where $b'$ and $\liste b1r$ are positive rational numbers.
   We claim now that we can find positive rational numbers
   $c'$ and $\liste c1r$ such that the divisor
   $$
      D=P_B + c'C' +\sum_{i=1}^{r}c_i C_i
   $$
   satisfies
   \begin{equation}\label{desired-inequ}
      D\cdot C'>0 \quad\mbox{and}\quad
      D\cdot C_i<0 \mbox{ for } i =1,\dots,r
      \,.
   \end{equation}
   These inequalities then tell us that
   $D\in W_S$, but $D\not\in Z_S$.

   To prove the existence of $D$, we solve first
   the system of inequalities
   \begin{eqnarray*}
      D \cdot C' = N_D \cdot C' &<& 0 \\
      D \cdot C_1 = N_D \cdot C_1 &<& 0 \\
      \vdots                      & &   \\
      D \cdot C_r = N_D \cdot C_r &<& 0 \,.
   \end{eqnarray*}
   for the variables $c',\liste c1r$.
   Lemma~\ref{lemma-inverse-matrix} guarantees that the solutions are positive.
   We claim finally that upon
   replacing $c'$ with $c'=\frac14\min c_i$,
   the desired inequalities \eqnref{desired-inequ} hold.
   This latter fact follows from a case-by-case analysis using the
   fact that the connected components of $\Neg(D)$ are A-D-E
   curves (see Figure~\ref{fig-Dynkin} for the possible
   configurations of their components). We omit the details.
\end{proof}

\begin{proposition}\label{prop-zariski-in-weil}
   Let $X$ be a K3 surface, and let $\mathcal S\in\mathcal Z(X)$.
   We have
   $$
      \interior Z_{\mathcal S}\subset W_{\mathcal S}
   $$
   if and only
   if $C_1\cdot C_2\ne 1$ for all curves $C_1,C_2\subset S$.
\end{proposition}

\begin{proof}
   We show first that
   if there are two curves $C_i,C_j\in\mathcal S$ with
   $C_i\cdot C_j=1$, then
   there is a big divisor $E$, which belongs to
   $\interior{Z}_{\mathcal S}$ but not to $W_{\mathcal S}$.
   Setting $\mathcal S=\set{\liste C1r}$, assume that $C_1$ and
   $C_2$ are such curves.
   Choose an ample divisor $H$, and consider the big divisor
   $$
      D=H+\sum_{i=1}^{r} a_iC_i \,,
   $$
   where $\liste a1r$ are positive rational numbers. Its Zariski
   decomposition is of the form
   $$
      D=\underbrace{H+\sum_{i=1}^{r}a^*_iC_i}_{P_D}
      +\underbrace{\sum_{i=1}^{r}\left(a_i-a^*_i\right)C_i}_{N_D}\,,
   $$
   where the coefficients $a^*_i$ satisfy
   $$
      0=P_D\cdot C_i=H\cdot C_i +\sum_{j=1}^{r} a^*_jC_j\cdot
      C_i \qquad\mbox{for}\ i=1,\dots,r.
   $$
   Choosing the $a_i$ large enough,
   we may assume $0<a^*_i<a_i$ (see \cite{Bau}).
   Consider the divisor
   $$
      D^*=H+a^*_1C_1+\sum_{i=2}^{r}a_iC_i
      = P_D+\sum_{i=2}^r (a_i-a_i^*)C_i
      \,.
   $$
   Then we have $P_{D^*}=P_D$, so $\Null(P_{D^*})=\Null(P_D)$, but
   $$
      \Neg(D^*)=\set{\liste C2r}\subsetneq\set{\liste C1r}=\Neg(D)
      \,.
   $$
   Therefore, by \cite[Prop.~1.7]{BKS},
   the divisor $D^*$ lies on the boundary of the
   Zariski chamber $\Sigma_{P_D}$. We have
   $$
      D^*\cdot C_1=P_D\cdot C_1+\sum_{i=2}^r(a_i-a_i^*)C_i\cdot C_1
      \ge(a_2-a_2^*)C_2\cdot C_1=a_2-a_2^*>0 \,.
   $$
   There exists an $\epsilon>0$, such that
   $\left(D^*+\epsilon C_1\right)\cdot C_1>0$, and the divisor
   $D^*+\epsilon C_1$ lies in the interior of $\Sigma_{P_D}$
   by \cite[Prop.~1.8]{BKS}.
   Setting $E=D^{*}+\epsilon C_1$ we are done.

   Conversely, let $D\in\interior{Z}_{\mathcal{S}}$ be a
   big divisor with Zariski decomposition  $D=P_D+N_D$. Then
   $N_D=\sum_{i=1}^{r} a_iC_i$ with positive rational numbers
   $a_i$ and $(-2)$-curves $C_i$. Thanks to negative definiteness
   and the condition in the proposition we have
   $C_i\cdot C_j=0$ for $i\neq j$. Therefore:
   $$
      D\cdot C_i=N_D\cdot C_i=a_i\,C_i^2=-2a_i<0
      \qquad \mbox{for } i=1,\dots,r.
   $$
   For any $(-2)$-curve $C$ different from the $C_i$, we clearly
   have $D\cdot C\ge 0$. We claim that $D\cdot C>0$. In
   fact, if we had $D\cdot C=0$, then
   $P_D\cdot C=0$. So $C\in\Null(P)$, but
   $C\not\in\Neg(D)$, and then, again by
   \cite[Prop.~1.7]{BKS}, $D$ would lie on the boundary of the
   chamber $Z_S$.
\end{proof}

\begin{figure}[t]
   \begin{flushleft}
      \leftskip=0.1\textwidth
      \ADEdiagrams

      \vspace{2\baselineskip}
      \caption{\label{fig-Dynkin}
         The Dynkin diagrams corresponding to A-D-E curves on a K3
         surface.}
   \end{flushleft}
\end{figure}

\section{Examples}\label{sect-example}

\paragraph{K3 surfaces with differing decompositions.}

   We analyze now in detail
   examples of K3 surfaces where the
   two chamber decompositions differ.
   Specifically, we consider smooth quartic surfaces in $\P^3$
   containing a
   hyperplane section that decomposes into two lines and an
   irreducible conic. It is easy to see -- for instance as in
   \cite[Lemma~2.2b]{Bau97a} -- that such surfaces exist.
   Fix such a quartic surface $X$, and let $L_1,L_2,C$ be two
   lines and an irreducible conic such that $L_1+L_2+C$ is a
   hyperplane section of $X$.
   In the N\'eron-Severi vector space
   $\NS_\R(X)$ we consider the subspace
   $$
      V \eqdef\vecspan{L_1,L_2,C}\subset\NS_\R(X)
   $$
   spanned by the classes of
   $L_1$, $L_2$, and $C$.
   The intersection form is given on $V$ by the matrix
   \begin{equation}\label{L1L2C-matrix}
      \matr{-2 & 1 & 2 \\
         1 & -2 & 2 \\
         2 & 2 & -2}
         \,.
   \end{equation}
   We now show:

\begin{proposition}\label{prop-L1L2C-surface}
   \begin{itemize}\compact
   \item[(i)]
      The intersection $\mathcal C\eqdef\BigCone(X)\cap V$ is
      the interior of the cone generated by the classes of
      $L_1$, $L_2$, and $C$.
   \item[(ii)]
      The curves $L_1$, $L_2$, and $C$ are the only $(-2)$-curves
      in $V$.
   \item[(iii)]
      The intersection $\Nef(X)\cap V$
      consists of the classes $aC+b_1L_1+b_2L_2$ such that the
      real numbers $a,b_1,b_2$ satisfy the inequalities
      $$
         b_1+b_2\ge a,\quad
         2a+b_2\ge 2b_1,\quad
         2a+b_1\ge 2b_2\,.
      $$

   \item[(iv)]
      The cone $\mathcal C$
      decomposes into
      five Zariski chambers and into five simple Weyl
      chambers. These two decompositions do not coincide.
   \end{itemize}
\end{proposition}

\begin{figure}
   \begin{center}\footnotesize
      \FigureQuarticChambers
      \caption{\label{fig-quartic-chambers}
         The Weyl chamber decomposition (left picture) and the
         Zariski chamber decomposition (right picture) of the
         cone $\mathcal C$ on the quartic surfaces from
         Proposition~\ref{prop-L1L2C-surface}. The pictures show
         the intersection of the (3-dimensional) cone~$\mathcal
         C$ with the hyperplane in $V$ passing through the
         classes of $L_1$, $L_2$, and $C$. The labels in the
         chambers indicate in the left picture the curves that
         are intersected negatively and in the right picture the
         support of the Zariski chambers. The unlabelled chamber
         in the center is the nef chamber, i.e., the
         chamber whose interior is the ample cone and whose
         closure is the nef cone.
         }
   \end{center}
\end{figure}

\begin{proof}
   (i) As $\BigCone(X)$ is the closure of the effective cone, the
   issue is to see that if a numerical class $D=aC+b_1L_1+b_2L_2\in
   V$ is effective, then all coefficients $a,b_1,b_2$ are
   non-negative. Note that if we had $V=\NS_\R(X)$, then this
   would follow from
   \cite[Theorem~6.1]{Kov}. In our situation it can be seen as
   follows:
   One checks first that the divisor $C+2L_1+2L_2$ is
   nef. From the fact that its intersection with $D$ must be
   non-negative we get then that $a\ge 0$.
   Proceeding in the same way with the nef divisors $C+L_1$ and
   $C+L_2$ we get $b_2\ge 0$ and $b_1\ge 0$ respectively.

   (ii) Suppose that $F$ is a $(-2)$-curve in $V$.
   Then by (i) the numerical
   class of $F$ lies in the closure of $\mathcal C$.
   Writing
   $F=aC+b_1L_1+b_2L_2$ for this class,
   we have
   $$
      (aC+b_1L_1+b_2L_2)F=F^2=-2<0 \,.
   $$
   As $F$ is irreducible and the divisor in brackets
   is effective, this can only happen if
   $F$ is one of the curves $C,L_1,L_2$.

   (iii)
   A class $aC+b_1L_1+b_2L_2$ is nef if and only if it meets the
   curves $C,L_1,L_2$ non-negatively. These three
   conditions yield the
   asserted inequalities.

   (iv)
   The intersection matrix \eqnref{L1L2C-matrix}
   has
   exactly four negative definite prinicipal submatrices,
   corresponding to the four
   divisors
   $$
      L_1,\ L_2,\ L_1+L_2,\ C \,.
   $$
   Therefore,
   by Lemma~\ref{lemma-neg-def},
   there are exactly five Zariski chambers supported on the
   curves $L_1,L_2,C$, and hence
   by Theorem~\ref{thm-bijection}
   also exactly five Weyl chambers.
   The fact that the decompositions do not coincide is of course
   a consequence of Theorem~\ref{thm-equality-criterion}.
   Alternatively, it can be verified directly by computing the
   chambers from the shape of $\Nef(X)\cap V$, i.e., using the
   inequalities given in (iii).
\end{proof}

\begin{remark}\rm
   The two decompositions of $\mathcal C$ are shown in
   Figure~\ref{fig-quartic-chambers}. Using
   Theorem~\ref{introthm-inclusions} we get also information
   about the chamber decomposition of the cone $\BigCone(X)$
   (which might be bigger than $\mathcal C$). In particular, we
   have the following inclusions and non-inclusions of chambers
   on $X$:
   \begin{itemize}
   \item[(i)]
      $\interior Z_\emptyset=W_\emptyset$, as always,
   \item[(ii)]
      $\interior Z_{\set{L_1}}\subsetneq W_{\set{L_1}}$
      and
      $\interior Z_{\set{L_2}}\subsetneq W_{\set{L_2}}$,
   \item[(iii)]
      $\interior Z_{\set{C}}\subset W_{\set{C}}$,
   \item[(iv)]
      $\interior Z_{\set{C}}=W_{\set{C}}$,
      if and only if there is no $(-2)$-curve $C'\subset X$ with
      $C'\cdot C=1$.
   \item[(v)]
      $\interior Z_{\set{L_1,L_2}}\not\subset W_{\set{L_1,L_2}}$
   \item[(vi)]
      $W_{\set{L_1,L_2}}\subset Z_{\set{L_1,L_2}}$, if
      any only if there is no $(-2)$-curve $C'\subset X$ such
      that $C'\cdot L_1=0$ and $C'\cdot L_2=1$ or conversely.
   \end{itemize}
\end{remark}

   The situation becomes particulary transparent when
   one has quartics as above whose Picard number is exactly
   three.
   We will show that
   that surfaces with this property exist:

\begin{proposition}\label{prop-thequartic}
   There exist smooth quartic surfaces $X\subset\P^3$
   of Picard number three
   having a hyperplane section of the form $L_1+L_2+C$ such that
   $L_1$ and $L_2$ are lines and $C$ is a smooth conic.
\end{proposition}

   On such surfaces one has
   $V=\NS_\R(X)$, and then the
   conditions in (iv) and (vi) are fulfilled by
   Proposition~\ref{prop-L1L2C-surface}. Therefore, in that case
   \begin{itemize}
   \item
      $\interior Z_{\set{L_1}}\subsetneq W_{\set{L_1}}$
      and
      $\interior Z_{\set{L_2}}\subsetneq W_{\set{L_2}}$,
   \item
      $\interior Z_{\set{C}}= W_{\set{C}}$,
   \item
      $W_{\set{L_1,L_2}}\subsetneq Z_{\set{L_1,L_2}}$.
   \end{itemize}
   The picture in Figure~\ref{fig-quartic-chambers}
   describes then the whole big cone of $X$.

\begin{proof}[Proof of Proposition~\ref{prop-thequartic}]
   Consider the \textit{K3 lattice}
   $$
      \Lambda_{K_3}=U^3\oplus(-E_8)^2
   $$
   where $U$ is the lattice $\Z^2$ with the bilinear form
   given by the matrix
   $$
      \matr{0 & 1\\ 1 & 0}
   $$
   and $-E_8$ is the lattice $\Z^8$ with the bilinear form
   given by
   $$
      \matr{
         -2&1&0&0&0&0&0&0 \\
         1&-2&1&0&0&0&0&0 \\
         0&1&-2&1&0&0&0&0 \\
         0&0&1&-2&1&0&0&0 \\
         0&0&0&1&-2&1&0&1 \\
         0&0&0&0&1&-2&1&0 \\
         0&0&0&0&0&1&-2&0 \\
         0&0&0&0&1&0&0&-2
         }
   $$
   For any K3 surface $X$, there is an isomorphism
   $\sigma:H^2(X,\Z)\to\Lambda_{K3}$.
   The 21-dimensional \textit{period space}
   $$
      \Omega=\sset{\C\cdot x\in\P(\Lambda_{K_3}\tensor\C)\with\bilin xx=0,\
         \bilin x{\bar x}>0}
   $$
   is a fine moduli space for marked K3 surfaces, i.e., for pairs
   $(X,\sigma)$ consisting of a K3 surface $X$ and an
   isormorphism
   $\sigma:H^2(X,\Z)\to\Lambda_{K3}$
   (see e.g.\ \cite{BPV} for the facts mentioned here).

   Let now $X_0\subset\P^3$ be a smooth quartic surface
   admitting a hyperplane section of the form $L_1+L_2+C$
   as considered above, and let
   $\lambda_1,\lambda_2,\gamma\in\Lambda_{K3}$ be the lattice
   vectors corresponding to the classes of
   $L_1,L_2,C$ under a fixed isomorphism
   $H^2(X,\Z)\isomto\Lambda_{K3}$. The point $x_0\in\Omega$
   corresponding to $X_0$ is contained in the intersection
   $$
      \Omega'=\Omega\cap\lambda_1^\perp\cap\lambda_2^\perp\cap\gamma^\perp
      \ .
   $$
   Consider now a small deformation $X$ of $X_0$
   corresponding to
   a point $x\in\Omega'$ close to $x_0$.
   As $\Lambda_{K3}$ is a countable set,
   the generic such $x$ is not contained in any intersection
   $$
      \Omega'\cap\alpha^\perp
   $$
   where $\alpha\in\Lambda_{K_3}$ is a lattice vector different
   from $\lambda_1,\lambda_2,\gamma$. But this implies that the
   Picard group of $X$ is generated over $\Q$ precisely by the
   classes of $L_1,L_2$ and $C$. The Riemann-Roch theorem implies
   that these $(-2)$-classes are on the deformed surface still
   represented by effective divisors. Therefore, when
   $x$ is close enough to $x_0$, they are still represented by
   irreducible curves.
\end{proof}

\paragraph{K3 surfaces with coinciding decompositions.}

   If a K3 surface does not contain any $(-2)$-curves at all --
   for instance if its Picard group is one-dimensional --
   then of course the Zariski chamber decomposition coincides
   with the Weyl chamber decomposition. More substantial examples
   are given by the following proposition. The double
   covering construction in its proof was
   suggested to us by T.~Szemberg.

\begin{proposition}\label{prop-K3-coinciding}
   There exist K3 surfaces $X$ of Picard number three that contain
   three $(-2)$-curves $F_1,F_2,C$ such that
   $$
      F_1\cdot F_2=0,\quad F_1\cdot C=F_2\cdot C=2 \,,
   $$
   and such that there are no other $(-2)$ curves on $X$.

   The decomposition of the big cone of $X$ into Zariski chambers
   is the same as the decomposition
   into simple Weyl chambers. It consists of five chambers.
\end{proposition}

\begin{proof}
   Let $B\subset\P^2$ be a sextic curve that has two ordinary
   double points $x_1$ and $x_2$
   and no other singularities, and consider the
   blow-up $f:Y=\Bl_{x_1,x_2}(\P^2)\to\P^2$ at these two points. On
   the blow-up let $B'$ be the proper transform of $B$, i.e.,
   $$
      B'=f^*B-2E_1-2E_2 \,,
   $$
   where $E_1$ and $E_1$ are the exceptional divisors over $x_1$
   and $x_2$ respectively. The line bundle $\O_Y(B')$ is then
   2-divisible, so that we can consider the double covering
   $$
      g:X\to Y
   $$
   that is ramified over $B'$.
   As $K_{X}=g^*(K_Y+\frac12 B')=0$ and
   $h^1(\O_X)=0$ (cf.~\cite[Sect.~22]{BPV}),
   the surface $X$ is a K3 surface.
   One checks that the pullbacks
   $$
      F_1=g^*E_1,\quad
      F_2=g^*E_2,\quad
      C=g^*(H-E_1-E_2)
   $$
   are $(-2)$-curves on $X$.
   These curves generate a subspace of the
   N\'eron-Severi vector space of dimension three, on which the
   intersection form is given by the matrix
   \begin{equation}\label{F1F2C-matrix}
      \matr{
         -2 & 0 & 2 \\
         0 & -2 & 2 \\
         2 & 2 & -2
         }
   \end{equation}
   Arguing as in the proof of Proposition~\ref{prop-thequartic},
   i.e., replacing $X$ with a small deformation in the
   period space, we obtain a K3 surface of Picard number three
   with curves $F_1,F_2,C$ intersecting as in
   \eqnref{F1F2C-matrix}
   that generate the N\'eron-Severi group
   over $\Q$.
   For simplicity, let us
   denote the deformed surface again by $X$.

   We show next that there are no other $(-2)$-curves on $X$.
   Suppose that $D$ is a $(-2)$-curve on $X$, and write in
   numerical equivalence
   $$
      D=aC+b_1F_1+b_2F_2
   $$
   with rational numbers $a,b_1,b_2$.
   We can now proceed as in the proof of
   Proposition~\ref{prop-L1L2C-surface}(i)
   to show that $a\ge 0$ and
   $b_1,b_2\ge 0$. (In fact, intersect $D$ with the nef divisors
   $C+F_1$, $C+F_2$, and $C+F_1+F_2$.)
   One sees then as in the proof of
   Proposition~\ref{prop-L1L2C-surface}(ii) that $F_1$, $F_2$,
   and $C$ are
   the only $(-2)$-curves on $X$.
   The intersection matrix
   \eqnref{F1F2C-matrix}
   has exactly four negative definite principal submatrices,
   corresponding to the divisors
   $$
      F_1,\ F_2,\ F_1+F_2,\ C \ .
   $$
   With Lemma~\ref{lemma-neg-def} we conclude that
   there are exactly five Zariski chambers, and
   Theorem~\ref{thm-equality-criterion} implies that the chamber
   decompositions coincide.
\end{proof}

\section*{Appendix}
\renewcommand\thesection{A}
\setcounter{satz}{0}

   The following useful
   lemma from \cite{BKS} is used
   several times in
   the present paper. While probably well-known, a
   somewhat technical proof was included in \cite{BKS} for lack of a
   reference. Here we give a much simpler argument, which is
   inspired by the beginning of the proof of
   \cite[Theorem~3.2]{PanRos}.

\begin{lemma}\label{lemma-inverse-matrix}
   Let $S$ be a negative definite $(r\times r)$-matrix over the
   reals such that $s_{i,j}\ge 0$ for $i\ne j$. Then all
   entries of the inverse matrix $S\inverse$ are $\le 0$.
\end{lemma}

\begin{proof}
   It is enough to show that if $a,b\in\R^r$ are vectors with
   $a=S\inverse b$ such that
   $b_i\le0$ for all $i$, then $a_i\ge 0$ for all $i$.
   To prove this claim, write $a=p-q$, where $p_i=\max(a_i,0)$.
   Then $p_i\ge 0$, $q_i\ge 0$, and $p_iq_i=0$ for all $i$.
   Using these relations along with
   the hypothesis that $s_{i,j}\ge 0$ for $i\ne j$,
   we get
   $$
      p^tSq=\sum_{i,j}p_is_{i,j}q_j=\sum_{i\ne j}p_is_{i,j}q_j\ge 0
   $$
   and hence
   $$
      p^tSq-q^tSq>0 \qquad\mbox{if }q\ne 0\,,
   $$
   because $S$ is negative definite.
   On the other hand, we have
   $$
      p^tSq-q^tSq=(p-q)^tSq=a^tSq=(Sa)^tq=b^tq\le 0 \,,
   $$
   so we conclude that $q=0$.
   This implies that $a_i\ge 0$ for all $i$, as
   claimed.
\end{proof}

\frenchspacing

config: arxiv


TITLE: Continuous map example
QUESTION [0 upvotes]: I am trying to understand the distinction between continuous maps between varieties and morphisms between varieties, and I believe a concrete example illustrating the distinction will help. What is an example of a continuous map $\pi:A\rightarrow B$ where $A,B$ are varieties and $\pi$ is not a morphism?

REPLY [0 votes]: How about $A=B$ being the affine line over $\Bbb C$, and $\pi:z\mapsto\overline z$?

REPLY [0 votes]: One easy source of examples is $\Bbb A^1_k$. Mapping the generic point to the generic point and permuting the closed points will produce a continuous map, and most examples of this fail to be morphisms. For instance, you can swap $0$ and $1$ and leave all other points invariant, which cannot be the result of a morphism.


TITLE: A poker deck has $52$ cards, $13$ ranks ${A,2,...,10,J,Q,K}$ and $4$ suits {diamond, club, heart, spade}.
QUESTION [0 upvotes]: A poker deck has $52$ cards, $13$ ranks $\{A,2,...,10,J,Q,K\}$ and $4$ suits {diamond, club, heart, spade}. If $5$ cards are dealt, what is the probability that there is an ace of diamonds?
Here is my thinking: event over the sample space, which is event of getting
(ace of diamonds)/ $\begin{pmatrix} 52 \\ 5 \end{pmatrix}$.
then for the event: _ _ _ _ _. We got 5 chances which is $\begin{pmatrix} 5 \\ 1 \end{pmatrix}$, then for the ace of diamonds $\begin{pmatrix} 52 \\ 1 \end{pmatrix} $since there is only one ace of a diamond from the $52$ cards. so $\frac{ \begin{pmatrix} 5 \\ 1 \end{pmatrix} \begin{pmatrix} 52 \\ 1 \end{pmatrix}} {\begin{pmatrix} 52 \\ 5 \end{pmatrix}}$? is this right?
What about if we just count of the p(ace) and P(a diamond)?   Thank you in advance.

REPLY [4 votes]: Another way of looking at the problem, which might be even easier to digest, is that the probability that you do get an ace of diamonds is very simply $1$ minus the probability that you don't get the ace of diamonds.
The number of five card hands total is $\binom{52}{5}$, each of which are equally likely to have occurred (which allows us to calculate probabilities by taking ratios).  The number of five card hands which don't contain the ace of diamonds is $\binom{51}{5}$.
The probability is then:
$$1 - \dfrac{\binom{51}{5}}{\binom{52}{5}}$$
This generalizes easily to the other problems you seem interested in, the probability that within your five card hand you have at least one ace.  The probability that you get at least one ace is $1$ minus the probability you got no aces.  The number of hands with no aces would be $\binom{48}{5}$, so the probability of at least one ace is:
$$1 - \dfrac{\binom{48}{5}}{\binom{52}{5}}$$
Similarly, the probability of getting at least one diamond is $1 - \dfrac{\binom{39}{5}}{\binom{52}{5}}$


TITLE: Where does $f_n(x)=\frac{x^n}{1+x^{2n}}$ converge uniformly?
QUESTION [1 upvotes]: I'm given the function
\begin{equation}
f_n(x)=\frac{x^n}{1+x^{2n}},
\end{equation}
where I can assume $f_n:[0,\infty)\to \mathbb{R}$. I need to determine and show the sets over which the convergence is uniform.
Clearly for $x=1$ the pointwise limit is $\frac{1}{2}$, because $f_n(1)=\frac{1}{2}$ for all $n$.
For $x \in [0,1)$, it converges pointwise to $0$, which doesn't seem too hard to show.
For $x \in (1,\infty)$ however, I believe it also converges to $0$ but I am not sure. The numerator is growing but the denominator is growing faster. Is it growing fast enough for the pointwise limit to go equal zero? Any way to see this easily?
Finally, I need to find the intervals over which this convergence is uniform. Any help here please? Thanks!

REPLY [2 votes]: Yes you're right $f_n$ converges to $0$ for $x>1$ since in this case
$$\frac{x^n}{1+x^{2n}}\sim_\infty x^{-n}$$
and since the limit function isn't continuous on $1$ then we haven't the uniform convergence on $[0,+\infty)$. However we have the uniform convergence on every interval $[0,a]$ where $0<a<1$. Indeed,
$$|f_n(x)|\le x^n\le a^n,\quad \forall x\in[0,a]$$
We have the same result on the interval $[a,+\infty)$ where $a>1$. In fact by taking the derivative of $f_n$ we can see that it's decreasing on the interval $[a,+\infty)$ so
$$\sup_{x\ge a}|f_n(x)|=f_n(a)\sim_\infty a^{-n}\xrightarrow{n\to\infty}0$$


TITLE: Number of finite simple groups of given order is at most $2$ - is a classification-free proof possible?
QUESTION [20 upvotes]: This Wikipedia article states that the isomorphism type of a finite simple group is determined by its order, except that:

$L_4(2)$ and $L_3(4)$ both have order $20160$
$O_{2n+1}(q)$ and $S_{2n}(q)$ have the same order for $q$ odd, $n > 2$

I think this means that for each integer $g$, there are $0$, $1$ or $2$ simple groups of order $g$.
Do we need the full strength of the Classification of Finite Simple Groups to prove this, or is there a simpler way of proving it?

REPLY [5 votes]: There are many mathematicians outside finite group theory who asked whether important infinite fragments of the classification were possible without the entire classification.  I believe the favorite questions has always been : Can you prove the finiteness of the sporadics without the full classification?
There is a good chance that third generation proof technology will reduce the entanglement between different portions of the classification because one knows the unipotent primes during earlier arguments where current methods only reveal semi-simple structure.  There has been one remarkable success in this direction :
Theorem (Altinel, Borovik, Cherlin).  A simple group of finite Morley rank containing an infinite elementary abelian 2-group is a Chevalley group over an algebraically closed field of characteristic 2.
There is no known proof that simply groups of finite Morley rank even have an involution, much less that groups with odd characteristic looking Sylow 2-subgroup are also algebraic.  In consequence, there is a conjecture by Borovik that basically proposes one might classify finite simple groups who's 2-rank vastly exceeds the p-rank for any other p prime.
The final proof of [ABC] weighs in around 500 pages, but any finite analog would require many thousands of pages to deal with issues like twisted groups of Lie type and alternating groups, even assuming you find some trick for avoiding all the sporadics.
In short, there are an awful lot of interesting results that depend upon the full CFSG for the foreseeable future because only funky asymptotic fragments look even vaguely realistic as stand alone results and even those look extremely difficult.


text:
\begin{document}
\title{\bf Depth  in a pathological  case }

\author{ Dorin Popescu}

\thanks{The  support from  grant ID-PCE-2011-1023 of Romanian Ministry of Education, Research and Innovation is gratefully acknowledged.}

\address{Dorin Popescu,  Simion Stoilow Institute of Mathematics of Romanian Academy, Research unit 5,
 P.O.Box 1-764, Bucharest 014700, Romania}
\email{dorin.m.popescu@gmail.com}
\maketitle
\begin{abstract} Let $I$ be  a  squarefree monomial ideal of a polynomial algebra over a field minimally generated by $f_1,\ldots,f_r$ of  degree $ d\geq 1$, and a set $E$ of monomials  of degree $\geq d+1$. Let $J\subsetneq I$ be a squarefree monomial ideal generated in degree $\geq d+1$.  Suppose that  all squarefree monomials of $I\setminus (J\cup E)$ of degree $d+1$ are some least common multiples of $f_i$. If $J$ contains all least common multiples of two of $(f_i)$ of degree $d+2$ then $\depth_SI/J\leq d+1$ and Stanley's Conjecture holds for $I/J$.

 \noindent
  {\it Key words } : Monomial Ideals,  Depth, Stanley depth.\\
 {\it 2010 Mathematics Subject Classification: Primary 13C15, Secondary 13F20, 13F55,
13P10.}
\end{abstract}

\section*{Introduction}
Let $K$ be a field and $S=K[x_1,\ldots,x_n]$ be the polynomial $K$-algebra in $n$ variables. Let  $I\supsetneq J$ be  two    monomial ideals of $S$ and suppose that  $I$ is generated by some monomials of degrees $\geq d$   for some positive integer $d$.  After  a multigraded isomorphism we may assume either that $J=0$, or $J$ is generated in degrees $\geq d+1$.

 Suppose that $I \subset S$ is minimally generated by some  monomials $f_1,\ldots,f_r$ of degrees $d$,  and a  set $E$  of  monomials of degree $\geq d+1$.
 Let  $B$ (resp. $C$) be the set of   squarefree monomials of degrees $d+1$  (resp. $d+2$) of $I\setminus J$.   Let $w_{ij}$ be the least common multiple of $f_i$ and $f_j$, $i<j$ and set $W$ to be the set of all $w_{ij}$.
By \cite[Proposition 3.1]{HVZ} (see  \cite[Lemma 1.1]{P}) we have $\depth_S I/J\geq d$. It is easy to see that if $d=1$, $E=\emptyset$ and $B\subset W$ then $\depth_SI/J=d$ (see for instance \cite[Lemma 1.8]{P} and \cite[Lemma 3]{AP}). Attempts to extend this result were made in \cite[Proposition 1.3]{PZ}, \cite[Lemma 4]{AP}. However \cite[Example 1]{AP} (see here Example \ref{e2}) shows that for $d=2$, $E=\emptyset$ and $B\subset W$ it holds $\depth_SI/J=d+1=3$.

 If $B\cap (f_1,\ldots,f_r)\subset W$ we call $I/J$ a {\em pathological case}.
 It is the purpose of this paper  to show the following theorem.

\begin{Theorem} \label{t}   If  $B\cap (f_1,\ldots,f_r)\subset W$  and $C\cap W=\emptyset$ then $\depth_S I/J \leq d+1$.
\end{Theorem}

In particular, if $C\cap W=\emptyset$ then the so called Stanley's Conjecture  holds in the pathological case (see Corollary \ref{c}).
But why is important this pathological case? The methods used in \cite{PZ1}, \cite{AP}, \cite{P1} to show a weak form of Stanley's Conjecture when $r\leq 4$ (see \cite[Conjecture 0.1]{P1}) could be applied only   when $B\cap (f_1,\ldots,f_r)\not\subset W$, that is when $I/J$ is not  pathological. Thus the above theorem solves partially one of the  obstructions to prove this weak form. We believe that the condition $C\cap W=\emptyset$ could be removed from the above theorem. The proof of Theorem \ref{t} relies on Lemmas \ref{d'}, \ref{m} and Examples \ref{e4}, \ref{e}, \ref{e3} found after many computations with the Computer Algebra System SINGULAR \cite{DGPS}.

The above theorem hints a possible positive answer to the following question.

\begin{Question} Let $i\in [r-1]$. Suppose that $E=\emptyset$ and every squarefree monomial from $I\setminus J$ of degree $d+i$ is a least common multiple of $i+1$ monomials $f_j$. Then is it $\depth_SI/J\leq d+i$?
\end{Question}

We owe thanks to A. Zarojanu and a Referee, who noticed  some mistakes and a  gap in   Section 2 of  some previous versions of this paper.

\section{Depth and Stanley depth}

Suppose that $I$ is minimally generated by some squarefree monomials $f_1,\ldots,f_r$ of degree $ d$  for some $d\in {\mathbb N}$ and a set $E$ of some squarefree monomials of degree $\geq d+1$.  Let $C_3$ be the set of all $c\in C\cap (f_1,\ldots,f_r)$ having all degree $(d+1)$ divisors from $B\setminus E$ in $W$. In particular each monomial of $C_3 $ is the least common multiple of at least three of the $f_i$.

Next  lemma is closed to \cite[Lemma 1.1]{PZ}.
\begin{Lemma} \label{d} Suppose that $E=\emptyset$ and $\depth_SI/(J,b)=d$ for some $b\in B$. Then $\depth_SI/J\leq d+1$.
\end{Lemma}
\begin{proof} If there exists no $c\in C$ such that $b|c$ then we have $\depth_SI/J\leq d+1$ by \cite[Lemma 1.5]{PZ}.
Otherwise, in the exact sequence
$$0\to (b)/J\cap (b)\to I/J\to I/(J,b)\to 0$$
the first term has depth $\geq d+2$ because for a multiple $c\in C$ of $b$ all the variables of $c$ form a regular system. By hypothesis  the last term has depth $\geq d$ and so the middle one has depth $d$ too using  the Depth Lemma.
\hfill\ \end{proof}

We recall the following example from \cite{AP}.
\begin{Example} \label{e2}  {\em Let $n=5$, $r=5$, $d=2$, $f_1=x_1x_2$, $f_2=x_1x_3$, $f_3=x_1x_4$,  $f_4=x_2x_3$, $f_5=x_3x_5$ and $I=(f_1,\ldots,f_5)$,
$J=(x_1x_2x_5,x_1x_4x_5, x_2x_3x_4,x_3x_4x_5).$
It follows that  $B=\{x_1x_2x_3,x_1x_2x_4,x_1x_3x_4,x_1x_3x_5,x_2x_3x_5\}$ and so $s=|B|=r=5$. Note that $B\subset W$. A computation with SINGULAR when char $K=0$ gives $\depth_SI/J=\depth_SS/J=3$ and $\depth_SS/I=2$. Since depth depends on the characteristic of the field it follows in general only that $\depth_SS/J\leq 3$, $\depth_SS/I\leq 2$ using \cite[Lemma 2.4]{AE}. In fact $\depth_SI/J\leq d+1=3$ using \cite[Proposition 2.4]{Sh} because $q=|C|=2<r=5$. Note that choosing any $b\in B$ we  have $\depth_SI/(J,b)=2$ because the corresponding $s'<r $ and we may apply \cite[Theorem 2.2]{P}. But then $\depth_SI/J\leq 3$ by Lemma \ref{d}.}
\end{Example}
\begin{Example} \label{e4}  {\em In the above example set $I'=(f_1,\ldots,f_4)$, $J'=J\cap I'=$\\
$(x_1x_2x_5,x_1x_4x_5,x_2x_3x_4)$. Note that we have an injection $I'/J'\to I/J$ and so $\depth_SI'/J'>2$ because otherwise we get $\depth_SI/J=2$ which is impossible. Given $B',W'$ for $I'/J'$ we see that $B'\not\subset W'$  since $x_2x_3x_5\in B'\setminus W'$, that is $I'/J'$ is not anymore in the pathological case even this was the case of $I/J$.   We have $I/(J,f_5)\cong I'/(J',x_1x_3x_5,x_2x_3x_5)$. Using SINGULAR when char $K=0$ we see that $\depth_SI'/(J',x_1x_3x_5)=\depth_SI'/(J',x_1x_3x_5,x_2x_3x_5)=2$, $\depth_SI'/(J',x_2x_3x_5)=3$. It follows that always $\depth_SI'/(J',x_1x_3x_5),\depth_SI'/(J',x_1x_3x_5,x_2x_3x_5)\leq 2$ using \cite[Lemma 2.4]{AE}. These inequalities are in fact equalities because $I'$ is generated in degree $2$. Thus we cannot apply Lemma \ref{d} for $I',J'$, $b=x_2x_3x_5$ but we may apply this lemma for $I',J'$, $b'=x_1x_3x_5$ to get $\depth_SI'/J'\leq 3$.}
\end{Example}

\begin{Lemma} \label{d'} Suppose that $r>1$ and $\depth_SI/(J,f_r)=d$. Then $\depth_SI/J\leq d+1$.
\end{Lemma}
\begin{proof}  Let $B\cap (f_r)=\{b_1,\ldots,b_p\}$.
 As  $I/(J,f_r)$ has a squarefree, multigraded free resolution we see that only the
 components of squarefree degrees of
$$\Tor_S^{n-d}(K, I/(J,f_r))\cong H_{n-d}(x; I/(J,f_r))$$
are nonzero, the last module being the Koszul homology of $I/(J,f_r)$.
 Thus we may find
$$z =
\sum^{r-1}_{i=1} y_i f_i e_{[n]\setminus \supp f_i} \in K_{n-d}(x; I/(J,f_r)),$$
 $y_i \in K$
inducing a nonzero element in $H_{n-d}(x; I/(J,f_r))$.  Here we set $\supp f_i=\{t\in [n]:x_t|f_i\}$ and $e_A = \wedge_{j\in A} e_j$ for a
subset $A\subset [n]$.
We have
$$\partial z=\sum_{b\in B} P_b(y)be_{[n]\setminus \supp b}=\sum_{i=1}^p P_{b_i}(y)b_ie_{[n]\setminus \supp b_i},$$
where $P_b$ are linear homogeneous polynomials in $y$. Note that  $P_b(y)=0$ for all $b\not\in\{b_1,\ldots,b_p\}$. Choose $j\in \supp f_r$ and consider
$$z_j= \sum^{r-1}_{i=1, f_i\not\in (x_j)} y_i f_i e_{[n]\setminus  (\{x_j\}\cup\ \supp f_i)} \in K_{n-d-1}(x; I/J).$$
In $\partial z_j$ appear only terms of type $ue_A$, $j\not\in A$ with $|A|=n-d-1$ and $u=\Pi_{i\in [n]\setminus (A\cup \{j\})}x_i$. Thus terms of type $b_ie_{[n]\setminus \supp b_i}$, $i\in [p]$ are not present in $\partial z_j$ because $b_i\in (x_j)$. It follows that $\partial z_j=0$ and so $z_j$ is a cycle. Note that a cycle of $K_{n-d-1}(x; I/J)$ could contain also terms of type $ve_{A'}$ with $|A'|=n-d-1$ and $v=\Pi_{i\in [n]\setminus A'}x_i\in B$, but $z_j$ is just a particular cycle.

Remains to show that we may find $j$ such that $z_j$ is a nonzero cycle.
Suppose that $y_m\not =0$ for $m\in [r-1]$ and  choose  $j\in\supp f_r\setminus\supp f_m$. It follows that $z_j$ is a nonzero cycle because $y_mf_me_{[n]\setminus  (\{x_j\}\cup\ \supp f_m)}$ is present in $z_j$.
 Thus $\depth_SI/J\leq d+1$ by \cite[Theorem 1.6.17]{BH}.
\hfill\ \end{proof}

\begin{Remark}\label {r'} {\em Applying the above lemma to Example \ref{e4} we see that $\depth_SI/J\leq 3$ because $\depth_SI/(J,f_5)=2$.}
\end{Remark}

  Let $P_{I\setminus J}$  be the poset of all squarefree monomials of $I\setminus J$  with the order given by the divisibility. Let $ P$ be a partition of  $P_{I\setminus J}$ in intervals $[u,v]=\{w\in  P_{I\setminus J}: u|w, w|v\}$, let us say   $P_{I\setminus J}=\cup_i [u_i,v_i]$, the union being disjoint.
Define $\sdepth  P=\min_i\deg v_i$ and  the  {\em Stanley depth} of $I/J$ given by $\sdepth_SI/J=\max_{ P} \sdepth  P$, where $ P$ runs in the set of all partitions of $P_{I\setminus J}$ (see  \cite{HVZ}, \cite{S}).  Stanley's Conjecture says that  $\sdepth_S I/J\geq \depth_S I/J$.

In Example \ref{e2} we have $s=5<q+3=7$ and so it follows that $\depth_SI/J\leq d+1$ by \cite[Theorem 1.3]{P0} (see also \cite[Theorem 2]{AP}). Next example follows \cite[Example 1.6]{P1} and has $s=q+r$, $\sdepth_SI/J=d+2$ but $\depth_SI/J=d$.

\begin{Example}\label{e} {\em Let $n=12$, $r=11$, $f_1=x_{12}x_1$, $f_2=x_{12}x_2$, $f_3=x_{12}x_3$, $f_4=x_{12}x_4$, $f_5=x_{12}x_5$, $f_6=x_{12}x_6$, $f_7=x_6x_7$, $f_8=x_6x_8$, $f_9=x_6x_9$, $f_{10}=x_6x_{10}$, $f_{11}=x_6x_{11}$, $J=(x_7,\ldots,x_{11})(f_1,\ldots,f_5)+(x_1,\ldots,x_{5})(f_7,\ldots,f_{11})+f_6(x_9,\ldots,x_{11})$, $I=(f_1,\ldots,f_{11})$. We have $B=\{w_{ij}:1\leq i<j\leq 5\}\cup \{w_{kt}:6< k<t\leq 11\}\cup \{w_{i6}:i\in [8],i\not=6\}$, that is $s=|B|=27$.
 Let $c_1=x_6w_{12}$,   $c_2=x_6w_{23}$,  $c_3=x_6w_{34}$,  $c_4=x_6w_{45}$,  $c_5=x_6w_{15}$, $c_6=x_8w_{67}$, $c_7=x_9w_{78}$, $c_8=x_{10}w_{89}$, $c_9=x_{11}w_{9,10}$, $c_{10}=x_7w_{10,11}$,  $c_{11}=x_7w_{8,11}$, $c'_{13}=x_4w_{13}$, $c'_{14}=x_5w_{14}$,  $c'_{24}=x_6w_{24}$,
   $c'_{25}=x_3w_{25}$,  $c'_{35}=x_6w_{35}$. These are all monomials of $C$, that is $q=|C|=16$ and so $s=q+r$. The intervals $[f_i,c_i]$, $i\in [11]$ and $[w_{13},c'_{13}]$, $[w_{14},c'_{14}]$, $[w_{24},c'_{24}]$, $[w_{25},c'_{25}]$, $[w_{35},c'_{35}]$ induce a partition $P$ on $I/J$ with sdepth $4$.

We claim that $\depth_SS/J=2$. Indeed, let
$J'=(x_7,\ldots,x_{11})(f_1,\ldots,f_5)+(x_1,\ldots,x_{5})(f_7,\ldots,f_{11})=(x_{12},x_6)(x_7,\ldots,x_{11})(x_1,\ldots,x_{5})$. By \cite[Theorem 1.4]{Ap} we get $\depth_SS/J'=2=d$. Set $J_1=J'+(x_{12}x_6x_9)$, $J_2=J_1+(x_{12}x_6x_{10})$. We have $J=J_2+(x_{12}x_6x_{11})$. In the exact sequences
$$0\to (x_{12}x_6x_9)/(x_{12}x_6x_9)\cap J'\to S/J'\to S/J_1\to 0,$$
$$0\to (x_{12}x_6x_{10})/(x_{12}x_6x_{10})\cap J_1\to S/J_1\to S/J_2\to 0,$$
$$0\to (x_{12}x_6x_{11})/(x_{12}x_6x_{11})\cap J_2\to S/J_2\to S/J\to 0$$
the first terms have depth $\geq 5$. Applying the Depth Lemma by recurrence we get our claim.

Now we see that $\depth_SS/I=6$. Set $I_j=(f_1,\ldots,f_j)$ for $6\leq j\leq 11$. We have $I=I_{11}$, $I_6=x_{12}(x_1,\ldots,x_6)$ and $\depth_SS/I_6=6$. In the exact sequences
$$0\to (f_{j+1})/(f_{j+1})\cap I_j\to S/J_j\to S/I_{j+1}\to 0,$$
$6\leq j<11$ we have $(f_{j+1})\cap I_j=f_{j+1}(x_{12},x_7,\ldots,x_{j})$ and so $\depth_S(f_{j+1})/(f_{j+1})\cap I_j=12-(j-5)\geq 7$
for $6\leq j<11$. Applying the Depth Lemma by recurrence we get $\depth_SS/I_{j+1}=6$ for $6\leq j<11$ which is enough.

Finally using the Depth Lemma in the exact sequence
$$0\to I/J\to S/J\to S/I\to 0$$
 it follows $\depth_SI/J=2=d$.   }
 \end{Example}

The following lemma is the key in the proof of Theorem \ref{t} and its proof is given in the next section.
\begin{Lemma} \label{m} Suppose that  $E=\emptyset$, $C\subset  C_3$, $C\cap W=\emptyset$ and Theorem \ref{t} holds for $r'<r$. Then $\depth_SI/J\leq d+1$.
\end{Lemma}

\begin{Proposition} \label{p} Suppose that   $C\cap (f_1,\ldots,f_r)\subset  C_3$, $C\cap W=\emptyset$ and  Theorem \ref{t} holds for  $r'<r$. Then $\depth_SI/J\leq d+1$.
\end{Proposition}
\begin{proof}
Suppose that $E\not =\emptyset$, otherwise  apply Lemma \ref{m}. Set $I'=(f_1,\ldots,f_r)$, $J'=J\cap I'$. In the exact sequence
$$0\to I'/J'\to I/J\to I/(I',J)\to 0$$
the last term is isomorphic to something generated by $E$ and so its depth is $\geq d+1$. The first term satisfies the conditions of Lemma \ref{m} which gives $\depth_SI'/J'\leq d+1$. By the Depth Lemma we get $\depth_SI/J\leq d+1$ too.
\hfill\ \end{proof}

{\bf Proof of Theorem \ref{t}}

Apply induction on $r$. If $r<5$ then $B\cap (f_1,\ldots,f_r)\subset W$ implies   $|B\cap (f_1,\ldots,f_r)|<2r$ and so $\sdepth_SI/J\leq d+1$ and even $\depth_SI/J\leq d+1$ by \cite[Proposition 2.4]{Sh} (we may also apply \cite[Theorem 0.3]{P1}). Suppose that  $r\geq 5$. Since all divisors of a monomial $c\in C\cap (f_1,\ldots,f_r)$ of degrees  $d+1$ are in $B$, they are also in $W$ by our hypothesis.   Thus  $C\cap (f_1,\ldots,f_r)\subset C_3$ and we may apply Proposition \ref{p} under induction hypothesis.  $\hfill\ \square$

\begin{Corollary} \label{c} Suppose that $B\cap (f_1,\ldots,f_r)\subset W$ and  $C\cap W=\emptyset$. Then $\depth_SI/J\leq \sdepth_SI/J$, that is the Stanley Conjecture holds for $I/J$.
\end{Corollary}
\begin{proof} If $\sdepth_SI/J=d$ then apply \cite[Theorem 4.3]{P}, otherwise apply Theorem \ref{t}.
\hfill\ \end{proof}

\section{Proof of Lemma \ref{m}.}

 We may suppose that $B\subset W$ because each monomial of $B$ must divide a monomial of $C$, otherwise we get $\depth_SI/J\leq d+1$ by \cite[Lemma 1.5]{PZ}. Then we may suppose that $B\subset \cup_i \supp f_i$ and we may reduce to the case when $[n]=\cup_i \supp f_i$ because then $\depth_SI/J=\depth_{\tilde S} (I\cap {\tilde S})/(J\cap {\tilde S})$ for ${\tilde S}=K[\{x_t:t\in \cup_i \supp f_i\}]$.

  On the other hand, we may suppose that for each $i\in [r]$ there exists $c\in C$ such that $f_i|c$, otherwise we may apply again \cite[Lemma 1.5]{PZ}. Since  $c\in C_3$, let us say $c$ is the least common multiple of $f_1,f_2,f_3$ we see that at least,let us say, $w_{12}\in B$. Then  $f_i\in (u_1)$, $i\in [2]$ for some monomial $u_1=(f_1f_2)/w_{12}$ of degree $d-1$.

 We may assume that $f_i\in (u_1)$ if and only if $i\in [k_1]$ for some $2\leq k_1\leq r$.
Set $U_1=\{f_1,\ldots,f_{k_1}\}$.
We also assume that
$$\{u_i:i\in [e]\}=\{u:u=gcd(f_i,f_j), \deg u=d-1, i\not = j\in [r]\},$$
and define
$$U_i=\{f_j:f_j\in (u_i), j\in [r]\}$$
for each $i\in [e]$.
 Since  each $f_t\in U_i$  divides a certain $c\in C$ we see from our construction that  there exist $f_p,f_l\in U_i$ such that $w_{tp},w_{tl}\in B$.  Note that if $|U_i\cap U_j|\geq 2$ then we get $u_i=u_j$ and so $i=j$. Thus $|U_i\cap U_j|\leq 1$ for all $i,j\in [e]$, $i\not =j$.

  Suppose that  $w_{ij}\in J$ for all $i\in [k_1]$ and for all $j>k_1$,
 let us say $f_i=u_1x_i$ for $i\in [k_1]$.  Set $I'=(f_1,\ldots,f_{k_1})$, $J'=I'\cap J$  and
 $\hat S=K[\{x_i:i\in [k_1]\cup \supp u_1\}]$. Then
 $\depth_SI'/J'=\depth_{\hat S} I'\cap \hat S/J'\cap \hat S=\deg u_1+\depth_{\hat S} (x_1,\ldots,x_{k_1})\hat S/(J':u_1)\cap \hat S=d$. Thus $\depth_SI/J=d$ by the Depth Lemma applied to the exact sequence
$$0\to I'/J'\to I/J\to I/(I',J)\to 0,$$
since the last term has depth $\geq d$  being generated by squarefree monomials of degree $\geq d$. In particular, $\depth_SI/J=d$  if $e=1$.
If $e>1$ we may assume that for each $i\in [e]$ there exists $j\in [e]$ with $U_i\cap U_j\not =\emptyset$.
\begin{Example}{\em Back to Example \ref{e} note that we may take $u_1=x_{12}$, $u_2=x_6$ and $U_1=\{f_1,\ldots,f_6\}$, $U_2=\{f_6,\ldots,f_{11}\}$.}
\end{Example}
\begin{Lemma}\label{l2} Suppose that $e\geq 2$ and $f_r\in U_e$. Let $I'$ be the ideal generated by all $f_k\in U_e\setminus \{f_r\}$. If
$\depth_SI/(J,I',f_r)\geq d+1$ and
there exists $t$ with $f_t\in (\cup_{i=1}^{e-1} U_i)\setminus U_e$ such that $w_{rt}\in B\setminus I'$ then $\depth_SI/(J,I')=d+1$.
\end{Lemma}
\begin{proof}
Using the Depth Lemma applied to the exact sequence
$$0\to (f_{r})/(f_{r})\cap (J,I')\to I/(J,I')\to I/(J,I',f_r)\to 0$$
 we see that it is enough to show that
 $\depth_S (f_{r})/(f_{r})\cap (J,I')=  d+1$.
 By our hypothesis the squarefree monomials from $(f_{r})\setminus (J,I')$ have the form $w_{rt'}$ for some $t'$ with
 $f_{t'}\in (\cup_{i=1}^{e-1} U_i)\setminus U_e$ and   $w_{rt'}\in B\setminus I'$.

 Next we will describe the above set of monomials.
 If there exists no $U_{l}$ containing $f_t$, $f_{t'}$ then their contributions to $(f_{r})/(f_{r})\cap (J,I')$ consist in two different monomials $w_{rt}$,  $w_{rt'}$. Otherwise, we must have $f_r=x_kx_pv$, $f_t=x_kx_mv$ and $f_{t'}=x_px_mv$  for some different $k,m,p\in [n]$ and one monomial $v$ of degree $d-2$. Thus $w_{rt}= w_{rt'}$ and the contributions of $f_t$, $f_{t'}$ consist in just one monomial.  Let $A$ be the set of all $f_k \in  (\cup_{i=1}^{e-1} U_i)\setminus U_e$ such that $w_{rk}\in B\setminus I'$  and  define an equivalence relation on $A$ by $f_t\sim f_{t'}$ if $f_t, f_{t'}\in U_i$ for some $i\in [e-1]$.
For some $f_t$ from an equivalence class of $A/\sim$ we have $w_{rt}=x_{\gamma_t}f_{r}$ for one $\gamma_t\in [n]$.  Let $\Gamma$ be the set of all these variables $x_{\gamma_t}$ for which $w_{rt}\not\in (J,I')$. For two $x_{\gamma_t}$, $x_{\gamma_{t'}}$ corresponding to different classes we have
$w_{tt'}=x_{\gamma_t}x_{\gamma_{t'}}f_{r}$ since $f_t$, $f_{t'}$ are not in the same equivalence class. Thus $w_{tt'}\in J$ because otherwise $w_{tt'}\in C$ which is impossible by our hypothesis. Let $Q\subset K[\Gamma]$  be the ideal  generated by all squarefree quadratic monomials. The multiplication by $f_{r}$ gives a bijection between $K[\Gamma]/Q$ and $(f_{r})/(f_{r})\cap (J,I')$ because each squarefree monomial of $(B\cap(f_r))\setminus I'$ has the form $w_{rt}=f_rx_{\gamma_t}$ for some $t$, $x_{\gamma_t}$ being in $\Gamma$. Then $\depth_S (f_{r})/(f_{r})\cap (J,I')=d+\depth_{K[\Gamma]}K[\Gamma]/Q=d+1 $ since the variables of $f_r$ form a regular sequence for $f_{r})/(f_{r})\cap (J,I')$ (in the squarefree frame).
 Note that if $A/\sim$ has just one class of equivalence containing some $f_t$ with $w_{tr}\in B$ then $|\Gamma|=1$, $Q=0$
and also it holds $\depth_{K[\Gamma]}K[\Gamma]/Q=1$.
\hfill\ \end{proof}

\begin{Remark}\label{l2'} {\em In the notations of the above lemma suppose that $w_{rt}\in (J,I')$ for all $t$ with $f_t\in (\cup_{i=1}^{e-1} U_i)\setminus U_e$.
Then there exists no squarefree monomial of degree $d+1$ in $(f_r)\setminus (J,I')$ and so $\depth_S(f_r)/(f_r)\cap (J,I')=d$. It follows that $\depth_SI/(J,I')=d$ too.}
\end{Remark}

\begin{Lemma} \label{l3} Suppose that $e\geq 2$.  If
$\depth_SI/(J,(U_e))\leq  d+1$ then $\depth_SI/J\leq d+1$.
\end{Lemma}
\begin{proof} Suppose that $U_e\supset \{f_{k+1},\ldots,f_r\}$ for some $k\leq r$.
 Let $I'_k$ be the ideal generated by all $f_t\in U_e\setminus \{f_{k+1},\ldots,f_r\}$.
 We  claim that $\depth_S I/ (J,I'_k)\leq   d+1$. Apply induction on $r-k$, the case $k=r-1$ being done in Lemma \ref{l2} and Remark \ref{l2'}.
 Assume that $r-k>1$ and note that $I'_{k+1}=(I'_k,f_{k+1})$. By induction hypothesis we have $\depth_S I/(J,I'_{k+1})\leq d+1$. If $\depth_S I/(J,I'_{k+1})=d$
 then we get $\depth_SI/(J,I'_k)\leq d+1$ by Lemma \ref{d'}. If $\depth_S I/(J,I'_{k+1})=d+1$ we get again  $\depth_SI/(J,I'_k)\leq d+1$ by Lemma \ref{l2} and Remark \ref{l2'}. This proves our claim.

 Now choose $r-k$ maxim, let us say  $U_e= \{f_{k+1},\ldots,f_r\}$. Then $I'_k=0$ and so we get $\depth_SI/J\leq d+1$.
\hfill\ \end{proof}

\begin{Example} \label{e3} {\em Let $n=6$, $r=8$, $d=2$, $f_1=x_1x_2$, $f_2=x_1x_3$, $f_3=x_1x_4$,  $f_4=x_2x_3$, $f_5=x_3x_5$, $f_6=x_2x_6$, $f_7=x_3x_6$,   $f_8=x_4x_6$, and $I=(f_1,\ldots,f_8)$,
$$J=(x_1x_2x_5,x_1x_3x_6,x_1x_4x_5,x_1x_4x_6, x_2x_3x_4, x_2x_5x_6,  x_3x_4x_5, x_3x_5x_6, x_4x_5x_6).$$
It follows that $U_1=\{f_1,\ldots,f_3\}$, $U_2=\{f_2,f_4,f_5,f_7\}$, $U_3=\{f_6,f_7,f_8\}$, $U_4=\{f_1,f_4,f_6\}$, $U_5=\{f_3,f_8\}$, $u_1=x_1$, $u_2=x_3$, $u_3=x_6$, $u_4=x_2$, $u_5=x_4$.  Let ${\tilde S}=K[x_1,\ldots,x_5]$, ${\tilde J}\subset {\tilde I}\subset {\tilde S}$ be the corresponding ideals given in Example \ref{e2}. For  $I'=(U_3)$ we have $I/(J,I')\cong {\tilde I}S/({\tilde J},x_6{\tilde I})S$. Thus  $\depth_SI/(J,I')=\depth_{\tilde S}{\tilde I}/{\tilde J}\leq 3=d+1$ by Example \ref{e2}  and so using Lemma \ref{l3} we get $\depth_SI/J\leq 3$ too.}
\end{Example}

\begin{Proposition}\label{pr} If $\cap_{i\in [e]}U_i\not=\emptyset$ then $\depth_SI/J\leq d+1$.
\end{Proposition}
\begin{proof} As we have seen $\depth_SI/J=d$ if $e=1$. Assume that $e>1$ and let us say $f_r\in \cap_{i\in [e]}U_i$. Set $I'=(U_e)$. In $I/(J,I')$ we have $(e-1)$ disjoint $U'_i=U_i\setminus \{f_r\}$, $i\in [e-1]$. It follows that $\depth_SI/(J,I')=d$ and so $\depth_SI/J\leq d+1$ by Lemma \ref{l3}.
\hfill\ \end{proof}

{\bf Proof of Lemma \ref{m}.}

 If $\cap_{i\in [e]}U_i\not=\emptyset$ then apply the above proposition. Otherwise,  suppose that
$f_r\in U_j$ if and only if $1\leq j<k$ for some $1<k\leq e$. Set $I'=(U_{k},\ldots,U_e)$. Applying again the above proposition we get $\depth_SI/(J,I')\leq d+1$.
Set $L_i=\cup_{j\geq i} U_j$.  Since  $I'=(L_{k}) $ and $\depth_SI/(J,L_{k})\leq d+1$  we see that   $\depth_SI/(J,L_{k+1})\leq d+1$ by Lemma \ref{l3}. Using by recurrence Lemma \ref{l3} we get  $\depth_SI/J=\depth_SI/(J,L_{e+1})\leq d+1$ since $L_{e+1}=\emptyset$.
 $\hfill\ \square$

It is not necessary  to assume in  Proposition \ref{pr} that $C\cap W=\emptyset$ because anyway this follows as shows the following lemma.

\begin{Lemma} \label{l4} If $\cap_{i\in [e]}U_i\not=\emptyset$ then $C\cap W=\emptyset$.
\end{Lemma}
\begin{proof} Clearly we may suppose that $e>1$. Let $f_r\in \cap_{i\in [e]}U_i$. Suppose that $w_{tt'}\in C$ for some $t\in U_i$, $t'\in U_j$. Then  $i\not =j$ and let us say $f_t=u_ix_1$, $f_{t'}=u_jx_2$ and $f_r=u_ix_3=u_jx_4$. It follows that $u_i=x_4v$, $u_j=x_3v$ for some monomial $v$ and so $f_t=x_1x_4v$, $f_{t'}=x_2x_3v$, $f_r=vx_3x_4$. Note that $f_tx_2\in B$ and so must be of type $w_{tt''}$ for some $t''\in [r]$. It follows that $f_t,f_{t''}\in U_k$ for some $k\in [e]$. By hypothesis $f_r\in U_k$ and so $k=i$ because otherwise $|U_k\cap U_i|>1$ which is impossible. Since $f_tx_2\in B$ we see that $x_2\not\in \supp u_i$ and it follows that $f_{t''}=x_2x_4v$. Therefore, $w_{t't''}=x_2x_3x_4v\in B$ and so $f_{t'},f_{t''}\in U_p$ for some $p$.  This is not possible because otherwise one of $|U_p\cap U_i|$, $|U_p\cap U_j|$, $|U_i\cap U_j|$ is  $\geq 2$. Contradiction!
\hfill\  \end{proof}


TITLE: Find stationary distribution decomposable Markov chain
QUESTION [0 upvotes]: Again a probability exercise:
Let $X=U \cup V$ be the finite state space of a Markov chain, where $U$ and $V$ are disjoint subsets of $X$ and $p_{ij}=0$ if both $i,j \in U$ or both $i,j \in V$. Otherwise, they are positive. Find a stationary distribution?
I'm assuming that it has multiple distributions but I wrote down the system that I'm trying to solve and either I'm really tired or I should be thinking of smth else: if $\pi$ is the stationary distribution then I'm trying to find we have $\pi_{j} = \sum_{i \in V} {\pi_{i} p_{ij}}$ for all $j \in U$ and also $\pi_{j} = \sum_{i \in U} {\pi_{i} p_{ij}}$ for all $j \in V$... Can someone please help again?
Thanks!

REPLY [1 votes]: The wording of this exercise seems a bit strange to me, since
every finite state space Markov chain has a stationary distribution.

Added: Let $P$ be the transition matrix of your chain. Any stationary distribution $\pi$ for $P$ is also stationary for $P^2$. But under $P^2$, the sets $U$ and $V$ are disjoint, recurrent communicating classes with unique stationary distributions
$\pi_U$ and $\pi_V$ (of lengths $|U|$ and $|V|$).
Let $\tilde\pi_U$ and $\tilde\pi_V$ be vectors of length $|X|$ obtained by padding
$\pi_U$ and $\pi_V$ with zeros in the appropriate places. Then the set of all  stationary distributions for $P^2$ is
$$\{\alpha\tilde\pi_U+(1-\alpha)\tilde\pi_V,\, 0\leq \alpha\leq 1\}.$$
The only one of these that is also stationary for $P$ is ${1\over 2}\tilde\pi_U+ {1\over 2}\tilde\pi_V$.


TITLE: One such group of interest among 10
QUESTION [2 upvotes]: It's known that birth months are uniformly distributed. A class is divided into 10 groups of 5 students. A group that all five members were born in different months is our interest. What is the probability that there is one such group of interest among 10 groups?
I thought it would be 1/12 (1 of the months out of 12) then multiplied by 1/11 (as all 5 months have to be unique) all the way to 1/8, however i dont believe this is correct. Please let me know how to solve this.

REPLY [4 votes]: As Henry said in the comments:
$P(5$ randomly chosen ppl are all in different months$) = \frac{11}{12}\times \frac{10}{12}\times \frac{9}{12}\times \frac{8}{12} =  \frac{55}{144}.$
Define $Y$ to be the number of interesting groups out of $10$ groups.
$Y\sim B\left(10,\frac{55}{144} \right).$
$P(Y=1)=\ldots$
$P(Y\geq 1) = \ldots$
depending on what you want.


TITLE: Left multiplication by invertible matrix.
QUESTION [2 upvotes]: If M is an $n\times m$ matrix of rank $n$ and $A\in GL_n$, that is, $A$ is an invertible $n\times n$ matrix, then why is the row reduced echelon form of $M$ is the same as the row reduced echelon form of $AM$?

REPLY [1 votes]: I will not answer your question directly, but rather give you hints by referring to the relevant definitions and theorems:

First, be sure to understand that the row reduced echelon form of a matrix is unique (Refer for example to Matrices and linear transformations by Cullen, theorem 2.18), but a matrix in row reduced echelon form might be row equivalent to more than one matrix (equivalence relation - so several matrices may belong to the same row equivalent class):— so $M$ and $AM$ will generally not be the same matrix, but they could be row equivalent to the same matrix in row reduced echelon form.
The matrix $M$ is row equivalent to a matrix in row reduced echelon form (say this matrix is $N$) iff there exists an invertible matrix $P$ such that $PM=N$. Also, every $n \times m$ matrix over a field is row equivalent to a matrix in row reduced echelon form (so we are guaranteed that such a $P$ and $N$ exist). For the theory related to these statements please refer to, for example, section 1.8 and 1.9 in the same text as mentioned above (this is all about row equivalence, series of elementary operations, elementary matrices, etc. it helps to know this).
Now take $AM$: if $PM=N$ with $N$ in row reduced echelon form, can you find an invertible matrix $Q$ such that $QAM=N$? By finding $Q$ you will prove that $M$ and $AM$ have the same row reduced echelon form.

I hope that is enough guidance? Ask if you want clarification on something specific.


TITLE: Fibonacci induction proof?
QUESTION [0 upvotes]: The Fibonacci Numbers $(f_n)$ are defined $f_1=f_2=1$, and $f_n=f_{n-1}+f_{n-2} ,\,\,\,\forall n \geq2$.
Prove that for every integer $n \geq 1$,
$$f_1 +f_2 +···+f_n =f_{n+2}−1$$

REPLY [0 votes]: $f_1 +f_2 +···+f_n =f_{n+2}−1 \to f_1 +f_2 +···+f_n =f_n + f_{n+1}−1$, so:
$f_1 +f_2 +···+f_{n-1} =f_{n+1}−1$.
Repeat until $f_1=f_3-1$, which is true ($f_3=2$).


TITLE: Can one infer independence by simple reasoning/intuition?
QUESTION [7 upvotes]: From my recent experience in probability, it feels as though independence is something we "discover" from the system via the equation:
$$P(A)*P(B)=P(A\cap B)$$
Could one ever conclude independence from the "system" by intuition? Is it wise to conclude independence for events that are "seemingly" independent? What would be some interesting examples where this would fail.

REPLY [0 votes]: An interesting example is as follows
Suppose there are two fair six faced dice, say I and II. Both dice are rolled together, let $X$ denotes the face shown by die I and $Y$ denotes the face shown by die II.
Define $Z:= X+Y$ and $U := $ remainder when $Z$ is divided by $6 $.
Surprisingly in the above construction random variables $U$ & $X$ are independent.
For example, let $A= \{X= 1\}  $  & $B=\{ U= 1\} =
\{ (X,Y)= (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)\} $
Then $\mathbb{P}(A)= \dfrac{1}{6}$,  $\mathbb{P}(B)= \dfrac{1}{6}$ and $\mathbb{P}(A\cap B)= \mathbb{P}(\{ (X,Y)= (1,6)\})=  \dfrac{1}{36}= \mathbb{P}(A) * \mathbb{P}(B)$.


TITLE: Partial Derivative.
QUESTION [0 upvotes]: Can I write the right hand side of the following equation ?
$$\frac{\partial}{\partial x}[\frac{\partial f(y)}{\partial y}]=\frac{\partial^2 f(y)}{\partial y\partial x}.$$
Does this operator follow the property ?

REPLY [1 votes]: If $f$ carries an open interval of $\mathbb{R}$ to $\mathbb{R}$, then it makes no sense to find a partial derivative of $f$; by definition a partial derivative of a map is the directional derivative of the map in the direction of a standard unit vector, provided that the map is a scalar field.
If $n \geq 2$, if $f$ carries some set $A$ open in $\mathbb{R}^{n}$ to $\mathbb{R}$, and if $f$ is twice differentiable on $A$, then we have
$$
D_{i}D_{j}f = D_{j}D_{i}f
$$
on $A$ for all $1 \leq i,j \leq n$;
this is an elementary result in analysis.
In particular, if $n := 2$, then $D_{1}D_{2}f = D_{2}D_{1}f$ on $A$; in the language of applied math, we may write this as
$$
\frac{\partial^{2}f}{\partial x \partial y} = \frac{\partial^{2}f}{\partial y \partial x},
$$
which is of course valid on $A$ too.


TITLE: A sequence of polynomials with degree going to $\infty$
QUESTION [0 upvotes]: Let $(\mathcal p_n)$ be a sequence of polynomials which converges uniformly on $[0,1]$ to some function $\mathcal f$, assume that $\mathcal f$ is not a polynomial. Prove the $\lim \limits_{n \to \infty}\deg(\mathcal p_n)=\infty,$ where deg$(\mathcal p)$ denotes the degree of a polynomial $\mathcal p$.
It is well known that space of polynomials is dense in the space of continuous functions. Intuition,If the degree of polynomial is finite then space of polynomial would be closed which no longer dense in the space of $C[0,1]$. Is it correct intuition? Give me ideas.

REPLY [3 votes]: Indeed, suppose that all your polynomials have degree at most $d$. Then the subspace $S$ of $C[0,1]$ of polynomials of degree at most $d$ is closed (as is any other finite dimensional subspace, for that matter), so any sequence contained in $S$ which converges in $C[0,1]$ has limit in $S$.


TITLE: How to Determine the Area of Integration of the Solid: $S=\{(x,y,z) \in \mathbb{R}^3: x^2+y^2+z^2 \le 4; \frac{1}{\sqrt{3}}x \le y \le \sqrt{3}x\}$
QUESTION [0 upvotes]: I am trying to determine the area of integration of the solid represented by:
$$S=\{(x,y,z) \in \mathbb{R}^3: x^2+y^2+z^2 \le 4; \frac{1}{\sqrt{3}}x \le y \le \sqrt{3}x\}$$
From this I gathered that there are three regions represented:
$$
x^2+y^2+z^2=4 \rightarrow \text{ sphere of radius 2 } \\
y=\frac{1}{\sqrt{3}}x \rightarrow \text{ plane parallel to the z axis } \\
y = \sqrt{3}x \rightarrow \text{ plane parallel to the z axis }
$$
With this, I have tried the following points in the conditions of $S$ to determine which area of the solid I will be working with:
$$
(2, 0, 0) \rightarrow \text{ belonging to the sphere } \\
(0, 2, 0) \rightarrow \text{ belonging to the sphere } \\
(0, 0, 2) \rightarrow \text{ belonging to the sphere } \\
$$
From which only the point $(0, 0, 2)$ fulfills all three conditions. How do I proceed from here?

REPLY [1 votes]: Looking from the +z axis, two planes look like this, and together with sphere they bound the red area.

Insight solution: I can now observe how volume slices (that extend in z direction, blue area on picture) change depending on  $r$ (distance from center). Express the volume $dV$ of each slice as a function of $r$ and I would remain with single integral.

$\int_{0}^{R}dV=\int_{0}^{R}(r*\alpha)*h(r)dr$.
Here $\alpha$ is angle of red area, and $h(r)$ is the height of each slice that depends on $r$. $r$ is the variable of integration and $R$ is sphere radius $=2$.

Longer solution: Use double integral with variables of integration $r$ and $\phi$. As height of the volume in question is constant with $\phi$ it resolves as constant factor $\alpha=\phi2-\phi1$ used above.
$\int_{0}^{R}\int_{\phi1}^{\phi2}h(r, \phi)d\phi dr= \int_{0}^{R}h(r)\int_{\phi1}^{\phi2}d\phi dr = \int_{0}^{R}h(r)(\phi2-\phi1)dr $


TITLE: Convergence radius of a power series $\sum_{k=0}^\infty a_kx^{2k}$ with $\lim\limits_{n \rightarrow \infty}|a_n|^{\frac{1}{n}}=\frac{1}{r}$
QUESTION [1 upvotes]: I need help for the following task:
Let {$a_n$}$_{n\in\mathbb{N}}$ be a sequence of real numbers such that $\lim\limits_{n \rightarrow \infty}|a_n|^{\frac{1}{n}}=\frac{1}{r}$ with $r$ being a positive real number.
Show that the power series $\sum_{k=0}^\infty a_kx^{2k}$, $x \in \mathbb{R}$ has radius of convergence $\sqrt{r}$.

I mean this is the formula we have to work with. If we have $x^n$, then $a_n=1$ and the radius is $1$. But in this situation I don't get the result that I should have and I don't know what I did wrong.

REPLY [1 votes]: Let us apply the root test to the proposed power series:
\begin{align*}
\lim_{n\to\infty}|a_{n}x^{2n}|^{1/n} = x^{2}\times\lim_{n\to\infty}|a_{n}|^{1/n} = \frac{x^{2}}{r} <  1 \Rightarrow |x| < \sqrt{r}
\end{align*}
and we are done.
Hopefully this helps !


TITLE: Finding points $X=(x,y)$ and directions where directional derivative of $f(x,y)=3x^2+y^2$ is maximum. The points are to be taken on $x^2+y^2=1$
QUESTION [1 upvotes]: Let the directional vector be $d=(a,b)$ so that $a^2+b^2=1$
Directional derivative of $f$ along $d$ is $ f'(X,d)=\nabla f.d=(6x,2y).(a,b)$
$=6xa+2yb=6\sin \theta\sin\phi+2\cos\theta\cos\phi=4+2\cos(\theta-\phi)$, where let $x=\cos\theta,a =\cos\phi$
Hence, $f'(X,d)|_{max}=6$, which occurs when $\theta -\phi =2n\pi\implies$that is, when $x=a,y=b$
Is this correct? I am confused because I saw the solution online and $f'(X,d)|_{max}=8$.
Can someone please help me understand how it is possible? Thanks in advance.

REPLY [1 votes]: You basically try to solve the following problem: Find the max value of $g(a,b,x,y) = 6ax+2by$ such that $a^2+b^2=1=x^2+y^2$. Simple. Apply the CS inequality: $|g(a,b,x,y)|^2 \le (36a^2+4b^2)(x^2+y^2)= 36a^2+4b^2= 4+32a^2\le 4+32=36\implies |g| \le \sqrt{36}=6$. This is the max value you are seeking and it is attained when $a = \pm 1, b = 0\implies g = \pm 6x=6\implies x =\pm 1, y=0$. Note that if $a = 1, x = 1$, and $a = -1, x = -1$.


TITLE: If $(f,g)$ is measurable than $f$ and $g$ are measurable?
QUESTION [1 upvotes]: Suppose $f : (Z,z) \rightarrow (X,x)$ and $g : (Z,z) \rightarrow (Y,y)$. If $(f,g)=(Z,z) \rightarrow (XxY, x⊗y)$ are measurable then $f$ and $g$ are measurable? If so how can I show that?
My approach to this question is to take a measurable set $B \times C$ in $x⊗y$ and than consider inverse and so on... But I can't have the answer.

REPLY [1 votes]: We know that
$$
z\ni(f,g)^{-1}(B\times C)=f^{-1}(B)\cap g^{-1}(C)
$$
for all $B\in x$ and $C\in y$. To show that, say, $f$ is measurable, you simply choose $C$ wisely.


text:
\section{Background}
\label{sec:background}
\subsection{Gaussian Sketches for Least Squares Optimization}
Consider an unconstrained least squares optimization problem
\begin{equation} x_{LS} = \text{arg}\min\limits_{x\in\R^d} \, \norm{Ax - y}_2^2
\label{eq:ls}
\end{equation}
where the data matrix $A\in\R^{n\times d}$ and the data vector $y\in\R^n$ are fixed and known. It is assumed that $n>d$. Throughout this project, we will assume that rank$(A)=d$, so that $A^TA$ is invertible. The classical Gaussian sketch is defined as
\begin{equation}
    \widehat{x} = \text{arg}\min\limits_{x\in\R^d} \, \norm{SAx - Sy}_2^2
    \label{eq:sketchls}
\end{equation}
where $A$ and $y$ are replaced by low dimensional projections $SA$ and $Sy$, where $S\in\R^{m\times n}$ has i.i.d. entries $S_{ij}\sim\frac{1}{\sqrt{m}}\N(0,1)$. Note that the sketching matrix $S$ satisfies $\E[S^TS]=I_n$.
\begin{lemma}
\label{lem:ls}
The classical Gaussian sketch $\widehat{x}$ satisfies
\begin{align}
\label{eq:sketch_solution}
    \widehat{x} &= x_{LS} + (SA)^\dagger Sy^\perp
\end{align}
where $(SA)^\dagger = (A^TS^TSA)^{-1}A^TS^T$ is the pseudoinverse of $SA$ and $y^\perp=y-Ax_{LS}$ such that $A^Ty^\perp=0$. $\norm{y^\perp}_2^2$ is called the residual error. Further, $Sy^\perp$, and the columns of $SA$ are independent multivariate Gaussians, and
\begin{equation}
\label{eq:sketch_dist}
    \widehat{x} \sim \N\left(x_{LS}, \frac{1}{m}\norm{y^\perp}_2^2 (A^TS^TSA)^{-1}\right) \mid SA
\end{equation}
\end{lemma}
Here the notation $\mid SA$ means conditioning on the random variable $SA$. Also Lemma \ref{lem:ls} assumes that $A^TS^TSA$ is invertible, which is true with probability 1, if $A^TA$ is invertible \cite{gupta2018matrix}.

This property allows us to look at the least squares optimization problem as a statistical estimation problem of estimating the mean of a multivariate Gaussian from its sample, conditioned on the matrix $SA$. Note that this property holds irrespective of the distribution of the data, and is a property of the Gaussian sketch method. In particular this means that $\widehat x$ is an unbiased estimator of $x_{LS}$ since $\E[\widehat x]=x_{LS}$.

The expected error of the classical Gaussian sketch can be evaluated exactly. This is known from earlier works as well \cite{ahfock2017statistical}, and is stated below again.

\begin{lemma}
\label{lem:gaussian_sketch}
The prediction error of the classical Gaussian sketch $\widehat x$ is:
\begin{equation}
\label{eq:sketch_error}
    \E\left[\norm{A(\widehat{x}-x_{LS})}_2^2 \right] = \frac{d}{m-d-1} \norm{y^\perp}_2^2
\end{equation}
\end{lemma}

\subsection{James-Stein Shrinkage Estimator}
Suppose we are given $X \sim \mathcal{N}(\theta,I_d)$ where $\theta\in\mathbb{R}^d$ is an unknown mean vector. We want to design an estimator $\widehat{\theta}(X)$ such that it minimizes the expected squared error $R(\widehat{\theta},\theta)=\mathbb{E}\lVert\widehat{\theta}-\theta\rVert_2^2$. The maximum likelihood estimator for $\theta$ is $\widehat{\theta}(X)=X$ which is an unbiased estimator, i.e. $\E\,\widehat\theta=\theta$. However, from \cite{stein1956} and \cite{james1961}, we have that the estimator given by
\begin{equation}
\label{eq:shrink_I}
    \widehat{\theta}(X) = \left(1-\frac{d-2}{\lVert X\rVert_2^2} \right) X
\end{equation}
results in a lower variance than the maximum likelihood estimator for $d>2$ for any value of $\theta$. In statistical language, $\widehat{\theta}(X)=X$ is an inadmissible estimator and (\ref{eq:shrink_I}) strictly dominates the former. This is a "shrinkage estimator" (as it shrinks the estimate towards 0), now known as the James-Stein estimator.

This estimator has been very useful in statistics, see \cite{efron1977} for example. A simple proof and explanation of this result can be found in \cite{yung1999explaining} and \cite{samworth2012stein}. There are also several improvements and generalizations of this estimator \cite{efron1976, hansen2008generalized, hoff2012shrinkage, shao1994improving} to other shrinkage estimators. Particularly, we look at the generalization to the case where $X\sim\N(\theta,\Sigma)$.
\begin{lemma}
\label{lem:stein}
Suppose $X\sim\N(\theta,\Sigma)$ where $\theta\in\R^d$ is the unknown mean with $d>2$. $\Sigma$ and a single sample $X$ are known. Then consider the estimator for $\theta$ as
\begin{equation}
\label{eq:shrink_Sigma}
    \widehat\theta(X) = \left(1- \frac{d-2}{X^T\Sigma^{-1}X} \right)X
\end{equation}
Then, its expected normalized squared error is given by
\begin{equation}
\label{eq:shrink_error}
    \E[(\widehat\theta - \theta)^T\Sigma^{-1} (\widehat\theta - \theta)] = d - (d-2)^2 \E\left[ \frac{1}{X^T\Sigma^{-1}X} \right]
\end{equation}
\end{lemma}

If we used the estimator $\widehat\theta(X)=X$, then we would have $\E[(X-\theta)^T \Sigma^{-1}(X-\theta)]=d$ and the estimator shown above achieves strictly lesser expected error for any value of $\theta$ when $d>2$. However, this is a biased estimator and the expected squared error will be lower when the true value of $\theta$ is close to zero, as compared to $\theta$ far from zero. More generally, we can also shrink the estimator towards any other point $\theta_0$ and derive an estimator $\theta_0 + \left(1- (d-2)/X^T\Sigma^{-1}X \right)(X-\theta_0)$, so that the error is smaller when $\theta$ is close to $\theta_0$.

The above section motivates us to use the shrinkage estimator to estimate the true least squares solution $x_{LS}$ from the solution $\widehat x$ of one sketched least squares problem.


TITLE: How are Superposition and Entanglement related?
QUESTION [2 upvotes]: How are Superposition and Entanglement related? I don't know much of Quantum Mechanics. I am CSE student and got started with  this Quantum Computing. It is interesting! If anybody can help me on this, without digging deep into mathematical details.

"The principle of quantum superposition refers to pure states of a quantum system. One considers a primary beam of quantal entities that passes into a primary beam splitter or quantum analyser that has multiple output channels. A beam has a pure state when every one of its quantal entities passes into one and the same output channel. A primary beam with a pure state is passed into another, different, secondary beam splitter or quantum analyser. Then the emerging quanta are probabilistically in its several output channels. These several emergent, intermediate, beams are respectively pure with respect to the secondary analyser. They are then passed to respective copies of the secondary analyser, arranged so as to bring them together into a single reconstituted beam. That beam is then passed to a copy of the primary analyser. In general, it will be probabilistically split into the several output channels." - Wikipedia
I did not get that! Any help is appreciated.

REPLY [2 votes]: Superposition & Measurement
The two analyzers measure different properties, or observables, of the beam quanta.
Prior to measurement, a quantum state may be in a superposition of many values for an observable, so there is a probability that you will measure any of these values. When you measure an observable, the quantum state "collapses" so that a repeated measurement gives the same result with probability 1.
However these analyzers are measuring two non-commuting observables. What this means is that the quantum states representing definite values of these observables - "eigenstates" - cannot be the same; a eigenstate of observable A is a superposition of eigenstates for observable B, and vice versa.
So if a beam with quanta in eigenstate A=1 is measured by the first analyzer it takes the A=1 channel. But when the second analyzer measures B, the state of each quanta is in a superposition of B-eigenstates and can take multiple channels. Each quanta collapses to different B-eigenstates. Upon measuring A again at the first analyzer, the quanta are now in superpositions of A-eigenstates and can take channels other than A=1. At the end of this process each of the quanta is in one of the A-eigenstates.
(FYI this picture of quantum collapse, while it describes sequences of measurements very well, is probably incorrect. The more complicated picture is that the state doesn't collapse, instead it becomes entangled with the measuring device and all the superposition states decay away thanks to decoherence)
Entanglement
Sets of quanta can be in a superposition. They are entangled if that superposition cannot be factored into superpositions of the individual quanta, which results in a correlation between measurements. For instance, if observable A of two quanta can be 0 or 1 with equal probability, you might measure {0,0}, {1,0}, {0,1}, and {1,1} all with equal probability - these quanta are not entangled because the results are factorable into "A1 can be 0 or 1" and "A2 can be 0 or 1", with no relation between them. An entangled example would be that observable A for the two quanta is measured as {1,0} or {0,1} with equal probability. These are entangled because while measuring either quanta can give A=0 or A=1, you can't measure {0,0} or {1,1}. When quanta are entangled, measuring one allows you to better (or even completely) predict the result of the other measurement.

REPLY [1 votes]: My 1st answer described what is a coherent superposition. Now, in short about entanglements. Taking two quantum systems, e.g. one quantum fowl described by
$(\text i) \ |\psi_1 \rangle = \frac {|QT\rangle \ + \ |QD\rangle}{\sqrt {2}}$
and the other quantum fowl described by
$(\text {ii}) \ |\psi_2 \rangle = \frac {|QT\rangle \ - \ |QD\rangle}{\sqrt {2}}$
where $|QT\rangle$ is the state quantum turkey and a $|QD\rangle$ is the state quantum duck, the joint system of the two fowls should be
$(\text {iii}) \ |\Psi\rangle = |\psi_1\rangle |\psi_2\rangle = \frac {|QT_1\rangle |QT_2\rangle \ - \ |QT_1\rangle |QD_2\rangle \ + \ |QD_1\rangle |QT_2\rangle \ - \ |QD_1\rangle |QD_2\rangle}{2}.$
That means you have four possibilities, you can find that both fowls are turkeys, or the fowl 1 is quantum turkey and fowl 2 quantum duck, etc. each possibility with probability 1/4.
Well, an entanglement is not separable as a product of two independent states, $|\psi_1\rangle |\psi_2\rangle$. E.g. the state of the two systems may be
$(\text {iv}) \ |\Psi \rangle = \frac {|QT_1\rangle |QT_2\rangle \ - \ |QD_1\rangle |QD_2\rangle}{\sqrt {2}}.$
The uncommon situation with this entanglement is that if you check which type is the one fowl and find turkey, the other fowl will be with certainty also turkey. But if you find that the 1st checked fowl is duck, the other fowl will be with certainty duck. So are the entanglements. Checking of which type is one quantum object, you already know something about which is the type of the other object.


TITLE: odd logical structures
QUESTION [2 upvotes]: How you find contrapositive and converse of these sentences.

Only if John chops down the tree, will he be a lumberjack.
You can't win if you don't fight.
All people that root for the Ducks are from Oregon.

The logical has me thrown at where the if is happening so I can switch it around from p implies q to q implies p. Please help

REPLY [1 votes]: HINT
We have to rewrite 1) as :

"Only if John chops down the tree, will he be a lumberjack"

as :

if John will be a lumberjack, then he chops down the tree.

This one has the "logical form" : $p \rightarrow q$; thus, its converse ($q \rightarrow p$) will be :

if John chops down the tree, then he will be a lumberjack,

while its contrapositive ($\lnot q \rightarrow \lnot p$)will be :

if John does not chop down the tree, then he will not be a lumberjack.

For 2), we simply have :

if you don't fight, then you can't win.

Thus, converse and contrapositive must be straightforward.

For 3), assuming that we have to "analyze" it without predicate logic, I agree with you :

if a person roots for the Ducks, then he is from Oregon.

Again, having reduced it to the standard "logical form" : $p \rightarrow q$, we have only to apply the above formulae to get converse and contrapositive.