text:
\begin{document}
\maketitle
\begin{abstract}
We prove that a terminal three-dimensional del Pezzo fibration
has no fibers of multiplicity
$\ge 6$.
We also obtain a rough classification possible configurations of singular points
on multiple fibers and give some examples.
\end{abstract}

\section{Introduction}
Throughout this paper a \textit{weak del Pezzo fibration} is
a projective morphism $f\colon X\to Z$ with connected fibers
from a threefold $X$ with
terminal singularities to a smooth curve $Z$
such that $-K_X$ is $f$-nef and $f$-big near a general fiber.
If additionally $-K_X$ is $f$-ample, we say that $f\colon X\to Z$
is a \textit{del Pezzo bundle}.
(We do not assume that $X$ is $\QQ$-factorial nor $\rho(X/Z)=1$).
The main reason to study del Pezzo fibrations
comes from the three-dimensional birational geometry,
namely the class of del Pezzo bundles with $\QQ$-factorial singularities and
relative Picard number one is one of three possible outcomes of
the minimal model program for threefolds of negative Kodaira dimension.

Our main result is the following.
\begin{theorem}
\label{main}
Let $f\colon X\to Z$ be a weak del Pezzo fibration and let $f^*(o)=m_oF_o$
be a special fiber of multiplicity $m_o$.
Then $m_o\le 6$.
Moreover, all the cases $1\le m_o\le 6$ occur.
Furthermore,
let $\B(F_o)=\basket{r_1,\dots,r_n}$ be the basket of singular
points of $X$ at which $F_o$ is not Cartier.
Then, in the case $m_o\ge 2$, there are only the following
possibilities:
\begin{enumerate}
\item
$m_o= 2$, $\B(F_o)=\basket{8}$,
$\basket{2,6}$,
$\basket{4,4}$,
$\basket{2, 2, 4}$, or
$\basket{2, 2, 2, 2}$,
\item
$m_o= 3$, $\B(F_o)=\basket{9}$,
$\basket{3, 3, 3}$, or
$\basket{3,6}$,
\item
$m_o= 4$, $\B(F_o)=\basket{2, 4, 4}$,
\item
$m_o= 5$, $\B(F_o)=\basket{5,5}$,
\item
$m_o= 6$, $\B(F_o)=\basket{2, 3, 6}$.
\end{enumerate}
The possible types of
singularities in $\B(F_o)$,
the local behavior of $F_o$ near singular points, and
the possible types of
a general fiber
are collected in Table \xref{table}.
\end{theorem}

\begin{warning}
In the statement of Theorem \ref{main}
and Table \xref{table} we do not assert that
the basket $\B(F_o)$ contains all the singularities along
$F_o$. It is possible that $F_o$ is Cartier at some non-Gorenstein points
(see Example \ref{ex-not-basket}).
\end{warning}

\begin{table}[h]
\renewcommand{\extrarowheight}{5pt}
\renewcommand{\arraystretch}{1.3}
\caption{}
\label{table}

\begin{tabular}{|l|l|l|l|l|c|}
\hline
type&$m_o$&$\B(F_o)=$&$(b_1,\dots,b_n)$&$q_i$&$K_{F_g}^2$
\\[-10pt]
&&$(r_1,\dots,r_n)$&&&
\\
\hline\hline
\type{2, 3, 6}&
$6$&
$\basket{ 2, 3, 6}$
&
$(1,\pm 1,\pm 1)$
&
$q_i\equiv -1$
&
6
\\ \hline
\type{5, 5}&
$5$
&
$\basket{5, 5}$
&$b_1^2+b_2^2\equiv 0$
&
$q_i\equiv -1$
&
5
\\ \hline
\type{2, 4, 4}&
$4$
&
$\basket{2, 4, 4}$
&
$(1,\pm 1,\pm 1)$
&
$q_i\equiv -1$
&
4,\, 8
\\ \hline
\type{3, 3, 3}&
$3$
&
$\basket{3, 3, 3}$
&
$(\pm 1,\pm 1,\pm 1)$
&
$q_i\equiv -1$
&
3,\, 6,\, 9
\\ \hline
\type{2, 2, 2, 2}&
$2$
&
$\basket{2, 2, 2, 2}$
&
$(1,1,1,1)$
&
$q_i\equiv 1$
&
even
\\ \hline
\type{3, 6}&
$3$
&
$\basket{3, 6}$
&
$(\pm 1,\pm 1)$
&
$q_i\equiv 4$
&
3,\, 6,\, 9
\\ \hline
\type{9}&
$3$
&
$\basket{9}$
&
$b_1=\pm 2q_1/3$
&
$q_1=3$ or $6$
&
$\equiv q_1/3\mod 3$
\\ \hline
\type{2, 2, 4}&
$2$
&
$\basket{2, 2, 4}$
&
$(1,1,\pm 1)$
&
$q_i\equiv r_i/2$
&
odd
\\ \hline
\type{4, 4}&
$2$&$\basket{4, 4}$
&
$(\pm 1,\pm 1)$
&
$q_i\equiv r_i/2$
&
even
\\ \hline
\type{2, 6}&
$2$& $\basket{2, 6}$
&
$(1, \pm 1)$
&
$q_i\equiv r_i/2$
&
even
\\ \hline
\type{8}&
$2$&$\basket{8}$
&
$(\pm 1)$ or $(\pm 3)$
&
$q_i\equiv r_i/2$
&
odd
\\ \hline
\end{tabular}

\end{table}

The idea of the proof is easy. In fact,
it is sufficient to compute dimensions of
linear systems $\dim |lF_o|$
by using the orbifold Riemann-Roch formula \cite{Reid-YPG1987}.
The main theorem is proved in \S\S \ref{sect-prep} -- \ref{sect-proof}.
In \S \ref{sect-exam} we give some examples.
In fact, it will be shown that
all cases in Table \ref{table}
except possibly for cases \type{2,6} and \type{8} occur.
Finally, in \S \ref{sect-45} we discuss
fibers of multiplicity $5$ and $6$.

\par\medskip\noindent
\textbf{Notation in Table \ref{table}.}
The number $b_k$ in the fourth column
is the weight which appears in a singularity
$\frac{1}{r_k}(1,-1, b_k)\in \B(F_o)$,
$q_k$ in the fifth column is an integer such that $F_o\sim q_kK_X$ near $P_k\in \B(F_o)$.
$F_g$ in the final column denotes a general fiber of $f$.
We say that the fiber $f^*(o)=m_oF_o$
is of type \type{r_1,\dots,r_n} if $\B(F_o)=\basket{r_1,\dots,r_n}$.

We also say that the fiber $F_o$
is \textit{regular} if it is of type \type{2,3,6},
\type{5,5}, \type{2,4,4}, \type{3,3,3}, or
\type{2,2,2,2}. Otherwise, $F_o$ is said to be \textit{irregular}.

\par\medskip\noindent
\textbf{Acknowledgments.}
Much of this
work was done at the Research Institute for Mathematical
Sciences (RIMS), Kyoto University in February 2008.
The second author would like to
thank RIMS for invitation to work there, for
hospitality and stimulating working environment.

\section{Preliminaries}
\label{sect-prelim}
\begin{discussion}
\textbf{Terminal singularities \cite{Mori-1985-cla}, \cite{Reid-YPG1987}.}
Let $(X,P)$ be a three-dimensional terminal singularity
of index $r$ and let $D$ be a Weil $\QQ$-Cartier divisor on $X$.
\end{discussion}

\begin{lemma}[{\cite[Corollary 5.2]{Kawamata-1988-crep}}]
\label{lemma-Kawamata-index-divisors}
In the above notation,
there is an integer $i$ such that $D\sim iK_X$ near $P$.
In particular, $rD$ is Cartier.
\end{lemma}

\begin{discussion}
\label{not-01}
Notation as above.
There is a deformation $X_\lambda$ of $X$ such that
$X_\lambda$ has only cyclic quotient singularities
$(X_\lambda,P_{\lambda,k})\simeq \frac {1}{r_k}(1,-1,b_k)$,
$0<b_k<r_k$, $\gcd(b_k, r_k)=1$.
Thus, to every theefold $X$ with terminal singularities,
one can associate
a collection $\B=\basket{(r_{P,k},b_{P,k})}$, where
$P_{\lambda,k}\in X_{\lambda,k}$ is a singularity
of type $\frac 1{r_{P,k}}(1,-1,b_{P,k})$.
This collection is called
the \textit{basket} of singularities of $X$. By abuse of
notation, we also will write $\B=\basket{r_{P,k}}$ instead of
$\B=\basket{(r_{P,k},b_{P,k})}$.
The index of $(X,P)$ is the least common multiple
of indices of points $P_{\lambda,k}$.
For any Weil divisor $D$, $\B(D)\subset \B$ denotes the collection of
points where $D$ is not Cartier.

Deforming $D$ with $(X,P)$
we obtain Weil divisors $D_{\lambda}$ on $X_\lambda$.
Thus we have a collection of numbers $q_k$ such that
$0\le q_k<r_k$ and
$D_{\lambda}\sim q_kK_{X_\lambda}$ near $P_{\lambda,k}$.

\end{discussion}

\begin{discussion}
\textbf{Orbifold Riemann-Roch formula \cite{Reid-YPG1987}.}
Let $X$ be a threefold with terminal singularities
and let $D$ be a Weil $\QQ$-Cartier divisor on $X$.
Then
\begin{multline}
\label{eq-RR}
\chi(D)=
\frac1{12}D\cdot (D-K_X)\cdot (2D-K_X)+
\\
+\frac1{12}D\cdot c_2(X)+\chi(\OOO_X)+\sum_{P\in \B} c_P(D),
\end{multline}
where
\begin{equation}
\label{eq-RR-cP-def}
c_P(D)=-q_P\frac{r_P^2-1}{12r_P}+\sum_{j=1}^{q_P-1}\frac{\ov{b_Pj}(r_P-\ov{b_Pj})}{2r_P},
\end{equation}
$q_P$ is such as in \ref{not-01}, and
$\overline{\phantom{X}}$ denotes the smallest residue $\mod r_P$.

Assume that
$D^2\equiv 0$. Then
\begin{equation}
\label{eq-RR-D2=0}
\chi(D)=
\frac1{12}D\cdot K_X^2+\frac1{12} D\cdot c_2(X)+\chi(\OOO_X)
+\sum c_P(D).
\end{equation}
We have (see, e.~g., \cite[proof of 2.13]{Alexeev-1994ge})
\begin{equation}
\label{eq-RR-cPK-K}
c_P(-K)=\frac{r_P^2-1}{12r_P}-\frac{b_P(r_P-b_P)}{2r_P},
\qquad
c_P(K)=-\frac{r_P^2-1}{12r_P}.
\end{equation}
\end{discussion}

\begin{construction}[Base change]
\label{base-change}
Let $f\colon X\to Z$ be a weak del Pezzo fibration
and let $f^*(o)=m_oF_o$
be a special fiber of multiplicity $m_o$.
Regard $f\colon X\to (Z,o)$ as a germ. Let
$(\CC,0)\simeq (Z',o')\to (Z,o)\simeq (\CC,0)$ is given by $t \mapsto t^{m_o}$
and let $X'$ be the normalization of $X\times_Z Z'$. We obtain the
following commutative diagram:
\begin{equation}
\label{eq-base-change}
\xymatrix{
X'\ar[r]^{\pi}\ar[d]_{f'}&X\ar[d]^{f}
\\
Z'\ar[r]&Z
}
\end{equation}
Here $f'$ is a weak del Pezzo fibration with
special fiber $F_o'=f'^{*}o'=\pi^{*}F_o$ of multiplicity $1$ and
$\pi$ is a $\muu_{m_o}$-cover which is
\'etale outside of the set $M$ of points where
$F_o$ is not Cartier.
Hence there is a $\muu_{m_o}$-action on
$X'$ such that $X=X'/\muu_{m_o}$ and the action is free
outside of $M$.

Conversely, let $f'\colon X'\to Z'\ni o'$ be a weak del Pezzo fibration
with central fiber of multiplicity $1$. Assume that $f$
equipped with an equivariant $\muu_{m_o}$-action such that
the action on $X'$ is \'etale in codimension two.
If the quotient $X'/\muu_{m_o}$ has only terminal singularities,
then $X'/\muu_{m_o}\to Z'/\muu_{m_o}$ is a weak del Pezzo fibration
with special fiber of multiplicity $m_o$.
\end{construction}

\begin{proposition}
\label{cor-point}
Let $f\colon X\to Z$ be a weak del Pezzo fibration. Let $f^*(o)=m_oF_o$
be a special fiber of multiplicity $m_o$.
There is a point $P\in F_o$ such that the
index of $F_o$ at $P$ is divisible by $m_o$.
\end{proposition}
\begin{proof}
Regard $f\colon X\to (Z,o)$ as a germ and apply
Construction \ref{base-change}.
It is sufficient to show that
$\muu_{m_o}$ has a fixed point on $F_o'$ (see Lemma \ref{lemma-Kawamata-index-divisors}).

First we consider the case of del Pezzo bundle, i.e., the case where $-K_X$ is ample.
Let $\gamma$ be the log canonical threshold of $(X',F_o')$
and let $W'\subset X'$ be a minimal center of log canonical
singularities of $(X',\gamma F_o')$ (see \cite[\S 1]{Kawamata1997}).

Assume that $\dim W'\le 1$.
Let $H$ be a general hyperplane section of $X$
passing through
$\pi(W')$ and let $H':=\pi^*H$.
For $0<\varepsilon\ll 1$, the pair $(X',\, \gamma F_o'+\varepsilon H')$
is not LC along $\pi^{-1}\pi(W')$ and LC outside.
Therefore for some $0<\delta\ll \epsilon$
the pair $(X',\, (\gamma-\delta)F_o'+\varepsilon H')$
is not KLT along $\pi^{-1}\pi (W')$ and KLT outside.
Moreover, $W'$ is a minimal LC center for $(X',\, (\gamma-\delta)F_o'+\varepsilon H')$.
Recall that
any irreducible component of
the intersection of two LC centers
is also an LC center  \cite[Proposition 1.5]{Kawamata1997}.
Hence $W'$ is the only LC center in its
neighborhood. Since the boundary
$(\gamma-\delta)F_o'+\varepsilon H'$ is $\muu_{m_o}$-invariant,
all the $gW'$ for $g\in \muu_{m_o}$ are also
centers of log canonical singularities for
the pair $(X',\, (\gamma-\delta)F_o'+\varepsilon H')$.
On the other hand, the locus $\muu_{m_o}W'$
of log canonical singularities
for the pair
$(X',\, (\gamma-\delta)F_o'+\varepsilon H')$ is connected,
see \cite[\S 5]{Shokurov-1992-e-ba}, \cite[17.4]{Utah}.
Hence $\muu_{m_o}W'$ is irreducible and so $W'$ is
$\muu_{m_o}$-invariant. If $W'$ is a point, we are done.
Otherwise $W'$ is a smooth rational curve
\cite[Th. 1.6]{Kawamata1997},
\cite{Kawamata-1997-Adj}.
But any cyclic group acting on $\PP^1$ has a fixed points.

Assume that $\dim W'=2$,
that is, $W'=\down{\gamma F_o'}$ and the pair $(X',\gamma F_o')$ is PLT.
By the inversion of adjunction \cite[3.3]{Shokurov-1992-e-ba}, \cite[17.6]{Utah}
and Connectedness Lemma \cite[\S 5]{Shokurov-1992-e-ba}, \cite[17.4]{Utah}
the surface
$W'$ is irreducible, normal and has only KLT singularities.
Hence $W'$ is a KLT log del Pezzo surface.
In particular, $W'$ is rational.
Then the assertion follows by Lemma \ref{lemma-claim-fp}
below.

Now we consider the general case.
We apply $\muu_{m_o}$-equivariant MMP in the category
$\muu_{m_o}$-threefolds (i.e., threefolds with terminal singularities
and such that every $\muu_{m_o}$-invariant Weil divisor is $\QQ$-Cartier,
see e.g. \cite[0.3.14]{Mori-1988}).
Let $X_1\to X'$ be a $\muu_{m_o}$-equivariant $\QQ$-factorialization.
Run $\muu_{m_o}$-equivariant MMP over $Z'$:
\[
X_1 \dashrightarrow X_2\dashrightarrow \cdots \dashrightarrow
X_N.
\]
These maps induce a sequence maps
\[
X_1/\muu_{m_o} \dashrightarrow X_2/\muu_{m_o}\dashrightarrow \cdots \dashrightarrow
X_N/\muu_{m_o}.
\]
where each step is either $K$-negative divisorial contraction or a flip
(both are not neseccarily extremal).
Hence, on each step the quotient $X_i/\muu_{m_o}$ has only terminal singularities
and the action of $\muu_{m_o}$ on $X_i$ is free in codimension two.
On the last step $X_N$ is either a del Pezzo bundle over $Z'$
with $\rho^{\muu_{m_o}}(X_N/Z')=1$ or a $\QQ$-conic bundle over a surface $S$ and
$S/Z'$ is a rational curve fibration.
In both cases $\muu_{m_o}$ has a fixed point on $X_N$.
We prove the existence of fixed point on $X_i$ by a descending induction on $i$.
So we assume that $X_{i+1}$ has a fixed point, say $P$.
If $\psi_i\colon X_{i} \dashrightarrow X_{i+1}$ is a flip, we may assume that
$P$ is contained in the flipped curve $C_{i+1}\subset X_{i+1}$.
In this case $\muu_{m_o}$
acts on
a connected closed subset of
the flipping curve $C_{i}\subset X_{i}$.
Since $C_{i}$ is a tree of rational curves, $\muu_{m_o}$
has a fixed point on $C_{i}$.
Similar argument works in the case where $\psi_i\colon X_{i} \to X_{i+1}$
is a contraction of a
$\muu_{m_o}$-invariant
divisor to a curve.
Thus we may assume that $\psi_i\colon X_{i} \to X_{i+1}$
is a divisorial contraction that contracts a
$\muu_{m_o}$-invariant
divisor $E\subset X_i$
to $P$.
Let $\gamma$ be the log canonical threshold of $(X_i,E)$
and let $W_i\subset X_i$ be a minimal center of log canonical
singularities of $(X_i,\gamma E)$.
As in the del Pezzo bundle case above,
considering the action of $\muu_{m_o}$ on $W_i$ we find
a fixed point. This proves our proposition.
\end{proof}

\begin{lemma}
\label{lemma-claim-fp}
Let $S$ be a rational surface.
Then any action of a finite
cyclic group on $S$ has a fixed point.
\end{lemma}

\begin{proof}
Let $\muu_m$ be the cyclic group acting on $S$.
Replacing $S$ with its normalization and the minimal
resolution, we may assume that $S$ is smooth.
Since $S$ is rational, $H^i(S,\CC)=0$ if $i$ is odd.
Then the assertion follows by the Lefschetz fixed point formula.
\end{proof}

\section{Preparations}
\label{sect-prep}
\begin{notation}
\label{not}
Let $f\colon X\to Z$ be a weak del Pezzo fibration.
Compactify $X$ and $Z$ and resolve $X$ only above the added points of $Z$.
Thus we may assume that both $X$ and $Z$ are projective.
Let $F_g$ be a general fiber and let $f^*(o)=m_oF_o$
be a special fiber of multiplicity $m_o$.
Write $m_o=m\alpha$, where $m$ and $\alpha$ are positive integers
and put $D:=\alpha F_o$. Then $m_oF_o=mD=f^*(o)$.
\end{notation}

\begin{computation}
By a variant of J. Koll\'ar's Higher Direct Images Theorem
(see \cite[1-2-7]{KMM}, \cite{Nakayama1986}), one has that
$R^i f_*\OOO_X(K_X-jD)$ is torsion free for all $i$.
But its restriction to the general fiber $F_g$ is zero for $i \neq2$
because $-K_{F_g}$ is nef and big.
Hence $R^i f_*\OOO(K_X-jD)=0$ for $i\neq 2$.
Further, the
Leray spectral sequence yields
\[
H^q(X,K_X-jD)=H^{q-2}(Z,R^2 f_*\OOO(K_X-jD))=0
\]
for $q-2 \neq1$
and $j \gg0$ because $R^2 f_* \OOO(K_X-jD)$ is very negative.
By Serre duality
\[
H^{3-q}(X,jD)\simeq H^q(X,K_X-jD)^{\vee} =0
\]
for $q \neq 3$ and $j \gg0$.

Finally, $H^{i}(X,jD)=0$ for all $i>0$, $j> j_0\gg 0$. We also have
\[
H^0(X,j f^*(o)+lD)\simeq H^0(X,j f^*(o))
\]
for $l=0,\dots,m-1$.
Put $j_1:=\down{j_0/m}$ and
\[
\Theta_l:=\frac{1}{mj_1 }h^0(X,
j_1 f^*(o))-\frac{1}{mj_1 +l} h^0(X,j_1 f^*(o)+lD).
\]
Thus for $l=0,\dots,m-1$ we have
\begin{equation}
\label{eq-RR-e1}
\Theta_l=\frac{l}{mj_1 (mj_1 +l)}h^0(X,j_1 f^*(o))=\frac{l(j_1 -p_a+1)}{mj_1 (mj_1 +l)},
\end{equation}
where $p_a$ is the genus of $Z$. On the other hand, by
\eqref{eq-RR-D2=0}
\begin{equation}
\label{eq-RR-e2}
\Theta_l=-\frac{1}{mj_1 +l}\sum_{P\in \B} c_P(lD)+\frac{l}{mj_1 (mj_1 +l)}\chi(\OOO_X).
\end{equation}
Comparing \eqref{eq-RR-e1} and \eqref{eq-RR-e2} we get
\begin{equation}
\label{eq-RR-e3}
m\sum_{P\in \B} c_P(lD)=-l,\qquad l=0,\dots,m-1.
\end{equation}
\end{computation}

\begin{computation}
Denote
\[
\begin{array}{lll}
\Delta_a&:=&\chi(\OOO_X(-K-aF_o))-\chi(\OOO_X(-K-(a+1)F_o)),
\\[9pt]
\delta_a&:=&\sum\limits_{P\in \B} c_P(-K-aF_o)- \sum\limits_{P\in \B} c_P(-K-(a+1)F_o).
\end{array}
\]
As above, for $a=0,\dots,m_o-2$, the following equality holds
\begin{multline*}
\Delta_a=
\frac{13}{12}K^2\cdot F_o
+\frac{1}{12}F_o\cdot c_2(X)+
\sum_{P\in \B} c_P(-K-aF_o)-
\\
-\sum_{P\in \B} c_P(-K-(a+1)F_o)
=
\frac{13}{12m_o}K^2\cdot F_g
+\frac{1}{12m_o}F_g\cdot c_2(X)+\delta_a.
\end{multline*}
Since $K^2\cdot F_g=K_{F_g}^2$ and
$F_g\cdot c_2(X)=c_2(F_g)=12-K_{F_g}^2$, we have
\begin{equation}
\label{eq-Delta}
\Delta_a=\frac{K_{F_g}^2+1}{m_o}+\delta_a.
\end{equation}
\end{computation}

\begin{discussion}
\textbf{Some computations.}
Let $(X,P)$ be a cyclic quotient singularity of type
$\frac 1r (a,-a,1)$, let $D$ be a Weil divisor on $X$,
and let $m$ be a natural number. We have $D\sim q K_X$ for some $0\le q < r$. Denote
\begin{equation}
\label{eq-RR-su-def}
\Xi_{P,m}:=\sum_{l=1}^{m-1} c_P(lD).
\end{equation}
We also will write $\Xi_P$ or $\Xi$ instead of $\Xi_{P,m}$ if no confusion is likely.
By definition
\begin{equation}
\label{eq-S1}
\Xi_{P,m}=\sum_{l=1}^{m-1}
\left(-
\ov{ql}\frac{r^2-1}{12r}+ \sum_{j=1}^{\ov{ql}-1}\frac{\ov{bj}(r-\ov{bj})}{2r}\right).
\end{equation}
We compute $\Xi$ in some special situation:
\end{discussion}

\begin{lemma}
\label{lemma-Sp}
Let $s:=\gcd(r,q)$.
Write
$r=sm$ and $q=sk$ for some $s,\, k\in \ZZ_{>0}$
\textup(so that $\gcd(m,k)=1$\textup).
Then
\begin{equation}
\label{eq-sum-f}
\Xi_{P,m}=-\frac{m^2-1}{24m}r.
\end{equation}
\end{lemma}
\begin{proof}
By our assumption $\gcd(m,k)=1$ the parameter $\ov{ql}=\ov{skl}$
runs through all the values $sl$, $l=1,\dots,m-1$. Hence,
\begin{equation*}
\Xi=-\sum_{l=1}^{m-1}
sl\frac{r^2-1}{12r}+\sum_{l=1}^{m-1} \sum_{j=1}^{sl-1}\frac{\ov{bj}(r-\ov{bj})}{2r}.
\end{equation*}
Since $\ov{bj}(r-\ov{bj})=\ov{bj'}(r-\ov{bj'})$ for $j+j'=r$, we have
\[
\sum_{j=1}^{sl-1}\frac{\ov{bj}(r-\ov{bj})}{2r}
=
\sum_{j=r-sl+1}^{r-1}\frac{\ov{bj}(r-\ov{bj})}{2r}.
\]
Therefore,
\begin{multline*}
\Xi=-
\frac{m(m-1)s}{2}
\frac{r^2-1}{12r}+
\frac{m-1}{2}
\sum_{j=1}^{r-1}\frac{\ov{bj}(r-\ov{bj})}{2r}-
\frac12 \sum_{l=1}^{m-1}\frac{\ov{bsl}(r-\ov{bsl})}{2r}=
\\
-\frac{(m-1)r}{2}
\frac{r^2-1}{12r}+
\frac{m-1}{2}
\sum_{j=1}^{r-1}
\frac{\ov{bj}(r-\ov{bj})}{2r}
-\frac12 \sum_{l=1}^{m-1}\frac{sl(r-sl)}{2r}
=\frac{m-1}{2}c_P(rK)\\
-\frac{s}{4} \sum_{l=1}^{m-1}l
+\frac{s^2}{4r} \sum_{l=1}^{m-1} l^2=
-\frac{sm(m-1)}{8}+\frac{s^2}{24r}(m-1)m(2m-1)=
\\
=
-\frac{m-1}{8}
\left(r- \frac{s}{3}(2m-1)\right)=
-\frac{m-1}{24}
\left(r+\frac{r}{m}\right).
\end{multline*}
(We used $sm=r$ and $c_P(rK)=0$.) This proves our lemma.
\end{proof}

\begin{lemma}
\label{lemma-su-div}
If $m=m_1m_2$, where $m_1D$ is Cartier, then
\[
\Xi_{P,m}=m_2\Xi_{P,m_1}.
\]
\end{lemma}
\begin{proof}
Follows by \eqref{eq-RR-su-def} because $c_P(tD)$
is $r$-periodic.
\end{proof}

\section{Proof of Theorem \ref{main}}
\label{sect-proof}
Notation as in \ref{not}.
Near each singular point $P\in X$ of index $r_P$
we write
\[
D\sim q_PK_X.
\]
Then $mq_PK_X\sim mD$ is Cartier near $P$. Hence,
\begin{equation}
\label{eq-RR-div1}
mq_P \equiv 0 \mod r_P.
\end{equation}
From \eqref{eq-RR-e3} we have
\begin{equation}
\label{eq-RR-su}
\sum_{P\in \B} \Xi_{P,m}=-\sum_{l=1}^{m-1} \frac lm
=-\frac{m-1}{2}.
\end{equation}

\begin{proposition}
\label{prop-Dprime}
Notation as above. If $m$ is prime, then
we have one of the following possibilities:
\begin{enumerate}
\renewcommand
\labelenumi{(\arabic{section}.\arabic{equation}.\arabic{enumi})}
\renewcommand
\theenumi{(\arabic{section}.\arabic{equation}.\arabic{enumi})}
\item
\label{main-m2-b8}
\quad $m= 2$, $\B(D)=\basket{8}$,
\item
\label{main-m2-b26}
\quad $m= 2$, $\B(D)=\basket{2,6}$,
\item
\label{main-m2-b44}
\quad $m= 2$, $\B(D)=\basket{4,4}$,
\item
\label{main-m2-b224}
\quad $m= 2$, $\B(D)=\basket{2, 2, 4}$,
\item
\label{main-m2-b2222}
\quad $m= 2$, $\B(D)=\basket{2, 2, 2, 2}$,
\item
\label{main-m3-b9}
\quad $m= 3$, $\B(D)=\basket{9}$,
\item
\label{main-m3-b333}
\quad $m= 3$, $\B(D)=\basket{3, 3, 3}$,
\item
\label{main-m3-b36}
\quad $m= 3$, $\B(D)=\basket{3,6}$,

\item
\label{main-m5-b55}
\quad $m= 5$, $\B(D)=\basket{5,5}$,

\item
\label{m5-b10}
\quad $m=5$, $\B(D)=\basket{10}$,
\item
\label{m11-b11}
\quad $m=11$, $\B(D)=\basket{11}$.

\end{enumerate}.
\end{proposition}

\begin{proof}
By \eqref{eq-RR-div1} we have $mq_P\equiv 0 \mod r_P$ and
$r_P\equiv 0\mod m$ for all $P\in \B(D)$
(otherwise $q_P\equiv 0 \mod r_P$ and $P\notin \B(D)$).
Put $s_P:=r_P/m$. Then $q_P=s_Pk_P$ for some $k_P\in \ZZ_{>0}$.
Since $\gcd(k_P,q_P)=1$,
the assumption
of Lemma \xref{lemma-Sp} holds for each point $P\in \B(D)$.
Combining \eqref{eq-sum-f} with \eqref{eq-RR-su} we obtain
\[
(m+1)\sum_{P\in \B} r_P=12m.
\]
Hence, $m\in \{2,\, 3,\, 5,\, 11\}$.
Using the fact
$r_P\equiv 0\mod m$ we get the statement.
\end{proof}

\begin{proposition}
\label{prop-Dprime-1}
Cases \xref{m5-b10} and \xref{m11-b11} do not occur.
In particular,
the assertion of Theorem \xref{main} holds if $m_o$ is prime.
\end{proposition}
\begin{proof}

Consider the case \ref{m11-b11}. Since $\gcd(q,m)=1$,
there is $0<l<r=m$ such that $ql\equiv 1 \mod m$.
Then by \eqref{eq-RR-e3} and \eqref{eq-RR-cPK-K} we have
\[
-\frac l{11}=c_P(lD)=c_P(K)=-\frac {r^2-1}{12r}=-\frac {10}{11},
\]
so $l=q=10$. Then again by \eqref{eq-RR-e3} and \eqref{eq-RR-cPK-K}
\[
-\frac 1{11} =c_P(D)=c_P(-K)=\frac {r^2-1}{12r}-\frac{b(r-b)}{2r}=
\frac {10}{11}-\frac{b(11-b)}{22}.
\]
Hence, $b(11-b)=22$ and $b$ cannot be coprime to $11$, a contradiction.

Consider the case \ref{m5-b10}.
Since $mq=5q\equiv 0\mod r=10$, $q$ is even.
There is $0<l<5$ such that $ql\equiv 2 \mod r$.
Then by \eqref{eq-RR-e3} we have
\[
-\frac l{5}=c_P(lD)=c_P(2K)=-\frac {2(r^2-1)}{12r}+\frac{b(r-b)}{2r}
=\frac {b(10-b)-33}{20}.
\]
Thus $b(10-b)+4l=33$, $b\in \{3, \, 7\}$, $l=3$, and $q=4$.
Again by \eqref{eq-RR-e3}
\begin{equation*}
-\frac 15=c_P(D)=-\frac {4(r^2-1)}{12r}+
\sum_{j=1}^3\frac{\ov{bj}(r-\ov{bj})}{2r}=
-\frac {33}{10}+
\sum_{j=1}^3\frac{\ov{3j}(10-\ov{3j})}{20}=
-\frac {3}{5},
\end{equation*}
a contradiction. This proves our lemma.
\end{proof}

\begin{corollary}
\label{cor-p}
For every prime divisor $d$ of $m_o$
we have $d\in \{2,\, 3,\, 5\}$.
\end{corollary}
\begin{proof}
Apply Propositions \ref{prop-Dprime} and \ref{prop-Dprime-1} with $D=\frac{m_o}{d}F_o$.
\end{proof}

Let $P_i$ be points of $\B(F_o)$.
Let $P=P_1$ be a point in $\B(F_o)$
whose index $r_{P_1}$ is divisible by $m_o$ (see Proposition \ref{cor-point}).
For short, below we will write $r_i$, $b_i$, $q_i$, etc
instead of $r_{P_i}$, $b_{P_i}$, $q_{P_i}$, respectively.

\begin{corollary}
$m_o$ is not divisible by $m \in \{16, 27, 25,
10,15, 12, 18\}$.
\end{corollary}

\begin{proof}
Let $d=2$, $3$ or $5$ be a prime divisor of $m_o$
and let $D=\frac {m_o}{d}F_o$.
Then $dD=f^*(o)$ and $D$
is not Cartier at $P_1$.
In this case, by
Propositions \ref{prop-Dprime} and \ref{prop-Dprime-1} the index of $(X,P_1)$
is at most $9$, a contradiction.
\end{proof}

\begin{corollary}
If $m_o$ is not prime, then $m_o \in \{4,\, 6,\, 8,\, 9\}$.
\end{corollary}

\begin{lemma}
\label{main-m6-b236}
If $m_o= 6$, then $\B(F_o)=\basket{2, 3, 6}$.
Moreover, $\gcd(r_P,q_P)=1$ for all $P\in \B(F_o)$.
\end{lemma}

\begin{proof}
Take $D=3F_o$.
Then $2D\sim f^*(o)$ but $D$ is not Cartier at $P_1$.
Hence $(X,P_1)$ is of index 6 and for $D$
we are in the case \ref{main-m2-b26},
that is, $\B(3F_0)=\basket{2,\, 6}$.
At all points $P_i\notin \B(3F_0)$ the divisor $3F_o$
is Cartier.
Similarly, take $D'=2F_o$. Then for
$D$ we get the case \ref{main-m3-b36},
that is, $\B(2F_o)=\basket{3,\, 6}$.
Hence $\B(F_o)$ contains three points
$P_1$, $P_2$, $P_3$ of indices $6$, $2$, $3$,
respectively, and in all other points
both $D'=2F_o$ and $D=3F_o$ are Cartier.
Hence $F_o=D-D'$ is Cartier outside of $P_1$, $P_2$, $P_3$
and $\B(F_o)=\basket{2, 3, 6}$.
\end{proof}

\begin{lemma}
\label{main-m4-b244}
If $m_o= 4$, then $\B(F_o)=\basket{2, 4, 4}$.
Moreover, $\gcd(r_P,q_P)=1$ for all $P\in \B(F_o)$.
\end{lemma}

\begin{proof}
Clearly, $2F_o$ is Cartier at all points of index $2$.
Hence $\B(2F_o)$ contains no such points
and for $\B(2F_o)$ we are in the case \ref{main-m2-b8}
or \ref{main-m2-b44}.
For all points $P_i\notin \B(2F_o)$
the divisor $2F_o$ is Cartier at $P_i$.
Hence, $q_{i}=r_{i}/2$.

Assume that $\B(2F_o)=\basket{8}$.
Let $P\in \B(2F_o)$.
Since $4F_o$ is Cartier, $4q_P\equiv 0 \mod 8$
(but $2q_P\not \equiv 0 \mod 8$).
By Lemma \ref{lemma-Sp} and \ref{lemma-su-div} we have
\[
\Xi_{P_1,4}=-\frac{5}{4},
\qquad \Xi_{P_j,8}=4\Xi_{P_j,2}=-\frac {r_{j}}4 ,
\quad j\neq 1.
\]
Therefore, by \eqref{eq-RR-su} the following holds $\sum_{i\neq 1} r_i=1$,
a contradiction.

Hence $\B(2F_o)=\basket{4,4}$.
At both points $P_i\in \B(2F_o)$
we have $F_o\sim \pm K_X$ near $P_i$.
Again by Lemma \ref{lemma-Sp} and \ref{lemma-su-div}
\[
\Xi_{P_i,4}=-\frac{5}{8},\quad i=1,\, 2
\qquad \Xi_{P_j,4}=2\Xi_{P_j,2}=-\frac {r_{j}}8,
\quad j\neq 1, \, 2.
\]
Therefore, by \eqref{eq-RR-su} we have $\sum\limits_{i\neq 1,\, 2} r_i=2$
and there is only one solution $\B(F_o)=\basket{4,4,2}$.
\end{proof}

\begin{corollary}
$m_o\neq 8$
\end{corollary}
\begin{proof}
Indeed, if $m_o=8$, then for $\B(2F_o)$
there is only one possibility from Lemma \ref{main-m4-b244}.
This contradicts Proposition \ref{cor-point}.
\end{proof}

\begin{lemma}
$m_o\neq 9$.
\end{lemma}
\begin{proof}
Assume that $m_o=9$. Take $D:=3F_o$.
Then $3D\sim f^*(o)$ but $D$ is not Cartier at $P_1$.
Hence, $\gcd(q_{1},r_{1})=1$, $(X,P_1)$ is of index $9$ and for $D$
we are in the case \ref{main-m3-b9},
that is, $\B(D)=\basket{9}\subset \B(F_o)$.
In all points $P_i\in \B(F_o)$, $P_i\neq P_1$ the divisor $D=3F_o$
is Cartier.
Hence by Lemma \ref{lemma-Sp} and \ref{lemma-su-div} we have
\[
\Xi_{P_1,9}=-\frac{10}{3},
\qquad \Xi_{P_i,9}=3\Xi_{P_i,3}=-\frac {r_{i}}3, \quad i\neq 1.
\]
Therefore, by \eqref{eq-RR-su}
\[
-4=\sum \Xi_{P_i,m}=-\frac{10}{3}-\frac 13 \sum_{i\neq 1} r_{i},
\qquad r_{i}= 2.
\]
This contradicts $r_{i}\equiv 0\mod 3$.
\end{proof}

\begin{computation}
The last lemma finishes the proof of Theorem \ref{main}.
It remains to compute values $b_k$, $q_k$, and $K_{F_g}^2$
in Table \ref{table}.

First we compute the possible values of $q_i$.
We may assume that $1\le q_i<r_i$.
In regular cases (\type{2,3,6}, \type{5,5}, \type{3,3, 3},
\type{2,4,4}, \type{2,2,2,2}) we have $\gcd (q_i, r_i)=1$
(see Lemmas \ref{main-m6-b236} and \ref{main-m4-b244})
and $m_o\ge r_i$ for all $i$.
Take $1\le l\le m_o-1$ so that $q_il\equiv 1\mod r_i$.
Then by \eqref{eq-RR-cPK-K} and \eqref{eq-RR-e3}
the following equality holds
\[
\sum_i c_{P_i}(lF_o)=\sum_i c_{P_i}(K)=
-\sum_i \frac{r_i^2-1}{12r_i}=-\frac{l}{m_o}.
\]
From this we immediately obtain $l\equiv q_i\equiv -1 \mod r_i$
for all $i$.

If $m_o=2$ (cases \type{4\times 2},
\type{2,2,4}, \type{4,4}, \type{2,6},
\type{8}), then $2F_o$ is Cartier.
Hence $q_i=r_i/2$. It remains to consider only cases
\type{9} and \type{3,6}.
In case \type{9}, since $3F_o$ is Cartier,
we have $q:=q_1=3$ or $6$.
If $q=3$, then by \eqref{eq-RR-e3} we have
\[
-1=3c_P(F_o)=3c_P(3K)=- \frac {40}{6}
+\frac{b(9-b)}{6}+\frac{\ov{2b}(9-\ov{2b})}{6}.
\]
Hence, $34=b(9-b)+\ov{2b}(9-\ov{2b})$ and
$5b^2\equiv 2\mod 9$. This immediately implies $b\equiv \pm 2$.
Similarly, if $q=6$, then $b^2\equiv -2\mod 9$
and $b\equiv \pm 4$.

Finally consider the case \type{3,6}.
Then by \eqref{eq-RR-cPK-K} and \eqref{eq-RR-cP-def}
\[
c_{P_1}(F_o)=
\begin{cases}
-2/9&\text{if $q_1=1$}
\\
-1/9&\text{if $q_1=2$}
\end{cases}
\qquad
c_{P_2}(F_o)=
\begin{cases}
-5/9&\text{if $q_1=2$}
\\
-1/9&\text{if $q_1=4$}
\end{cases}
\]
The equality $c_{P_1}(F_o)+c_{P_2}(F_o)=-1/3$
(see \eqref{eq-RR-e3}) holds only if $q_1=1$, $q_2=4$.

\begin{corollary}
The fiber $F_o$ is regular if and only if $q_i\equiv -1\mod r_i$
for all $i$. In particular,
for regular $F_o$ near each point
$P\in F_o$
where $F_o$ is not Cartier we have $K_X+F_o\sim 0$ .
\end{corollary}

\end{computation}

\begin{computation}
Now we find the possible values of $b_i$.
In all cases except for \type{5,5} and \type{9}
the relations $\gcd(r_i,q_i)=1$ is sufficient to get the conclusion.
The case \type{9} was treated above.
Consider the case \type{5,5}. Then by \eqref{eq-RR-cPK-K} and
\eqref{eq-RR-e3} we have $10=b_1(5-b_1)+b_2(5-b_2)$.
Hence $b_1^2+b_2^2\equiv 0\mod 5$.
\end{computation}

\begin{computation}
To obtain the possible values for $K_{F_g}^2$
we use \eqref{eq-Delta} with $a=0$.
Since $\Delta_a$ is an integer, it is sufficient to
compute $\delta_0=c_P(-K)-c_P(-K-F_o)$.
Table \ref{table1}
gives all values of $\delta_0$.
For example, if $F_o$ is regular, then
$q_P\equiv -1\mod r_P$ for all $P$ and
$\delta_0=\sum c_P(-K)=\sum c_P(F_o)$. So by
\eqref{eq-RR-cPK-K} and \eqref{eq-RR-e3}
we have $\delta_0=-1/m_o$.
Assume that $q_P=r_P/2$ (and all the $r_p$ are even). Then
\[
\delta_0 =\sum_{P\in \B}\left( c_P(-K)- c_P\left(\frac{r_P-2}{2}K\right)\right).
\]
Hence by \eqref{eq-RR-cPK-K} and \eqref{eq-RR-cP-def}
\begin{itemize}
\item[]
$r_P=2$ $\Longrightarrow$ $\delta_0= c_P(-K)= -1/8$,
\item []
$r_P=4$ $\Longrightarrow$ $\delta_0= c_P(-K)- c_P(K)= 1/4$,
\item []
$r_P=6$ $\Longrightarrow$ $\delta_0= c_P(-K)- c_P(2K)= 5/8$,
\item []
$r_P=8$ $\Longrightarrow$ $\delta_0= c_P(-K)- c_P(3K)=
1$ or $0$ if $b_P=1$ or $3$, respectively.
\end{itemize}
This immediately gives the values of $\delta_0$ in cases \type{2,2,4},
\type{4,4}, \type{2,6}, and \type{8}. Cases \type{3,6} and \type{9}
are similar.
\end{computation}

\begin{table}[h]
\renewcommand{\extrarowheight}{5pt}
\renewcommand{\arraystretch}{1.3}
\caption{}
\label{table1}

\begin{tabular}{|p{50pt}||c|c|c|c|c|c|c|c|c|c|c|}
\hline
 &regular &\type{3,6}
&\type{9}
&\type{2,2,4}&\type{4,4}&\type{2,6}
&\type{8}
\\[5pt]
\hline

$\delta_0$&$-\frac1{m_o}$
&$\frac23$& $\frac{6-q_1}9$&
$0$&$\frac12$&$\frac12$&$\frac{3-|b_1|}{2}$
\\[5pt] \hline
\end{tabular}

\end{table}

\section{Examples}
\label{sect-exam}
In this section we construct some examples of del Pezzo bundles
with multiple fibers. We
use notation of Construction \ref{base-change}.
We start with regular case.
\begin{proposition}
\label{prop-examples}
Let $f'\colon X'\to Z'\ni o'$ be a Gorenstein del Pezzo bundle.
Assume that the central fiber $F_o':=f'^{-1}(o')$
has only Du Val singularities.
Assume also that the cyclic group
$\muu_{m_o}$ acts on $X'$ and $Z'$ so that
\begin{enumerate}
\item
the action on $Z'$ is free outside of $o'$,
\item
$f'$ is $\muu_{m_o}$-equivariant,
\item
the action on $F_o'$ is free in codimension one,
\item
the quotient $F_o:=F_o'/\muu_{m_o}$ has only
Du Val singularities.
\end{enumerate}
Then $f\colon X=X'/\muu_{m_o}\to Z=Z'/\muu_{m_o}$
is a del Pezzo bundle with regular central fiber of multiplicity
$m_o$ and, moreover, $F_o\sim -K_X$ near each point $P\in X$.
\end{proposition}
\begin{proof}
In notation of Construction \ref{base-change} it is sufficient to
show that $X$ has only terminal singularities.
Since $X'$ has only terminal singularities and
the action of $\muu_{m_o}$ is free outside of
a finite number of points $P'_k$ lying on $F_o'$,
the quotient $X$ is smooth outside of $\pi(P_k')\in F_o$.
By the inversion of adjunction \cite[17.6]{Utah} the
pair $(X,F_o)$ is PLT near $F_o$. Since $F_o$ is Gorenstein,
the divisor $K_X+F_o$ is Cartier. Hence the
pair $(X,F_o)$ is canonical near $F_o$ and so
$X$ has only terminal singularities.
\end{proof}

Now we apply Proposition \ref{prop-examples} to construct concrete
examples.
\begin{example}
Let $F_o'$ be a del Pezzo surface of degree $d:=K_{F_o'}^2$
with at worst Du Val singularities.
Assume that the group $\muu_{m_o}$, $m_o\ge 2$
acts on $F_o'$ freely in codimension one
and so that the quotient $F_o:=F_o'/\muu_{m_o}$ has again only
Du Val singularities. Clearly, $F_o$
is del Pezzo surface and $m_oK_{F_o}^2=d$.
Hence, $d\ge m_o\ge 2$.
For $d=2$, $3$, $4$, and $8$,
according to \cite{Hidaka1981} there is an embedding
\[
\begin{array}{llll}
F_o'&\subset& \PP:=\PP(1,1,1,2) &\text{if $d=2$}
\\[5pt]
F_o'&\subset& \PP:=\PP^3 &\text{if $d=3$}
\\[5pt]
F_o'&\subset& \PP:=\PP^4 &\text{if $d=4$}
\\[5pt]
F_o'&\subset& \PP:=\PP^3\quad &\text{if $d=8$}
\end{array}
\]
Moreover, if $d=2$, $3$, $8$, then $F_o$ is a (weighted) hypersurface
of degree $4$, $3$, $2$, respectively and if $d=4$, then $F_o'$
is an intersection of two quadrics.
The action of $\muu_{m_o}$ on $F_o'$ induces the action on $\PP$.
We fix a linearization of this action
and take semi-invariant coordinates $x_i$ in $\PP$.
Now we define $\muu_{m_o}$-equivariant del Pezzo bundle $f'\colon X'\to Z'$.
If $F_o'$ is smooth, we can take $X'=F_o'\times \CC_t$.
In general case, $X'$ is embedded into $\PP\times \CC_t$, $Z'=\CC_t$
and $f'$ is the projection, where
$t$ is a coordinate in $\CC$ with $\wt t=1$.
Consider for example the case
$d\le 3$ (case $d=4$ is similar).
Let $\phi=\phi(x_1,x_2,x_3,x_4)$
be the defining
equation of $F_o'$ and let $\gamma_k$ be all monomials of weighted degree
$d$.
For each $\gamma_k$, let
$n_k$ be the smallest positive integer such that
$ n_k \equiv -\wt \gamma_k \mod m_o$. Then the polynomial
$\psi(x_1,\dots,x_4; t):=\phi+ \sum c_k t^{n_k} \gamma_k$,\ $c_k\in \CC$
is $\muu_{m_o}$-semi-invariant. Let $X'=\{\psi=0\}\subset \PP\times \CC_t$.
By Betrtini's theorem, for sufficiently general constants $c_k$,
fibers $F_t'$ of $f'$ over $t\neq 0$ are
smooth del Pezzo surfaces.
Hence we can apply Proposition \ref{prop-examples}
and get a del Pezzo bundle with a regular fiber of multiplicity $m_o$.

Note that the map $F_o'\to F_o$ is \'etale outside of $\Sing F_{o}$.
Hence there is a surjection $\pi_1(F_o\setminus \Sing F_{o})\twoheadrightarrow \muu_{m_o}$.
Conversely, assume that $F_o$ is a del Pezzo surface with Du Val singularities
such that $\pi_1(F_o\setminus \Sing F_{o})\twoheadrightarrow \muu_{m_o}$.
Then there is an \'etale outside of $\Sing F_o$ cyclic $\muu_{m_o}$-cover
$\upsilon\colon F_o'\to F_o$. Since $K_{F_o'}=\upsilon^*K_{F_o}$,
$F_o'$ is also a del Pezzo surface with Du Val singularities.
The fundamental groups of smooth loci of Du Val del Pezzo surfaces
are described in \cite{Miyanishi-Zhang-1988}, \cite{Miyanishi-Zhang-1993}.
For example, from \cite{Miyanishi-Zhang-1988} we have the following
examples with $\rho(F_o)=1$ (we do not list all the possibilities):
\par\medskip\noindent
\setlongtables\renewcommand{\arraystretch}{1.3}
\begin{longtable}{|c|l|c|c|c|c|l|}
\hline
$K_{F_o}^2$&$\Sing F_o$&$m_o$& $K_{F_o'}^2=K_{F_g}^2$&$\rho(F_o')$&$F_o'$, \ $\Sing F_o'$&type
\\[5pt]
\hline
\endfirsthead
\hline
$K_{F_o}^2$&$\Sing F_o$&$m_o$& $K_{F_o'}^2=K_{F_g}^2$&$\rho(F_o')$&$F_o'$, \ $\Sing F_o'$&type
\\[5pt]
\hline
\endhead
\hline
\endlastfoot
\hline
\endfoot
$1$& $A_1A_2A_5$&$6$&$6$&$4$&smooth&\type{2,3,6}
\\
$1$& $2A_4$&$5$&$5$&5&smooth&\type{5,5}
\\
2&$A_12A_3$&4&8&2&$\PP^1\times \PP^1$&\type{2,4,4}
\\
1&$A_3D_5$&4&4&4&$A_2$&\type{2,4,4}
\\
3&$3A_2$&3&9&1&$\PP^2$&\type{3,3,3}
\\
2&$A_2A_5$&3&6&3&$A_1$&\type{3,3,3}
\\
1&$A_8$&3&3&5&$A_2$&\type{3,3,3}

\\
4&$2A_1A_3$&2&8&1&$\PP(1,1,2)$&\type{2,2,2,2}
\\
3&$A_1A_5$&2&6&2&$A_2$&\type{2,2,2,2}
\\
2&$A_7$&2&4&3&$A_3$&\type{2,2,2,2}
\\
1&$D_8$&2&2&3&$D_5$&\type{2,2,2,2}
\end{longtable}
\end{example}

\begin{example}
In some cases we can give more explicit construction.
As was mentioned above, if $ F_o'$ is smooth, we can
take $X'=Z'\times F_o'$.
Consider the following cases:
\begin{itemize}
\item
$F_o'=\PP^2$, $\muu_3$ acts on $\PP^2_{x,y}$
by $x\mapsto \epsilon x$, $y\mapsto \epsilon^{-1} y$
(here $x$, $y$ are non-homogeneous coordinates on $\PP^2$ and $\epsilon^3=1$).
Then $\PP^2/\muu_3$ is a toric del Pezzo surface of degree 3
having three
singular points of type $A_2$.
The quotient $f\colon X\to Z$ is a del Pezzo bundle
with special fiber of type \type{3,3,3}.
\item
$F_o'=\PP^1\times \PP^1$, $\muu_2$ acts on $\PP^1_x\times \PP^1_y$
by $x\mapsto - x$, $y\mapsto - y$.
Then $\PP^1\times \PP^1/\muu_2$
is a del Pezzo surface of degree 4
having four
singular points of type $A_1$.
The quotient $f\colon X\to Z$ is a del Pezzo bundle
with special fiber of type \type{2,2,2,2}.
\item
$F_o'=\PP^1\times \PP^1$, $\muu_4$ acts
by $x\mapsto y$, $y\mapsto - x$.
Then $\PP^1\times \PP^1/\muu_4$
is a del Pezzo surface of degree 2
having two points of type $A_3$ and one point of type $A_1$.
The quotient $f\colon X\to Z$ is a del Pezzo bundle
with special fiber of type \type{2,4,4}.
\end{itemize}
\end{example}

Now we give some examples of irregular multiple fibers.
\begin{example}
Recall that any smooth del Pezzo surface of degree $1$
can be realized as a weighted hypersurface of degree $6$ in $\PP=\PP(1,1,2,3)$.
Let \[
\phi(x_1,x_2,y,z)=a_1x_1^6+a_2x_2^6+y^2(b_1x_1^2+b_2x_2^2) +c z^2,\qquad a_i, b_j, c\in \CC^*
\]
be a polynomial of weighted degree $6$,
where $x_1$, $x_2$, $y$, $z$ are coordinates in $\PP$ with
$\wt x_i=1$, $\wt y=2$, $\wt z=3$.
Consider the hypersurface $F_o'\subset \PP$
given by $\phi=0$.
By Bertini's theorem, for sufficiently general $a_i, b_j, c$, the surface
$F_o'$ is smooth outside of $P':=(0:0:1:0)$.
Consider the subvariety $X'$ in $\PP\times \CC_t$ given by
$\phi+ty^3=0$ and let $f'\colon X'\to Z'=\CC$ be the natural projection.
Since $F_o'$ is the scheme fiber of the projection
$f'\colon X'\to Z'$, the variety $X'$ is smooth outside of $P'$.
We identify $F_o'$ with the fiber over $t=0$.
Then $f'$ is a del Pezzo bundle of degree $1$ having a unique
singular point of type $\frac12(1,1,1)$ at $P'$.

Now let $\muu_2$ acts on $\PP\times \CC$ and $X'$ by
\[
(x_1,x_2,y,z; t) \longmapsto (x_1,-x_2,-y,-z; -t).
\]
The locus of fixed points $\Lambda$ consists of
the line $L:=\{x_1=y=t=0\}$ and two isolated points
$P':=(0:0:1:0; 0)$ and $P_1:=(1:0:0:0; 0)$.
Then $F_o'\cap \Lambda=\{P', Q_1, Q_2\}$, where
$Q_1\neq Q_2$ are points given by $x_1=y=a_2x_2^6+z^2=t=0$.
Let $f\colon X=X'/\muu_2\to Z=Z'/\muu_2$
be the quotient of $f'$. Since the action of $\muu_2$ on $X'$
is free in codimension one, $-K_X$ is $f$-ample and $F_o:=F_o'/\muu_2$
is a fiber of multiplicity $2$.
We show that $X$ has only terminal singularities.
By the above, $X$ is smooth outside of images of $P'$, $Q_1$, $Q_2$.
Since the $(X',Q_i)$ are smooth points, quotients $(X',Q_i)/\muu_2$
are terminal of type $\frac12(1,1,1)$.
Consider the affine chart
$\{y\neq 0\}\simeq \CC^4_{x_1', x_2',z',t}/\muu_2(1,1,1,0)$
containing $P'$.
Here $X'$ is given by the equation $\phi(x_1',x_2',1,z')+t=0$
and the action of $\muu_2$ on $\PP$ induces the following action of
$\muu_4$:
\[
(x_1',x_2',z',t) \longmapsto (\mt{i}x_1',-\mt{i}x_2',\mt{i}z',-t),\quad \mt{i}=\sqrt{-1}.
\]
Thus the quotients $(X',P')/\muu_2$ is a
terminal cyclic quotient of type $\frac14(1,-1,1)$.
Therefore, $f\colon X\to Z$ is a del Pezzo bundle
with special fiber of type \type{2,2,4}.
\end{example}

\begin{example}
As above let $\PP=\PP(1,1,2,3)$ and let
\[
\phi(x_1,x_2,y,z)=a_1x_1^6+a_2x_2^6+c y^3,\qquad a_i, c\in \CC^*
\]
be a $\muu_2$-invariant polynomial of weighted degree $6$.
Consider the hypersurface $F_o'\subset \PP$
given by $\phi=0$.
Again for sufficiently general $a_i, c$, the surface
$F_o'$ is smooth outside of $P'':=(0:0:0:1)$.
Consider the subvariety $X'$ in $\PP\times \CC_t$ given by
$\phi+tz^2=0$ and let $f'\colon X'\to Z'=\CC$ be the natural projection.
Then $f'$ is a del Pezzo bundle of degree $1$ having a unique
singular point of type $\frac13(1,1,-1)$ at $P''$.
Now let $\muu_3$ acts on $\PP\times \CC$ and $X'$ by
\[
(x_1,x_2,y,z; t) \longmapsto (x_1,\epsilon x_2,\epsilon y,\epsilon z; \epsilon t),
\qquad \epsilon :=\exp(2\pi\mt{i}/3).
\]
The only fixed point on $X'$ is $P''$.
As above, one can check that $(X',P'')/\muu_3$
is a terminal point of type $\frac19(-1,2,1)$.
Therefore, $X/\muu_3\to Z'/\muu_3$ is a del Pezzo bundle
with special fiber of type \type{9}.
\end{example}

\begin{example}
\label{ex-not-basket}
Let $\PP:=\PP(1,1,1,2,2)$, let $x_1,x_2,x_3,y_1,y_2$
be coordinates,
and let $X'\subset \PP\times \CC$ be subvariety given by
\[
\begin{cases}
c_1y_1^2+c_2y_2^2&=a_1x_1^4+a_2x_2^4+a_3x_3^4
\\
ty_2&=b_1x_1^2+b_2x_2^2+b_3x_3^2,
\end{cases}
\]
where $t$ is a coordinate on $\CC$ and $a_i$, $b_j$, $c_k$
are sufficiently general constants. By Bertini's theorem
$X'$ is smooth outside of $\{x_1=x_2=x_3=0\}\subset \Sing \PP$.
It is easy to check that $X'\cap \Sing \PP$ consists of two points
\[
\{P_1',\, P_2'\}=\{t=x_1=x_2=x_3=0,\ c_1y_1^2+c_2y_2^2=0\}
\]
and these points are
terminal of type $\frac12(1,1,1)$.
The projection $X'\to \CC$ is a del Pezzo bundle of degree $2$.
Define the action of $\muu_2$ by
\[
(x_1,x_2,x_3,y_1,y_2; t) \longmapsto (x_1,x_2,-x_3,y_1,-y_2; -t).
\]
There are four fixed points
\[
\{Q_1',\dots, Q_4'\}=\{t=x_3=y_2=0,\ c_1y_1^2=a_1x_1^4+a_2x_2^4, \ b_1x_1^2+b_2x_2^2=0\}.
\]
The quotient $f\colon X'/\muu_2\to \CC/\muu_2$ is a del Pezzo bundle
of type \type{2,2,2,2}. Note however that the image $P$ of
$\{P_1',\, P_2'\}$ on $X'/\muu_2$ is a point of type $\frac12(1,1,1)$
and $F_o$ is Cartier at $P$ (i.e., $P\notin \B(F_o)$).
\end{example}
\begin{example}
In the above notation define another action of $\muu_2$:
\[
(x_1,x_2,x_3,y_1,y_2; t) \longmapsto (x_1,x_2,-x_3,-y_1,-y_2; -t).
\]
Then the quotient $f\colon X'/\muu_2\to \CC/\muu_2$ is a del Pezzo bundle
of type \type{4,4}.
\end{example}

\begin{example}
\label{ex-I36}
Let $\PP:=\PP(1,1,1,1,2)$, let $x_1,x_2,x_3,x_4,y$
be coordinates,
and let $X'\subset \PP\times \CC$ be subvariety given by
\[
\begin{cases}
a_1x_1^2+a_2x_2^2+a_3x_1x_2+a_4x_3x_4=ty
\\
b_1x_1^3+b_2x_2^3+b_3x_3^3=x_4y
\end{cases}
\]
where $t$ is a coordinate on $\CC$ and $a_i$, $b_j$
are sufficiently general constants. Then the variety
$X'$ is smooth outside of the point $P'=\{x_1=x_2=x_3=x_4=0\}$
and $P'\in X'$ is of type $\frac12(1,1,1)$.
The projection $X'\to \CC$ is a del Pezzo bundle of degree $3$.
Define the action of $\muu_3$ by
\[
(x_1,x_2,x_3,x_4,y; t) \longmapsto
(\omega^{-1}x_1,\omega^{-1}x_2,\omega^{}x_3,x_4,y; \omega^{}t).
\]
There are two fixed points $\{t=x_1=x_2=x_3=x_4y=0\}$
and quotients of these points are of types $\frac16(1,1,-1)$ and
$\frac13(,1,-1)$.
Hence the quotient $f\colon X'/\muu_3\to \CC/\muu_3$ is a del Pezzo bundle
of type \type{3,6}.
\end{example}

\section{On del Pezzo bundles with fibers of multiplicity $\ge 5$.}
\label{sect-45}
\begin{notation}
\label{not-56}
Let $f\colon X\to Z\ni o$ be the germ of a del Pezzo bundle
and let $m_oF_o=f^*(o)$ be a fiber of
multiplicity $m_o$.
In this section we assume that $m_0\ge 5$, i.e., $F_o$ is of type \type{2,3,6}
or \type{5,5}.
\end{notation}

\begin{conjecture}
\label{conj-non-Gor}
In notation of \xref{not-56}
$f$ is a quotient
of a Gorenstein del Pezzo bundle by a cyclic
group acting free in codimension $2$ on $X$.
\end{conjecture}

\begin{proposition}
\label{prop-non-Gor}
Notation as in \xref{not-56}.
If either
\begin{enumerate}
\item
$\B(F_o)=\B$, that is, each point $P\in F_o$
where $F_o$ is Cartier is Gorenstein on $X$, or
\item
a general member $S\in |-K_X|$ has only Du Val singularities
\textup(Reid's general elephant conjecture\textup),
\end{enumerate}
then \xref{conj-non-Gor} holds.
\end{proposition}

\begin{proof}
Assume that (i) holds.
By Table \ref{table}
near each singular point
$K_X+F_o\sim 0$.
Apply Construction \ref{base-change}.
Then $F_o'=\pi^*F_o$ is Cartier.
Since $\pi$ is \'etale in codimension one, $K_{X'}+F_o'\sim 0$.
Hence, $X'$ is Gorenstein.

Now assume that (ii) holds.
Then $\varphi\colon S\to Z$ is an elliptic fibration with
Du Val singularities. We have $K_S=(K_X+S)|_S\sim 0$.
Let $\mu \colon \tilde S\to S$ be the minimal resolution.
Since $S$ has only Du Val singularities, $K_{\tilde S}\sim 0$.
In particular, $\psi\colon \tilde S\to Z$ is a minimal elliptic fibration.
By Kodaira's canonical bundle formula
$\psi$ has no multiple fibers \cite[Th. 12]{Kodaira-1964}.
Since $\psi^*o$ has a component of multiplicity $\ge 5$,
for $\psi^*o$ we have only one possibility $\tilde E_{8}$
in the classification of singular fibers
\cite[Th. 6.2]{Kodaira-1963}.
More precisely, $\Supp (\psi^*o)$ is a tree
of smooth rational curves with self-intersection number $-2$
and the dual graph $\Gamma$ is the following:
\[
\label{eq-graph-tildeE8}
\xymatrix@R=0.8pc{
\stackrel{1}\circ\ar@{-}[r]
&\stackrel{2}\circ\ar@{-}[r]
& \stackrel{3}\circ\ar@{-}[r]
&\stackrel{4}\circ\ar@{-}[r]
&\stackrel{5}\circ\ar@{-}[r]
&\stackrel{6}\circ\ar@{-}[r]
&\stackrel{4}\circ\ar@{-}[r]
&\stackrel{2}\circ
\\
&&&&&\stackunder{3}\circ\ar@{-}[u]
}
\]
Further we consider the case $m_o=6$ (the case $m_o=5$ is similar).
It is easy to see that the curve $S\cap F_o$ is irreducible and
correspond to the central vertex $v$ of $\Gamma$.
Then $\Gamma \setminus \{v\}$ has three connected components corresponding to
points of types $A_1$, $A_2$ and $A_5$ on $S$.
Therefore, $\B(F_o)=\B$.
\end{proof}

\begin{proposition}
In notation of \xref{not-56}, assume that $F_o$ is irreducible.
let $f_{\an}\colon X_{\an}\to Z_{\an}$
be the analytic germ near $F_o$.
Then
$X_{\an}$ is $\QQ$-factorial over $Z_{\an}$,
$\rho (X_{\an}/ Z_{\an})=1$, and $\rho(F_o)=1$.
\end{proposition}

\begin{warning}
Here the $\QQ$-factoriality condition of
$X_{\an}$ means
that every global Weil divisor of the total germ $X_{\an}$ along
$F_o$ is $\QQ$-Cartier, not that every analytic local ring of $X_{\an}$
is $\QQ$-factorial.
\end{warning}

\begin{proof}
Let $q\colon \hat X_{\an}\to X_{\an}$ be a $\QQ$-factorialization over
$Z_{\an}$. Run the MMP over $Z_{\an}$. So, we have the following diagram
\[
\xymatrix{
&\hat X_{\an}\ar[dl]_{q}\ar@{-->}[dr]&
\\
X_{\an}\ar[d]_{f_{\an}}&&\bar X_{\an}\ar[d]^{\bar f_{\an}}
\\
Z_{\an}&&\bar Z_{\an}\ar[ll]_{g_{\an}}
}
\]
Here $\bar X_{\an}$ is $\QQ$-factorial over $\bar Z_{\an}$ and
$\rho (\bar X_{\an}/ \bar Z_{\an})=1$.
Note that $\hat X_{\an} \dashrightarrow \bar X_{\an}$
is a composition of flips and divisorial contractions that
contract divisors to curves dominating $Z_{\an}$.
Let $\bar F_o$ be the proper transform of $F_o$ on $\bar X_{\an}$.
There are two possibilities:

1) $\bar Z_{\an}$ is a surface. Then $g_{\an}$ is a rational curve fibration
with $\rho(\bar Z_{\an}/Z_{\an})=1$. Let $C:=\bar f_{\an} (\bar F_o)$.
Since $\bar X_{\an}$ has only isolated singularities, $\bar F_o=\bar f_{\an}^*(C)$.
Further, $g_{\an}^*(o)=nC$ for some $n\in \ZZ_{>0}$ and
$\bar f_{\an}^*g_{\an}^*(o)=n\bar f_{\an}^*C=n\bar F_o$.
So, $n=m_o$.
By the main result of \cite{Mori-Prokhorov-2008}
the surface $\bar Z_{\an}$ has only Du Val singularities.
Therefore, $m_o=n\le 2$, a contradiction.

2) $\bar Z_{\an}$ is a curve. Then $g_{\an}$ is an isomorphism
and $\bar f_{\an}\colon \bar X_{\an}\to \bar Z_{\an}$ is a del Pezzo bundle
with central fiber $\bar F_o$ of multiplicity $\bar m_o=m_o\ge 5$.
By Table \ref{table} the degree of the generic fiber of $\bar f_{\an}$
(and $f_{\an}$)
is equal to $m_o$.
This means that degrees of generic fibers of $\bar f_{\an}$
and $f_{\an}$ coincide. In particular, the MMP $\hat X_{\an}\dashrightarrow \bar X_{\an}$
does not contract any divisors. Hence,
$\rho(\hat X_{\an}/Z_{\an}) =\rho(\bar X_{\an}/Z_{\an})=1$.
This implies that $q$ is an isomorphism and $\rho(X_{\an}/Z_{\an})=1$.
The last assertion follows from the exponential exact sequence and vanishing
$R^1 f_{\an *}\OOO_{X_{\an}}=0$.
\end{proof}

\begin{proposition}
Notation as in \xref{not-56}.
Conjecture \xref{conj-non-Gor}
holds under the additional assumption that $F_o$ has only log terminal singularities.
\end{proposition}
\begin{proof}
Assume that $F_o$ has only log terminal singularities.
By Table \ref{table} near each point $P\in \B(F_o)$ we have $K_X+F_o\sim 0$.
By Adjunction $F_o$ has only Du Val singularities at these points.
In points $P\notin \B(F_o)$ the divisor $F_o$ is Cartier.
Hence $F_o$
has only singularities of type T \cite{Kollar-ShB-1988}.
By Noether's formula \cite[Prop. 3.5]{Hacking-Prokhorov-2005}
\[
K_{F_o}^2+\rho(F_o)+\sum_{P\in F_o} \mu_P=10.
\]
Since points in
$\B(F_o)$ correspond to distinct points on
$X$, we have $\sum_{P\in \B(F_o)} \mu_P\ge 8$.
Hence, $K_{F_o}^2=1$, $\rho(F_o)=1$, and $\B(F_o)=\B$.
Now the assertion folows by Proposition \ref{prop-non-Gor}.
\end{proof}


TITLE: Integers expressible as the sum of its digits each raised to $p \in\mathbb N$
QUESTION [1 upvotes]: Let a number n $ \in\mathbb Z,$ be expressible as the sum of its digits each raised to  p $ \in\mathbb N$ . How to prove or disprove that there are infinite such numbers for a particular p?

REPLY [2 votes]: For more on this, see https://oeis.org/A023052 and references there.


TITLE: What is the value of this expression: $\log(\tan 6^\circ) + \log (\tan 12^\circ) + \log (\tan 18^\circ)+\cdots+\log(\tan 84^\circ)$
QUESTION [1 upvotes]: I am not able to solve this problem. Please try it to solve. The log base is 10 and angles are in degree.
Which of the following option is true for the given expression?
(A) Whole number
(B) An irrational number
(C) An negative number

REPLY [0 votes]: We know $\tan(x^\circ) \tan(90^\circ-x^\circ)=1 \\\tan(x^\circ)=\cot(90^\circ-x^\circ)$
$$6,12,18,\ldots,84\\\text{number of terms}=\frac{\text{last}-\text{first}}{\text{step}}+1=\frac{84-6}{6}+1=13+1=14$$ those are coupled $$6+84=90\\12+78=90\\18+72=90\\\vdots$$so $$\log(\tan 6^\circ) + \log (\tan 12^\circ) + \log (\tan 18^\circ)+\cdots+\log(\tan 84^\circ)=\\
\log(\tan 6^\circ\cdot\tan 12^\circ\cdot\tan 18^\circ\cdots\tan 84^\circ)
=\\\log(\underbrace{\tan 6^\circ\cdot\tan 84^\circ\cdot\tan 12^\circ\cdot\tan 78^\circ\tan 18^\circ\cdot\tan 72^\circ\cdots}_{14 \text{terms}})=\\
\log(\underbrace{1\cdot1\cdot1\cdots}_{7 \text{ terms}})=\\\log(1)=0$$


TITLE: Uncertainty principle in Quantum mechanics
QUESTION [1 upvotes]: The Uncertainty principle says that "△x△p>h/2"; we cannot precisely obtain both position $x$ and momentum $p$ simultaneously.
Is this because the uncertainty is the natural characteristic or it is because we do not know additional values? (ex. like additional 11 dimensions in superstring theory.)

REPLY [0 votes]: "△x△p>h/2" is a simple consequence of the fundamental principle of using wavefunctions ("Amplitudes") to determine the probability of finding a particle.
A plane wave is evenly spread over all space and is the eigenfunction of one precisely known value of p.
In order to get anything other than such complete indeterminacy of position x, one must add several plane waves with different p, forming a wave packet which tails out at the spacial extremes and becomes more and more localised at one value of x, the more different p are added to the superposition.
In the limit you get an infinitely narrow wave packet (Dirac impulse), which is the eigenfunction of a precisely known value of x, which contains all possible p values (p is completely indeterminate).
Reality always lies in between these two extreme situations, and △x△p>h/2 follows from a Fourier analysis of the wave superposition (see e.g. Schiff: Quantum Mechanics).


TITLE: Show two probability measures are equal
QUESTION [2 upvotes]: Let $P$ and $Q$ be two probability measures on a measure space $(\Omega,\mathcal{F})$. Show that if $P(A) = Q(A)$ for all $A\in \mathcal{F}$ such that $P(A) \leq 1/2$, then $P=Q$. Show that the result fails if we replace the constraint $P(A) \leq 1/2$ by the more stringent $P(A) < 1/2$.
Is my proof for part 1 correct? Also I would like some tips/ideas for how to approach part 2
PART 1
Suppose $P$ and $Q$ are two probability measures on a measurable space such that $P(A) = Q(A)$ whenever $P(A) \leq 1/2$. We want to show $P$ and $Q$ agree on all $A\in \mathcal{F}$ such that $P(A)>1/2$. Let $A\in \mathcal{F}$ such that $P(A)> 1/2$. then $$
P(A^C) = 1 - P(A) < 1/2 $$ Therefore $P(A^C) = Q(A^C)$. Since we also have $Q(A^C) = 1-Q(A)$, then $1-Q(A) = 1-P(A)$ implies $Q(A)=P(A)$.
PART 2

REPLY [2 votes]: Your proof for the first part seems fine.
Hint for the second part:
Note that $P$ and $Q$ can only possibly disagree on a set $A$ with $P(A) = \frac{1}{2}$. Furthermore, it would be nice if $P$ does not take any values smaller than $\frac12$ other than $0$.
Does this help?


TITLE: co-variance matrix of discrete multivariate random variable
QUESTION [0 upvotes]: Let $\mathcal{S} = \{(\delta,0,0),(0,\delta,0),(0,0,\delta)\}$. If we consider a random variable $S$ defined over $\mathcal{S}$ such that
\begin{equation}
S =
    \begin{cases}
    s, & \text{ if } s\in\mathcal{S}\\
    0, & \text{otherwise}.
  \end{cases}
\end{equation}
Then $\mathbb{P}(S=s) = 1/m$, where $m = |S|$ cardinality of $\mathcal{S}$,
What is $\mathbb{E}(S^{T}AS)$?
My attempt: $\mathbb{E}(S^{T}AS) = trace{(A\Sigma)} + \mu^{T}A\mu$, where $\mu$ is the expected value of $S$ and $\Sigma$ is the covariance matrix. In this case how to find $\Sigma$?

REPLY [1 votes]: Since $|\mathcal{S}| = 3$ the easiest way will be by direct enumeration:
\begin{align*}
\mathbf{E}[S^T A S] & = \frac{1}{3} \bigg( (\delta,0,0)^T A (\delta, 0,0) +
(0,\delta,0)^T A (0,\delta, 0) +
(0,0,\delta)^T A (0,0,\delta) \bigg)
\end{align*}
If $A = (a_{i,j})$ with $1 \leq i \leq 3, 1 \leq j \leq3$ then
\begin{align*}
(\delta,0,0)^T A (\delta, 0,0) & =a_{1,1}\delta^2 \\
(0,\delta,0)^T A ( 0,\delta, 0) & =a_{2,2}\delta^2 \\
(0,0,\delta)^T A ( 0,0,\delta) & =a_{3,3}\delta^2
\end{align*}
Hence
$$
\mathbf{E}[S^T A S] = \delta^2(a_{1,1} + a_{2,2} + a_{3,3}) = \delta^2 \text{Tr}(A)
$$

If you are really keen to solve the problem using the covariance matrix, then let us denote $S = (s_1,s_2,s_3)$ for the random element of $\mathcal{S}$ and then the covariance matrix $\Sigma$ is given by
\begin{align*}
\Sigma_{i,j} \, \colon & = \mathbf{E}[S_iS_j] - \mathbf{E}[S_i]\mathbf{E}[S_j] \\
& = \delta^2\mathbf{1}_{i,j} -  \mathbf{E}[S_i]\mathbf{E}[S_j] \\
& = \delta^2\mathbf{1}_{i,j} - \left(\frac{\delta}{3} \times \frac{\delta}{3} \right) \\
& =\delta^2\left( \mathbf{1}_{i,j} - \frac{1}{9}\right)
\end{align*}
where $\mathbf{1}_{i,j} = 1$ if $i = j$ and is $0$ otherwise. That $\mathbf{E}[S_iS_j] =\mathbf{1}{i,j}$ follows since only one element of $S$ is non-zero at any one time. Hence we have
$$ \Sigma = \delta^2 \left( I - \frac{1}{9} J \right)$$
where $I$ is the $3 \times 3$ identity matrix, and $J$ is the $3 \times 3$ matrix with all entries equal to 1.


TITLE: How to prove $f(x)=x^4$ is concave up by definition?
QUESTION [3 upvotes]: I know $f(x)=x^4$ is concave up, by calculating its second derivative. However, how to prove that $f(x)=x^4$ is concave up by definition, say $f(\frac{x_1+x_2}{2})<(1/2)f(x_1)+(1/2)f(x_2)$ for all $x_1,~x_2$? I'd tried binomial theorem, but can't get anything.

REPLY [6 votes]: As an intermediate step, use the inequality
$$ \frac{x^2 + y^2}{2} \ge \left(\frac{x + y}{2}\right)^2 $$
which is relatively easy to prove.
This gives
$$ \frac{x^4 + y^4}{2} = \frac{(x^2)^2 + (y^2)^2}{2} \ge \left(\frac{x^2 + y^2}{2}\right)^2 \ge \left(\left(\frac{x + y}{2}\right)^2\right)^2 = \left(\frac{x + y}{2}\right)^4. $$

Alternatively, using the "usual" definition of convexity, we have
$$ [(1 - \lambda)x^2 + \lambda y^2] - [(1 - \lambda)x + \lambda y]^2 = (1 - \lambda)\lambda (x - y)^2 \ge 0. $$
Being a quadratic form, this is an easy enough sum of squares to compute. One can do it by hand even. Then we do what we did above:
$$ (1 - \lambda)x^4 + \lambda y^4 \ge [(1 - \lambda)x^2 + \lambda y^2]^2 \ge [(1 - \lambda)x + \lambda y]^4 . $$


TITLE: How to express $\cos(\frac{x}{k})$ and $\sin(\frac{x}{k})$ in terms of $\cos(x)$ and $\sin(x)$, respectively?
QUESTION [4 upvotes]: How can we express $\cos(\frac{x}{k})$ ($k \in \mathbb{N}$) in terms of $\cos(x)$?
And $\sin(\frac{x}{k})$ in terms of $\sin(x)$?
Edit
Maybe this another question helps. Is there a $T_n(x)$ inverse? (Link don't say anything about this)
If $\cos(x)=T_k(\cos(\frac{x}{k}))$, is there $T^{-1}_n(x)$ that $\cos(\frac{x}{k})=T^{-1}_k(\cos(x))$ is at least one solution for the question? What is the $T^{-1}_n(x)$ formula?

REPLY [1 votes]: For any positive integer $k$, the trigonometric function $\cos ky$ is a polynomial in $\cos y$.  For example, with $k = 2$,
$$
\cos 2y = 2\cos^2 y - 1.
$$
The polynomial here is $T_2(u) = 2u^2 - 1$.  In order to express $\cos y$ in terms of $\cos 2y$ (this is equivalent to expressing $\cos \frac{x}{2}$ in terms of $\cos x$ via the substitution $y = 2x$), we have to invert the polynomial function $T_2$.
In other words, you have to solve $v = 2u^2 - 1$ for $u$.  This doesn't have a unique solution, but we can describe both possible solutions:
$$
u = \pm \sqrt{\frac{1 + v}{2}}.
$$
This is the well-known "half-angle formula for cosine"
$$
\cos \frac{x}{2} = \pm \sqrt{\frac{1 + \cos x}{2}}.
$$
For larger $k \in \mathbb{N}$, you have to solve the higher degree polynomial equations
$$
v = T_k(u)
$$
for $u$ in order to express $\cos \frac{x}{k}$ in terms of $\cos x$.


TITLE: Definition of outer product
QUESTION [5 upvotes]: I am trying to understand the concept of outer product in quantum mechanics. I read "Quantum Computing explained" of David MacMahon.

I can understand the transition in (3.12):
$$(|\psi\rangle \langle \phi | )|\chi\rangle \rightarrow |\psi\rangle \langle \phi |\chi\rangle  $$
But how to get $(\langle \phi | \phi | \chi \rangle ) | \psi \rangle$ ?
Why it is possible to get through such steps?

$|\psi\rangle \langle \phi | \chi\rangle  $
$|\psi\rangle \langle \phi | \phi | \chi\rangle  $
$\langle \phi | \phi | \chi\rangle |\psi\rangle  $

REPLY [3 votes]: I'm thinking that it's a typo, and all the author wanted in the last term was to write
$$
(\langle \phi|\chi\rangle)\,|\psi\rangle.
$$
The proof uses that you have a kind of associativity in the first equality $(|\psi\rangle\langle\phi|)\,|\chi\rangle= |\psi\rangle\,\langle\phi|\chi\rangle$ which I think is brought out of the blue if you introduce bras and kets out of nowhere.
The equality is obvious if you notice that kets as simple column vectors in $\mathbb C^n$, and bras are their adjoints (conjugate transpose). In that setting your equality is
$$
(\psi\phi^*)\,\chi=\psi\,(\phi^*\chi)=(\phi^*\chi)\,\psi,
$$
where the associativity is that of the product of matrices.


TITLE: How does composition affect eigendecomposition?
QUESTION [8 upvotes]: What relationship is there between the eigenvalues and vectors of linear operator $T$ and the composition $A T$ or $T A$? I'm also interested in analogous results for SVD.

REPLY [3 votes]: Denote the eigenvalues of $T,A,TA$ by $\lambda_1,\lambda_2, \ldots, \lambda_n,
\mu_1,\mu_2, \ldots, \mu_n, \nu_1,\nu_2, \ldots, \nu_n$, respectively. What relations exist between them ? There are several easy constraints :
(1) We must have $\prod_{k=1}^{n} \lambda_k\mu_k=\prod_{k=1}^{n}\nu_k$ because ${\sf det} (TA)={\sf det} (T) {\sf det} (A)$.
(2) We must have $k_{\nu} \geq {\sf max }(k_{\lambda},k_{\mu})$, where  we denote by $k_{\lambda},(k_{\mu},k_{\nu})$ the number of indices $i$ such that $\lambda_i$ (or $\mu_i,\nu_i$ respectively), is zero (this is because ${\sf rank}(AT) \leq {\sf min}({\sf rank}(A),{\sf rank}(T))$ and ${\sf rank}(T)=n-k_{\lambda}$).
(3) If one of $A,T$ or $AT$ is a homothety, then $\lbrace \nu_{k} \rbrace_{1 \leq k \leq n}=\lbrace \lambda_k\mu_k \rbrace_{1 \leq k \leq n}$.
I think there are no other constraints besides the three above.
I can actually prove this for a "generic enough" example : assume $n=2$, and $\lambda_1\mu_1\lambda_2\mu_2=\nu_1\nu_2$ (to ensure (1)), and $\lambda_1 \neq \lambda_2, \mu_1 \neq \mu_2,  \nu_1 \neq \nu_2$ (to ensure (3)).
Then consider the two matrices
$$
P=\bigg(
\begin{matrix}
(\nu_1-\lambda_1\mu_2) & (\nu_1-\lambda_1\mu_1)\\
(\nu_1-\lambda_2\mu_2) & (\nu_1-\lambda_2\mu_1)\\
\end{matrix}
\bigg),
$$
$$
Q=\bigg(
\begin{matrix}
(\nu_1-\lambda_2\mu_1)(\nu_1-\lambda_2\mu_2) & -(\nu_1-\lambda_1\mu_1)(\nu_1-\lambda_1\mu_2) \\
(\nu_2-\lambda_2\mu_1)(\nu_1-\lambda_2\mu_2) & -(\nu_2-\lambda_1\mu_1)(\nu_1-\lambda_1\mu_2) \\
\end{matrix}
\bigg)
$$
We have
$$
{\sf det}(P)=\nu_2(\lambda_2-\lambda_1)(\mu_2-\mu_1),
$$
$$
{\sf det}(Q)=\mu_1(\lambda_2-\lambda_1)(\nu_2-\nu_1)(\nu_1-\lambda_1\mu_2)(\nu_1-\lambda_2\mu_2)
$$
so that $P$ and $Q$ are both inversible whenever  $\mu_1\neq 0,\nu_2\neq 0, \nu_1 \not\in \lbrace \lambda_1\mu_1, \lambda_1\mu_2 \rbrace$. If we put
$$
L={\sf diag}(\lambda_1,\lambda_2), M={\sf diag}(\mu_1,\mu_2), N={\sf diag}(\nu_1,\nu_2)
$$
then $QLPM=NQP$ (GP can check this), so that $LPMP^{-1}=Q^{-1}NQ$. If we put $T=L$, $A=PMP^{-1}$, then $T$ has eigenvalues $\lambda_1,\lambda_2$, $A$ has eigenvalues $\mu_1,\mu_2$, $TA$ has eigenvalues $\nu_1,\nu_2$.


text:
\section{Optimizers}
\label{sec:opt}

In this section, we introduce six optimizers to solve the following constrained problem:
\begin{equation}
    \min\limits_{\bm{x} \in \Omega} \left\{ f(\bm{x}) ~~\mbox{s.t.}~~ \bm{g}(\bm{x})\geq 0 \right\},
    \label{eq:opt_prob}
\end{equation}
where $f: \mathbb{R}^d \mapsto \mathbb{R}$ is the objective function, $\bm{g}:\mathbb{R}^d \mapsto \mathbb{R}^m$ gives the inequality constraints which are expensive to evaluate and $\bm{x} \in \Omega \subset \mathbb{R}^d $ is the vector of design variables.
Two of the optimizers are included in the Isight software, which is mandatory to use because of CMDO and PMDO frameworks. We developed an Isight interface for the SEGOMOE python tool-box in which the other optimizers are implemented.

\subsection{Bayesian optimizers}
This section introduces the \textit{constrained Bayesian optimization} (CBO) framework \cite{MockusBayesianmethodsseeking1975,JonesEfficientglobaloptimization1998} that aims to solve the optimization problem~\eqref{eq:opt_prob} with a minimal number of calls.
To do so, one uses \textit{Gaussian Process} (GP) (also known as Kriging) \cite{RasmussenGaussianprocessesmachine2006,Krigestatisticalapproachbasic1951} trained with a pre-computed \textit{design of experiments} (DoE) (i.e. set of designs evaluated on the objective and constraints functions) of $l$ points.
GPs are then used to provide, with a cheap computational cost, a prediction $\mu_s^{(l)}: \mathbb{R}^d \mapsto \mathbb{R}$ and an associated uncertainty $\sigma_s^{(l)}: \mathbb{R}^d \mapsto \mathbb{R}$ for each point of $\bm{x} \in \Omega$ where $s : \mathbb{R}^d \mapsto \mathbb{R}$ can be either $f$ and $g_i$ for a given constraint component $i$.
Concerning the objective function, these information are combined in an acquisition function  $\alpha_f^{(l)}: \mathbb{R}^d \mapsto \mathbb{R}$ \cite{frazier2018tutorial,Bartoliadaptivemodeling2019,WangMaxvalueentropysearch2017} coding the trade-off between exploration of the highly uncertain domain that can hide a minimum and exploitation of the minimum of the GP prediction.
For the constraints, these information are joined to produce a feasibility criterion $\alpha_{\bm{g}}^{(l)}: \mathbb{R}^d \mapsto \mathbb{R}^m$ \cite{frazier2018tutorial,lam2015multifidelity,priem2019use} which is generally explicit.
The point $\bm{x}^{(l+1)}$, solving the constrained maximization trade-off sub-problem:
\begin{equation}
    \bm{x}^{(l+1)}=\arg\min\limits_{\bm{x} \in \Omega} \left\{ \alpha_f^{(l)}(\bm{x}) ~~\mbox{s.t.}~~ \bm{x}\in \Omega_{\bm{g}}^{(l)} \right\},
    \label{eq:vst_prob}
\end{equation}
where $\Omega_{\bm{g}}^{(l)}$ is the approximated feasible domain defined by the feasibility criterion $\bm{\alpha}_{\bm{g}}^{(l)}$, is thus iteratively added to the DoE until a maximum number of iterations max\_nb\_it is reached.
The solution provided by CBO to the problem~\eqref{eq:opt_prob} is eventually the best point in the DoE (i.e with the minimal feasible value of $f$).
The main steps of CBO are finally summarized in Algorithm~\ref{alg:BO}.
\begin{algorithm}[ht!]
     \begin{algorithmic}[1]
        \INPUT{Objective and constraints functions, initial DoE for objective and constraints, a maximum number of iterations max\_nb\_it\;}

        \FOR{$l = 1$ \TO \mbox{max\_nb\_it}}
            \STATE {Build the surrogate models using GPs\;}
            \STATE {Find $\bm{x}^{(l+1)}$ a solution of the enrichment maximization sub-problem\;}
            \STATE {Evaluate the objective and constraints functions at $\bm{x}^{(l+1)}$\;}
            \STATE {Update the DoE\;}
        \ENDFOR
        \OUTPUT{The best point found in the DoE\;}
    \end{algorithmic}
    \caption{The constrained Bayesian optimization framework.}
    \label{alg:BO}
\end{algorithm}

In this context, \citet{Bartoliadaptivemodeling2019} implemented a CBO python tool-box based on the \textit{super efficient global optimization} (SEGO) algorithm of \citet{SasenaExplorationmetamodelingsampling2002} which has been enhanced by the use of \textit{mixture of experts} (MOE) \cite{hastieelements2005, bettebghor2011surrogate}, the \textit{kriging with Partial Least Squares} (KPLS) for high dimensional problems~\cite{BouhlelImprovingkrigingsurrogates2016}, additional acquisition functions (e.g. WB2S \cite{Bartoliadaptivemodeling2019}) for highly multimodal objective functions, supplementary feasibility criteria (e.g. the \textit{upper trust bound} (UTB) \cite{lam2015multifidelity,priem2019use}) for non-linear constraint functions and a multiprocessing ability to speed-up the optimization process.
The tool-box \cite{Bartoliadaptivemodeling2019} is thus named the \textit{super efficient global optimization with mixture of experts} (SEGOMOE) tool-box.
We also developed an interface of the python tool-box to allow its use in the Isight software in which BRAC is implemented.

In this paper, the four following variants of SEGO (available with the SEGOMOE tool-box) are tested to show their common ability on industrial MDO problems:
\begin{itemize}
    \item SEGO \cite{SasenaExplorationmetamodelingsampling2002}: it can be seen as a basic implementation of the super efficient global optimization framework, it was developed for standard constrained optimization problems.
    \item SEGO-UTB \cite{priem2019use}: it includes a decreasing upper trust bound in the SEGO framework. In the context of this paper and to encourage the exploration of the feasible domain, we used an exponential decrease for the constraints learning rate, see \cite[Fig.  3]{priem2019use}.
    \item SEGOMOE \cite{Bartoliadaptivemodeling2019}: it combines the use of multiple models (local GPs) with the SEGO framework in order to mitigate  the high non-linearity of the targeted optimization problem.
    \item SEGOMOE-UTB: similarly to SEGO-UTB, it combines the use of the UTB strategy and the SEGOMOE framework. Again, we choose to work with an exponential decrease for the constraints learning rate.
\end{itemize}
We note that for all the four variants we used the WB2S acquisition function combined with the KPLS models as the dimension of the design space is larger than 12.

\subsection{Isight optimizers}

The Isight software \cite{vanisight2010} is delivered with many ready-to-use constrained optimizers.
In this paper, we focus on two of the popular derivative-free Isight constrained optimizers: the \textit{evolutionary optimization algorithm} (Evol) \cite{schwefelevolutionsstrategie1975} and a variant of the Pointerdog algorithm (Pointer-2)  \cite{vanisight2010}.

Evol is an evolutionary strategy that mutates iteratively the best known point
 by adding a normally distributed  perturbation to the design variables.
The standard deviation of the normal distribution is adapted during the optimization process to solve the regarded problem
with a minimal number of evaluations.
This simple algorithm is enhanced by different features such as: (1) a repeat calculation check to ensure that all the points computed are different, (2) a standard deviation expansion process when the same point is always evaluated, (3) a consecutive variable search allowing the exploration in a single canonical direction of the design space, and (4) a parallel execution accelerating the optimization process.

The Pointer-2 strategy is an optimization framework based on an Isight proprietary algorithm that managed a set of well-known optimizers: Evol \cite{schwefelevolutionsstrategie1975}, the Hooke-Jeeves direct search method \cite{hookedirect1961}, the \textit{non-linear programming by quadratic Lagrangian} (NLPQL) algorithm \cite{schittkowskinlpql1986}, the downhill simplex algorithm \cite{neldersimplex1965}, the \textit{multi-function optimization system tool} (MOST) technique \cite{vanisight2010} and the \textit{multi-objective particle swarm optimization} (MOPSO) algorithm \cite{coellohandling2004}.
The choice of the set of optimizers used in the optimization process is made with respect to a classification system using the information available on the problem (i.e. often given by the user).
The best optimizer, and its settings, are then updated thanks to the information collected all along the optimization process.
Furthermore, the optimization method can be used in two ways.
First, the optimizers and their settings are selected to have the higher improvement leading to the best solution in the shortest time.
On the contrary, one searches a robust solution to uncertainties which leads to a slower optimization process due to a higher number of calls.
The Pointer-2 strategy periodically performs a surrogate based optimization to speed up the convergence.
In fact, the Pointer-2 is able to solve a wide range of constrained optimization problems and thus allows non-specialist to solve an optimization problem only providing the objective and constraints functions, the design variables, and a targeted optimization time.

config: arxiv


TITLE: Comparing fields with same degree
QUESTION [4 upvotes]: Two part question: Are the fields $\mathbb{Q} (\sqrt[3]{2}, i \sqrt{3})$ and $\mathbb{Q} (\sqrt[3]{2}, i, \sqrt{3})$ identical in algebraic structure? I have in notes that they both have degree of 6 over $\mathbb{Q}$.
How do I show explicitly that $\mathbb{Q} ( i \sqrt{3})$ is only degree 2 over $\mathbb{Q}$. The usual trick is to adjoin the real roots and then adjoin the complex root, but it's a different story when it's not just $i$ by itself.
Edit: I'm starting to mistrust that the degree of the two extensions are identical.

REPLY [2 votes]: For starters, neither $i$ nor $\sqrt{3}$ is in $\mathbb{Q}(\sqrt[3]{2},i\sqrt{3})$. And I think you're right to mistrust that the degrees are equal since $|\mathbb{Q}(\sqrt[3]{2},i,\sqrt{3})/\mathbb{Q}|=12$.


TITLE: How can I calculate the argument of amplification factor?
QUESTION [0 upvotes]: For example, I have an amplification factor of upwind scheme for hyperbolic conservation law, $$\lambda(k)=1-\nu(1-e^{-ik{\triangle}x})$$magnitude of which is
$$|\lambda|^2=1-4\nu(1-\nu)\sin^2\frac{1}{2}k{\triangle}x$$
I want to derive the argument of this amplification factor which is
$$\arg\lambda=-\tan^{-1}\left[\frac{\nu\sin k{\triangle}x}{(1-\nu)+\nu\cos k{\triangle}x}\right].$$
My text book(Numerical Solution of Partial Differential Equation : An introduction by Motron and Mayers) just gives this without explanation.

REPLY [0 votes]: We have
$$\lambda(k) = 1 - \nu (1 - e^{-ik \Delta x}) = 1 - \nu[1 - (\cos(k \Delta x) - i \sin(k \Delta x))]$$
$$ \lambda(k) = (1 - \nu  + \nu \cos(k \Delta x)) + i \nu \sin(k \Delta x)) $$
Therefore
$$\tan(\arg [\lambda(k)]) = \frac{\nu \sin(k \Delta x)}{1 - \nu  + \nu \cos(k \Delta x)}$$


TITLE: How to use the properties of the logarithmic function
QUESTION [1 upvotes]: I'm coding the game asteroids. I want to make a levels manager who can create a infinity number of level increasing in difficulty.
My levels have as parameters :

The number of asteroids on the board;
The average radius of the asteroids on the board;
The minimum radius an asteroid has to have to exists.

However, I can't use a linear function to calculate the parameters in function on the level because each parameter has their restrictions.

The number of asteroids on the board increase from $2$ to a maximum of $20$.
The average radius of the asteroids increase from $30$ to a maximum of $50$.
The minimum radius of the asteroids decrease from $15$ to a minimum of $5$.

How can I use the properties of the logarithmic function to be able to generate a minimum of $100$ level?
Thanks for your help.

REPLY [1 votes]: Finally, I use this method :
$$f(l):=m\left( \frac{m}{M} \right)^{\frac{l-1}{1-N}}$$
I calculate it by resolving the ODE :
$$\frac{d}{dl}f(l)=kf(l)$$
with $f(1)=m$ and $f(N)=N$.


TITLE: Let consider a disjoint union of intervals from $(0,1)$ such that there is no two points at distance $\frac{1}{12}$ in the union.
QUESTION [0 upvotes]: Let   consider a  disjoint    union  of  intervals from $(0,1)$   such  that   there  is  no   two  points at   distance $\frac{1}{12}$  in   the  union. Prove  that the   sum  of   the  lenght  of   the   interval  is lower than $\frac{1}{2}$. How   to   start?  Thank  you!  I  considered  two consecutive   intervals $(a,b)$  and $(c,d)$. Then $ a+ 1/12 \geq b$ and $  a+ 1/12 \leq c$

REPLY [2 votes]: Let $S$ be the set of elements $x \in (0,1)$ not in one of the intervals
We specify a set $T$ as follows:

Define the subintervals $A_i \doteq (\frac{i}{12}, \frac{i+1}{12})$; $i=0,1,\ldots, 11$
For each $i$ even let $T \cap A_i$ be the set of points $x: x+\frac{1}{12} \in S$ and for $i$ odd let $T \cap A_i$ be the set of points $x: x-\frac{1}{12}  \in S$.

Then note the following: On the one hand, $\mu(S) = \mu(T)$. On the other hand, every point in $\cup_{i=0}^{11} A_i$ is either in $S$ or $T$.
As $\mu(\cup_{i=0}^{11} A_i)=1$ it follows that
$$1= \mu(\cup_{i=0}^{11} A_i)= \mu(S \cup T) \le \mu(S) + \mu(T) = 2\mu(S),$$
so $\mu(S)$ must be at least $\frac{1}{2}$.


TITLE: How to express the Euclidean basis vector $(0, 1)$ under this new basis?
QUESTION [1 upvotes]: Changing our basis. Suppose we are working in the $\mathbb{R}^2(\mathbb{R})$ vector space under the standard Euclidean basis and dot product.  Suppose we then transform the standard basis vectors $e_1 = (1,0) \mapsto (1/\sqrt{2}, 1/\sqrt{2}) = d_1$ and $e_2 = (0,1) \mapsto (-1/\sqrt{2}, 1/\sqrt{2})= d_2$.  Note that this is an orthogonal change in basis (in fact a $90$ degree rotation), so norms should be preserved.
How to re-express $e_2$ under the new $d_i$ basis? Now let us consider $e_2$ under the new $d_i$ basis.  It seems to me $e_2 = (1, 1)$ under this new basis, since
$$
1(d_1) + 1(d_2) = e_2
$$
Yielding a contradiction. But doesn't this then imply that
$$
||e_2|| \ne 1
$$
which is false? To see why, consider the expansion of $||e_2||$:
$$
||e_2|| = \sqrt{e_2 \cdot e_2} = \sqrt{(1,1) \cdot (1,1)} = \sqrt{1*1 + 1*1} = \sqrt{2} \ne 1
$$
So what I have done wrong? Note that under the standard basis, $||e_2|| = 1$ since
$$
||e_2|| = \sqrt{e_2 \cdot e_2} = \sqrt{(0,1) \cdot (0, 1)} = \sqrt{0*0+1*1} = 1
$$
...so this is clearly not right.

REPLY [1 votes]: Your basis change is
$$d_1=\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2,$$
$$d_2=-\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2.$$
Solving for $e_2$ requires to do
\begin{eqnarray*}
d_1+d_2&=&\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2
+\left(-\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2\right),\\
&=&\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2
-\frac{1}{\sqrt 2}e_1+\frac{1}{\sqrt 2}e_2,\\
&=&\frac{2}{\sqrt 2}e_2.
\end{eqnarray*}
Then
$$e_2=\frac{\sqrt 2}{2}d_1+\frac{\sqrt 2}{2}d_2.$$
Also, here we can see $\|e_2\|=1$


TITLE: What is Gödel's pairing function on ordinals?
QUESTION [11 upvotes]: I find many references to Gödel's pairing function on ordinals but I have not found a definition.  What is it?

REPLY [6 votes]: Asaf and Joel have answered the question. Let me add a remark that expands the fact that it helps us prove that $\kappa\times$ and $\kappa$ have the same size. All the claims here can be verified rather easily. Multiplication and exponentiation are in the ordinal sense.
It is customary to write $\Gamma(\alpha,\beta)$ for the order type of the predecessors of $(\alpha,\beta)$ under the order than Asaf denotes $\prec$. For example, $\Gamma(\omega,\omega\cdot2)=\omega^2+\omega$.
An ordinal $\alpha$ is (additively) indecomposable iff $\alpha\gt 0$ and whenever $\beta,\gamma\lt\alpha$, then $\beta+\gamma\lt \alpha$. One can easily check that the indecomposable $\alpha$ are precisely those of the form $\omega^\beta$. Say that $\alpha$ is multiplicatively indecomposable iff $\alpha>0$ and $\beta\gamma\lt \alpha$ whenever  $\beta,\gamma\lt\alpha$. Then $\alpha$ is multiplicatively indecomposable iff it is $1$ or has the form $\omega^{\omega^\beta}$.
Ok, we can now state the remark; unfortunately I would not know who to credit for this observation, I think of it as folklore: An ordinal $\alpha$ is multiplicatively indecomposable iff it is closed under Gödel pairing, that is, $\Gamma(\beta,\gamma)\lt\alpha$ whenever $\beta,\gamma\lt\alpha$. In particular, $\Gamma(\kappa,\kappa)=\kappa$ for any infinite cardinal $\kappa$, which of course implies that $\kappa\times\kappa$ and $\kappa$ have the same size. Also, if $\kappa$ is uncountable, then there are $\kappa$ ordinals $\alpha$ below $\kappa$ such that $\Gamma(\alpha,\alpha)=\alpha$. Of course, all of this works well in $\mathsf{ZF}$ and all the definitions involved are absolute.

I prefer a different approach when verifying that $\kappa\times\kappa$ and $\kappa$ have the same size, one that (again) is absolute and goes through in $\mathsf{ZF}$, but only requires the use of additively indecomposable ordinals: One first checks that there is a (recursive) bijection $h:\omega\times\omega\to\omega$ with $h(0,0)=0$. Then, given ordinals $\alpha,\beta$, use their Cantor's normal form to write them as
$$ \alpha= \omega^{\alpha_1}n_1 + \omega^{\alpha_2}n_2 + \dots + \omega^{\alpha_k}n_k $$
and
$$ \beta= \omega^{\alpha_1}n'_1 + \omega^{\alpha_2}n'_2 + \dots + \omega^{\alpha_k}n'_k $$
where $\alpha_1 \gt \alpha_2 \gt \dots \gt \alpha_k$ are ordinals, and $n_1,\dots,n_k, n'_1,\dots,n'_k$ are natural numbers. (Note that these representations are not unique, but at least one of $n_i$ and $n_i'$ is non-zero iff $\alpha_i$ appears as an exponent in the canonical form of $\alpha$ or $\beta$).
Now set
$$ H(\alpha,\beta)=\omega^{\alpha_1}h(n_1,n'_1)+\omega^{\alpha_2}h(n_2,n'_2)+\dots+ \omega^{\alpha_k}h(n_k,n'_k). $$
Then $H$ is a bijection between $\alpha\times\alpha$ and $\alpha$ whenever $\alpha$ is indecomposable. And an easy inductive argument, appealing to the explicit proof of Schröder-Bernstein, allows us to use $H$ to argue that there is, provably in $\mathsf{ZF}$, a class function that assigns to each infinite ordinal $\alpha$ a bijection between $\alpha\times\alpha$ and $\alpha$.
(Of course, the existence of this class function can also be argued from $\Gamma$, using that there are $\kappa$ ordinals $\alpha$ below $\kappa$ with $\Gamma(\alpha,\alpha)=\alpha$, but this second approach is somewhat easier.)
I found this argument a while ago, but then saw that Levy gives essentially the same approach in his textbook on set theory. Again, I am not sure who to credit for this construction, it seems to go back to Gerhard Hessenberg's 1906 book, "Grundbegriffe der Mengenlehre".


TITLE: some particular extensions of $SL_2(5)$
QUESTION [1 upvotes]: With GAP software, I can calculate
\begin{gather}\dim H^2(SL(2,5),GF(2)^9)=1
\end{gather}
So, if I understand correctly, there is exactly one non-split extension of $SL(2,5)$ by an elementary abelian $2$-group of dimension $2^9$. Let's say I want to know the character table of this extension. Then I look for all perfect groups $K$ of size $120\cdot2^9$. Unfortunately, GAP seems that doen't help me.
gap> NrPerfectGroups(120*2^9);
fail

The GAP's library PerfectGroup relies on the book "Perfect Groups" of Holt and Plesken, but I didn't find information neither there.
Edit: I was trying to get the information about a perfect group $K$ with the following properties: there is $M \unlhd K$ an elementary abelian group of order $2^9$ such that $M=C_K(M)$, $K/M$ si isomorphic to $SL(2,5)$ and acts irreducibly on $M$. Hence what I calculated with GAP is the cohomology group $H^2(SL(2,5),M)$ where $M$ is an irreducible faithful $SL(2,5)$ module.

REPLY [4 votes]: If we take the 9-dimensional indecomposable module listed on the ATLAS webpages (as indicated in a commentary), we can construct the corresponding nonsplit extension in GAP as follows:
gap> LoadPackage("atlasrep");;
gap> DisplayAtlasInfo("2.A5");
Representations for G = 2.A5:    (all refer to std. generators 1)
-----------------------------
 1: G <= Sym(24)                             rank 8, on cosets of 5 < 2.D10
 2: G <= Sym(40)                             rank 16, on cosets of 3 < 2.S3
 3: G <= GL(8,2)
 4: G <= GL(9,2)
[...] Rest deleted

so we want representation 1 for permutations and representation 4 for the module:
gap> perms:=AtlasGenerators("2.A5",1).generators;
[ (1,2,5,4)(3,6,8,7)(9,13,11,14)(10,15,12,16)(17,19,18,20)(21,24,23,22),
  (1,3,2)(4,5,8)(6,9,10)(7,11,12)(13,16,17)(14,15,18)(19,21,22)(20,23,24) ]
gap> mats:=AtlasGenerators("2.A5",4).generators;
[ <an immutable 9x9 matrix over GF2>, <an immutable 9x9 matrix over GF2> ]
gap> gp:=Group(perms);;
gap> mo:=GModuleByMats(mats,GF(2));;
gap> coh:=TwoCohomologyGeneric(gp,mo);;
gap> coh.cohomology; # indeed dimension 1
[ <an immutable GF2 vector of length 144> ]

Now form the extension:
gap> gp:=FpGroupCocycle(coh,coh.cohomology[1]);
<fp group of size 61440 on the generators [ F1, F2, F3, F4, m1, m2, m3, m4,  m5, m6, m7, m8, m9 ]>
gap> gp:=Image(IsomorphismPermGroup(gp));
<permutation group of size 61440 with 13 generators>

Should you like the full library of perfect groups that I computed (which will be in the future GAP 4.12 release) send me a private email and I'm happy to provide you with a link.


text:
(* Copyright 2021 (C) Mihails Milehins *)

section\<open>Further properties of natural numbers\<close>
theory CZH_Sets_Nat
  imports CZH_Sets_Sets
begin

subsection\<open>Background\<close>

text\<open>
The section exposes certain fundamental properties of natural numbers and
provides convenience utilities for doing arithmetic within the type \<^typ>\<open>V\<close>.

Many of the results that are presented in this sections were carried over
(with amendments) from the theory \<open>Nat\<close> that can be found in the main
library of Isabelle/HOL.
\<close>

notation ord_of_nat (\<open>_\<^sub>\<nat>\<close> [999] 999)

named_theorems nat_omega_simps

declare One_nat_def[simp del]

abbreviation (input) vpfst where "vpfst a \<equiv> a\<lparr>0\<rparr>"
abbreviation (input) vpsnd where "vpsnd a \<equiv> a\<lparr>1\<^sub>\<nat>\<rparr>"
abbreviation (input) vpthrd where "vpthrd a \<equiv> a\<lparr>2\<^sub>\<nat>\<rparr>"

subsection\<open>Conversion between \<^typ>\<open>V\<close> and \<open>nat\<close>\<close>

subsubsection\<open>Primitive arithmetic\<close>

lemma ord_of_nat_plus[nat_omega_simps]: "a\<^sub>\<nat> + b\<^sub>\<nat> = (a + b)\<^sub>\<nat>"
  by (induct b) (simp_all add: plus_V_succ_right)

lemma ord_of_nat_times[nat_omega_simps]: "a\<^sub>\<nat> * b\<^sub>\<nat> = (a * b)\<^sub>\<nat>"
  by (induct b) (simp_all add: mult_succ nat_omega_simps)

lemma ord_of_nat_succ[nat_omega_simps]: "succ (a\<^sub>\<nat>) = (Suc a)\<^sub>\<nat>" by auto

lemmas [nat_omega_simps] = nat_cadd_eq_add

lemma ord_of_nat_csucc[nat_omega_simps]: "csucc (a\<^sub>\<nat>) = succ (a\<^sub>\<nat>)"
  using finite_csucc by blast

lemma ord_of_nat_succ_vempty[nat_omega_simps]: "succ 0 = 1\<^sub>\<nat>" by auto

lemma ord_of_nat_vone[nat_omega_simps]: "1 = 1\<^sub>\<nat>" by auto

subsubsection\<open>Transfer\<close>

definition cr_omega :: "V \<Rightarrow> nat \<Rightarrow> bool"
  where "cr_omega a b \<longleftrightarrow> (a = ord_of_nat b)"

text\<open>Transfer setup.\<close>

lemma cr_omega_right_total[transfer_rule]: "right_total cr_omega"
  unfolding cr_omega_def right_total_def by simp

lemma cr_omega_bi_unqie[transfer_rule]: "bi_unique cr_omega"
  unfolding cr_omega_def bi_unique_def
  by (simp add: inj_eq inj_ord_of_nat)

lemma omega_transfer_domain_rule[transfer_domain_rule]:
  "Domainp cr_omega = (\<lambda>x. x \<in>\<^sub>\<circ> \<omega>)"
  unfolding cr_omega_def by (auto simp: elts_\<omega>)

lemma omega_transfer[transfer_rule]:
  "(rel_set cr_omega) (elts \<omega>) (UNIV::nat set)"
  unfolding cr_omega_def rel_set_def by (simp add: elts_\<omega>)

lemma omega_of_real_transfer[transfer_rule]: "cr_omega (ord_of_nat a) a"
  unfolding cr_omega_def by auto

text\<open>Operations.\<close>

lemma omega_succ_transfer[transfer_rule]:
  includes lifting_syntax
  shows "(cr_omega ===> cr_omega) succ Suc"
proof(intro rel_funI, unfold cr_omega_def)
  fix x y assume prems: "x = y\<^sub>\<nat>"
  show "succ x = Suc y\<^sub>\<nat>" unfolding prems ord_of_nat_succ[symmetric] ..
qed

lemma omega_plus_transfer[transfer_rule]:
  includes lifting_syntax
  shows "(cr_omega ===> cr_omega ===> cr_omega) (+) (+)"
  by (intro rel_funI, unfold cr_omega_def) (simp add: nat_omega_simps)

lemma omega_mult_transfer[transfer_rule]:
  includes lifting_syntax
  shows "(cr_omega ===> cr_omega ===> cr_omega) (*) (*)"
  by (intro rel_funI, unfold cr_omega_def) (simp add: nat_omega_simps)

lemma ord_of_nat_card_transfer[transfer_rule]:
  includes lifting_syntax
  shows "(rel_set (=) ===> cr_omega) (\<lambda>x. ord_of_nat (card x)) card"
  by (intro rel_funI) (simp add: cr_omega_def rel_set_eq)

subsection\<open>Elementary results\<close>

lemma ord_of_nat_vempty: "0 = 0\<^sub>\<nat>" by auto

lemma set_vzero_eq_ord_of_nat_vone: "set {0} = 1\<^sub>\<nat>"
  by (metis elts_1 set_of_elts ord_of_nat_vone)

lemma vone_in_omega[simp]: "1 \<in>\<^sub>\<circ> \<omega>" unfolding \<omega>_def by force

lemma nat_of_omega:
  assumes "n \<in>\<^sub>\<circ> \<omega>"
  obtains m where "n = m\<^sub>\<nat>"
  using assms unfolding \<omega>_def by clarsimp

lemma omega_prev:
  assumes "n \<in>\<^sub>\<circ> \<omega>" and "0 \<in>\<^sub>\<circ> n"
  obtains k where "n = succ k"
proof-
  from assms nat_of_omega obtain m where "n = m\<^sub>\<nat>" by auto
  with assms(2) obtain m' where "m = Suc m'"
    unfolding less_V_def by (auto dest: gr0_implies_Suc)
  with that show ?thesis unfolding \<open>n = m\<^sub>\<nat>\<close> using ord_of_nat.simps(2) by blast
qed

lemma omega_vplus_commutative:
  assumes "a \<in>\<^sub>\<circ> \<omega>" and "b \<in>\<^sub>\<circ> \<omega>"
  shows "a + b = b + a"
  using assms by (metis Groups.add_ac(2) nat_of_omega ord_of_nat_plus)

lemma omega_vinetrsection[intro]:
  assumes "m \<in>\<^sub>\<circ> \<omega>" and "n \<in>\<^sub>\<circ> \<omega>"
  shows "m \<inter>\<^sub>\<circ> n \<in>\<^sub>\<circ> \<omega>"
proof-
  from nat_into_Ord[OF assms(1)] nat_into_Ord[OF assms(2)] Ord_linear_le
  consider "m \<subseteq>\<^sub>\<circ> n" | "n \<subseteq>\<^sub>\<circ> m"
    by auto
  then show ?thesis by cases (simp_all add: assms inf.absorb1 inf.absorb2)
qed

subsection\<open>Induction\<close>

lemma omega_induct_all[consumes 1, case_names step]:
  assumes "n \<in>\<^sub>\<circ> \<omega>" and "\<And>x. \<lbrakk>x \<in>\<^sub>\<circ> \<omega>; \<And>y. y \<in>\<^sub>\<circ> x \<Longrightarrow> P y\<rbrakk> \<Longrightarrow> P x"
  shows "P n"
  using assms by (metis Ord_\<omega> Ord_induct Ord_linear Ord_trans nat_into_Ord)

lemma omega_induct[consumes 1, case_names 0 succ]:
  assumes "n \<in>\<^sub>\<circ> \<omega>" and "P 0" and "\<And>n. \<lbrakk> n \<in>\<^sub>\<circ> \<omega>; P n \<rbrakk> \<Longrightarrow> P (succ n)"
  shows "P n"
  using assms(1,3)
proof(induct rule: omega_induct_all)
  case (step x) show ?case
  proof(cases \<open>x = 0\<close>)
    case True with assms(2) show ?thesis by simp
  next
    case False
    with step(1) have "0 \<in>\<^sub>\<circ> x" by (simp add: mem_0_Ord)
    with \<open>x \<in>\<^sub>\<circ> \<omega>\<close> obtain y where x_def: "x = succ y" by (elim omega_prev)
    with elts_succ step.hyps(1) have "y \<in>\<^sub>\<circ> \<omega>" by (blast intro: Ord_trans)
    have "y \<in>\<^sub>\<circ> x" by (simp add: \<open>x = succ y\<close>)
    have "P y" by (auto intro: step.prems step.hyps(2)[OF \<open>y \<in>\<^sub>\<circ> x\<close>])
    from step.prems[OF \<open>y \<in>\<^sub>\<circ> \<omega>\<close> \<open>P y\<close>, folded x_def] show "P x" .
  qed
qed

subsection\<open>Methods\<close>

text\<open>
The following methods provide an infrastructure for working with goals of the
form \<open>a \<in>\<^sub>\<circ> n\<^sub>\<nat> \<Longrightarrow> P a\<close>.
\<close>

lemma in_succE:
  assumes "a \<in>\<^sub>\<circ> succ n" and "\<And>a. a \<in>\<^sub>\<circ> n \<Longrightarrow> P a" and "P n"
  shows "P a"
  using assms by auto

method Suc_of_numeral =
  (
    unfold numeral.simps add.assoc,
    use nothing in \<open>unfold Suc_eq_plus1_left[symmetric], unfold One_nat_def\<close>
  )

method succ_of_numeral =
  (
    Suc_of_numeral,
    use nothing in \<open>unfold ord_of_nat_succ[symmetric] ord_of_nat_zero\<close>
  )

method numeral_of_succ =
  (
    unfold nat_omega_simps,
    use nothing in
      \<open>
        unfold numeral.simps[symmetric] Suc_numeral add_num_simps,
        (unfold numerals(1))?
      \<close>
  )

method elim_in_succ =
  (
    (
      elim in_succE;
      use nothing in \<open>(unfold triv_forall_equality)?; (numeral_of_succ)?\<close>
    ),
    simp
  )

method elim_in_numeral = (succ_of_numeral, use nothing in \<open>elim_in_succ\<close>)

subsection\<open>Auxiliary\<close>

lemma two: "2\<^sub>\<nat> = set {0, 1\<^sub>\<nat>}" by force

lemma three: "3\<^sub>\<nat> = set {0, 1\<^sub>\<nat>, 2\<^sub>\<nat>}" by force

lemma four: "4\<^sub>\<nat> = set {0, 1\<^sub>\<nat>, 2\<^sub>\<nat>, 3\<^sub>\<nat>}" by force

lemma two_vdiff_zero[simp]: "set {0, 1\<^sub>\<nat>} -\<^sub>\<circ> set {0} = set {1\<^sub>\<nat>}" by auto
lemma two_vdiff_one[simp]: "set {0, 1\<^sub>\<nat>} -\<^sub>\<circ> set {1\<^sub>\<nat>} = set {0}" by auto

text\<open>\newpage\<close>

end

subset_name: curated


TITLE: Convergence of $a_n=d(u_n,\Bbb{Z})$ where $u_n=\bigl(\frac{1+\sqrt{5}}{2}\bigr)^n$.
QUESTION [2 upvotes]: I would like to  study the sequence defined by
  $$
a_n=d(u_n,\Bbb{Z}).
$$
Where $u_n=\bigl(\frac{1+\sqrt{5}}{2}\bigr)^n$ and $d(u_n,\Bbb{Z})=\inf\{d(u_n,x):x\in\Bbb{Z}\}$.

I do not really have ideas, here is some 'though',
$$a_{n+1}-a_n=d(u_{n+1},\Bbb{Z})-d(u_n,\Bbb{Z})=\inf(\{d(u_{n+1},x):x\in\Bbb{Z}\})-\inf(\{d(u_{n},x):x\in\Bbb{Z}\}).$$
Now by theorem I have $\vert d(u_{n+1},\Bbb{Z})-d(u_n,\Bbb{Z})\vert\le\Vert u_{n+1}-u_n\Vert$.
Let us compute $\Vert u_{n+1}-u_n\Vert=\Vert\bigl(\frac{1+\sqrt{5}}{2}\bigr)^{n+1}-\bigl(\frac{1+\sqrt{5}}{2}\bigr)^n\Vert=\Vert\bigl(\frac{1+\sqrt{5}}{2}\bigr)^n\frac{\sqrt{5}-1}2\Vert.$

After that I am not sure what I have to do. Any ideas?

REPLY [4 votes]: Hint: Show that
$$\left(\frac{1+\sqrt{5}}{2}\right)^n+\left(\frac{1-\sqrt{5}}{2}\right)^n$$
is an integer. The second term has limit $0$.
Remark: There are many ways to show that our expression $\alpha^n+\beta^n$ is an integer. One way is by induction, using the fact that
$$\alpha^{n+1}+\beta^{n+1}=(\alpha+\beta)(\alpha^n+\beta^n)-\alpha\beta(\alpha^{n-1}+\beta^{n-1}),$$
and $\alpha+\beta$ and $\alpha\beta$ are integers.
Another way is to show that each of $\alpha^n$ and $\beta^n$, and therefore their sum, satisfies the recurrence $x_{n+1}=x_n+x_{n-1}$. Then from the fact that $\alpha^0+\beta^0$ and $\alpha^1+\beta^1$ are integers, we can conclude that $\alpha^n+\beta^n$ is an integer for all $n$. By the way, these are the Lucas numbers, close relatives of the Fibonacci numbers.
Whenever $\frac{1+\sqrt{5}}{2}$ has a problem, its conjugate $\frac{1-\sqrt{5}}{2}$ is ready to pitch in and help.


TITLE: Does the measurability of a set depend on a specific measure or is it "universal"?
QUESTION [4 upvotes]: I know that a measurable space is a tuple $(X,\Sigma)$, where $X$ is a set and $\Sigma$ is a $\sigma$-algebra of (measurable) subsets of $X$. A measure space, in contrast is a triple $(X,\Sigma,\mu)$, where $X$ is a set, $\Sigma$ is a $\sigma$-algebra of (measurable) subsets of $X$, and $\mu$ is a measure.
The obvious main difference is that a measurable space does not require a specific measure. Does this mean that measurability as a concept is completely independent of any specific measure? Are $\sigma$-algebras always measurable by any measure, or are they considered measurable if at least one valid measure exists? If measurability is a concept independent of specific measures, why is there a correspondence between, for example, Borel $\sigma$-algebras and the Borel measure, or Lebesgue $\sigma$-algebras and the Lebesgue measure?
Perhaps someone can clarify and point me to some resources that make this clear. Thank you!

REPLY [5 votes]: See the answers you have received allready. I would like to add that so-called outer measures induce $\sigma$-algebras. This in such a way that the the outer measure restricted to it is a measure on it. That might explain the use of terminology like "Lebesgue measure". We start with what we allready call "the Lebesgue outer measure", and the $\sigma$-algebra induced by it is called "Lebesgue $\sigma$-algebra". So the notion of measuring sets then precedes the notion of $\sigma$-algebra. See here and/or here for more information.


TITLE: Given a square matrix where $a_{11}=c\neq 0$ and $a_{ij}=0$ otherwise, can we find a matrix B such that B and A+B have no common eigenvalues?
QUESTION [0 upvotes]: Given a matrix where $a_{11}=c\neq 0$ and $a_{ij}=0$ otherwise, can we find a matrix B such that B and A+B have no common eigenvalues?
If instead the matrix had its nonzero entry component at $a_{1n}$ (or $a_{n1}$) an easy example would be to take the matrix with ones on the subdiagonal (or superdiagonal) and zeros elsewhere. This matrix would be nilpotent and thus its only eigenvalue would be zero, but whereas the sum matrix would be invertible and therefore have no zero eigenvalues.
Returning the original question, for the $n=2$ case, we can take the permutation matrix which has eigenvalues $\pm1$ and the sum matrix would be the Fibonacci matrix which has irrational eigenvalues. I'm not sure how to create a general example and see the result in a simple way. I'd imagine the construction wouldn't be too different from this case, but I also would think there is an easier way to do this. I considered looking at characteristic polynomials but didn't see anything enlightening there.
Thanks!

REPLY [1 votes]: If $p(\lambda)$ is the characteristic polynomial of $B$ and $q(\lambda)$ is the characteristic polynomial of the matrix $C$ obtained from $B$ by removing the first row and column, then the characteristic polynomial of $A+B$ is
$p(\lambda) + c q(\lambda)$.  So all you need to do is ensure that $q$ and $p$ share no roots.  That will almost always be the case.  For a simple example, you might take $A$ to be the companion matrix for a polynomial with nonzero constant term $p(0)$, in which case $p(\lambda) = \lambda q(\lambda) + p(0)$.


TITLE: Perfect numbers less than 10 000
QUESTION [0 upvotes]: Im trying to find perfect numbers less that 10 000. I was told that the best way to do this is by using maple, but I don't know how to use latex except the basics like graphing. Can someone help me do this in maple? thanks.

REPLY [0 votes]: In Maple,
seq(`if`(`+`(numtheory:-divisors(i)[])=2*i,i,NULL),i=1..9999);

                    6, 28, 496, 8128

Or, if you accept some background theory as given,
restart:

i:=1:
k:=numtheory:-mersenne([i]):
T:=k*((k+1)/2):
L:=[]:
while T < 10000 do
   L:=[L[],T];
   i:=i+1;
   k:=numtheory:-mersenne([i]);
   T:=k*((k+1)/2);
end do:
L;

                   [6, 28, 496, 8128]