text:
\begin{document}
\maketitle
\begin{abstract}
It is well known that the Lie-algebra structure on quantum algebras gives rise to a Poisson-algebra structure on classical algebras as the Planck constant goes to 0. We show that this correspondance still holds in the generalization of super-algebra introduced by Scheunert, called $\ep$-algebra. We illustrate this with the example of Number Operator Algebras, a new kind of object that we have defined and classified under some assumptions.
\end{abstract}
\eject

\section{Introduction}
In \cite{S} Scheunert has introduced the concept of $\ep$-algebra, which seems to be the widest generalization of super-algebra. The sign appearing in super-algebra is generalized in $\ep$-algebra by a so-called ``commutation factor''. We will define this concept in section 2, and quickly review some basic facts, most of which can be found in \cite{S} or \cite{KN1}. However, we will emphasize on free $\ep$-modules, since they do not behave as simply as super-modules, something that is not always clearly stated. We will define three different notions of rank for these modules and give a sufficient condition for an $\ep$-algebra to have invariant basis number, in the sense of Cohn (see \cite{C}).

Although the existence of $\ep$-algebras is quite well known, their study has not been undertaken with the same energy as the study of super-algebras mainly for two reasons : they are much too general and they have not proved to be useful for physics yet. Nevertheless, the particular case where the gradation is over $\ZZ$ and $\ep=(-1)^d$, with $d$ an (anti-)symmetric bilinear form seems to be the first logical step away from super-symmetry if ever this step has to be made. This is precisely this kind of $\ep$-algebras we were naturally led to while studying ``number operator algebras''. The latter are algebras we expect to play a role in the quantization procedure of the equations of motion of a system of harmonic oscillators (see \cite{Bes}). We define them and give a classification theorem in section 3.

Interestingly enough, every number operator algebra depends on a single constant $h$ (which plays the role of the Planck constant). It is then natural to try to interpret the algebras at $h\not=0$ (quantum algebras) as deformations of those at $h=0$ (classical algebras) as we shall see in section 4.

In section 5, we show that except in two special cases that happen only for a finite number of degrees of freedom (thus, not in field theory) and which deserve to be studied on their own, the classical algebras are endowed with an $\ep$-Poisson structure coming from the $\ep$-Lie structure on the quantum algebras. This $\ep$-Poisson structure can be used to formulate ``classical'' equations of motion for the system of harmonic oscillators under consideration.

Throughout the article, we denote by $K$ a field, and by $K\langle X\rangle$ the free algebra generated by a set $X$ over $K$. All rings and algebras are unital, and morphisms preserve unit.

\section{$\ep$-algebra}
\subsection{Basic definitions}
The following definition comes from \cite{Bou} (chap. III, p. 46. See also p. 116). Note that, in contrast with \cite{S} and \cite{KN1}, we do not assume that $G$ is a group.

\begin{definition}
Let $(G,+)$ be a commutative monoid. A commutation factor on $G$ with values in $K$ is a mapping $\ep : G\times G\rightarrow K$ such that :
\bea
\ep(g,h)\ep(h,g)=1 \label{c1}\\
\ep(g+h,k)=\ep(g,k)\ep(h,k)\label{c2}
\eea
\end{definition}
>From (\ref{c1}) and (\ref{c2}) one sees at once that $\ep(g,h+k)=\ep(g,h)\ep(g,k)$. Of course, the range of $\ep$ is included in $K^\times$, the group of invertible elements of $K$. Moreover, from (\ref{c1}), we have $\ep(g,g)^2=1$ for any $g\in G$. >From (\ref{c2}) we also have $\ep(0,g)=\ep(g,0)=1$. In particular $\ep(0,0)=1$. If $g$ has an opposite $g'$, we see from (\ref{c2}) that $\ep(g',k)=\ep(g,k)^{-1}$. Thus, if $G$ is a semi-group, we can extend $\ep$ uniquely to the group obtained from $G$ by symmetrization. It is obvious that this extension is a commutation factor.

We now define a map $p$ (for ``parity''), from $G$ to $\ZZ/2\ZZ$ by $p(g)=0$ if $\ep(g,g)=1$ and $p(g)=1$ if $\ep(g,g)=-1$. Since $\ep(g+h,g+h)=\ep(g,g)\ep(g,h)\ep(h,g)\ep(h,h)$ $=\ep(g,g)\ep(h,h)$, and $\ep(0,0)=1$, $p$ is a monoid homomorphism from $(G,+)$ to $(\ZZ/2\ZZ,+)$. Let us call $G_0=Ker(p)$, and $G_1=G\setminus G_0$. The elements in $G_0$ are called ``even'', the ones in $G_1$ are called ``odd''.

Note that in order to have $G_1\not=\emptyset$ we must impose a condition on $G$. Indeed, since $\ep(n.g,g)=\ep(g,g)^n$ every odd element must be of even order. For instance, if $G=\ZZ/p\ZZ$ with $p$ odd then $G_1=\emptyset$ for every $\epsilon$ defined on $G$.

\smallbreak
Now for an example : let $G=\ZZ\times\ZZ$, take $q\in K^\times$, and set $\ep_q((k,l),(m,n))=q^{lm-kn}$. More generally, for any $G$ consider a monoid homomorphism $\phi : G\times G\rightarrow(\ZZ,+)$ which is anti-symmetric, then for any $q\in K^\times$, $\ep=q^\phi$ is a commutation factor. Note that if $q=\pm 1$, $\phi$ may either be taken symmetric or anti-symmetric.

\begin{definition}
Let $A$ be a $K$-algebra. $A$ is called a $G$-graded $K$-algebra iff there exist $K$-subspaces $(A_g)_{g\in G}$, such that :
\bea
A=\bigoplus_{g\in G}A_g\\
A_gA_h\subset A_{g+h}
\eea
\end{definition}

An element in $A_g$ for some $g$ is called homogenous. In the sequel, we use the following notation : if $a\in A_g$, we write $\bar a=g$. More generally, if $x$ is a homogenous element in any $G$-graded object, we will write $\bar x$ for the grade of $x$.

We also introduce the notation ``$\forall_h$'' that will mean ``for all homogenous''. For example, $\forall_h x\in A$ means ``for all homogenous $x$ in $A$''.

A couple $(A,\ep)$ where $A$ is a $G$-graded $K$-algebra, and $\ep$ is a commutation factor on $G$ with values in $K$ is called an ``$\ep$-algebra''. We will also say ``$A$ is an $\ep$-algebra''.

If $G$ is a semi-group, it can be embedded into a group $G'$. Since $\ep$ extends to a commutation factor $\ep'$ on $G'$, any $\ep$-algebra can be considered as an $\ep'$-algebra with $A_g=\{0\}$ for $g\in G'\setminus G$. Thus, we can always work with groups instead of semi-groups.

Let us look at an example. Let $M_q$ be Manin's quantum plane, that is : $M_q=K\langle x,y\rangle/\langle xy-qyx\rangle$, where $\langle xy-qyx\rangle$ is the two-sided ideal generated by $xy-qyx$, with $q\in K^\times$. Since $\{y^kx^l|k,l\in\NN\}$ is a $K$-basis of $M_q$, one sees at once that there is a unique $\NN\times\NN$-grading

 such that $\bar x=(1,0)$, $\bar y=(0,1)$, endowing $(M_q,\ep_q)$ with the structure of an $\ep_q$-algebra.
\smallbreak
Throughout the rest of this section, $A$ will be a fixed $\ep$-algebra.

$\forall_h$ $x,y\in A$, we define the $\ep$-commutator by : $$[x,y]_\ep=xy-\ep(\bar x,\bar y)yx$$
It is then extended to non-homogenous elements by linearity. We say that $x$ and $y$ $\ep$-commute iff $[x,y]_\ep=0$. The $\ep$-center of $A$, $Z_\ep(A)$, is the set $\{x\in A|\forall y\in A, [x,y]_\ep=0\}$. $A$ is said to be $\ep$-commutative iff $Z_\ep(A)=A$. For instance, $M_q$ is $\ep_q$-commutative.

There is a super-algebra $\tilde A$ naturally associated with $A$ : it is defined by $\tilde A_0=\bigoplus_{g\in G_0}A_g$, and $\tilde A_1=\bigoplus_{g\in G_1}A_g$. If $A$ is $\ep$-commutative, it is not true in general that $\tilde A$ is super-commutative. However, it is easy to see that in an $\ep$-commutative algebra, every $x\in\tilde A_1$ is nilpotent.

In the next two definitions, we assume the characteristic of $K$ to be $\not=2,3$.
\begin{definition}
Let $V$ be a $G$-graded $K$-space, and $[.,.]$ be a bilinear map from $V\times V$ to $V$, such that $\forall_h$ $x,y,z\in V$ :
\be
[x,y]=-\ep(\bar y,\bar x)[y,x]
\ee
\be
\ep(\bar z,\bar x)[x,[y,z]]+\ep(\bar y,\bar z)[z,[x,y]]+\ep(\bar x,\bar y)[y,[z,x]]=0
\ee
$V$ is called an $\ep$-Lie algebra.
\end{definition}

For example, $(A,[.,.]_\ep)$ is an $\ep$-Lie algebra.

\begin{definition}
Let $\{.,.\} : A\times A\rightarrow A$ be a $K$-bilinear map. $\{.,.\}$ is called an $\ep$-Poisson bracket, and $A$ is an $\ep$-Poisson algebra, if and only if :
\begin{enumerate}
\item $(A,\{.,.\})$ is an $\ep$-Lie algebra
\item $\forall_h\ x,y,z\in A,\ \{x,yz\}=\{x,y\}z+\ep(\bar x,\bar y)y\{x,z\}$
\end{enumerate}
\end{definition}

\subsection{$\ep$-modules}

They are just $G$-graded $A$-modules. More precisely, a left module $M$ over $A$ is called a left $\ep$-module over $A$ if, and only if, there exists a decomposition $M=\bigoplus_{g\in G}M_g$ as $K$-space, such that $A_gM_h\subset M_{g+h}$. Right $\ep$-modules are similarly defined.

Given an $\ep$-module over $A$, a super-module $\tilde M$ over $\tilde A$ is defined by $\tilde M_0=\bigoplus_{g\in G_0}M_g$ and $\tilde M_1=\bigoplus_{g\in G_1}M_g$.

If $A$ is $\ep$-commutative, there exists on any left $\ep$-module a canonical right $\ep$-module structure compatible with it, given by :
\be
\forall_h a\in A, \forall_h m\in M,\quad m.a=\ep(\bar m,\bar a)a.m\label{rs}
\ee

Let $M$, $N$ be two left $\ep$-modules over $A$. For all $\gamma\in G$, we define a homomorphism of grade $\gamma$ from $M$ to $N$ to be a $K$-linear map $f$ such that\footnote{if $G$ is not a semi-group, the same homomorphism may have several grades}
\be
\forall g\in G,\quad f(M_g)\subset N_{g+\gamma}\label{h1}
\ee
and furthermore :
\be
\forall_h a\in A, \forall_h m\in M,\quad f(a.m)=\ep(\gamma,\bar a)f(m)\label{h2}
\ee
We denote by  $\Hom^\gamma_A(M,N)$ the set of those morphisms. For two right modules, a homomorphism of grade $\gamma$ has to fulfill (\ref{h1}), but (\ref{h2}) is replaced by
\be
f(m.a)=f(m).a\label{h3}
\ee
One sees that with these definitions, if $A$ is $\ep$-commutative, any grade $\gamma$ left homomorphism is automatically a grade $\gamma$ right homomorphism for the right structure (\ref{rs}).

We will be mainly concerned with grade 0 homomorphisms over an $\ep$-commuta\-tive $A$.

A free $\ep$-module is just a free graded module, that is a graded module in which a homogenous basis exists. Note that if $M$ is free as an ungraded module, it need not be free as a graded module, even if $A$ is commutative. An important exception is when $M$ is a super-module over a super-commutative algebra. In this case, given an ungraded basis one can find a graded one (see \cite{D} p 19).

We will be only concerned with finite free $\ep$-modules. There are three natural notions of rank for them.

Let us recall that a ring $R$ is said to have ``invariant basis number'' (IBN) if (see \cite{C}) : $$R^m\simeq R^n\Rightarrow m=n$$

It is equivalent to saying that for any two matrices $P\in{\cal M}_{m,n}(R)$ and $Q\in{\cal M}_{n,m}(R)$ : $$(PQ=I_m \mbox{ and } QP=I_n)\Rightarrow m=n$$ If there exists a ring homomorphism from $R$ to $S$, we can extend it to a ring homomorphism of matrix algebras, thus $S$ has IBN $\Rightarrow$ $R$ has IBN.
\smallbreak

Let ${\bf n} : G\rightarrow\NN$ be a map such that the set $\mbox{Supp}({\bf n})=\{g\in G|{\bf n}(g)\not=0\}$ is finite. We define the (right, say) module $F=A^{\bf n}$ to be the direct sum $\bigoplus_{g\in \mbox{Supp}({\bf n})}A^{{\bf n}(g)}$. If $\{e_g^i|g\in \mbox{Supp}({\bf n}), 1\leq i\leq {\bf n}(g)\}$ is the canonical basis, we define a grading by $\bar e_g^i=g$.

\begin{definition}
The integer $n=\sum_g{\bf n}(g)$ is called the ``total rank'' of $F$.

The couple $(p,q)$ where $p=\sum_{g\in G_0} {\bf n}(g)$, $q=\sum_{g\in G_1}{\bf n}(g)$, denoted by $p|q$, is called the ``super-rank'' of $F$. ${\bf n}$ is called the ``$\ep$-rank'' of $F$.
\end{definition}

$F$ is the canonical (right) free $\ep$-module over $A$ of $\ep$-rank ${\bf n}$. If two canonical free $\ep$-modules over $A$ that are isomorphic must have the same $\ep$-rank, we say that $A$ has invariant $\ep$-rank, and the $\ep$-rank is uniquely defined for any free $\ep$-module over $A$. The properties of having ``invariant super-rank'' and ``invariant total rank'' are similarly defined. Of course $A$ has invariant $\ep$-rank $\Rightarrow$ $A$ has invariant super-rank $\Rightarrow$ $A$ has invariant total rank.

Elements of $\Hom^\gamma(A^{\bf m},A^{\bf n})$ may be represented by matrices in ${\cal M}_{n,m}(A)$. For more details see \cite{KN1} (but note that their matrices must act on the right of row vectors, which represent elements of a left module, whereas if we prefer working with right modules, we must take column vectors, with matrices acting on the left, and this is what we will do).

\begin{theorem}\label{ibn}
Let $A$ be an $\ep$-algebra and $B$ be an algebra having IBN. Denote by $H$ the set of algebra homomorphisms from $A$ to $B$. Then :
\begin{enumerate}
\item  If $H\not=\emptyset$, $A$ has invariant total rank.
\item  If $\exists\pi$, $\pi\in H$, $\pi(A_g)=\{0\}$ for all $g\in G_1$, then $A$ has invariant super-rank.
\item If $\exists\pi$, $\pi\in H$, $\pi(A_g)=\{0\}$ for all $g\not=0$, then $A$ has invariant $\ep$-rank
\end{enumerate}
\end{theorem}
\dem\newline
The first assertion comes from the trivial fact that a homogenous basis is also an ungraded basis.
\smallbreak
For the second assertion we take 2 homogenous bases $\{e_\rho,u_\sigma|1\leq\rho\leq m, 1\leq\sigma\leq k\}$ and $\{f_\alpha,v_\beta|1\leq\alpha\leq n, 1\leq\beta\leq l\}$ of a free $\ep$-module over $A$, such that the $e_\rho$ and the $f_\alpha$ are even, and the $u_\sigma$, $v_\beta$ are odd.
We write : $f_\alpha=\sum_\rho e_\rho a_{\rho,\alpha}+\sum_\sigma u_\sigma b_{\sigma,\alpha}$, and $v_\beta=\sum_\rho e_\rho c_{\rho,\beta}+\sum_\sigma u_\sigma d_{\sigma,\beta}$.

Of course we also have : $e_\rho=\sum_\mu f_\mu a'_{\mu,\rho}+\sum_\nu v_\nu b'_{\nu,\rho}$, and $u_\sigma=\sum_\mu f_\mu c'_{\mu,\sigma}+\sum_\nu v_\nu d'_{\nu,\sigma}$.

We can suppose without loss of generality that :
$$a_{\rho,\alpha}\in\bigoplus_{g|g+\bar e_\rho=\bar f_\alpha} A^g$$
and
$$b_{\sigma,\alpha}\in\bigoplus_{g|g+\bar u_\sigma=\bar f_\alpha}A^g$$
and similarly for the other coefficients. We have the matrix equalities :
\be
\matrix{ &\matrix{ {\scriptstyle m} &{\scriptstyle k}}\cr
\matrix{{\scriptstyle n}\cr {\scriptstyle l}}&\left(\matrix{a'&c'\cr b'&d'}\right)}\matrix{ &\matrix{ {\scriptstyle n} &{\scriptstyle l}}\cr
\matrix{{ }_m\cr { }_k}&\left(\matrix{a&c\cr b&d}\right)}=\matrix{ &\matrix{ & }\cr
\matrix{ \cr }&\left(\matrix{I_n&0\cr 0&I_l}\right)}\label{mat1}
\ee
and
\be
\matrix{ &\matrix{ {\scriptstyle n} &{\scriptstyle l}}\cr
\matrix{{ }_m\cr { }_k}&\left(\matrix{a&c\cr b&d}\right)}\matrix{ &\matrix{ {\scriptstyle m} &{\scriptstyle k}}\cr
\matrix{{\scriptstyle n}\cr {\scriptstyle l}}&\left(\matrix{a'&c'\cr b'&d'}\right)}=\matrix{ &\matrix{ & }\cr
\matrix{ \cr }&\left(\matrix{I_m&0\cr 0&I_k}\right)}\label{mat2}
\ee
Since the parity map is a monoid homomorphism, the coefficients of $b$, $b'$, $c$, $c'$ are odd. Thus the images under $\pi$ of (\ref{mat1}) and (\ref{mat2}) are :
\be
\left(\matrix{\pi(a')&0\cr 0&\pi(d')}\right)\left(\matrix{\pi(a)&0\cr 0&\pi(d)}\right)=\left(\matrix{I_n&0\cr 0&I_l}\right)
\ee
and
\be
\left(\matrix{\pi(a)&0\cr 0&\pi(d)}\right)\left(\matrix{\pi(a')&0\cr 0&\pi(d')}\right)=\left(\matrix{I_m&0\cr 0&I_k}\right)
\ee
Therefore $\pi(a')\pi(a)=I_n$ and $\pi(a)\pi(a')=I_m$. Since $B$ has IBN, we get $m=n$. In the same way, we find $k=l$.
\smallbreak
For the last assertion, we take two bases of $\ep$-rank ${\bf m}$ and ${\bf n}$. Let us suppose first that $G$ is a group.

As before, the change of bases gives the following matrix equations
\be
\left(\matrix{ & & \cr  & a_{g,h} & \cr & & }\right)\left(\matrix{ & & \cr  & a'_{g,h} & \cr & & }\right)=I_m
\ee
\be
\left(\matrix{ & & \cr  & a'_{g,h} & \cr & & }\right)\left(\matrix{ & & \cr  & a_{g,h} & \cr & & }\right)=I_n
\ee
where the $a_{g,h}$ and $a'_{g,h}$ are block matrices. They are indexed by the elements of $\mbox{Supp}({\bf m})\cup\mbox{Supp}({\bf n})$. Their coefficients are of grade $g-h$. The block $a_{g,h}$ has ${\bf m}(g)$ rows and ${\bf n}(h)$ columns, and $a'_{g,h}$ has ${\bf n}(g)$ rows and ${\bf m}(h)$ columns. $I_m$ and $I_n$ are the identity matrices of rank $m=\sum_g{\bf m}(g)$ and $n=\sum_g{\bf n}(g)$.

Taking the image under $\pi$ makes all the off-diagonal block matrices vanish. We thus find for all $g\in\mbox{Supp}({\bf m})\cup\mbox{Supp}({\bf n})$ : $\pi(a_{g,g})\pi(a'_{g,g})=I_{{\bf m}(g)}$ and $\pi(a'_{g,g})\pi(a_{g,g})=I_{{\bf n}(g)}$. Since $B$ has IBN, we see that ${\bf m}$ and ${\bf n}$ must coincide on their support. Thus, they are equal.

In the general case the coefficients of the matrices $a_{g,h}$ and $a_{g,h}'$ belong to $\bigoplus_{k\in G|k+h=g}A^k$, and if $g\not=h$, we have $k\not=0$, and $\pi(A^k)=0$. The result follows.

\hfill QED.

\smallbreak
It is well known that commutative algebras have invariant total rank, and that super-commutative algebras have invariant super-rank (cf. \cite{D}). But the following example shows that the invariance of the $\ep$-rank is not always true, even for an $\ep$-commutative $A$.

Consider the algebra $A=K\{x,y,x^{-1},y^{-1}\}$ generated by variables $x$,\ldots,$y^{-1}$ such that $x$ and $x^{-1}$ anti-commute with $y$ and $y^{-1}$. A $K$-basis for $A$ is given by $\{x^ky^l|k,l\in\ZZ\}$, a $\ZZ/2\ZZ\times\ZZ/2\ZZ$-grading is uniquely defined by $\bar x=(1,0)$, $\bar y=(0,1)$. Finally, $A$ is $\ep$-commutative with $\ep((k,l),(m,n))=(-1)^{kn+lm}$. But $x$ and $y$ are two homogenous bases of $A$ considered as a module over itself, with different $\ep$-rank.
\smallbreak
Nevertheless, there are many cases in which the conditions of theorem \ref{ibn} are met. For instance, an $\ep$-commutative $A$ can sometimes be written $A=C\oplus I$, where $I$ is a two-sided ideal, and $C$ is a commutative sub-algebra. Then of course the hypothesis of theorem \ref{ibn} is satisfied by the quotient map. An example of this situation is when $G=\NN^k$, or more generally, the semi-group of positive elements of some ordered group. Then we can take $C=A_0$ and $I=\bigoplus_{g\not=0}A_g$. Another situation that we shall meet is when $C=K.1$, in which case $A$ can be viewed as a non-unital algebra to which a unit has been added.

\subsection{Tensor products}

In this paragraph, $A$ is $\ep$-commutative, all modules are considered as $A-A$-bimodules through (\ref{rs}), and all bases are homogenous.

Let $V_1,\ldots,V_n$ be $\ep$-modules over $A$. Their tensor product $V_1\otimes\ldots\otimes V_n$ is defined using the general construction given in \cite{Bou} (III, p. 65-69). In particular, note that for $a\in A$, $v_i\in V_i$, one has :
$$v_1\otimes\ldots\otimes v_i.a\otimes v_{i+1}\otimes\ldots v_n=v_1\otimes\ldots\otimes v_i\otimes a.v_{i+1}\otimes\ldots\otimes v_n$$
Note also that the commutativity isomorphism $V_1\otimes V_2\simeq V_2\otimes V_1$ should be defined by $v_1\otimes v_2\mapsto \ep(\bar v_1,\bar v_2)v_2\otimes v_1$.

We then define the $\ep$-tensor algebra of $V$ : $T_\ep(V)=\bigoplus_{n\in\NN}V^{\otimes n}$. The gradation $\overline{v_1\otimes\ldots\otimes v_n}=\bar v_1+\ldots+\bar v_n$ turns it into an $\ep$-algebra.

In particular, if $V$ is a free $\ep$-module with basis $\{x_1,\ldots,x_n\}$, $A_\ep\langle x_1,\ldots,x_n\rangle:=T_\ep(V)$ is the free $\ep$-algebra over $A$ on the generators $\{x_1,\ldots,x_n\}$.

In general, all usual constructions carry over, provided one puts in an epsilon term each time two factors are exchanged. For instance, one can define the $\ep$-antisymmetric algebra $\Lambda_\ep(V)=T_\ep(V)/I$, where $I$ is the two-sided ideal generated by the elements of the form $v\otimes w+\ep(\bar v,\bar w)w\otimes v$. Note that if any generator is odd this algebra is of infinite rank, thus we cannot use it to define a determinant. If we had chosen not to put the epsilon factor in the definition of $\Lambda_\ep(V)$, we would have had an algebra of finite rank with the top exterior product of rank one, and with basis $x_1\wedge\ldots\wedge x_n$, as usual. However, this construction is not functorial when there are odd elements, so the usual determinant cannot be defined. Nevertheless, the generalization of the determinant to super-algebra, known as the Berezinian, can be extended to $\epsilon$-algebra, as it is shown in \cite{KN1}.

\underline{Remark} : For the reader acquainted with these matters, we mention here that Bergman's diamond lemma, which is a most useful result, is valid in this setting for graded reduction systems, that is to say systems $\{(w_\sigma,f_\sigma)|\sigma\in S\}$ where $f_\sigma$ is homogenous of grade $\bar w_\sigma$. This is an immediate application of \cite{Ber}, section 6.

\section{Number Operator Algebras}
In \cite{Bes} we have introduced number operator algebras.

\begin{definition}
Let $K$ be a field of characteristic 0 endowed with an involutive automorphism $\tau$, $B$ a non-trivial $K$-algebra (that is $B\not=0$, $B\not=K$), $Z(B)$ the centre of $B$, and let $C^+=\{\aix|i\in\II\}$, $C^-=\{\ai|i\in\II\}$, and $N=\{N_i|i\in\II\}$ be 3 sets indexed by $\II$, with the $\ai$'s and $\aix$'s in $B$, and $N_i$'s in $B/Z(B)$. $(B,C^+,C^-,N)$ is said to be a number operator algebra if, and only if :

\item{(i)} $B$ is generated by $C^+\cup C^-$.
\item{(ii)} One uniquely defines an anti-involution $J$ on $B$ by setting $J(\ai)=\aix$.
\item{(iii)} $\forall i,j\in\II$, $[N_i,\ajx]=\delta_{ij}\ajx$ and $[N_i,\aj]=-\delta_{ij}\aj$.
\end{definition}

\underline{Remarks} :
\begin{itemize}
\item{}In the physical case $K=\CC$ and $\tau$ is the complex conjugation.
\item{}We shall say ``$B$ is a n.o.a.'' rather than using the lengthy expression ``$(B,C^+,C^-,N)$ is a number operator algebra''.
\item{}We say that $B$ is of type $\alpha$ when $\alpha$ is the cardinal of $\II$.
\end{itemize}
We go on with a few more definitions. From now on we fix a n.o.a. $B$, with all its features, $C^+$, $C^-$, etc\ldots

Let us call $L$ the free $K$-algebra generated by $C^+\cup C^-$, and $\pi$ the canonical morphism from $L$ onto $B$. If the kernel of $\pi$ is generated by elements of degree two or less, we say that $B$ is quadratically presented.

Let $\SS_\II$ be the group of one-one mappings from $\II$ to $\II$ leaving all elements invariant except for a finite number. Every $\sigma$ in $\SS_\II$ naturally gives rise to an algebra automorphism of $L$, denoted by $\sigma^*$, such that for all $i\in\II$, $\sigma^*(\ai)=a_{\sigma(i)}^{}$, and $\sigma^*(\aix)=a_{\sigma(i)}^+$.

If every such $\sigma^*$ induces an automorphism of $B$, we say that $B$ is symmetric.

It is possible to classify all n.o.a. of infinite type which are symmetric and quadratically presented (see \cite{Bes}).

\begin{theorem}\label{clas}
Let $B$ be symmetric and quadratically presented. If $\II$ is infinite, then there exists an $h\in K^+\setminus\{0\}$ ($K^+$ is the sub-field of elements of $K$ that are invariant under $\tau$) such that Ker$(\pi)$ is generated by one of the following sets :
\item{(a)} $\{\ai^2,\aix^2,\ai\aj+\aj\ai,\aix\ajx+\ajx\aix,\ai\ajx+\ajx\ai,\ai\aix+\aix\ai-h|i,j\in\II,i\not=j\}$ (Fermionic case)
\item{(a')} $\{\ai^2,\aix^2,\ai\aj-\aj\ai,\aix\ajx-\ajx\aix,\ai\ajx-\ajx\ai,\ai\aix+\aix\ai-h|i,j\in\II,i\not=j\}$ (Pseudo-Fermionic case)
\item{(c)} $\{\ai\aj-\aj\ai,\aix\ajx-\ajx\aix,\ai\ajx-\ajx\ai,\ai\aix-\aix\ai-h|i,j\in\II, i\not=j\}$ (Bosonic case)
\item{(c')} $\{\ai\aj+\aj\ai,\aix\ajx+\ajx\aix,\ai\ajx+\ajx\ai,\ai\aix-\aix\ai-h|i,j\in\II, i\not=j\}$ (Pseudo-Bosonic case)
\end{theorem}

We denote by $\hat \fer^h$, $\fer^h$, $\bos^h$ and $\hat\bos^h$, respectively, the algebras corresponding to each of these four cases. Actually we should call them $\hat\fer^{h,\II}$, etc\ldots but we assume that $\II$ is fixed and thus no confusion can be made.

It should be noted that although these four kinds of algebras are generally not isomorphic to each other, this can still happen for $\hat\fer^h$ and $\fer^h$, at least when $\II$ is countable (Brauer-Weyl isomorphism).

Nevertheless, these four cases are distinct as symmetric n.o.a. : that is to say, there exists no isomorphism $\phi$ between any two of them, such that $\phi$ commutes with the anti-involution and with the action of $\SS_\II$, and such that $\phi$ send any number operator to a number operator. Of course, all this can be formulated in terms of categories.

What is the situation inside each of the four cases ?
We see that the algebras only depend on a parameter $h\in K^+$. Consider $B^h$ and $B^{h'}$, two n.o.a. of one of the four species (either two algebras of fermions, or two algebras of pseudo-fermions, or etc\ldots). Then one can show that these algebras are isomorphic as symmetric number operator algebras iff $\exists\lambda\in K^+$ such that $h'=\lambda\tau(\lambda)h$. When such a relation exists between $h$ and $h'$, let us define the mapping $\phi_\lambda$, given by $\phi_\lambda(\ai)=\lambda\ai'$, $\phi_\lambda(\aix)=\tau(\lambda)\aix'$, where the elements with a prime are in $B^{h'}$ and the others are in $B^h$. $\phi_\lambda$ is easily found to have all the required properties for an isomorphism of symmetric number operator algebras.

In the physical case, we see that the four families of theorem 2 depend on a non-zero real constant, and furthermore in each of the four families there are exactly two isomorphism classes in the category of symmetric n.o.a., one for $h>0$ and the other for $h<0$, which makes 8 isomorphism classes as a whole. The isomorphisms $\phi_\lambda$ clearly correspond to a rescaling of the units.

\underline{Remark} : When $\alpha$ is finite, a classification can be done under a supplementary hypothesis, namely the ``confluence hypothesis''. In this case we have two more families of algebras, which have the following presentation :
\begin{itemize}
\item{(b)} $\{\ai^2,\aix^2,\ai\aj,\aix\ajx,\ai\ajx,\ai\aix+\sum_{k\in\II}\akx\ak-h|i,j\in\II,i\not=j\}$
\item{(b')} $\{\ai^2,\aix^2,\ai\aj,\aix\ajx,\ajx\ai,\aix\ai+\sum_{k\in\II}\ak\akx-h|i,j\in\II,i\not=j\}$
\end{itemize}

Of course, these two kinds of algebras are isomorphic as algebras (and are isomorphic to matrix algebras), but once again not as number operator algebras (except when $n=1$, in which case only two kinds of algebras remain\penalty 1000 : bosonic and fermionic). We call them ${\cal E}^h$ and ${\cal E'}^{h}$. In the physical case, there are also two isomorphism classes in each case $(b)$ or $(b')$, distinguished by the sign of $h$, in the category of number operator algebras. Particles corresponding to such algebras would follow an exclusion principle even more severe than Pauli's : only one such particle could exist at a given time, regardless of its state.

\section{Classical limit of Number Operator Algebras}

We quickly recall the definition of formal deformations of algebras. See \cite{Ger} or \cite{Gui} for more details.

Let $B$ be a $K$-algebra whose multiplication is seen as a bilinear map $\mu : B\times B\rightarrow B$. Let $K[[\hbar]]$ stand for the algebra of formal series in $\hbar$, and $\tilde B$ stand for the $K[[\hbar]]$-algebra of formal series with coefficients in $B$.

A ``formal deformation of $(B,\mu)$'' is a $K[[\hbar]]$-algebra structure $\tilde\mu$ on $\tilde B$, such that the canonical map $\tilde B/\hbar\tilde B\rightarrow B$ is an algebra isomorphism. These data are equivalent to the existence of a sequence of bilinear maps $\mu_n: B\times B\rightarrow B$, with $\mu_0=\mu$, such that :
\be
\forall x,y,z\in B,\, \forall n\geq 1,\, \sum_{p+q=n}(\mu_p(\mu_q(x,y),z)-\mu_p(x,\mu_q(y,z)))=0\label{st}
\ee

$\tilde\mu$ is then defined by setting, for all $x,y\in B\subset\tilde B$ :
\be
\tilde\mu(x,y):=\sum_n\mu_n(x,y)\hbar^n\label{stst}
\ee

and extending to formal series in the obvious way.

It is sometimes possible to ``fix the parameter'', i.e. to replace everywhere $\hbar$ by a constant $h\in K$. In particular, this is the case when the right-hand side of (\ref{stst}) is a polynomial for every $x,y\in B$. By doing so, a new algebra structure $\mu^h$ is defined on $B$. We shall say that $(B,\mu^h)$ is a deformation of $(B,\mu)$.

Let $B^h$ be a n.o.a. If we replace $h$ by $0$ in the presentation of $B^h$, we get a new algebra $B^0$ that we call the ``classical limit'' of $B^h$. Let us see what these algebras look like in the cases $(a)$, $(a')$, $(c)$ and $(c')$ of theorem \ref{clas}.

It is easy to see that $\hat\fer^0$ is the exterior algebra $\Lambda[\ai,\aix|i\in\II]$ over variables $\ai$ and $\aix$, and that ${\cal A}^0$ is the symmetric (or polynomial) algebra $S[\ai,\aix|i\in\II]$ over the same variables. The other two have no names (however $\fer^0$ is a particular case of the generalized Grassman algebras of \cite{OK}), but we have the following algebra isomorphisms :

$$\fer^0\simeq\tens_\II\Lambda[a,a^+]\qquad\hat \bos^0\simeq\hattens_\II S[a,a^+]$$
where $\hattens$ means ``graded tensor product''.

Let us have a closer look at $\bos^h$ and $\bos^0$. Fix a total ordering $<$ on $\II$. Take two tuples of indices $\JJ=(j_1,\ldots,j_r)$ and $\KK=(k_1,\ldots,k_s)$ such that $j_1\leq\ldots\leq j_r$, $k_1\leq\ldots\leq k_s$, and set $\ajjx:=a_{j_1}^+\ldots a_{j_r}^+$ and $\akk:=a_{k_1}^{}\ldots a_{k_s}^{}$. If $\JJ$ or $\KK$ is empty we set $a_\emptyset^{}=a_\emptyset^+=1$. ${\cal A}^h$ has a basis $T$ of the form : $\{\ajjx\akk|\JJ$ and $\KK$ are any ordered tuples of indices$\}$. But $\bos^0$ has a basis of the same form and we can use it to identify $\bos^0$ and $\bos^h$ as vector spaces. Now take any two elements of $T$, multiply them in $\bos^h$, and write the result in terms of basis vectors : we can see it as a polynomial in $h$ and take the coefficients to define $\mu_n$ as in the formula below :
$$\mu(x,y)=:\sum_{n\geq 0}\mu_n(x,y)h^n$$
where $\mu$ is the multiplication of $\bos^h$.

We can then extend $\mu_n$ by bilinearity. We claim that the $\mu_n$ just defined fulfil the conditions (\ref{st}) and that $\mu_0$ is actually the multiplication of $\bos^0$. To show it we should use the concepts of reduction systems and confluence (see \cite{Ber}). The reader acquainted with these matters will see it to be an easy consequence of the confluence of the reduction system $S=\{(\ai\aj,\aj\ai),(\aix\ajx,\ajx\aix),(\ai\ajx,\ajx\ai),(\aj\aix,\aix\aj),(\ai\aix,\aix\ai+h)|i,\penalty -100 j\in\II,$ $i<j\}$ for any value of $h$. We refer to \cite{Gui} (Chap. 3, theorem 2.6.2.) for more details.

We can thus define a formal deformation $\tilde\mu$ of $\bos^0$, and if we set the constant to $h$, we obviously have $(\bos^0,\mu^h)\simeq (\bos^h,\mu)$.

We can do the same with $\hat\bos^h$, $\hat\fer^h$ and $\fer^h$.

\underline{Remark} : Let us say a little word about the case $(b)$ ($(b')$ being symmetrical). ${\cal E}^h$ is a deformation of ${\cal E}^0$, which is a $(\alpha+1)^2$-dimensional local algebra. We do not know if ${\cal E}^0$ has ever been considered. Since it is not an $\ep$-Poisson algebra, we will not study it in the next section. Nevertheless, it has an interesting structure that we plan to study in another article.

\section{$\ep$-Poisson structures}
Let us begin by the example of $\bos^h$, which is well known.

Since $\bos^0$ is commutative, the formula
$$\{x,y\}:=\mu_1(x,y)-\mu_1(y,x)$$
defines a Poisson bracket on $\bos^0$ (see \cite{Gui})\footnote{To follow usual conventions, one should divide out by $i$ in the physical case}. It can also be written in the heuristic form :
$$\{x,y\}=\lim_{\hbar\rightarrow 0}{1\over\hbar}(\tilde\mu(x,y)-\tilde\mu(y,x))$$
This last formula is useful to see that all the properties of the Poisson bracket come from the corresponding properties of the normal (commutator) bracket of $\tilde\mu$, to the first order in $\hbar$. $\bos^0$ is thus a Poisson algebra.
\smallbreak

$\hat\fer^h$, $\fer^h$ $\bos^h$ and $\hat\bos^h$, as well as their classical limits are naturally $\ZZ^{(\II)}$-graded (where $(\II)$ means direct sum over $\II$). It comes from the gradation on the free algebra $L$ uniquely defined by $\bar\ai^+=p_i$, $\bar\ai=-p_i$, where $p_i$ is the element of $\ZZ^{(\II)}$ defined by $p_i(j)=\delta_{ij}$.

We call it the gradation by the number of particles. Since the ideals of definition of quantum as well as classical algebras are homogenous with respect to this gradation, it goes to the quotient in both cases.

Let us define bilinear maps from $\ZZ^{(\II)}\times\ZZ^{(\II)}$ to $\ZZ$ :
$$d_{a}(p,q):=(\sum_{i\in\II}p(i))(\sum_{i\in\II}q(i))$$
$$d_{a'}(p,q):=\sum_{i\in\II}p(i)q(i)$$
$$d_c(p,q):=0$$
$$d_{c'}(p,q)=\sum_{i,j\in\II \atop i\not=j}p(i)q(j)$$
We then define commutation factors on $\ZZ^{(\II)}$ by $\ep_l(p,q)=(-1)^{d_l(p,q)}$, with $l=a$, $a'$, $c$ or $c'$.

\begin{theorem}
$\hat\fer^0$ is an $\ep_a$-commutative $\ep_a$-algebra, $\fer^0$ is $\ep_{a'}$-commutative, $\bos^0$ is $\ep_c$-commutative (that is to say commutative), and $\hat\bos^0$ is $\ep_{c'}$-commutative.

Moreover, they have invariant $\ep$-rank.
\end{theorem}
\dem\newline
Let us examine for instance the case of $\hat\bos^0$ : take a basis element $\aiix\ajj$, $p$ its grading, and $\ak$ a generator. We see that $\ak$ anti-commutes with everything, except with $\ak$ and $\akx$, to which it commutes. Thus we find $\aiix\ajj\ak=(-1)^{\sum_{i\not=k}p_i}\ak\aiix\ajj$. It is of course the same for $\akx$, and by iteration, we find the commutation rules for two basis elements. Then it is easy to extend the result to homogenous elements.

Now the map $\pi : \hat\bos^0\rightarrow K$ given by the projection on the basis element $1$, with respect to the basis $T$, is an algebra homomorphism. Thus $\hat\bos^0$ has invariant $\ep$-rank, by theorem 1.

The three other cases are similar.\hfill QED.
\smallbreak

\begin{theorem}
Let $B^h$ be one of the algebras of Theorem 2 (or more generally, any formal deformation of an $\ep$-commutative algebra), and call $\ep$ the corresponding commutation factor. Let us define :
$$\forall_h\ x,y\in B^0,\quad \{x,y\}_\ep=\mu_1(x,y)-\ep(\bar x,\bar y)\mu_1(y,x)$$
Then, $(B^0,\{.,.\}_\ep)$ is an $\ep$-Poisson algebra.
\end{theorem}
\dem\newline
If we write $[.,.]_\ep$ for the $\ep$-commutator in $(B^0,\mu^h)$, $\ep$-commutativity of $\mu_0$ implies :
$$[x,y]_\ep=\hbar\{x,y\}_\ep+{\cal O}(\hbar^2)$$
where ${\cal O}(\hbar^2)$ stands for terms of order $\geq\hbar^2$. Thus, it is clear that $\{.,.\}_\ep$ defines an $\ep$-Poisson algebra structure on $B^0$.\hfill QED.

\underline{Remark} : Let us go back to the physical case. If we define :
$$p_i:={1\over\sqrt 2}(\ai+\aix)$$
$$q_i:={1\over i\sqrt 2}(\aix-\ai)$$
and
$$H_i:=hN_i$$
we find that all the equations for a system of $\ep$-classical harmonic oscillators are satisfied :
$$\{p_i,p_j\}_\ep=\{q_i,q_j\}_\ep=0$$
$$\{p_i,q_j\}_\ep=\delta_{ij}$$
$$\{H_i,p_j\}_\ep=\delta_{ij}q_j$$
$$\{H_i,q_j\}_\ep=-\delta_{ij}p_j$$


TITLE: Deriving AM-GM in a way that reflects my intuition.
QUESTION [2 upvotes]: One of my core intuitions with inequalities (especially the olympiad kind in which this is very explicit, but this has more profound influences on how I understand the relationship between inequalities) is that in a homogenous inequality, on average, terms that are more mixed ($x^2yz$) are smaller than their less-mixed counterparts ($x^4$).
With this bit of intuition, many inequalities seem trivially true, like am-gm:
$$x^3+y^3+z^3\geq xyz + xyz +xyz$$
Or this one:
$$x^2y + y^2z + z^2x\geq xyz+xyz+xyz$$
And a generalisation of part of this idea: Murihead's Inequality.
I want to find a proof (by which I mean not particularly formal derivation) of AM-GM, that adequately captures 'mixed terms being smaller on average' as being the reason for the inequality. I think this would satisfy some innate and inexplicable urge.
There may not exist a satisfying answer to this question (though I would certainly no one if I saw it), but hopefully, I have communicated the kind of thing I'm looking for.

REPLY [3 votes]: I think there is a proof satisfying your intuition. I will talk just about the common AM-GM for three numbers $x^3+y^3+z^3\geq 3xyz$ but you can straightforwardly generalize the idea for more variables.
Very intuitive rearrangement inequality can be generalized for three ordered sequences of nonnegative numbers. For $a_1 \geq a_2 \geq a_3\geq 0$ and $b_1 \geq b_2 \geq b_3\geq 0$ and $c_1 \geq c_2 \geq c_3\geq 0$ the generalized rearrangement inequality states that
$$a_1b_1c_1 + a_2b_2c_2 + a_3b_3c_3 \geq a_{\sigma(1)}b_{\tau(1)}c_1 + a_{\sigma(2)}b_{\tau(2)}c_2 + a_{\sigma(3)}b_{\tau(3)}c_3$$
for any permatutions $\sigma, \tau:\{1,2,3\}\rightarrow\{1,2,3\}$.
Since the AM-GM inequality is symmetric, we may assume that $x\geq y\geq z$. Then AM-GM is an immediate consequence of the generalized rearrangement inequality for three ordered sequences $x\geq y\geq z\geq 0$, $x\geq y\geq z\geq 0$, $x\geq y\geq z\geq 0$ since on the LHS of AM-GM we have the greatest possible sum of products and on the RHS we have some "mixture" of products.

You may wonder how to (intuitively) prove rearrangement inequality and its generalization. If you read the paragraph called Intuition on Wikipedia, it should be obvious why rearrangement inequality holds for two sequences. Nevertheless I will present slightly unusual proof with something I call adding level by level (someone told me it is actually Abel's summation but I admit I was lazy to check it).
I will prove the following inequality (all numbers are nonnegative)
$$\color{red}{a_1b_1} + \color{fuchsia}{a_2b_2} + \color{maroon}{a_3b_3} \geq \color{blue}{a_1b_2} + \color{lime}{a_2b_3} + \color{green}{a_3b_1}$$
and I believe you will understand how it works for any other permutation. I start with $b_3$-level (the first line which is actually equality), the second line is $b_2$-level and the third is $b_1$-level.
\begin{align}
\color{red}{a_1b_3}\phantom{-x_y)} + \color{fuchsia}{a_2b_3}\phantom{-x_y)} + \color{maroon}{a_3b_3} &\geq \color{blue}{a_1b_3}\phantom{-x_y)} +\color{lime}{a_2b_3}  +\color{green}{a_3b_3} \\
\color{red}{a_1(b_2-b_3)} + \color{fuchsia}{a_2(b_2-b_3)} \phantom{+x_yx_y}  &\geq \color{blue}{a_1(b_2-b_3)}\phantom{+x_yx_y} +\color{green}{a_3(b_2-b_3)} \\
\color{red}{a_1(b_1-b_2)}\phantom{+ a_2(b_2-b_3) +x_yx_y}   &\geq \phantom{a_1(b_2-b_3) + x_yx_y+} \color{green}{a_3(b_1-b_2)} \\
\end{align}
Just observe that we get the desired inequality after summing all these three "levels".
The idea for three sequences is exactly the same! Assume we want to prove this inequality
$$\color{red}{a_1b_1c_1} + \color{fuchsia}{a_2b_2c_2} + \color{maroon}{a_3b_3c_3} \geq \color{blue}{a_1b_2c_3} + \color{lime}{a_2b_3c_1} + \color{green}{a_3b_1c_2}.$$
Again observe that it is just a sum of these three inequalities that follow from the rearrangement inequality for two sequences:
\begin{align}
\color{red}{a_1b_1c_3}\phantom{-x_y)} + \color{fuchsia}{a_2b_2c_3}\phantom{-x_y)} + \color{maroon}{a_3b_3c_3} &\geq \color{blue}{a_1b_2c_3}\phantom{-x_y)} +\color{lime}{a_2b_3c_3}\phantom{x}  +\color{green}{a_3b_1c_3} \\
\color{red}{a_1b_1(c_2-c_3)} + \color{fuchsia}{a_2b_2(c_2-c_3)} \phantom{+x_yx_yx_y}  &\geq \phantom{x_yx_y++}  \color{lime}{a_2b_3(c_2-c_3)} + \color{green}{a_3b_1(c_2-c_3)} \\
\color{red}{a_1b_1(c_1-c_2)}\phantom{+ a_2b_2(c_2-c_3) +x_yx_yx_y}   &\geq \phantom{x_yx_y++} \color{lime}{a_2b_3(c_1-c_2)}\\
\end{align}


TITLE: How is the total mass of Andromeda determined?
QUESTION [0 upvotes]: I am currently attempting to use Doppler shifts on both the advancing and receding sides of the galaxy to determine the mass. I am not sure exactly how I would convert the advancement and recession velocities to tangential velocities, which I need to determine the mass contained within the galaxy.
Also, how would I account for the dark matter halo? As far as I understand, the Doppler shift method only works for the farthest visible sides of the galaxy.

REPLY [2 votes]: You don't need to worry about any spherically symmetric distribution of matter, dark or otherwise, that lies outside the radii you have measurements for. It has no effect (see the shell theorem).
Your velocities tell you something about the mass interior to your velocity measurements.
Unless you have lots of measurements as a function of radius, there's not much you can do without making simplifying assumptions. I'll assume you don't.
In which case what you could do is take two measurements on opposite sides and take half the difference as the velocity at their average radius. You need to correct this by dividing by the sine of the estimated inclination of the disc to the line of sight (where 90 degrees would be seeing the disc "edge on"). You also need to correct your measurement by dividing by the sine of an azimuth angle of the location of your velocity measurement in the disc (where 90 or 270 degrees would be an object at the Galaxy "limb").
Then, assuming that the mass interior to your radius, $M_r$ is either symmetrically distributed or centrally concentrated, you can use the shell theorem again to argue that
$$\frac{v^2}{r} = \frac{GM_r}{r^2}$$
NB. This approximation may be quite poor if a large fraction of the mass is in a disk.
If this is too simple a treatment, then you need to indicate exactly what data you have and what way you wish to model the mass distribution.


TITLE: Integrating $ x \phi(x, \mu, \sigma) $
QUESTION [0 upvotes]: I am trying to compute the following integral:
$$\int x \cdot \phi (x) dx $$
... where $\phi(x)$ is the normal density function with mean $\mu$ and stdev $\sigma$.
Owen (1980), see https://en.wikipedia.org/wiki/List_of_integrals_of_Gaussian_functions, states the solution as:
$$\int x \phi(a+bx) dx = - \frac{1}{b^2}(\phi(a+bx)+a \Phi(a+bx)) + C $$
In order to test out the solution, I have tried to compute the integral for a couple of example values using the built in integrate function in R. One example is:
mu <- 10
sigma <- 5
x <- 15

tst <- function(x, mu, sigma){
  return(x * dnorm(x, mu, sigma))
}

integrate(tst, mu = mu, sigma = sigma, lower = -Inf, upper = x, abs.tol = 0)$value
> 7.203594

... where the result of roughly 7.2 makes intuitively good sense. However, when I use Owen's function, I get a completely different result. Again, in R (where I have assumed that $a = \mu$ and $b = \sigma$):
-(1/sigma^2) * (dnorm(mu+sigma*x)+mu*pnorm(mu+sigma*x))
> -0.4

I am probably missing something or using Owen's integral improperly since I would expect to get the same results using the integrate function in R and Owen's formula. Can you help me identify why the two are not equal?

REPLY [0 votes]: We can first transform the integral to evaluate the expression easier,
\begin{equation}
I=\int \phi(a+bx)xdx=\int \phi(y)\frac{y-a}{b}dx\\
\end{equation}
Using the fact that $x\phi(x)=-\phi'(x)dx$
\begin{eqnarray}
I=\frac{1}{b}\int \phi(y)\frac{y-a}{b}dy=\frac{1}{b^2}\int \phi(y)ydx-\frac{a}{b^2}\int \phi(y)ydx\\=\frac{-\phi(y)}{b^2}-\frac{a}{b^2}\Phi(y)=\frac{-\phi(a+bx)}{b^2}-\frac{a}{b^2}\Phi(a+bx)+C
\end{eqnarray}
which is Owen's result.
When do a numerical test sticking to the original integral parametrization we see that they coincide in first having full $\infty$ boundaries and a truncated boundary, $x=-2$. Actually when we evaluate Owen's integral at $\infty$ and $a=10,b=5$ we see the result $\frac{-a}{b^2}$ which is indeed -0.4.
fx<-function(x,a,b){x*dnorm(a+b*x)}
 b=5
 a=10
 integrate(fx,-Inf,Inf,a=a,b=b)
-0.4 with absolute error < 7.8e-07
 integrate(fx,-15,15,a=a,b=b)
-0.4 with absolute error < 3.3e-05
fI<-function(x,a,b){-1/b^2*(dnorm(a+b*x)+a*pnorm(a+b*x))}
fI(Inf,a,b)
[1] -0.4
integrate(fx,-Inf,-2,a=a,b=b)
-0.2159577 with absolute error < 5.8e-05
fI(-2,a,b)
[1] -0.2159577


TITLE: Complement of all finite subsets of $\mathbb{R}$
QUESTION [1 upvotes]: I was skimming through the topology book recommended through this question and I came across a question that I apparently solved incorrectly. It's question 2.5 on page 12.
Repeated here:

Let $X$ be $\mathbb{R}$, and let $\Omega$ consists [sic] of the empty set and complements of all finite subsets of $\mathbb{R}$. Is $\Omega$ a topological structure?

And my answer is "no", because I can posit some finite intersection of infinite sets and come up with a finite set, which is no longer in $\Omega$. For example, for some $a,b \in \mathbb{R}$ and $a<b$, two infinite sets are $X_1 = (-\infty,b)$ and $X_2 = (a,\infty)$ and obviously, $X_1, X_2 \in \Omega$ but $X_1 \cap X_2 \notin \Omega$.
But it seems it is a topology, because in the line following the question, they state:

The space of Problem 2.5 is denoted by $\mathbb{R}_{T_1}$ and called the line with $T_1$-topology.

So where am I going wrong?

REPLY [3 votes]: The problem with your argument is that not all infinite subsets of $\mathbb R$ are complements of finite sets. To see that we in fact have a topology, let $X_1$ and $X_2$ be finite subsets of $\mathbb R$ and observe that $(\mathbb R - X_1) \cap (\mathbb R - X_2) = \mathbb R - (X_1 \cup X_2)$ which is again the complement of a finite set. The union property should be obvious.


TITLE: Probability that 3 people meet at the same place
QUESTION [2 upvotes]: $3$ people arrive at a place separately at a random time between $9$ am and $10$ am. Each person remains there for $15$ minutes. What is the probability that all $3$ meet at the same time?

I understand how this works if it's only $2$ people. If $9:00$ is $0$ and $10:00$ is $1$, then it can be graphed $2d$. There would be a $x$ and a $y$ for each person. Since one person has to be within $\dfrac14$ of an hour to another person, then it can be said that $x- \dfrac14 < y < x + \dfrac14$.
Here is a visual representation of that work:

So then the probability would be $\dfrac7{16}$ for two people case
But then we turn to $3$ people. If I were to use a similar approach then I could say that
$$y - \dfrac14 < x < y + \dfrac14$$
$$z - \dfrac14 < x < z + \dfrac14$$
But then I don't know what this graph shape would look like? Or do I need more inequalities (two with $y$ in the middle, two with $z$ in the middle?) Bit confused from here.

REPLY [0 votes]: Using Geometry.

The combined grey areas cover the two person case.
and the area is  $1 - (\frac {3}{4})^2$
Now we need to exclude the cases when z is early.
If $x,y$ meet up at some time in the light grey area, then it is impossible for z to be early.  If $x,y$ meet up at some time in the dark grey area, then we have this pair of triangular prisms to exclude.
And there is also a pair of triangular prisms to exclude when z is late, and these excluded region are symmetrical.
Exclude 4 triangular prisms each $\frac 14 \times 34 \times  34$
$\frac 7{16} - (\frac 34)(\frac 14)(\frac 34)\cdot 2 = \frac 5{32}$
Using calculus:
WLOG assume $x\ge y\ge z$
$\frac {\int_9^{9.25}\int_{9}^x\int_{x}^z \ dy\ dz\ dx+\int_{9.25}^{10}\int_{x-0.25}^x\int_{z}^x \ dy\ dz\ dx}{\int_8^9\int_{8}^x\int_{z}^x \ dy\ dz\ dx}$


TITLE: Showing $\int_0^\infty \frac{x\sin(2x)}{x^2+3}dx=\frac{\pi}{2}e^{-2\sqrt3}$
QUESTION [2 upvotes]: Things are about to get real, prepare your mind!

$$f(z)=\frac{ze^{2iz}}{z^2+3}$$
Now this has two singular points, $\pm\sqrt3i$
$$f(z)=\frac{ze^{2iz}}{(z-(-\sqrt3i))(z-\sqrt3i)}$$ and hence both poles are of order $1$.
We can find $\operatorname{Res}_{z=\sqrt3i}f(z)=\frac{(\sqrt3i)e^{2i(\sqrt3i)}}{(\sqrt3i)+\sqrt3i}=e^{-2\sqrt3}/2$
Now I want to show that:
$$\int_0^\infty \frac{x\sin(2x)}{x^2+3}dx=\frac{\pi}{2}e^{-2\sqrt3}$$
Apparently this is pretty much immediate from my work above it, how?

REPLY [2 votes]: We first note that the integrand $\frac{x\sin 2x}{x^2+3}$ of the integral of interest $I$ is an even function around $x=0$.  Therefore we can write
$$I=\frac12 \int_{-\infty}^{\infty}\frac{x\sin 2x}{x^2+3}$$
Next, let $C_R$ be a closed contour that is comprised of (i) that part of the real axis $|x|\le R$ and (ii) a semi-circle in the upper-half plane such that $|z|=R$ and $\text{Im}(z)>0$.  We then note that
$$\lim_{R\to \infty}\text{Im}\left(\oint_{C_R} \frac{z\,e^{ i2z}}{z^2+3}dz\right)=\int_{-\infty}^{\infty}\frac{x\sin 2x}{x^2+3}dx+\lim_{R\to \infty}\int_{|z|=R,\text{Im}(z)>0}\frac{z\,e^{ i2z}}{z^2+3}dz \tag 1$$
We can easily see that the limit of last integral on the right-hand side of $(1)$  vanishes by using Jordan's Lemma.
Therefore, we have that

$$\begin{align}
\bbox[5px,border:2px solid #C0A000]{I=\frac12\text{Im}\left(2\pi i\text{Res}\left(\frac{z\,e^{ i2z}}{z^2+3},z=i\sqrt{3}\right)\right)=\frac{\pi}{2}e^{-2\sqrt{3}}}
\end{align}$$


TITLE: Is it possible to solve a differential equation without analytical term?
QUESTION [0 upvotes]: I have been confused by this differential equation for several days. I have an equation like this:
$$my''(t) = cV'(y(t)) - n y'(t)$$
Where $m,c,n$ are known parameters. And $V(y(t))$ is a two dimensional coordinate data set $(y(t_1),V_1),(y(t_2),V_2),...,(y(t_i),V_i)$ and I don't know the exact analytical expression of $V(t)$. So, is it possible to solve this differential equation? By what kind of method or algorithm?
I know maybe I could solve this equation by fitting the data set $V(y(t))$, and obtain an analytical expression, then substitute it into the equation and solve it. But that maybe inaccurate and sometimes it is difficult to fit the curve. So is it possible to solve this kind of equation that contains a term without analytical expression in other methods?

REPLY [0 votes]: I will give you idea on how to solve it numerically:
First, there are 3 different options:

Finite-Difference method
Finite-Element method
Finite-Volume  method

2 and 3 are actually the same but they are so much more complicated than finite-differences and that should work fine.
I will  make an example with an easier equation (heat equation):
$$\dfrac{\partial y}{\partial t} =\lambda \cdot \dfrac{\partial^2 y}{\partial x^2} $$
First, let's define the space on which we want to solve our equation. In this case, you should think about a two dimensional grid. x on the first axis and t on the second axis. (y is not an axis, its the value at a given point).
This grid will be divided into nodes and we solve our equations for these nodes. E.G. the derivative at a node can be calculated using the values of the nodes around it. Depending on the taylor expansion, your terms might look different and have different advantages but I wont talk about that here because that's to complicated.
Furthermore, we want a grid with constant distances on the x and t direction ($\Delta t$, $\Delta x$)
Then, we will write down a taylor expansion for every term that we cannot express (we dont want to have any derivatives). Furthermore, the indices look like this:
$$y^n_{i,k,..}$$ $n$ is the time index, and $i,k$ are the spacial indices. (We only need i here, because we only got x).
Furtermore, we will solve our equation for one timestep and once we solved that, we can solve for the next time step (we dont want to solve all the time_steps at the same time).
We will end up with something like this (evaluated at $n+1$ and $i$)
$$\dfrac{\partial y}{\partial t}\Big|_{n+1, i} \approx \dfrac{y^{n+1}_i-y^n_i}{\Delta t}$$
$$\dfrac{\partial^2 y}{\partial x^2}\Big|_{n+1, i} \approx \dfrac{y^{n+1}_{i+1}-2y^{n+1}_{i}+y^{n+1}_{i-1}}{\Delta x}$$
Plugging this back into our initial equation:
$$\dfrac{y^{n+1}_i-y^n_i}{\Delta t} = \lambda \dfrac{y^{n+1}_{i+1}-2y^{n+1}_{i}+y^{n+1}_{i-1}}{\Delta x}$$
We want to write every unknown on the LHS(Assuming that we solved for timestep n and want to solve for timestep n+1):
$$\dfrac{y^{n+1}_i-y^n_i}{\Delta t} = \lambda \dfrac{y^{n+1}_{i+1}-2y^{n+1}_{i}+y^{n+1}_{i-1}}{\Delta x}$$
$$\Rightarrow y^{n+1}_i-y^n_i = \dfrac{\lambda \Delta t}{\Delta x} \left( y^{n+1}_{i+1}-2y^{n+1}_{i}+y^{n+1}_{i-1} \right)$$
$$\Rightarrow y^{n+1}_i \cdot \left( 1 + 2 \dfrac{\lambda \Delta t}{\Delta x}\right)  - \dfrac{\lambda \Delta t}{\Delta x}y^{n+1}_{i+1}-\dfrac{\lambda \Delta t}{\Delta x}y^{n+1}_{i-1}=  y^n_i$$
This equation is really nice because its linear and that's all we need to have at the end. It is an equation that brings together the nodes at the next timesteps and values that we calculated at before. Let's say you want to solve for timestep 2 (and you know values for timestep 1 (also known as initial conditions)) you can write that down for all your nodes in your grid for timestep 2:
Now there is still a problem, because we cannot actually use that equation for ALL our nodes because it takes into account the node left and right of a node. If we talk about boundary nodes, this would not work.
So e.g. We take a grid with 4 nodes in x direction (labeled 1 to 4), and want to solve for timestep 2:
$$\Rightarrow y^{2}_2 \cdot \left( 1 + 2 \dfrac{\lambda \Delta t}{\Delta x}\right)  - \dfrac{\lambda \Delta t}{\Delta x}y^{2}_{3}-\dfrac{\lambda \Delta t}{\Delta x}y^{2}_{1}=  y^1_2$$
$$\Rightarrow y^{2}_3 \cdot \left( 1 + 2 \dfrac{\lambda \Delta t}{\Delta x}\right)  - \dfrac{\lambda \Delta t}{\Delta x}y^{2}_{4}-\dfrac{\lambda \Delta t}{\Delta x}y^{2}_{3}=  y^1_3$$
Our boundary conditions might be: $$y^n_1 = 1; y^n_4 = 3$$.
You can write this into a matrix and solve it.
To summarize, this is the easiest method but also pretty complicated and I need to advise against it for your problem because it will get really complicated with the taylor expansion but I think that's the way to go.


TITLE: What will be quantity of wine at the end in the vessel?
QUESTION [1 upvotes]: There are two vessels containing 12 liter of water the other contains 6 liters of wine respectively.One liter of mixture was taken out from each vessel and poured into the other and if this process is to be done four times. Find how much liters of wine will be at the end in second vessel?

REPLY [1 votes]: Hint:
We know that the two vessels will always have respectively 12L and 6L of mixture. Let $v_i(n)$ the percentage of wine in vessel $i$ after $n$ times.
We have $v_1(0)=0$ and $v_2(0)=1$.
And the to get the percentage of wine we will have after the next time, here is how I'd go (here is for the first vessel, it goes the same for the second):
Since I remove one liter and replace it with one liter from the second vessel, the concentration of wine will be 11/12 what I had before in the first vessel, plus 1/12 what I have in the second one.
This is indeed how concentration works when you mix two liquids that don't react with each other (we suppose they don't of course).
Finally you can come up with a formula:
$$
\begin{cases}
v_1(n+1) = \frac{11}{12}v_1(n)+\frac1{12}v_2(n)\\
v_2(n+1) = \frac56v_2(n)+\frac16v_1(n)\\
\end{cases}
$$
You compute the values one after the other from here, or try and calculate an explicit forumla. This is not so easy, but this answer could set you on the way.

Solution:
(hover the mouse to temporarily reveal spoilers, click on them to display them permanently)
Diagonalizing the matrix describing our sequences as in the linked answer, I get:

Eigenvalues $1$ and $\frac34$, eigenvectors $\begin{bmatrix}1 \\1\end{bmatrix}$ and $\begin{bmatrix}1 \\-2\end{bmatrix}$

Thus the following equality:

$$\begin{bmatrix}\frac{11}{12} & \frac1{12}\\\frac16 & \frac56\end{bmatrix} = \frac13 \begin{bmatrix}1 & 1\\1 & -2\end{bmatrix} \begin{bmatrix}1 & 0\\0 & \frac34\end{bmatrix} \begin{bmatrix}2 & 1\\1 & -1\end{bmatrix}$$

Putting that to the $n$th, we finally have:

$$\forall n,\begin{cases}v_1(n) = \frac13 \left(1-\left(\frac34\right)^n\right)\\v_2(n) = \frac13 \left(1+2\left(\frac34\right)^n\right)\end{cases}$$

And for your special case,

$v_1(4) = \frac{175}{768} \approx 22.8\%$ and $v_2(4) = \frac{209}{384} \approx 55.4\%$

This means that the answer to your very problem is :

There are $3.27$ liters of wine ($209/384$ of $6$ liters) contained in the mixture of the second vessel after the fourth trade.

To be sure that your answer is correct, compute the amount of wine in each vessel, and add them: it should always be $6$, since we didn't add or remove wine.


TITLE: $L_1\subset L_p$?
QUESTION [1 upvotes]: I am trying to check whether the implication $\forall p>1\quad f\in L_p(X,\mu)\Rightarrow f\in L_1(X,\mu)$ is true when $\mu(X)<\infty$. By $L_p(X,\mu)$ I mean the space of Lebesgue integrable functions $f:X\to\mathbb{C}$ with respect to measure $\mu$.
That holds for $p=2$ thanks to the inequality $|f(x)|\leq\frac{1}{2}(|f(x)|^2+1)|$ and the fact that constant functions are Lebesgue-integrable on a domain of finite measure. Does it hold for any $p>1$? If it does, how can it be proved? $\infty$ thanks!
To specify the details of where this question has come to my mind, I add that I am trying to prove the completeness of normed space $L_p(X,\mu)$, where functions are identified as a class of equivalence when they are equal almost everywhere and $\|f\|:=(\int_X |f|^p d\mu)^{1/p}$, by trying to adapt this proof for the case $p=2$.

REPLY [2 votes]: Hint: use the Hölder's inequality with $f$ and $x\to 1$.


text:
\section{Descent and reductions}\label{sec:descent}
The goal in this section is to prove the main theorem of this article. Given $h \geq 1$ and $\mathcal L \in F$ we write $V_{h+1,\mathcal L}$ for the unique two-dimensional representation of $G_{\mathbb Q_p}$ such that $D_{\st}^{\ast}(V_{h+1,\mathcal L}) \cong D_{h+1,\mathcal L}$ where $D_{h+1,\mathcal L}$ is as in Section \ref{subsec:filtered-phiN}. Write $\overline V$ for the semi-simple reduction modulo $\mathfrak m_F$ of $V$. Let $\mathbb Q_{p^2}$ be the unramified quadratic extension of $\Qp$, $\chi$ the unramified quadratic character of $G_{\mathbb Q_{p^2}}$, and $\omega_2$ a niveau 2 fundamental character of $\mathbb Q_{p^2}$. Note that $\operatorname{Ind}_{G_{\mathbb Q_{p^2}}}^{G_{\mathbb Q_p}}(\omega_2^{h}\chi)$ has determinant $\omega^h$, where $\omega$ is the cyclotomic character, and its restriction to inertia is $\omega_2^h \oplus \omega_2^{ph}$.

\begin{thm}\label{thm:main-thm-in-text}
Assume $h \geq 2$, $p \neq 2$, and if $p = 3$ then $h \geq 3$. Then, if $\mathcal L$ satisfies
$$
v_p(\mathcal L^{-1}) > {h-1\over 2} - 1 + v_p((h-1)!),
$$
then $\overline V_{h+1,\mathcal L} \cong \operatorname{Ind}_{G_{\mathbb Q_{p^2}}}^{G_{\mathbb Q_p}}(\omega_2^{h}\chi)$.
\end{thm}

\begin{rmk}\label{eqn:small-p}
Our contribution toward Theorem \ref{thm:main-thm-in-text} is limited to $h \geq 4$ and $p = h = 3$. We could also include $p=3$ and $h=2$ if we weaken the bound to $v_p(\mathcal L^{-1}) > {h-1\over 2} + v_p((h-1)!)$. In fact, if we use the weaker bound then we do not need to give references to prior work, either. The proof and further details are given in Section \ref{subsec:theorem-proof} below.
\end{rmk}

We plan to take the matrix of $\varphi$ acting on $\mathcal M_{h+1,\mathcal L}=\underline{\mathcal M}_{\mathcal O_F}(D_{h+1,\mathcal L}) \otimes_{\mathcal O_F} S_F$ as in Theorem \ref{thm:initialSF-basis} and replace it with a $\varphi$-conjugate defined over $\mathfrak S_{\Lambda}$ when $v_p(\mathcal L^{-1})$ satisfies the bound in theorem. This defines a Kisin module $\mathfrak M$ for $V_{h+1,\mathcal L}$ that allows us to calculate the reduction $\overline{V}_{h+1,\mathcal L}$. Throughout, we assume without further comment that:
\begin{align}\label{eqn:blanket-assumptions}
p &\neq 2 \text{ and } h \geq 2; \\
v_p(\mathcal L^{-1}) &> {h-1\over 2} - 1 + v_p((h-1)!).\nonumber
\end{align}
We will clarify result-by-result where we need to limit to $h\geq 3$ or $h \geq 4$. Also, fix $z = \sum x_j E^j$ as in Theorem \ref{thm:initialSF-basis}. Note that by \eqref{eqn:blanket-assumptions} $v_p(\mathcal L^{-1}) \geq -1$ so the estimates \eqref{eqn:xj-estimate} in Theorem \ref{thm:initialSF-basis} hold.

\subsection{Preparing for descent}\label{subsec:Rm-basis}
Consider the ring
$$
R_2 = \{ f = \sum a_i u^i \in F[\![u]\!] \mid i + 2v_p(a_i) \rightarrow \infty \text{ as $i \rightarrow \infty$}\}.
$$
Thus $R_2$ is the $F$-Banach algebra of series converging on $|u| \leq p^{-1/2}$. We equip $R_2$ with the valuation $v_{R_2}(\sum a_i u^i) = \inf_i \{i + 2v_p(a_i)\}$. The canonical map $\mathcal O_F \hookrightarrow R_2$ factors through $S_F$ since $v_{R_2}(E^p/p) = p - 2 > 0$.  Finally, given $v \in \R$, we define additive subgroups $H_v^{\circ} \subseteq H_v \subseteq R_2$ by
\begin{align*}
H_v &= \{ f \in R_2 \mid v_{R_2}(f) \geq v\}; &
H_v^{\circ} &= \{ f \in R_2 \mid v_{R_2}(f) > v\}.
\end{align*}
Our first lemma, concerning some entries of the matrix in Theorem \ref{thm:initialSF-basis}, is straightforward so we omit the proof (compare with \cite[Lemma 5.1.1]{BergdallLevin-BLZ}).

\begin{lemma}\label{lemma:lambda-assertion}
Let $\lambda_- = \prod_{n\geq 0} \varphi^{2n+1}(E)/p$ and $\lambda_{++} = \varphi(\lambda_-)$ be as in Theorem \ref{thm:initialSF-basis}. Then,
\begin{enumerate}[label=(\alph*)]
\item $\lambda_- \in 1 + H_{p-2}$ and $\lambda_{++} \in 1 + H_{p^2-2}$;\label{enum:lambda-assertion:1plus}
\item $\lambda_-, \lambda_{++}\in R_2^\times$;\label{enum:lambda-assertion:unit}
\item $v_{R_2}(\lambda_-^{\pm 1}) = 0 = v_{R_2}(\lambda_{++}^{\pm 1})$.\label{enum:lambda-assertion:valuation}
\end{enumerate}
\end{lemma}
We also prepare estimates for $z$. Note that by \eqref{eqn:blanket-assumptions} the estimate \eqref{eqn:xj-estimate} becomes
\begin{equation}\label{eqn:xs-assumption}
v_p(x_j) > v_p((h-1)!) - v_p(j!) - j - 1 \geq - j - 1.
\end{equation}
Recall, we write $a_p = \varpi^{h-1} + \varpi^{h+1}$. Thus, $v_{p}(a_p) = \frac{h-1}{2}$.
\begin{lemma}\label{lemma:z-estimates}
For $z = \sum_{j=1}^{h-1} x_j E^j$ as above, and $\nu = -1 + \varphi(z)(a_p - p^h z)$, we have
\begin{multicols}{2}
\begin{enumerate}[label=(\alph*)]
\item $p^h z \in H_{h-1}^{\circ}$;\label{enum:z-estimates:p^hz}
\item $\varphi(z) \in H_{-2}^{\circ}$;\label{enum:z-estimates:phiz}
\item $\nu \in -1 + H_{h-3}^{\circ}$;\label{enum:z-estimates:nu}
\item If $h \geq 3$, then $\nu \in R_2^\times$.\label{enum:z-estimates:nu-unit}
\end{enumerate}
\end{multicols}
\noindent Furthermore, if $p =3$ and $h =3$, then $\phz(z) \in H^{\circ}_{-1}$ and $\nu \in -1 + H_{h-2}^{\circ} = -1 + H^{\circ}_{1}$.
\end{lemma}
\begin{proof}
First, $v_{R_2}(E^j) = j$. By the ultrametric inequality and \eqref{eqn:xs-assumption}, we see
$$
v_{R_2}(z) > \inf\{2(-j-1) + j \mid 1 \leq j \leq h-1\} = -1-h.
$$
Part \ref{enum:z-estimates:p^hz} follows because $v_{R_2}(p^h) = 2h$. For \ref{enum:z-estimates:phiz},  note $v_{R_2}(\varphi(E)^j) = 2j$. Thus, using \eqref{eqn:xs-assumption},
\begin{equation*}
v_{R_2}(\varphi(z)) > \inf \{2(-j-1) + 2j \mid 1 \leq j \leq h-1 \} = -2.
\end{equation*}
Continuing, $\varphi(z)p^h z \in H_{h-3}^{\circ}$ by parts \ref{enum:z-estimates:p^hz} and \ref{enum:z-estimates:phiz} and, since $v_{R_2}(a_p) = h-1$, we have  $\varphi(z)a_p \in H_{h-3}^{\circ}$. This proves \ref{enum:z-estimates:nu}. Finally, part \ref{enum:z-estimates:nu-unit} follows from the geometric series and part \ref{enum:z-estimates:nu}.

Finally, suppose $p = h = 3$. By the argument for (c) above, it suffices to show $\varphi(z) \in H_{-1}^{\circ}$. We note $v_{R_2}(\varphi(E)^j - E(0)^j) \geq p + 2j - 2$ for any $j$. Thus, by \eqref{eqn:xs-assumption}
\begin{equation}\label{eqn:phi-diff}
v_{R_2}(\phz(z) - \varphi(z)(0))>  p + 2j - 2  - 2( j +1)> p - 4 = -1.
\end{equation}
But, by Example \ref{exam:rmk-p=3-h=3-L} we have $\varphi(z)(0) = z(0) = \frac{-1}{2\mathcal L}\left(\frac{1}{\mathcal L} + 1\right)$. Since $v_p(\mathcal L^{-1}) > 0$, \eqref{eqn:phi-diff} then implies so $v_{R_2}(\varphi(z)) > -1$ as we wanted.
\end{proof}

We now write $\mathcal M_2 = \mathcal M_{h+1,\mathcal L} \otimes_{S_F} R_2 \cong \underline{\mathcal M}_{\mathcal O_F}(D_{h+1,\mathcal L}) \otimes_{\mathcal O_F} R_2$. Thus, $\mathcal M_2 \in \Mod_{R_2}^{\varphi,\leq h}$. We also introduce some notation. Given $A \in \Mat_d(R_2)$ and $C \in \GL_d(R_2)$ we write $
C \ast_\varphi A = C\cdot A \cdot \varphi(C)^{-1}$. Thus, if $\{e_1,e_2\}$ is a basis of $\mathcal M_2$ and $A$ is the matrix of $\varphi_{\mathcal M_2}$ in that basis, then $C\ast_\varphi A$ is the matrix of $\varphi_{\mathcal M_2}$ in the basis $\{e_1',e_2'\}$ is given by $(e_1',e_2') = (e_1,e_2)C^{-1}$.

\begin{prop}\label{prop:descent-almostready} Assume $h \geq 4$ or $p=h=3$.   Then, there exists a basis of $\mathcal M_2$ in which the matrix of $\varphi_{\mathcal M_2}$ is $\begin{smallpmatrix} G & -1 \\ E^h & 0 \end{smallpmatrix}$, where $
G \in (a_p - p^h z)\left({\lambda_-\over \lambda_{++}}\right)^h + H_h^{\circ}$.
\end{prop}
\begin{proof}
By Theorem \ref{thm:initialSF-basis}, there is a basis $\{e_1,e_2\}$ of $\mathcal M_2$ such that $\varphi_{\mathcal M_2}((e_1,e_2)) = (e_1,e_2)A$, where
\begin{equation*}
A = \begin{pmatrix} (a_p - p^h z)\left({\lambda_-\over \lambda_{++}}\right)^h & -1 + \varphi(z)(a_p - p^h z) \\ E^h & E^h \varphi(z) \left({\lambda_{++}\over \lambda_-}\right)^h \end{pmatrix} = \begin{pmatrix} \mu & \nu \\ E^h & \eta \end{pmatrix},
\end{equation*}
where $\nu$ is as in Lemma \ref{lemma:z-estimates} and $\mu$ and $\eta$ are defined by the equality. Assume for now just that $h \geq 3$. Then, by Lemma \ref{lemma:z-estimates}\ref{enum:z-estimates:nu-unit}, $\nu \in R_2^\times$. Making a change of basis on $\mathcal M_2$, we replace $A$ by (note that $\mu\eta = (1+\nu)E^h$)
\begin{equation*}
A' = \begin{pmatrix} 1 & 0\\ -\eta/\nu & 0 \end{pmatrix} \ast_\varphi A = \begin{pmatrix} \mu + {\nu\varphi(\eta)\over \varphi(\nu)} & \nu \\ -E^h \nu^{-1} & 0 \end{pmatrix}.
\end{equation*}
Since $v_{R_2}(\nu +1) > 0$ by Lemma \ref{lemma:z-estimates}\ref{enum:z-estimates:nu}, we have $\nu(0) \in \Lambda^\times$. Thus $\nu_0 = \nu/\nu(0) \in 1 + (H_{h-3}^{\circ} \cap uR_2)$. By \cite[Lemma 4.1.1]{BergdallLevin-BLZ}, we have $\varphi^k(\nu_0) \in 1 + H_{h-3+m_k}$ where $m_k \rightarrow \infty$ as $k \rightarrow \infty$. Thus, the infinite product $\nu_+ = \prod_{n \geq 0} \varphi^{2n}(\nu_0)$ converges in $R_2$. Set $\nu_- = \varphi(\nu_+)$, so $\nu_{\pm} \in 1 + H_{h-3}^{\circ} \subseteq R_2^\times$. We now change basis on $\mathcal M_2$ again to get a matrix $A''$ for $\varphi_{\mathcal M_2}$ given by
\begin{equation*}
A'' = \begin{pmatrix}  {-1\over \nu(0)} {\nu_- \over \nu_+} & 0 \\ 0 & {\nu_+ \over \nu_-}\end{pmatrix} \ast_{\varphi} A' = \begin{pmatrix} G & -1 \\ E^h & 0 \end{pmatrix},
\end{equation*}
where
\begin{equation}\label{eqn:F-defn}
G = \left(\mu + {\nu\varphi(\eta)\over \varphi(\nu)}\right) {\nu_-^2 \over \nu_+ \nu_{++}}
\end{equation}
and $\nu_{++} = \varphi(\nu_-)$.

To complete the argument, we justify $G \in \mu + H_h^{\circ}$. We already know $\nu_-^2/\nu_+\nu_{++} \in 1 + H_{h-3}^{\circ}$. The same is true for $\nu/\varphi(\nu)$. So,
\begin{equation}\label{eqn:nuphi(eta)/phi(nu)}
v_{R_2}\left({\nu\varphi(\eta)\over \varphi(\nu)}\right) \geq v_{R_2}(\varphi(\eta)) \geq v_{R_2}(\varphi(E)^h\varphi^2(z)),
\end{equation}
where we used Lemma \ref{lemma:lambda-assertion} to remove $\lambda_-$ and $\lambda_{++}$ from the estimate. We note $v_{R_2}(\varphi(E)^h) = 2h$ and $v_{R_2}(\varphi^2(z)) \geq v_{R_2}(\varphi(z)) > -2$, by Lemma \ref{lemma:z-estimates}\ref{enum:z-estimates:phiz} (cf.\ \cite[Lemma 4.1.1]{BergdallLevin-BLZ}). Thus from \eqref{eqn:nuphi(eta)/phi(nu)} we deduce that $v_{R_2}(\nu\varphi(\eta)/\varphi(\nu)) > 2h - 2 = 2(h-1)$.  We also note that $a_p-p^hz \in H_{h-1}$. Thus, $\mu \in H_{h-1}$ and so, returning to the definition \eqref{eqn:F-defn} of $\mu$ and $G$, we see
\begin{equation*}
G \in \bigl(\mu + H_{2(h-1)}^{\circ}\bigr)\cdot (1 + H_{h-3}^{\circ}) \subseteq \mu + H_{2h-4}^{\circ}+  H^{\circ}_{2(h-1)} = \mu + H_{2h-4}^{\circ}.
\end{equation*}

Now, if $h \geq 4$, then $2h-4 \geq h$ and so $G \in \mu + H_h^{\circ}$.  This completes the proof except if $p = h = 3$. In that case, Lemma \ref{lemma:z-estimates} shows $\nu \in -1 + H^{\circ}_1$, rather than $-1 + H_0^{\circ}$, from which we deduce
\begin{equation*}
G \in \bigl(\mu + H_{4}^{\circ}\bigr)\cdot (1 + H_{1}^{\circ}) \subseteq \mu + H_3^{\circ} = \mu + H_h^{\circ}
\end{equation*}
anyways. This completes the proof.
\end{proof}

\subsection{Descent}\label{subsec:descent}
To descend to $\mathfrak S_{\Lambda}$, we use the algorithm from \cite[Section 4]{BergdallLevin-BLZ}. Write $T_{\leq d} : R_2 \rightarrow F[u]$ for the ``truncation'' operation $T_{\leq d}(\sum a_i u^i) = \sum_{i\leq d} a_i u^i$ and $T_{>d}(f) = f - T_{\leq d}(f)$. In the next two proofs, we will use the following principle:\ if $f \in R_2$ and $v_{R_2}(T_{\leq d}(f)) > d$ then $f \in \mathfrak m_F[u]$.

\begin{prop}\label{prop:descent-prop}
Suppose that $G \in R_2$ such that
\begin{enumerate}[label=(\alph*)]
\item $G \in H_{h-1}$;\label{enum:descent-prop:g-ass}
\item $T_{>h}(G) \in H_{h-1}^{\circ}$;\label{enum:descent-prop:error-ass}
\item $T_{\leq h}(G) \in \mathfrak m_F[u]$.\label{enum:descent-prop:integral-ass}
\end{enumerate}
Then, given $A = \begin{smallpmatrix} G & -1 \\ E^h & 0 \end{smallpmatrix}$, there exists $C \in \GL_2(R_2)$  and $P \in \mathfrak m_F[u]$ such that $C \ast_{\varphi} A = \begin{smallpmatrix} P & -1 \\ E^h & 0 \end{smallpmatrix}$.
\end{prop}
\begin{proof}
Since $E^h \in u^h + H_{h+1}$, the assumption \ref{enum:descent-prop:g-ass} implies that
\begin{equation*}
A \in \begin{pmatrix} 0 & -1 \\ u^h & 0 \end{pmatrix} + \begin{pmatrix} H_{h-1} & 0 \\ H_{h + 1} & 0 \end{pmatrix}.
\end{equation*}
In the notation of \cite[Section 4.3]{BergdallLevin-BLZ}, set $a = 0$, $b = h$, $a' = {h \over 2} - {p-1 \over 2}$ and $b' = {h \over 2} + {p-1 \over 2}$, and $(c_0,c_h) = (-1,1)$. Since $h-1-a' = {h\over 2} - 1 + {p-1\over 2} \geq 1$, we see $A$ is $\gamma$-allowable with $\gamma = 1$ in the sense of \cite[Definition 4.3.1]{BergdallLevin-BLZ}. The error of $A$, in the same definition, is $\varepsilon = v_{R_2}(T_{>h}(G)) - a'$. By \cite[Theorem 4.3.7]{BergdallLevin-BLZ}, with $R=R_2$ in {\em loc.\ cit.}, there exists $C \in \GL_2(R_2)$ such that $C|_{u=0} = I$ and
\begin{equation*}
C \ast_{\varphi} A = \begin{pmatrix} P & -1 \\ f & 0 \end{pmatrix}
\end{equation*}
where $P$ and $f$ are polynomials of degree at most $h$ and $P$ satisfies, because of assumption \ref{enum:descent-prop:error-ass},
\begin{equation*}
v_{R_2}(P - T_{\leq h}(G)) \geq \varepsilon + a' + 1 =  v_{R_2}(T_{>h}(G)) +1 > h.
\end{equation*}
Since $P-T_{\leq h}(G)$ has degree at most $h$, we deduce $P-T_{\leq h}(G) \in \mathfrak m_F[u]$. By assumption \ref{enum:descent-prop:integral-ass}, we have $P\in \mathfrak m_F[u]$ as well. Finally, $f = \det(C\ast_{\varphi} A) = rE^h$ for some $r \in R_2^\times$. But, $f$ and $E^h$ are both polynomials of degree at most $h$ with the same constant term, since $C|_{u=0} = I$, and thus $f = E^h$.
\end{proof}
We now verify the $G$ from Proposition \ref{prop:descent-almostready} satisfies the hypothesis of Proposition \ref{prop:descent-prop}.
\begin{lemma}\label{lemma:Gintegral-properties}
Let $G \in (a_p - p^hz)\left({\lambda_-\over \lambda_{++}}\right)^h + H_{h}^{\circ}$. Then,
\begin{enumerate}[label=(\alph*)]
\item $G \in H_{h-1}$,\label{enum:Gintegral-prop:G}
\item $T_{>h}(G) \in H_{h-1}^{\circ}$, and\label{enum:Gintegral-prop:>hG}
\item $T_{\leq h}\left(G \right) \in \mathfrak m_F[u]$.\label{enum:Gintegral-prop:<hG}
\end{enumerate}
\end{lemma}
\begin{proof}
First, the conclusions depend only on $G \bmod H_h^{\circ}$, so we suppose $G = (a_p - p^h z)\left({\lambda_-\over \lambda_{++}}\right)^h$. Part \ref{enum:Gintegral-prop:G} follows from Lemmas \ref{lemma:lambda-assertion} and \ref{lemma:z-estimates}. For part \ref{enum:Gintegral-prop:>hG}, we first have, by Lemma \ref{lemma:lambda-assertion}\ref{enum:lambda-assertion:1plus}, that $a_p\left({\lambda_-\over \lambda_{++}}\right)^h \in a_p + a_p H_{p-2}$. So, $T_{>0}\big(a_p\left({\lambda_-\over \lambda_{++}}\right)^h\big) \in H_{h+p-3} \subseteq H_h$. On the other hand, by Lemma \ref{lemma:z-estimates}\ref{enum:z-estimates:p^hz} we have $p^hz \in H_{h-1}^{\circ}$. Thus we've shown in fact $T_{>0}(G) \in H_{h-1}^{\circ}$.

Finally, we consider part \ref{enum:Gintegral-prop:<hG}. Since $E = u+p$, any $f \in S_\Lambda$ can be written $f= \sum\limits_{n = 0}^\infty \alpha_n \frac{E ^ n}{p^{\lfloor \frac n p  \rfloor}}$ with $\alpha_n \in \Lambda$. Let $f = {\lambda_-\over \lambda_{++}} \in S_\Lambda$, in particular. Since $v_p (a_p) = \frac{h -1}{2} > \lfloor \frac h  p  \rfloor$ unless $ p = h =3$ (or $p=2$, which we have excluded in \eqref{eqn:blanket-assumptions}), we see immediately that $T_{\leq h} (a_p f^h) \in \fm_F[u]$ except when $h = p = 3$. When $h = p =3$, though, $f^h = f^p$ is of the form $\alpha'_0 + \sum\limits_{n = 1}^\infty p \alpha'_n  \frac{E ^ n}{p^{\lfloor \frac n p  \rfloor}}$ with $\alpha_i' \in \Lambda$; thus  $T_{\leq h} (a_p f^h) \in \fm_F[u]$ in every case.

It remains to show that $T_{\leq h} (p ^h z f^h ) \in \fm_F[u]$ as well. Since $z= \sum_{j < h} x_jE^j$, it suffices to show  $h + v_p (x_j) -  \lfloor \frac {n-j} p  \rfloor>  0$ for $n \leq h$ and $j < h$. This follows from \eqref{eqn:xs-assumption}.
\end{proof}

\subsection{Proof of Theorem \ref{thm:main-thm-in-text}}\label{subsec:theorem-proof}
Finally, we give the proof of the main theorem:
\begin{quote}
{\it
Assume that $h \geq 2$, $p \neq 2$, and if $p = 3$ then $h \geq 3$. Then, if $\mathcal L$ satisfies
$$
v_p(\mathcal L^{-1}) > {h-1\over 2} - 1 + v_p((h-1)!),
$$
then $\overline V_{h+1,\mathcal L} \cong \operatorname{Ind}_{G_{\mathbb Q_p^2}}^{G_{\mathbb Q_p}}(\omega_2^{h}\chi)$.
}
\end{quote}
\begin{proof}[Proof of Theorem \ref{thm:main-thm-in-text}]

First if $p \geq 5$ and $h < 4$ then $h < p-1$. Thus the result follows from \cite[Theorem 4.2.4.7(iii)]{BreuilMezard-Multiplicities} when $h$ is odd and \cite[Theorem 5.0.5]{GuerberoffPark-Semistable} when $h$ is even.

Now we assume that either $h \geq 4$ or $p = h = 3$. Then, applying Proposition \ref{prop:descent-almostready}, Lemma \ref{lemma:Gintegral-properties}, and Proposition \ref{prop:descent-prop}, we deduce that there exists a basis of $\mathcal M_2$ in which the matrix of $\varphi_{\mathcal M_2}$ is given by $A = \begin{smallpmatrix} P & - 1 \\ E^h & 0 \end{smallpmatrix}$ and $P \in \mathfrak m_F[u]$. Define $\mathfrak M = \mathfrak S_{\Lambda}^{\oplus 2}$ with the matrix of $\varphi$ being given by $A$. Clearly $\mathfrak M$ is a Kisin module over $\mathfrak S_{\Lambda}$ of height $\leq h$, and
$$
\mathfrak M \otimes_{\mathfrak S_{\Lambda}} R_2 \cong \mathcal M_2 = \underline{\mathcal M}_{\mathcal O_F}(D_{h+1,\mathcal L})\otimes_{\mathcal O_F} R_2
$$
as $\varphi$-modules over $R_2$. Thus, by Proposition \ref{prop:reductions-first} we deduce $\mathfrak M = \mathfrak M(T)$ for some lattice $T \subseteq V_{h+1,\mathcal L}$. Furthermore, $\mathfrak M \otimes_{\mathfrak S_{\Lambda}} \mathbb F[u^{-1}]$ is a $\varphi$-module over $\mathbb F(\!(u)\!)$ with Frobenius given by $\begin{smallpmatrix} 0 & - 1 \\ u^h & 0 \end{smallpmatrix}$. This shows, in particular, that $\overline V_{h+1,\mathcal L}$ is the same for any $\mathcal L$ satisfying \eqref{eqn:blanket-assumptions} (see \cite[Corollary 2.3.2]{BergdallLevin-BLZ}).

Let $V_{h+1,\infty}$ be as in the introduction. By \cite[Corollary 5.2.2]{BergdallLevin-BLZ}, for  $V_{h+1,\infty}$ there exists a Kisin module $\M'$  so that  $M':= \mathfrak M'  \otimes_{\mathfrak S_{\Lambda}} \mathbb F[u^{-1}]$ has Frobenius also given by $\begin{smallpmatrix} 0 & - 1 \\ u^h & 0 \end{smallpmatrix}$ and $M'$ determines $\overline V_{h+1,\infty} \cong \operatorname{Ind}_{G_{\mathbb Q_p^2}}^{G_{\mathbb Q_p}}(\omega_2^{h}\chi)$. Therefore, $\overline V_{h+1,\mathcal L} \cong \overline V_{h+1,\infty} \cong \operatorname{Ind}_{G_{\mathbb Q_p^2}}^{G_{\mathbb Q_p}}(\omega_2^{h}\chi)$.
\end{proof}

\begin{rmk}\label{remark:final-remark}
We return to Remark \ref{eqn:small-p}. Suppose we replace \eqref{eqn:blanket-assumptions} with
\begin{equation}\label{eqn:p3-assumptions}
v_p(\mathcal L^{-1}) > {h-1\over 2} + v_p((h-1)!).
\end{equation}
This has the impact of scaling $z$ by a $p$-adic unit multiple of $p$, thus increasing $v_{R_2}(z)$ by $2$ throughout our estimates in Section \ref{subsec:Rm-basis}. The reader may check that Proposition \ref{prop:descent-almostready} holds with these new estimates, and so the proof goes through for all $h \geq 2$ and $p \geq 3$ under the assumption \eqref{eqn:p3-assumptions}.
\end{rmk}

config: arxiv


TITLE: Need a mathematical name for a particular conversion.
QUESTION [0 upvotes]: I can convert 12 into 2 $\def\sf{\sqrt 3}\sf$
But I don't know the correct mathematical name for this conversion. can anyone tell me the exact mathematical name for this process and link me to the details of the same.?

REPLY [3 votes]: The operation that turns the number $12$ into $2\sqrt3$ is the square root. One writes $\sqrt{12}=2\sqrt3$.
I gather that you're asking how we get the formal expression $2\sqrt3$ from $\sqrt{12}$. In elementary algebra textbooks, $2\sqrt3$ is called the "simplest radical form" or the "simplified form" of the radical $\sqrt{12}$. There's a little more detail at Wikipedia.


TITLE: Find all solutions of $y^2 dx + (e^x - y)dy = 0$
QUESTION [1 upvotes]: I need to find all solutions of:
$y^2 dx + (e^x - y)dy = 0$
So I take:
$M = y^2$
$N = e^x - y$
And I see that:
$\frac{M}{dy} = 2y$
$\frac{N}{dx} = e^x$
I'm therefore looking for an integrating factor:
$$ \displaystyle e^{\int \frac{1}{y^2}(e^x - 2y) dy} = e^{\int \frac{e^x - 2y}{y^2}dy} = e^{-2ln(y) - \frac{e^x}{y}}$$
And at this point I think it might be too complicated and I probably do something wrong.

REPLY [0 votes]: Divide equation by $y$.
Then easy get that integrating factor of $y^2 dx + (e^x - y)dy = 0$ is
$$\mu=\frac{e^{-x}}{y}$$
and all solutions
$$\ln y-e^{-x}y=C$$


TITLE: Covariance of maxima
QUESTION [5 upvotes]: Let $X_1,\dots,X_n$ and $Y_1,\dots,Y_n$ be random variables. Is it true that
  \begin{align*}
\text{Cov}(\max_i X_i, \max_j Y_j) \leq \sum_{i, j}|\text{Cov}(X_i,Y_j)|
\end{align*}

My intuition as to why this may be true is from the previous question: variance of maximum
I believe the identity $2\text{Cov}(X,Y)=E(X-X')(Y-Y')$ should be relevant, where $(X',Y')$ is an independent copy of the couple $(X,Y)$, but I'm not sure how to proceed, since we can no longer write the expected value as an integral of the tail probability (since $(X-X')(Y-Y')$ is not necessarily positive).
If needed, we may also assume that all variables are positively correlated with each other, and that they are each marginally distributed as bernoulli coin flips.
But hopefully the more general statement above is true on its own.

REPLY [4 votes]: The general form of the conjecture is false, due to the (often forgotten?) fact that dependent variables can be uncorrelated (i.e. zero covariance).  This makes the RHS $=0$ and the rest is not difficult.  Here is a simple counter-example:

$n=2$ and $X_1, X_2$ i.i.d. uniform $\in \{-1, 0, 1\}$ with prob $\frac13$ for each choice.
$Y_i = |X_i|$
$E[X_i Y_i] = \frac13 ( -1 + 0 + 1) = 0 = E[X_i]E[Y_i]$ so $Cov(X_i, Y_i) = 0$ i.e. they are uncorrelated.  Thus RHS $=0$ because each term in the sum is either independent or uncorrelated.

Meanwhile these tables show $\max X_i$ (top table) and $\max Y_i$ (bottom table) for each of the $9$ equi-probable choices of $(X_1, X_2)$:
maxX  -1   0   1
 -1   -1   0   1
  0    0   0   1
  1    1   1   1

maxY  -1   0   1
 -1    1   1   1
  0    1   0   1
  1    1   1   1

From these tables:

$E[\max X_i] = \frac19 (5 - 1) = \frac49$
$E[\max Y_i] = \frac89$
$E[\max X_i \cdot \max Y_i] = E[\max X_i] = \frac49$
so $Cov(\max X_i, \max Y_i) = \frac49 - \frac49 \frac89 = \frac{4}{81} > 0 = $ RHS.

Further guess: I think adding "each variable positively correlated with each other" won't help, because a very small perturbation to above can (my guess) make the positive covariances on the RHS arbitrarily small, not enough to overwhelm the $\frac{4}{81}$ covariance on the LHS.

UPDATE: Here is a coin-based counter-example.  The trick we now exploit is that pairwise independence does not equal mutual independence.  And pairwise independence is sufficient to make RHS $= 0$.

$n = 2$ and $X_1, X_2$ i.i.d. uniform $\in \{0,1\}$ with prob $\frac12$ for each choice.
$Y_1 = Y_2 = \max Y_j = (X_1 \neq X_2)$, i.e. $(X_1 ~~~\text{xor}~~~ X_2)$
As is well known, $\forall (i,j) \in \{1,2\}^2: X_i, Y_j$ are (pairwise) independent, so $Cov(X_i, Y_j) = 0$ and RHS $=0$.

Here are all four possibilities:
X1  X2  maxX  Y1=Y2=maxY
 0   0    0           0
 0   1    1           1
 1   0    1           1
 1   1    1           0

$E[\max X_i] = \frac34$
$E[\max Y_i] = \frac12$
$E[\max X_i \cdot \max Y_i] = \frac12$
$Cov(\max X_i, \max Y_i) = \frac12 - \frac12 \frac34 = \frac18 > 0 =$ RHS.