text:
\begin{document}
\begin{abstract}
We show that  a characteristic $0$
 model $X_R\to \Spec R$, with Picard number $1$ over a geometric generic point,   of a  K3 surface in characteristic $p\ge 3$, essentially kills  all automorphisms (Theorem~\ref{cor:entropy2}). We show that there is an  explicitely constructed automorphism on a supersingular  K3
surface in characteristic $3$, which has positive entropy, the logarithm of a Salem number of degree $22$
 (Theorem~\ref{NoLiftPosEntropy}). In particular it does not lift to characteristic $0$. In addition, we show that in any large characteristic,  there is an automorphism of a supersingular K3  which has positive entropy and does not lift to characteristic $0$ (Theorem~\ref{thm:beta}).
\end{abstract}
\maketitle

\section{Introduction}\label{intro}

Let $X$ be a K3 surface over an algebraically closed field $k$ of characteristic $p>0$. A classical theorem \cite{Del81} asserts that the formal universal deformation space $\hat S$  of $X$  is unobstructed, and is formally smooth of dimension $20$ over $W(k)$, the ring of Witt vectors of $k$.  Moreover, the closed formal  subscheme of $\hat S$ parametrizing the locus $\hat \Sigma(X, L)$  over which a given line bundle $L$ on $X$ lifts, is  a  hypersurface, flat over $W(k)$. The aim of our article is to understand  conditions for automorphisms of $X$  to be or not to be liftable to a proper  model $X_R\to \Spec R$ of $X$, where $R$ is a discrete valuation ring such that $\Spec R\to \hat S$ dominates $\Spec W(k)$.  Said in words, we study conditions on automorphisms of $X$   to lift to characteristic $0$, or not.  One motivation for this study is the observation that the crystalline classes of graphs of automorphisms on a positive characteristic K3 surface obey the Fontaine-Mazur  $p$-adic variational Hodge conjecture as expressed in
\cite[Conj.~1.2]{BlEsKe14} (see Remark~\ref{rmk:hodge}).

\medskip

 Our main results are Theorem~\ref{cor:entropy2}, Theorem~\ref{NoLiftPosEntropy} and Theorem~\ref{thm:beta}. Simplified versions are Theorems~\ref{IntroThm2},~\ref{IntroThm3},~\ref{IntroThm4} explained below. For the discussion in the introduction, we assume for simplicity that $p\ge 3$.

\medskip

Recall one has  a natural  injective specialization  homomomorphism
$$ \iota: {\rm Aut}^e\, (X_{\bar K}/\bar K) \to {\rm Aut}\, (X/k)\,\, $$
where  $K$ is the field of fractions of $R$ and ${\rm Aut}^e(X_{\bar K}/\bar K) \subset {\rm Aut}(X_{\bar K}/\bar K)$ is the subgroup of automorphisms which lift to some proper model $X_R\to {\rm Spec} \ R$.
We say  $f \in {\rm Aut}\, (X/k)$ is not geometrically liftable to characteristic $0$  if it  is not in the image
of $\iota$
  (see Section~\ref{sec:notations} for details).

\medskip

In the complex case, for any K3 surface $M$ and any given line bundles $L_i$ ($1 \le i \le d \le 19$) on $M$  for which  the  $ L_i,  \ i=1, \ldots, d$ are part of a $\Z$-basis of ${\rm Pic}\, (M)$,  there is  a smooth proper small deformation
$$\psi : (\sM, \sL_1, \ldots , \sL_d) \to \Delta$$
of $(M, L_1, \ldots , L_d)$, where $\Delta$ is the analytic disc,  such that ${\rm Pic}\, (\sM_t) = \langle (\sL_1)_t, \ldots , (\sL_d)_t \rangle \simeq \Z^d$ for  a point $t \in \Delta,$ generic  in the  complex analytic sense.
The proof goes via a study of the period map.
 In particular, if $d=1$ and $L_1$ is ample primitive, then $\psi : \sM \to \Delta$ is also projective and ${\rm Pic}\, (\sM_t) = \Z (\sL_1)_t$ for generic $t$. As a consequence,  the specialization homomorphism has very small image even if ${\rm Aut}\, (M)$ is very large. In other words, interesting automorphisms disappear on the generic fiber $\sM_t$  (see eg. \cite{Og03}).

\medskip

Our first aim is to show the analogous results Theorems ~\ref{IntroThm1}, ~\ref{IntroThm2}  on  liftings  from characteristic $p$ to characteristic $0$  (see  Theorems~\ref{thm:one},  ~\ref{cor},
~\ref{cor:entropy2} for more precise statements):

\begin{theorem} \label{IntroThm1} If $p\ge 3$,
there is a discrete valuation ring $R$, finite over the ring of Witt vectors $W(k)$, together with a
projective model $X_R\to \Spec R$, such that the Picard rank of $X_{ \bar K}$ is $1$.
\end{theorem}
The proof is given in \cite[App.~A]{LieOls11}. The proof in {\it loc.cit.} relies on  \cite{Ogu79}, \cite{Ogu83} and on the properties of the stack parametrizing deformations of a K3 surface together with line bundles.  In Section~\ref{sec:lifting}, we sketch another proof, relying on
 \cite{Del81}. Theorem~\ref{cor}  has its own interest.

As an immediate but remarkable consequence of Theorem~\ref{IntroThm1}, we obtain:

\begin{theorem} \label{IntroThm2} If $p\ge 3$, there is a projective model $X_R\to \Spec R$
such that no subgroup $G \subset {\rm Aut}\, (X)$, except for
$G = \{ {\rm id}_X \}$ is geometrically liftable to $X_R\to \Spec R$, unless
${\rm Pic}\, (X) = \Z \cdot  H$ with  self-intersection number $(H^2) = 2$.
\end{theorem}
See also Theorem ~\ref{cor:entropy2} 2) for the exceptional case. We also note that when $X$ is not supersingular, Theorem~\ref{IntroThm2} is in  sharp contrast to the model constructed by Lieblich and Maulik \cite{LieMau11}, to show the  Kamawata-Morrison Cone Conjecture for K3 surfaces in positive characteristic.  We prove Theorem~\ref{IntroThm2} in Section~\ref{sec:Nolifting}.

\medskip

The second aim of our article is to show the richness of automophisms of supersingular K3 surfaces of Artin invariant $1$, in  view of  the non-liftability problem. Supersingular K3 surfaces of Artin invariant $1$ are unique up to isomorphisms, for each field $k$ of characteristic $p >0$. They are the most special K3 surfaces (see Section ~\ref{sec:NonLiftable} for a brief review). We denote them by $X(p)$. Recently,  several interesting aspects of automorphisms of $X(p)$ for various $p$ were studied (\cite{DK09}, \cite{DK09-2}, \cite{Sh13}, \cite{KS12} and references therein). The notion of entropy is classical in the complex case, and is of topological nature (see e.g. \cite{Og14} and
references therein); it has been  introduced   in  \cite{EsnSri13} in positive characteristic. The positivity of entropy is  a numerical  measure of complexity or richness of automorphisms in any characteristic (see Section \ref{ss:K3}). We note here that an automorphism of positive entropy is necessarily of infinite order, but it is a stronger constraint as  there are many automorphisms of infinite order with null-entropy.

\medskip

Our next main results are Theorems ~\ref{IntroThm3} and  ~\ref{IntroThm4}, showing the richness of automorphisms of supersingular K3 surfaces of Artin invariant $1$:

\begin{theorem} \label{IntroThm3}
There is an $f \in {\rm Aut}\, (X(3))$ of positive entropy such that for all $n\in \Z\setminus \{0\}$,   $f^n$
is  not geometrically  liftable to characteristic zero. The entropy of $f$ is the logarithm of a Salem number of degree $22$.  \end{theorem}

This is the  first explicit example of an automorphism  of positive entropy which can  never be lifted to characteristic zero. In characteristic $2$, (non-explicit) examples have been constructed in \cite{BC13} (see \cite[Thm.~4.2]{EOY14} for a slight  clarification).
See Theorem~\ref{NoLiftPosEntropy} for the precise statement and Section~\ref{sec:Salem} for the definition of Salem numbers.

\medskip

Recall that there are  non-projective  complex K3 surfaces with an automorphism for which  the entropy  is the logarithm of a Salem number of degree $22$ (\cite{Mc02}). However,  the  entropy of an automorphism of a  projective K3 surface over a field of characteristic zero is either zero or the logarithm of Salem number of degree $\le 20$ (\cite{Mc02}, \cite{Mc13}).
In particular, the Salem number we construct in Theorem~\ref{IntroThm3} can not be the Salem number associated to the entropy of a projective K3 surface in characteristic $0$.

Our construction is entirely based on a result of Kondo and Shimada \cite{KS12} and is mildly supported by a  {\tt Mathematica} computation. It would be nicer if one could find a more conceptual reason for the existence.

\medskip

Finally we show:
\begin{theorem}\label{IntroThm4}
For $p$ large, there is an automorphism  of $X(p)$, of positive entropy, which is not geometrically  liftable to characteristic $0$.
\end{theorem}

See also Theorem~\ref{thm:beta} for a more general statement.
Our construction is based on Jang's result \cite[Thm.~3.3]{Jan14} together
with
 a  result on the Mordell-Weil groups of elliptic fibrations
due to Shioda \cite{Sh90}.
In the proof, one shows a way to construct  an automorphism of positive entropy out of  those of null-entropy. This might have  an  interest on its own.
\footnote{Since the construction presented in this article has been written, using  Shioda's work on the existence of different elliptic structures with rich N\'eron-Severi groups  on the surfaces $X(p)$ , it is shown in \cite[Thm.~1.1]{EOY14} that for almost all $p$, $X(p)$ carries an automorphism, the entropy of which is the logarithm of a Salem number of degree $22$.}
We prove Theorem~\ref{IntroThm3}, Theorem~\ref{IntroThm4} in Sections~\ref{sec:NonLiftable},
 ~\ref{sec:EllipticK3}.

\medskip

\noindent
{\bf Ackowledgement.} \\[.2cm]
We thank  Takao Iohara for kindly accepting to check again our {\tt Mathematica}  computation  using  {\tt Maxima}. We thank Gabriel F\"urstenheim for asking  us for concrete non-liftable automorphisms, which prompted the starting point of our work.  We thank Luc Illusie and Arthur Ogus for a friendly and  interesting discussion on the articles \cite{Del81} and \cite{Ogu79}.  We thank Matthias Sch\"utt for carefully reading   the first version,
Junmeyong Jang for mentioning \cite{Jan14} to us, which enabled us to find Theorem ~\ref{IntroThm4}, and  Olivier Benoist for mentioning \cite{LieOls11} to us. Finally we thank the referee for a friendly and careful reading of our article which enabled us to improve it.

\section{Lifting automorphisms of K3 surfaces: notations and formulation of the  problems} \label{sec:notations}
\noindent
We introduce some notations and formulate the main questions addressed in our article.
\subsection{Models and Lifts} \label{ss:models}
 Let $M$ be a  proper   variety defined over a perfect characteristic $p>0$ field $k$, and $R$ be a discrete valuation ring (in the sequel abbreviated as DVR)  with residue field $k$ and  field  of fractions  $K = {\rm Frac}\, (R)$. A {\it model of $M$ over $\Spec R$} is a proper flat morphism  of schemes $M_R\to \Spec R$ lifting $M \to \Spec k$. If $M$ is smooth, then  a lift  $M_R\to \Spec R$   is a model if and only if $M_R\to \Spec R$ is smooth. We call a model $M_R\to \Spec R$ a {\it lift to characteristic zero} if $K$ is of characteristic zero. If $M_R\to \Spec R$ is a model of $X$,  and $L\supset K$ is any field extension, we say that $M_L=M_R\otimes_R L$  is a {\it lift} of $M$ to $L$.
 If $K$ has characteristic $0$, we say $M_L$ is a lift of $M$ {\it to characteristic} $0$. We call a lift $M_R\to \Spec R$ to characteristic zero a {\it projective model} of $M$ if $M_R\to \Spec R$ is projective.

\subsection{Automorphisms} \label{ss:auto}

Let $M$ and $S$ be schemes and $\varphi : M \to S$ be a morphism.
We denote by ${\rm Aut}\, (M/S)$ the {\it group of automorphisms of}  $M$ over $S$.  When $\varphi$ is flat projective, ${\rm Aut}\, (M/S)$  is the group of $S$-points of a group scheme representing the ${\rm Aut}(-/S)$-functor, but we will just use the abstract group ${\rm Aut}\, (M/S)$. If $S$ is the spectrum of a ring $R$, we also write ${\rm Aut}(M/R)$. Finally for $S=\Spec k$, we write ${\rm Aut}(M)$ instead of ${\rm Aut}(M/k)$ if there is no danger of confusion.

\subsection{K3 surfaces and automorphisms.} \label{ss:K3}
A {\it K3 surface} $V$  {\it over a field} $F$ is a smooth projective geometrically irreducible $2$-dimensional variety defined over $F$ such that $H^1(V, {\mathcal O}_V) = 0$ and the dualizing sheaf $\omega_V$ is trivial, i.e., $\omega_V \simeq {\mathcal O}_V$. As  a smooth proper scheme  of dimension $2$ over a field is necessarily projective, we stick to the notion of 'projective' surface rather than 'algebraic' surface.

\medskip

 {\it Thoughout this article,   $X $ is a K3 surface over an algebraically closed field   $k$  of characteristic
$p >0$}.  A model $X_R\to \Spec R$ of a K3 surface has the property that $X_K/K$ is a $K3$ surface.

\medskip

 The {\it  N\'eron-Severi group  ${\rm NS}\, (X)$} of $X$  is isomorphic to
${\rm Pic}\, (X)$, and it is a free $\Z$-module
of finite rank. The rank is called the {\it  Picard number} of $X$ and denoted by $\rho(X)$. We have $1 \le \rho(X) \le 22$.   The Hodge index theorem implies that the  N\'eron-Severi group ${\rm NS}\, (X)$
is an even  hyperbolic lattice with respect to the intersection from $(*, **)$,  i.e., $(*, **) \in  {\rm Sym}^2 ({\rm NS}(X)^\vee)$,
 of signature $(1, \rho(X) -1)$ on ${\rm NS}\, (X) \otimes_{\Z} \R$. In  addition,
$(x^2) := (x, x) \in 2 \Z$ for all $x \in {\rm NS}\, (X)$, as by the Riemann-Roch theorem $x^2=2( \chi(X, x) -\chi(X, \mathcal{O}_X))$.  We denote the group of isometries of $({\rm NS}\, (X), (*, **))$ by ${\rm O}\, ({\rm NS}\, (X))$.

\medskip

The action by pull-back of line bundles $L\mapsto f^*L$ defines a contravariant representation
$${\rm Aut}\, (X) \to {\rm O}\, ({\rm NS}\, (X)).$$ Let $f \in {\rm Aut}\, (X)$. The   {\it spectral radius} of $f^* \in {\rm O}\, ({\rm NS}\, (X))$, denoted by ${\rm sp}\, (f)$, is the maximum of the absolute values of eigenvalues of $f^* \otimes {\rm id}_{\C} \vert_{{\rm NS}\, (X) \otimes \C}$. Here and hereafter for a complex number $\alpha = a + b\sqrt{-1}$ ($a, b \in \R$) the absolute value $\vert \alpha \vert$ of $\alpha$ is the non-negative real number
$$\vert \alpha \vert = \sqrt{a^2 + b^2}.\,\, $$
 One has  ${\det}\, f^* = \pm 1$. Thus, ${\rm sp}\, (f) \ge 1$. One defines $f$ to be of {\it positive entropy} (resp. of {\it null-entropy}) if
${\rm sp}\, (f) > 1$ (resp. ${\rm sp}\, (f) =1$). The entropy of $f$ is defined by
$$h(f) = \log\, {\rm sp}\, (f)\,\, .$$
This definition is in fact equivalent to the one obtained by first defining the entropy as the natural logarithm of the maximum of absolute values of the eigenvalues of $f^*$ acting of the $\ell$-adic cohomology ring $\oplus_i H^i(X, \mathbb{Q}_\ell)$,  with respect to all complex embeddings of the eigenvalues, as it is shown in \cite{ EsnSri13}  that regardless of the choice of the complex embedding, this maximum is taken on the N\'eron-Severi group.
This is also consistent with  the notion of entropy for complex projective K3 surfaces. Note that if $f$ is of positive entropy, then $f$ is of infinite order, while the converse is not true in general. Let $G \subset {\rm Aut}\, (X)$ be a subgroup. We call $G$ {\it of null-entropy} (resp. {\it of positive entropy}) if all the elements of $G$  are of null-entropy (resp. some element of $G$ is of positive entropy).

 \subsection{Specializations.}  \label{ss:sp}
 Let $X_R\to \Spec R$ be a smooth proper  morphism.
 Recall  (\cite[X,~App.]{SGA6}) that  one has a {\it specialization homomorphism }
 $sp: \Pic(X_{\bar K})\to \Pic(X)$ {\it on the Picard group},  which is defined as follows:  any $\sL_{\bar K}\in \Pic(X_{\bar K})$ is defined over a finite extension $L\supset K$, $L\subset \bar K$, so $\sL_{\bar K}=\sL_L\otimes_L \bar K$. Let $R_L\subset L$ be the ring of integers. The restriction homomorphism $\Pic(X_{R_L})\to \Pic(X_L)$ is an isomorphism as $X_{R_L}$ is smooth. So $\sL_L=\sL_{R_L} \otimes L$. Then the specialization of $\sL_{\bar K}$ is $\sL_{R_L}\otimes k$.

 The specialization factors through the N\'eron-Severi group $sp_{NS}: NS(X_{\bar K})\to NS(X_k)$ and through the N\'eron-severi group modulo torsion $sp_{NS/{\rm torsion}}:  NS(X_{\bar K})/{\rm torsion} \to NS(X_k)/{\rm torsion}$. Then $sp_{NS/{\rm torsion}}$ is injective, as $sp$ is compatible with the  injections  $NS(X_{\bar K})/{\rm torsion}\to H^2(X_{\bar K}, \Q_\ell(1)), \  NS(X)/{\rm torsion} \to H^2(X, \Q_\ell(1))$ defined by the Chern class  and the specialization
 $H^2(X_{\bar K}, \Q_\ell(1))\to H^2(X, \Q_\ell(1))$
  on $\ell$-adic cohomology, which is an
 isomorphism
  by the smooth proper base change theorem (\cite[V,~Thm.~3.1]{SGA4.5}).

 \medskip

 Let $X_R\to \Spec R$ be a  model of a K3 surface $X$.
 Then  one has a {\it restriction homomorphism  } ${\rm Aut}(X_R/R)\to {\rm Aut}(X)$.
 Let us define the subset ${\rm Aut}^e (X_{\bar K}/{\bar K})\subset {\rm Aut}(X_{\bar K}/{\bar K})$ consisting of those automorphisms which lift to some model $X_R\to {\rm Spec} \ R$. (Here {\it e}  stands  for extendable).   It is clearly a subgroup, where the group law is define after  finite base change.  Then the restriction homomorphism yields a {\it specialization homomorphism}
 \ga{}{\iota: {\rm Aut}^e(X_{\bar K}/\bar K)\to {\rm Aut}(X/k) . \notag}
Moreover, $sp$ is equivariant under $\iota$.
   In addition,  as automorphisms are recognized on the associated formal scheme, and $H^0(X, T_{X/k})=0$, the specialization homomorphism
$\iota$ is injective (see \cite[~Lem.~2.3]{LieMau11}) .

 We call  $f$  {\it geometrically liftable} if
it is in the image of the specialization homomorphism $\iota$.
One  similarly defines  geometric liftability of a  subgroup $G\subset {\rm Aut}(X)$.

 \begin{remark}\label{biraut} It is natural to ask whether the subgroup ${\rm Aut}^e(X_{\bar K}/\bar K) \subset {\rm Aut}(X_{\bar K}/\bar K)$  is a strict subgroup.  We  give an explicit example for which  ${\rm Aut}^e(X_{\bar K}/\bar K) \not= {\rm Aut}(X_{\bar K}/\bar K)$ in
Theorem~\ref{NoLiftTwoDegreeTwo}.
  Any $f \in {\rm Aut}\, (X_{\bar K}/\bar K)$ extends uniquely to $\tilde{f} \in {\rm Bir}\, (X_{R_L}/R_{L})$, the group of birational automorphisms, where $L$, with $K \subset L \subset \bar{K}$,  is a field of definition of $f$.   As $X$ is regular, if $\tilde{f}$ is not regular, then there is a $1$-dimensional subscheme $C\subset X$ such that $\tilde{f}$ is well defined as a morphism $\tilde{f}: X_{R_L}\setminus C\to X_{R_L}$, but the morphism does not necessarily extend to $X_{R_L}$. Finally this implies that for any field extension $L'\supset L$, $L'\subset \bar K$, the base changed morphism   $\tilde{f}\otimes R_{L'}:  X_{R_{L'}}\setminus C\to X_{R_{L'}}$ does not extend either.
 See \cite[Prop.~4.1]{LieMat14} for related phenomena. \end{remark}

\section{Automorphisms of even hyperbolic lattices} \label{sec:Salem}

We call a polynomial $P(x) \in \Z [x]$ a {\it Salem polynomial} if it is irreducible, monic,  of even degree $2d \ge 2$ and the complex zeroes of $P(x)$ are of the form ($1 \le i \le d-1$):
$$a > 1\,\, ,\,\, 0 < a^{-1} < 1\,\, ,\,\, \alpha_i, \overline{\alpha}_i \in S^1 :=
\{z \in \C\, \vert\, \vert z \vert = 1\} \setminus \{\pm 1\}\,\, .$$

\begin{proposition}\label{salem} Let $r$ be a positive integer and $L = (\Z^r, (*,**) \in {\rm Sym}^2 (\Z^r)^{\vee})$ be a hyperbolic lattice, i.e., the bilinear form $(*,**)$ is non-degenerate of signature $(1, r-1)$. Let $C := \{x \in L \otimes \R\, \vert\, (x^2) > 0\}$. Then $C$ has exactly two connected components, say $C^0$ and $-C^0$. Let $f \in {\rm Aut}\, L$ and assume that $f(C^0) \subset C^0$. Then, the characteristic polynomial of $f$ is the product of cyclotomic polynomials and at most one Salem polynomial. In particular,  when $f(C^0) \subset C^0$,
 the characteristic polynomial of $f$ is the product of cyclotomic polynomials if and only if $f$ is of null-entropy  if and ony if  $f$ is quasi-unipotent,
  i.e., all the eigenvalues of $f^n$ are $1$ for some positive integer $n$.
\end{proposition}
\begin{proof} This is well-known and essentially due to McMullen \cite{Mc02}. See also \cite{Og10}.
\end{proof}
One way for an automorphism $f$  of an hyperbolic lattice to perserve $C^0$ is to fulfill $f(e)=e$ for a non-zero isotropic vector $e$.
\begin{remark}\label{entropy}
 For $f$ as in Proposition~\ref{salem}, we define (by a slight abuse of notation)  $f$  to be  of positive entropy (resp. of null entropy) if ${\rm sp}\, (f) > 1$ (resp. ${\rm sp}\, (f) =1$). Thus $f$ is of positive entropy (resp. of null entropy) if and only if the characteristic polynomial of $f$ has a  Salem factor (then exactly one)  (resp. only cyclotomic factors).
\end{remark}

\begin{proposition}\label{nullentropy} Let  $L$ be as in Proposition~\ref{salem} and $f$ be in ${\rm Aut}(L)$.
  Assume that there is $e \in L \setminus \{ 0\}$
such that $f(e) = e$ with $(e^2) :=(e,e) = 0$. Then the characteristic polynomial of $f$ is the product of cyclotomic polynomials.
\end{proposition}

\begin{proof}

We may assume without loss of generality that $e$ is primitive
in the sense that $e$ is a part of $\Z$-basis of $L$. By the assumption, $f$ acts on the flag  $\Z e \subset (\Z e)^{\perp}$ and hence induces an automorphism $\bar f$ of
$$\overline{L} := (\Z e)^{\perp}/\Z e\,\, .$$
The bilinear form of $L$ induces a bilinear form of
$\overline{L}$ of signature $(0, r-2)$, i.e., $\overline{L}$ is  negative definite or $\{0 \}$.
If  $ \overline{L} \neq 0$,
 the eigenvalues of $\bar f$ on $\overline{L}$ are of absolute value $1$. Here we use the well-known fact that eigenvalues of a real orthogonal matrix are of absolute value $1$. Combining this with $f(e) = e$, we find that the eigenvalues of $f$ are of absolute value $1$ except perhaps one eigenvalue counted with multiplicities. Note that $\det\, f = \pm 1$, as
an automorphism of free $\Z$-module $L$ of finite positive rank. Hence the last  eigenvalue is also of absolute value $1$. Since $f(C^0) \subset C^0$ as $f(e)= e$, this implies the result by Proposition~\ref{salem}.
\end{proof}

\section{Lifting to characteristic $0$  K3 surfaces with Picard number one} \label{sec:lifting}

Let $M$ be a complex projective  K3 surface with a primitive ample line bundle $H$. Recall   $H$ is said to be {\it primitive}  if it is a part of $\Z$-basis of the finitely generated free $\Z$-module ${\rm Pic}\, (M) = {\rm NS}\, (M)$.
 It is well-known that any generic  fiber ${\mathcal X}_t$ of the Kuranishi family $ \kappa: ({\mathcal X}, {\mathcal H}) \to {\mathcal K}$ of $(X, H)$ is of Picard number $1$. Here 'generic' is to be understood in the complex analytic sense.  In particular, restricting $\kappa$ to the germ of a smooth curve $\sB \subset \mathcal{K}$ containing a generic point, the kernel of the Gau{\ss}-Manin connection on $H^2$ on $\sB$ is spanned by the first de Rham Chern class of a generator of ${\rm NS}\,(\mathcal{X}_t)$.

  \medskip

In \cite[App.~A]{LieOls11},
   M. Lieblich and M. Olsson,
   prove the analogous result on Picard  rank $1$ lifts to characteristic $0$ of characteristic $p\ge 3$ K3 surfaces.

\begin{theorem} \label{thm:one}
Let $X$ be a K3 surface defined over
 an algebraically  closed field   $k$  of characteristic
$p >0$, where $p>2$ if $X$ is  Artin-supersingular.
Then there is a discrete valuation ring $R$, finite over the ring of Witt vectors $W(k)$, together with a
projective model $X_R\to \Spec R$, such that the Picard rank of $X_{ \bar K}$ is $1$, where $ K={\rm Frac}\,(R)$ and $\bar K\supset K$ is an algebraic closure.
\end{theorem}
Their proof  (in $p \ge 3$) relies on \cite{Ogu79} and \cite{Ogu83} and properties of stacks for pairs of K3 surfaces together with line bundles.

\medskip

 For a line bundle $L$ on $X$, we denote by $c_1^{\rm Hodge}(L)\in H^1(X, \Omega^1_X)$ its Hodge Chern class.
\begin{proposition} \label{existence}
Let $X$ be a K3 surface  defined over
 an algebraically  closed field   $k$  of characteristic
$p >0$,
with $p>2$  if     $X$ is  Artin-supersingular.  Then there is an ample primitive   line bundle $L$ such that $c_1^{\rm Hodge}(L)\neq 0$.

\end{proposition}

\begin{proof}
Unfortunately, we can not prove it directly, this is the reason for the restriction on $p$.  If $X$ is not Artin-supersingular, then by \cite[Prop.~10.3]{GK00}, $$c_1^{{\rm Hodge}}: \Pic(X)/p\Pic(X)  \to H^1(X, \Omega^1_{X/k})$$
is injective.  Else, due to our assumption, it is Shioda-supersingular,
thus by
\cite[Prop.~11.9]{GK00}, $c_1^{{\rm Hodge}}$ is not identically zero.  Once one line bundle $M$ fulfills $c_1^{{\rm Hodge}}(M)\neq 0$, then given any ample line bundle $H$, $0\neq c_1^{{\rm Hodge}}(M)=c_1^{{\rm Hodge}}(M+mpH)$ for any integer $m$, and for $m$ large, $M+pmH$ is ample.  The $\Z$-module
$\Q\cdot ( M+pmH)\cap NS(X)$  of rank $1$ has a generator $L$ such $(M+mH)=aL$ where $a \in \N\setminus \{0\}$.
 Then $L$
is primitive, ample, and fulfills $c_1^{\rm Hodge}(L)\neq 0$.

\end{proof}

In fact, one can prove Theorem~\ref{thm:one} by using \cite{Del81} and Proposition~\ref{existence} only. Indeed, for $L$ as in Proposition~\ref{existence}, one shows (\cite[Prop.~2.2]{Ogu79} and \cite[Lem.~4.3]{LieMau11}) that the formal hypersurface $\Sigma(X,L)\subset \hat S={\rm Spf} \ W[[t_1,\ldots, t_{20}]]$, which is defined as the solution to the deformation functor of the pair $ (X,L)$, while $\hat S$ is defined as the solution of the deformation functor of $X$, is formally smooth.

Let $\hat X$ be the formal universal K3 surface over $\hat S$. Set $\hat Y=\hat X \times_{\hat S} \Sigma(X,L)$ and let $\sL$ be the formal universal line bundle on $\hat Y$ lifting $L$ on $X$. Then the de Rham class $c_1^{DR}(\sL)\in H^2_{DR}(\hat Y/\Sigma(X,L))$ is non-zero as its restriction in $H^1(X, \Omega^1_X)$ is non-zero.

  Then one shows, using  the injectivity of the crystalline Chern class map ${\rm Pic}(X)\to H^2_{DR}(\hat X_W/W)$ (\cite[2.10]{Del81})
\begin{theorem} \label{cor}
The kernel of the Gau{\ss}-Manin connection
\ga{}{\nabla: H^2(\Omega^{\ge 1}_{\hat Y/\hat \Sigma}) \to
\Omega^1_{\hat \Sigma /W}  \otimes H^2_{DR}(\hat Y/\hat \Sigma) \notag}
is spanned by $c_1^{DR}(\sL)$ over $W$.

\end{theorem}
From Theorem~\ref{cor}  one easily deduces Theorem~\ref{thm:one}.

\begin{remark} \label{rmk:follow}
Given $X$ as in Theorem~\ref{thm:one}, and $L\in \Pic(X)$ as in Proposition~\ref{existence}, then  $X_R\to \Spec R$  of  Theorem~\ref{thm:one} is constructed in such a way   that $L$ lifts to $X_R$. \end{remark}
\section{No lifting of automorphisms}\label{sec:Nolifting}

 In this section, applying Theorem ~\ref{thm:one}, we construct a projective model
 $X_R\to \Spec R$
 of a K3 surface $X$, with $K={\rm Frac}(R)$  of characteristic zero, for   which almost all automorphisms of $X$ are not geometrically liftable.

\begin{theorem} \label{cor:entropy2}
Let $X$ be a K3 surface defined over
 an algebraic closed field  $k$ of  characteristic
$p >0$,  where $p>2$ if $X$ is  Artin-supersingular.
\begin{itemize}
\item[1)]  Assume  that either the Picard number of $X$ is $\ge 2$ or that ${\rm Pic}\, (X) = \Z \cdot  H$ and $H^2\neq 2$.
Then there is a DVR $R$, finite over $W(k)$,  together with a projective model $X_R\to \Spec R$ of $X\to \Spec k$ such that no subgroup $G \subset {\rm Aut}\, (X)$, except for
$G = \{ {\rm id}_X \}$,  is geometrically liftable to $X_R\to \Spec R$;
\item[2)]  Assume  that
${\rm Pic}\, (X) = \Z \cdot  H$ and  $(H^2) = 2$.   Then, for any projective   model $X_R\to \Spec R$ with $R$ finite over $W(k)$,   ${\rm Aut}^e(X_R/R)={\rm Aut}(X_R/R)$,
the specialization homomorphism $\iota: {\rm Aut}(X_{\bar K}) \to {\rm Aut}(X)$ is an isomorphism, and ${\rm Aut}(X)=\Z/2$.

\end{itemize}
\end{theorem}

\begin{proof}  First we prove 2).
 By replacing $H$ by $-H$ if necessary, we may assume that $H$ is ample.
By Proposition ~\ref{existence} or, if $p \neq 2$, simply by $H^2(X, \Omega^2_{X/k}) \ni c_1^{{\rm Hodge}}(H)^{\cup 2}=$ residue class of $2$ in $k$, which is  thus non-zero,
 $c_1^{\rm Hodge}(H) \not= 0$ and $H$ extends to a line bundle
$H_R$ for any projective lift
$X_R \to {\rm Spec}\, R$ to characteristic $0$. By \cite{SD74}, $h^0(X, H) = 3$, $h^i(X, H) = 0$ ($i \ge 1$), and $H$ is globally generated.  Strictly speaking,  $p \not= 2$ is assumed  in  \cite{SD74}. However, since $H$ is an ample generator, $h^0(H) \ge 3$ by the Riemann-Roch theorem. Moreover, any element in $\vert H \vert$ is irreducible and reduced. (Indeed, if  $C + D \in |H|$ for some non-zero effective divisors $C, D$, with possibly $C = D$, then $C \in |nH|$ and $D \in |mH|$ for some positive integers $n$, $m$ by ${\rm Pic}\, (X) = \Z\cdot  H$.
However, then
$2 \le n+m =1$, a contradiction.) So, we can apply \cite[Prop.~2.6, Thm.~3.1]{SD74}, which is characteristic free, to our $X$, to conclude that  $\vert H \vert$ defines a finite surjective morphism $\varphi : X \to \P^2_k$ of degree $2$. This is also separable even for $p=2$, as $X$ has no non-zero vector field by \cite{RS81}. We denote the covering involution by  $\iota\in {\rm Aut}\, (X)$. Since $h^i(X, H) = 0$ ($i \ge 1$), it follows that $H^0(X_R, H_R)$ is a rank $3$  free module over $R$, which satisfies base change. It thus defines a finite  surjective morphism $\varphi_R: X_R\to \P(H^0(X_R, H_R)^\vee)\cong \P^2_R$, of degree $2$, the specialization  of which over ${\rm Spec}\, k$ is $\varphi : X \to \P^2_k$.  We denote by   $\iota \in {\rm Aut}\, (X_R/R)$  the covering involution of $\varphi_R$, which exists as $\varphi_R$ is finite. Then  $\iota_R$ specializes to $\iota$.
 Hence, in the exceptional case ${\rm Pic}\, (X) = \Z \cdot  H$ with $(H^2) = 2$, the involution $\iota \in {\rm Aut}\, (X)$ lifts as an automorphism to any projective lift to characteristic $0$, in particular, $\iota$ is  geometrically liftable to any projective lift to characteristic zero.   More precisely, ${\rm Pic}\, (X_{\bar K}) = \Z \cdot  H_{\bar K}$, $(H_{\bar K}^2) = 2$ and ${\rm Aut}\, (X_{\bar K}) = \langle \iota_{\bar K} \rangle \simeq \Z/2$, where $\bar K$ is an algebraic closure of $K = {\rm Frac}\, (R)$ and $\iota_{\bar K}$ is the covering involution
of the morphism $\varphi_{\bar K} : X_{\bar K} \to \P_{\bar K}^2$ given by $|H_{\bar K}|$.  This finishes the proof of 2).
\end{proof}
\begin{proof} We prove 1).
The following lemma  ought to  be well-known to the experts:
\begin{lemma}\label{trivial}
Let $Z$ be a complex projective K3 surface, with Picard group ${\rm Pic}\, (Z)$  generated by an ample class $H$. Assume that $(H^2) \not= 2$. Then ${\rm Aut}\, (Z) = \{\rm id_Z \}$.
\end{lemma}

We do not know whether Lemma~\ref{trivial}  holds also for a K3 surface $X$ with $\rho(X) = 1$ in characteristic $p \ge 3$. However, if $X$ is a K3 surface defined over $\bar{\F}_p$, with $p\ge 3$, the positive solution to the Tate conjecture (\cite{MPe13}, also \cite{Ben14} with references therein) implies that  $\rho(X)$ is  even (\cite{Ar74}).

\begin{proof} We give a proof for the reader's  convenience.

Since $H^0(Z, T_Z) = 0$ and ${\rm Aut}\, (Z)$ preserves the ample generator $H$, it follows that ${\rm Aut}\, (Z)$ is of dimension $0$ and also a closed algebraic subgroupscheme of the affine groupscheme ${\rm Aut}\, (\P (H^0(Z, mH)^\vee))$ for large $m >0$. It follows that ${\rm Aut}\, (Z)$ is a finite group.

Let $T(Z)$ be the transcendental lattice of $Z$, that is, the orthogonal complement of ${\rm NS}\, (Z)$ in $H^2(Z, \Z(1))$. Thus the representation ${\rm Aut}\, (Z) \to {\rm O}\, ({\rm NS}\, (Z)) \times {\rm O}\,  (T(Z))$ has image in $\{1\}\times {\rm O}\,  (T(Z))$. On the other hand, since $Z$ is projective and ${\rm Aut}\, (Z)$ is a finite group,  by Nikulin \cite{Ni79}, the image of the natural map $G \to {\rm O}\, (T(Z))$
is a cyclic group of finite order $N$,  where  $N$ is exactly the order of  the image of the  representation ${\rm Aut}\, (Z) \to {\rm GL}\, (H^0(Z, \Omega_Z^2)) = \C \omega_Z,$ and  the Euler function $\varphi (N) := [\Q (e^{2\pi i/N}) : \Q]$ of $N$
is a divisor
of the rank of the transcendental lattice $T(Z)$.  In our case, $T(Z)$  is of rank $21 = 22 -1$.  In particular, it is an odd number. Hence $N = 1$ or $2$. As $({\rm NS}(X)\oplus T(Z))\otimes_{\Z} \Q=H^2(Z, \Q(1))$ and ${\rm Aut}\, (Z)$ stabilizes  the lattice
${\rm NS}( Z)\oplus T( Z)\subset H^2( Z, \Q(1))$, the image of the representation ${\rm Aut}\, (Z)\to GL(H^2(Z, \Q(1)))$ is cyclic of order $1$ or $2$, as well as the image of the representation $G\to GL(H^2( Z, \Z(1))$.
 On the other hand, by the global Torelli theorem for complex projective K3 surfaces (\cite{PS71}), the action of ${\rm Aut}\, (Z)$ on $H^2(Z, \Z(1))$ is faithful. Thus ${\rm Aut}\, (Z)$ is either $\{{\rm  id}_Z \}$ or cyclic  of order $2$.

 So far, we did not use $(H^2) \not= 2$. Set $2d = (H^2) \ge 2$.  Then  by \cite[Cor.~1.6.2]{Ni79-2}, we have
$$\Z/2d \simeq {\rm NS}\, (Z)^*/{\rm NS}\, (Z) \simeq T(Z)^*/T(Z)\,\, ,$$
as $H^2(Z, \Z)$ is free and unimodular  and ${\rm NS}\, (Z)$ is primitive in $H^2(Z, \Z)$. Here $(-)^*$ means ${\rm Hom}_{\Z}((-), \Z)$  and $T(Z)=(\Z\cdot H)^\perp$.

As $\iota^*H$ is ample, and  $H$ is the unique ample generator of ${\rm Pic}\, (Z)$, one concludes that $\iota^*(H)=H$, thus
$\iota^*={\rm id} $  on ${\rm NS}\, (Z)^*/{\rm NS}\, (Z)$.
Hence $\iota^* = {\rm  id}$ on $T(Z)^*/T(Z)$ as well. On the other hand, the involution $\iota$ satisfies $\iota^* \omega_Z = -\omega_Z$ by Nikulin's result above. Thus
$\iota^* = - {\rm id}$ on $T(Z)^*/T(Z)$. Thus ${\rm id} = -{\rm id}$ on $\Z/2d$. Hence $2d = 2$ as claimed.  \end{proof}

\begin{lemma}\label{trivial2}
Let $X$ be as in Theorem ~\ref{cor:entropy2}.  Assume that ${\rm Pic} (X) = {\rm NS}\, (X)$
is not isomorphic to $\Z\cdot  H$ with self-intersection number $(H^2) = 2$.  Then, there is an ample primitive line bundle $L$ such that $c_1^{\rm Hodge}(L) \not= 0$ and $(L^2) \not= 2$.
\end{lemma}

\begin{proof} By Proposition ~\ref{existence}, there is an ample primitive line bundle $L_0$ such that $c_1^{\rm Hodge}(L_0) \not= 0$. If $(L_0^2) \not= 2$, then we may take  $ L=L_0$. In particular, if $\rho(X) = 1$, then we are done, as we exclude the case ${\rm Pic}\, (X) = \Z\cdot L$ with $(L^2) = 2$.

So, we may assume without loss of generality that $\rho(X) \ge 2$
and $(L_0^2) = 2$. Since $L_0$ is primitive and $\rho(X) \ge 2$, we can choose a line bundle $M$ such that
$\{L_0, M\}$  is   part of  a $\Z$-basis
of ${\rm Pic}\, (X)$. Replacing $M$ by $M + nL_0$ with large integer $n$, we may further assume, without loss of generality, that $M$ is also ample. Note here that $\Z \langle L_0, M \rangle = \Z \langle L, M +nL_0 \rangle$ and $L_0$ and $M$ remain part of free $\Z$-basis under this replacement. Now consider $L= pM + L_0$. Then $L$ is ample, as $M$ and $L_0$ are ample. $L$ is also primitive, as $L_0$ and $M$ form part of free $\Z$-basis of $\Z$-module ${\rm Pic}\, (X)$. Moreover,
$$(L^2) = p^2(M^2) + 2p(M.L_0) + (L_0^2) > 2\,\, ,$$
by $(M^2) > 0$, $(M.L_0) >0$ and $(L_0^2) >0$ by the ampleness. Thus $(L^2) \not= 2$.  Moreover,
$$c_1^{\rm Hodge}(L) = pc_1^{\rm Hodge}(M) + c_1^{\rm Hodge}(L_0) = 0 + c_1^{\rm Hodge}(L_0) = c_1^{\rm Hodge}(L_0) \not= 0\,\, ,$$
in the $k$-vector space $H^1(X, \Omega_X^1)$. So,
$L = pM +L_0$ satisfies all the requirements.
\end{proof}

Now we are ready to finish the proof of Theorem~\ref{cor:entropy2} 1).
We take the model $X_R\to \Spec R$ of Theorem~\ref{thm:one}, Remark~\ref{rmk:follow}, applied to the ample primitive line bundle $L$ in Lemma ~\ref{trivial2}.  Assume that $G$ is geometrically liftable to $X_R\to \Spec R$. Then $G$
 has to stabilize $\Pic(X_{\bar K})$, thus it fixes the  polarisation $\sL_{\bar K}$. Hence $G$ is a finite group. So we may  assume that there is an abstract field isomorphism $\bar K\rightarrow \C$ and $X_{\bar K}$ is a complex projective K3 surface, say $Z$, with Picard group ${\rm Pic}\, (Z)$  generated by an ample class $H_Z$ with $(H_Z^2) \not= 2$, and $G$ is now a group of automorphisms of $Z$. Then $G = \{{\rm id}_{X}\}$
by Lemma ~\ref{trivial}.

\end{proof}

\section{Non-liftable automorphism of positive entropy}\label{sec:NonLiftable}

The aim of this section is to construct an example  of an automorphism of a supersingular  K3 surface over an algebraically closed  field
$k$ of characteristic $p \ge 3$,
which is not geometrically liftable to any projective model.  As far as we are aware of, this is  the first such example.
 Our construction is based on the work by Kondo-Shimada \cite{KS12},
and is (mildly) computer supported. The characteristic $p$ will be equal to $3$.

\medskip

Recall that ${\rm det}\, {\rm NS}\, (X) = -p^{2 \sigma_0}$ for a supersingular K3 surface defined over $k$. The value $\sigma_0$ is called the {\it Artin invariant} of $X$. Artin \cite{Ar74} proved that $1 \le \sigma_0 \le 10$ and Ogus \cite{Ogu79} proved the uniqueness of a supersingular K3 surface over $k$ with $\sigma_0 = 1$,  up to isomorphisms. Then  this K3 surface  is isomorphic to the Kummer K3 surface ${\rm Km}\, (E \times_k E)$ associated to the product abelian surface $E \times_k E$ of any supersingular elliptic curve $E/k$. Tate and Shioda (\cite{Sh75}) proved that the Fermat quartic K3 surface is supersingular if and ony if $p \equiv 3\ {\rm mod}   \ 4$. There are several other descriptions of supersingular K3 surfaces of Artin invariant $1$ (see e.g. \cite{Sh13}).

\medskip

{\it From now until the end of this section, $X$ is a supersingular K3 surface, defined over $k$ of characteristic $3$, with Artin invariant $1$.} As remarked above, $X$ is isomorphic to the Fermat quartic K3 surface.
 We denote  by $q : X \hookrightarrow \P^3$ the projective embedding and set $H=q^*\sO_{\P^3}(1)$.

 \medskip

In \cite{KS12},  Kondo and Shimada prove the following  statements, which are crucial for our construction:
\begin{itemize}
\item[(i)]$X$ has two  globally generated line bundles $L_i$ ($i = 1$, $2$) of degree $2$ (${\sL_{m_i}}$ in their notation  \cite[p.~19]{KS12}).  They are not ample. The linear system associated to $H^0(X, L_i)$ induces a well defined morphism $\varphi_i: X\to \P^2$ which is generically finite $2:1$, but not finite.  The Galois involution of the function field extension extends as  an automorphism $\tau_i \in {\rm Aut}\, (X/\P^2)$. Indeed, it clearly extends as an automorphism in ${\rm Aut}(Y_i/\P^2)$ where $\varphi_i: X\to Y_i\to \P^2$ is the Stein factorization, and on the other hand, $X$ is the minimal desingularization of  the surface $Y_i$.

\item[(ii)] Let ${\rm Aut}\, (X, H)$ be the automorphism group of $X$ induced by the projective linear automorphisms of $\P^3$ under $q$. It is known that ${\rm Aut}\, (X, H)$ is a finite group but  of huge order (see. e.g. \cite{DK09}, \cite{Mu88}). In ${\rm Aut}\, (X, H)$, there is a special element $\tau \in {\rm Aut}\, (X, H)$  of order $28$ (\cite[Ex.~3.4]{KS12}).
\end{itemize}

They prove the following beautiful description of the automorphism group of $X$:

\begin{theorem} \label{KondoShimada}
${\rm Aut}\, (X) = \langle \tau_1, \tau_2, {\rm Aut}\, (X, H) \rangle$.
\end{theorem}

In the course of the proof, they  work with an explicit $\Z$-basis $\sB$ of ${\rm NS}\, (X)$, consisting of $22$ lines
among the 112 lines on $X$  (\cite[Lem. 6.3]{SSL10}) and compute the (right, hence covariant) representation of $(\tau_i)_* \vert_{ {\rm NS}\,(X)}$, $\tau_* \vert_{ {\rm NS}\,(X)}$ on ${\rm NS}\, (X)$. They actually write explicitely  the matrices  of $(\tau_i)_* \vert_{ {\rm NS}\,(X)}$ and $\tau_* \vert_{ {\rm NS}\,(X)}$  in the basis $\sB$.  We denote them by $A_1$, $A_2$, $T$  respectively, of which explicit forms are in  Tables 5.4, 5.5, 3.3 in \cite{KS12}. These forms are important  for the proof of Theorem \ref{KondoShimada}.

Suppose one has a model $X_R\to {\rm Spec} \ R$ on which $L_i$ lift to $L_{i, R}$. Then, as $H^0(X_R, L_{i, R})\otimes_R k=  H^0(X, L_i)$, one has base change $Y_{i, R}\otimes_R k=Y_i$ for the Stein factorization
$\varphi_{i,R}: X_R\to Y_{i, R}\to \P^2_R$
of the well defined morphism $\varphi_{i, R}: X_R\to \P^2_R$ associated to $H^0(X_R, L_{i, R})$.
The Galois involution of the function field extension induced by $\varphi_{i, R}$ extends as an automorphism
$\tau_{Y_{i, R}} \in {\rm Aut}(Y_{i,R}/\P^2_R)$, which induces a birational automorphism  $\tau_{i,R}\in
{\rm Bir}(X_R/\P^2_R)$. One says that $\tau_i$ {\it lifts to } $X_R$ if   $\tau_{i,R}\in {\rm Aut}(X_R/\P^2_R)\subset
{\rm Bir}(X_R/\P^2_R)$.

\begin{theorem} \label{NoLiftTwoDegreeTwo}
There is a projective model $X_R\to \Spec \ R$  of $X = X(3)$ with $R$ of characteristic $0$ such that ${\rm Aut}^e(X_{\bar K}) \not= {\rm Aut}\, (X_{\bar K})$.  Here $K = {\rm Frac}(R)$ and $\bar K$ is an algebraic closure
of $K$.
\end{theorem}

\begin{proof}  Let $H$ be an ample line bundle on $X$. By \cite[App.~A]{LieOls11}, there is a model $X_R\to \Spec \ R$ such that $L_1$, $L_2$ and $H$ lift.  This property is compatible with further base change $R\subset R_L$ where $L\supset K$ is a field extension with $L\subset \bar K$.
This model is projective. Let us denote by $f_i$ the restriction of $\tau_{i,R}$ to $X_{\bar K}$. Then $f_i \in {\rm Aut}(X_{\bar K})$ as $X_{\bar K}$ is a minimal smooth projective surface. It suffices to show that one of $f_i$ ($i =1$, $2$) is not in ${\rm Aut}^e(X_{\bar K})$. Assume to the contrary that both $f_i$ ($i =1$, $2$) are in ${\rm Aut}^e(X_{\bar K})$.
 Then   $ f_i^*\omega = -\omega,$ where $\omega$ is a non-zero  global $2$-form of $X_{\bar K}$. Then  $(f_1 \circ f_2)^*\omega = \omega$.  Thus $(f_1\circ f_2)^*$ has one  eigenvalue equal to one on de Rham cohomology $H^2_{DR}(X_{\bar K}/{\bar K})$, thus, by the comparison theorem, on $\ell$-adic cohomology $H_{{\rm \acute{e}t}}^2(X_{\bar K}, \Q_\ell(1))$ as well, thus by  (\cite[V,~Thm.~3.1]{SGA4.5}), on $H_{{\rm \acute{e}t}}^2(X, \Q_\ell(1))$ as well.

On the other hand, using the explicit forms of
$A_1$ and $A_2$  in Tables 5.4, 5.5 in \cite{KS12}, and {\tt Mathematica  (all we need here are Dot command, CharacteristicPolynomial command, Factor command}), we find that the characteristic polynomial of $(\tau_2 \circ \tau_1)^* \vert_{ {\rm NS}\,(X)}  = \tau_{1*} \circ \tau_{2*}|_{NS(X)}$, i.e., of $A_1A_2$, is
$$(1 + x + x^2) (1 - 11 x + 10 x^2 - 9 x^3 + 9 x^4 - 10 x^5 + 15 x^6 -
   23 x^7 + 19 x^8 - 14 x^9 + 14 x^{10} - 14 x^{11} + 19 x^{12}$$
$$ - 23 x^{13} +
   15 x^{14} - 10 x^{15} + 9 x^{16} - 9 x^{17} + 10 x^{18} - 11 x^{19}
+ x^{20})\,\,,$$
of which  $1$ is not a zero,  a contradiction.
\end{proof}

\begin{remarks} \label{NoLiftAlg}
 On the model of Theorem~\ref{NoLiftTwoDegreeTwo}, one of the $\tau_i$ does not lift, which means that a small modification occurs on the special fiber.  In fact one can say the following.
Given a  projective model $X_R\to {\rm Spec}  \ R$, with ample line bundle $L_R$, then $\tau_i$  always lifts to an automorphism $\tau_i^0$  of $X_R^i:=X_R\setminus \Sigma_i$, where $\Sigma_i$  is  the exceptional  locus of $X\to Y_i$.
So one can always define the line bundle $L_{i, R}= ( \tau_i^0)^*L_R|_{X_R^i} \in {\rm Pic} (X_R^i)={\rm Pic}(X_R)$. Then $\tau_i$ lifts to $X_R$ if and only if
$L_{i,R}$ is ample, which is equivalent to
$L_{i, R}\otimes_R k= \tau_i^* (L_R\otimes_R K)$.

\end{remarks}
The next theorem gives the (first)  explicit example of a positive entropy  automorphism of  a K3 surface which
is not geometrically liftable  to characteristic zero.
\begin{theorem} \label{NoLiftPosEntropy} The automorphism
$f := \tau_1 \circ \tau \circ \tau_2 \circ \tau \in {\rm Aut}\, (X)$
\begin{itemize}
 \item[(1)] is not geometrically liftable to any projective model $X_R\to \Spec R$ of characteristic zero  as well as its power $f^n$
($n \in \Z \setminus \{0\}$);
 \item[(2)] has positive entropy $h(f)$ equal to the logarithm of a Salem number  $a$ of degree $22$;
 \item[(3)] $h(f)$ is not the entropy of any automorphism on any  projective K3 surface  in characteristic $0$;
 \item[(4)] numerically
$$a =  26.9943 \ldots\,\, ,\,\, h(f) = \log\, 26.9943 \ldots\,\, .$$
\end{itemize}
\end{theorem}

\begin{proof} Using the explicit forms  of $A_1$, $A_2$, $T$ in Tables 5.4, 5.5, 3.3 in \cite{KS12}, and {\tt Mathematica (all we need here are Dot command, CharacteristicPolynomial command, Factor command, and NSolve command)}, we find that the characteristic polynomial  $P$ of
$f_* \vert_{ {\rm NS}\,(X))}$,
i.e., of $A_1TA_2T$, is
$$1 - 27 x + 4 x^3 + 3 x^4 + 24 x^5 + 15 x^6 - 7 x^7 + x^8 - 14 x^9 -
 2 x^{10} - 5 x^{11} - 2 x^{12}$$
$$ - 14 x^{13} + x^{14} - 7 x^{15} + 15 x^{16} +
 24 x^{17} + 3 x^{18} + 4 x^{19} - 27 x^{21} + x^{22}\,\, .$$
This is an irreducible Salem polynomial of degree $22$.
The fact that this is irreducible is checked by {\tt Factor command}. Then this is a Salem polynomial by Proposition~\ref{salem}. Indeed, it is either a Salem polynomial or a cyclotomic polynomial of degree $22$. But {\tt NSolve command} shows that one of  the zeroes of the above polynomial is approximately $26.9943$. Hence it is a Salem polynomial of degree $22$ with Salem number approximately $26.9943$. This shows (2).

By \cite[Lem.~2]{Smy14}, if  $\lambda$ is a Salem number of degree $d$, then $\lambda^n$ is a Salem number of the same degree $d$ for all $n \in \N \setminus \{0\}$.  Applying this to the eigenvalue $\lambda$  of $f$ or  of $f^{-1}$ which is a Salem number, one concludes that $f^n$,  for all $n\in \Z\setminus \{0\}$,  has entropy the logarithm of a Salem number of degree $22$.

As in characteristic $0$, an automorphism  $g$ always stabilizes $NS(X)\subset H^2(X_{\bar K}, \Q_\ell(1))$ and the Picard rank is at most $20$, the logarithm of the absolute value of a root of $P$  can not be
the entropy of $g$.  This shows (3).

If $f^n, \ n\in \Z$ was lifting to an automorphism $g$ on  $X_R\to \Spec  R$, then the specialization $\iota: NS(X_{\bar K})\hookrightarrow NS(X)$  (see Section~\ref{ss:K3}) would be $g$ equivariant, thus, as the Picard rank of $X_{\bar K}$ is at most $20$, the minimal polynomial of $g$ could not have degree $22$. This shows (1) and finishes the proof.
\end{proof}
\begin{remarks} \label{rmk:hodge}
1) As discussed in \cite[Conj.~1.2]{BlEsKe14}, one expects that the rational crystalline cycle class of an algebraic cycle, expressed as a de Rham class on a model in characteristic $0$, is the cycle class of an algebraic cycle on the model, if and only if it is in the right level of the Hodge filtration.  For the cycle class  $c$ of the graph of an automorphism $f$ on a K3 surface, the conjecture is verified (as written in  \cite[Cor.~2.5]{Ogu79}). Indeed,
$$c\in F^2H^4_{DR}(X_R\times_R X_R/R)=F^2H^4(\hat X_R\times_R \hat X_R/R)$$
if and only if $f^*$ acting on $H^2_{DR}(\hat X_R/R)=\varprojlim_n H^2_{DR}(X_n/(R/ \langle \pi^n \rangle))$ respects the Hodge filtration.  Here $\pi$ is the uniformizer of $R$ and $X_n=X\otimes_R R_n, \ R_n= R/\langle \pi^n\rangle$.
Clearly, if $f$ lifts, then $c$ is the cycle class of the graph and lies in $F^2H^4(\hat X_R\times_R \hat X_R/R)$.
Let us now assume that $c\in F^2H^4(\hat X_R\times_R \hat X_R/R)$.
 The obstruction to lifting $f_n$ on $X_n$ to $f_{n+1}$ on $X_{n+1}$  lies in $H^1(X_n, f_n^*T_{X_n/ R_n} \otimes \pi^n|_{X_n})$ and is identified with the action of $f_n^*$ in  ${\rm Hom} (H^1(f^*(\Omega^1_{X_n/R_n} )  ), \pi^n \otimes H^2(\sO_{X_n}))$, thus dies. One constructs  in  this way a prosystem of lifts $\varprojlim_n f_n$, thus, a formal scheme $\varprojlim_n \Gamma_n$, where $\Gamma_n\subset X_n\times_{R_n} X_n$ is the graph of $f_n$, thus, by  \cite[Chap.III,~Thm.~5.4.5]{EGA3},  a projective scheme $\Gamma_R\subset X_R\times_R X_R$ which lifts the graph of $f$ and thus defines the lift.

So the test whether or not  an automorphism lifts to characteristic $0$ is of $p$-adic nature.
On the other hand,  the test we develop  in  Theorem~ \ref{NoLiftPosEntropy} relies on the degree of an  algebraic integer.   It is of course very specific to our situation, nonetheless it is intriguing.

 \medskip

 2) The Salem number we define in Theorem~ \ref{NoLiftPosEntropy} does not come from  the entropy  of an automorphism on a  projective K3 surface in characteristic $0$. One could perhaps speculate that there is a projective model $V_R\to \Spec R$ of a higher dimensional smooth projective variety $V_K$ in characteristic $0$, with an automorphism $f_R$ of $V_R/R$, such that its entropy is reached on the class of a $1$-cycle, the support of which,  by specialization,  lies on the K3 surface considered in Theorem~ \ref{NoLiftPosEntropy}, as a higher codimensional cycle on $V_R\otimes_R k$.  Though we do not have any computation going in this direction, this would just be nice.

\end{remarks}

\section{Lifting of automorphisms of supersingular  K3 surfaces of Artin invariant $1$ in large characteristic}
 \label{sec:EllipticK3}
Throughout this section, $k$ is
 an algebraically closed  field of  characteristic
$p  \ge 3$
and $X=X(p)$
as in Section~\ref{sec:NonLiftable}. So $X(p)$ is a supersingular K3 surface of Artin invariant $1$, and is uniquely defined up to isomorphism with this property.

Using Ogus' crystalline Torelli theorem \cite{Ogu83}, J. Jang \cite[Thm.3.3]{Jan14} proved the following theorem:
\begin{thm} \label{thm:jang}
The image of the representation of ${\rm Aut}(X)$ in the linear automorphism of the one dimensional vector space $k\cdot \omega= H^0(X, \Omega^2_{X/k})$ is a cyclic group of cardinality $p+1$.

\end{thm}
We denote by  $h$ an element in ${\rm Aut}(X)$ such that $h^*\omega=\xi_{p+1} \omega$ where $\xi_{p+1}$ is a primitive $(p+1)$-th root of unity in $k$.

\begin{remark} \label{rmk:jang}
Let $M$ be a projective K3 surface over a characteristic $0$ field $K$. Then the image of ${\rm Aut}(M)$
 in the linear automorphism of the one dimensional vector space $K\cdot \omega= H^0(M, \Omega^2_{M/K})$
is a cyclic group of order $\le 66$ (\cite{Ni79}). In fact Nikulin considered the image of a finite subgroup, but, by the finiteness of the pluri-canonical representation in characteristic $0$ (Ueno-Deligne,  \cite[Thm.~14.10]{Ue75}),  the proof extends to the whole automorphism group.
\end{remark}

Jang deduces from Theorem~\ref{thm:jang}  and Remark~\ref{rmk:jang}  the following:

\begin{corollary} \label{cor:jang}
If $p\ge 67$, then $h$ is not geometrically liftable to characteristic $0$.

\end{corollary}

The aim of this section is to construct an element $\tau \in {\rm Aut}(X)$ {\it of positive entropy} which is not geometrically liftable to characteristic $0$.  (Recall $X=X(p)$).

\begin{definition} \label{def:number}
We define $\beta$ to be the least common multiple of the natural numbers $n$ such that the value of the Euler function $\varphi(n)$ is smaller or equal to  $22$.

\end{definition}
\begin{thm} \label{thm:beta}
If $p+1\ge  67 \beta$, then there is an automorphism  $\tau $ in  $ {\rm Aut}(X(p))$, of positive entropy, which is not geometrically  liftable to characteristic $0$.

\end{thm}
In order to prove the theorem, we first state the following lemmata.

For an endomorphism $\theta$ of  a free  $\Z$-module  of finite type, we denote by ${\rm tr} (\theta)$ its trace with values in $\Z$.

\begin{lemma}\label{trace} Let $X$ be a K3 surface over any field.
Let $f \in {\rm Aut}\, (X)$ such that  $$\vert {\rm tr}\, (f^*  |_{{\rm NS}\, (X)}) \vert \ge 23.$$ Then $f$ is of positive entropy. Conversely, if $f$ is of positive entropy, then there is a positive integer $N$ such that
$\vert {\rm tr}\, ((f^{n})^{*} \vert _{{\rm NS}\, (X)}) \vert \ge 23$ for all
integers $n$ such that $n \ge N$.
\end{lemma}
\begin{proof} Recall that ${\rm NS}\, (X)$ is of rank $\rho \le 22$. Let $\alpha_i$ ($1 \le i \le \rho$) be the eigenvalues of $f^* \vert_{ {\rm NS}\, (X)}$. If $f$ is not of positive entropy, then $\alpha_i$ are cyclotomic integers by
Remark~\ref{entropy}. Hence
$$\vert {\rm tr}\, (f^* \vert _{{\rm NS}\, (X)}) \vert = \vert \sum_{i=1}^{\rho} \alpha_i \vert \le
\sum_{i=1}^{\rho} \vert \alpha_i \vert = \rho < 23\,\, .$$
Hence $f$ is of positive entropy if $\vert {\rm tr}\, (f^* \vert_{{\rm NS}\, (X)}) \vert \ge 23$. Assume that $f$ is of positive entropy. Then, after renumbering,  $\alpha_1$ is a Salem number $a >1$, $\alpha_2$ is $1/a$ and all other $\alpha_k$ are of absolute value $1$ by Remark~\ref{entropy}. Then
$${\rm tr}\, ((f^{N})^{*} \vert _{{\rm NS}\, (X)}) = a^{N} + \frac{1}{a^N} + \sum_{k=3}^{\rho} \alpha_k^{N}\,\, ,$$
which is an integer, in particular, real. Hence
$${\rm tr}\, ((f^{N})^{*} \vert _{{\rm NS}\, (X)}) = a^{N} + \frac{1}{a^N} + \sum_{k=3}^{\rho} {\rm Re}\,(\alpha_k^{N})\,\, .$$
Since $\vert \alpha_k^N \vert = \vert \alpha_k \vert^N = 1$, it follows that
$$-18 = \sum_{k=3}^{\rho} -1 \le \sum_{k=3}^{\rho} {\rm Re}\,(\alpha_k^{N}) \le \sum_{k=3}^{\rho} 1 = 18\,\, .$$
On the other hand, since $a > 1$, it follows that
$$\lim_{N \to \infty} a^{N} + \frac{1}{a^N} = +\infty\,\, .$$
Hence, there is $N$ such that ${\rm tr}\, ((f^{n})^{*} \vert_{ {\rm NS}\, (X)}) \ge 23$, hence $\vert {\rm tr}\, ((f^{n})^{*} \vert_{ {\rm NS}\, (X)}) \vert \ge 23$, for all integers $n$ such that $n \ge N$.
\end{proof}

\begin{lemma}\label{positiveentropy}
Let $X=X(p)$. Then
there is $f \in {\rm Aut}\, (X)$ such that $\vert {\rm tr}\, (f^* \vert_{ {\rm NS}\,(X)}) \vert > 23$,  and $f^*\omega=\omega$. In particular, $f$ is of positive entropy.
\end{lemma}

\begin{proof}
Recall that $X(p)={\rm Km}(E\times_k E)$ where $E$ is a supersingular elliptic curve.
The projections $pr_i: E\times_k E\to E, \ i=1,2$  descend to elliptic fibrations $\varphi_i: X\to \P^1\cong E/\langle \pm 1 \rangle$ (thus with section) and with exactly $4$ singular fibers of type $I_0^*$.
We choose  a zero-section of $\varphi$ and denote it by $O$. We identify it with its image $O \subset X$. Let ${\rm MW}\, (\varphi_i)$ be the Mordell-Weil group of $\varphi_i$. It acts  on the generic fiber $X_\eta$ of $\varphi$ by translation, thus is an abelian  subgroup of ${\rm Aut}(X)$.

One has (\cite{Sh90})

\ga{}{{\rm rank} \ {\rm MW}(\varphi_i)={\rm rank} \ {\rm NS}(X) -2 -4\times 4= 4. \notag
}

For each $i = 1$, $2$, choose $f_i \in {\rm MW}\, (\varphi_i)$ such that $f_i$ is of infinite order. The N\'eron-Severi class $e_i$ of a closed fiber of $\varphi_i$ is stable under $f_i$, i.e.,  $f_i^*(e_i) = e_i$  for $i =1$, $2$, $e_1$,  $e_2$ are of self-intersection $0$,  and they are linearly independent in the hyperbolic lattice ${\rm NS}\, (X)$. Then, by \cite[Thm.~ 3.1]{Og09}, $f_3 := f_1^{n_1} \circ f_2^{n_2} \in
{\rm Aut}\, (X)$ is of positive entropy for large positive integers $n_1$, $n_2$.
Note that $f_i^*\omega=\omega$ as $f_i$ is a translation automorphism of an elliptic fibration. Thus $f_3^*\omega=\omega$ as well.
Hence $f = f_3^N$ for large $N$
is a required solution.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{thm:beta}]
If $h$ is of positive entropy, we are done. So we assume $h$ has null-entropy. This means, the characteristic polynomial of $h^*|_{{\rm NS}(X)}$ is the product of cyclotomic polynomials of degree $\le 22$. Set $g:= h^\beta$. As ${\rm rank} \ {\rm NS}(X)= 22$,  $g^*|_{{\rm NS}(X)}$ is unipotent, so  there is a basis of ${\rm NS}\, (X) \otimes \Q$ in which $g^* |_{{\rm NS}(X)\otimes \Q}  $ is represented by the Jordan canonical form:
$$g^* |_{{\rm NS}(X)\otimes \Q}  = J := J(r_1, 1) \oplus \cdots \oplus J(r_s, 1)\,\, .$$
Set
$s = {\rm max}\, \{r_i\}_{i=1}^{s} -1$ and choose $f$ as in Lemma~\ref{positiveentropy}.

 If $s=0$, then $g^*|_{{\rm NS}(X)}={\rm Id}$. Hence $(f\circ g)^*|_{{\rm NS}(X)} =f^*|_{{\rm NS}(X)}$ and $(f\circ g)^*\omega=\xi_m \omega$ with some $m\ge 67$. Thus $f\circ g \in {\rm Aut}(X)$ is of positive entropy and not geometrically liftable to characteristic $0$.

Assume $s>0$.

In the same basis, $f^*|_{{\rm NS}(X)\otimes \Q}$ is represented by a matrix $A = (A_{ij})$ where
$A_{ij}$ is $r_i \times r_j$ matrix located at the $(i, j)$-block. Then
$${\rm tr}\, ((f \circ g^N)^* \vert _{{\rm NS}\,(X)}) = {\rm tr}\, (J^NA) = \sum_{i=1}^{s}{\rm tr}\, (J(r_i, 1)^NA_{ii}).$$
We want to estimate each summand of the formula above.
For instance, if $r_1 = 4$, then
$$J_4 := J(4, 1) = I_4 + R_4\,\, ,\,\, R_4 := \left(\begin{array}{rrrr}
0 & 1 & 0 & 0\\
0 & 0  & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0
\end{array} \right)\,\, ,$$
and $R^4 = 0$. Thus, by the binomial expansion
$$J_4^N = I_4 + N
\left(\begin{array}{rrrr}0 & 1 & 0 & 0\\
0 & 0  & 1 & 0\\
0 & 0 & 0 & 1\\
0 & 0 & 0 & 0\end{array} \right) + \frac{N(N-1)}{2}
\left(\begin{array}{rrrr}
0 & 0 & 1 & 0\\
0 & 0  & 0 & 1\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\end{array} \right) + \frac{N(N-1)(N-2)}{6}
\left(\begin{array}{rrrr}
0 & 0 & 0 & 1\\
0 & 0  & 0 & 0\\
0 & 0 & 0 & 0\\
0 & 0 & 0 & 0\end{array} \right)\,\, .$$
Hence, for a $4 \times 4$-matrix $A_4 := (a_{ij})$ of rational entries,
we have
$${\rm tr}\, (J_4^NA_4) = {\rm tr}\, (A_4) + N(a_{21} + a_{32} + a_{43}) + \frac{N(N-1)}{2}(a_{31} + a_{42}) + \frac{N(N-1)(N-2)}{6}a_{41}\,\, ,$$
which is a polynomial of $N$ of degree $\le 3 = 4-1$, with
rational coefficients depending on $A_4$ (and independent of $N$).
For exactly the same reason,  from the expansion of $J(r, 1)^N$ as above,
we find that
 ${\rm tr}\, ((f \circ g^N)^* \vert_{ {\rm NS}\,(X)})$
is of the form
$${\rm tr}\, ((f \circ g^N)^* \vert_{ {\rm NS}\,(X)}) = a_sN^s + a_{s-1}N^{s-1} + \cdots
+ a_1 N + {\rm tr}\, A\,\, .$$
Here  $a_k$ are rational numbers depending only on $A$ (and independent of $N$). So, if it  happened that $\vert {\rm tr}\, ((f \circ g^{k_n})^* \vert_{ {\rm NS}\,(X)}) \vert \le 22$ for some sequence of positive integers
$$k_1 < k_2 < k_3 < \cdots < k_n < \cdots\,\, \to +\infty\,\, ,$$
 one would have
$$a_s = a_{s-1} = \cdots = a_{1} = 0\,\, ,$$
and hence
$$\vert {\rm tr}\, ((f \circ g^N)^* \vert _{{\rm NS}\,(X)}) \vert = \vert {\rm tr}\, (A) \vert$$
for all positive integer $N$.  But
$$\vert {\rm tr}\, (A) \vert = \vert {\rm tr}\, (f^* \vert_{ {\rm NS}\,(X)})) \vert \ge 23\,\, ,$$
by our choice of $f$, a contradiction to $\vert {\rm tr}\, ((f \circ g^{k_n})^* \vert_{ {\rm NS}\,(X)})\vert \le 22$.
Hence, there are only finitely many positive integers $\ell$ such that $\vert {\rm tr}\, ((f \circ g^{\ell})^* \vert_{ {\rm NS}\,(X))}) \vert \le 22$. Thus, there is a positive integer $M$ such that $\vert {\rm tr}\, ((f \circ g^N)^* \vert _{{\rm NS}\,(X)})) \vert \ge 23$ for all integers $N$ such that $N \ge M$.

Let us choose $N$ such that $N\ge M$ and $( N, m )=1$. Define $\tau:= f\circ g^N$. Then $|{\rm tr} (\tau^*|_{{\rm NS}(X)})| \ge 23$, and $\tau^*\omega=\xi_m^N \omega$.  Note $\xi_m^N$ is a primitive $m$-th root of unity with $m\ge 67$.  Thus $\tau\in {\rm Aut}(X)$ is of positive entropy and not geometrically liftable to characteristic $0$.

\end{proof}

\begin{remark}\label{independentoflift}
In \cite{EOY14} we show that various elliptic structures on $X(p)$, for
 $p \ge 11$ and $p \not= 13$, with a rich Mordell-Weil group force the existence of autmorphisms with entropy the logarithm of a degree $22$ Salem number.
\end{remark}

config: arxiv


TITLE: $y^5 =(x+2)^4+(e^x)(ln y)−15$ finding $\frac{dy}{dx}$ at $(0,1)$
QUESTION [2 upvotes]: Unsure what to do regarding the $y^5$.
Should I convert it to a $y$= function and take the $5$ root of the other side. Then differentiate? Any help would be great thanks.

REPLY [1 votes]: Unfortunately we cannot rearrange this to find $y$. Fortunately we can still use implicit differentiation to find a derivative. Take the derivative of all terms w.r.t. $x$, using the chain rule and treating $y$ as a function of $x$ - we obtain
$$
5y^4\frac{\mathrm{d}y}{\mathrm{d}x} = 4(x+2)^3 + \mathrm{e}^x\ln y +\frac{\mathrm{e}^x}{y}\frac{\mathrm{d}y}{\mathrm{d}x}.
$$
Note that we used the product rule for the $\mathrm{e}^x\ln y$ term. Now the trick is to rearrange to solve for the derivative, and then substitute $(x,y) = (0,1)$. Can you take it from here?


TITLE: Why do archery arrows tilt downwards in their descent?
QUESTION [67 upvotes]: In the movies, arrows shot into the air rotate so that during the descent, the arrow head hits ground first. What is the source of this angular momentum?  It would seem that the bow string exerts a force directly in line with the arrow.

REPLY [0 votes]: The arrow ideally will fly out like a drag race car with the parachute deployed!
 A well designed arrow should have these properties.
1- Sharp and proportionally heavy point to accept a large momentum and deliver it as kinetic energy E= mv^2/2
2- long and balanced stem to accommodate a big arch and maintain separation between the tip and the fletching.
3- an aerodynamically well designed fletching to keep the arrow straight and also provide the small amount of drag needed to keep the end of arrow always steady behind the tip and prevent the stem from getting disoriented.
The reason the arrow fly aligned in its arc is the small drag created by fletching always position it aligned with its track, same as badminton ball.
 In aerodynamic design this scheme of installing the rudder and elevators at the end of fuselage for stability is common practice! All airplanes have it as empennage or tail end!


TITLE: Maximum principle - prove that function is increasing/decreasing
QUESTION [2 upvotes]: $P$ is an $n$-degree polynomial . For $r>0$ let's define:
$$M(r,P)=\sup_{|z|=r} |P(z)| $$
Prove that function $r\mapsto M(r,P)$ is increasing and function $r\mapsto\frac{M(r,P)}{r^n}$ is decreasing.
I think I should use the principle of maximum. Have you got any hints?

REPLY [1 votes]: I will answer the first part of the question, and give a hint for the second.
By the maximum modulus principle, $\left|P(z)\right|$ achieves no maximum in the domain $|z|<r$ (I'm assuming $n>0$, so $P$ isn't constant). However, as a continuous function, it does a achieve a maximum in the compact domain $|z|\le r$. This implies that $\max_{|z|\leq r}\left|P(z)\right| = \max_{|z|=r}\left|P(z)\right| = M(r,P)$, and it isn't achieved anywhere in $|z|<r$.
In particular, for all $|z_0| < r$ we have
$$\left|P(z_0)\right| \leq \max_{|z|\leq |z_0|}\left|P(z)\right| < \max_{|z|=r}\left|P(z)\right| = M(r,P)$$
Put differently, for $r<R$ and all $|z|=r$ we have $\left|P(z)\right|<M(R,P)$. A direct consequence is that $M(r,P)<M(R,P)$.
For the second part, as mentioned in comments, we can develop
$$\frac{M(r,P)}{r^n} = \sup_{|z|=r}\left|\frac{1}{z^n}P(z)\right| = \sup_{\left|\frac{1}{z}\right|=\frac{1}{r}}\left|Q\left(\frac{1}{z}\right)\right|$$
for a different polynomial $Q(z)$ (which?). We can now use part (a) to conclude (how?).


TITLE: The undecidability of validity and satisfiability of first-order logic
QUESTION [0 upvotes]: I'm new to mathematical logic and I'm recently confused by some concepts about first-order logic validity and satisfiability, here are some of my problems:
I learned that validity and also unsatisfiability is semi-decidability, does that mean one can use method like resolution reasoning and eventually get a 'yes' once the statement is actually valid not matter how long it will take, but may get nothing (the mothod does not stop) if the statement is not valid?
I learned that satisfiability is undecidable and even not semi-decidable, because once it is it will lead to that the whole problem is decidable, I can understand it from this perspective but does that mean satisfiability is harder than validity? If it is so then it confuses me a lot, cause I think to prove satisfiability one just needs to find a model and an explanation but to prove validity one needs to check all models and explanations, which seems to me that satisfiability is easier than validity.
Is there something wrong about my understanding? Thanks!

REPLY [0 votes]: Validity and satisfiability are either equally complicated or incomparably complicated, depending on how we compare complexities (i.e. many-one reductions or Turing reductions); basically, they're "dual" to each other, since $\varphi$ is valid iff $\neg\varphi$ is not satisfiable. Granted, in a sense semidecidability (or recursive enumerability, or computable enumerability) is more natural than "co-semidecidability" (or co-r.e.-ness, or co-c.e.-ness), but this doesn't really translate to a genuine "simpler than" comparison.
Now what about model checking? After all, a sentence is satisfiable iff it has a model, so shouldn't we be able to positively test for satisfiability by searching through all possible models? Well, this works for finite models. However, most models aren't finite: there are lots of sentences which are satisfiable but have no finite models. How are we supposed to search through all possible infinite models?
The fact that we're allowing infinite models as well as finite models makes both satisfiability and validity seem impossibly hard. This is where the highly nontrivial completeness theorem comes in, and I've written more about this here.
Incidentally, if we do restrict attention to finite models only, the situation reverses: satisfaction is semidecidable but not co-semidecidable, and validity is co-semidecidable but  not semidecidable (see Trakhtenbrot's theorem). Finite model theory is a very different thing from infinite model theory!


TITLE: If the norm of an ideal is prime then the ideal is prime
QUESTION [6 upvotes]: Let $K$ be a number field and $D = \mathcal{O}_K$ its ring of integers. Let $I$ be a non-zero integral ideal of $D$ and suppose that the norm $N(I)$ is a (rational) prime. I want to prove that $I$ is a prime ideal in $D$.
Since $D$ is a Dedekind domain, $I$ has a unique factorisation into prime ideals, so write $I = \mathfrak{p}_1\cdots\mathfrak{p}_n$. Then, since the norm is completely multiplicative, we have
$$N(I) = N(\mathfrak{p}_1)\cdots N(\mathfrak{p}_n) = p$$
from some prime $p \in \Bbb Z$. In particular, this means that $p \mid N(\mathfrak{p}_i)$ for some $i \in \lbrace 1, \dots, n\rbrace$, and the product of the remaining norms is equal to $1$.
Does this mean that $N(I) = N(\mathfrak{p}_i)$ for some $i$ and hence that $I = \mathfrak{p}_i$?

REPLY [6 votes]: Note that the norm of any proper ideal is never $1$.
Thus $N(I) = N(\mathfrak{p}_1)\cdots N(\mathfrak{p}_n) = p$ implies in particular that $n=1$, so $I=\mathfrak{p_1}$.
You write "$N(I) = N(\mathfrak{p}_i)$, [...] hence $I=\mathfrak{p}_i$". The fact alone that two ideals have the same norm does not necessarily imply that they are equal. If one ideal divides the other ideal (as in this case), it does.
Alternatively, $N(I) = |D/I|$, so $D/I$ is a finite ring of order $p$. The only such ring is $\Bbb Z/(p)$, thus $D/I$ is an integral domain.


TITLE: Why are the two calculations of $ \int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x $ give two distinct answers?
QUESTION [4 upvotes]: One of the calculations:
$$
\begin{aligned}
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x &=\int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \cdot \frac{1}{\sqrt{x}} \mathrm{~d} x \\
&=2 \int \frac{1}{\sqrt{1-(\sqrt{x})^{2}}} \mathrm{~d}(\sqrt{x})=2 \arcsin \sqrt{x}+C .
\end{aligned}
$$
The other:
$$
\int \frac{1}{\sqrt{x(1-x)}} \mathrm{d} x = \int \frac{d\left(x-\frac{1}{2}\right)}{\sqrt{\left(\frac{1}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}}} = \arcsin \frac{x-1 / 2}{1 / 2}+C=\arcsin(2x-1)+C
$$

REPLY [6 votes]: The answers are actually the same.
$\begin{align*}
\theta = 2\arcsin \sqrt x &\implies \sin \frac \theta 2 = \sqrt x \\
&\implies \sin^2 \frac \theta 2 = x \\
&\implies \frac{1-\cos \theta}{2} = x \\
&\implies \cos \theta = 1-2x \\
&\implies \sin(\theta+\pi/2) = 2x-1 \\
&\implies \theta = \arcsin(2x-1) - \pi/2
\end{align*}$
And $\pi/2$ gets absorbed into $+C$

REPLY [0 votes]: They are both correct. Let $y=2 \arcsin \sqrt x$. Then $x=(\sin (\frac  y 2))^{2}$. So $2x-1=2(\sin (\frac  y 2))^{2}-1=\sin (y+\frac {\pi} 2)$. So $y=\sin (2x-1)-\frac {\pi} 2$.


TITLE: "Distribution functions should have disappeared long ago"
QUESTION [13 upvotes]: In "Mathematical Foundations of the Calculus of Probability" by Jacques Neveu, the author says the following about distribution functions:

These functions, which are in fact of very little
  practical use (except in certain questions where the order structure of the
  real line plays a predominant role), should have disappeared a long time
  ago to the benefit of the ensemble definition of the notion of probability.

This sentiment (that they should have disappeared long ago) is also expressed by Erhan Ҫinlar in "Probability and Stochastics".
I have two closely related questions:

What is the justification for this view? I don't really see how you would come to this conclusion. How are distribution functions impractical?
Is this a commonly held sentiment among experienced mathematicians?

For the record: The distribution function of a RV $X$ is the function $x\mapsto \mu(-\infty\,..x]$, where $\mu$ is the distribution of $X$.

REPLY [1 votes]: I (think) the argument being made is how distribution functions only give you information about the probabilities of sets of the form $$X^{-1}( (-\infty, a])\,,$$ whereas the real source of interest is the values of the probability measure for sets of the form $$X^{-1}(A)$$ for any (Borel)-measurable set $A$.
I base this inference on the parenthetical comment from your quotation:

except in certain questions where the order structure of the real line plays a predominant role

In those certain cases I imagine that sets of the form $(-\infty, a]$ (and their inverse images under a random variable $X$) are more important than arbitrary Borel-measurable sets $A$.
This is, however, just a guess -- although I read Neveu's book a year or so ago, I don't quite remember what else (if anything) he mentioned about the topic. I have not yet read Cinlar's book, although I hope to do so in the future.


text:
\section{Conclusion}
\label{sec:conclusion}

In this paper, we have described a \revision{red}{regularized} homogeneous interior-point algorithm and its implementation in Tulip, an open-source linear optimization solver.
\revision{blue}{Our solver is written in Julia, and leverages some of the language's features to propose a flexible and easily-customized implementation.}
Most notably, \revision{blue}{Tulip's} algorithmic framework is fully disentangled from linear algebra implementations \revision{blue}{and the choice of arithmetic}.

The performance of the code has been evaluated on \revision{blue}{generic instances from H. Mittelmann's benchmark} testset, on \revision{blue}{two sets} of structured instances for which we developed specialized linear algebra routines, \revision{blue}{and on numerically problematic instances using higher-precision arithmetic.
The computational evaluation has shown three main results.
First, when solving generic LP instances, Tulip is competitive with open-source IPM solvers that have a Julia interface.
Second, when solving structured problems, the use of custom linear algebra routines yields a tenfold speedup over generic ones, thereby outperforming state-of-the-art commercial IPM solvers on larger and denser instances.
}
These results demonstrate the benefits of being able to seamlessly integrate specialized linear algebra within an interior-point algorithm.
\revision{blue}{Third, in a development context, Tulip can be conveniently used in conjunction with higher-precision arithmetic, so as to alleviate numerical issues.}

\revision{blue}{
Finally, future developments will consider the use of iterative methods for solving linear systems, the development of more general structured linear algebra routines and their multi-threaded implementation, and more efficient algorithmic techniques for solving problems in extended precision.
Because of the way in which Tulip has been designed, all those developments do not require any significant rework of the code structure.}


TITLE: How to compute the value of the discriminant of $\mathcal{O}_K$
QUESTION [2 upvotes]: I found a $\mathbb{Z}$-basis $\left\{1, \sqrt{2}, \sqrt{-1}, (\sqrt{2}+\sqrt{-2})/2 \right\}$ for ring of integers $\mathcal{O}_K$ of some number filed $K$ and I need to compute the discriminant of $\mathcal{O}_K$. I know that the formula is
$$\det(Tr(a_ia_j), 1 \leq i, j \leq n)$$
where $a_i$ are just numbers in the basis $\left\{1, \sqrt{2}, \sqrt{-1}, (\sqrt{2}+\sqrt{-2})/2 \right\}$. I found that computing each $Tr(a_ia_j)$ is quite tedious to do it by hand and I feel like there should be an easier way to do it?

REPLY [4 votes]: This is not tedious as you think, since $Tr(a+bi+c\sqrt{2}+di\sqrt{2})=4a$.
The first 3 elements of your basis are then pairwise orthogonal. Moreover, $1$ and $i$ are also orthogonal to $(\sqrt{2}+i\sqrt{2})/2$, so really you just have 5 easy computations to do...
It is then easy to compute the matrix you are interested in. We get $$\pmatrix{4 &0 &0 & 0\cr0 &8 & 0 & 4\cr0 & 0& -4& 0\cr 0 & 4 & 0&0  },$$
which has determinant $4\times (-4)\times (-16)=2^8$.
Side remark. Taking the field of fractions, I see that your number field is $K=\mathbb{Q}(i,\sqrt{2})=\mathbb{Q}(\zeta_8)$, and a integral basis and the discriminant of cyclotomic fields are well-known( so you could just use the basis $\zeta_8^k, k=0,...,3$ instead.) The result I found is consistent with this fact.


TITLE: Effects of a non-Lorentz-invariant vacuum state
QUESTION [3 upvotes]: I'm here asking about real or though experiments (i.e., physical effects) where, at least in principle, one can see some consequence of a non-Lorentz-invariant vacuum state in an otherwise Poincare invariant theory.
Let me develop the question. Assume a theory in which the Hamiltonian (that closes Poincare algebra with the rest of generators) $H$ acts non-trivially on the vacuum state $|0>$, where 'non-trivially' simply means:
$$H|0> = E \, |0>$$
with $E$ a positive constant. Since the Hamiltonian transforms as the temporal component of a 4-vector, the vacuum state is not Lorentz invariant. Therefore, the theory is not Lorentz invariant (it is usually claimed that this is an additional condition besides the Poincare algebra). However, I'm not able to see any consequence of this fact. I think that this does not affect any cross-section or decay rate.
I think that one can redefine the Hamiltonian so that $H'=H-E$ and $H'|0>=0 \,$ (this does not affect the Poincare algebra if one also redefines the boost generators properly). I know this seems obvious (this redefinition is usually done in canonical quantization when one doesn't adopt normal ordering), but I've just read this in this forum:
https://physics.stackexchange.com/a/8360/10522

However, in special relativity, energy is the time component of a
  4-vector and it matters a great deal whether it is zero or nonzero. In
  particular, the energy of the empty Minkowski space has to be exactly
  zero because if it were nonzero, the state wouldn't be
  Lorentz-invariant: Lorentz transformations would transform the nonzero
  energy (time component of a vector) to a nonzero momentum (spatial
  components).

One has a family of vacuum states related by Poincare transformations which are unitary, I don't think this is a problem... what do you think?
Added:
1) I'm not thinking about a Poincare invariant Lagrangian with a potential of the form $(A^2(x)-v)^2$, where $A_{\mu}(x)$ is a vectorial field that acquires a vacuum expectation value $v$. Assume that every field has zero vev.
2) I'm looking for Lorentz violating effects instead of vacuum energy effects unless you argue that these vacuum energy effects (Lamb shift, spontaneous emission, etc.) break Lorentz symmetry.

REPLY [2 votes]: In axiomatic quantum field theory, it is assumed that there is a unique Poincare invariant (projective) state. An arbitrary normalized representative $|0\rangle$ of this state is called the vacuum state. Poincare invariance implies that for all $x$, the state $e^{x\cdot P/\hbar}|0\rangle$ is a multiple of $|0\rangle$. This implies that $P|0\rangle=p|0\rangle$ for some $p$. Redefining $P$ as $P-p$ gives another representation of the Poincare group in which the vacuum state has zero 4-momentum, as usually assumed.
On the other hand, Rn Maimon considers a situation not covered by axiomatic QFT, where a state $|0\rangle$ he calles (in my view mistakenly) the vacuum state is not in the domain of the translation group. Thus $e^{x\cdot P/\hbar}|0\rangle$ is (fro nonzero $x$) not a state vector in the physical Hilbert space (but a proper distribution), the vacuum expectation value of the momentum vector does not exist, and the usual covariance arguments break down. As I understand him, he takes this to be the situation in an infinite universe with finite local momentum density but infinite total energy. However, in such a situation, Poincare invariance is no longer relevant. Indeed, in quantum gravity, there doesn't seem to be an observer-independent notion of a vacuum state, due to the Unruh effect. Instead opne has a family of Hadamard states that, taken together, replace the vacuum state of flat QFT. This family as a whole is indeed Poincare invariant, even invariant under the group of volume-preserving diffeomorphisms.


TITLE: A question about symmetric bilinear forms
QUESTION [1 upvotes]: If b is an indefinite symmetric bilinear form is it nondegenerate?
And conversely if b is nondegenerate is it positive/negative definite or indefinite?
How can i start to prove this?
Note:Edited and changed the question

REPLY [3 votes]: $b$ is indefinite iff its matrix representation $B$ has both positive and negative eigenvalues; $b$ is nondegenerate iff $B$ has no zero eigenvalues. So,

an indefinite $b$ can be degenerate (e.g. $B=\operatorname{diag}(1,0,-1)$);
a nondegenerate $b$ can be positive definite, negative definite or indefinite (examples: consider $B=1,\ B=-1$ and $B=\operatorname{diag}(1,-1)$.)


TITLE: Transforming a sum to an integral: why does it work?
QUESTION [3 upvotes]: The problem at hand has been discussed in loads of previous questions (1, 2, 3)), and my version can be stated as follows.
Consider the sum $$\sum_{\mathbf k} \ln(1+e^{-(\alpha+\beta \varepsilon_k)}) \ .$$ We are summing over, say, two dimensional k-space lattice $\mathbf k = \frac{2\pi}{L}(n_x,n_y),$ where the $n_x,n_y$ run through the positive integers. I want to understand how we can write this as an integral. Our energy is given by $\varepsilon_k = \frac{\hbar^2 k^2}{2m}$.

REPLY [2 votes]: Why does it work? By assuming the separation distance between the points in the discrete space is negligibly small compared to the total volume, we can make use of the definition of the Riemann integral.

Say we are summing over discrete points $\mathbf k$ in k-space $$\sum_\mathbf{k}\ f(\mathbf k)\ .$$
If we want to rewrite this as an integral we have to assume that the
separation between the points in k-space $\Delta k$ is negligible in
comparison to the volume of the k-space $V$.
In the following we can then use the definition of the Riemann
integral
\begin{align}
\sum_{\mathbf k} f(\mathbf k) &= \frac{1}{\Delta k} \sum_{\mathbf k}
f(\mathbf k) \Delta k \\
&\equiv \frac{1}{\Delta k}  \int_{\mbox{all space}} f(\mathbf k) d\mathbf k
\ ,
\end{align}
where, in the last step we used our assumption that our seperation
distance is negligibly small compared to the volume of the whole
space.


TITLE: Combination problem: retrieve four kinds of objects with restriction
QUESTION [1 upvotes]: How many ways are there to select (combinations) $n$ objects from $a$ identical  objects of first kind, $b$ identical  objects of second kind, $c$ identical  objects of third kind and  $d$ identical objects of fourth kind?
P.S. : Since the maximum number of any kind is restricted it is different from the original stars and bars problem where there is no restriction.

REPLY [1 votes]: I figured out the answer. The question same as,
The total no of non negative  integral solutions of x+y+z+w=n , where,
$0\le x\le a,0\le y\le b\ ,\ 0 \le z\le c\ $ and $\ 0\le w\le d$
changing variables as,
x1=a-x,y1=a-y,z1=a-z and w1=a-w,
we get the answer to be,
number of solutions of the equation x1+y1+z1+w1=a+b+c+d-n
answer= $$\binom{a+b+c+d-n+4-1}{4-1}$$


TITLE: Yoneda extensions in derived categories
QUESTION [9 upvotes]: If given an abelian category $\mathcal{A}$, we can consider the bounded derived category $D^b(\mathcal{A})$. For two objects $A,B \in \mathcal{A}$, we know that there is a natural identification between
$$\text{Ext}_{\mathcal{A}}^i(A,B)$$
and
$$\text{Hom}_{D^b(\mathcal{A})}(A,B[i])$$
using Yoneda extensions. This is proven in Verdier's thesis (see also group of Yoneda extensions and the EXT groups defined via derived category).
I wondered whether one in general can identify the morphisms in the derived category with Yoneda extensions in the following way:
Given two complexes $E^\bullet, F^\bullet \in D^b(\mathcal{A})$, can we identify the set of morphisms $$\text{Hom}_{D^b(\mathcal{A})}(E^\bullet, F^\bullet[i])$$
with sequences of morphisms in the derived category $D^b(\mathcal{A})$
$$F^\bullet \to Z_{i-1}^\bullet \to \dots \to Z_0^\bullet \to E^\bullet $$
such that the above sequence breaks into distinguished triangles, i.e. there exist distinguished triangles
$$F^\bullet \to Z_{i-1}^\bullet \to G_{i-1}^\bullet, \quad G_{i-1}^\bullet \to Z_{i-2}^\bullet \to G_{i-2}^\bullet, \dots$$
In the case that $E^\bullet,F^\bullet$ are complexes concentrated in degree 0 (i.e. objects in $\mathcal{A}$), this is precisely the Yoneda extension with the $Z_j^\bullet, G_j^\bullet$ also complexes concentrated in degree 0.
For $i=1$ and $E^\bullet, F^\bullet$ arbitrary, this also holds true since a morphism $E^\bullet \to F^\bullet[1]$ can be completed to a distinguished triangle $$G^\bullet \to E^\bullet \to F^\bullet[1]$$ which we can rotate to obtain $$F^\bullet \to G^\bullet \to E^\bullet.$$ I am curious about the general case.

REPLY [4 votes]: There is such a sequence, but it's not very interesting.
Given an element of $\text{Hom}_{D^b(\mathcal{A})}(E,F[i])$, then in the same way you describe, this gives a distinguished triangle
$$F\to Z_{i-1}\to E[-i+1].$$
Then you can just take the distinguished triangles $E[-i+1]\to0\to E[-i+2]$,  $E[-i+2]\to0\to E[-i+3]$, $\dots$, up to $E[-1]\to0\to E$, which can be spliced to give a sequence
$$F\to Z_{i-1}\to Z_{i-2}\to\dots\to Z_0\to E,$$
where $Z_{i-2}=\dots= Z_0=0$.
This might seem like cheating, but I don't think it is. In the case of  (a nonzero element of) $\text{Ext}_{\mathcal{A}}^i(A,B)$, the only reason you can't have zero objects in your sequence
$$B \to Z_{i-1} \to Z_{i-2}\to\dots \to Z_0 \to A$$
is that you insist that the $Z_j$ are not just objects of $D^b(\mathcal{A})$, but of $\mathcal{A}$. If you drop this requirement, then you just have too much freedom.


TITLE: GNS construction: Finding a representation
QUESTION [1 upvotes]: I am applying the GNS construction to the algebra $M_{N \times N}$ of all $N\times N$ matrices on $\mathbb{C}$. I have the state $\omega$ given by
$$\omega(A)  = Tr(\rho A),$$
for $A\in M_{N\times N}$, with $\rho = \rho^* \in M_{N\times N}$ and $\rho = diag(\lambda_1, \ldots, \lambda_N)$, $\sum_i \lambda_i =1, \ \lambda_i \geq 0$.
I have deduced (please correct me if I'm wrong), but the kernel of the above state: $\ker(\omega):= \{A\in M_{N \times N} : \forall B \in M_{N \times N} \implies \omega(B^*A) = \omega(A^*B)=0 \}$ is the set:
$$\ker(\omega) = \{A\in M_{N \times N} : A^2 = 0\}.$$
By the GNS construction, we then look to the quotient space given by $M_{N \times N}/ \ker (\omega)$. Let $\xi_A$ denote the equivalence class of $A$, then we need to construct a representation $\pi_{\omega}$ such that we have:
$$\pi_{\omega}(A) \xi_B = \xi_{AB}.$$
I'm a little unsure of how to do this, but I have two ideas:

$\pi_{\omega}(A)B = AB,$
$\pi_{\omega}(A)B = \sqrt{A}B.$

I'm not sure about the accuracy of either of these answers, but I need a little guidance on this technique.

REPLY [2 votes]: You are looking for the subspace
$$
J=\{A:\ \omega(A^*A)=0\},
$$
which is not the kernel of $\omega$. If $\lambda_j>0$ for all $j$, this will be just $\{0\}$. You have
$$
0=\omega(A^*A)=\sum_j \lambda_j (A^*A)_{jj}=\sum_{\lambda_j>0}\lambda_j (A^*A)_{jj}.
$$
So, for each $\lambda_j>0$, the $j,j$ entry of $A^*A$ has to be zero. This is equivalent to $A$ having its $j^{\rm th}$ row and column equal to zero. So, if $F=\{j:\ \lambda_j>0\}$ and $m=|F|$, then
$$
M_N(\mathbb C)/J=M_m(\mathbb C).
$$
The representation is indeed
$$
\pi_\omega(A)B=AB,
$$
but this needs to be understood in the quotient. That is, as you wrote yourself, if $\xi_B$ denotes the class of $B$ in $M_N(\mathbb C)/J$, then
$$
\pi_\omega(A)\xi_B=\xi_{AB}.
$$


TITLE: Generating Function: Why is $G(0)=P(X=0)$?
QUESTION [1 upvotes]: $G$ is the generating function:
$$G(s)=E\left[s^X\right]=\sum_{i=0}^{\infty}P(X=i)s^i$$
But the textbook claims that $G(0)=P(X=0)$. Why?
$$G(0)=\sum_{i=0}^{\infty}P(X=i)(0)^i=0$$ but this is not $P(X=0)$, this is wrong?

REPLY [6 votes]: The definition for $G(s)$ and result of $G(0)$ are both correct.
Since $ 0^0 = 1$, and $0^i = 0$ for $i \in \mathbb(R) \neq 0$, we have
$$ G(0) = \sum_{i=0}^{\infty} P(X=i)(0)^i = P(X=i) \cdot 1 = P(X=i)$$


TITLE: Quadratic fit using sum of least squares without matrices
QUESTION [1 upvotes]: I've been hunting around for examples using the sum of least squares + partial derivative method to fit a polynomial to a set of points but am completely stuck.  All the examples I've found involve matrices, but I am unfamiliar with how to use matrices.
Full disclosure, this is a homework problem for a multivariable calculus course.  Here's what I've tried:
Points
$(-2,4)$, $(-1,2)$, $(0,1)$, $(1,1)$, and $(2,2)$ so n=5
My method:
$$f(a,b,c) = \sum{(y_i-(ax_i^2+bx_i+c))^2}$$
$$f(a,b,c) = \sum{y_i^2}-2a\sum{y_ix_i^2}=2b\sum{y_ix_i}-2c\sum{y_i}+a^2\sum{x_i^4}+2ab\sum{x_i^3}+2ac\sum{2_i^2}+2bc\sum{x_i}+b^2\sum{x_i^2}+5c^2$$
Find the partial derivatives for a, b, c and set to 0
$$f_a = -2\sum{y_ix_i^2}+2a\sum{x_i^4}+2b\sum{x_i^3}+2c\sum{x_i^2}$$
$$f_b = -2\sum{y_ix_i}+2a\sum{x_i^3}+2c\sum{x_i}+2b\sum{x_i^2}$$
$$f_c = -2\sum{y_i}+2a\sum{x_i^2}+2b\sum{x_i}+10c$$
Calculate all the sums:
$$\sum{x_i^4} = 34$$
$$\sum{x_i^3} = 0$$
$$\sum{x_i^2} = 10$$
$$\sum{x_i} = 0$$
$$\sum{y_i} = 10$$
$$\sum{y_ix_i} = -5$$
$$\sum{y_ix_i^2} = 27$$
But when I plug the sums in and set the partial deriatives to 0, I get the wrong answer:
$$68a+20c = 54$$
$$20b = -10$$
$$20a = 20$$
Which gives:
$$y = x^2-\frac{1}{2}x - \frac{7}{10}$$
I checked over the equations and I think they look right...but clearly they're not :/
Thank you!

REPLY [0 votes]: eep this was a simple algebra error!
Going to leave this up just in case anyone in the future is hunting around for an example of a polynomial fit using least squares without matrices.


TITLE: Coordinate transform
QUESTION [1 upvotes]: Can anyone see what transformation $$r\to f(r)$$ transforms $$\exp(2\phi(r))(dr^2+r^2d\theta^2)$$ to
$$df^2+\sinh^2(f)d\theta^2$$?
I there a systematic way to attack such a problem -- rather than just hoping that  I somehow spot it?
Thank you!

REPLY [1 votes]: Well, $\phi$ can't be very creative. Since the curvature of this surface must be invariant under the change of coordinates, with a bit of computation we'll get (from the latter) $K=-1$ and (from the former) $K=-e^{-2\phi(r)}\big(\phi''(r)+\frac1r\phi'(r)\big)$. Thus, $\phi$ must satisfy the ODE
$$\phi''(r)+\frac1r\phi'(r)=e^{2\phi(r)}\,.\tag{$\star$}$$
(This should look somewhat familiar if you've computed curvature of a conformally flat metric before.)
It also makes us think we're looking at some hyperbolic metric, of course.
Starting at the other end of the problem, you need
\begin{align*} f'(r) &= e^{\phi(r)} \\
\sinh f(r) &= re^{\phi(r)}\,,
\end{align*}
and we can actually integrate patiently to obtain
$$\cosh f(r) = \frac{Cr^2+1}{Cr^2-1}\,,$$
so $$f(r) = \log\left(\frac{(Cr+1)^2}{Cr^2-1}\right)\,.$$
Amazingly, if we now set $\phi(r) = \log f'(r)$, we do in fact (!) get a solution of the ODE ($\star$) above.


TITLE: Probability: random elevator stops
QUESTION [1 upvotes]: I have a elevator probability problem but with a twist (instead of people exiting the elevator, it is the elevator stopping on each level). Require some help understanding and completing the probability questions.
Problem:
4 people go into the elevator of a 5-storey shophouse. Assume that each of them exits the building uniformly at random at any of the 5 levels and independently of each other. N is the random variable which is the total number of elevator stops.

Describe the sample space for this random process.

Answer: {1,2,3,4,5}
Understanding: Sample space represents all the possible outcomes of the random event. Hence, the elevator can only stop at these 5 floors. Just like throwing a die and randomly getting a number, the sample space for that is {1,2,3,4,5,6}.

Let $X_i$ be the random variable that equals to 1 if the elevator stops at floor i and 0, otherwise. Find the probability that the elevator stops at both floors $a$ and $b$ for $a$, $b$ $\in$ {1,2,3,4,5}. Find $EX_aX_b$.

Question: Do they mean the elevator stop at two consecutive floors or two different floors? Can a and b be the same, a smaller than b and vice versa?

Prove the independence of $EX_1X_2$.
Calculate $EN^2$ (represented as ($X_1 + ...+ X_5)^2 = \sum_{(i,j)}  X_aX_b$ where the sum is over all ordered pairs (a,b) of numbers from {1,2,3,4,5} and the linearity of expectation. Find the variance of N.
Determine the distribution of N. That is, determine the probabilities of events N = i. Compute $EN and EN^2$ directly by using the laws of expectation.

I got stuck from question 2 onwards with thinking of the possible ways that this can be done. It would be superb if someone can shed some light on them. Many thanks.

REPLY [1 votes]: You did not specify the events $X_a$ and $X_b$, but from the context I assume it is meant that
$$ X_a = 1\{\text{elevator stops at floor } a\}. $$
You have to find $\mathbb{E}X_aX_b$, which is equal to the probability that the elevator stops at both floor $a$ and $b$. In the question it is only stated that $a,b \in \{1,\dots,5\}$ but it is not specified that they can't be equal, meaning we should also answer the question in the event $a = b$. Moreover, it does not have to be the case that $a$ and $b$ are two consecutive floors, since in that case they would have stated that $a \in \{1,\dots,5\}$ and $b = a+1$. So you should just read this question as: given two floors, what is the probability that the elevator stops at these two floors (or, in case $a = b$: given this floor, what is the probability that the elevator stops at this floor).
As mentioned above, we have two different scenario's: $a$ and $b$ different floors or $a = b$ and we only consider one floor. I will now give a hint on answering the question for both scenario's:
$a = b$:
in the case $a = b$ we have to find $\mathbb{P}(\text{ elevator stops at floor a})$. Since the only way the elevator stops at this floor is if there are people exiting the elevator at this floor, we have that:
\begin{align*}
\mathbb{P}(\text{elevator stops at floor a}) &=
\mathbb{P}(\geq 1 \text{ people exit the elevator at floor a}) \\
&= 1 - \mathbb{P}(\text{no people exit at floor a})
\end{align*}
and we can compute this last probability, working with binomial random variables.
$a \neq b$:
in the case $a \neq b$ we have to find $\mathbb{P}(\text{ elevator stops at both floor a and b})$. A hint to find this probability is to use the rule: $\mathbb{P}(A \cap B) = \mathbb{P}(A) + \mathbb{P}(B) - \mathbb{P}(A \cup B)$ and use the results from the case $a = b$.

REPLY [1 votes]: 1)
Your answer only gives the set of values that can be taken by $N$ as outcomes. That is not enough in this situation.
You could go for $\Omega=\{0,1\}^5$ where e.g. $\omega=(1,0,0,1,0)$ is the outcome that the elevator stops on the floors $1$ and $4$.
Then for $i\in\{1,2,3,4,5\}$ we prescribe random variable $X_i:\Omega\to\mathbb R$ by $\omega=(\omega_1,\omega_2,\omega_3,\omega_4,\omega_5,)\mapsto\omega_i$.
Then the event "the elevator stops at floor $i$" can be recognized as the event $\{X_i=1\}$.
2)
I would rather avoid $a,b$ and choose for $i,j$ instead. Note that $X_iX_j=1$ if and only if the elevator stops at floor $i$ and stops at floor $j$. This with $\mathbb EX_iX_j=P(X_iX_j=1)$ because $X_iX_j$ only takes values in $\{0,1\}$.
If $i=j$ then $X_iX_j=X_i^2=X_i$ so in that case we must find $P(X_i=1)$.
This can be done by:$$P(X_i=1)=1-P(X_i=0)=1-\left(\frac45\right)^4$$
Observe here that $\{X_i=0\}$ is the event that all passengers exit on a floor different from $i$.
If $i\neq j$ then we go for calculation:$$P\left(X_{i}X_{j}=1\right)=P\left(X_{i}=1\wedge X_{j}=1\right)=1-P\left(X_{i}=0\vee X_{j}=0\right)=$$$$1-P\left(X_{i}=0\right)-P\left(X_{j}=0\right)+P\left(X_{i}=0\wedge X_{j}=0\right)=1-2\cdot\left(\frac{4}{5}\right)^{4}+\left(\frac{3}{5}\right)^{4}$$
3)
I don't know what is meant by "prove the independence of $\mathbb EX_1X_2$" since $\mathbb EX_1X_2$ is a real number. They might mean something like: "check that $\mathbb EX_iX_j$ is not depending on $(i,j)$ if $i\neq j$.
4)
$N^2=\sum_{i=1}^5\sum_{j=1}^5X_iX_j$ and this equality enables you to find an expression for $\mathbb EN^2$ by means of linearity of expectation and symmetry.
5)
Finding $P(N=i)$ for $i=1,2,3,4,5$ might turn out to be the hardest part of the question, but something must be left for you.
If you don't manage to do it yourself then please let me know.

edit (concerning determination of distribution of $N$).
Number the floors with $1,2,3,4,5$ using index variable $j$ for
them.
Number the persons with $1,2,3,4$ using index variable $i$ for them.
Let $Y_{i,j}=1$ if person $i$ leaves the elevator at floor $j$
and let $Y_{i,j}=0$ otherwise.
Then $Y_{j}:=\sum_{i=1}^{4}Y_{i,j}$ equalizes the number of persons
that leave the elevator at floor $j$.
Here $Y_{1,j},Y_{2,j},Y_{3,j},Y_{4,j}$ are iid and have Bernoulli
distribution with parameter $\frac{1}{5}$.
Consequently $Y_{j}$ has binomial distribution with parameters $n=4$
and $p=\frac{1}{5}$.
Next to that we have: $$\sum_{j=1}^{5}Y_{j}=4\tag1$$
$Y:=\left(Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right)$ is a random vector where
$Y_{j}$ denotes the number of persons that leave the elevator at
floor $j$.
$Y$ has $5$-nomial distribution with parameters $n=4$ and probabilities
$\left(p_{1},p_{2},p_{3},p_{4},p_{5}\right)=\left(\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5},\frac{1}{5}\right)$.
This means that: $$P\left(Y=\left(r_{1},r_{2},r_{3},r_{4},r_{5}\right)\right)=5^{-4}\frac{4!}{r_{1}!r_{2}!r_{3}!r_{4}!r_{5}!}\tag2$$
for nonnegative integers $r_{1},\dots,r_{5}$ that satisfy $r_{1}+r_{2}+r_{3}+r_{4}+r_{5}=4$
We can define $N$ as the cardinality of the random set $S=\left\{ j\in\left\{ 1,2,3,4,5\right\} \mid Y_{j}>0\right\} $.
Based on $(1)$ we find:

$N=1\iff\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 4,0\right\} $
$N=2\iff\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 2,0\right\} \text{ or }\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 3,1,0\right\} $
$N=3\iff\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 2,1,0\right\} $
$N=4\iff\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 1,0\right\} $

This enables you to find the distribution of $N$.
As an example let us find $P\left(N=3\right)$.
In order to get $\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 2,1,0\right\} $
first observe on base of $(2)$ that: $$P\left(Y_{1}=2,Y_{2}=1,Y_{3}=1\right)=5^{-4}\frac{4!}{2!1!1!0!0!}=12\cdot5^{-4}$$
Then realize that $Y=(2,1,1,0,0)$ is just one of $5\binom{4}{2}=30$ possibilities to arrive
at $\left\{ Y_{1},Y_{2},Y_{3},Y_{4},Y_{5}\right\} =\left\{ 2,1,0\right\} $
so that finally we arrive at: $$P\left(N=3\right)=30\cdot12\cdot5^{-4}=360\cdot5^{-4}$$