text:
\begin{document}

\title{On positivity and base loci of vector bundles}

\author[Bauer, Kov\'acs, K\"uronya, Mistretta, Szemberg, Urbinati]{Thomas Bauer,
  S\'andor J Kov\'acs, Alex K\"uronya, Ernesto Carlo Mistretta, Tomasz Szemberg, Stefano
  Urbinati}

\address{TB: Fachbereich Mathematik und Informatik, Philipps-Universit\"at Marburg, Hans-Meerwein-Stra{\ss}e, Lahnberge, 35032 Marburg, Germany}
\address{SJK: University of Washington, Department of Mathematics, Seattle, WA 98195, USA}
\address{AK: Budapest University of Technology and Economics, Mathematics Institute, Egry J\'ozsef u. 1, H-1111 Budapest, Hungary}
\address{ECM: Universit\`a degli Studi di Padova, Dipartimento di Matematica, ROOM 619, Via Trieste 63, 35121 Padova, Italy}
\address{TSz: Pedagogical Univarsity Cracow, Podchor\c a\.zych 2, 30-084 Krak\'ow, Poland}
\address{SU: Universit\`a degli Studi di Padova, Dipartimento di Matematica, ROOM 608, Via Trieste 63, 35121 Padova, Italy}

\email{TB: tbauer@mathematik.uni-marburg.de}
\email{SJK: skovacs@uw.edu}
\email{AK: alex.kuronya@math.bme.hu}
\email{ECM: ernesto.mistretta@unipd.it}
\email{TSz: tomasz.szemberg@gmail.com}
\email{SU: urbinati.st@gmail.com}

\date{\usdate\today}

\maketitle
\thispagestyle{empty}

\silentremark
{\center{\sandor{060914}: \small\tt\bf PLEASE!! When you add something, try to follow the
  notational conventions already in the paper. I corrected a bunch of
  inconsistencies, and I don't want to do it again. Please look at the rest of the
  paper how things are denoted. E.g., intersection numbers with ``cdot'' but without
  enclosing parantheses. There were three different notations in different parts!}}

\silentremark{\sandor{060914}: I removed ``Viehweg'' from ``weakly positive'' everywhere
  I found, because I don't believe there is any other notion of weak positivity. Same
  for Miyaoka and AEA.}

\section{Introduction}

The aim of this note is to shed some light on the relationships among  some notions of positivity for vector bundles that arose
in recent decades.

Positivity properties of line bundles  have  long played a major role in projective geometry;
they have once again become a center of attention recently, mainly in relation with advances in birational geometry,
especially in the framework of the Minimal Model Program. Positivity of line bundles has often
been studied in conjunction with numerical invariants and various kinds of asymptotic base loci (see
for example \cite{ELMNP} and \cite{BDPP}).

At the same time, many positivity notions have been introduced for vector bundles of higher rank,
generalizing some of the properties that hold for line bundles.  While the situation in rank one is well-understood, at least
as far as the interdepencies between the various positivity concepts is concerned, we are quite far from an analogous state of
affairs for vector bundles in general.

In an attempt to generalize bigness for the higher rank case, some positivity properties have been put forward
by Viehweg (in the study of fibrations in curves, \cite{vie}), and Miyaoka (in the context of surfaces, \cite{Miy83}), and are known to
be different from the generalization given by using the tautological line bundle on
the projectivization of the considered vector bundle (cf. \cite{PAG}).  The
differences between the various definitions of bigness are already present in the
works of Lang concerning the Green-Griffiths conjecture (see \cite{lang}).

Our purpose  is to study  several of the positivity notions studied for vector
bundles with some notions of asymptotic base loci that can be defined on the
variety itself, rather than on the projectivization of the given vector bundle.  We
relate some of the different notions conjectured to be equivalent with the help of
these base loci, and we show that these can help simplify the various relationships
between the positivity properties present in the literature.

In particular, we define augmented and restricted base loci $\mathbb{B}_+ (E)$ and
$\mathbb{B}_- (E)$ of a  vector bundle $E$ on the variety $X$, as generalizations of the corresponding
notions studied extensively for line bundles.  As it turns out, the asymptotic base loci defined here behave well with
respect to the natural map induced by the projectivization of the vector bundle $E$,
as shown in Section~\ref{asvect}.

The relationship between these base loci with the positivity notions appearing in the
literature goes as follows:

\begin{theorem}

  Let $X$ be a smooth projective variety and $E$ a vector bundle on $X$. Then:

\begin{enumerate}

\item $E$ is \emph{ample} if and only if ${\mathbb{B}_+ (E)}  = \emptyset$;

\item $E$ is \emph{nef} if and only if ${\mathbb{B}_- (E)}  = \emptyset$;

\item $E$ is \emph{pseudo-effective} if and only if ${\mathbb{B}_- (E)} \neq X$;
  \ilabel{changed}
  \silentremark{\sandor{060914}: changed ``$\emptyset$'' to ``$X$'' in \iref{changed}}

\item $E$ is
  \emph{weakly positive} if and only if $\overline{\mathbb{B}_- (E)} \neq X$ (see
  Section~\ref{restricted});

\item $E$ is \emph{V-big} if and only if ${\mathbb{B}_+ (E)} \neq X$ (see section
  \ref{augmented});

\item Assume that $E$ is a nef vector bundle. Then $E$ is
  \emph{almost everywhere ample} if and only if ${\mathbb{B}_+ (E)} \neq X$ (cf.\
  Section~\ref{badaea}).

\end{enumerate}

\end{theorem}

The paper is organized as follows: in sections \ref{def} and \ref{asvect} we give the
definition and basic properties of the asymptotic base loci for vector bundles, and
relate these loci with the ones on the projectivizations. In Section~\ref{line} we recall the various positivity properties for line bundles
and their relationship with asymptotic base loci. Section~\ref{restricted} is devoted to a  study of  positivity properties of vector bundles
related to the restricted base locus, while   Section~\ref{augmented} is given over to an investigation of connection between
positivity properties of  vector bundles related and  augmented base loci.
In sections \ref{sectionaea} and \ref{badaea} we study almost everywhere ampleness
and relate it to V-bigness.

\section{Definitions and first properties}
\label{def}

\begin{convention}\label{conv}
  Throughout the paper we are working with vector bundles of finite rank, but for
  various reasons we find it  more convenient to work with the associated sheaf of sections
  which is a locally free coherent $\sO_X$-module. We will follow the usual abuse of
  terminology and while exclusively using this associated sheaf, we will still call it a
  vector bundle. If, rarely, we want to refer to a vector bundle and mean a vector
  bundle we will call it the \emph{total space} of the vector bundle.

  We will also work with line bundles, which of course refers to a locally free sheaf
  of rank $1$. For a line bundle $L$ we will denote by $c_1(L)$ the associated Weil
  divisor on $X$.
\end{convention}

With that convention fixed we are making the following notation that we will use
through the entire paper:

\begin{no}
  Let $X$ be a smooth projective variety over the complex numbers, and $E$ a vector
  bundle (i.e., according to \ref{conv} really a locally free sheaf) over $X$. For a
  point $x\in X$, $E_x=E\otimes_{\sO_X}\sO_{X,x}$ denotes the stalk of $E$ at the
  point $x$ and $E(x)=E\otimes_{\sO_X} \kappa(x)$ where $\kappa(x)$ is the residue
  field at $x$. Clearly, $E(x)$ is the fiber of the total space of $E$ over the point
  $x$. In particular, $E(x)$ is a vector space of dimension $r=\rk E$.
\end{no}

\begin{definition}

  We define the \emph{base locus} of $E$ (over $X$) as the subset
\[
\mathrm{Bs} (E) := \{ x \in X ~|~ H^0(X, E) \to E(x) ~\textrm{is not surjective} \}\ ,
\]
and the \emph{stable base locus} of $E$ (over $X$) as
\[
\mathbb{B} (E) := \bigcap_{m>0} \mathrm{Bs}(\Sym^m E)\ .
\]
\end{definition}

\begin{remark} The assertions below follow immediately from the definition:
  \begin{enumerate}
  \item As $\mathrm{Bs} (E) = {\rm Bs} (\mathrm{Im} (\bigwedge^{\rk E} H^0(X, E) \to
    H^0(X, \det E) ))$, these loci are closed subsets, and carry a natural scheme
    structure.
  \item For any positive integer $c>0$, $\mathbb{B}(E) = \mathbb{B}(\Sym^c E)$, and
    the same holds for $\mathbb{B}_-$ and $\mathbb{B}_+$. \ilabel{base-of-sym}
  \end{enumerate}
\end{remark}

\begin{remark}
The rank of the natural linear map $H^0(X, E) \to E(x)$ induces a stratification of $X$ into locally closed subsets.
\end{remark}

\begin{definition}

  Let $r = p/q \in \mathbb{Q}_{>0}$ be a positive rational number, and $A$ a line
  bundle on $X$.  We will use the following notation:
  \begin{align*}
    \mathbb{B}(E + rA) &:= \mathbb{B}(\Sym^q E \otimes A^{p}), \quad\text{and}\\
    \mathbb{B}(E - rA) &:= \mathbb{B}(\Sym^q E \otimes A^{-p}) \ .
  \end{align*}

  Note that if $r=p'/q'$ is another representation of $r$ as a fraction, then
  $q'p=p'q$, hence
  $$
  \Sym^{q'}(\Sym^q E \otimes A^p)\simeq \Sym^{q'q} E \otimes A^{q'p}\simeq
  \Sym^{q'q} E \otimes A^{p'q}\simeq \Sym^{q}(\Sym^{q'} E \otimes A^{p'}),
  $$
  and therefore, by \iref{base-of-sym}, $\mathbb{B}(\Sym^q E \otimes
  A^p)=\mathbb{B}(\Sym^{q'} E \otimes A^{p'})$ and hence $\mathbb{B}(E + rA)$ is
  well-defined. A similar argument shows that $\mathbb{B}(E - rA)$ is also
  well-defined.

Let $A$ be an ample line bundle on $X$,
we define the \emph{augmented base locus} of $E$ as
\[
\mathbb{B}_+^A (E):= \bigcap_{r \in \mathbb{Q}^{>0}} \mathbb{B}(E - r A ) \ ,
\]
and the \emph{restricted base locus} of $E$ as
\[
\mathbb{B}_-^A (E):= \bigcup_{r \in \mathbb{Q}^{>0}} \mathbb{B}(E + r A )\ .
\]

\end{definition}

\begin{remark}

The definitions above yield the following properties:

\begin{enumerate}

\item The loci \( \mathbb{B}_+^A (E)\) and \( \mathbb{B}_-^A (E)\) do not depend on
  the choice of the ample line bundle $A$, so we can write
  \( \mathbb{B}_+ (E)\) and \( \mathbb{B}_- (E)\) for the augmented and restricted base locus of $E$, respectively.

\item For any $r_1 > r_2 > 0$ we have \( \mathbb{B} (E + r_1 A) \subseteq \mathbb{B}
  (E + r_2 A) \) and \( \mathbb{B} (E - r_2 A) \subseteq \mathbb{B} (E - r_1 A) \).

\item In particular, for any $\eps >0$ we have \( \mathbb{B} (E + \eps A) \subseteq
  \mathbb{B} (E) \subseteq \mathbb{B} (E - \eps A) \).

\item Therefore we have that
  $$
    \mathbb{B}_+ (E):= \bigcap_{q \in \mathbb{N}}
    \mathbb{B}(E - (1/q) A ) \qquad \text{and} \qquad
    \mathbb{B}_- (E):= \bigcup_{q \in \mathbb{N}} \mathbb{B}(E + (1/q) A ).
  $$

\item It follows that $\mathbb B_+(E)$ is closed, but even for line bundles, the
  locus $\mathbb{B}_- (E)$ is not closed in general: Lesieutre \cite{lesieutre}
  proved that this locus can be a proper dense subset of $X$, or a proper dense
  subset of a divisor of $X$.

\end{enumerate}

\end{remark}

\section{Asymptotic invariants for vector bundles}
\label{asvect}

In the following sections we will relate  augmented and restricted base loci for
vector bundles to various positivity notions found in the literature. In order to
achieve  a better understanding of these positivity properties and the relations
between them, it is necessary to investigate the dependence of
asymptotic base loci for vector bundles,  and the corresponding loci of the tautological quotient line
bundles on the appropriate projectivizations.

Let $E$ be a vector bundle on a smooth projective variety $X$, $\pi \colon
\mathbb{P}(E) \to X$ the projective bundle of rank one quotients of $E$, and
$\sO_{\mathbb{P}(E)}(1)$ the universal quotient of $\pi^* E$ on $\mathbb{P}(E)$.
Then we immediately have
\[
\pi (\mathbb{B} (\sO_{\mathbb{P}(E)}(1))) \subseteq \mathbb{B} (E) \ .
\]
In fact, if the evaluation map $H^0(X, E) \otimes \sO_X \to E$ is surjective over a
point $x \in X$, then the map
\[
H^0(\mathbb P(E), \sO_{\mathbb P(E)}(1)) \otimes\sO_{\mathbb{P}(E)} = H^0(X, E) \otimes \sO_{\mathbb{P}(E)} \to \pi^* E
\twoheadrightarrow \sO_{\mathbb{P}(E)}(1)
\]
is surjective over any point $y \in
\mathbb{P}(E)$ such that $\pi(y) = x$, and a similar argument applies to
$\Sym^m E$.

More precisely, we have $\pi (\mathrm{Bs} (\sO_{\mathbb{P}(E)}(1))) = \mathrm{Bs} (E)$: if a
point $x \in X$ does lie in $\mathrm{Bs} (E)$, then the image of the map $H^0(X, E)
\to E(x)$ is contained in some hyperplane $H \subset E(x)$, where the  hyperplane $H$
corresponds to a point $y \in \pi^{-1} (x)$  contained in $\mathrm{Bs}(\sO_{\mathbb{P}(E)}(1))$.

It is not clear whether the inclusion $\pi(\mathbb{B} (\sO_{\mathbb{P}(E)}(1)))
\subseteq \mathbb{B} (E) $ of  stable loci is strict in general.  However, as we
will show right below, some useful connections  rely on properties
of  augmented and restricted base loci, which exhibit  a more predictable
behavior with respect to the map $\pi$.

\begin{proposition}
\label{bmin}

Let $E$ be a vector bundle on a smooth projective variety $X$,
$\pi \colon \mathbb{P}(E) \to X$ the projective bundle of one dimensional quotients of $E$,
and $\sO_{\mathbb{P}(E)}(1)$ the universal quotient of $\pi^* E$ on $\mathbb{P}(E)$.
Then
\[
\pi(\mathbb{B}_- (\sO_{\mathbb{P}(E)}(1))) =  \mathbb{B}_- (E) ~.
\]

\end{proposition}

\begin{proof}

  Let us fix $H \in \mathrm{Pic} (X)$, a sufficiently ample line bundle such that
  $A:= \sO_{\mathbb{P}(E)}(1) \otimes \pi^* H$ is very ample on $\mathbb{P}(E)$.
  Then
  \[
  \mathbb{B}_- (\sO_{\mathbb{P}(E)}(1)) = \bigcup_{a \in \mathbb{N}}
  \mathbb{B}(\sO_{\mathbb{P}(E)}(a) \otimes A)= \bigcup_{a \in \mathbb{N}}
  \Big(\bigcap_{b\in \mathbb{N}} \mathrm{Bs}(\sO_{\mathbb{P}(E)}(ab) \otimes A^{b})
  \Big)
  \]
  and
  \[
  \mathbb{B}_- (E) = \bigcup_{a\in \mathbb{N}} \Big(\bigcap_{b\in \mathbb{N}}
  \mathrm{Bs} (\Sym^{ab} E \otimes H^{b}) \Big).
  \]

  $(\subseteq)$ The easier inclusion is \( \pi(\mathbb{B}_- (\sO_{\mathbb{P}(E)}(1)))
  \subseteq \mathbb{B}_- (E).  \) In order to show this, suppose that $x \in X$ and
  that $x \notin \mathbb{B}_- (E)$.  Then for any integer $a >0$ there exists a $b>0$
  such that the vector bundle $\Sym^{ab} E \otimes H^b$ is generated by its global
  sections at $x \in X$.  Then for all $a > 0$ the line bundle $\sO(2(a-1)b) \otimes
  A^{2b} = \sO (2ab) \otimes \pi^* H^{2b}$ which is a quotient of $\pi^* (\Sym^{2ab}
  E \otimes H^{2b})$ is generated by its global sections (defined over the whole
  space $\mathbb{P} (E)$) on any point of the fibre $\pi^{-1} (x)$, so the fibre
  $\pi^{-1} (x)$ is contained in the complement of $\mathbb{B}_-
  (\sO_{\mathbb{P}(E)}(1))$.

  $(\supseteq )$ Let us show now that \( \pi(\mathbb{B}_- (\sO_{\mathbb{P}(E)}(1)))
  \supseteq \mathbb{B}_- (E) \): Let $x \in \mathbb{P}(E)$ be a point such that $x
  \notin \pi(\mathbb{B}_- (\sO_{\mathbb{P}(E)}(1))) $.  Then for any $a>0$ there
  exists a $b>0$ such that $\sO_{\mathbb{P}(E)}(2(a-1)b) \otimes A^{b}$ is generated on
  any point $y \in \pi^{-1} (x)$ by its global sections (defined on the whole
  $\mathbb{P}(E)$).

  Then the line bundle
  \begin{align*}
    L := \sO_{\mathbb{P}(E)} (2ab) \otimes \pi^* H^{2b} &\simeq \sO_{\mathbb{P}(E)} (2(a-1)b)
  \otimes \sO_{\mathbb{P}(E)} (b) \otimes \pi^* H^{b} \otimes \sO_{\mathbb{P}(E)} (b)
  \otimes \pi^* H^{b} \simeq\\
  &\simeq \Big (\sO_{\mathbb{P}(E)}(2(a-1)b) \otimes A^{b} \Big ) \otimes A^{b}
  \end{align*}
  is the product of a line bundle which is generated by global sections (on
  $\mathbb{P}(E)$) at any point of the fiber $\mathbb P_x := \pi^{-1} (x) = \mathbb{P}(E(x))$
  with a very ample line bundle, so its global sections (on $\mathbb{P}(E)$) define a
  closed immersion of $\mathbb P_x$ into a projective space.

    In other words, the linear system $H^0(\mathbb{P}(E), L)$ defines a rational
    map \hbox{ $\varphi \colon \mathbb{P}(E) \dashrightarrow
      \mathbb{P}(H^0(\mathbb{P(E)}, L)) =\mathbb{P}^N$} which is a regular immersion
    on $\mathbb P_x$.  Then in particular, for $m \gg 0$ the multiplication map $\Sym^m
    H^0(\mathbb{P(E)}, L) \to H^0(Y , L|_{Y})$ is surjective.

  It follows that the map $\pi_* (\Sym^m H^0(\mathbb{P}(E), L) \otimes
  \sO_{\mathbb{P}(E)}) \to \pi_*(L^{m})$ is surjective at the point $x \in X$.  As
  $\pi_*(L^{m}) = \Sym^{2abm} E \otimes H^{2bm}$ we may conclude that for any $a>0$
  and $m$ large enough the vector bundle $\Sym^{2abm} E \otimes H^{2bm}$ is generated
  at $x$ by its global sections, hence $x \notin \mathbb{B}_- (E)$.
\end{proof}

The analogous claim  holds for  augmented base locus, with a similar proof.

\begin{proposition}
  \label{bplus}

  Let $E$ be a vector bundle on a smooth projective variety $X$, with the same
  notations as in Proposition \ref{bmin}, we have
  \[
  \pi(\mathbb{B}_+ (\sO_{\mathbb{P}(E)}(1))) =  \mathbb{B}_+ (E) ~.
  \]

\end{proposition}

\begin{proof}

  Let $H \in \mathrm{Pic} (X)$ be a sufficiently ample line bundle
  such that $A:= \sO_{\mathbb{P}(E)}(1) \otimes \pi^* H$ is very ample on
  $\mathbb{P}(E)$.  Then
  \[
  \mathbb{B}_+ (\sO_{\mathbb{P}(E)}(1)) = \bigcap_{a>0}
  \mathbb{B}(\sO_{\mathbb{P}(E)}(a) \otimes A^{-1})= \bigcap_{a>0} \Big(\bigcap_{b>0}
  \mathrm{Bs}(\sO_{\mathbb{P}(E)}(ab) \otimes A^{-b}) \Big)
  \]
  and
  \[
  \mathbb{B}_+ (E) = \bigcap_{a>0} \Big(\bigcap_{b>0} \mathrm{Bs} (\Sym^{ab} E
  \otimes H^{-b}) \Big).
  \]

  In order to show that \( \pi(\mathbb{B}_+ (\sO_{\mathbb{P}(E)}(1))) \subseteq
  \mathbb{B}_+ (E), \) observe that if $\Sym^{ab}E \otimes H^{-b}$ is globally
  generated at a point $x \in X$, then $\pi^* \Sym^{ab}E \otimes \pi^*H^{-b}$ is
  generated at all points in $\pi^{-1}(x)$, hence $\sO_{\mathbb{P}(E)} (ab) \otimes
  \pi^*H^{-b} = \sO_{\mathbb{P}(E)} ((a+1)b) \otimes A^{-b} $ is globally generated
  at all points in $\pi^{-1}(x)$.

  To show the other inclusion, set $U = X \setminus \pi(\mathbb{B}_+
  (\sO_{\mathbb{P}(E)}(1)))$ and observe that $(\sO_{\mathbb{P}(E)} (ab) \otimes
  A^{-b})$ is generated by its global sections at the points of $\pi^{-1} (U)$ for
  $a$ and $b$ sufficiently large.  Let us consider $b>0$ a sufficiently large
  positive integer, $a = (b-1)k >0$ a sufficiently large multiple of $b-1$, and set
  $c:= ((a-1)b +1)/(b-1) = kb-1 = a + k-1$.  Finally, let $L$ be the line bundle
  \begin{multline*}
    L:= \sO_{\mathbb{P}(E)} (c (b-1)) \otimes \pi^*H^{-(b-1)} = \sO_{\mathbb{P}(E)}
    ((a-1)b +1) \otimes \pi^*H^{-(b-1)}\simeq \\
    \simeq \Big( \sO_{\mathbb{P}(E)} (ab) \otimes \big( \sO_{\mathbb{P}(E)} (-1)
    \otimes \pi^*H^{-1} \big)^{b} \Big) \otimes (\sO_{\mathbb{P}(E)} (1) \otimes
    \pi^*H) \simeq (\sO_{\mathbb{P}(E)} (ab) \otimes A^{-b}) \otimes A.
  \end{multline*}

  Now for $b$ and $k$ large enough $L$ is the product of the very ample line bundle
  $A$ with a line bundle which is generated by global sections on $\pi^{-1} (U)$, so
  it is very ample on the open subset $\pi^{-1} (U)$. Furthermore, we have that
  $\pi_* (L) = \Sym^{c(b-1)} E \otimes H^{-(b-1)}$ and so we can apply the same
  argument as in the proof of Proposition~\ref{bmin} to finish the proof.
\end{proof}

\section{Positivity properties for line bundles}
\label{line}

We recall here how  augmented and restricted base loci are involved with various
positivity notions of line bundles, as well as  loci defined by negative curves:

\begin{de}
  Let $L$ be a line bundle on a smooth projective variety $X$.  Fix an ample line
  bundle $A$ and a rational number $\eps>0$. We define

  \begin{enumerate}
  \item ${T_\eps^A=\overline{\set{ x \with \exists\, C\subseteq X \textrm{ curve on $X$
            s.t. } x \in C, c_1(L)\cdot C<\eps\cdot c_1(A)\cdot C}}}$
    to be the  \emph{non-AEA locus of $L$ with respect to $A$ and $\eps$};

    \silentremark{\sandor{060914}: I added ``$\exists\, C\subseteq X$'' here, because
      it seemed missing.\\ I also added here and elsewhere ``$c_1(\ )$'' to change a
      line bundle into a divisor...}

  \item  $\displaystyle{\mathbb{T}(L):=\bigcap_{\eps>0} T_\eps^A}$ to be the \emph{stable non-AEA locus of $L$}, and

  \item  $ ~ \mathbb{T}^0(L):=\set{x \with x \in C \textrm{ such that  }  L\cdot C < 0}$ the \emph{negative locus of $L$}.
  \end{enumerate}
\end{de}

\smallskip

\silentremark{\vskip-48pt\sandor{060914}: When labeling an ``item'', please use
  ``ilabel'' and refer to it with ``iref''. That will add the theorem number as well.
  (See sk-preamble.tex for definition).}

\begin{propdef}
\label{defline}
For  a line bundle $L$ on the  variety $X$ we have the following.
\begin{enumerate}

\item
\ilabel{nonnef-restricted}
$ \mathbb{T}^0(L) \subseteq \mathbb{B}_- (L)$, the inequality can be strict (cf.
\cite[Remark 6.3]{BDPP}).

\item
\ilabel{nonample-augmented}
 $ \mathbb{T}(L) \subseteq \mathbb{B}_+ (L)$.

\item $L$ is \emph{ample} iff $\mathbb{B}_+ (L) = \emptyset$.

\item $L$ is \emph{semiample} iff $\mathbb{B}(L) = \emptyset$.

\item $L$ is \emph{nef} iff $\mathbb{B}_- (L) = \emptyset$ iff $ \mathbb{T}^0(L) =
  \emptyset$.

\item
\ilabel{big}
$L$ is \emph{big} iff $\mathbb{B}_+ (L) \neq X$.

\item
\ilabel{psef}
$L$ is \emph{psef} (pseudo-effective) iff $\mathbb{B}_- (L) \neq X$.

\item
\ilabel{anef}
$L$ is \emph{almost nef} iff  $\mathbb{T}^0(L)$  is  contained
in a countable union of proper closed subsets of $X$.

 \item
   \ilabel{aea}
   $L$ is \emph{AEA} (almost everywhere ample) if $ \mathbb{T} (L) \neq X$.

\item
\ilabel{wpos} $L$ is \emph{weakly positive} if $\overline{\mathbb{B}_- (L)} \neq X$.

\end{enumerate}

\end{propdef}

\begin{proof}

  Points (\ref{defline}.\ref{nonnef-restricted}-\ref{big}) are well-known statements.
  The claims \iref{anef}, \iref{aea} and \iref{wpos} are the definitions of respective notions
  according to \cite{BDPP}, \cite{Miy83} and \cite{vie}, respectively.  The only
  statement in need of a proof is \iref{psef}.

  Note that a line bundle is pseudo-effective precisely if  its numerical equivalence class lies in the closure of the
  effective cone in the real N\'eron-Severi group. Hence the line bundle  $L$ is psef if and only if  $\forall m>0
  ~ L+ (1/m) A$ is effective, or, equivalently, if  $\forall m>0 ~\mathbb{B} (mL + A) \neq X$. Therefore
  $\mathbb{B}_-(L) \neq X$ as it is contained in a countable union of proper closed
  subsets of $X$.  Conversely,  if $\mathbb{B}_-(L) \neq X$, then the class of $L$ is a
  limit of effective classes.
\end{proof}

\begin{remark}

  Positivity properties related to asymptotic base loci are best  summarized in the form of a table.

  \begin{center}

    \begin{tabular}{|c|c|c|c|c|}

      \hline

      & $\mathbb{B}_- (L)$ & ${\overline{\mathbb{B}_- (L)}  }$  & $ \mathbb{B} (L) $
      & $\mathbb{B}_+ (L)$ \vphantom{$\overline{B}^{\text{\normalsize B}}$}\\
      \hline
      $=\emptyset $& nef & nef &  semiample & ample \\
      \hline
      $\neq X $ & pseudo-effective & weakly positive & effective &  big \\

      \hline

    \end{tabular}

  \end{center}

\medskip

\end{remark}

\begin{remark}

  See Section~\ref{badaea} and in particular Remark \ref{remark-augm} for more
  details about non-AEA loci and their relationship with the augmented base loci.

  Furthermore, as we mentioned earlier,  Lesieutre \cite{lesieutre} proved
  that there exist line bundles which are pseudo-effective but not weakly positive. In
  particular, $\mathbb{B}_- (L)$ is not necessarily closed.
\end{remark}

\begin{proposition}

\label{anef-psef}
A line bundle $L$ is almost nef if and only if it is pseudo-effective.

\end{proposition}

\begin{proof}
  One implication is obvious by \iref{nonnef-restricted}.  The other implication
  follows from \cite{BDPP}, as if $L$ is not pseudo-effective then there exists a
  reduced irreducible curve $C \subseteq X$, such that $c_1(L)\cdot C <0$ and $C$
  moves in a family covering all $X$, so $ \mathbb{T}^0(L)$ cannot be contained in a
  countable union of proper (Zariski) closed subsets.
\end{proof}

The following theorem will be proved in Section~\ref{sectionaea}:

\begin{theorem}

A line bundle $L$ is big if and only if it is AEA.

\end{theorem}

\silentremark{\sandor{060914}: I changed ``AEN'' to ``AEA'' assuming that it was a
  typo.}

A recent result of Lehmann \cite{brian} gives a characterization of the relationship
between the non-AEA locus and the diminished base locus. We will use the following
when describing all the relationships.

\begin{de} Let $X$ be a smooth projective variety over $\mathbb{C}$ and let $D$ be a
  pseudo-effective $\R$-divisor on $X$. Suppose that $\phi : Y \to X$ is a proper
  birational map from a smooth variety $Y$ . The movable transform of $L$ on $Y$ is
  defined to be
  $$
  \phi_{\mov}^{-1}(D) := \phi^*D - \sum_{E ~ \phi-exc} \sigma_E(\phi^*D)\cdot E.
  $$
  For a pseudo-effective line bundle $L$  define $\phi_{\mov}^{-1}(L)$ as the line
  bundle associated to $\phi_{\mov}^{-1}(c_1(L))$ on $Y$.
\end{de}

Note that the movable transform is not linear and is only defined for
pseudo-effective divisors.

\begin{remark}
  In the above, $\sigma_E$ is the asymptotic multiplicity function introduced by
  Nakayama \cite[Section III.1]{Nak04}.
  If $X$ is a smooth projective variety, $L$ a pseudo-effective $\R$-divisor, $E$ a
  prime divisor on $X$, then
  \[
  \sigma_E(L) := \lim_{\epsilon\to 0^+} \inf \{ \mult_E{L'}\mid L' \geq 0 \text{ and
  } L'\sim_{\R} L+\epsilon A\}\ ,
  \]
  where $A$ is an arbitrary but fixed ample divisor.
\end{remark}

\begin{remark}
  Following \cite[Definition 1.2]{brian}, we call an irreducible curve $C$ on $X$ to
  be a $\mov^1$-curve, if it deforms to cover a codimension one subset of $X$.
\end{remark}

\begin{thm}[\cite{brian}] Let $X$ be a smooth projective variety over $\C$ and $D$ a
  pseudo-effective $\R$-divisor. $D$ is not movable if and only if there is a
  $\mov^1$- curve $C$ on $X$ and a proper birational morphism $\phi: Y \to X$ from a
  smooth variety $Y$ such that $$\phi_{\mov}^{-1}(D) \cdot \bar{C} < 0,$$ where
  $\bar{C}$ is the strict transform of a generic deformation of $C$.
\end{thm}

The following reformulation is easy to see.

\begin{thm}
  Let $X$ be a smooth projective variety and $D$ a pseudo-effective $\R$-divisor.
  Suppose that $V$ is an irreducible subvariety of $X$ contained in $\mathbb{B}_-(L)$
  and let $\psi : X' \to X$ be a smooth
  birational model resolving the ideal sheaf of $V$.  Then there is a birational
  morphism $\phi: Y \to X'$ from a smooth variety $Y$ and an irreducible curve
  $\bar{C}$ on $Y$ such that $\phi^{-1}_{\mov}(\psi^*D)\cdot \bar{C} < 0$ and $\psi
  \circ \varphi (\bar{C})$ deforms to cover $V$ .

\end{thm}

\begin{remark} Let $L$ be any line bundle on $X$ smooth projective, then
$$\mathbb{B}_-(L)= \bigcup_{f: Y \to X} f(\mathbb{T}^0(f_{\mov}^{-1}(L))).$$

\end{remark}

There are several examples for which the loci $\mathbb{T}^0(L)$ and $\mathbb{B}_-(L)$
do not coincide, and in some cases the difference is divisorial.

\begin{qsn}
Is it true that $\mathbb{B}_-(L)$ is contained in a proper closed subset of $X$
(i.e., $L$ is weakly positive) if and only if the same holds for $\mathbb{T}^0(L)$?
\end{qsn}

\begin{remark}
In \cite[Question 7.5]{BDPP} the authors ask if for a vector bundle $E$, $\mathbb{B}_-(\sO_{\mathbb{P}(E)}(1))$
doesn't dominate $X$ if and only if neither does $\mathbb{T}^0(\sO_{\mathbb{P}(E)}(1))$.
\end{remark}

\section{Restricted base loci for vector bundles}
\label{restricted}

Here we explore the connections between  the positivity properties of a vector bundle
and  the associated asymptotic base loci.  We will start recalling some classical
definitions for vector bundles. Note that these definitions do sometimes appear
slightly differently in the literature, but we will try to follow and indicate specific
selected references each time.

\begin{definition}

  Let $E$ be a vector bundle on a smooth projective variety $X$, $\pi \colon
  \mathbb{P}(E) \to X$ the projective bundle of one dimensional quotients of $E$, and
  $\sO_{\mathbb{P}(E)}(1)$ the universal quotient of $\pi^* E$ on $\mathbb{P}(E)$.
  We say that $E$ is

  \begin{enumerate}

  \item \emph{nef} if $\sO_{\mathbb{P}(E)}(1))$ is a nef line bundle, {i.e.,} if
    $\mathbb{B}_- (\sO_{\mathbb{P}(E)}(1)) = \emptyset$;

  \item \emph{almost nef} if $\pi (\mathbb{T}^0(\sO_{\mathbb{P}(E)}(1)))$ is
    contained in a countable union of proper closed subsets of $X$ (cf. \cite{BDPP});

  \item \emph{pseudo-effective} if $\mathbb{B}_- (E) \neq X$ (cf. \cite{BDPP});

  \item \emph{weakly positive} if $\overline{\mathbb{B}_- (E)} \neq X$ (cf.
    \cite{vie}); \ilabel{wp}

  \end{enumerate}

\end{definition}

\begin{proposition}

A vector bundle $E$ is nef if and only if $\mathbb{B}_- (E)= \emptyset$.

\end{proposition}

\begin{proof}

This follows directly from Proposition~\ref{bmin}.
\end{proof}

\begin{proposition}

  A vector bundle $E$ is pseudo-effective if and only if $\sO_{\mathbb{P}(E)}(1)$ is
  pseudo-effective and $\pi (\mathbb{B}_- (\sO_{\mathbb{P}(E)}(1))) \neq X$.

\end{proposition}

\begin{proof}
  This follows immediately from Proposition~\ref{bmin}.
\end{proof}

\begin{remark}

  The proposition above is the same as \cite[Proposition 7.2]{BDPP}. Observe that the
  locus $L_{\nonnef}$ in \cite{BDPP} is what we call $\mathbb{B}_- (L)$ here (we are
  in the smooth projective case).

\end{remark}

The following proposition is immediate.

\begin{proposition}

  A vector bundle $E$ is almost nef if and only if there exists a countable union $T
  = \bigcup_{i \in \mathbb{N}} T_i$ of proper closed subsets of $X$, such that for
  any curve $C \subseteq X$ not contained in $T$ the restriction $E|_{C}$ is a nef
  vector bundle.

\end{proposition}

\begin{remark}

  It follows from the definitions and propositions above that
  \[
  E ~\textrm{weakly positive} \Rightarrow E ~\textrm{pseudo-effective} \Rightarrow E
  ~\textrm{almost nef} ~.
  \]
  We have seen that the first implication is not an equivalence, while it is an open
  question whether, for vector bundles, being almost nef is equivalent to being
  pseudo-effective, as in the line bundle case cf.\ \cite[Question~7.5]{BDPP}.

\end{remark}

\begin{question}
Does $E$ being {almost nef} imply that $E$ is {pseudo-effective}?
\end{question}

If $E$ is almost nef, then the line bundle $\sO_{\mathbb{P}(E)}(1)$ is almost nef,
hence pseudo-effective.  In order to have that $E$ is pseudo-effective, one needs to
show that $\mathbb{B}_- (\sO_{\mathbb{P}(E)}(1))$ does not dominate $X$.

\section{Augmented base loci for vector bundles}
\label{augmented}

\begin{definition}

  Let $E$ be a vector bundle on a smooth projective variety $X$, $\pi \colon
  \mathbb{P}(E) \to X$ the projective bundle of one dimensional quotients of $E$, and
  $\sO_{\mathbb{P}(E)}(1)$ the universal quotient of $\pi^* E$ on $\mathbb{P}(E)$.

\begin{enumerate} We say that

\item $E$ is \emph{ample} if $\sO_{\mathbb{P}(E)}(1)$ is ample on $\mathbb{P}(E)$;

\item $E$ is \emph{L-big} if $\sO_{\mathbb{P}(E)}(1)$ is big on $\mathbb{P}(E)$; and

\item $E$ is \emph{V-big} (or \emph{Viehweg-big}) if there exists an ample line
  bundle $A$ and a positive integer $c>0$ such that $\Sym^c E \otimes A^{-1}$ is
  weakly positive, {i.e.,} such that $\overline{\mathbb{B}_- (\Sym^q E \otimes
    A^{-1})} \subsetneq X$ (cf.\ \iref{wp}).

\end{enumerate}

\end{definition}

\begin{proposition}
  A vector bundle $E$ is ample if and only if $\mathbb{B}_+ (E)= \emptyset$.
\end{proposition}

\begin{proof}
Follows directly from Proposition~\ref{bplus}.
\end{proof}

\begin{remark}

  It is well-known that a line bundle $L$ is big if and only if $\mathbb{B}_+ (L)
  \neq X$, equivalently,  if there exist $A$ ample and a positive integer $c>0$ such that
  $L^{\otimes c} \otimes A^{-1}$ is pseudo-effective.  Next we will prove that the
  same equivalences hold for V-bigness for vector bundles of arbitrary rank.

\end{remark}

\begin{theorem}
  \label{thm:V-big}

  Let $E$ be a vector bundle on a smooth projective variety $X$.  Then the following
  properties are equivalent:

\begin{enumerate}

\item $E$ is V-big.
\ilabel{one}

\item There exist an ample line bundle $A$ and a positive integer $c>0$ such that
$\Sym^c E \otimes A^{-1}$ is pseudo-effective.
\ilabel{two}

\item $\mathbb{B}_+ (E) \neq X$.
\ilabel{three}

\end{enumerate}

\end{theorem}

\begin{proof}

  The implication $\iref{one} \Rightarrow \iref{two}$ is clear; let us consider
  $\iref{two} \Rightarrow \iref{three}$.  Suppose there exist $A$ ample and $c>0$
  such that $\Sym^c E \otimes A^{-1}$ is pseudo-effective, {i.e.,}
 \[
 \mathbb{B}_-(\Sym^c E \otimes A^{-1}) \neq X \ .
 \]
 Then
\[
\bigcup_{q>0} \mathbb{B} (E -(1/c)A + (1/q)A) =
\bigcup_{q>0} \mathbb{B} (E - \frac{q-c}{qc}A) \neq X ~.
\]
Now for $q\gg0$ and $\frac{q-c}{qc} > \frac{1}{2c}$, one has
\[
\mathbb{B}_+ (E) \,\subseteq\, \mathbb{B} (E - \frac{1}{2c}A)
\,\subseteq\, \mathbb{B} (E - \frac{q-c}{qc}A) \,\subsetneq\, X  \ ,
\]
hence the validity of the desired implication.

Next we verify that $\iref{three} \Rightarrow \iref{one}$.
If $E$ satisfies $\mathbb{B}_+ (E) \neq X$,
then there exists $q>0$ such that $\mathbb{B} (E - \frac{1}{q}A) \subsetneq X$ is a closed proper subset.
Consequently,

\[
\overline{\mathbb{B}_- (\Sym^q E \otimes A^{-1})} \,\subseteq\, \mathbb{B} (\Sym^q E
\otimes A^{-1}) \,\subsetneq\, X \qedhere
\]
\end{proof}

\begin{corollary}
  If $E$ is V-big, then it is  L-big as well.
\end{corollary}

\begin{proof}
  Theorem~\ref{thm:V-big} yields $\mathbb{B}_+ (E) \neq X$, therefore
  $\mathbb{B}_+ (\sO_{\mathbb P(E)}(1)) \neq \mathbb P(E)$ via Proposition~\ref{bplus}. As a consequence
  $\sO_{\mathbb P(E)}(1)$ is big on $\mathbb P(E)$, equivalently,  $E$ is L-big.
\end{proof}

\begin{remark}\label{L-big-vs-V-big}
  L-big vector bundles are not necessarily V-big, as the example of
  $\sO_{\P^1}\oplus\sO_{\P^1}(1)$ on $\P^1$ shows (see \cite[p.24]{KellyThesis}).

  The key difference between V-big and L-big vector bundles is that being L-big means
  that $\sO_{\P(\she)}(1)$ is ample with respect to an open set $V\subseteq \P$
  whereas $\she$ is V-big if we can take $V$ to be of the form $V=\pi^{-1}(U)$,
  $U\subseteq X$ open.
\end{remark}

\begin{remark}

  A vector bundle $E$ on a variety $X$ satisfying $\mathbb{B}_+ (E) \neq X$ is also
  called \emph{ample with respect to an open subset} (cf. \cite[Chapter
  3]{KellyThesis}).

\sideremark{\sandor{060914}: this should be properly defined. I'll do this later.}

\end{remark}

\begin{remark}
\label{diverio}

In the case where $E = \Omega_X$ is the cotangent sheaf of a variety $X$, the
definitions and Proposition~\ref{bplus} imply the following inclusion $\mathbb{B}_+
(\Omega_X) \supseteq DS(X, T_X)$, where $DS(X, T_X)$ is the Demailly-Semple locus.

The work of Diverio and Rousseau \cite{divrous} therefore provides examples of
complex projective varieties of general type $X$ where $\Omega_X$ is a semistable
L-big vector bundle with a big determinant, which is nevertheless not V-big.

\end{remark}

\section{Almost Everywhere Ampleness}
\label{sectionaea}

The notion of almost everywhere ampleness was first defined by Miyaoka in the context of his work on vector bundles on surfaces; the definition
goes through in all dimensions verbatim.

\begin{definition}[\cite{Miy83}]
  Let $X$ be a smooth projective variety,  $E$ a rank $r$ vector bundle on $X$.
  Consider the projectivized bundle $\P=\P(E)$ with projection morphism $\pi:\P\to X$ and
  tautological bundle $\sO_\P(1)$. We say that $E$ is \emph{almost everywhere ample}
  (AEA for short), if there exists an ample line bundle $A$ on $X$, a Zariski closed
  subset $T\subset \P$, whose projection $\pi(T)$ onto $X$ satisfies $\pi(T)\neq X$,
  and a positive number $\eps>0$ such that
   $$
   c_1(\sO_\P(1))\cdot C\ge \eps\cdot \pi^*(c_1(A))\cdot C
   $$
   for all curves $C\subset \P$ that are not contained in $T$.
\end{definition}

\begin{comment}
  It is possible that the scale-invariant version will turn out to be a more useful
  definition: $E$ is AEA if the condition in the definition is fulfilled for
  $\Sym^mE$ for some $m>0$.  This changes nothing about our results for line bundles,
  but might give a better Kodaira lemma.
\end{comment}

For line bundles, this notion coincides with bigness:

\begin{proposition}\label{proposition-AEA-big}
   For a line bundle $L$ on a smooth projective variety $X$, the
   following are equivalent:
   \begin{enumerate}
   \item $L$ is AEA, i.e, there is an ample line bundle $A$ on $X$, a number
     $\eps>0$, and a proper Zariski closed subset $T\subset \P$ such that
     $$
     c_1(L)\cdot C\ge \eps\cdot c_1(A)\cdot C
     $$
     for all curves $C\subset \P$ not contained in $T$.
     \ilabel{onec}
   \item For every ample line bundle $A$ on $X$, there is an $\eps>0$ and a proper
     Zariski closed subset $T\subset X$ such that
     $$
     c_1(L)\cdot C\ge \eps\cdot c_1(A)\cdot C
     $$
     for all curves $C\subset X$ not contained in $T$.
     \ilabel{twoc}
   \item
     $L$ is big.
     \ilabel{threec}
   \end{enumerate}
\end{proposition}

\begin{proof}
  Assume \iref{threec}, and let $A$ be any ample line bundle.  Then, by Kodaira's lemma,
  there is a positive integer $m$ such that we can write
   $$
   mc_1(L) = c_1(A)+F ,
   $$
   where $F$ is an effective divisor.  Taking $T$ to be the support of $F$, it
   follows for every curve $C\subset X$ not contained in $T$ that
   $$
   mc_1(L)\cdot C=c_1(A)\cdot C+F\cdot C\ge c_1(A)\cdot C
   $$
   and this implies \ref{twoc} with $\eps:=1/m$.

   Obviously \ref{twoc} implies \iref{onec}, so let us assume condition \iref{onec}
   and show that it implies \iref{threec}.  A curve $C\subset X$ such that $c_1(L)\cdot
   C<\eps\cdot c_1(A)\cdot C$ cannot be a movable curve (in the sense of
   \cite[Sect.~11.4.C]{PAG}), since these cover all of $X$ (by \cite[Lemma
   11.4.18]{PAG}), whereas $T\ne X$.  So $L$ must have positive intersection with all
   movable curves.  This implies that $L$ lies in the dual of the cone of movable
   curves $\Mov(X)$, which by the theorem of Boucksom-Demailly-Paun-Peternell
   \cite{BDPP} is the pseudo-effective cone $\overline{\Eff}(X)$.  In order to
   conclude that $L$ is big -- and thus to complete the proof -- it is therefore
   enough to show that $L$ lies in the interior of that cone.

   The assumption that $L$ be AEA says that
   $$
   (c_1(L)-\eps c_1(A))C\ge 0 \qquad\mbox{for all } C\not\subset T \ .
   $$
   Therefore, writing $c_1(L)-\eps c_1(A)=(c_1(L)-\frac\eps2 c_1(A))-\frac\eps2
   c_1(A)$, we see that $c_1(L)-\frac\eps2 c_1(A)$ is AEA as well. Moreover, every
   class in the open set
   $$
   c_1(L)-\frac\eps2 c_1(A)+\Amp(X)
   $$
   is AEA, and $c_1(L)$ lies in this open set.
\end{proof}

\begin{proposition}\label{proposition-AEA}
  Let $E$ be a vector bundle on a smooth projective variety $X$, let $\P=\P(E)$. If
  $E$ is AEA on $X$, then so is $\sO_\P(1)$ on $\P$.
\end{proposition}
\begin{proof}
  For $E$ to be AEA means that for every ample line bundle $A$ on $X$, there exists a
  Zariski-closed subset $T\subseteq X$, and $\eps >0$ such that
  \[
  c_1(\sO_\P(1))\cdot C \geq \eps (\pi^*c_1(A)\cdot C)
  \]
  for all irreducible curves not contained in $T$.

  Since $\sO_\P(1)$ is $\pi$-ample, the line bundle $\pi^*A\otimes \sO_\P(m)$ is
  ample for all $m\geq m_0\gg 0$ by \cite[Proposition 1.7.10]{PAG}.  According to
  Proposition~\ref{proposition-AEA-big}, $\sO_\P(1)$ is AEA if and only if it is big,
  therefore it suffices to prove the AEA property for $\sO_\P(k)$ for some large $k$.
  This means in particular, that we are allowed to work with $\Q$-divisors as well.

  Let $m_0$ be, as above, a positive integer such that $\pi^*A\otimes\sO_\P(m_0)$ is
  ample.  We will prove that $\sO_\P(1)$ is AEA on $\P$ with closed subset
  $T\nsubseteq \P$, and a suitable $\eps'>0$.  We need that
  \[
  c_1(\sO_\P(1))\cdot C \geq \eps'(c_1(\pi^*A\otimes\sO_\P(m_0))\cdot C) \ ,
  \]
  or equivalently,
  \[
  c_1(\sO_\P(1))\cdot C \geq \frac{\eps'}{1-\eps'm_0}(\pi^*c_1(A)\cdot C)
  \]
  for all curves not contained in $T$. By our assumption on $E$, this holds whenever
  \[
  \eps' < \frac{\eps}{1+\eps m_0}\ . \qedhere
  \]
\end{proof}

\begin{corollary}
  If $E$ is AEA, then it is L-big.
\end{corollary}
\begin{proof}
  Immediate from Proposition~\ref{proposition-AEA-big} and
  Proposition~\ref{proposition-AEA}.
\end{proof}

\section{The bad AEA locus in the line bundle case}
\label{badaea}

Consider a line bundle $L$, and fix an ample line bundle $A$ and a number $\eps>0$.
We defined the \textit{non-AEA locus} of $L$ with respect to $A$ and $\eps$ as the
subvariety
$$
T_\eps^A=\mbox{closure}(\bigcup\set{C\with C\mbox{ curve on $X$ with $c_1(L)\cdot
    C<\eps\cdot c_1(A)\cdot C$}})\ .
$$
The AEA assumption on $L$ simply  means that there exists an $\eps>0$ such that
$T^A_\eps\ne X$.  For $\eps<\delta$ we have $T_\eps\subset T_\delta$, so that we can
 express the AEA condition equivalently as saying that the intersection
${\mathbb{T}}(L):=\bigcap_{\eps>0} T_\eps^A$ is not all of $X$.

\begin{remark}\label{remark-augm}
   It is immediate that
   $$
       \mathbb{T}(L)\subseteq \mathbb{B}_+(L) \ .
   $$
   In fact, by the noetherian property there are positive real numbers $\eps_0$ and
   $\delta_0$ such that

   \[
   \mathbb{T}(L) = T_{\eps}^A ~~\forall \eps \leqslant \eps_0 \textrm{ and }
   \mathbb{B}_+ (L) = \mathbb{B}(L - \delta A) ~~\forall \delta \leqslant \delta_0.
   \]

   Now choose $\eps < \min (\eps_0 , \delta_0)$, then
   \[
   \mathbb{T}(L) = T_{\eps} = \overline{\set{ x \with x \in C \mbox{ curve on $X$
         s.t. } c_1(L)\cdot C<\eps\cdot c_1(A)\cdot C}}.
   \]
   If $C$ is a curve such that $(L - \eps A)C <0$ then $C \subseteq \mathbb{B}(L -
   \eps A) = \mathbb{B}_+ (L)$ which is a closed set.

\end{remark}

\begin{remark}

  A line bundle $L$ is ample if and only if $ \mathbb{T}(L) = \emptyset$. In general
  the inclusion
  \silentremark{\sandor{060914}: should we mention Seshadri constants here?}
  \[
  \overline{\set{ x \with x \in C \mbox{ curve on $X$ s.t. } c_1(L)\cdot C \leqslant
      0}} \subseteq \mathbb{T}(L)
  \]
  is strict, as shown by a strictly nef non ample line bundle $L$, where the first
  set is empty but the second one is not.  Examples of line bundles that are strictly
  nef (and even big) and non ample have been first given by Mumford (cf.
  \cite{har70}), and a complete description can be found in \cite{urbin}.

\end{remark}

\begin{remark} A few words on the relationship between $\mathbb{T}(L)$ and $\mathbb{B}_\pm(L)$.
  We'll show here that $\mathbb{T}(L)\neq B_+(L)$ in general. A bit more precisely,
  we will try to understand the relationship of $\mathbb{T}(L)$ to the augmented and
  restricted base loci of $L$ when $\dim X=2$. Recall that
\[
 \mathbb{B}_-(L) = \bigcup_{m=1}^{\infty} \mathbb{B} (L+\frac{1}{m}A)
\]
for any integral ample divisor $A$ on $X$.

Let $D$ be a big divisor on a smooth projective surface $X$ with Zariski
decomposition $D=P_D+N_D$. Then \cite[Examples 1.11 and 1.17]{ELMNP} tell us that
\begin{eqnarray*}
  \mathbb{B}_+(\sO_X(D)) & = & \Null(P_D) \equ \bigcup_{C}\st{C\subseteq X \text{ irred
    }\,|\, P_D\cdot C=0}\ , \\
  \mathbb{B}_-(\sO_X(D)) & = & \Supp N_D\ .
\end{eqnarray*}
\end{remark}

\begin{example} Here we present an  example where $T(L)\neq \mathbb{B}_{\pm}(L)$.
  Let $X$ be a surface that carries a big divisor $D$ and an irreducible curve
  $C\subseteq X$ satisfying $C\subseteq \Supp N_D$ and $D\cdot C>0$. Then
  $C\not\subseteq T(L)$, but $C\subseteq \mathbb{B}_-(\sO_X(D))\subseteq
  \mathbb{B}_+(\sO_X(D))$.  In this case we have
   \[
   T(\sO_X(D)) \neq \mathbb{B}_-(\sO_X(D)),\mathbb{B}_+(\sO_X(D)).
   \]
   Surfaces carrying such $D$ and $C$ exist by \cite{BauFun10}: Consider a K3 surface
   $X$, on which the Zariski chamber decomposition does not coincide with the Weyl
   chamber decomposition. The latter is by \cite[Theorem~1]{BauFun10} the case if and
   only if there are $(-2)$-curves on $X$ having intersection number 1. For a
   concrete example one can, as done in \cite[Section~3]{BauFun10}, take a smooth
   quartic surface $X\subset\mathbb P^3$ that has a hyperplane section of the form
   $H=L_1+L_2+Q$, where $L_1$ and $L_2$ are lines and $Q$ is a smooth conic. Then the
   divisors of the form
   $$
      D=H+a_1L_1+a_2L_2
   $$
   with $a_1\ge 1$ and $a_2\ge 1$ have $L_1$ and $L_2$ in the support of the negative
   part of their Zariski decomposition, but one can find $a_1,a_2$ such that $D\cdot
   L_1>0$ and $D\cdot L_2<0$ (for instance $a_1=2,a_2=4$). (In the notation of
   \cite{BauFun10}, $D$ lies in the Zariski chamber $Z_{\{L_1,L_2\}}$, but in the
   Weyl chamber $W_{\{L_2\}}$.)
\end{example}

\begin{example}
  In general $T(\sO_X(D))$ is not contained in $B_-(\sO_X(D))$ either, where $L=\sO_X(D)$ for a suitable integral Cartier  divisor $D$.
  To see this,
  take a surface where all negative curves have self-intersection $-1$. Then the
  intersection form of the negative part of the Zariski decomposition of any big
  divisor is $-\Id$, in other words, no two curves in it can intersect. Consequently,
  \[
  T(\sO_X(D)) \equ B_+(\sO_X(D))\ .
  \]
  This can be seen as follows: let $C\subseteq B_+(\sO_X(D))=\Null (P_D)$ be an
  irreducible curve. Since $P_D$ is big and nef, the intersection form on $P_D^\perp$
  is negative definite, which under the given circumstances means that $(C\cdot
  C')=0$ for every irreducible curve $C\neq C'$ coming up in $N_D$. Therefore
  \[
  D\cdot C \equ P_D\cdot C + N_D\cdot C \equ 0 + (\leq 0) \equ (\leq 0)\ .
  \]
  Consequently, $C\subseteq T(\sO_X(D))$.

  Take a non-stable (in the sense if \cite[Definition 1.22]{ELMNP} big divisor $D$ on
  $X$, then $B_-(\sO_X(D))\subsetneq B_+(\sO_X(D)) = T(\sO_X(D))$.
\end{example}

\medskip
The following lemma is well-known to experts working in the area, but for lack of an
adequate reference we include it here.

\begin{lemma}
  Let $D$ be a big divisor on a smooth projective surface, $C\subseteq X$ irreducible
  curve, $D\cdot C=0$. Then $(C^2)<0$.
\end{lemma}
\begin{proof}
  Let $D=P_D+N_D$ denote the Zariski decomposition of $D$. If $C\subseteq \Supp N_D$,
  then it must have negative self-intersection, since the intersection form on $N_D$
  is negative definite. Assume $C$ is not in $N_D$.  Then
  \[
  0 \equ D\cdot C \equ P_D\cdot C +N_D\cdot C \equ (\geq 0) + (\geq 0) \ ,
  \]
  since $P_D$ is nef, $C$ is effective, $N_D$ is effective with no common components
  with $C$. This can only happen if
  \[
  P_D\cdot C \equ N_D\cdot C \equ 0\ .
  \]
  Therefore, $C$ is orthogonal to the big and nef divisor $P_D$, hence we must
  have $(C^2)<0$.
\end{proof}

\section{V-big vs. AEA}

Let $X$ be a smooth projective variety and $\she$ a vector bundle on $X$.  There
exist two non-equivalent definitions for bigness in the literature: V-big and L-big
vector bundles.  It is known that V-bigness implies L-bigness and that the converse
does not hold if $\rk\she\geq 2$ (cf.\ Remark~\ref{diverio}).

Throughout this section we will point out some differences (for example a different
Kodaira's lemma) between L-big and V-big vector bundles, and compare V-bigness and
almost everywhere amplenessq.  In particular we will show that these positivity
properties coincide for nef vector bundles.  V-big vector bundles are also called
\emph{ample with respect to an open set} \cite[Chapter 3]{KellyThesis}.

We have seen that if $E$ is V-big, then $E$ is also AEA cf.\
Remark~\ref{remark-augm}.

\begin{question}
  Does $E$ being AEA imply that $E$ is V-big?

\noindent
We will see that this is the case if $E$ is nef.

\end{question}

\begin{remark}
  As pointed out in \cite[Lemma 3.44]{KellyThesis}, a vector bundle on a projective
  curve is ample with respect to an open set exactly if it is ample.
\end{remark}

Next we will show that a strong form of Kodaira's lemma is valid for vector bundles
that are ample with respect to an open set.

\begin{lemma}[(Kodaira's lemma for vector bundles)]
  Let $X$ be a smooth projective variety, $\she$ a vector bundle, and $\sha$ an ample
  line bundle on $X$. Then the following are equivalent.
  \begin{enumerate}
  \item $\she$ is ample with respect to an open subset.
    \ilabel{oneb}
  \item $\Sym^m\she$ contains an ample vector bundle of the same rank for some $m>0$.
    \ilabel{twob}
  \item There exists $m>0$ and an injective morphism
    \[
    \bigoplus^{\rk\Sym^m\she}\sha \hookrightarrow \Sym^m\she\ ,
    \]
    which is an isomorphism over an open subset.
    \ilabel{threeb}
\end{enumerate}
\end{lemma}
\begin{proof}
  The equivalence of \iref{oneb} and \iref{threeb} is the content of \cite[Lemma
  3.42]{KellyThesis}; \iref{threeb} obviously implies \iref{twob}, and
  \iref{twob} implies \iref{oneb} holds if being ample with respect to an open set is
  scale-invariant.
  \sideremark{\sandor{060914}: this needs to be rewritten after including the defn of
    ample wrt an open set}
\end{proof}

There is a useful characterization of ampleness with respect to an open subset in terms of $\sO_{\P(\she)}(1)$.

\begin{lemma}(cf.\ Proposition~\ref{bplus})\label{lemma-equiv} With notation as
  above, $\she$ is ample with respect to the dense open set $U\subseteq X$ precisely
  if $\sO_{\P(\she)}(1)$ is ample with respect to $\pi^{-1}(U)\subseteq \P(\she)$.
\end{lemma}

\begin{comment}
 With notation as above, recall that  a vector bundle $\she$ is called \emph{L-big},
 if the line bundle $\sO_{\P(\she)}(1)$ on $\P(\she)$ is big.
\end{comment}
\medskip

Here we have the following weaker version of the Kodaira lemma.

\begin{lemma}

  \label{lemma-PAG}
  Let $\she$ be a vector bundle using the notation above.
\begin{enumerate}

\item Assume that $\HH{0}{X}{\Sym^m\she}\neq 0$ for some $m>0$.  Then for any ample
  line bundle $\sha$ on $X$ and any $k>0$, ${\Sym^k\she}\otimes\sha$ is L-big.
  \ilabel{oned}

\item Assume that for some $m>0$ and some $x \in X$ the vector bundle $\Sym^m\she$ is
  generated at $x$ by its global sections $\HH{0}{X}{\Sym^m\she}$.  Then for any
  ample line bundle $\sha$ on $X$ and any $k>0$, ${\Sym^k\she}\otimes\sha$ is V-big.
  \ilabel{twod}

\item Conversely, assume that $\she$ is L-big. Then for any line bundle $\shl$ on
  $X$,
  \[
  \HH{0}{X}{\Sym^m\she\otimes \shl}\neq 0
  \]
  for all $m\gg 0$.
  \ilabel{threed}

\item Assume that $\she$ is V-big. Then for any line bundle $\shl$ on $X$, \(
  {\Sym^m\she\otimes \shl} \) is generically generated by its global sections for all
  $m\gg 0$.
  \ilabel{fourd}

\end{enumerate}

\end{lemma}

\begin{proof}
  To prove \iref{oned}, assume that $\HH{0}{X}{\Sym^m\she}\neq 0$ for some $m>0$.
  This means that $\HH{0}{\P(\she)}{\sO_{\P(\she)}(m)}\simeq \HH{0}{X}{\Sym^m\she}
  \neq 0$, hence $\sO_{\P(\she)}(1)$ is $\Q$-effective. Then
  \[
  \sO_{\P(\she\otimes\sha)}(1)\simeq \sO_{\P(\she)}(1)\otimes\pi^*\sha\ .
  \]
  By \cite[Proposition 1.7.10]{PAG}, the $\Q$-divisor $a c_1(\op) +
  \pi^*c_1(\shl)$ is ample for $0<a\ll 1$. This implies that
  \[
  c_1(\op)+\pi^*c_1(\shl) \equ (1-a) c_1(\op) + (a c_1(\op)+\pi^*c_1(\shl))
  \]
  can be written as the sum of an effective and an ample divisor, hence it is big.

  To prove \iref{twod}, assume that for some $m>0$ and some $x \in X$ the vector
  bundle $\Sym^m\she$ is generated at $x$ by its global sections.

  Then $\mathbb{B}({\she}) \neq X$ and hence $\mathbb{B}_- ({\she}) \subset
  \mathbb{B}({\she}) \neq X$. Therefore ${\she}$ is weakly positive and for all $H$
  ample \[\mathbb{B}_+ ({\she} + (1/m) H) = \bigcap \mathbb{B}({\she} + (1/m)H -
  (1/n)H) \subseteq \mathbb{B}({\she}) \neq X.\]

  \iref{threed} is a reformulation of the Kodaira lemma on $\P(\she)$ (see
  \cite[Lemma 2.2.6]{PAG}).

  Finally, for \iref{fourd}, assume that $\she$ is V-big, $H$ an ample line bundle,
  and $\shl$ any line bundle on $X$.  Using notations of the first chapters we have
  that for $m$ sufficiently large $H - (1/m)L$ is ample, and as ${\she}$ is V-big we
  have
  \[
  \mathbb{B}(\Sym^{mk}{\she} +L) = \mathbb{B}(\Sym^k {\she} + (1/m) L) \subseteq
  \mathbb{B} (\Sym^k {\she} - (H - (1/m) L)) \neq X
  \]
  for $k$ big enough.
\end{proof}

\begin{remark}
  A line bundle is (L-)big if and only if it is ample with respect to an open set.
  We have seen in \ref{L-big-vs-V-big} that there exist simple vector bundles that
  are L-big, but not V-big, and hence not ample with respect to an open set.
\end{remark}

\begin{comment}
  It follows from Lemma~\ref{lemma-equiv} and the previous remark that a vector
  bundle, which is ample with respect to an open set is necessarily L-big.
\end{comment}

\begin{proposition}
  Let $X$ be a smooth projective variety, $\she$ a vector bundle on $X$ containing an
  ample vector bundle $\sha$ of the same rank. Then
  \begin{enumerate}
  \item $\she$ is almost everywhere ample with respect to the closed subset $T= \Supp
    ({\she}/A)$.
  \item $\she$ is ample with respect to the dense open set $X\setminus T$.
  \end{enumerate}
\end{proposition}

\sideremark{\sandor{060914}: This proof is a mess. I will clean this up after adding
  the def of ample wrt an open set.}

\begin{proof}

  Let $\shh$ be an ample line bundle on $X$, and $\shl=\sO_{\P(\she)}(1)$ and
  $C\subseteq \P(\she)$ an irreducible curve. If $C$ is contained in a fibre that
  maps to a point away from $T$, then
  \[
  \shl \cdot C \equ \pi^*\sha\cdot C\ ,
  \]
  which is good. If $C$ is contained in a fibre mapping to $T$, then we do not care
  about the intersections numbers at all.

  We may now assume that $C$ is not contained in a fibre of $\pi$. Let $B\deq
  \pi(C)\subseteq X$.

  By restricting everything to $B$ via base change along $B\hookrightarrow X$, we may
  assume that $X$ is a curve, and $\pi|_C:C\to X$ is a dominant morphism.  Consider
  the short exact sequence of sheaves
  \[
  0\lra \sha\lra\she\lra \shq\lra 0\ ,
  \]
  where $\shq\deq \she/\sha$ is a torsion sheaf on $X$ with support $T$.

  The vector bundle map $\pi^*\she\to\op$ is surjective, hence $\pi^*\sha$ maps
  surjectively onto a sub-line-bundle $\shb$ of $\op$. Since $\pi^*\sha$ is ample, so
  is the quotient $\shb$.

\end{proof}

\begin{comment}
  It is known that a direct sum of line bundles is L-big if and only if at least one
  of the summands is big; a direct sum of line bundles is V-big if all summands are
  big. What is the situation with AEAness?
\end{comment}

\sideremark{\sandor{060914}: This remark is a bit confusing. It needs a rewrite.}

\begin{remark}
  In the case of a line bundle $\shl$, the largest open subset over which the
  evaluation map
  \[
  \HH{0}{X}{\shl}\otimes\sO_X\lra \shl
  \]
  is surjective is the complement of the stable base locus $X\setminus
  \mathbb{B}(\shl)$.

  An L-big vector bundle $E$ is V-big if $\mathbb{B}_+(\op)$ is contained in a union
  of fibres over a proper Zariski closed subset of $X$.
\end{remark}

When $\shl$ is a line bundle on a surface, then it is immediate from the
intersection-theoretic characterizations that
\[
\mathbb  B_+(L) \equ \mathbb T(P_D)\ .
\]
Since we would need something along these lines on $\P(E)$, which in interesting cases has dimension at least three, the above observations can only serve as a pointer what kind of statements we would like to prove in higher dimensions.

\begin{lemma}\label{lemma-nef}
  Let $X$ be an irreducible projective variety, $\shl$ a nef line bundle on $X$. Then
  \[
  \mathbb B_+(\shl) \equ \mathbb T(\shl)\ .
  \]
\end{lemma}

\sideremark{\sandor{060914}: This proof needs a facelift!!}

\begin{proof}
  We have already observed in Remark~\ref{remark-augm} that $T(\shl)\subseteq
  B_+(\shl)$ in general.

  \begin{comment}
    According to the main result of \cite{Nak} (see also \cite[Corollary
    5.6]{ELMNP2}),
  \[
  B_+(\shl) \equ \bigcup_{V\subseteq X,\ (\shl^d\cdot V)=0}V\ ,
  \]
  where $V$ runs through all positive-dimensional irreducible subvarieties of $X$
  restricted to which $\shl$ ceases to be big; while
  \[
  \mathbb{T}(\shl) = T_{\eps} = \overline{\bigcup_{C\subseteq X,\ (\shl\cdot C)< \eps
      c_1(A)\cdot C } C } \textrm { for } 0< \eps \ll 1 ,
  \]
  where $C\subseteq X$ means an irreducible curve, and $A$ is a fixed ample divisor
  on $X$.  We have shown that $T(\shl)\subseteq B_+(\shl)$ in general.
  \end{comment}

To prove the other direction in the case where $L$ is nef,
we use  the main result of \cite{Nak} (see also \cite[Corollary 5.6]{ELMNP2}):
\[
 B_+(\shl) \equ \bigcup_{V\subseteq X,\ (\shl^d\cdot V)=0}V\ .
\]
We have to show that if
$V\subseteq X$ is an irreducible subvariety for which
  $\shl|_V$ is not big, then $V$ is contained in  $T(\shl)$.
We will show that $V$ is covered by curves $C$ satisfying
  $\shl\cdot C < \eps A \cdot C$ for $\eps$ small enough.

Assume first that $V$ is
  smooth, and apply \cite{BDPP} (the
  pseudo-effective cone is the dual of the cone of moving curves).
Then
  $\shl|_V$ not big implies that it is not in the interior of
the pseudo-effective  cone,
hence there must exist a real 1-cycle $ 0 \neq C \in N_1(V)$ limit of moving
  curves $C_n\subseteq V$ with $\shl|_{V}\cdot C=0$.
As $A|_{V}$ is ample on $V$,
  then $A|_{V} \cdot C >0$,
so $\lim \frac{\shl|_{V}\cdot C_n}{A|_{V} \cdot C_n} =0$.
  Thus, for $n$ sufficiently large, $\shl \cdot C_n < \eps A \cdot C_n$.  The class
  of $C_n$ however covers $V$, which implies $V\subseteq \mathbb{T}(\shl)$.

  If $V$ is not smooth, then let $\mu:V'\to V \subseteq X$ be a resolution of
  singularities. Since $\shl|_V$ was pseudo-effective and not big to begin with, the
  same applies to $\mu^*(\shl|_V)$. Using the argument as above, there exist moving
  curves $C_n$ on $V'$ such that their limit is a non zero 1-cycle $C \in N_1(V')$
  such that $\mu^*(\shl|_V) \cdot C =0$, now let $A$ be an ample divisor on $X$, then
  $\mu^*(A)$ is big on $V'$, so $\mu^*(A) \cdot C >0$. Then as above $\lim
  \frac{\mu^*(\shl|_{V}) \cdot C_n}{\mu^*(A) \cdot C_n} =0$.  Thus, for $n$
  sufficiently large, $\mu^*(\shl|_{V}) \cdot C_n <\eps \mu^*(A) \cdot C_n$, hence by
  projection formula $\shl \cdot \mu_*(C_n) < \eps A \cdot \mu_*(C_n)$.  And the
  class of $\mu_*(C_n)$ covers $V$, which implies $V\subseteq \mathbb{T}(\shl)$.
\end{proof}

\begin{proposition}
  \label{aeavbig}
  Let $\she$ be a nef vector bundle on an irreducible projective variety $X$. Then
  $\she$ is AEA if and only if it is V-big.
\end{proposition}

\begin{proof}
  Both AEA and V-bigness imply that $\shl\deq \sO_\P(1)$ is a big and nef line
  bundle.  By the previous lemma, $B_+(\shl)=T(\shl)$, hence
  \[
  qE \text{ is V-big } \Leftrightarrow \pi(B_+(\shl))\neq X \Leftrightarrow
  \pi(T(\shl))\neq X \Leftrightarrow E \text{ is AEA}\ .\qedhere
  \]
\end{proof}

\begin{comment}
\silentremark{\sandor{060914}: Do we need this here?}
  Let now $D$ be a Cartier divisor on an irreducible projective variety $X$, and
  assume it has a Zariski decomposition; more precisely, I'd like to assume that
  there exists a unique minimal effective divisor $E$ on $X$ such that
  \[
  P \deq D-E
  \]
  is nef, and $P|_V$ is not big for any irreducible component $V$ of $E$. Following
  the argument of \cite[Example 1.11]{ELMNP}, we would get that
  \[
  B_+(\sO_X(D)) \equ \Null (P) \equ T(\sO_X(D))
  \]
  by Lemma~\ref{lemma-nef}.

\end{comment}


TITLE: Show that $x\in \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$ if and only if $x\in A_k$ for infinitely many sets $A_k$
QUESTION [0 upvotes]: Let $x\in \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$ then $x\in \bigcup_{k=1}^{\infty}A_k$ and $x\in \bigcup_{k=2}^{\infty}A_k$ and so on so that $x\in \bigcup_{k=n}^{\infty}A_k$, but since $x\in \bigcup_{k=n}^{\infty}A_k\subseteq \bigcup_{k=1}^{\infty}A_k\space \forall n\in\mathbb{N}$ $\implies x\in \bigcup_{k=1}^{\infty}A_k \implies\space x\in A_k$ for infinitely many sets $A_k$.
Conversely, Let $x\in A_k$ for infinitely many sets $A_k$. Let $x\in A_1\implies x\in\bigcup_{k=1}^{\infty}A_k$. Now, suppose $x\notin A_1$ but $x\in A_2\implies x\in\bigcup_{k=2}^{\infty}A_k$. Continue in this manner, so that $x\notin A_k, \forall k\in\{1,...,n-1\}$ but $x\in A_n$ and $x\in\bigcup_{k=n}^{\infty}A_k $. Now, $\bigcup_{k=n}^{\infty}A_k\subseteq...\subseteq  \bigcup_{k=2}^{\infty}A_k\subseteq\bigcup_{k=1}^{\infty}A_k$. So, $x\in \bigcup_{k=n}^{\infty}A_k\implies x\in \bigcup_{k=1}^{\infty}A_k\implies x\in \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k$

REPLY [1 votes]: Use more words and fewer formulae:
Suppose $x \in \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$.
We want to show that $x$ is in $A_k$ for infinitely many $k$ so we assume $x \in A_k$ only for $A_k$ among $A_{n_1}, A_{n_2}, \ldots, A_{n_m}$, $m$ many sets.
Now if $n_0 \in \Bbb N$ is picked to be larger than all indices $\{n_1, \ldots, n_m\}$ we know that $x \notin \bigcup_{k=n_0}^\infty A_k$ as no set with index $\ge n$ can contain $A_k$ by assumption, so this $n_0$ contradict $x \in \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$, so our assumption that $x$ is in only finitely many $A_k$ was false. This shows one inclusion.
To show the reverse, assume that for some $x$ the set $N_x=\{k\mid x \in A_k\}$ is infinite. To see that $x \in  \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$, let $n \in \Bbb N$ be arbitrary. Then $N_x$ contains some $k \ge  n$ or $N_k$ would have been finite (atmost size $n$ in fact). So we have some $k$ (depending on $n$) such that $k \ge n$ and $x \in A_k$. This shows that $x \in \bigcup_{k=n}^\infty A_k$ for this $n$ and as $n$ was arbitrary, $x \in \bigcap_{n=1}^\infty \bigcup_{k=n}^\infty A_k$ as required.


TITLE: Construct a Mobius map that maps the strip $\{z\in \mathbb C:0<\Im (z)<1\}$ onto the area between the circles $|z-1|=1$ and $|z-2|=2$.
QUESTION [2 upvotes]: Construct a Mobius map that maps the strip $\{z\in \mathbb C:0<\Im
 (z)<1\}$ onto the area between the circles $|z-1|=1$ and $|z-2|=2$.

(The question earlier asked for a map from $|z-1|<1$ onto $|z|>2$ for which is got $g(z)=\frac 2{z-1}$ but I can't really see how to use this)
I can't get anywhere useful, I just have pages of algebra getting me nowhere.
Any help is appreciated, thank you

REPLY [1 votes]: For me, it's easier to move a pair of parallel lines to a desired destination than doing it for a pair of nested circles touching at one point. Therefore I suggest first constructing the map in the other direction, and then inverting the resulting Möbius transformation to obtain the map in the required direction.
One transforms a pair of nested circles touching at $p$ into a pair of parallel lines by applying a Möbius transformation that maps $p$ to $\infty$. Here, the two circles touch at $0$, and the simplest Möbius transformation mapping $0$ to $\infty$ is $\rho \colon z \mapsto 1/z$. $\rho$ maps $\mathbb{R}\cup \{\infty\}$ to itself, and the two circles intersect the real axis at right angles, so the image of each of the two circles under $\rho$ is a straight line intersecting $\mathbb{R}$ at a right angle, i.e. a line of the form $\{ w : \operatorname{Re} w = c\}$.
The smaller circle intersects $\mathbb{R}$ at $0$ and at $2$, and $\rho(2) = \frac{1}{2}$, so the image of $\{ z : \lvert z-1\rvert = 1\}$ under $\rho$ is $\bigl\{ w : \operatorname{re} w = \frac{1}{2}\bigr\}$. The larger circle intersects $\mathbb{R}$ at $0$ and at $4$, so the image of $\{ z : \lvert z-2\rvert = 2\}$ is $\bigl\{ w : \operatorname{Re} w = \frac{1}{4}\bigr\}$.
The point $3$ lies between the two circles ($\lvert 3-1\rvert > 1$ and $\lvert 3-2\rvert < 2$), and $\frac{1}{4} < \rho(3) = \frac{1}{3} < \frac{1}{2}$, so the region between the two circles is mapped to the strip $\bigl\{ w : \frac{1}{4} < \operatorname{Re} w < \frac{1}{2}\bigr\}$.
We now have a vertical strip of width $\frac{1}{4}$, and we want a horizontal strip of width $1$, thus we need a scaling by a factor of $4$, and a rotation by $\pm \frac{\pi}{2}$. Then we need a translation to move the strip to the required position. All that is easily achieved by
$$\sigma \colon w \mapsto i(4w-1),$$
and hence a Möbius transformation doing the ultimate goal is $(\sigma\circ \rho)^{-1}$. All other such transformations are obtained by composing it with an automorphism of the strip.
To find $(\sigma\circ \rho)^{-1}$, we isolate $z$ in
$$w = i\biggl(\frac{4}{z} - 1\biggr),$$
yielding
$$z = \frac{4}{1-iw} = \frac{4i}{w+i} =: g(w).$$
So you did get the (well, a, it's not unique) right map, and you just have miscalculated when trying to verify.
Let's make a couple of checks. The image of the line $\{w : \operatorname{Im} w = 0\}$ under $g$ is the circle through $g(\infty) = 0$, $g(0) = 4$, and $g(1) = \frac{4}{1-i} = \frac{4(1+i)}{2} = 2(1+i)$. It is easily seen that this is the circle $\{ z : \lvert z-2\rvert = 2\}$.
The image of the line $\{ w : \operatorname{Im} w = 1\}$ is the circle through $g(\infty) = 0$, $g(i) = \frac{4}{1-i^2} = 2$, and $g(2+i) = \frac{4}{1 - i(2+i)} = \frac{4}{2(1-i)} = 1+i$, which is $\{ z : \lvert z-1\rvert = 1\}$.
A point between the two lines is $\frac{i}{3}$, and $g(i/3) = \frac{4}{1-i^2/3} = \frac{4}{4/3} = 3$ lies between the two circles. Hence $g$ maps the strip $\{ w : 0 < \operatorname{Im} w < 1\}$ to the region between the two circles, as desired.


TITLE: How to find the equation of a line which passes through 10 known points
QUESTION [0 upvotes]: I know how to find the equation of a line which passes through two known points. But here, I have 10 points $(x,y)$ through which my line passes through. I know I can draw a graph and find then find the equation of the line, but is there a better way or online tool to do the task? The values are given below

REPLY [0 votes]: If all 10 points lie along a line, then you can pick any two points, find the equation for the line through those, and that line will pass through the other eight points. This is because "two points determine a line."
You should know, though, that 10 arbitrarily chosen points don't necessarily all lie along a line. Indeed the points in your question do not lie along a line:

You can get a line that approximates the points as nearly as possible by "linear regression". In R, you can get the intercept and slope of the regression line with the lm function (for "linear model"):
> mydata = read.csv("~/tmp/known-points.csv")
> lm(mydata)

Call:
lm(formula = mydata)

Coefficients:
(Intercept)          ADC
   17.29737      0.07425


text:
(* Title: ETTS/Manual/ETTS_CR.thy
   Author: Mihails Milehins
   Copyright 2021 (C) Mihails Milehins
*)

theory ETTS_CR
  imports ETTS_Examples
begin
end

subset_name: curated


TITLE: Order type in finite sets
QUESTION [3 upvotes]: This is a general question regarding power (cardinal number)  and type (ordinal number) of a set. These definitions are taken from Kolmogorov and Fomin(1970). I can see why given power there are (uncountably) many sets with different types with this power in the infinite case. For instance, $\aleph_0$ corresponds to the usual order $\omega$ of $\mathbb{N}$ which is
$$
1,2,3,\dots,
$$
but another order type can be written as
$$
1,3,5,\dots,2,4,6\dots.
$$
However, it is claimed that in the finite case there is a unique type for given power. I do not understand this because we can follow the same arguments. For instance take a set with power $2n$, usual order type is
$$
1,2,,\dots,2n
$$
in which $2n$ is the maximal element. However, another order type
$$
1,3,5,\dots,2n-1,2n,\dots 2
$$
in which $2$ is the maximal element. I do not see why this argument should not hold. Thanks for any help

REPLY [2 votes]: Order type does not care about naming of the elements. i.e. the order
$$1,2,\dots,2n \>\>\>\>\text{ and }\>\>\>\>1,3,5,\dots,2n-1,2n,\dots 2$$
are the same. Just create the natural bijection (mapping least element to least, second least to second least etc.) between the sets, and it is straight forward to check that it is an order preserving bijection.
Your question should really be why the order types of
$$
1,2,3,\dots
\>\>\>\>\text{ and }\>\>\>\>
1,3,5,\dots,2,4,6\dots
$$
are different.
Answer: The easy way to see this is to observe that in $1,2,3,\dots$ for each element there is a finite amount of elements which is less than that element. However in $
1,3,5,\dots,2,4,6\dots$ the element $2$ has an infinite amount of elements which is less than it. Thus any order preserving bijection between $1,2,3,\dots$ and $1,3,5,\dots,2,4,6\dots$ has an impossible task to map any element in the domain to the element $2$ in the co-domain.
Especially if we map least element to least, second least to second least etc. we will not create a function which is onto.
Formal proof that there is no order preserving bijection: Assume $f$ is an order preserving bijection from $1,2,3,\dots$ to $1,3,5,\dots,2,4,6\dots$ . Now assume that $f(n)=2$. As $1,3,5,\ldots,2n+3$ are all numbers less than 2 (in the right order type) we see that $f^{-1}(1),f^{-1}(3),\ldots,f^{-1}(2n+1)$ are all distinct numbers which are less than $n$. However these consitute a set of $n$ different positive integers which are less than $n$, which is a contradiction, since there are only $n-1$ positive integers which are less than $n$.


TITLE: Double slit experiment setup
QUESTION [2 upvotes]: This is a better form of a previous question I asked: Double slit experimental procedure
Basically, I want to know what constitutes a measurement of which slit a particle goes through in the double slit experiment.  For the following three setups:

Two slits.
Two slits with devices that assign the particles perpendicular spins.
Two slits with devices that assign the particles perpendicular spins, and where the screen can tell the spin of the electron.

In the first, we have interference.  In the third, we have no interference because we have measured which slit the electron passed through.  In the second, is there interference or no interference?

REPLY [4 votes]: Basically, I want to know what constitutes a measurement of which slit
  a particle goes through in the double slit experiment

Simple: whatever setup you use, if that setup can give you the which-way information, it is a measurement and you will not get interferences. If the setup is useless, you will get interferences. The details have no importance.
So your second setup is not a measurement if you do nothing with the spin tags. If you use them to recover the which-way info, then it is a measurement.
Why is it so simple? Because the fundamental idea is that any slit-observing device must by definition couple with the particle (else it cannot detect it). This means particle and observing device gets entangled; the superposition of trajectories is now linked to the superposition of the device yes/no states. In other words, the device yes/no states interfere together, along with the trajectories.
Using the device to get the which-way information means measuring it in a specific yes or no state. But the entanglement is such that when the device states do not interfere any more, the particle paths do not interfere either. That's all there is to it, and there is no way to cheat with this, whatever detection trick you may come up with.


TITLE: $\|f\|_\infty = \sup \{|f(x)| : x \in \mathbb R^n\}$
QUESTION [2 upvotes]: Let $f$ be a  continuous on the measure space $\mathbb R^n,$ $L,$ $\lambda.$ Prove that $\|f\|_\infty = \sup \{|f(x)| : x \in \mathbb R^n\}$
I got "$\leq $" easily, but I have problems  using the continuity of $f$ to show it can not be $<$ strictly.

REPLY [1 votes]: The definition of the $L^{\infty}$ norm I'm assuming you're using is the following:
$$
\|f\|_{\infty}=\inf\{C\geq 0;\,\,|f(x)| \leq C \text{ almost everywhere}\}.
$$
Let $\alpha = \sup\{|f(y)|;y\in \mathbb{R}^{n}\} \in [0,\infty]$. Since $|f(x)|\leq \alpha$ for all $x\in\mathbb{R}^{n}$, then $\|f\|_{\infty} \leq \alpha$. Suppose $\|f\|_{\infty}<\alpha$ (assume for the moment that $\alpha<\infty$). Then, there would be a constant $0\leq C<\alpha$ and a full measure set $X \in L$ (meaning that $\lambda(\mathbb{R}^n\setminus X)=0$) such that
$$
|f(x)| \leq C
$$
for  $x\in X$. Since $\alpha$ is the supremum, there is a $y_0\in \mathbb{R}^{n}$ such that $|f(y_0)|>\frac{\alpha+C}{2}>C$ (because $(\alpha+C)/2 < \alpha$). From continuity, of $f$, there is a $\delta>0$ such that if $y\in B_{\delta}(x_0) = \{y\in \mathbb{R}^{n};|y-y_0|<\delta\}$, then
$$
|f(y)- f(y_0)|<\frac{\alpha-C}{2},
$$
which means that if $y\in B_{\delta}(y_0)$,
$$
|f(y)|>|f(y_0)| - \frac{\alpha-C}{2}>C.
$$
Then, $B_{\delta}(y_0) \subset \mathbb{R}^{n}\setminus X$, which contradicts the fact that $\lambda(\mathbb{R}^{n}\setminus X)=0$.
If $\alpha = \infty$ and $\|f\|_{\infty}<\infty$, then the result is simpler, as you can find a point  $y_0$ where $|f(y_0)| > C+1$ (say), and use continuity to find a ball where $f$ is greater than $C$ in absolute value, reaching the same contradiction.


TITLE: Why doesn't the speed of the wind have an effect on the apparent frequency?
QUESTION [9 upvotes]: A boy is standing in front of stationary train. The train blows a horn of $400Hz$ frequency . If the wind is blowing from train to boy at speed at $30m/s$, the apparent frequency of sound heard by the boy will be?
The answer: The frequency remains the same at $400Hz$
MY QUESTION:
Why doesn't the speed of the wind have an effect on the apparent frequency?

REPLY [3 votes]: The Doppler Shift formula $f'=f(\frac{v \pm v_{obs}}{v \mp v_{source}})$ only works if the wind (or the medium that the sound is moving in) is constant. Therefore, if the wind is moving at a constant speed, change the reference frame so that the wind is stationary.
In your case, change the reference frame so that it is moving from the train to the boy at $30m/s$. This way, the wind is stationary in the new reference frame. Then, calculate the speed of both the observer and source in the new reference frame: $f'=f(\frac{v \pm v_{obs}}{v \mp v_{source}})=f(\frac{343m/s + 30m/s}{343+ 30m/s})=f=400Hz$
Note: If the windspeed changes, the frequency will be different as the reference frame will be changing all the time.


TITLE: why the name "relaxation" in the SOR Gauss-Seidel iterative method?
QUESTION [1 upvotes]: The question is that simple. It is in the subject.
Why the name "relaxation" in the SOR Gauss-Seidel iterative method?
I have googled for it without luck.
Moreover, when $\omega$ is used in Gauss-Seidel is named "relaxation" and when $\omega$ is used in Jacboi is named: "weight" weighted Jacobi method
So, something got to be special about Gauss-Seidel to deserve such a name.
I observed that there is a "relaxation" technique explained in
this thesis   back from 1948. I do not see a connection between
relaxation in that context of mechanics (PDEs) and the relaxation asked here which is in the context of linear systems such as Jacobi and Gauss-Seidel.
Hence, I do not consider this as a duplicate questions.
Thanks.

REPLY [1 votes]: Being curious about the same question, I read the related part of the thesis:

Southwell considered a framework having elastic members and frictionless joints. Rigid constraints were imposed at the joints such that the movements of the joints were prevented but the ends of the members were left free to turn. All joints were constrined in this manner and specified forces applied. Since the constraints were rigid they sustrained the whole force at every joint.
Then one constrint was relaxed so that one joint could move slowly (so that equilibrium is maintaned and vibrations are not excited) through a specified distance in a specified direction. If the initial force on the constrint had a component in the direction of travel, this component of the force was relieved and the strain energy stored in the member. All joints but one having remained fixed, it was possible to calculate how much force was transfered.
Another constraint was relaxed and a similar operation performed. Systematically proceedings whith each constraint the amount of total strain energy was increased with each step until the forces remaining on the constraints were so small as to be negligible. Since each constraint was "relaxed" in turn, the name "Relaxation Methods" was given to the procedure.

So for iterative method in solving linear equation system $Ax=b, (Ax^k-b=r^k)$, my understanding is that:

Each component of the unknown (i.e., $x_i$) is regarded as a joint constraint;
Each component of the residual (i.e., $r_i$) is regarded as a force on the corresponding joint;
Relaxing "joint" $x_i$ means $x_i$ is allowed to be updated, in order to diminish the corresponding "force" $r_i$. This is possible since only $x_i$ is relaxed while all the rest are still fixed (All joints but one having remained fixed, it was possible to calculate how much force was transferred).
Other $x_i$ are relaxed in turn similarly, until the residual $r_i$ on all joints are so small to be negligible.

So far the analogy above is satisfactory to me, although I don't know the details of the mechanics/physics involved.


TITLE: When does a Lagrangian dynamical system have an equivalent Hamiltonian description?
QUESTION [9 upvotes]: Let a Lagrangian dynamical system with $n$ degrees of freedom and configuration space $\mathbb{R}^n$
(i.e. phase space $\mathbb{R}^{2n}$), which is described by  $L=L(q_{i},\dot{q}_{i},t)$, $i=1,2,...,n$.
It is well known from classical mechanics, that a sufficient condition for a Lagrangian dynamical system to be equivalently described as a Hamiltonian system, is the Hessian of the Lagrangian w.r.t. the generalized velocities to be non-degenerate:
$$
\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0
$$
Now the question is the following: It is frequently mentioned in the literature of classical mechanics (but I have not seen a direct proof of this), that the above sufficient condition is independent of the choice of generalized coordinates and depends only on the dynamical system itself (i.e. only on the Lagrangian function and the underlying configuration space). What kind of direct proof could be provided for this ?
P.S.: the motivation for posting this question was a student's (not my student) question on whether the property of a Lagrangian system to be Hamiltonian as well, is a purely geometrical property. I am not an expert in the subject but I think that essentially, it suffices to answer the question for the case that the configuration space is $\mathbb{R}^n$. In the sense that the independence of the property of the particular coordinate system used, means that in the general case, it will be a property of the underlying differentiable structure on the configuration space manifold.

REPLY [4 votes]: Here's what I have done:
$\bullet$ Let the Lagrangian $L(q_{i},\dot{q}_{i},t)$, which under the point transformations
$$
\{q_{i}\}\leftrightsquigarrow\{Q_{i}\}
$$
given by the invertible relations $Q_{i}=Q_{i}\big(q_{j},t\big)\Leftrightarrow q_{j}=q_{j}\big(Q_{i},t\big)$,  $\ \ i,j=1,2,...,n$, (i.e. $\det\Big|\frac{\partial Q_{j}}{\partial q_{i}}\Big|\neq 0$), can be written as:
$$
L(q_{i},\dot{q}_{i},t)\stackrel{}{\leftrightsquigarrow}L(Q_{i},\dot{Q}_{i},t)
$$
$\bullet$ Differentiating the point transformations $Q_{i}=Q_{i}\big(q_{j},t\big)$, we get that:
$$
\frac{dQ_i}{dt}\equiv\dot{Q}_i=\frac{\partial Q_i}{\partial q_j}\dot{q}_i+\frac{\partial Q_i}{\partial t}\ \ \ \ \ \Rightarrow \ \ \ \ \ \frac{\partial\dot{Q}_i}{\partial\dot{q_j}}=\frac{\partial Q_i}{\partial q_j} \ \ \ \ \ \ \ \ \ \ \ \ \ (1)
$$
$\bullet$ On the other hand, differentiating the Lagrangian we get that:
$$
\frac{\partial L}{\partial\dot{q}_i}=\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i} \Rightarrow
$$
$$
\Rightarrow \frac{\partial^2 L}{\partial\dot{q}_j\partial\dot{q}_i}=\frac{\partial}{\partial\dot{q}_j}\Big(\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\Big)= \sum_{m=1}^{n}\frac{\partial}{\partial\dot{Q}_m}\Big(\sum_{k=1}^{n}\frac{\partial L}{\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\Big)\frac{\partial\dot{Q}_m}{\partial\dot{q}_j}=
$$
$$
=\sum_{k,m=1}^{n}\frac{\partial^2 L}{\partial\dot{Q}_m\partial\dot{Q}_k}\frac{\partial\dot{Q}_k}{\partial\dot{q}_i}\frac{\partial\dot{Q}_m}{\partial\dot{q}_j}\stackrel{(1)}{\Rightarrow}
$$
$$
\stackrel{(1)}{\Rightarrow}
\frac{\partial^2 L}{\partial\dot{q}_j\partial\dot{q}_i}=\sum_{k,m=1}^{n}\frac{\partial^2 L}{\partial\dot{Q}_m\partial\dot{Q}_k}\frac{\partial Q_k}{\partial q_i}\frac{\partial Q_m}{\partial q_j}
$$
Thus:
$$
\Big[\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big]=\Big[\frac{\partial Q_{j}}{\partial q_{i}}\Big]^{t}
\Big[\frac{\partial^{2}L}{\partial\dot{Q}_{i}\partial\dot{Q}_{j}}\Big]\Big[\frac{\partial Q_{j}}{\partial q_{i}}\Big]
$$
where $[..]$ stands for the respective Jacobian and hessian matrices and $[..]^t$ stands for the transpo-se matrix. The last relation clearly tells us that the non-vanishing of the hessian determinant is invari-ant under the point transformations, i.e.
$$
\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}\partial\dot{q}_{j}}\Big|\neq 0 \Leftrightarrow \det\Big|\frac{\partial^{2}L}{\partial\dot{Q}_{i}\partial\dot{Q}_{j}}\Big|\neq 0
$$
which concludes the proof.
P.S.: We can easily see why the above, is a sufficient condition for a Hamiltonian description to exist: Generalized momenta $p_{i}$, are defined by
\begin{equation} \label{7.55}
p_{i}=\frac{\partial L}{\partial\dot{q}_{i}}=p_{i}\big(q_{j},\dot{q}_{j},t\big), \ \ \   i,j=1,2,...,n
\end{equation}
If the Jacobian of the generalized momenta w.r.t. the generalized velocities, or equivalently the Hessian of the
Lagrangian w.r.t. the generalized velocities
\begin{equation}   \label{7.57}
\det\Big|\frac{\partial p_{i}}{\partial\dot{q}_{j}}\Big|=\det\Big|\frac{\partial^{2}L}{\partial\dot{q}_{i}
\partial\dot{q}_{j}}\Big|\neq 0
\end{equation}
is non-zero, then, due to the inverse function theorem in $\mathbb{R}^{n}$, the above relations can be solved w.r.t.
the generalized velocities:
\begin{equation} \label{7.58}
\dot{q}_{j}=\dot{q}_{j}\big(q_{i},p_{i},t\big), \ \ \   i,j=1,2,...,n
\end{equation}
Then, the Hamiltonian function is defined through a Legendre transform:
\begin{equation} \label{7.59}
H=H\big(q_{i},p_{i},t\big)=\sum_{i=1}^{n}p_{i}\dot{q}_{i}-L\big(q_{i},\dot{q}_{i},t\big)
\end{equation}
with $i=1,2,...,n$. The Hamiltonian function  $H:\mathbb{R}^{2n+1}\rightarrow\mathbb{R}$, is a function,
of generalized coordinates, generalized momenta and time.


TITLE: In Nuclear Magnetic Resonance what is the Rotating Frame?
QUESTION [0 upvotes]: My understanding is that there's an initial precession frequency ($\omega_0$) about the direction of magnetization ($z$). Then when you excite the nucleus into the spin state with the radio frequency pulse, the direction of Nucleus's Magnetic Moment rotates orthogonality ($x$) to the pulse direction ($y$) changing the bulk magnetization.
This bulk magnetization is still precessing about the direction of magnetization ($z$), so is the rotating frame just accounting for that precession?
I'm pretty sure the rotation ($\omega_1$) about the direction of pulse magnetization (that is about $x$) doesn't have anything to do with the frequency of the rotating frame. My understanding is that precession only relates to the pulse length.
I'm just not sure what the rotating frame is or how it relates to the measurement of induced current from free induction decay.
Edit: I posted this question in several locations, so here's a summary of the answers I got.

The Initial Precession Frequency (ω0) is the Rotating Frame.
The Pulse Rotation (ω1) is in the Rotating Frame.
I'm still not clear on how Bulk Magnetization develops after a pulse sequence.

REPLY [0 votes]: The rotating frame is, as you say, just to account for the rotation about the $z$ axis. More precisely, it's a choice of frame that "undoes" the rotation. This also allows us to think of the RF magnetic field as simply a pulsed static magnetic field along a particular direction (say $x$). From this, we can easily see how the RF magnetic field simply rotates the magnetization in the $xz$ plane.
This also answers the question regarding $\omega_1$, since it is clear that the precession amount only depends on the pulse length and the strength of the pulsed RF field. $\omega_0$ was already taken into account in transforming to the rotating frame, and in a sense is no longer relevant.


TITLE: Function with 2 unknowns and one needs to be solved
QUESTION [2 upvotes]: I have a homework question stating:

Find the set of values of $k$ for which $f(x)=3x^2-5x-k>1$ for all $x\in\mathbb R$.

This question has really confused me because I looked at the answer and the value of $k$ is meant to be $k<-\frac{37}{12}$ and I'm afraid I don't know how to get there. May I please have some help?

REPLY [2 votes]: Hint
Consider the function
$$f(x)=3x^2-5x-k-1$$
Its derivative $f'(x)=6x-5$ is zero when $x=\frac 56$ and you want that $f\left(\frac{5}{6}\right) >0$.
Then, ....

REPLY [2 votes]: Rearranging gives $3x^2-5x-k-1>0$, i.e. the resulting quadratic in $x$ has no real roots. This means that the discriminant $b^2-4ac$ must be negative:
$$(-5)(-5)-4\cdot3(-k-1)<0$$
$$25+12(k+1)<0$$
$$k+1<-\frac{25}{12}$$
$$k<-\frac{37}{12}$$
as expected.

REPLY [0 votes]: Hint: Completing the square we get $$\left(x-\frac{5}{6}\right)^2-\left(\frac{25}{36}+\frac{k}{3}+\frac{1}{3}\right)>0$$


TITLE: Using SUVAT equations of kinematics when using an accelerating frame
QUESTION [0 upvotes]: I have a conceptual question arising from the following problem, which relates to the application of the constant translational acceleration in a straight line (SUVAT) equations of kinematics of a mass point.

A bullet of mass $m$ is ﬁred horizontally with speed u into a wooden block of mass $M$ at rest on a horizontal surface. The coefficient of friction between the block and the surface is $\mu$. While the bullet is moving through the block, it experiences a constant force of resistance to its motion of magnitude $R$, where $R > (M+m)\mu g$. The bullet moves horizontally in the block and does not emerge from the other side of the block.
(i) Show that the magnitude of the deceleration a of the bullet relative to the block while the bullet is moving through the block is given by

$$a=\frac{R}{m}+\frac{R-(M+m)\mu g}{M} \tag{1}.$$
This part of the problem is easy to solve using the SUVAT equations. However, it is the second part of the problem that is puzzling me. This part is as follows.

(ii)  Show that the common speed of the block and bullet when the bullet stops moving through the block, which is denoted by $v$, satisﬁes
  $$av=\frac{Ru-(M+m)\mu gu}{M} \tag{2}.$$

Let us see the two conflicting solutions which I worked out for this part of the problem. The speed $v$ can be calculated using the fact that the final speed of the bullet relative to the accelerating block is zero, as $$0=u-at_c \tag{3},$$
since $u$ would be same for both the stationary frame and the accelerating-block frame, and $t_c$ denotes the time when $v$ is reached. Now consider the acceleration of the block (of mass $M$) in the stationary horizontal surface frame $S$ given by $$a_{MS}=\frac{R-(M+m)g\mu}{M} \tag{4}.$$ The speed $u$ is then given by
$$v=v_{MS}=(a_{MS})(t_c)=\left(\frac{R-(M+m)g\mu}{M}\right) \left(\frac{u}{a}\right) \tag{5}.$$
On the other hand, if we consider the acceleration of the bullet (of mass $m$) relative to the stationary surface frame $S$, applying the SUVAT again yields $$v = v_{mS}=u-a_{mS}t_c \tag{6},$$ thus implying that, $$v=u-\left(\frac{R}{m} \right) \left(\frac{u}{a}\right) \tag{7},$$ which does not seem to co-incide with the previously calculated solution.
Why is it that I get a different answer when considering $a_{mS}$ alone and what conceptual errors are present in my work?

REPLY [1 votes]: This is an excellent kinematics and dynamics question on the basic application of Newton's laws of motion.
The solution presented to both problems in the question is accurate. The question pertains to an alternate solution to the second problem, which evokes the question of whether the alternate solution is consistent with the prior analysis. Consider the two solutions $v = \frac{R - \mu (m + M) g}{M} \cdot \frac{u}{a}$ and $v = u - \frac{R}{m} \cdot \frac{u}{a}$. Setting the right hand sides of both solutions to be equal results in the equation $$a - \frac{R}{m} = \frac{R - \mu (m + M) g}{M},$$ which implies by doing some algebra, including some relative acceleration manipulations, that, $$M(a -\frac{R}{m}) = M(-a_{mM} + a_{mS}) \\
= M a_{MS} = R - \mu (m + M) g,$$
where we denote by $a_{mM} := - a < 0$, $a_{mS}$ and $a_{MS}$ the relative accelerations of the bullet w.r.t. the block, the bullet w.r.t. the stationary surface and the block w.r.t. the stationary surface, respectively. The last equation is consistent with our analysis.  In other words, given these two solutions which were arrived at by applying kinematics and Newton's dynamics equations, equating them yields the consistent Newton's second law of motion for the block.


TITLE: Suppose we have a 1D smooth manifold that locates on the x-y plane, and we have some information about it. Can we identify the manifold uniquely?
QUESTION [3 upvotes]: Suppose we have a 1D smooth manifold that locates on the x-y plane, and we know the number of intersections for any straight line on the x-y plane with the manifold. Can we recover its shape uniquely?

REPLY [1 votes]: First of all, one has to clarify what the "number of intersections" means: Even for (smooth) simple closed curves in the plane, the intersection of a line with the curve can be infinite: Both countably infinite and of cardinality of continuum. A more precise version of your question then is:

Suppose that $A, B$ are 1-dimensional smooth submanifolds of ${\mathbb R}^2$ such that for every affine line $L\subset  {\mathbb R}^2$ the sets  $A\cap L$ and $B\cap L$ have the same cardinality. Does it follow that $A=B$?

The answer to the question posed in this form is negative if we allow noncompact submanifolds. The simplest examples are given by Archimedean spirals, which, in polar coordinates $(r,\theta), r>0, \theta\in (0,\infty)$, are written as
$$
r= b\theta, b> 0.
$$
Each spiral has countably infinite intersection with every affine line in the plane, but, (for instance) choosing different values of the parameter $b$, one obtains different spirals (which are not even congruent to each other by a Euclidean rigid motion).
However, one can prove that if $A$ and $B$ are compact smooth 1-dimensional submanifolds, then your question indeed has positive answer. Let me know if you are interested in the proof.

Edit. Here is how to handle simple closed curves.
Definition. Given a smooth curve $C$ in the plane and a (straight) line $L$ in the plane, define $n_C(L)$ to be the cardinality of the intersection $C\cap L$.
In general, this cardinality is infinite (either countably-infinite or cardinality of continuum). In the special case, when $C$ is a simple closed real-analytic curve, the function $n_C$ takes only finite values.   I will need definitions.
Definition. A line $L$ is said to meet $C$ transversally at a point $p\in C\cap L$, if the line $L$ is not tangent to $C$ at $p$.
A line $L$ is said to be transversal to $C$ if it meets $C$ transversally at every point $p\in C\cap L$ (this intersection could be empty).
From now on, I will only consider only smooth simple closed curves $C$ in the plane.
Important properties of lines meeting $C$ transversally are:

$n_C(L)<\infty$.

If $L_1$ is another line sufficiently close to $L$ (one can state this closeness for instance in terms of coefficients of linear equations defining lines), then $L_1$ also meets $C$ transversally and $n_C(L)=n_C(L_1)$.

Definition. Suppose that $L$ is a line tangent to a smooth simple curve $C$ at a point $p$. Then the intersection of $L$ and $C$ at $p$ is called nondegenerate or the first order of contact if the following holds. Choose Cartesian $xy$- coordinates in the plane so that $L$ is the $x$-axis and $p=(0,0)$. Then a small arc of $C$ containing $p$ is the graph of a smooth function $y=f(x)$. The 1-st order of contact means that $f'(0)=0$ (the tangency condition) and $f''(0)\ne 0$.
In particular, for first order of contact at $p$, a small arc of $C$ containing $p$ lies entirely on one side of $L$ (I'll call this the $C$-side).
Here is an important property of nondegenerate intersections:
Lemma 1. Suppose that $L$ is a line which intersects $C$ at points $p_1,...,p_n$ (thus, $n_C(L)=n$), such that the intersections are transversal at $p_1,...,p_{n-1}$ and the intersection at $p_n$ is the 1st order of contact. Then, for every line $L_1\ne L$ parallel to $L$ and sufficiently close to $L$, $n_C(L_1)= n\pm 1$. Moreover, it is $+1$ if and only if $L_1$ lies on the $C$-side of $L$.
Theorem. Suppose that $C_1, C_2$ are distinct (smooth simple closed) planar curves. Then there is a line $L$ such that
$n_{C_1}(L)\ne n_{C_2}(L)$.
Proof.
Lemma 2. Let $A_1, A_2$ be smooth arcs in $R^2$ such that for every point $p\in A_1$ the tangent line $L_p$ to $A_1$ at $p$ is also tangent to $A_2$ at some point. Then either $A_1$ is a straight line segment or $A_1\subset A_2$.
Suppose that $L$ is a line which meets $C_2$ transversally, while it meets $C_1$ as in Lemma 1. Assume that $n_{C_1}(L)=n_{C_2}(L)=n$ (otherwise, we would be done). Then, replacing $L$ with a different line $L_1$, parallel to $L$ and sufficiently close to $L$, we obtain:
$$
n_{C_1}(L_1)= n\pm 1, n_{C_2}(L_1)= n.
$$
This would imply  $n_{C_1}(L_1)\ne n_{C_2}(L_1)$! It remains to find a line $L$ intersecting $C_1, C_2$ as described above.
Here is a proof of existence of such $L$ in the case of real-analytic curves $C_1, C_2$. (The real-analyticity assumption cuts down on the number of technicalities in the proof, which is already quite long and does not even includes proofs of Lemmata 1 and 2, but is not really essential.)
For each $p\in C_1$, let $L_p$ denote the line tangent to $C_1$ at $p$. Since $C_1$ is real-analytic, there is a finite subset $E_1\subset C_1$, such that for all $p\in C_1\setminus E_1$,  the line $L_p$ has 1st order of contact with $C_1$ at $p$. It is a priori, possible, that for some $p\in C_1\setminus E_1$, the line $L_p$ is tangent to $C_1$ at another point $q\ne p$. Lemma 2 (applied to pairs of arcs $A_1, A_2$ in $C_1$) implies that the subset $E_2\subset C_1$ of such points $p$ is finite (I again use real-analyticity assumption here): Since $p\ne q$ the option $A_1\subset A_2$ is impossible, while no arc of $C_1$ can be a straight-line segment, since $C_1$ is assumed to be real-analytic.
Now, consider points $p\in C_1\setminus (E_1\cup E_2)$. Applying Lemma 2 again (now, to pairs arcs
$A_1\subset C_1, A_2\subset C_2$) and using real-analyticity one more time, we see that there is a finite subset $E_3\subset C_1$ such that for all $p\in C_1\setminus (E_1\cup E_2\cup E_3)$, the line $L_p$ meets $C_2$ transversally. Indeed, the case of an arc in $C_1$ which is a straight-line segment is ruled out as above, while $A_1\subset A_2$ would imply (by real-analyticity of $C_1, C_2$!) that $C_1=C_2$, which we assumed is not the case.
Thus, for every $p\in C_1\setminus (E_1\cup E_2\cup E_3)$, the line $L_p$ meets $C_2$ transversally, while it meets $C_1$ as in Lemma 1. As was noted above, this proves existence of  a line $L$ such that $n_{C_1}(L)\ne n_{C_2}(L)$. qed


TITLE: Tangent space of a flag manifold?
QUESTION [2 upvotes]: I am studying differential geometry and now I am trying to find the tangent space to a flag manifold $F(a_1,a_2,...,a_k, \mathbb{R^n}).$
Any hint would be greatly appreciated!

REPLY [0 votes]: Let $\mathbb{V}$ be a finite dimensional real (Euclidean) or complex (Hermitian) vector space. Let $G(r ; \mathbb{V})$ denote the Grassmann manifold of $r$ dimensional linear subspaces of $\mathbb{V}$, which is
$$
G(r ; \mathbb{V}):=\{\text { all subspaces } E \subset \mathbb{V} \mid \operatorname{dim}(E)=r\}.
$$
Let $a_{\bullet}$ be an increasing of positive integers
$$
a_{\bullet}=\left(1 \leq a_1<a_2<\cdots<a_k<\operatorname{dim}(\mathbb{V})\right).
$$
Define the flag manifold $F\left(a_{\bullet} ; \mathbb{V}\right)$ by
$$
F\left(a_{\bullet} ; \mathbb{V}\right):=\left\{\left(E_1, \ldots, E_k\right) \in G\left(a_1 ; \mathbb{V}\right) \times G\left(a_2 ; \mathbb{V}\right) \times \cdots \times G\left(a_k ; \mathbb{V}\right) \mid E_i \subset E_{i+1}\right\}.
$$
Let $\left(E_1, \ldots, E_k\right)$ be a point of $F\left(a_{\bullet} ; \mathbb{V}\right)$, and the tangent space to $F\left(a_{\bullet} ; \mathbb{V}\right)$ at this point will be as subspace of
$$
\operatorname{Hom}\left(E_1, E_1^{\perp}\right) \oplus \operatorname{Hom}\left(E_2, E_2^{\perp}\right) \oplus \cdots \oplus \operatorname{Hom}\left(E_k, E_k^{\perp}\right).
$$
To show this we only need basic knowledge of linear algebra.


TITLE: What is the relevance of potential energy as a concept?
QUESTION [0 upvotes]: As far as I understand, for an object to have a fixed quantity of potential energy, it must be in current equilibrium and have no net force applied to it (otherwise the potential would convert to kinetic). For example, I picture a ball on a hill as a classical example.
And as far as I know, potential energy also requires some external perturbation to convert it into kinetic energy (i.e. someone pushing the ball).
But what can be said about potential energy where two fundamental forces are pushing directly against each other? i.e. the ball has rolled to the ground, but it still has a gravitational force vector pushing it downwards, however it is balanced by an equal and opposite electromagnetic force vector. The potential is still there because the force vector exists, but it would be almost impossible to do anything useful with it because of the nature of the equilibrium.
So here is the conclusion I'm making that I'd like to check the correctness of:
Potential energy is just a measure of the force vector applied by a fundamental force on an object, and the amount of "perturbation" that one has to do to turn that energy into kinetic energy is completely arbitrary, thus some potential energy is very functional (requires only a small push to get the ball rolling) whereas other equivalent potential energy would require even more work done on the system to turn the potential into kinetic, thus rendering it "unharnessable".
Edit
I've just realized that since force fields aren't homogeneous, a force vector at a given point wouldn't adequately represent potential energy. But aside from that correction, the underlying inquiry I have still remains.

REPLY [1 votes]: As the name suggests, when something has Potential Energy, it just means that it has the potential to do some work.
The amount of Potential Energy that any body possesses when associated with a force depends on the position of that body. For example, the higher you go, more is your potential to drop and do some 'work'. The work that the body does only depends on its starting and end position. It has nothing to do with the trajectory.
It does not matter if the body is in equilibrium or not. A book kept on your table is also in equilibrium, its weight and normal force cancel out each other. But, it still possesses Potential  Energy because it is at a height from the floor, which is the reference frame that we have chose to have zero potential.
You can apple the same logic to any type of potential energy, gravitational, electric, spring etc.
Force fields don't have to be homogeneous. If the field does some work by moving a body from one point to another in such a way that it gains potential energy, the potential energy can be evaluated as a function between the two end points.