text:
\begin{document}

\title[Contractions of 2-Fanos]{Extremal contractions of 2-Fano fourfolds}

\author{Florian Schrack}
\address{Mathematisches Institut\\ Universität Bayreuth\\ 95440 Bayreuth\\ Germany}
\email{florian.schrack@uni-bayreuth.de}

\subjclass[2010]{Primary 14J45; Secondary 14E30}
\keywords{2-Fano manifolds, fourfolds, extremal contractions, Mori theory}

\date{June 12, 2014}

\begin{abstract}
We consider extremal contractions on smooth Fano fourfolds whose second Chern character is positive. We show that such contractions can neither be of fiber type nor contract a divisor to a point.
\end{abstract}

\maketitle
\section{Introduction}
A Fano manifold~$X$ is called \emph{2-Fano}, if its second Chern character $\ch_2(X) := \ch_2(T_X)$ is positive, i.e., $\ch_2(X).S > 0$ for all $S\in \overline{NE}_2(X)\setminus\{0\}$. We recall that for a vector bundle~$E$, its second Chern character~$\ch_2(E)$ is given by
\[ \ch_2(E) = \tfrac{1}{2}\bigl(c_1^2(E)-2c_2(E)\bigr). \]

2-Fano manifolds have first been introduced by de Jong and Starr in \cite{dJS06} and \cite{dJS07} in order to obtain a natural sufficient condition for the technical notion of \emph{rational simple connectedness}. The property of rational simple connectedness is used in~\cite{dJHS11} to prove a generalization to the surface case of the Graber--Harris--Starr theorem on the existence of sections in families of rationally connected varieties over curves \cite{GHS03}.

The classification of 2-Fano manifolds is still very much an open problem. In dimension~$3$, as shown by Araujo--Castravet in~\cite{AC12}, the only 2-Fano manifolds are $\bP_3$ and $Q_3$. For $n:=\dim X \ge 4$, \cite{AC12} gives a classification of 2-Fano manifolds with index $\ge n-2$, but for arbitrary index, the overall picture remains unclear.

All known examples of $2$-Fanos seem to have second Betti number one, so a natural question to ask is whether it is possible for a $2$-Fano manifold~$X$ to have $b_2(X) \ge 2$.

The present paper studies this question in dimension~$4$, i.e., we let $X$ be a $2$-Fano fourfold and assume that $b_2(X)\ge 2$. Then, by Mori theory, there exists an extremal contraction
\[
f\colon X \to Y,
\]
where $Y$ is a normal projective variety with $1\le \dim Y \le 4$.

We first study the cases where $\dim Y < 4$. In section~\ref{sec:conic} we  exclude the case $\dim Y = 3$ by showing that a 2-Fano manifold $X$ of arbitrary dimension cannot have an extremal contraction to a variety of dimension $\dim X - 1$ (cf.~Proposition~\ref{prop:conicbund}). Sections \ref{sec:p2fib} and \ref{sec:p3q3fib} deal with the cases $\dim Y = 2$ and $\dim Y = 1$, respectively. We thus arrive at the following theorem:
\begin{thm}
Let $X$ be a smooth 2-Fano fourfold and $f\colon X \to Y$ the contraction of an extremal ray in~$\overline{NE}(X)$. Then $f$ is birational.
\end{thm}
Now in the birational case, the situation is more complicated, so we cannot give a complete result. Nevertheless, we do some calculations for blow-ups in section~\ref{sec:bir} and can thus rule out one important case of birational contractions:
\begin{thm}
Let $X$ be a smooth 2-Fano fourfold and $f\colon X \to Y$ a birational contraction of an extremal ray in~$\overline{NE}(X)$. Then $f$ cannot contract a divisor to a point.
\end{thm}
\subsection*{Acknowledgments}
I would like to thank Thomas Peternell for bringing the subject at hand to my attention and for many useful discussions and suggestions.

\section{Conic bundles}\label{sec:conic}
Let $Z$, $U$ be smooth varieties and $g\colon Z \to U$ a conic bundle (i.e., a proper morphism such that each fiber of~$g$ is isomorphic to a conic in~$\bP_2$). Then there exists an effective divisor $\Delta_g \subset U$ such that $Z_u := g^{-1}(u)$ is smooth if and only if $u\not\in\Delta_g$.
\begin{lem}\label{lem:conicbund}
In the situation given above, for any smooth proper curve $C\subset U$ such that $S:=g^{-1}(C)$ is smooth, we have
\[ \ch_2(T_Z|_S) = -\tfrac{3}{2} \deg \Delta_g|_C \le 0.\]
\end{lem}
\begin{proof}
Since $g$ is flat, we have $N^*_{S|Z}=(g|_S)^*N^*_{C|U}$ and thus $\ch_2(N^*_{S|Z}) = (g|_S)^*\ch_2(N^*_{C|U})$, which is zero because $\dim C=1$.
So from the exact sequence
\[
0 \longrightarrow T_S \longrightarrow T_Z|_S \longrightarrow N_{S|Z} \longrightarrow 0,
\]
we obtain
\[
\ch_2(T_Z|_S) = \ch_2(S).
\]
This shows that we can assume from now on that $U=C$ and $Z=S$.

We now consider the rank-3 vector bundle $E:= g_*\sO_Z(-K_Z)$ and the projective bundle
\[
\pi\colon \bP(E) \to U.
\]
An easy calculation shows that $Z$ can be embedded into~$\bP(E)$ as a divisor
\[
Z \in \lvert \sO_{\bP(E)}(2) + \pi^*(\det E^* -K_U ) \rvert.
\]
Equivalently, $Z$ is given by a section
\[
s \in H^0(S^2E\otimes\det E^*\otimes \sO_U(-K_U)) \subset \Hom(E^*, E\otimes\det E^*\otimes \sO_U(-K_U)).
\]
From this, it follows that
\[
\Delta_g \in \lvert \det E^* -3K_U\rvert.
\]
On the other hand, by \cite[Lem.~4.1]{dJS06}, we have
\[
\ch_2(\bP(E)) = {\tfrac{3}{2}}\xi^2 + \pi^*c_1(E^*).\xi,
\]
where $\xi := c_1(\sO_{\bP(E)}(1))$. It follows that
\[
\begin{split}
\ch_2(Z) &= \bigl(\ch_2(\bP(E)) - \tfrac{1}{2}(2\xi + \pi^*c_1(E^*)+\pi^*c_1(U))^2\bigr)|_Z\\
&= \bigl(-\tfrac{1}{2}\xi^2-\pi^*c_1(E^*).\xi-2\pi^*c_1(U).\xi\bigr)\bigl(2\xi+\pi^*c_1(E^*)+\pi^*c_1(U)\bigr)\\
&= -\tfrac{3}{2}c_1(E^*)-\tfrac{9}{2}c_1(U) = -\tfrac{3}{2} \deg \Delta_g.
\end{split}
\]
Since $Z$ and $U$ are smooth, the general fiber of~$g$ is smooth, so $\Delta_g$ is effective, i.e., $\deg\Delta_g\le 0$.
\end{proof}
\begin{prop}\label{prop:conicbund}
Let $X$ be a smooth projective variety with $n:=\dim X \ge 2$ and $f\colon X \to Y$ the contraction of an extremal ray on~$X$. Suppose $\dim Y = n-1$. Then there exists a smooth surface $S\subset X$ such that $\ch_2(X).S \le 0$.
\end{prop}
\begin{proof}
There exists a subvariety $A\subset Y$ of codimension $\ge 2$ such that if we let $U:=Y\setminus A$, then $U$ is smooth and
\[
g := f|_{f^{-1}(U)}\colon f^{-1}(U) \to U
\]
is equidimensional. By~\cite[Thm.~3.1]{And85} (cf.~\cite[Prop.~2.2]{Kac97}), this implies that $g$ is a conic bundle. Now if we take $n-2$ general ample divisors $H_1$, $\dots$, $H_{n-2}$ on $Y$, then $H_1\cap \dots \cap H_{n-2}$ is contained in~$U$ and both $H_1\cap \dots \cap H_{n-2}$ and
\[
S:=g^{-1}(H_1\cap \dots \cap H_{n-2})
\]
are smooth. Then we have $\ch_2(X).S=\ch_2(T_{f^{-1}(U)}|_S)\le 0$ by Lemma~\ref{lem:conicbund}.
\end{proof}

\section{$\bP_2$-fibrations}\label{sec:p2fib}
We now consider extremal contractions
\[
f\colon X \to Y
\]
where $X$ is a 2-Fano fourfold and $Y$ is a surface. We will show in this section that the above situation cannot occur.

We start by observing that a general fiber~$F$ of~$f$ is a smooth surface with $N_{F|X}\cong \sO_{F}^{\oplus2}$, so $-K_F=-K_X|_F$ is ample and $\ch_2(F) = \ch_2(X).S > 0$. In other words, $F$ is 2-Fano, and thus $F \cong \bP_2$.

By~\cite[Cor.~(1.4)]{AW97}, $Y$ is smooth. So, since $X$ is Fano, $Y$ must be a blow-up of either~$\bP_2$ or a Hirzebruch surface. In particular, there exists a base point free linear system on~$Y$ whose general element~$\ell$ is isomorphic to~$\bP_1$.

If we consider the preimage $X_\ell := f^{-1}(\ell)$, then $X_\ell$ is a smooth threefold with the property that $\ch_2(X_\ell).S > 0$ for any $S\in \overline{NE}_2(X_\ell)\setminus\{0\}$.

We now apply the following lemma:
\begin{lem}\label{lem:p2fib}
Let $g\colon Z \to T$ be a surjective morphism from a smooth projective threefold~$Z$ to a smooth curve~$T$.
Assume that
\begin{enumerate}
\item $-K_Z$ is $g$-ample,
\item $\ch_2(Z).S > 0$ for all irreducible surfaces $S\subset Z$ and
\item a general fiber of~$g$ is isomorphic to~$\bP_2$.
\end{enumerate}
Then any singular fiber~$F_{\mathrm{sing}}$ of~$g$ can (as a divisor) be decomposed as
\[
F_{\mathrm{sing}} = F_1 + F_2 + F_3,
\]
where $F_i\cong \bF_1$ for $i=1$, $2$, $3$ and $F_i\cap F_j$ is a line for $i\ne j$ (either a fiber of the ruling or the exceptional section of~$\bF_1$).

More precisely, there is a finite sequence of birational morphisms
\[ Z = Z_0 \stackrel{\alpha_1}{\longrightarrow} Z_1 \stackrel{\alpha_2}{\longrightarrow}
\dots \stackrel{\alpha_s}{\longrightarrow} Z_s,\]
where $Z_s$ is a $\bP_2$-bundle over~$T$ and each~$\alpha_i$ is given as the composition
\[ \alpha_i = \gamma_i\circ\beta_i,\]
where $\beta_i$ is the blow-up of a line in a smooth fiber of~$Z_i$ over $T$ and $\gamma_i$ is the blow-up of a non-trivial fiber of~$\beta_i$.

\end{lem}
\begin{proof}
The result is a corollary of the more general classification of Del~Pezzo fibrations over curves carried out by T.~Fujita in~\cite{Fuj90}. In our special case, the proof is simplified considerably by ruling out most cases using the positivity assumption on the second Chern character as we will do in the following.

We let $F_{\mathrm{sing}}$ be a singular fiber of~$g$. Then by~\cite[Thm.~5]{ARM12}, $F_{\mathrm{sing}}$ is reducible. So, as a divisor, we can decompose $F_{\mathrm{sing}}$ as
\[
F_{\mathrm{sing}} = \sum_{j=1}^r m_jF_j,
\]
where $r\ge2$, the $F_j$ are the irreducible components of~$F_{\mathrm{sing}}$ and $m_j\ge1$. Then we have the following
\begin{lem}\label{lem:compextr}
For any irreducible component $F_j$ of $F_{\mathrm{sing}}$, there exists an irreducible curve $C_j\subset F_j$ with $C_j.F_j < 0$ whose numerical class~$[C_j]$ is extremal in~$\overline{NE}(Z/T)$.
\end{lem}
\begin{proof}
Since the fiber $F_{\mathrm{sing}}$ is connected, the irreducible component~$F_j$ meets another irreducible component~$F_k$ with $k\ne j$. Let $\tilde{C}\subset F_j$ be a curve meeting $F_k$ but not contained in any $F_i$ for $i\ne j$. Then clearly
\[
F_k.\tilde{C} > 0\qquad\text{and}\qquad F_i . \tilde{C} \ge 0\quad\text{for $i\ne j$}.
\]
Since $F_{\mathrm{sing}}.\tilde{C} = 0$, this implies
\[
F_j.\tilde{C} < 0.
\]
Now since $-K_Z$ is $g$-ample, by the relative cone theorem, $\overline{NE}(Z/T)$ is generated by the classes of finitely many irreducible extremal curves, so in particular, we can write
\[
\tilde{C} \equiv \sum_{l=1}^m \alpha_l \tilde{C}_l,
\]
where the $\tilde{C}_l$ are irreducible extremal curves and $\alpha_l >0$.
From $F_j.\tilde{C} < 0$, it then follows that there exists an $l\in\{1,\dots, m\}$ such that $F_j.\tilde{C}_l < 0$ and thus $C_j:=\tilde{C}_l \subset F_j$.
\end{proof}

By the relative contraction theorem, there exists for each~$j$ a relative Mori contraction of~$g$ over~$T$ contracting the $C_j\subset F_j$ constructed in the preceding lemma, i.e., a Mori contraction $\phi_j\colon Z \to Z'_j$, where $Z'_j$ is a normal projective variety, and a morphism $g'_j\colon Z'_j \to T$ such that $g_j'\circ \phi_j = g$.

If $\phi_j$ contracts any curve contained in a general fiber of~$g$, then $\phi_j = g$, thus $g$ is itself a Mori contraction and thus a $\bP_2$-bundle by~\cite[Thm.~3.5]{Mor82}. This is a contradiction to the reducibility of~$F_{\mathrm{sing}}$, so $\phi_j$ must be birational.

Thus, by~\cite[Thm.~3.3]{Mor82}, there is an irreducible divisor~$D_j$ on~$Z$ contained in a fiber of~$g$ such that $\phi_j$ contracts~$D_j$ and $\phi_j$ is an isomorphism on~$Z\setminus D_j$. For any $i\ne j$, any curve $C'\subset F_i$ with $C'\not\subset F_j$ satisfies $C'.F_j\ge 0$, so $C'$ is not numerically proportional to~$C_j$. In particular, $\phi_j$ cannot contract $F_i$. Thus we have shown that $D_j = F_j$, so again by~\cite[Thm.~3.3]{Mor82}, we obtain the following types of possible fiber components~$F_j$:
\begin{enumerate}
\item[(T1)] $F_j \cong \bP_2$ with normal bundle $N_{F_j|Z}\cong \sO_{\bP_2}(-k)$, $k\in\{1,2\}$,
\item[(T2)] $F_j \cong \bP_1\times\bP_1$ with $N_{F_j|Z} \cong \sO_{\bP_1\times\bP_1}(-1,-1)$,
\item[(T3)] $F_j \cong Q$, where $Q\subset\bP_3$ is a quadratic cone and $N_{F_j|Z}\cong \sO_{\bP_3}(-1)|_Q$,
\item[(T4)] $Z_j'$ is smooth and there exists a smooth curve $C_j'\subset Z_j'$ such that $\phi_j$ is the blow-up of~$C_j'$. Thus, $F_j \cong \bP(N_{C_j'|Z_j'}^*)$ and $N_{F_j|Z}\cong \sO_{\bP(N_{C_j'|Z_j'}^*)}(-1)$.
\end{enumerate}
We further note that in types (T1), (T2) and (T3), the fiber component~$F_j$ is mapped to a point by the contraction morphism~$\phi_j$.

By straightforward computations, we can obtain the intersection number $\ch_2(Z).F_j$ in each of these cases:
\begin{enumerate}
\item[(T1)] $\ch_2(Z).F_j = \tfrac{3}{2} + \tfrac{1}{2}k^2$,
\item[(T2)] $\ch_2(Z).F_j = 1$,
\item[(T3)] $\ch_2(Z).F_j = 1$,
\item[(T4)] $\ch_2(Z).F_j = -\tfrac{1}{2} \deg N_{C_j'|Z_j'}$.
\end{enumerate}
By assumption, we have $\ch_2(Z).F_j>0$ for all~$j$. Furthermore, since
$\ch_2(Z).F_{\mathrm{sing}} = \ch_2(\bP_2) = \tfrac{3}{2}$, we have
\[
\sum_{j=1}^r m_i \ch_2(Z).F_j = \tfrac{3}{2}.
\]
Combining this with the above calculations, we see that only the following possibilities remain (up to re-numbering the~$F_j$):
\begin{enumerate}
\item[(P1)] $r=2$, $m_1 = 1$, $F_1$ is of type~(T4) with $\deg N_{C_1'|Z_1'} = -1$, and $m_2 \in \{1, 2\}$ with $F_2$ of type (T2), (T3) or (T4).
\item[(P2)] $r=3$, $m_1 = m_2 = m_3 = 1$, and each $F_j$ is of type~(T4) with $\deg N_{C_j'|Z_j'} = -1$.
\end{enumerate}
We now show that possibility~(P1) cannot occur. We assume that a fiber~$F_{\mathrm{sing}}$ of~$g$ satisfies~(P1). We consider the contraction~$\phi_1\colon Z\to Z_1'$ constructed above and let $F_{\mathrm{sing}}'=\phi_1(F_{\mathrm{sing}})$. We know that $Z_1'$ is smooth, which implies that $g_1'$ is flat. Since $F_{\mathrm{sing}}'$ is irreducible, we can apply~\cite[Thm.~5]{ARM12} to conclude that $F_{\mathrm{sing}}'\cong\bP_2$. But now $\phi_1$ is the blow-up of a smooth curve $C_1'\subset F_{\mathrm{sing}}'$, so in particular, $F_2$, which is the strict transform of $F_{\mathrm{sing}}'$ in~$Z$, is isomorphic to~$F_{\mathrm{sing}}'\cong\bP_2$, contradicting (T2), (T3) and (T4).

It remains to study possibility~(P2) in greater detail. We let $F_{\mathrm{sing}}=F_1+F_2+F_3$ be a fiber of~$g$ that satisfies~(P2). We first note that since $-K_Z$ is $g$-ample, we have
\[ F_i.F_j.(-K_Z) \ge 0\quad\text{for $i\ne j$} \]
and since $F$ is connected, we can for each~$j$ even find an $i_0\ne j$ such that
\[ F_{i_0}.F_j.(-K_Z) > 0. \]
Together with $F_j.(F_1+F_2+F_3) = 0$, this implies that
\begin{equation}\label{eq:negselfint}
F_j^2.(-K_Z) < 0 \quad\text{for $j=1$, $2$, $3$.}
\end{equation}

By applying Kodaira vanishing and using~(T4) and the adjunction formula, we obtain
\begin{equation}\label{eq:vaneuler}
\chi(N_{F_j|Z}) = \chi(K_{F_j} - K_Z|_{F_j}) = 0.
\end{equation}
Riemann-Roch yields
\[ \chi(N_{F_j|Z}) = \tfrac{1}{2}(F_j|_{F_j})^2-\tfrac{1}{2}K_{F_j}.F_j|_{F_j} + \chi(\sO_{F_j}) = \tfrac{1}{2}F_j^2.(-K_Z) + \chi(\sO_{F_j}).\]
Applying \eqref{eq:negselfint} and \eqref{eq:vaneuler}, we obtain
\[ \chi(\sO_{F_j}) > 0. \]
Since obviously $h^2(\sO_{F_j}) = h^0(K_{F_j}) = 0$, we conclude that $h^1(\sO_{F_j}) = 0$ and thus the curve $C_j'\subset Z_j'$ from~(T4) must be isomorphic to~$\bP_1$.

We now let $\tilde{F}_j=\phi_j(F_{\mathrm{sing}})$. Then we have the following normal bundle sequence:
\begin{equation}\label{eq:normbund}
0 \longrightarrow N_{C_j'|\tilde{F}_j} \longrightarrow N_{C_j'|Z_j'} \longrightarrow N_{\tilde{F}_j|Z_j'}|_{C_j'} \longrightarrow 0.
\end{equation}
But now $N_{\tilde{F}_j|Z_j'}$ is trivial, so from $\deg N_{C_j'|Z_j'}=-1$, we infer that $N_{C_j'|\tilde{F}_j}\cong\sO_{\bP_1}(-1)$ and hence
\[ N_{C_j'|Z_j'} \cong \sO_{\bP_1}(-1) \oplus \sO_{\bP_1}. \]
So in particular, $F_j \cong \bF_1$.

We consider again the contraction $\phi_1\colon Z \to Z_1'$, which is the blow-up of a smooth curve $C_1'\subset Z_1'$. If we let $F_2'$ and $F_3'$ be the images of $F_2$ and $F_3$ via $\phi_1$, respectively, we can assume that $C_1'\subset F_2'$. Then $F_2$ is mapped isomorphically onto $F_2'$ by~$\phi_1$.

We now want to apply Lemma~\ref{lem:compextr} to~$Z_1'$ in order to find a contraction morphism $\psi_2\colon Z_1' \to Z_2''$ which contracts~$F_2'$. In order to do this, we have to show that $-K_{Z_1'}$ is $g_1'$-ample. Since $\overline{NE}(Z/T)$ is generated by finitely many curves, the same is true for~$\overline{NE}(Z_1'/T)$. So it suffices to show that $-K_{Z_1'}.C > 0$ for any curve $C$ contained in a fiber of~$g_1'$. But now since $\phi_1$ is the blow-up of~$C_1'$, we have
\[ K_Z = \phi_1^*K_{Z_1'} + F_1, \]
so $-K_{Z_1'}.C > 0$ is certainly true for all irreducible $C \ne C_1'$ since such $C$ can be obtained as the image of a curve $\tilde{C}\subset Z$ with $\tilde{C} \not\subset F_1$. But now also $-K_{Z_1'}.C_1' > 0$, which one can see as follows: Let $C_e \subset Z$ be the exceptional section of $F_1 \to C_1'$. Then $\phi_{1*}C_e = C_1'$ and $K_{F_1}|_{C_e} \cong \sO_{\bP_1}(-1)$, hence by adjunction
\[ F_1.C_e = -K_Z.C_e - 1 \ge 0 \]
and thus $-K_{Z_1'}.C_1' > 0$ as in the previous case.

So by Lemma~\ref{lem:compextr}, we obtain a relative contraction morphism $\psi_2\colon Z_1' \to Z_2''$ contracting $F_2'$ as desired. We let $g_2''\colon Z_2'' \to T$ be the induced morphism. Since the exceptional divisor of $\psi_2$ is $F_2' \cong \bF_1$ we conclude by~\cite[Theorem~3.3]{Mor82} that $Z_2''$ is smooth and $\psi_2$ is the blow-up of a smooth curve~$C_2''\subset Z_2''$. By the same argument as in the discussion of possibility~(P1) above, we see that the fiber of $g_2''$ over $g(F_{\mathrm{sing}})\in T$ must be isomorphic to~$\bP_2$. If we consider the normal bundle sequence for $C_2''\subset \psi_2(\phi_1(F_{\mathrm{sing}})) \subset Z_2''$ (cf.~sequence~\eqref{eq:normbund}), we can conclude from $\bF_1 \cong F_2' \cong \bP(N_{C_2''|Z_2''})$ that $C_2''$ must be a line in~$\bP_2$.

We conclude by showing that $C_1'$ must be a fiber of $F_2'\to C_2''$. We first note that by adjunction, we have
\[
K_{Z_1'}.C_1' = \deg K_{C_1'} - \deg N_{C_1'|Z_1'} = -1.
\]
Furthermore, obviously $F_3'=\phi_1(F_3)$ is the strict transform of $\psi_2(F_3')\cong \bP_2$, so $F_3'\cong\bP_2$. Since $F_3\cong\bF_1$ is the blow-up of $F_3'$ in $C_1'\cap F_3'$, we conclude
that $F_3'.C_1'=1$ and thus $F_2'.C_1'=-1$ since $(F_2'+F_3').C_1'=0$.
Finally, since $K_{Z_1'} = \psi_2^*K_{Z_2''}+F_2'$, it follows that $\psi_2^*K_{Z_2''}.C_1' = 0$, from which we conclude that $\psi_2(C_1')$ must be a point since $-K_{Z_2''}$ is obviously $g_2''$-ample.
\end{proof}

We now apply Lemma~\ref{lem:p2fib} to~$X_\ell$. We get a finite sequence
\[ X_\ell = Z_0 \stackrel{\alpha_1}{\longrightarrow} Z_1 \stackrel{\alpha_2}{\longrightarrow}
\dots \stackrel{\alpha_s}{\longrightarrow} Z_s\]
of birational morphisms as described in the lemma, where $Z_s \to \ell\cong\bP_1$ is a $\bP_2$-bundle. We now apply to~$Z_s$ the following lemma:

\begin{lem}\label{lem:projbundp1}
Let $E$ be a vector bundle on~$\bP_1$ of rank~$r\ge 2$. Then there exists a rank-$2$ quotient bundle $F$ of $E$ such that the surface
\[ S := \bP(F) \subset \bP(E) \]
satisfies $\ch_2(\bP(E)).S \le 0$.
\end{lem}
\begin{proof}
By Grothendieck, there exist integers $a_1 \le \dotsb \le a_r$ such that
\[ E \cong \sO_{\bP_1}(a_1) \oplus \dotsb \oplus \sO_{\bP_1}(a_r). \]
We consider the quotient
\[ E \twoheadrightarrow \sO_{\bP_1}(a_1) \oplus \sO_{\bP_1}(a_2) =: F \]
given by the projection onto the first two summands of~$E$. Then we have a natural inclusion $S:=\bP(F) \subset \bP(E)$ and if we let $\zeta := \sO_{\bP(E)}(1)|_{\bP(F)}$, the normal bundle is given by
\[ N_{S|\bP(E)} \cong \zeta \otimes \pi^*{E^*/F^*} \cong \zeta \otimes \pi^*\bigl(\sO_{\bP_1}(-a_3) \oplus \dotsb \oplus \sO_{\bP_1}(-a_r)\bigr),\]
where $\pi\colon S=\bP(F) \to \bP_1$ denotes the natural projection. We now consider the sequence
\[ 0 \longrightarrow T_S \longrightarrow T_{\bP(E)}|_S \longrightarrow N_{S|\bP(E)} \longrightarrow 0. \]
Since $F$ has rank~$2$, we have $\ch_2(T_S) = 0$ by~\cite[Prop.~4.3]{dJS06}, so we obtain from the above sequence:
\[ \begin{split}
\ch_2(\bP(E)).S &= \ch_2(N_{S|\bP(E)})\\
&= \tfrac{1}{2}\zeta^2\cdot(r-2) + \zeta.\pi^*\sO(-a_3-\dotsb-a_r)\\
&= \tfrac{1}{2}(r-2)(a_1+a_2)-a_3-\dotsb-a_r \le 0 \qedhere
\end{split} \]
\end{proof}

We let $S_s := S\subset Z_s$ be the surface constructed in Lemma~\ref{lem:projbundp1}. We denote by $S_k$, $k=0$, $\dotsc$, $s-1$, the strict transform of~$S$ inside~$Z_k$.
\begin{claim}\label{cl:line}
For any $k=0$, $\dotsc$, $s$, the intersection of $S_k$ with any smooth fiber of~$Z_k$ over~$\ell$ is a line in~$\bP_2$. Furthermore, $\ch_2(Z_k).S_k \le 0$.
\end{claim}
To prove Claim~\ref{cl:line}, we argue by induction over~$k$. The case $k=s$ follows from Lemma~\ref{lem:projbundp1}. Now we assume by induction that Claim~\ref{cl:line} is true for some~$k\ge1$. We consider the birational morphism $\alpha_k\colon Z_{k-1}\to Z_k$, which by Lemma~\ref{lem:p2fib} can be decomposed as
\[ \alpha_k = \gamma_k\circ\beta_k, \]
where $\beta_k\colon \tilde{Z}_k \to Z_k$ is the blow-up of a line $L_1$ contained in a smooth fiber of~$Z_k$ over~$\ell$ and $\gamma_k\colon Z_{k-1}\to\tilde{Z}_k$ is the blow-up of a one-dimensional fiber $L_2\cong\bP_1$ of~$\beta_k$. We let $\tilde{S}_k$ be the strict transform of $S_k$ in~$\tilde{Z}_k$ and consider different cases.

The first case is that $L_1 \not\subset S_k$. Then $L_1 \cap S_k$ is a simple point by the induction hypothesis, so in particular $\tilde{S}_k$ is the blow-up of~$S_k$ in the point~$L_1 \cap S_k$. Furthermore, by~\cite[Lem.~4.13(i)]{AC12},
\[ \ch_2(\tilde{Z}_k).\tilde{S}_k = \ch_2(Z_k).S_k-\tfrac{3}{2} < 0. \]
Now if we blow up $L_2$ in this case, we either have $L_2 \not\subset\tilde{S}_k$, whence
\[ \ch_2(Z_{k-1}).S_{k-1} \le \ch_2(\tilde{Z}_k).\tilde{S}_k < 0 \]
by~\cite[Lem.~4.13(i)]{AC12}, or we have $L_2 \subset \tilde{S}_k$. But then $L_2$ is a $(-1)$-curve in~$\tilde{S}_k$ by the above considerations, so
\[ \ch_2(Z_{k-1}).S_{k-1} = \ch_2(\tilde{Z}_k).\tilde{S}_k - \tfrac{3}{2} + 1 < 0 \]
by~\cite[Lem.~4.13(ii)]{AC12}.

It remains to consider the case $L_1\subset S_k$. Then $L_1$ is a fiber of~$S_k$ over~$\ell$, so by~\cite[Lem.~4.13(ii)]{AC12},
\[
\ch_2(\tilde{Z}_k).\tilde{S}_k = \ch_2(Z_k).S_k - 1 < 0.
\]
But now since $L_1\subset S_k$, the blow-up morphism~$\beta_k$ maps $\tilde{S}_k$ isomorphically onto~$S_k$, so in particular $L_2 \not\subset \tilde{S}_k$. This implies by~\cite[Lem.~4.13(i)]{AC12} that
\[ \ch_2(Z_{k-1}).S_{k-1} \le \ch_2(\tilde{Z}_k).\tilde{S}_k < 0. \]
Thus we have inductively constructed an irreducible (smooth) surface $S \subset X_\ell = Z_0$ with $\ch_2(X_\ell).S \le 0$ (even $\ch_2(X_\ell).S < 0$ if $X_\ell$ is not a $\bP_2$-bundle). Now since $N_{X_\ell|X}=(f|_{X_\ell})^*N_{\ell|Y}$, we have $\ch_2(N_{X_\ell|X})=0$, so it follows that
\[ \ch_2(X).S \le 0, \]
therefore $X$ is not $2$-Fano.

\section{$\bP_3$- and $Q_3$-fibrations}\label{sec:p3q3fib}
In this section we deal with the case of a smooth projective fourfold~$X$ admitting an extremal contraction
\[f\colon X\to C\]
over a curve~$C$. Then if $X$ is 2-Fano, a general fiber~$F$ of~$f$ must also be 2-Fano (cf.~the beginning of section~\ref{sec:p2fib}), so either $F\cong\bP_3$ or $F\cong Q_3$ (where $Q_3 \subset \bP_4$ is a smooth quadric hypersurface) by~\cite[Thm.~1.2]{AC12}. In the first case we easily obtain:
\begin{prop}
In the above situation, assume that the general fiber of~$f$ is $\bP_3$. Then $X$ is not 2-Fano.
\end{prop}
\begin{proof}
Since the relative Picard number of~$f$ is $1$, every fiber of~$f$ is irreducible. By~\cite[Thm.~5]{ARM12}, this implies that $f$ is a $\bP^3$-bundle. Since $C$ is a curve, there exists a rank-4 vector bundle on~$C$ such that $X\cong \bP(E)$. The proposition now follows from~\cite[Cor.~4.6]{dJS06}.
\end{proof}

It remains to consider the case where $F\cong Q_3$:
\begin{prop}
In the above situation, assume that the general fiber of~$f$ is $Q_3$. Then $X$ is not $2$-Fano.
\end{prop}
\begin{proof}
Since the relative Picard number of~$f$ is $1$, every fiber of~$f$ is irreducible. By \cite[Prop.~21]{Ara09}, this implies that $f$ is a \emph{geometric quadric bundle,} i.e., there exists a rank-5 bundle $E$ on~$C$ such that $X$ embeds into~$\bP(E)$ as a divisor of relative degree~$2$. Now if we assume $X$ to be $2$-Fano, $-K_X$ is ample, so for $m \gg 0$, if we take two general divisors $H_1$, $H_2\in\lvert -mK_X \rvert$, the intersection
\[
S:= H_1\cap H_2
\]
is a smooth surface.

We will now calculate $\ch_2(X).S$. We denote by $\pi\colon \bP(E) \to C$ the natural projection and let $\xi := \sO_{\bP(E)}(1)$. Then
\[
-K_{\bP(E)} = 5\xi+\pi^*(\det E^* -  K_C)
\]
and
\[
\ch_2(\bP(E)) = \tfrac{5}{2}\xi^2+\pi^*c_1(E^*).\xi
\]
(cf.~\cite[Lem.~4.1]{dJS06}).

Since $f$ is a geometric quadric bundle, there exists a line bundle $L$ on~$C$ such that
\[
X \in \lvert 2\xi + \pi^* L \rvert.
\]
In particular,
\[
-K_X = \bigl(3\xi+\pi^*(\det E^* -  L - K_C)\bigr)|_X
\]
and
\[
\begin{split}
\ch_2(X) &= \bigl(\tfrac{5}{2}\xi^2+\pi^*c_1(E^*).\xi-\tfrac{1}{2}(2\xi+\pi^*L)^2\bigr)|_X\\
&= \bigl(\tfrac{1}{2}\xi^2+\pi^*(c_1(E^*)-2L).\xi\bigr)|_X.
\end{split}
\]
We thus obtain
\[
\begin{split}
\ch_2(X).S &= \ch_2(X).(-mK_X)^2\\
&= m^2\Bigl(
\bigl(\tfrac{1}{2}\xi^2+\pi^*(c_1(E^*)-2L).\xi\bigr).\bigl(3\xi+\pi^*(c_1(E^*) -  L - K_C)\bigr)^2
\Bigr)|_X\\
&= m^2
\bigl(
\tfrac{9}{2}\xi^4 +\pi^*(12c_1(E^*)-21L-3K_C).\xi^3
\bigr).(2\xi+\pi^*L)\\
&= m^2\bigl(9\xi^5 + \pi^*(24c_1(E^*)-\tfrac{75}{2}L-6K_C).\xi^4\bigr)\\
&= m^2(15\deg E^* -\tfrac{75}{2}\deg L - 6 \deg K_C).
\end{split}
\]
Since $X$ is given by a section
\[
s\in H^0(S^2E\otimes L)\subset \Hom(E^*,E\otimes L),
\]
the discriminant locus of~$f$ is a divisor
\[
\Delta_f\in \lvert 2\det E+5 L\rvert.
\]
We can thus rewrite the result of the above calculation as
\[
\ch_2(X).S=-\tfrac{15}{2}m^2(\deg \Delta_f+\tfrac{4}{5}\deg K_C).
\]
We now argue by contradiction: We assume that $\ch_2(X).S >0$. Then since $\Delta_f$ is effective, we must have $\deg K_C <0$, i.e., $C\cong \bP_1$, and so
\[
\deg\Delta_f\in\{0,1\}.
\]
By tensorizing $E$ with a suitable line bundle, we can assume
\[
0\le \deg E \le 4.
\]
In the case $\deg\Delta_f=0$, we thus obtain $\deg E = \deg L = 0$. We first show that the line bundle
\[
\xi \otimes \pi^*\sO_{\bP_1}(-1)
\]
cannot have a section: Suppose the contrary, then we get an effective Divisor
\[
D \in \lvert \xi - \pi^*\sO_{\bP_1}(1) \rvert
\]
on $X$. Intersecting $D$ with a general member of~$\lvert -mK_X \rvert$ gives a surface in~$X$ and we have
\[
\begin{split}
\ch_2(X).D.(-mK_X) &= m\xi^3.(\xi-\pi^*\sO_{\bP_1}(1)).(3\xi+\pi^*\sO_{\bP_1}(2))\\
&= m\bigl(3\xi^5 - \pi^*\sO_{\bP_1}(1).\xi^4\bigr) = -m,
\end{split}
\]
which contradicts $X$ being $2$-Fano. So $H^0\bigl((\xi\otimes\pi^*\sO_{\bP_1}(-1))|_X\bigr)=0$ and from the exact sequence
\[
0 \longrightarrow \xi^*\otimes\pi^*\sO_{\bP_1}(-1) \longrightarrow \xi\otimes\pi^*\sO_{\bP_1}(-1)  \longrightarrow (\xi\otimes\pi^*\sO_{\bP_1}(-1))|_X \longrightarrow 0
\]
we conclude that also $H^0(\xi\otimes\pi^*\sO_{\bP_1}(-1))=0$. But this means that
\[
H^0(E\otimes\sO_{\bP_1}(-1)) = 0.
\]
Since $E$ splits as a direct sum of line bundles, we conclude that $E\cong \sO_{\bP_1}^{\oplus 5}$ and thus $X\cong \bP_1 \times Q_3$ is not $2$-Fano.

In the case $\deg\Delta_f = 1$, we have $\deg E = 3$ and $\deg L = -1$, so we obtain
\[
-K_X = 3 \xi|_X,
\]
thus $X$ has index~$3$, so $X$ is not $2$-Fano by~\cite[Thm.~1.3]{AC12}.
\end{proof}

\section{The birational case}\label{sec:bir}
In this section we study the case of a birational extremal contraction
\[
f\colon X \to Y,
\]
where $X$ is a $2$-Fano fourfold. Unfortunately, we do not have a general classification result for this situation. We start by studying the blow-up of a smooth surface inside a smooth fourfold:
\begin{lem}\label{lem:blowupsurf}
Let $p \colon X \to Z$ be the blow-up of a smooth fourfold~$Z$ along a smooth surface $W \subset Z$. Then, for any smooth curve $C \subset W$, the smooth surface $S:= p^{-1}(C) \subset X$ satisfies
\[
\ch_2(X).S = - \tfrac{1}{2}\deg N_{W|Z}|_C.
\]
\end{lem}
\begin{proof}
Since $S$ is a $\bP_1$-bundle, we have $\ch_2(S)=0$ by~\cite[Prop.~4.3]{dJS06}.
The sequence
\[
0 \longrightarrow T_S \longrightarrow T_{X}|_S \longrightarrow N_{S|X} \longrightarrow 0.
\]
then implies that
\[ \ch_2(X).S = \ch_2(N_{S|X}). \]
Now if we denote by~$E$ the exceptional divisor of~$p$, we have an exact sequence
\[
0 \longrightarrow N_{S|E} \longrightarrow N_{S|X} \longrightarrow N_{E|X}|_S \longrightarrow 0.
\]
Since $N_{S|E} = (p|_S)^*N_{C|W}$, we have $\ch_2(N_{S|E}) = 0$, so it remains to calculate $\ch_2(N_{E|X}|_S)$. To this end, we observe that $N_{E|X} = \sO_E(-1)$, where $\sO_E(1)$ is the tautological line bundle on ~$E\cong \bP(N_{W|Z}^*)$. Thus we obtain
\[
\ch_2(N_{E|X}|_S) = \tfrac{1}{2} (\sO_E(-1))^2.((p|_E)^*\sO_W(C)) = \tfrac{1}{2} c_1(N_{W|Z}^*).c_1(\sO_W(C)).\qedhere
\]
\end{proof}

We can use this Lemma to obtain the following result:
\begin{prop}
Let $f\colon X \to Y$ be a birational divisorial extremal contraction which maps its exceptional divisor $D\subset X$ to a point in~$Y$. Then $X$ is not $2$-Fano.
\end{prop}
\begin{proof}
We assume that $X$ is $2$-Fano.

As in~\cite[Prop.~5]{Tsu06}, we can choose an extremal ray $\mathbb{R}_+[C_0]$ on $X$ such that $D.C_0 > 0$. Then by the proof of~\cite[Prop.~5]{Tsu06} (cf.~also \cite[Prop.~3.1]{Cas09}), the associated extremal contraction $g\colon X \to Z$ is either a conic bundle or the blow-up of a smooth surface inside a smooth Fano fourfold. We already showed in Proposition~\ref{prop:conicbund} that the conic bundle case cannot occur, so we conclude that $Z$ must be a Fano fourfold and $g$ is the blow-up of some smooth surface $W \subset Z$.

We denote by $E\subset X$ the exceptional divisor of~$g$. Then we have $E \cong \bP(N_{W|Z}^*)$ with the tautological line bundle $\sO_E(1) := \sO_{\bP(N_{W|Z}^*)}(1)$ and the natural projection map
\[
\pi\colon E \to W.
\]

In the case $\rho(X) \ge 3$ this situation has been studied in~\cite{Fuj12}. Fujita gives a very explicit description of the possible configurations that can occur, but for our purposes it is sufficient to notice that from his Thm.~1.1 it easily follows that $\det N_{W|Z}$ is always ample, which contradicts our Lemma~\ref{lem:blowupsurf}.

So for the rest of the proof we can assume that $\rho(X)=2$. We consider the scheme-theoretic intersection
\[ V := D \cap E.\]
\begin{claim}\label{cl:red}
$V$ is a reduced section of~$\pi$, i.e., $\pi|_V$ is an isomorphism.
\end{claim}
If we assume Claim~\ref{cl:red} to be true, we immediately obtain that $D$ and $E$ intersect transversally and that $g|_D\colon D \to g(D)=: D'$ is an isomorphism. Now since $f$ is the contraction of an extremal ray, we have
\begin{equation}\label{eq:relpic}
\operatorname{Im}(\Pic X \to \Pic D) \cong \bZ
\end{equation}
(cf.~\cite[Thm.~2.1]{And85}). This implies in particular that $V=E|_D$ is an ample Cartier divisor on~$D$ and thus that $W$ is an ample Cartier divisor on $D' = g(D)$.

Since $\rho(X) = 2$, it follows that $\rho(Z) = 1$, so that $D'$ is an ample divisor on~$Z$. The adjunction formula now gives
\begin{equation}\label{eq:dbladj}
\sO_W(K_W) = (\omega_{D'}\otimes \sO_{D'}(W))|_W = \sO_{D'}(K_Z+D'+W)|_W.
\end{equation}

But since $Z$ and $W$ are smooth, we also have
\begin{equation}\label{eq:adjform}
\sO_W(K_W) = \sO_W(K_Z) \otimes \det N_{W|Z}.
\end{equation}
Comparing \eqref{eq:dbladj} and \eqref{eq:adjform}, we obtain
\[ \det N_{W|Z} = \sO_{D'}(D'+W)|_W, \]
thus $\det N_{W|Z}$ is ample by our previous considerations. This contradicts Lemma~\ref{lem:blowupsurf}.

It remains to prove Claim~\ref{cl:red}. To do this, we need to obtain more information about~$W$. We first note that since $X$ is $2$-Fano, we have
\[ \det N_{W|Z}^*.C \ge 1\quad\text{for any curve $C\subset W$} \]
by Lemma~\ref{lem:blowupsurf}. Since $Z$ is Fano, this implies by~\eqref{eq:adjform} that $-K_W$ is ample. Furthermore, \eqref{eq:adjform} gives
\[ -K_W.C \ge 2 \quad\text{for any curve $C\subset W$.} \]
In particular, $W$ cannot contain any $(-1)$-curve, so we conclude that either $W \cong \bP_2$ or $W \cong \bP_1 \times \bP_1$.

The map $f$ induces a contraction on~$E$ contracting $V$ to a point. We now consider the reduction
\[ \tilde{V} := V_{\mathrm{red}} = (D\cap E)_{\mathrm{red}}. \]
The restriction $g|_{\tilde{V}}$ is then a finite morphism, so $\tilde{V}$ is a multisection of~$\pi$. Now let $\ell \subset W$ be a general line. Then $f|_{\pi^{-1}(\ell)}$ must be the contraction of the minimal section of the Hirzebruch surface $\pi^{-1}(\ell)$. This implies that $\tilde{V}$ is indeed a section of~$\pi$. Furthermore, since $N_{\tilde{V}|E}^*|_\ell$ is ample, it follows that $N_{\tilde{V}|E}^*$ is ample.

Thus it only remains to show that $V$ is reduced. As a divisor on~$E$, we can write
\[ V = \lambda \tilde{V} \qquad\text{for some $\lambda \ge 1$.}\]
Claim~\ref{cl:red} is then proved if we show that $\lambda = 1$. In order to show this, we first use Ando's result on the classification of the general fiber of the exceptional divisor of an extremal contraction \cite[Thm.~2.1]{And85}. By this result we have one of the following cases:
\begin{enumerate}
\item $D \cong \bP_3$,
\item $D \cong Q_3$, or
\item $D$ is a Del Pezzo variety such that if $A$ is an ample generator of~$\operatorname{Im}(\Pic X \to \Pic D) \cong \bZ$, then
\begin{equation}\label{eq:delpezzo}
\omega_D \cong (A^*)^{\otimes 2}.
\end{equation}
\end{enumerate}
In the first case ($D\cong \bP_3$), Tsukioka proves in~\cite[Lem.~2]{Tsu06} that $\lambda = 1$, but his argument also applies to the second case ($D\cong Q_3$): If $\lambda > 1$, then $V = D\cap E$ is singular in every point of~$\tilde{V}$, so in particular $T_D|_{\tilde{V}} \cong T_E|_{\tilde{V}}$ (observe that $D$ is smooth). This implies that $N_{\tilde{V}|D} \cong N_{\tilde{V}|E}$. This is a contradiction since $N_{\tilde{V}|D}$ is ample because $\rho(D)=1$, but $N_{\tilde{V}|E}$ is negative as seen above.

It remains to study the Del Pezzo case. Since by adjunction,
\[ \omega_D \cong \sO_D(K_X+D), \]
we have
\begin{equation}
\sO_D(K_X) \cong \sO_D(D)
\end{equation}
by~\eqref{eq:delpezzo}. We thus get
\begin{equation}\label{eq:drestv}
\sO_{\tilde{V}}(D) \cong \sO_{\tilde{V}}(K_X) \cong \sO_{\tilde{V}}(g^*K_Z+E) \cong \sO_E(-1)|_{\tilde{V}}\otimes \sO_W(K_Z).
\end{equation}
On the other hand, we have
\[
\sO_{\tilde{V}}(D) \cong \sO_{E}(V)|_{\tilde{V}} \cong \bigl(\sO_{E}(\tilde{V})|_{\tilde{V}}\bigr)^{\otimes \lambda},
\]
i.e., $\sO_{\tilde{V}}(D)$ is divisible by~$\lambda$.

Now since $\tilde{V}$ is a section of~$\pi$, it is given by an element in $H^0(\sO_E(1)\otimes \pi^*L)$ for some $L \in \Pic W$, or, equivalently, by a section
\[ s \in H^0(N_{W|Z}^*\otimes L) \]
without zeroes. The section $s$ induces a short exact sequence of vector bundles
\begin{equation}\label{eq:normseq}
 0 \longrightarrow L^* \longrightarrow N_{W|Z}^* \longrightarrow \det N_{W|Z}^*\otimes L \longrightarrow 0,
\end{equation}
from which we conclude that
\begin{equation}\label{eq:taut}
\sO_E(1)|_{\tilde{V}} \otimes L \cong N_{\tilde{V}|E} \cong \det N_{W|Z}^*\otimes L^{\otimes 2}.
\end{equation}
This implies together with~\eqref{eq:drestv} that
\begin{equation}\label{eq:donv}
 \sO_{\tilde{V}}(D) \cong \sO_W(K_Z) \otimes \det N_{W|Z} \otimes L^* \cong \sO_W(K_W) \otimes L^*.
\end{equation}
Since $\sO_D(-D)$ is ample, we can conclude that $\sO_W(-K_W)\otimes L$ is an ample line bundle.

We now first consider the case $W\cong\bP_2$. Then $-K_W = \sO_{\bP_2}(3)$. Since $Z$ is Fano and Lemma~\ref{lem:blowupsurf} holds, only two cases can occur by~\eqref{eq:adjform}:
\begin{enumerate}
\item $\sO_W(-K_Z) \cong \sO_{\bP_2}(1)$ and $\det N_{W|Z}^* \cong \sO_{\bP_2}(2)$, or
\item $\sO_W(-K_Z) \cong \sO_{\bP_2}(2)$ and $\det N_{W|Z}^* \cong \sO_{\bP_2}(1)$.
\end{enumerate}
If we let
\[
L \cong \sO_{\bP_2}(a),
\]
then we have $a\ge -2$ because $L- K_W$ is ample as shown above. We already showed above that $N_{\tilde{V}|E}^* \cong \det N_{W|Z}\otimes (L^*)^{\otimes 2}$ is ample, which yields
$\deg N_{W|Z}-2a > 0$, i.e., $a < \tfrac{1}{2}\deg N_{W|Z}$. In the two cases stated above, this means the following:
\begin{enumerate}
\item In this case, we have $-2 \le a < -1$, so $a=-2$ and thus by~\eqref{eq:donv}
\[ \sO_{\tilde{V}}(D) \cong \sO_{\bP_2}(1). \]
Since $\sO_{\tilde{V}}(D)$ is divisible by~$\lambda$, we get $\lambda=1$.
\item Here we must have $-2 \le a < -\tfrac{1}{2}$, so $a\in\{-2, -1\}$. If $a=-2$, we get $\lambda=1$ as in the first case. If $a=-1$, we get
\[ \sO_{E}(-1)|_{\tilde{V}} \cong \sO_{\tilde{V}} \]
by~\eqref{eq:taut}. This is a contradiction because
\[ \sO_{E}(-1)|_{\tilde{V}} \cong \sO_E(E)|_{\tilde{V}} \cong \sO_D(E)|_{\tilde{V}} \]
is ample by~\eqref{eq:relpic}
\end{enumerate}

It remains to consider the case $W\cong \bP_1\times\bP_1$. Then $-K_W \cong \sO_{\bP_1\times\bP_1}(2,2)$ and thus by~\eqref{eq:adjform}, we must have
\[ \sO_W(-K_Z) \cong \det N_{W|Z}^* \cong \sO_{\bP_1\times\bP_1}(1,1) \]
(again using Lemma~\ref{lem:blowupsurf} and the fact that $Z$ is Fano). If we let
\[
L \cong \sO_{\bP_1\times\bP_1}(a,b)
\]
and use again the fact that $L-K_W$ is ample, we have $a$, $b\ge -1$. From the ampleness of~$N_{\tilde{V}|E}^*$ we obtain $-1-2a > 0$ and $-1-2b > 0$, so $a=b=-1$ and thus by~\eqref{eq:donv}
\[
\sO_{\tilde{V}}(D) \cong \sO_{\bP_1\times\bP_1}(1,1),
\]
from which we again conclude $\lambda=1$.
\end{proof}

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config: arxiv


TITLE: Non trivial homomorphism from $SU(2)$ to the diffeomorphism group of the circle
QUESTION [5 upvotes]: Is there a non trivial homomorphism $f: SU(2) \to \operatorname{Diff}(S^1)$?
(From the comments) By a previous question, we know that there is no nontrivial homomorphism $SU(2) \to O(2)$. Since $O(2)$ is a subgroup of $\operatorname{Diff}(S^1)$ (rotations are diffeomorphisms), maybe there is a nontrivial map in the larger group.

REPLY [3 votes]: Theorem. There are no nontrivial homomorphisms $f$ (continuous or not) from $SU(2)$ to $Homeo(S^1)$.
Proof. First of all, it is known that $SO(3)=SU(2)/(\pm 1)$ is a simple group (as an abstract group); a proof of this is a very nice exercise in elementary group theory and geometry. If you cannot prove it, see for instance Berger’s book “Geometry-I”, section 8.5. Thus, it suffices to show that each homomorphism $f$ has kernel consisting more than just of the center of $SU(2)$. Observe that $SU(2)$ contains the binary Icosahedral group $G$, a nontrivial $Z_2$-central extension of the
alternating group $A_5$. Thus, $f$ defines a homomorphism $f: G\to Homeo(S^1)$. I claim that each such homomorphism has to be trivial. Since $G$ is perfect, the image of $f$ is contained in the orientation-preserving subgroup of $Homeo(S^1)$. Now, if $g: S^1\to S^1$ is an orientation-preserving non-identity homeomorphism of finite order, it cannot have fixed points (prove first that each finite order homeomorphism of the interval has to be the identity map). Thus, each faithful topological action of $G$ (or its alternating quotient) on $S^1$ has to be fixed-point free, hence, defines a covering map $S^1\to S^1/f(G)$. However, the quotient $Y=S^1/f(G)$ has to be a topological circle, hence, by the basic covering theory, $G$ is isomorphic to the quotient $\pi_1(Y)/(i_*\pi_1(S^1))$, where $i_*$ is the monomorphism of fundamental groups induced by the covering map $S^1\to Y$. But such a quotient has to be cyclic. A contradiction. qed


TITLE: How many numbers are divisible by at least one element of $S$
QUESTION [1 upvotes]: I have a set S containing $N$ $(1 \le N \le 20)$ integers $a_i$ $(1 \le a_i \le 10^{18})$ and two numbers $a$ and $b$. I want to tell how many numbers between $a$ and $b$ are divisible by at least one of the numbers from set S. I can do it in $O(2^n \cdot n)$ time using inclusion-exclusion principle. Is there any faster way to do this?

REPLY [2 votes]: If $a$ and $b$ are very large numbers, then you can use the following.  I will assume that the $a_i$'s are relatively prime (to simplify).
Let $c=\prod_{i=1}^Na_i$.
Then, the number of integers less than $c$ which are divisible by $a_1$ is $\prod_{i=2}^Na_i$, the number of integers less than $c$ which are divisible by $a_1$ and $a_2$ is $\prod_{i=3}^Na_i$.
In general, using the inclusion/exclusion principle, we know that the number of integers relatively prime to $a_1,\cdots,a_n$ is given by (where $S=\{1,\cdots,N\}$)
$$
\prod_{i=1}^Na_i-\sum_{A\subseteq S,|A|=|S|-1}\prod_{i\in S}a_i+\sum_{A\subseteq S,|A|=|S|-2}\prod_{i\in S}a_i-\sum_{A\subseteq S,|A|=|S|-3}\prod_{i\in S}a_i+\cdots+(-1)^N
$$
These are the coefficients of the polynomial $\prod(a_i-x)$.  So, then, we can calculate the number of integers less than or equal to $c$ relatively prime to the $a_i$'s by substituting $1$ for $x$ and multiplying which takes $O(N)$.
In fact, any sequence of integers of length $c$ will have exactly this many integers relatively prime to the $a_i$'s.  So, you could calculate the number of integers less than $m$ which are relatively prime to the $a_i$'s where $1\leq m\leq c\leq 10^{360}$ and break up the interval between $b$ and $a$ into blocks of size $c$ and use a table to look up the rest.
This can give an algorithm that is around $O(N)$ (around because I'm ignoring bit complexity of multiplication) - but the constant is HUGE (at least $10^{360}$).


TITLE: Step from $\Big(\frac{d}{dx}\frac{dy}{dt}\Big)t$ to $\Big(\frac{d^2y}{dt^2}\frac{dt}{dx}\Big)t$
QUESTION [0 upvotes]: From this page between (2) and (3): https://math.dartmouth.edu/archive/m23s06/public_html/handouts/euler_eqns.pdf
$$\Big(\frac{d}{dx}\frac{dy}{dt}\Big)t=\frac{d^2y}{dt^2}\frac{dt}{dx}t$$
I always had difficulties with the chain rule, especially in Leibniz notation.
Where does the second derivative with respect to $t$ come from?

PDF transcribed with color for emphasis

$$\begin{align}  \frac{d^2 y}{dx^2} &=  \frac{d}{dx} \frac{dy}{dx}\\
 &=\frac{d}{dx}\left(\frac{dy}{dt} t\right)\\
 &=\color{red}{\left(\frac{d}{dx}\frac{dy}{dt}\right)t}+\frac{dy}{dt}\frac{dt}{dx}\\
 &=\color{red}{\frac{d^2
 y}{dt^2}\frac{dt}{dx}t}+\frac{dy}{dt}\frac{dt}{dx}\\ &=\frac{d^2
 y}{dt^2}t^2+\frac{dy}{dt}t \end{align}$$ Where we have again used the
   fact that $dx/dt = t$

REPLY [0 votes]: Perhaps, it'd help to view it as the below operator acting on $y$. Note that the $t$ at the end in your expression in the LHS and RHS is a just a multiplicative factor and not acted upon by the operator.$$ \frac{\mathrm{d}}{\mathrm{d} x} \left( \frac{\mathrm{d}}{\mathrm{d} t}\right)= \left(\frac{\mathrm{d }t}{\mathrm{d} x} \right)\frac{\mathrm{d}}{\mathrm{d} t}\left( \frac{\mathrm{d}}{\mathrm{d} t}\right) = \left(\frac{\mathrm{d }t}{\mathrm{d} x} \right)\frac{\mathrm{d^2}}{\mathrm{d} t^2}$$


TITLE: Constant random variable
QUESTION [3 upvotes]: How do I plot the cumulative distribution function and probability mass function of the constant random variable $X(\omega)=2$ for all $\omega$?

REPLY [2 votes]: The CDF is:
$$P(X\leq x) = \begin{cases}
  1 \quad \text{ if } x \geq 2 \\
  0 \quad \text{ if } x < 2 \\
\end{cases}$$
while the PMF is:
$$P(X = x) = \begin{cases}
  1 \quad \text{ if } x = 2 \\
  0 \quad \text{ if } x \not= 2 \\
\end{cases}$$

REPLY [2 votes]: If $P(X = c) = 1,$ then that equation is all that's necessary to show the PMF.
The CDF $F_X(x)$ is $0$ for $x < c$ and $1$ for $x \ge c.$
Here is a plot of the CDF for the case $c = 3.$


TITLE: A differentiable function whose Fourier series is not uniformly convergent.
QUESTION [1 upvotes]: We focus on $2\pi$-periodic functions.
1- Does there exist any differentiable function whose Fourier series is not uniformly convergent?
Even more,
2- Does there exist any differentiable function whose Fourier series con not be uniformly convergent on every set of positive measure in [$-2\pi,2\pi$]?

REPLY [1 votes]: Perhaps you already know the two things I have to say about this:  (i)  Dini's test shows that if $f$ is differentiable then the Fourier series fo $f$ converges pointwise to  $f$ (ii) it follows, and it's also easy to prove directly, that if $f$ is continuously differentiable then the Fourier series converges to $f$ uniformly. (Because $f'$ continuous implies $\sum n^2|c_n|^2<\infty$, hence $\sum|c_n|<\infty$.)
But if $f$ is just differentiable does that imply the Fourier series for $f$ converges uniformly? I haven't concocted an example, but surely not; if $f'$ exists but is not continuous then the differentiation is not uniform.


TITLE: If $A^m = 0$, then $\mbox{rank}(A) \leq \frac{m-1}{m}{n}$
QUESTION [4 upvotes]: Let $A$ be a $n \times n$ real matrix. Show that if $A^m = 0$, then
$\mbox{rank}(A) \leq \frac{m-1}{m}{n}$

My attempt:
If $m=1$, then $A=0$ so $\mbox{rank}(A)=0$.
If $m=2$, we have $\mbox{im}(A) \subset \ker(A)$ so $2\operatorname{rank}(A) \leq \dim \mbox{im}(A) + \dim \ker(A)=n$
For arbitrary $m$, I want to use induction. $B=A|_{\mbox{im}(A)}$ satisfies $B^{m-1}=0$ so $\mbox{rank}(B) \leq \frac{m-2}{m-1}\mbox{rank}(A)$. Thus $\dim \ker B \geq \mbox{rank}(A)- \frac{m-2}{m-1}\mbox{rank}(A) =\frac{1}{m-1}\mbox{rank}(A)$ by rank-nullity theorem. Thus $n=\dim \ker A + \mbox{rank}(A) \geq  \dim \ker B +\mbox{rank}(A)\geq \frac m {m-1}\mbox{rank}(A)$
Is this ok?

REPLY [1 votes]: Here's a solution using a Jordan canonical form for $A$ . . .

Let $A$ be a nonzero $n{\,\times\,}n$ matrix such that $A^m=0$ for some positive integer $m$, and let $r=\text{rank}(A)$.

Fix a Jordan canonical form for $A$.

Since $A$ is nonzero, there is at least one nonzero Jordan block.

Let $B_1,...,B_k$ be the sequence of nonzero Jordan blocks.

Let $n_i$ be the size of $B_i$ (i.e., $B_i$ is an $n_i{\,\times\,}n_i$ matrix), and let $N=\sum_{i=1}^k n_i$.

Then $B_i$ has rank $n_i-1$, hence $r=\sum_{i=1}^k (n_i-1)=N-k$.

Since $A^m=0$, we must have $B_i^m=0$ for all $i$, hence $m\ge n_i$ for all $i$.

For each $i$, let $x_i=m-n_i$, and let $X=\sum_{i=1}^k x_i$.

Then $X=\sum_{i=1}^k (m-n_i)=km-N$, hence
\begin{align*}
\Bigl(\frac{m-1}{m}\Bigr)n-r
&\ge
\Bigl(\frac{m-1}{m}\Bigr)N-r
\\[4pt]
&=
\Bigl(\frac{m-1}{m}\Bigr)N-(N-k)
\\[4pt]
&=
\frac{km-N}{m}
\\[4pt]
&=
\frac{X}{m}
\\[4pt]
&\ge
\;0
\\[4pt]
\end{align*}
so we have
$
r\le\Bigl({\Large{\frac{m-1}{m}}}\Bigr)n
$
as was to be shown.


TITLE: Define a relation on $\mathbb Z$ by $m\sim n$ if $\gcd(m,n)=m$. Show that this is NOT an equivalence relation
QUESTION [3 upvotes]: Define a relation on $\mathbb Z$ by specifying that $m\sim n$ if $\gcd(m, n) = m$. Show that this relation is NOT an equivalence relation.

I have shown that it passes the reflexive property and symmetric property, so that means it has to fail the transitive property, but I don't know how to show that.
Reflexive: $\gcd(m,m) = m$
Example: $\gcd(2,2) = 2$
Symmetric: $\gcd(m,n) = m$ and $\gcd (n,m) = m$
Example: $\gcd(2,4) = 2$ and $\gcd (4,2) = 2$

REPLY [0 votes]: Alternative approach:
Lemma 1: $a,b \in \Bbb{Z_{\neq 0}}, ~a ~| ~b \implies |a| \leq |b|$.
Proof:
$|a| > |b|$ makes it impossible for $a$ to be a divisor of $b$.
Lemma 2: gcd$(m,n) = m \iff m ~| ~n.$
For this Lemma, without loss of generality, assume that $m,n \in \Bbb{Z^+}$.  For either $m,n$ not positive, the analysis would be very similar.
$\implies$:
Assume gcd$(m,n) = m$.  This implies that the gcd, namely $m$, is a common divisor of both $m$ and $n$.  In particular, this implies that $m ~| ~n.$
$\impliedby$:
Assume that $m ~| ~n$.  Then, clearly, $m$ is a common divisor of both $m$ and $n$.  Further, it must be the greatest common divisor, because for any integer $d$ such that $|d| > |m|$ it is impossible for $d$ to be a divisor of $m$, by Lemma 1.  Therefore, for any integer $d > m$, such that $|d| > |m|$, $d$ can not be a common divisor of $(m,n)$.

Three questions need to be considered:

Does $a \sim a$? : Reflexive
Does $a \sim b \implies b \sim a$? : Symmetric
Does $\{ ~(a \sim b) \wedge (b \sim c) ~\} \implies (a \sim c)$? : Transitive

Use the Lemmas.

$\forall a \in \Bbb{Z} ~$ (including $a = 0$), $a ~| ~a \implies a \sim a.$

Given $a,b \in \Bbb{Z_{\neq 0}},$ suppose that $~a \sim b,~$ and $|a| \neq |b|$.  Such an occurrence is clearly possible, because any integer $b$, not an element of $\{-1, 0, 1\}$ will have $1$ as a divisor.  In this event, by Lemmas 1 and 2, you have that $|a| < |b|$.  This implies (again using Lemmas 1 and 2) that it is not the case that $b \sim a$.

Given $a,b,c \in \Bbb{Z},$
$[(a\sim b) \wedge (b\sim c)] \implies \{ ~a ~| ~b \wedge ~b ~| ~c ~\} \implies $
$a ~| ~c \implies (a \sim c).$

Therefore $(\sim)$ is reflexive and transitive, but not symmetric.


TITLE: logarithmic differentiation issue
QUESTION [0 upvotes]: Trying to understand a solution I was given to a problem I was told to use logarithmic differentiation on. $$ 1/x(x+1)(x+2) $$
and I know that $$log((ab)/c) = log(a) + log(b) - log(c)$$ So I tried to use that rule here and did: $$ln(a) - ln(b) - ln(c)$$
and got: $$ln(1) - ln(x(x+1)- ln(x(x+2)$$
which simplifies to: $$0 - ln(x^2+x)- ln(x^2+2x)$$
and then I look at the solution which gives:
$$y`=(1/(x(x+1)(x+2))) * (1/x + 1/(x+1) + 1/(x+2))$$
I'm just kind of confused on what I am doing wrong.

REPLY [1 votes]: You have that
$$\ln f(x)=\ln \frac{1}{x(x+1)(x+2)}=\ln 1-ln x- \ln (x+1)-\ln(x+2)$$ $$=-ln x- \ln (x+1)-\ln(x+2).$$
Taking derivatives:
$$\frac{f'(x)}{f(x)}=-\frac 1x-\frac{1}{x+1}-\frac{1}{x+2}.$$ Thus
$$f'(x)=\frac{-1}{x(x+1)(x+2)}\left(\frac 1x+\frac{1}{x+1}+\frac{1}{x+2}\right).$$
Note a difference in sign with your expected answer.


TITLE: How to model/simulate pressures and flows in a network of pipes
QUESTION [13 upvotes]: I'm having a hard time finding information on how to model/simulate this.  I attached a couple files, both of which show an example tank & pump network.  It's just nonsense that I made up for this post, but it shows what kind of system I'm trying to simulate.
(I just put it on the web instead:
pdf link,
vsd link)
I've been looking on the web to find information on how to do this, but I keep running into formulas that are way more complex than what I need to do.  I just want to know what the pressure and flow is at any node in the system, with 80% or so accuracy.  There's no need to consider fluid compression, and we can just use a constant for the resistance experienced through valves or narrowing pipes.  We don't need to consider friction of fluid against piping.  We can ignore heat.  It's important to note, again, that I don't need complete accuracy, it's much more the behavior of the simulated system that is important.
I've tried this going in both directions, that is following a flow path down from the tanks/sources and propogating pressure changes through the system, and I've also tried it directly from the node in question, and considering that each node has a number of source nodes in terms of pressures and flows.  Both of these simulated SW systems become highly erratic the further downstream from the pumps that you get, and the more on/off valves that I put into the system.
Even if you just told me where to look or what book to read, that would be a huge help.  If there's example source code out there somewhere that does this, I will buy you a 60 oz beer of your choice.
Thanks in advance.

(post edit 4/12)

Thanks for the answers so far, I've ended up spending quite a bit of time looking around for a solution that already exists, because I was just convinced that many have run into this problem before, and knew that it was solvable in a general sense.  Part of the reason I was so interested in a general solution is that I can rip through 12 more of these simulations that I have to do quickly and accurately if that's the case.
I've found another piece of software called EPAnet, which seems like a simpler solution than what was listed below, and I'm looking through their SDK at the moment and playing around with it.  I'll keep you posted, the few that are interested ... this might not be a big deal to most, but to the few that it matters to it will be huge, so I'll write up what I find.

(edit 4/13)

I've spent the last two days building a solid wrapper around the EPAnet programmer's toolkit, and making a scratch program that does some of the things I want to do.  (Incidentally, I've broken a new record for the # of times in a day I can make LabVIEW crash quickly and silently.)
It's really quite complex underneath, and I'm delighted by how much mathematical power I can stand on.  I don't think they've fleshed out all the issues, so I've been working around some of them.  But it's looking like it's 80% chance so far that it's going to work for what I want.

REPLY [1 votes]: Turns out, EPAnet can actually do what I want.  It's a little hairy to try to plug into, as I don't think any real updates have come for 5 years or more, but there's several commercial products that base their underlying engine on EPAnet, even if they extend or tweak it a bit.  This little fact gave me more confidence plowing forward that I wasn't jumping down a rabbit hole.


TITLE: Is the relation < necessarily a total order?
QUESTION [4 upvotes]: Let $M$ be a non-empty set, $+$ a binary operation on $M$, and $<$ a relation on M, satisfying the following:

For all $a, b, c \in M$, $(a + b) + c = a + (b + c)$

For all $a, b \in M$ with $a \ne b$, exactly one of $a < b$ and $b < a$ is true.

For all $a, b \in M$ with $a < b$, at least one of the equations $a + x = b$ and $x + a = b$ has a solution for $x$ in $M$.

For all $a, b \in M$ with $a < b$, neither of the equations $b + x = a$ nor $x + b = a$ have a solution for $x$ in $M$.

The question is: is it necessarily true that for all $a, b, c \in M$, that if $a < b$ and $b < c$ then $a < c$?

REPLY [3 votes]: Consider the oriented graph with vertex set $M$ in which edge $(a,b)$ exists if $a<b$.
Label each edge $(a,b)$ of this graph with the label $L$ if $x+a=b$ has solution.
Label each edge $(a,b)$ of this graph with the label $R$ if $a+x=b$ has solution.
Note each edge has at least one label.
Assume there is a triangle cycle $a\rightarrow b\rightarrow c\rightarrow a$. Such a cycle must contain two consegutive edges with the same label, so we can rotate it so that without loss of generality the first two arrows have the same label. without loss of generality the label is $L$.
We have $b=x+a$ and $c=y+b$ and so $c=y+x+a$ which means $a<c$ which is a contradiction, as we assumed $c<a$


TITLE: Basic question about higher derivative
QUESTION [1 upvotes]: A curve in the $xy$
  -plane is given parametrically by
  $$x=t^2+2t \\ y=3t^4+4t^3$$
  for all $t>0$. What is the value of $\frac{d^2y}{dx^2}$ at $(8,80)$ ?

I have the following doubt:
We know that $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ Similarly, can we say that
$$\frac{d^2y}{dx^2}=\frac{\frac{d^2y}{dt^2}}{\frac{d^2x}{dt^2}}$$

REPLY [0 votes]: The function $x(t) = t^2+2t$ has an inverse function when $t\gt 0$ (because $x(t)$ is continuous and monotonic when $t\gt0 $). So we can think of $t$ as a function of $x$.
We have: $${dy \over dx} = \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \tag{1}$$
To get the second derivative $\dfrac{d^2y}{dx^2}$, we differentiate (1) with respect to $x$ (keep in mind that $t$ is a function of $x$):
$$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left( \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\right)= \dfrac{d}{\color{magenta}{dt}}\left( \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\right)\dfrac{\color{magenta}{dt}}{dx}$$
But
$$\dfrac{d}{dt}\left( \dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}\right) = \dfrac{\dfrac{dx}{dt}\dfrac{d}{dt}\left(\dfrac{dy}{dt}\right)-\dfrac{dy}{dt}\dfrac{d}{dt}\left( \dfrac{dx}{dt}\right)}{\left( \dfrac{dx}{dt}\right)^2}= \dfrac{\dfrac{dx}{dt}\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}\dfrac{d^2x}{dt^2}}{\left( \dfrac{dx}{dt}\right)^2}$$
$$\dfrac{dt}{dx} = \dfrac{1}{\dfrac{dx}{dt}}$$
So we find:
$$\dfrac{d^2y}{dx^2}=\dfrac{\dfrac{dx}{dt}\dfrac{d^2y}{dt^2}-\dfrac{dy}{dt}\dfrac{d^2x}{dt^2}}{\left( \dfrac{dx}{dt}\right)^3}$$


text:
\begin{document}

\author[J. Campbell]{James T. Campbell} \address[J. T.
Campbell]{Dept. Math. Sci.\\Dunn Hall 373\\
  University of Memphis\\Memphis, TN 38152}
\email{jcampbll@memphis.edu}

\author[J. Collins]{Jared T. Collins}\address[J. T. Collins]{Dept. Math. and Comp. Sci.\\Freed-Hardeman University\\Henderson TN 38340}
\email{jtcollins@fhu.edu}

\author[S. Kalikow]{Steven Kalikow}\address[S. Kalikow]{Dept. Math. Sci.\\Dunn Hall 373\\
  University of Memphis\\Memphis, TN 38152}
\email{skalikow@memphis.edu}

\author[R. King]{Raena King}\address[R. King]{Dept. Math. and Comp. Sci \\Christian Bros. University\\ Memphis, TN, 38104} \email{rking2@cbu.edu}

\author[R. McCutcheon]{Randall McCutcheon} \address[R. McCutcheon]{Dept. Math. Sci.\\Dunn Hall 373\\
  University of Memphis\\Memphis, TN 38152}
\email{rmcctchn@memphis.edu}

\begin{abstract}

  Given a measure-preserving transformation $T$ of a probability space $(X, \mathcal B, \mu)$ and a finite measurable partition $P$ of $X$, we show how to construct an Alpern tower of any height whose base is independent of the partition $P$. That is, given $N \in \N$,  there exists a Rohlin tower of height $N$, with base $B$ and error set $E$, so that $B$  is independent of $P$, and $T(E) \subset B$. \bigskip

\textit{2010 Mathematics Subject Classification}: 28D05, 37M25, 60A10 \medskip

\textit{Keywords:} Rohlin tower, Alpern tower, independent sets, measure-preserving transformation, probability space.
\end{abstract}

\date{April 30, 2015}

\maketitle

\section{Introduction and Statement of Results}
It has long been known that, given an ergodic invertible probability measure preserving system, a Rohlin tower may be constructed with base independent of a given partition of the underlying space(\cite{Roh:52}, \cite{Roh:65}). In \cite{Alp:79}, meanwhile, S. Alpern proved a `multiple' Rohlin tower theorem (see \cite{EP:97} for an easy proof) whose full statement we will not give, but which has the following corollary of interest:

\begin{thm}\label{thm:alp}
Let $N \in \N$ and \epso be given. For any ergodic invertible measure-preserving transformation $T$ of a Lebesgue probability space $(X, \mathcal B, \mu)$, there exists a Rohlin tower of height $N$ with base $B$ and error set $E$ with $\mu(E) < \eps$, so that $T(E) \subset B$.
\end{thm}

A {\em Rohlin tower of height} $N$ {\em with base} $B$ {\em and error set} $E$ is characterized by the collection of sets
$\{B, TB, \dots, T^{N-1}B, E\}$ forming a partition of $X$. If in addition $T(E) \subset B$, we shall say {\bf Alpern Tower}.
It is our goal to show that for ergodic transformations on $(X, \mB, \mu)$, given a finite measurable partition $\mathbb P$ of $X$, an Alpern tower may be constructed with base $B$ independent of $\mathbb P$. Precisely:

\begin{bigthm}
\label{mainthm}
Let $(X, \mathcal B, \mu)$ be a Lebesgue probability space, and suppose $\mathbb P$ is a finite measurable partition of $X$.   For any ergodic invertible measure-preserving transformation $T$ of $X$, $N \in \N$,
there exists a Rohlin tower of height $N$ with base $B$ and error set $E$ such that $T(E) \subset B$ and $B$ is independent of $\mathbb P$.
\end{bigthm}

We do not specify the size of the error set; but the process of constructing our tower makes it clear that the error set may be made arbitrarily small.

\section{Proof of main result}

For the remainder of the paper, $(X, \mathcal B, \mu)$ will be a fixed Lebesgue probability space and  $T:X \to X$ will be an invertible ergodic measure-preserving transformation on $X$. All mentioned sets will be measurable and we will adopt a cavalier attitude toward null sets. In particular, ``partition'' will typically mean ``measurable partition modulo null sets''.
\begin{defn}
By a {\em tower over B} we will mean a set $B \subset X$, called the {\em base}, and a countable partition $B = B_1 \cup B_2 \cup \cdots$, together with their images  $T^iB_j$, $0 \le i < j$, such that the family $\{T^iB_j : 0 \le i < j\}$ consists in pairwise disjoint sets. If this family partitions $X$, we will say that the tower is {\em exhaustive}.
\end{defn}

If a tower over $B$ is exhaustive and $B = B_N \cup B_{N+1}$, we shall speak of an {\em exhaustive Alpern tower of height} $\{N, N+1\}$, as in such a case,
$\{B, TB, \ldots, T^{N-1}B, E=T^N B_{N+1}\}$ partitions $X$ with $T(E) \subset B$. So
we may re-phrase Theorem 1 as: \medskip

\noindent {\em {\bf Theorem \ref{mainthm}:} Let $(X, \mathcal B, \mu)$ be a Lebesgue probability space and suppose $\mathbb P$ is a finite measurable partition of $X$.   For any ergodic invertible measure-preserving transformation $T$ of $X$, $N \in \N$, one may find an exhaustive Alpern tower of height $\{N, N+1\}$ having base independent of $\mathbb P$.} \medskip

\noindent We require a lemma (and a corollary).

\begin{lem}\label{lem:m}
Let $M \in \N$ and let $\Pa = \{P_1, \dots, P_t\}$ be a partition of $X$ with $\mu(P_i) > 0$ for each $i$. There exists a set $S$ of positive measure so that if $x \in S$ with first return $n(x) = n$, say, then $|\{x, Tx, \dots, T^{n-1}x\} \cap P_i | \ge M, 1 \le i \le t$.
\end{lem}
\proof  For almost every $x$ we may find $K(x)$ so that for each $i$ between $1$ and $t$ we have $|\{x, Tx, \dots, T^{K(x) - 1}x\} \cap P_i| \ge M$. Since almost all of $X$ is the countable union (over $k \in \N$) of $\{x: K(x) = k\}$, there exists some fixed $K$ so that the set $A = \{x: K(x) \le K\}$ has positive measure. If $C \subset A$ has very small measure ($\mu(C) < 1/K$) then the average first-return time of $x \in C$ to $C$ is $\frac{1}{\mu(C)} > K$, so we can find $S \subset C$ with $\mu(S) > 0$  so that $S, TS, \dots, T^{K - 1}S$ are pairwise disjoint. \bs

\begin{cor}\label{cor:B}
 Let $M \in \N$ and $\Pa = \{P_1, \dots, P_t\}$ be a partition of $X$ with $\mu(P_i) > 0$ for each $i$. There is a tower having base
$S = S_{tM} \cup S_{tM+1} \cup \cdots$ where for each $x \in S_r$, $|\{x, Tx, \dots, T^{r - 1}x\} \cap P_i| \ge M$ for all $1 \le i \le t$.
\end{cor}
\proof Let $S$, $K$  be as in Lemma \ref{lem:m} and choose any $k \ge K$. \bs \medskip

We turn now to the proof of Theorem \ref{mainthm}. Fix a partition $\Pa = \{P_1, \dots, P_t\}$, an arbitrary natural number $N$, and $\epsilon >0$. Set $m_i = \mu(P_i)$, and assume (without loss of generality) that $0 < m_1 \le m_2 \le \dots \le m_t$. Select and fix $M > \frac{3N^3t}{m_1}$. Let $S$ be as in Corollary \ref{cor:B} for this $M$; hence $S = S_{tM} \cup S_{tM+1} \cup \cdots$.  (Some $S_i$ may be empty, of course.) For each non-empty $S_R$, partition $S_R$ by $\Pa$-name of length $R$. (Recall that $x, y$ in $S_R$ have the same $\Pa$-name of length $R$ if $T^ix$ and $T^iy$ lie in the same cell of \Pa for $0 \le i < R$.) Let $C$ be the base of one of the resulting columns; hence, every $x \in C$ has the same $\Pa$-name of length $R$ (for some $R\ge tM$), and the length $R$ orbit of each $x \in C$ meets each $P_i$ at least $M$ times.

Partition $C$ into pieces $C^{(1)}, C^{(2)}\dots$, $C^{(t)}$ whose measures will be determined later. Then partition each $C^{(i)}$ into $N$ equal measure pieces, $C^{(i)} = C^{(i)}_1 \cup C^{(i)}_2 \cup \dots \cup C^{(i)}_N$.

Now we fix $(R,C)$ and focus our attention on the height $R$ {\em column} over a single $C^{(i)}$ and its height $R$ {\em subcolumns} over $C^{(i)}_j$, $1\leq j\leq N$. We refer to the sets $T^rC^{(i)}$, $0 \le r < R$, as {\em levels} and to the sets $T^rC^{(i)}_j$ as {\em rungs}. We are going to build a portion of $B$ by carefully selecting some rungs from the subcolumns under consideration. As we move through the various subcolumns, we need to have gaps of length $N$ or $N+1$ between selections. Now to specifics. We want
to have our $\Ci$-selections form a ``staircase'' of height $N$ starting at level $N^2 - N$. That is, at height $(N-1)N$, the rung over $C^{(i)}_1$ is the only one selected; at height $N(N-1) + 1$, the rung over $C^{(i)}_2$ is the only one selected; etc., so that at height $N^2-1$, the rung over $C^{(i)}_N$ is the only one selected.

This is easy to accomplish. First, we select each base rung $C^{(i)}_j$, $j = 1, 2, \dots, N$ (i.e., the rungs in the zeroth level). Over $C^{(i)}_1$, we then select $N-1$ additional rungs with gaps of length $N$; that is, we select the rungs at heights $N$, $2N, \dots, (N-1)N$. Over $C^{(i)}_2$ we select $N-2$ rungs with gap $N$, then a rung with gap $N+1$.  We continue in this fashion, choosing one less gap of length $N$ and one more of length $N+1$ in each subsequent subcolumn. In the last subcolumn (that over $C^{(i)}_N$) we are thus choosing rungs with gaps of length $N+1$ a total of $N-1$ times.  See the left side of
Figure \ref{pic:bottom} for the case $N = 4$.

Now we perform a similar procedure moving down from the top, so as to obtain a staircase starting at height $R-(N^2-1)$.
Note that there are either $N$ or $N-1$ unselected rungs at the top of each subcolunm. See the right side of Figure \ref{pic:bottom}.

\begin{figure}[hbtp]

\caption[Bottom of Tower for $N = 4$]
   {Bottom, Top of Tower for $N = 4$}
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\label{pic:bottom}

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\end{figure} \medskip

Next, we want to select rungs through the middle of the tower so as to iterate the staircase pattern all the way up, except that we will skip certain levels (i.e. not select any of their rungs), continuing the staircase pattern where we left off with the following rung. As we want to match stride with the staircase already selected at the top, the total number of levels skipped in the middle section will be constrained to a certain residue class modulo $N$, and as we want the selected rungs to form a portion of an Alpern tower of height $\{N, N+1\}$, we cannot skip any two levels with fewer than $N$ levels between them.

Some terminology: an {\em appearance} of $P_j$ in \Ci\ is just a level of \Ci\ that is contained in $P_j$.
A {\em selection} of $P_j$ is just a selected rung in a subcolumn of \Ci\ that is contained in $P_j$. The {\em net skips} of $P_j$ in the tower over
\Ci\ is defined as \[ S_j(\Ci) = (\# \textnormal{ of appearances of $P_j$) } - (\# \textnormal{ of selections of $P_j$)}.\]
For example, looking at Figure \ref{pic:bottom}, one sees that $4$ zeroth level rungs are selected. So if the zeroth level belongs to $P_j$, the zeroth level
contribution to $S_j(\Ci)$ is $-3$ (one appearance and 4 selections).

Let $\delta = 2(N-1)(N-2)$ and choose $\gamma$ with
\begin{displaymath}{\delta\over m_1}+N > \gamma \ge {\delta\over m_1} \quad \mbox{and} \quad (t-1)\delta + \gamma \equiv R ~(\hspace{-2.2ex}\mod N).
\end{displaymath}
Over \Ci, we skip a quantity of ``middle'' levels belonging to each $P_j$ (for $j \neq i$) sufficient to ensure that $S_j(\Ci)=\delta$ for $j\neq i$ and $S_i(\Ci)=\gamma$. (Note that $P_j$ cannot have been skipped more than $\delta$ times in the outer rungs.) This is not delicate; one can just enact the selection greedily. That is to say, travel up the tower, beginning at level $N^2$, skipping rungs that belong to cells requiring additional skips whenever there's been no too-recent skip. Since each $P_j$ appears at least $M>{3N^3t\over m_1}$ times, and we need only $\gamma +(t-1)\delta\le {2N^2t\over m_1}$ net skips, we'll find all the skips we need.

We have not specified the relative masses of the bases of the columns $\Ci$. Set
\begin{equation}
\label{bi} b_j = \frac{\mu(P_j)(\gamma + (t-1)\delta) - \delta}{\gamma - \delta} \;
\end{equation}
and put $\mu(\Ci)= b_i \mu(C)$, $1\leq i\leq t$. Our choice of $\gamma$ ensures that $b_i\ge 0$ for each $i$, and one easily checks that $\sum b_i=1$, so this is coherent.

Let $B_{C}$ be the union of the rungs selected from the columns over $C$
(this includes each of the rungs selected from each of the $N$ subcolumns over $\Ci$, $1\le i\le t$)
and put $B=\bigcup_{C} B_{C}$. (Here $C$ runs over the bases of the columns corresponding to every $\Pa$-name of length $R$ for every $R\geq tM$.) It
is clear that $B$ forms the base of an Alpern tower of height $\{N, N+1\}$. It remains to show that $B$ is independent of $\Pa$, which we will do by constructing a set $A$, disjoint from $B$, such that both $A$ and $A\cup B$ can be shown to be independent of $\Pa$.

Here is how $A$ is constructed. Consider again the tower over $\Ci$. This tower had $R$ levels and $RN$ rungs, some of which were selected for the base $B$. We now choose $\gamma +(t-1)\delta$ additional rungs for the set $A$. For each $j\neq i$, $\delta$ of these rungs should be contained in $P_j$, with the remaining $\gamma$ contained in $P_i$. (We don't worry about gaps and whatnot; just choose any such collection of rungs disjoint from the family of $B$ selections.)
Denote the union of the these additional rungs (in all of the columns over $\Ci$, $1\leq i\leq t$) by $A_{C}$. Finally, put $A=\bigcup _{C} A_{C}$.

That $A\cup B$ is independent of $\Pa$ is a consequence of the fact that for each $\Ci$, the number appearances of $P_j$ in the column over $\Ci$ is
precisely the number of $B$-selections from $P_j$ plus the number of $A$-selections from $P_j$. Accordingly, the relative masses of the cells of $\Pa$ restricted to $A\cup B$ are equal to the relative frequencies of the appearances of the cells of $\Pa$ in the column over $\Ci$. Therefore, since the proportion of the column that is selected for $A\cup B$ is independent of $\Ci$ (in fact is always equal to ${1\over N}$), and since the columns over the various $\Ci$ exhaust $X$, $A\cup B$ is independent of $\Pa$ (in fact $\mu\big(P_j \cap (A\cup B)\big) = {1\over N} \mu(P_j)$, $1\leq j\leq t$).

That $A$ is independent of $\Pa$, meanwhile, is a consequence of equation (\ref{bi}). Fixing $C$ and recalling that $b_i = {\mu(\Ci)\over \mu(C)}$,
that there were $\delta$
$P_j$-rungs in the column over $\Ci$ selected for $A$, $i\neq j$, and that there were $\gamma$ $P_i$-rungs in the column over $\Ci$ selected for $A$,
the relative mass of $P_i$ among the $A$-selections in the tower over $C$ is
\[ r_i = \frac{b_i \gamma + (1-b_i)\delta}{\gamma + (t-1)\delta} .\]
But, solving for $\mu(P_i)$ in equation (\ref{bi}), one gets that
\[ \mu(P_i) = \frac{b_i \gamma + (1-b_i)\delta}{\gamma + (t-1)\delta} \]
as well. So the intersection of $A$ with the column over $C$ is independent of $\Pa$. That this is true for every $C$ gives independence of $A$ from $\Pa$ simpliciter. \bs

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  \href{http://www.ams.org/mathscinet-getitem?mr=#1}{#2}
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TITLE: Evaluate $\int_{-\infty}^{\infty}\sin(ax)\operatorname{sech}^4(b(x-c))dx$, for $a,b,c>0$
QUESTION [2 upvotes]: $\newcommand{\sech}{\operatorname{sech}}$
I am trying to compute the following integral:
$$\int_{-\infty}^{\infty}\sin(ax)\sech^4(b(x-c))dx; a,b,c>0$$
I cannot seem to tackle this integral. I have tried integration by parts, but this leads nowhere since we need to evaluate $\sin(ax)$ at the infinities. How should I be approaching this integral?

REPLY [2 votes]: $\newcommand{\sech}{\operatorname{sech}}$
$\newcommand{\res}{\operatorname{Res}}$
Put $x=z/b+c$ and use angle-addition:
$$
\int_{-\infty}^{\infty} \sin(ax)\sech ^4(b(x-c))\,dx
$$
$$
=\frac{1}{b}\int_{-\infty}^{\infty} \sin(a(z/b+c))\sech ^4(z)\,dz
$$
$$
=\frac{\sin(ac)}{b}\int_{-\infty}^{\infty} \cos(\frac{a}{b}z)\sech ^4(z)\,dz+\frac{\cos(ac)}{b}\int_{-\infty}^{\infty} \sin(\frac{a}{b}z)\sech ^4(z)\,dz
$$Now $\sech^4$ is improperly integrable (in fact, rapidly decaying) and even, so the second integral vanishes by symmetry. Note that
$$
\int_{-\infty}^{\infty} \cos(\frac{a}{b}z)\sech ^4(z)\,dz=\Re\left(\int_{-\infty}^{\infty} \exp(\frac{a}{b} i z)\sech ^4(z)\,dz\right)
$$ We will use the Residue Theorem. Let $f(z)$ denote the integrand. A pole of order four occurs at $z=\pi i/2$, so it makes sense to take our contour as the rectangle $\gamma$ (with sub-contours $\gamma_1,\gamma_2,\gamma_3,\gamma_4$) whose path is $$\gamma:-R\longrightarrow R \longrightarrow R+\pi i \longrightarrow -R+\pi i \longrightarrow -R.$$
The details are left to the reader (or to Mathematica or some such), but we have:
$$
\int_{\gamma_3}f(z)\,dz = -\exp(-\frac{a}{b}\pi) \int _{\gamma_1}f(z)\,dz;\\
\lim_{R\to \infty} \int_{\gamma_2}f(z)\,dz=\lim_{R\to \infty} \int_{\gamma_4}f(z)\,dz=0
$$
Computing the residue is tedious, but the end result is:
$$
\res(f(z),\pi i /2)
 = \frac{1}{6}  \lim_{z\to \pi i/2} \frac{d^3}{dz^3}\frac{\exp(\frac{a}{b} iz)(z-\pi i/2)^4}{\sech^4(z)}
$$
$$
= -\frac{i a \left(a^2+4 b^2\right) \exp(-\frac{a}{b}\pi/2)}{6 b^3}
$$ Then in the limit, we have
$$
\lim_{R\to\infty}\int _{\gamma_1}f(z)\,dz = \Re\left((1-\exp(-\frac{a}{b}\pi))^{-1}\cdot 2\pi i \cdot \left(\frac{-i a \left(a^2+4 b^2\right) \exp(-\frac{a}{b}\pi/2)}{6 b^3}\right)\right)
$$
$$=\frac{\pi  a \left(a^2+4 b^2\right) \text{csch}\left(\frac{\pi  a}{2 b}\right)}{6 b^3}
$$
So in summary,
$$
\int_{-\infty}^{\infty} \sin(ax)\sech ^4(b(x-c))\,dx =  \frac{\pi  a \left(a^2+4 b^2\right) \text{csch}\left(\frac{\pi  a}{2
   b}\right) \sin (a c)}{6 b^4}
$$

Note: originally, I took the contour to be the semicircle enclosing the upper half plane, with poles at $t=\pi i(k+1/2)$, $k\in\mathbb{N}^0$. Their sum is a geometric series but showing the integral vanished along the circular arc was slightly tricky because the integrand is large, of size $ \cosh(\frac{a}{b}R)\sec^4(R)$, near $\theta=\pi/2$. If one were careful with the approximation by bounding the size of this region (the integrand rapidly decreases elsewhere), you could do the problem the same way.