TITLE: Why are computers much smaller than they are fast?
QUESTION [0 upvotes]: In any medium there is a finite propagation speed at which information can spread. Special relativity provides an upper bound to that propagation speed, namely the speed of light in vacuum. Signals of light in media, or electric signals in wires come close/are similar to that theoretical maximum.
There is no fundamental difference between "transporting" and "processing" information, e.g., just sending a "bit" trough a wire is not fundamentally different from doing the same while negating the bit after every millimeter of the wire looking like 0101010101. As a more non-trivial example, consider adding two (very large) numbers whose digits are placed at every millimeter along the wire. To add the digits at one place we need the carry from the previous (lower-digit) additions, so the carry is a signal running through the wire which is processed in a non-trivial way. The theoretical maximum speed for this addition should still be the speed of light.
Now, consider the transistors modern computers are made of. The size of a modern CMOS transistor is about 5nm. I don't know what exactly the propagation delay through one of these transistors is, but I think it's supposed to be in the picosecond range, say 10ps. Please correct the numbers if they are very far off. Now, if we would densely pack transistors along a line to "process" the information on the way instead of merely "transporting" the signal, the effective speed of the processing would be 5e-9m/10e-12s=500m/s, which is much slower than the speed of light, by a factor of ~1e6.
So why is sending a signal through a transistor so much slower than sending a signal through a piece of wire? Are there any information processing technologies (maybe something based on light etc.) which provide a "transistor" element where the speed through is comparable to the speed of light? Of course, the "transistor" element should keep the ~speed-of-light propagation speed when downsized to a size comparable to a CMOS transistor (otherwise one might argue a 5nm CMOS transistor itself has ~speed-of-light when we put it in a 1mm box, i.e., pretend its size is 1mm).
I believe that part of the reason might be that transistors are not optimized for that since this is not the bottleneck in common computer architectures. There, signals do not always go from one transistor to a neighboring one but often are sent through "long" pieces of wire in between. Then the propagation delay on those long pieces of wire due to the speed of light might be of the same order as the propagation delay through the individual transistors, such that optimizing the latter does not yield much benefit. However, if modern computers were cellular automata, improving the propagation delay through a single transistor would provide a huge potential for speedup (up to ~1e6). Are there any areas (maybe graphics cards/parallel computing) where improving propagation delay of single transistors plays an important practical role? Are there any developments in transistor technology targeted at improving the propagation delay rather than the size?

REPLY [1 votes]: In CMOS every input to a transistor or gate is driven by a precious CMOS output.
So in all cases you end up with a rather high source impedance R (say 10 kOhm) driving a capacitive input C (say 1 fF).
The charging/discharging process takes time roughly equal to RC (10 ps). And before this time has passed the output of the transistor/gate being charged does not yet change appreciably. Only after RC does the output transition and the next stage will begin being charged/discharged for a duration of RC.
All these charging processes introduce a lot of delay in which signals cannot propagate further.


TITLE: When does Scott topology generated by specialization order induced by a sober space (X,$\tau$) equal the initial topology $\tau$?
QUESTION [7 upvotes]: Let X be a $T_{0}$ space. The specialization order ≤ on X is that if x is contained in cl{y}, then we call "x≤y". Obviously (X,≤) is a partially ordered set.
A sober space is a topological space such that every irreducible closed subset of X is the closure of exactly one point of X: that is, this closed subset has a unique generic point. It is not difficult to  see that specialization order induced by a sober space is a directed complete partial order (dcpo).
A subset O of a dcpo P is called Scott-open if it is an upper set and if it is inaccessible by directed joins, i.e. if all directed sets D with supremum in O have non-empty intersection with O. The Scott-open subsets of a dcpo P form a topology on P, the Scott topology. (https://en.wikipedia.org/wiki/Scott_continuity)
Let (X,$\tau$) be a sober space. The specialization order ≤ induced by a sober space makes (X,≤) a dcpo. Generally, the Scott topology on (X,≤) is finer than $\tau$. But when do these two topologies coincide with each other?
Two examples (sober c space and sober locally finitely compact space)
are given in a book "Non-Hausdorff topology and Domain theory". (http://www.cambridge.org/ca/academic/subjects/mathematics/geometry-and-topology/non-hausdorff-topology-and-domain-theory-selected-topics-point-set-topology)
Can you find more examples like this? or Can you characterize this kind of sober space by simple property?

REPLY [4 votes]: Here is a partial answer. In P. T. Johnstone, "Stone Spaces", pp. 292, 294, it is shown that Scott topologies of continuous posets are precisely all completely distributive complete lattices. Recall that a poset $P$ is continuous iff it is a dcpo and moreover the join map $\operatorname{Ideals}(P)\to P$ has a left adjoint.
I don't know a nice characterization of spaces whose specialization posets are continuous, but for those the answer is thus that their topology coincides with the Scott topology of the specialization iff open sets form a completely distributive lattice. And moreover, any completely distributive complete lattice is the topology of a sober space which is at the same time the Scott topology of its specialization order.


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\begin{document}

\setcounter{page}{1}

\title{  semi-order continuous operators on vector spaces}
\author[K. Haghnejad Azar]{Kazem Haghnejad Azar}
\author[M. Matin]{Mina Matin}
\author[R. Alavizadeh]{Razi Alavizadeh}
\address{Department  of  Mathematics  and  Applications, Faculty of  Sciences, University of Mohaghegh Ardabili, Ardabil, Iran.}
\email{\textcolor[rgb]{0.00,0.00,0.84}{haghnejad@uma.ac.ir}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{minamatin@uma.ac.ir}}
\email{\textcolor[rgb]{0.00,0.00,0.84}{ralavizadeh@uma.ac.ir}}
\subjclass[2010]{Primary 39B82; Secondary 44B20, 46C05.}
\keywords{ ordered vector space, pre-Riesz space, $\tilde o$-convergence, $\tilde o$-continuous, $o_{W_1}$-convergence, $o_{W_1}$-continuous.
\newline \indent $^{*}$Corresponding author}
\begin{abstract}
In this manuscript, we will study both $\tilde{o}$-convergence in (partially) ordered vector spaces and  a kind of convergence in a vector space $V$.
A vector space $V$ is called semi-order vector space (in short semi-order space), if there exist an ordered vector space $W$ and an operator $T$ from $V$ into $W$. In this way, we say that $V$ is semi-order space with respect to $\{W, T\}$.
	 A net $\{x_\alpha\}\subseteq V$ is said to be ${\{W,T\}}$-order convergent  to a vector $x\in V$ (in short we write $x_\alpha\xrightarrow {\{W, T\}}x$), whenever there exists a net $\{y_\beta\}$ in $W$  satisfying $y_\beta \downarrow 0$ in $W$ and for each $\beta$, there exists $\alpha_0$ such that $\pm (Tx_\alpha -Tx) \leq y_\beta$ whenever $\alpha \geq \alpha_0$. In this manuscript, we study and investigate some properties of  $\{W,T\}$-convergent nets and its relationships with other order convergence in partially ordered vector spaces.
Assume that $V_1$ and $V_2$ are semi-order spaces with respect to 	 $\{{W_1}, T_1\}$ and $\{W_2, T_2\}$, respectively. An operator $S$ from $V_1$ into $V_2$ is called semi-order continuous, if  $x_\alpha\xrightarrow {\{{W_1}, T_1\}}x$ implies $Sx_\alpha\xrightarrow {\{W_2, T_2\}}Sx$ whenever $\{x_\alpha\}\subseteq V_1$. We study some properties of this new
 classification of  operators.

\end{abstract}
\maketitle

\section{Introduction}
Let $W$ be a real vector space and $K$ be a cone in $W$, that is, $K$ is a wedge ($x,y \in K$ and $\lambda,\mu \geq 0$ imply $\lambda x + \mu y \in K$) and $K \cap (-K) = \{0\}$. In $W$ a partial order is defined by $x\leq y $ whenever $y-x \in K$. The space $(W, K)$ (or, loosely $W$) is then called a (partially) ordered vector space. A subspace $M \subseteq W$ is majorizing in $W$ if for every $x\in W$ there exists some $m\in M$ with $x\leq m$ (or, equivalently, if for each $x\in W$ there exists some $m\in M$ with $m\leq x$). A subspace $M\subseteq W$ is called directed if for every $x,y \in M$ there is an element $z\in M$ such that $x\leq z$ and $y\leq z$. An ordered vector space $W$ is directed if and only if $W_+ $ is generating in $W$, that is, $W = W_+ - W_+$.
An ordered vector space $W$ is called Archimedean if for every $x,y \in W$ with $nx\leq y$ for every $n\in \mathbb{N}$ one has $x\leq 0$.
The ordered vector space $W$ has the Riesz decomposition property (RDP) if for every $x_1 , x_2 , z \in K$ with $z\leq x_1 + x_2$ there exist $z_1 , z_2 \in K$ such that $ z = z_1 +z_2$ with $z_1 \leq x_1$ and $z_2 \leq x_2$. We call a linear subspace $M$ of an ordered vector space $W$ order dense in $W$ if for every $x\in W$ we have
\begin{equation*}
x = \inf \{z\in M : x \leq z\},
\end{equation*}
that is, the greatest lower bound of set $\{z\in M : x \leq z\}$ exists in $W$ and equals $x$, see page $360$ of \cite{1a}. Clearly, if $M$ is order dense in $W$, then $M$ is majorizing in $W$. Denote for a subset $M$ of $W$, the set of all upper bounds by $M^u = \{x\in W : x\geq m \ for \ all\ m \in M\}$.

   A subset $M$ of an ordered vector space $W$ is called solid if for every $x\in W$ and $y\in M$ the relation $\{\pm y\}^u \subseteq \{\pm x\}^u$ implies $x\in M$. A solid subspace $M$ of $W$ is called an ideal.  The elements $x,y \in W$ are called disjoint, in symbols $x \perp y$, if $\{\pm (x+y)\}^u = \{\pm (x-y)\}^u$. The disjoint complement of a subset $M\subseteq W$ is $M^d = \{x\in W \mid \forall y\in M: x \perp y\}$. A sequence $\{x_n\}\subseteq W$ is said to be disjoint, if for every $n \neq m$, $x_n \perp x_m$.  A linear subspace $M$ of an ordered vector space $W$ is called a band in $W$ if $M = M^{dd} $.\\
 Recall that a linear map $i : W_1 \rightarrow {{W_2}}$ between two ordered vector spaces is said to be bipositive if for every $x\in W_1$ one has $i(x)\geq 0$ if and only if $x\geq 0$. A partially ordered vector space $(W_1,K)$ is called pre-Riesz space if for every $x,y,z \in W_1$ the inclusion $\{x+y,x+z\}^u \subseteq \{y,z\}^u$ implies $x\in K$. Clearly, each vector lattice is pre-Riesz space, since the inclusion in definition of pre-Riesz space reduces to inequality $(x+y)\vee (x+z)\geq y \vee z$, so $ x+(y \vee z) \geq y \vee z$, which implies $x\geq 0$. By Theorem 4.3 of \cite{3},  ordered vector space $W_1$ is a pre-Riesz space if and only if there exist a vector lattice ${{W_2}}$ and a bipositive linear map $i: W_1\rightarrow {{W_2}}$ such that $i(W_1)$ is order dense in ${{W_2}}$. The pair $({{W_2}},i)$ (or, loosely ${{W_2}}$) is then called a vector lattice cover of $W_1$. The theory of pre-Riesz spaces and their vector lattice covers is due to van Haandel, see \cite{5}.\\
A net $\{x_\alpha\} $ in ordered vector space $W$ is said to be decreasing (in symbols, $x_\alpha \downarrow$), whenever $\alpha \geq \beta $ implies $x_\alpha \leq x_\beta$. For $x\in W$ the notation $x_\alpha \downarrow x $ means that $x_\alpha \downarrow $ and $\inf_\alpha \{x_\alpha\} = x$ both hold. The meanings of $x_\alpha \uparrow x$ are analogus.  We say that a net $\{x_\alpha\}\subseteq W$, $(o)$-converges to $x\in W$ (in symbols, $x_\alpha \xrightarrow{o}x)$, if there is a net $\{y_\alpha\}\subseteq W$ such that $y_\alpha \downarrow 0$ and for all $\alpha$ one has $ \pm(x_\alpha - x) \leq y_\alpha$. Let $W_1$ and $W_2$ be two ordered vector spaces. An operator $T:W_1\rightarrow W_2$ is said to be order continuous, if every net $\{x_\alpha\}\subseteq W_1$ with $x_\alpha \xrightarrow{o}0 $ implies $Tx_\alpha \xrightarrow{o}0$ in $W_2$. The collection of all order continuous operators between two ordered vector spaces $W_1$ and $W_2$, showed by $L_{oc}(W_1,W_2)$. From \cite{2} recall that the directed part of $L_{oc}(W_1,W_2)$ has been showed by $L_{oc}^\diamond (W_1,W_2) = L_{oc}(W_1,W_2)_+ - L_{oc}(W_1,W_2)_+$ where $W_1$ and $W_2$ are ordered vector spaces and $W_1$ is directed.  \\

Let $(W,K)$ be an ordered vector soace. For two elements $y,z \in K$ with $ y \leq z$ denote the according order interval by $ [y,z] = \{x\in W: y\leq x \leq z\}$. A set $M\subset W$ is called order bounded if there are $y,z \in W$ such that $M\subseteq [y,z]$.\\

\section{ $\tilde o$-continuous operators}
	Let $W$ be an ordered vector space. A net $\{x_\alpha\}\subseteq W$ is said to be $\tilde{o}$-convergent to $x\in W$ (in symbol, $x_\alpha \xrightarrow{\tilde{o}} x $) if  there exists a net $\{y_\beta\}\subseteq W$, possibly over a different index set, such that $y_\beta \downarrow 0$ in $W$ and for every $\beta$ there exists $\alpha_0$ such that $\pm(x_\alpha - x)\leq y_\beta$, whenever $\alpha \geq \alpha_0$.\\
It is clear that for each net $\{x_\alpha\}\subseteq W$, $x_\alpha \xrightarrow{o}x$ implies $x_\alpha \xrightarrow{\tilde{o}}x$, but by Example 1.4 of \cite{11}, the converse, in general not holds.

\begin{lemma}\label{eli}
	Let $W$ be an ordered vector space and $\{x_\alpha\}\subseteq W$. Then we have the following assertions.
\end{lemma}
\begin{enumerate}
	\item $x_\alpha \xrightarrow{\tilde o}x$ iff $(x_\alpha - x) \xrightarrow{\tilde o}0$.
	\item If $0\leq x_\alpha \xrightarrow{\tilde o}x$, then $0 \leq x$.
	\item If for each $\alpha$, $ x_\alpha \leq y$ and $x_\alpha \xrightarrow{\tilde o}x$, then $ x \leq y$.
	\item If $x_\alpha\xrightarrow{\tilde o}x$ and $x_\alpha \xrightarrow{\tilde o}y$, then $x = y$.

\item If $x_\alpha \xrightarrow{\tilde o}x$ and $y_\alpha \xrightarrow{\tilde o}y$, then $ \lambda x_\alpha + \mu y_\alpha \xrightarrow{\tilde o} \lambda x + \mu y$ for all $ \lambda , \mu  \in\mathbb{R}$.
\item If $x_\alpha \xrightarrow{\tilde o}x$, $z_\alpha \xrightarrow{\tilde o}z$ and $ x_\alpha \leq z_\alpha$ for all $\alpha$, then $ x \leq z$.
\end{enumerate}
\begin{proof}
	\begin{enumerate}
	\item By definition it is established.
	\item	Since $x_\alpha \xrightarrow{\tilde o}x$, therefore there exists net $\{y_\beta\}\subseteq W$ such that $y_\beta \downarrow 0$ and for every $\beta $ there exists $\alpha_0$ such that $ \pm (x_\alpha - x) \leq y_\beta$ whenever $ \alpha \geq \alpha_0$. Since  $x_\alpha - x \leq y_\beta$, then $0 \leq x_\alpha \leq x + y_\beta$. Therefore $0\leq y_\beta + x$ and $0 \leq \inf_\beta \{y_\beta\} + x = x$.
	\item By assumption there exists a net $\{y_\beta\}\subseteq W$ such that $y_\beta \downarrow 0$ and for every $\beta $ there exists $\alpha_0$ such that $ \pm (x_\alpha - x) \leq y_\beta$ whenever $ \alpha \geq \alpha_0$. We have $ x = x - x_\alpha + x_\alpha \leq y_\beta +y$ whenever $\alpha \geq \alpha_0$ and therefore $ x \leq \inf_\beta \{y_\beta\} + y$, and follows $ x \leq y$.
	\item We have $x - y = x - x_\alpha + x_\alpha - y \leq y_\beta + z_\gamma $ where $y_\beta \downarrow 0 , z_\gamma \downarrow 0$, therefore $x\leq y$. By the same way $ y \leq x$ and so $ x = y$.

	\item We know that if $y_\beta \downarrow 0$ and $x\leq y$, then for every $\lambda \geq 0$, $\lambda y_\beta \downarrow 0$ and $\lambda x \leq \lambda y$. Note that if $\lambda < 0$, then $\lambda (\pm x_\alpha) = -\lambda (\pm x_\alpha)$. Therefore the proof holds.
	\item According to the relationship $x - z = x - x_\alpha + x_\alpha - z \leq x - x_\alpha + z_\alpha - z$, the proof is obvious.\qedhere
	\end{enumerate}
\end{proof}

\begin{definition}
	Let $W_1$ and $W_2$ be two ordered vector spaces. An operator $T:W_1\rightarrow W_2$ is said to be
	\begin{enumerate}
		\item  $\tilde{o}$-continuous, if for every net $\{x_\alpha\}$ in $W_1$ with $x_\alpha \xrightarrow{\tilde{o}}0$ it follows that $T(x_\alpha)\xrightarrow{\tilde{o}}0$ holds in $W_2$.
		\item $\sigma$-$\tilde{o}$-continuous,
		 if for every sequence $\{x_n\}$ in $W_1$ with $x_n \xrightarrow{\tilde{o}}0$ it follows that $T(x_n)\xrightarrow{\tilde{o}}0$ holds in $W_2$.
	\end{enumerate}
We show the collection of all $\tilde{o}$-continuous ($\sigma$-$\tilde{o}$-continuous) operators between two ordered vector spaces $W_1,W_2$, by $L_{\tilde{o}c}(W_1,W_2)$ ($L_{\sigma \tilde{o}c}(W_1,W_2)$).
\end{definition}
Example 1.8 of \cite{11} shows that the classes of $\tilde o$-continuous operators and order continuous operators between two ordered vector spaces $W_1$ and $W_2$ are different.
\begin{example}\label{jhg}
		Let $B$ be a projection band of ordered vector space $W$ and $ P_{B}$
	the corresponding band projection.  Let $\{x_\alpha\}\subseteq W$ and $x_\alpha \xrightarrow{\tilde o}0$ in $W$. There exists a net $\{y_\beta\}\subseteq W$ that $y_\beta \downarrow 0$ in $W$. For each $\beta$ there is $\alpha_0$ such that $P_B x_\alpha \leq y_\beta$ for each $\alpha\geq \alpha_0$. It is clear that $y_\beta \xrightarrow{o}0$ in $W$ and there exists a net $\{z_\beta\}\subseteq B$ that $P_B(y_\beta)\leq z_\beta$ for each $\beta$. We have $P_B(x_\alpha) = P_B(P_B(x_\alpha))\leq P_B(y_\beta)\leq z_\beta$ whenever $\alpha \geq \alpha_0$.

\end{example}
\begin{theorem}\label{uyi}
	Let $T:W_1\rightarrow W_2$ be an operator between two ordered vector spaces.
	\begin{enumerate}
		\item If $0\leq T$ is order continuous, then $T$ is $\tilde{o}$-continuous.\label{ali}
		\item If $W_2$ is a Dedekind complete vector lattice and $T$ is $\tilde o$-continuous, then $T$ is order continuous.\label{alib}
		\item 	If $W_1$ is directed with RDP and $W_2$ is a Dedekind complete vector lattice. Then operator $T: W_1 \rightarrow W_2$ is order continuous iff $T$ is $\tilde o$-continuous.
	\end{enumerate}
\end{theorem}
\begin{proof}
	\begin{enumerate}
\item	Let $\{x_\alpha\}\subseteq W_1$ be a net such that $x_\alpha \xrightarrow{\tilde{o}}0$. There exists a net $\{y_\beta\}$ in $W_1$ such that $y_\beta \downarrow 0$ and for every $\beta$ there exists $\alpha_0$ such that $\pm x_\alpha \leq y_\beta$ whenever $\alpha \geq \alpha_0$. Due to $T$ being positive we obtain $\pm T(x_\alpha)\leq T(y_\beta)$. Since $T$ is positive and order continuous, hence by Lemma 7 of \cite{2}, $Ty_\beta \downarrow 0$ in $W_2$. It follows $T(x_\alpha)\xrightarrow{\tilde{o}}0$.
\item Let $\{x_\alpha\}\subseteq W_1$ be a net such that $x_\alpha \xrightarrow{o}0$ in $W_1$. It is clear that $x_\alpha \xrightarrow{\tilde o}0$ in $W_1$. By assumption $Tx_\alpha \xrightarrow{\tilde o}0$ in $W_2$. Since $W_2$ is a Dedekind complete vector lattice, $Tx_\alpha \xrightarrow{o}0$ in $W_2$ (see page 288 of \cite{11}). Hence $T$ is order continuous.
\item Let $T\in L_{oc}(W_1,W_2)$.	By Proposition 11 of \cite{2} we have $L_{oc}^\diamond (W_1,W_2) = L_{oc}(W_1,W_2)$. Therefore $T = T_1 - T_2 $ such that $T_1$ and $T_2$ are positive and order continuous. By \ref{ali}, $T_1 , T_2$ are $\tilde o$-continuous and therefore $T$ is $\tilde o$-continuous.\\
Conversely, it is clear by  \ref{alib}.
\end{enumerate}
\end{proof}

In this section we want to bring up two propositions similar to Theorem 13 and Proposition 20 of \cite{2}, respectively.
\begin{proposition}
	Let $W_1$ be a directed ordered vector space and $W_2$ be a pre-Riesz space with a vector lattice cover $(W_3,i)$. If $T\in L_{\tilde{o} c}(W_1,W_2)$, then $ioT \in L_{\tilde{o} c}(W_1,W_3)$.
\end{proposition}
\begin{proof}
	The proof has an argument similar to Theorem 13 of \cite{2}.
\end{proof}
\begin{proposition}
	Let $V_1$ and $V_2$ be two pre-Riesz spaces and $(W_1 , i_1)$, $(W_2 , i_2)$ be their vector lattice covers, respectively. Let a positive  operator $T: V_1 \rightarrow V_2$ has a positive linear extension $S: W_1 \rightarrow W_2$, i.e. $ Soi_1 = i_2 o T$. If $S\in L_{\tilde oc}(W_1,W_2)$, then $T\in L_{{\tilde oc}}
	(V_1,V_2)$.
\end{proposition}
\begin{proof}
	Let $\{x_\alpha\}$ be a net in $V_1$ with $x_\alpha \downarrow 0 $ in $V_1$. By Lemma 1(i) of \cite{2}, the infimum of the set $\{i_1 (x_\alpha): \alpha \in A\}$ exists in $W_1$ and equals $0$. It is clear that $i_1(x_\alpha)\xrightarrow{\tilde o}0$ in $W_1$. As $S$ is ${\tilde o}$-continuous, $S(i_1(x_\alpha))\xrightarrow{\tilde o} 0$. Therefore there exists a net $\{y_\beta\}\subseteq W_2$ such that $y_\beta \downarrow 0$ and for each $\beta$ there is an $\alpha_0$ such that $\pm S(i_1(x_\alpha)) \leq y_\beta$ whenever $\alpha \geq \alpha_0$. Clearly $i_2 (T(x_\alpha))\downarrow$. We have $0 \leq i_{2}(T(x_\alpha)) = S(i_{1}(x_\alpha))\leq y_\beta$ in $W_2$.  It is clear that $i_2 (T(x_\alpha))\downarrow 0$ in $W_2$ and by Lemma 1(ii) of \cite{2}, $T(x_\alpha)\downarrow 0$ in $V_2$.	By Lemma 7 of \cite{2}, $T$ is order continuous and therefore by Theorem \ref{uyi}, $T$ is $\tilde o$-continuous.
\end{proof}

\section{Order convergence in semi-order vector spaces }

A vector space $V$ is called semi-order vector space (in short semi-order space), if there exist an ordered vector space $W$ and an operator $T$ from $V$ into $W$. In this way, we say that $V$ is semi-order space with respect to $\{W, T\}$.
	 A net $\{x_\alpha\}\subseteq V$ is said to be ${\{W,T\}}$-order convergent (resp. $T(V)$-order convergent)  to a vector $x\in V$ in short we write $x_\alpha\xrightarrow {\{W, T\}}x$ (resp. $x_\alpha\xrightarrow {T(V)}x $), whenever there exists a net $\{y_\beta\}$ in $W$ (resp. $T(V)$)  satisfying $y_\beta \downarrow 0$ in $W$ (resp. $T(V)$)  and for each $\beta$, there exists $\alpha_0$ such that $\pm (Tx_\alpha -Tx) \leq y_\beta$ whenever $\alpha \geq \alpha_0$.\\
	  Whenever $V$ is a subspace of $W$ and $T$ is inclusion map, we use symbol $x_\alpha \xrightarrow{Wo}x$ in $V$ instead of symbol $x_\alpha  \xrightarrow{\{W,T\}}x$ in $V$.\\
In this section, we study some properties of  $\{W,T\}$-order convergent nets.

 Let $K^\prime$ be a cone in $W$. Obviously,  $K^\prime \cap T(V)= K^{\prime\prime}$ is a cone of $T(V)$. Then there exists $K\subseteq V$ with $T(K)=K^{\prime\prime}$. Now if $\ker T = \{0\}$,  then  $K$ is a cone of $V$. It means that if $V$ is a semi-order vector space with respect
  to $\{W,T\}$ that $\ker T = \{0\}$, then $V$ is an ordered vector space. \\

 We say that $V$ has order properties, when $T(V)$ has these order properties, for example  see the following definition for some of them.

\begin{definition}\label{puy}
Assume that $V$ is a semi-order space with respect to $\{W,T\}$.
\begin{enumerate}
\item For each $x\in V$, we define $x\geqslant_V 0$ whenever $Tx\geqslant 0$ ($\geqslant_V$ is named semi-order in $V$).
\item A subset $M$ of $V$ is $\{W,T\}$-order closed, $\{W,T\}$-order bounded in  $V$ whenever $T(M)$ is    order closed, order bounded in  $W$, respectively.
\item A subspace $B\subseteq V$ is $\{W,T\}$-order dense, $\{W,T\}$-ideal, $\{W,T\}$-band in  $V$ whenever $T(B)$ is   order dense, ideal, band in  $W$, respectively. The operator $P_B : V\rightarrow B$ defined via $P_B(x)=x_1$ where $P_{TB}(Tx)=Tx_1$ that $x_1 \in B$ and $P_{TB}: W\rightarrow TB$ is a band projection, is a band projection on $V$.
\item Let  a net $\{x_\alpha\}\subseteq V$. $x_\alpha\downarrow x$ whenever $Tx_\alpha\downarrow Tx$ in $W$.

\item A sequence $\{x_n\}\subseteq V$ is said to be  $\{W,T\}$-disjoint in $V$, if $\{Tx_n\}$ is disjoint  sequence in $W$.
\end{enumerate}

\end{definition}
For a semi-order space $V$, order convergence is depended to ordered vector space $W$ and  operator $T$.
  The following example, shows that ${\{W,T\}}$-order convergence depends on both of $W$ and $T$.

\begin{example}\label{elin}
		Consider the standard basis $\{e_n\}$ of $c_0$. $c_0$ is a subspace of $\ell^\infty$ and $e_n\xrightarrow{\{\ell^\infty o\} }0$ in $c_0$, but $\{e_n\}$ is not $I(c_0)$-order convergent to $0$ in $c_0$. Now set an operator $T$ from $c_0$ into $\ell^\infty$ with $Te_n=a_n$ whenever $a_n=(n,n,n,...n, 0,0,0,...)$,  the first $n$ places are occupied with $n$ and the rest is zero. It is clear that $\{e_n\}$ is not ${\{\ell^\infty,T\}}$-order convergent to zero.
\end{example}

\begin{lemma}
	Assume that  $V$ is a semi-order space  with respect to $\{W,T\}$. Then we have  the following assertions.
\end{lemma}
\begin{enumerate}
	\item $x_\alpha \xrightarrow{\{W, T\}}x$ in $V$ iff $(x_\alpha - x) \xrightarrow{\{W, T\}}0$ in $V$.
\item If $0\leq_V x_\alpha \xrightarrow{\{W, T\}}x$ in $V$, then $0 \leq_V x$.
\item If for each $\alpha$, $ x_\alpha \leq_V y$ in $V$ and $x_\alpha \xrightarrow{\{W, T\}}x$ in $V$, then $ x \leq_V y$.
\item If $x_\alpha\xrightarrow{\{W, T\}}x$ and $x_\alpha \xrightarrow{\{W, T\}}y$ in $V$, then $x = y$.
\item If $x_\alpha \xrightarrow{\{W, T\}}x$ and $y_\alpha \xrightarrow{\{W, T\}}y$ in $V$, then $ \lambda x_\alpha + \mu y_\alpha \xrightarrow{\{W, T\}} \lambda x + \mu y$ in $V$ for all $ \lambda , \mu \in\mathbb{R}$.
\item If $x_\alpha \xrightarrow{\{W, T\}}x$, $z_\alpha \xrightarrow{\{W, T\}}z$ and $ x_\alpha \leq_V z_\alpha$ for all $\alpha$ in $V$, then $ x \leq_V z$.
\end{enumerate}
\begin{proof}
By using of Definition \ref{puy} and the proof of  Lemma \ref{eli}, the proof is complete.
\end{proof}
\begin{theorem}
	\begin{enumerate}
\item	Let  $W$ be an order dense subspace of ordered vector space $U$. If $\{x_\alpha\}\subseteq V$ and $x_\alpha \xrightarrow{\{W, T\}}0$ in $V$, then $x_\alpha \xrightarrow{\{U, T\}}0$ in $V$.\label{rew}
\item If $V$ is order dense in $W$ and $x_\alpha \xrightarrow{T(V)}x$ in $V$, then $x_\alpha \xrightarrow{\{W,T\}}x$ in $V$.
\item Assume that  $V$  is semi-order space with respect to both $\{{{W_1}}, T\}$ and $\{W_2, T\}$ such that  ${{W_1}}$  is an ideal of vector lattice $W_2$. If $\{x_\alpha\}$ is $\{W_1,T\}$-order bounded in ${{V}}$, then $x_\alpha \xrightarrow{\{W_2, T\}}0$ in $V$ implies $x_\alpha\xrightarrow{\{{{W_1}}, T\}}0$ in $V$.
	\item Let $W$ be a vector lattice, $I$ be a $\{W,T\}$-ideal in $V$ and $\{x_\alpha\} \subseteq I$. $x_\alpha \xrightarrow{\{W,T\}}x$ in $V$ iff $x_\alpha \xrightarrow{\{W,T\mid_I\}}x$ in $I$.
	\item Let $(U,i)$ be a vector lattice cover of pre-Riesz space $W$ and $\{x_\alpha\}\subseteq V$. Then $x_\alpha \xrightarrow{{\{W,T\}}}0$ in $V$ implies $x_\alpha \xrightarrow{\{U, ioT\}}0$ in $V$.
\end{enumerate}
	\end{theorem}
\begin{proof}
\begin{enumerate}
\item  Let $\{x_\alpha\}\subseteq V$ and $x_\alpha \xrightarrow{\{W,T\}}x$ in $V$, therefore there exists a net $\{y_\beta\}\subseteq W$ that $y_\beta \downarrow 0$ in $W$. By Proposition 5.1 of \cite{3}, $y_\beta \downarrow 0$ in $U$. Hence the proof is complete.
\item By assumption and by Definition \ref{puy}, $T(V)$ is order dense in $W$. Therefore the proof is clear by \ref{rew}.
\item  Assume that   $\{x_\alpha\}\subseteq V$  and $x_\alpha \xrightarrow{\{W_2, T\}}0$ in $V$.
 Then there exists $\{y_\beta\}\subseteq W_2$ satisfying $y_\beta \downarrow 0$ and for each $\beta$ there exists $\alpha_0$ such that $ |Tx_\alpha|\leq y_\beta$ whenever $\alpha \geq \alpha_0$. By assumotion and by Definition \ref{puy}, there exists a $u\in {{W_1}}^+$ such that $|Tx_\alpha| \leq u$. Since  ${{W_1}}$ is an ideal of $W_2$, $\{u \wedge y_\beta\}\subseteq {{W_1}}$.  It is clear that  $u \wedge y_\beta \downarrow 0$  in ${{W_1}}$. It is obvious   that for each $\beta$ there exists $\alpha_0$ that $|Tx_\alpha | \leq u \wedge y_\beta$ whenever $\alpha \geq \alpha_0$. It follows that $x_\alpha \xrightarrow{\{{{W_1}}, T\}}0$ in $V$.
 \item  Let $\{x_\alpha\} \subseteq I$ and $x_\alpha \xrightarrow{\{W,T\}}x$ in $V$, that means that $T(x_\alpha)\xrightarrow{\tilde o}T(x)$ in $W$. By Definition \ref{puy}, $T(I)$ is an ideal in $W$. By Lemma 2.5 of \cite{4}, $T(x_\alpha)\xrightarrow{\tilde o}T(x)$ in $T(I)$. Hence $x\in I$.\\
 Conversely, it is clear that if $x_\alpha \xrightarrow{\{W,T\mid_I\}}x$ in $I$, then $x_\alpha \xrightarrow{\{W,T\}}x$ in $V$.
 \item 	Let $\{x_\alpha\}\subseteq V$ and $x_\alpha \xrightarrow{{\{W,T\}}}0$ in $V$. Then there exists a net $\{y_\beta\}\subseteq W$ such that $y_\beta \downarrow 0$ in $W$ and for each $\beta$ there exists $\alpha_0$ that $\pm (T(x_\alpha - x))\leq y_\beta$ whenever $\alpha \geq \alpha_0$. Since $W$ is order dense in $U$, therefore by Proposition 5.1 of \cite{3}, $y_\beta \downarrow  0$ in $U$. By Lemma 1 of \cite{2}, $i(y_\beta)\downarrow 0$ in $U$. Because $i$ is positive, we have $\pm i(T(x_\alpha - x))\leq i(y_\beta)$.
\end{enumerate}
\end{proof}

Assume that $V$ is a semi-order space with respect  to $\{W,T\}$. A set  $A\subseteq V$ is $\{W, T\}$-closed (resp. $T(V)$-closed) in $V$, if for any net $\{x_\alpha\}\subseteq A$ and $x\in V$ with $x_\alpha\xrightarrow{\{W, T\}}x$ (resp. $x_\alpha\xrightarrow{T(V)}x$) in $V$, one has $x\in A$.
\begin{proposition}
Let $T$ be onto, one-to-one and $A\subseteq V$. Then $A$ is $\{W,T\}$-order closed iff $T(A)$ is $\tilde{o}$-closed in $W$
\end{proposition}
\begin{proof}
	Let $\{x_\alpha\} $ be a net in $A$ and $x_\alpha \xrightarrow{\{W,T\}}x$ in $V$. It follows that  there exists a net $\{y_\beta\}\subseteq W$ such that $y_\beta \downarrow 0$ and for each $\beta$ there exists $\alpha_0$ that $\pm (Tx_\alpha - Tx) \leq y_\beta$ whenever $\alpha \geq \alpha_0$. Since $\{Tx_\alpha\}\subseteq T(A)$ and $T(A)$ is $\tilde o$-closed in $W$, therefore $Tx\in  T(A)$. It follows that $x\in A$.\\
 Conversely, let $\{T(x_\alpha)\}\subseteq T(A)$ and $Tx_\alpha\xrightarrow{\tilde o}y$. Since $T$ is onto, there exists $x\in V$ that $Tx = y$.  It is clear that $\{x_\alpha\}\subseteq A$, $x_\alpha \xrightarrow{\{W,T\}}x$ in $V$. By assumption $x\in A$. So $Tx \in T(A)$.
\end{proof}
\begin{proposition}
	Let $V$ be a semi-order space with respect to $\{W,T\}$ where $W$ is a vector lattice. Each $\{W,T\}$-disjoint and $\{W,T\}$-order bounded sequence $\{x_n\}\subseteq V$ is $\{W,T\}$-order null.
\end{proposition}
\begin{proof}
By Definition \ref{puy}, $\{Tx_n\}$ is order bounded and disjoint in $W$. Therefore by Corollary 3.6 of \cite{4}, $Tx_n \xrightarrow{uo}0$ in $W$. Since $\{Tx_n\}$ is order bounded in $W$, we have $Tx_n \xrightarrow{\tilde o}0$ in $W$. Hence there exists a sequence $\{y_m\}\subseteq W$ such that $y_m \downarrow 0$ in $W$ and for every $m$ there exists $n_0$ such that $\pm Tx_n \leq |Tx_n|\leq y_m$  whenever $n \geq n_0$. Thus $x_n \xrightarrow{\{W,T\}}0$ in $V$.
\end{proof}

\section {Semi-order continuous operators}

Assume that $V_1$ and $V_2$ are semi-order spaces with respect to 	 $\{{{W_1}}, T_1\}$ and $\{W_2, T_2\}$, respectively.
An operator $S$ from $V_1$ into $V_2$ is positive, when $x>_{V_1}0$ implies $Sx>_{V_2}0$

\begin{definition}
Assume that $V_1$ and $V_2$ are semi-order spaces with respect to 	 $\{{{W_1}}, T_1\}$ and $\{W_2, T_2\}$, respectively. An operator $S$ from $V_1$ into $V_2$ is called
\begin{enumerate}
	\item  semi-order continuous, if  $x_\alpha\xrightarrow {\{{{W_1}}, T_1\}}x$ implies $Sx_\alpha\xrightarrow {\{W_2, T_2\}}Sx$ whenever $\{x_\alpha\}\subseteq V_1$.
	\item  $\sigma$-semi-order continuous, if  $x_n\xrightarrow {\{{{W_1}}, T_1\}}x$ implies $Sx_n\xrightarrow {\{W_2, T_2\}}Sx$ whenever $\{x_n\}\subseteq V_1$.
\end{enumerate}
\end{definition}
It is obvious that an operator $S$ from $V_1$ into $V_2$ is called semi-order continuous if and only if $x_\alpha\xrightarrow {\{{{W_1}}, T_1\}}x$ implies $x_\alpha\xrightarrow {\{W_2, T_2S\}}x$ whenever $\{x_\alpha\}\subseteq V_1$.
Since semi-order continuity of an operator $S$ between two  semi-order spaces depends on ${{W_1}}$ and $W_2$, the collection of all semi-order continuous operators between two semi-order spaces $V_1$ and $V_2$ will be denoted by  $L_{o_{{{W_1}}W_2}}(V_1, V_2)$. Similarly, $ L_{\sigma o_{{{W_1}}W_2}}(V_1, V_2)$ will denote the collection of all operators from $V_1$ to $V_2$ that are $\sigma$-semi-order continuous. Whenever $S\in L_{o_{{{W_1}}W_2}}(V_1, V_2)$ (resp. $ L_{\sigma o_{{{W_1}}W_2}}(V_1, V_2)$), we say that, $S$ is $o_{W_1W_2}$-continuous (resp, $\sigma$-$o_{W_1W_2}$-continuous).\\
Here are some examples of semi-order continuous operators.

\begin{example}
\begin{enumerate}
	\item  Let $V$ be a pre-Riesz space and $W$ is its vector lattice cover that $W$ is Archimedean. The inclusion map $I : V \rightarrow W$ is $o_{V^\delta W^\delta}$-continuous ($V^\delta$, $W^\delta$ are Dedekind completions of $V$,$W$ respectively). Let $\{x_\alpha\}\subseteq V$ and $x_\alpha\xrightarrow{\{V^\delta o\}}0$ in $V$. Then by Theorem 2 of \cite{2}, $Ix_\alpha \xrightarrow{\{W^\delta o\}}0$ in $W$.
	\item Let $V$ be a semi-order space with respect to $\{W,T\}$, $B$ be a band of $V$ and $P_B:V\rightarrow B$ be a band projection. If $\{x_\alpha\}\subseteq V$ and $x_\alpha \xrightarrow{\{W,T\}}0 $ in $V$, then $Tx_\alpha\xrightarrow{\tilde o}0$ in $W$. By Definition \ref{puy}, $T(B)$ is a band in $W$. We consider the band projection $P_{TB}:W\rightarrow TB$. By Example \ref{jhg}, $P_{TB}(Tx_\alpha)\xrightarrow{\tilde o}0$ in $TB$. It is clear that $T|_B P_B = P_{TB}T$. Therefore $P_B(x_\alpha)\xrightarrow{\{T|_B(B)\}}0$ in $B$. So $P_B$ is $o_{W,TB}$-continuous.
	\item Let $f$ be a continuous functional on Banach lattice $E$. Since $\mathbb{R}$ is a $KB$-space so by Theorem 4.60 of \cite{1}, $c_0$ is not embeddable in $E$. By Theorem 4.63 of \cite{1}, there exist a $KB$-space $F$, lattice homomorphism $Q:E\rightarrow F$ and functional $g:F\rightarrow \mathbb{R}$ such that $ f = g o Q$. If  $I$ is identity map on $\mathbb{R}$, we have $Iof = g o Q$. If $\{x_\alpha\}\subseteq E$ and $x_\alpha\xrightarrow{\{F,Q\}}0$ in $E$, then $Q(x_\alpha)\xrightarrow{\tilde o}0$ in $F$. Since $F$ has order continuous norm, therefore $Q(x_\alpha)\xrightarrow{\|.\|}0$ in $F$. So $g(Q(x_\alpha))\xrightarrow{\|.\|}0$ and hence $g(Q(x_\alpha))\xrightarrow{\tilde o}0$ in $\mathbb{R}$. Therefore we have $f(x_\alpha)\xrightarrow{\{\mathbb{R}o\}}0$. Hence $f$ is  $o_{F,\mathbb{R}}$-continous.
\end{enumerate}
\end{example}
\begin{theorem}\label{ebadi}
	Let $V_1 , V_2$ be two pre-Riesz spaces, $(W_1,i_1), (W_2,i_2)$ be their vector lattice covers, respectively and $S:V_1 \rightarrow V_2$ be a positive operator.
	\begin{enumerate}
		\item If $S$ has extension positive and order continuous $\tilde{S} : W_1\rightarrow W_2$ that $i_2oS = \tilde{S}o i_1$, then $S\in L_{o_{{{W_1}}W_2}}(V_1, V_2)$.
		\item
		$ S\in L_{o_{{{W_1}}W_2}}(V_1, V_2)$ iff  $ x_\alpha \downarrow 0$ in ${{V_1}}$ implies $ Sx_\alpha \downarrow 0$ in $V_2$ for each net $\{x_\alpha\}\subseteq V_1$.
		\end{enumerate}

\end{theorem}
\begin{proof}
	\begin{enumerate}
	\item	 Let $\{x_\alpha\}\subseteq V_1$ and $x_\alpha \xrightarrow{\{W_1 , i_1\}}0$ in $V_1$. Then there exists a net $\{y_\beta\}\subseteq W_1$ such that $y_\beta \downarrow 0$ and for each $\beta$ there exists $\alpha_0$ that $\pm i_1(x_\alpha)\leq y_\beta$ whenever $\alpha \geq \alpha_0$. Since $\tilde{S}$ is order continuous and positive, therefore by Theorem 1.56 of \cite{1}, $\tilde{S}(y_\beta) \downarrow 0$ in $W_2$ and it is clear that $\tilde{S}(\pm i_1(x_\alpha))\leq
	\tilde{S}(y_\beta)$. Hence by assumption $S(x_\alpha)\xrightarrow{\{W_2,i_2\}}0$.
	\item

	Suppose that $0\leq S\in L_{o_{{{W_1}}W_2}}(V_1, V_2)$ and $\{x_\alpha\}\subseteq V_1$ with $x_\alpha \downarrow 0$ in ${{V_1}}$. Then by Lemma 1 of \cite{2}, $i_1(x_\alpha)\downarrow 0$ in $W_1$. It is clear that
	$x_\alpha \xrightarrow{\{{{W_1 ,i_1}}\}}0$ in $V_1$.
	By assumption, we have    $Sx_\alpha \xrightarrow{\{W_2,i_2\}}0$ in $V_2$. It follows that there exists a net $\{y_\beta\} \subseteq W_2$ satisfying, $y_\beta \downarrow 0$ in $W_2$ and for each $\beta$ there exists $\alpha_0$ such that $\pm i_2Sx_\alpha \leq y_\beta$ whenever $\alpha \geq \alpha_0$. It follows that $ i_2Sx_\alpha \leq \inf y_\beta =0$ for each $\alpha$ and so $ i_2Sx_\alpha \downarrow 0$ in $W_2$ and therefore by Lemma 1 of \cite{2}, $Sx_\alpha \downarrow 0$ in $V_2$.\\
	Conversely, let $\{x_\alpha\}\subseteq V_1$ and $x_\alpha \xrightarrow{\{W_1,i_1\}}0$ in $V_1$. There exists a net $\{y_\beta\}\subseteq W_1$ such that $y_\beta \downarrow 0$ in $W_1$ and for each $\beta$ there exists $\alpha_0$ such that $\pm i_1(x_\alpha)\leq y_\beta$ whenever $\alpha \geq \alpha_0$. Hence $i_1 x_\alpha \downarrow 0$ in $W_1$. By Lemma 1 of \cite{2}, $x_\alpha \downarrow 0$ in $V_1$. By assumption $Sx_\alpha\downarrow 0$ in $V_2$ and by Lemma 1 of \cite{2}, $i_2 S(x_\alpha)\downarrow 0$ in $W_2$. Therefore $S(x_\alpha)\xrightarrow{\{W_2,i_2\}}0$ in $V_2$.
	\end{enumerate}

\end{proof}

Assume that $V_1$ and $V_2$ are semi-order spaces with respect to 	 $\{{{W_1}}, T_1\}$ and $\{W_2, T_2\}$, respectively.
An operator $S:V_1\rightarrow V_2$ is semi-order bounded, if $S(A)$ is  $\{W_2,T_2\}$-bounded for each  $\{W_1,T_1\}$-bounded set $A\subseteq V_1$.

 If  $S:V_1\rightarrow V_2$  semi-order bounded, then we write that $S$ is  ${{W_1}}W_2$-order bounded (for short, $o_{{{W_1}}W_2}$-bounded).
 \\
We denote the collection of all  $o_{{{W_1}}W_2}$-bounded operators $S:V_1\rightarrow V_2$ by $L_{b_{{{W_1}}W_2}}(V_1, V_2)$.

\begin{theorem}\label{kat}
	Let $V_1, V_2$ be  subspaces of ordered vector spaces ${{W_1}}, W_2$, respectively, such that $W_2$ is  Archimedean Dedekind complete vector lattice. Then, $L_{o_{{{W_1}}W_2}}(V_1, V_2)$ is subspace of $ L_{b_{{{W_1}}W_2}}(V_1, V_2)$.
\end{theorem}
\begin{proof}
 Let $S:V_1 \rightarrow V_2$ be an $o_{W_1 W_2}$-continuous operator. First we consider $y \in {{W_1}}_+$ and  $A = V_1\cap [0,y]$.
 Let $I= \mathbb{N}\times A$ be an index set with the lexicographical order. Namely, $(n,x)>(m,z)$ if and only if either one the following holds true.\\
$(1)$ $n>m$,\\
$(2)$ $n=m$ and $x>z$.\\
It is easy to check that $I$ is a directed set, so we may consider a net indexed by $I$. Let us set $\varphi_{(n,x)}=\frac{1}{n}x$ for all $x\in A$. Then we have $0\leq\varphi_{(n,x)}\leq \frac{1}{n}x$. It follows that $\varphi_{(n,x)}$ is $\{W_1\}$-order convergent to zero.
By assumption,  $S\varphi_{(n,x)}$ is $\{W_2\}$-order convergent to zero.
Then there exists a net $(y_\beta)_{\beta}$ such that
$y_\beta\downarrow 0$ and for every $\beta$ there exists $(n,x)$ satisfying $ \pm S\varphi_{(m,z)}\leq y_\beta$ for all $(m,z)>(n,x)$.
Let us pick any $y_\beta$ and find corresponding index $(n,x)\in I$. Then, in particular, $\pm S\varphi_{(n+1,z)}\leq y_\beta$ for all $z\in A$. It follows that $-(n+1)y_\beta\leq Sz\leq (n+1)y_\beta$ for every
$z\in A$. Thus, $S$ is $o_{{{W_1}}W_2}$-bounded operator from $V_1$ into $V_2$.
\end{proof}

Let $V_1$ and $V_2$ be two semi-order spaces with respect to $\{W_1,T_1\}$ and $\{W_2,T_2\}$, respectively, where $W_1$ is directed. We define the directed part of $L_{o_{{{W_1}}W_2}}(V_1,V_2)$ by
$$L_{o_{{{W_1}}W_2}}^\diamond(V_1,V_2) := L_{o_{{{W_1}}W_2}}(V_1,V_2)_+ - L_{o_{{{W_1}}W_2}}(V_1,V_2)_+.$$

\begin{theorem}\label{po}
	Let $V_1$ and $V_2$ be two  pre-Riesz spaces with vector lattice covers $\{W_1,i_1\}$ and $\{W_2,i_2\}$,respectively, such that $V_1$ with a generating cone has the RDP and $V_2$, $W_2$ is Archimedean  Dedekind complete vector lattice. Then
	\begin{enumerate}
		\item $L_{b_{{{W_1}}W_2}}(V_1,V_2)$ is a Dedekind complete vector lattice.\label{iu}
		\item $L_{o_{{{W_1}}W_2}}(V_1,V_2)$ is a vector lattice.

		\item  $L_{o_{{{W_1}}W_2}}(V_1,V_2)$ is a band in $L_{b_{{{W_1}}W_2}}(V_1,V_2).$
		\end{enumerate}
\end{theorem}
\begin{proof}
	\begin{enumerate}
		\item By Theorem 8 of \cite{2}, $L_b(V_1,V_2)$ is a Dedekind complete vector lattice. Since $V_1$ and $V_2$ are majorizing in ${{W_1}}$ and $W_2$ respectively, it is clear that $T\in L_{b_{{{W_1}}W_2}}(V_1,V_2)$ iff $T\in L_b(V_1,V_2)$. Therefore $L_{b_{{{W_1}}W_2}}(V_1,V_2)$ is a Dedekind complete vector lattice.

		\item Let $T\in L_{o_{{{W_1}}W_2}}(V_1,V_2)$. By Theorem \ref{kat}, $T\in L_{b_{{{W_1}}W_2}} (V_1,V_2)$. Therefore by \ref{iu}, $|T|$ exists and belongs to $L_{b_{{{W_1}}W_2}} (V_1,V_2)$. Let $\{x_\alpha\}\subseteq V_1$ and $x_\alpha \downarrow 0$. By Theorem \ref{ebadi}$(2)$, it is enough to show that $|T|x_\alpha\downarrow 0$ in $V_2$. Since $|T|$ is positive and $V_2$ is a Dedekind complete, so there exists a $z\in V_2$ that $|T|x_\alpha \downarrow z$ in $V_2$. Let $z\neq 0$. We have $i_2 |T|(x_\alpha)\downarrow i_2z$ in $W_2$. There exists a net $\{y_\beta\}\subseteq W_2$ that $y_\beta\downarrow 0$ and for each $\beta$ there exists $\alpha_0$ that $\pm i_2 (|T|x_\alpha -z)\leq y_\beta$ whenever $\alpha \geq \alpha_0$. We have $\pm i_2 (Tx_\alpha) \leq i_2 (|T|x_\alpha)\leq y_\beta \pm i_2 z$. So $\pm i_2 (Tx_\alpha - z)\leq y_\beta$ whenere $\alpha\geq \alpha_0$. This is a contradiction with $Tx_\alpha \xrightarrow{\{W_2,i_2\}}0$.  Therefore $|T|x_\alpha \downarrow 0$ in $V_2$.
\item  By Theorem \ref{kat}, $L_{o_{{{W_1}}W_2}}(V_1,V_2)$ is a subspace in  $L_{b_{{{W_1}}W_2}} (V_1,V_2)$. Let $T\in L_{o_{{{W_1}}W_2}}(V_1,V_2)$, $S\in L_{b_{{{W_1}}W_2}}(V_1,V_2)$ with $|S|\leq |T|$ and $\{x_\alpha\}\subseteq V_1$ with $x_\alpha \xrightarrow{\{W_1, i_1\}}0$ in $V_1$. We have $|T|x_\alpha \xrightarrow{\{W_2,i_2\}}0$.  With loss of generality, we can assume $0\leq x_\alpha$ for each $\alpha$. By enequality $\pm i_2 S(x_\alpha)\leq i_2 |S|x_\alpha \leq i_2 |T|x_\alpha$, $Sx_\alpha \xrightarrow{\{W_2,i_2\}}0$. So $L_{o_{{{W_1}}W_2}}(V_1,V_2)$ is an ideal in $L_{b_{{{W_1}}W_2}}(V_1,V_2)$. To see that the ideal $L_{o_{{{W_1}}W_2}}(V_1,V_2)$ is a band, the proof has similar argument of Theorem 1.57 \cite{1}.

\end{enumerate}\end{proof}

config: arxiv


TITLE: inverse Laplace transform of the determinant
QUESTION [1 upvotes]: what is the inverse Laplace transform of the function
$x\mapsto\det x$? (where $x\in\mathbb{R}^{n\times n}$ is a symmetric metric, if necessary at all.)

REPLY [2 votes]: The one-variable inverse Laplace transform of $x$ is $\delta'(s)$ and that of $1$ is $\delta(s)$.
As a function on $\mathbb R^{n \times n}$, the multivariate inverse Laplace transform of
$\det(M) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{j=1}^n x_{j, \sigma(j)}$ is then
$$Y(s_{11}, \ldots, s_{nn}) = \sum_{\sigma \in S_n} \text{sgn}(\sigma) \prod_{j=1}^n \delta'(s_{j,\sigma(j)}) \prod_{k \ne \sigma(j)} \delta(s_{j,k})$$


TITLE: Dilation of an operator
QUESTION [2 upvotes]: Let $T\in B(\mathcal{H})$. Based on  https://en.wikipedia.org/wiki/Dilation_(operator_theory)
 dilation of the operator $T$  is an operator on a larger Hilbert space $K$, whose restriction to $\mathcal{H}$ composed with the orthogonal projection $P$ onto $\mathcal{H}$ is $T$.
I have several questions about this definition. Is there another version of this definition? Is the operator $P$ always projection?  I mean could one find a larger Hilbert space $K$ and operators $V, U$ on $K$, when $U$ is not an orthogonal projection, such that $UV|_{\mathcal{H}}= T$?

REPLY [1 votes]: Another version:
Theorem: Let $A$ be a bounded linear operator on a Hilbert space $\mathcal{H}$ with $\|A\|\le 1$. There there exists a Hilbert space $\mathcal{K}$, an isometry $V : \mathcal{H}\rightarrow\mathcal{K}$, and a unitary operator $U : \mathcal{K}\rightarrow\mathcal{K}$ such that
$$
 A^n = V^*U^n V \mbox{ for all } n \ge 0.
$$


TITLE: Why do we consider the viscous force to be tangential?
QUESTION [0 upvotes]: Is it that since it is a friction force, it needs to be tangential as other friction forces like kinetic friction and static friction forces are?

REPLY [1 votes]: If we orient a Cartesian coordinate system such that the xy plane coincides with the surface of interest, the viscous stress at the surface for a Newtonian fluid is given by:
$$\boldsymbol{\tau}=\sigma_{xz}\mathbf{i}+\sigma_{yz}\mathbf{j}+\sigma_{zz}\mathbf{k}\tag{1}$$where $$\sigma_{xz}=\eta\left(\frac{\partial v_x}{\partial z}+\frac{\partial v_z}{\partial x}\right)\tag{2}$$
$$\sigma_{yz}=\eta\left(\frac{\partial v_y}{\partial z}+\frac{\partial v_z}{\partial y}\right)\tag{3}$$$$\sigma_{zz}=2\eta\frac{\partial v_z}{\partial z}-\left(\frac{2}{3}\eta-\kappa\right)\left(\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}\right)\tag{4}$$where $\eta$ is the shear viscosity and $\kappa$ is the "dilatational viscosity."
From the non-slip boundary condition, we know that $v_x=v_y=v_z=0$ at the surface.  This enables us to reduce Eqns. 2-4 at the surface to:
$$\sigma_{xz}=\eta\frac{\partial v_x}{\partial z}\tag{5}$$
$$\sigma_{yz}=\eta\frac{\partial v_y}{\partial z}\tag{6}$$$$\sigma_{zz}=\left(\frac{4}{3}\eta+\kappa\right)\frac{\partial v_z}{\partial z}\tag{7}$$In general, since $\sigma_{zz}$ is not generally equal to zero, the viscous stress will not generally be tangent to the surface.  However, for the special case of an incompressible fluid, we know that $$\frac{\partial v_x}{\partial x}+\frac{\partial v_y}{\partial y}+\frac{\partial v_z}{\partial z}=0$$And then, from the zero slip boundary condition, it follows that $$\frac{\partial v_z}{\partial z}=0$$.  This means that for the special case of an incompressible fluid, $\sigma_{zz}=0$  Under these circumstances, the viscous stress will be tangent to the surface.


TITLE: Last digits of sum $n!+m!$
QUESTION [3 upvotes]: Please, give me a hint

Can the last four digits of the sum $n!+m!$ be 1990?

Or, in other words, can one find such $n,m$ that $m!+n! \mod 10^4 = 1990$?

REPLY [1 votes]: The remainder $k!\bmod 8$ is one of $1,2,6$, or (namely for all $k\ge 4$) $0$.
Since $x\equiv 1990\pmod{10}^4$ implies $x\equiv 6\pmod 8$, we conclude that wlog. $n=3$ and $m\ge 4$. So we need $m!\equiv 1984\pmod{10^4}$. Since that implies $m!\not\equiv 0\pmod 5$, we conclude $m\le 4$, but then $m!\le 43< 1984$. Hence no solution exists.


TITLE: Understanding Galois groups, too simple argumentation?
QUESTION [1 upvotes]: I am currently studying Galois Theory, and there I read the paper
Analyzing the Galois Groups of Fifth-Degree and Fourth-Degree Polynomials
In it the author states (page 26, case 1)
"Let's look at $f(x) = x - 1$ and $f(x) = x^2 + 3x + 2$. Both have coefficients and solutions that lie in $\mathbb Q$. When these conditions are met the only possible permutation on the rational numbers is the identity. Therefore, the Galois group for these types of polynomials is the trivial group $S_1$."
What does he mean "by these conditions met", this case gives no insight to me. The function $f$ has two root $-2$ and $-1$, so why I just can apply the identity permutation on them and not exchange them, like I do when I have a polynomial which has an imaginary root?

REPLY [2 votes]: The Galois Group is given by the set of $\mathbb{Q}$-automorphisms of the splitting field $\mathbb{Q}[x]/(\operatorname{f})$.
A $\mathbb{Q}$-automorphism is an automorphism that fixes $\mathbb{Q}$. (It is a map from the field to itself, which is an additive and multiplicative homomorphism.)
In the example you give, since the roots of $x^2+3x+2$ are already rational, we don't need to extend $\mathbb{Q}$ to be able to factorise $x^2+3x+2$. So the splitting field $\mathbb{Q}/(x^2+3x+2)$ is just $\mathbb{Q}$. Clearly, the only automorphism of $\mathbb{Q}$ that fixes $\mathbb{Q}$ is the identity. So the Galois Group is trivial.
What the author means is that any polynomial with rational coefficients, i.e. in $\mathbb{Q}[x]$, that factories over $\mathbb{Q}$, i.e. has $\mathbb{Q}$ as its splitting field, will have the trivial group as its Galois Group. The only automorphism from $\mathbb{Q}$ to $\mathbb{Q}$ that fixes $\mathbb{Q}$ is the identity!


TITLE: Find infimum and supremum of the function $f(x,y)=x^3+y^3$ in $\{(x,y) \in \mathbb R^2 : x^2+y^2 \le 1 \}$
QUESTION [0 upvotes]: Find infimum and supremum of the function $f(x,y)=x^3+y^3$ in $\{(x,y) \in \mathbb R^2 : x^2+y^2 \le 1 \}$.
From the Lagrange theorem: if the infimum and supremum exist then:
\begin{align} 3x^2&=\alpha x \tag{1}\\ 3y^2&=\alpha y\tag{2} \\ x^2+y^2 &\le 1 \tag{3}\end{align}
Then:
\begin{align} 3x^3&=\alpha x^2 \tag{1}\\ 3y^3&=\alpha y^2\tag{2} \\ x^2+y^2 &\le 1 \tag{3}\end{align}
S0: $$x^3+y^3 =\frac{\alpha}{3}(x^2+y^2) \le \frac{\alpha}{3}$$
But I don't know what I can do to find inf and sup.

REPLY [3 votes]: If $x^2 + y^2 \leq 1$ then $|x|, |y| \leq 1$, and therefore
$$|x^3 + y^3| \leq |x^3| + |y^3| = |x|^3 + |y|^3 \leq x^2 + y^2 \leq 1.$$
It follows that $-1 \leq x^3 + y^3 \leq 1$, for which the bounds are attained by $(\pm 1, 0)$.


TITLE: Using the Multinomial Theorem to Calculate a Finite Sum raised to an exponent
QUESTION [2 upvotes]: I know it's a simple question, but I keep getting different general formulas for the coefficients when I am trying to use the multinomial theorem for the following:
$$
\left(\sum_{k=0}^{M}\frac{(-x^2)^k}{4^kn^{k/2}k!(1+k)!}\right)^n
$$
Please help me to calculate it.
Thank you.

REPLY [1 votes]: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
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 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
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 \newcommand{\yy}{\Longleftrightarrow}$
$\ds{%
\bracks{\,\sum_{k = 0}^{M}{(-x^{2})^{k} \over 4^{k}n^{k/2}k!(1 + k)!}}^{n}:\
{\large ?}.\quad}$ Let $\ds{\quad z \equiv {-x^{2} \over 4n^{1/2}}\quad}$ and
$\ds{\quad a_{k} \equiv {1 \over k!\pars{1 + k}!}}$.

$$
\mbox{Then}\,,\quad
\bracks{\,\sum_{k = 0}^{M}{(-x^{2})^{k} \over 4^{k}n^{k/2}k!(1 + k)!}}^{n}
=
\pars{\sum_{k = 0}^{M}a_{k}z^{k}}^{n}\tag{1}
$$

\begin{align}
&\pars{\sum_{k = 0}^{M}a_{k}z^{k}}^{n}=
\sum_{k_{1} = 0}^{M}a_{k_{1}}z^{k_{1}}\sum_{k_{2} = 0}^{M}a_{k_{2}}z^{k_{2}}
\cdots\sum_{k_{n} = 0}^{M}a_{k_{n}}z^{k_{n}}\sum_{\ell = 0}^{nM}
\delta_{\ell,k_{1}\ +\ k_{2}\ +\ \cdots\ +\ k_{n}}
=
\sum_{\ell = 0}^{nM}A_{\ell}z^{\ell}
\end{align}
where

$$
\left.A_{\ell}
\equiv
\sum_{k_{1}, k_{2},\ldots,k_{n} = 0}^{M}
a_{k_{1}}a_{k_{2}}\ldots a_{k_{n}}\right\vert_{\sum\limits_{i\ =\ 1}^{n}k_{i}\ =\ \ell}
$$


TITLE: Non-homogeneous equation is also a difference equation
QUESTION [0 upvotes]: The problem is this:
The impulse response of a system is the output from this system
when excited by an input signal δ(k) that is zero everywhere, except at k = 0,
where it is equal to 1. Using this definition and the general form of the solution
of a difference equation, write the output of a linear system described by:
y(k) – 3y(k – 1) – 4y(k – 2) = δ(k) + 2δ(k – 1)

The initial conditions are: y(–2) = y(–1) = 0.
My question is: How can the particular solution be found using the method of undetermined coefficients if the non-homogeneous equation is also a difference equation?

REPLY [1 votes]: OK. So, I suppose you want to solve your difference equation for $k\geq 0$. If you set $k=0$ we easily get $y(0)=1$ and you could find $y(k)$ recursively for $k>0$, right? that's the point. To obtain the impulse response, you can see it as a response of an homogeneous equation with initial conditions different from zero. Then, check it out that the impulse response $h(k)$ to your problem is the same of this one:
$$
y(k) - 3y(k-1) - 4y(k-2)=0, \quad y(-2)=1/4, y(-1)=0
$$
Why? simply because they have the same solution for $k\geq 0$. Now its easy to find $h(k)$ right? The final solution is then $h(k)+2h(k-1)$.


TITLE: Proving pair consisting of a set and binary operation is a group and whether it is Abelian.
QUESTION [0 upvotes]: This question is in relation to Group Theory.
I am trying to determine which of the following pairs consisting of a set and a binary operation ($G$, *), is a group. And which are Abelian groups.
$1.$ $G=\mathbb R^3 , x.y:= x$ x $y$, the cross product.
$2.$ $G={x+\sqrt 2 y : x,y \mathbb Z}$ where  $a*b : a+b$, the usual addition
$3.$ $G={x+\sqrt 2 y : x,y \mathbb Q}$ where  $a*b : a+b$, the usual addition
$4.$ $G={x+\sqrt 2 y : x,y \mathbb Q}$ where  $a*b : a+b$, the usual multiplication
$5.$ $G=({x+\sqrt 2 y : x,y \mathbb Q})$ \{$0$} where  $a.b : a+b$, the usual multiplication
My understanding: I am new to this topic but I assume here i need to prove associativity here, that is that for any $3$ elements $a*(b*c)=(a*b)*c$
I would also need to show the existence of an identity element, that is $e*a=a*e=a$ where $e$ is the identity element.
And finally I would need to prove that there is closure under inverses, $a*a^{-1}=e$
In order to prove if these groups are abelian, I would need to show communtivity, where $a*b=b*a$
Is my understanding correct and if so how do i apply these axioms. A solution to one part would really help me to understand the work needed to be done to solve the rest.

REPLY [0 votes]: HINT
The most awkward result to prove in questions of this type is always associativity. However, here all but one of the operations is "the usual multiplication/addition" of reals and so you know these will be associative.  That leaves the vector product which is not associative, so just find an example of three vectors to show this. That will prove (1) is not a group and you then need do nothing further for that case.
Can you proceed now? If you show your work it will be easy to help you further.
P.S. Your understanding of the problem is correct except that you say you need to prove "closure under inverses" but what you actually must prove is the existence of inverses and also you must prove closure in general.
As an added help you should note that "the usual multiplication/addition" of reals is commutative and so that means that any of examples 2-5 which are groups will be abelian groups.


TITLE: Abstract definition of conjugate points
QUESTION [3 upvotes]: Let $S$ be a Cauchy hypersurface of a globally hyperbolic spacetime $(\mathcal{M},\mathcal{O},\mathcal{A},g,T)$ with unit normal vector field $n$. Define the exponential map on a neighborhood $U\subseteq \mathbb{R}\times S$ of $\{0\}\times S$ by $\exp(t,p)=c_p(t)$ where $c_p$ is the timelike geodesic which goes through $p$ with tangent vector $n_p$. A regular value of $\exp$ is called conjugate to $S$.
I'm having trouble relating this definition to the intuitive notion where conjugate points to $S$ are points where timelike geodesics starting at nearby points in $S$ intersect. I am sure this has to do with some mathematical theorem I'm not aware of. Can somebody help me understand this?

REPLY [0 votes]: In this context, I believe we are to consider $\exp(t,p)$ as a family of maps $\exp_t : S \to M$, and take a regular value to be a point $q \in \exp_t(S)$ such that the push-forward $\exp_{t*} : T_{\exp_t^{-1}(q)}S \to T_qM$ is a surjection. Here $t$ is to be considered merely a parameter which tells us which map we are dealing with. In other words, $q$ is a regular value we would have $T_qM = \exp_{t*}(T_{\exp_t^{-1}(q)}S).$ Since $S$ has co-dimension one, this is obviously only possible if the pre-image $\exp_t^{-1}(q)$ contains more than one point. At least this would recover, in some sense, the intuitive notion of conjugate points, and bears some resemblence to the standard definition.


TITLE: Infinite dimensional vector space of solutions
QUESTION [2 upvotes]: Possible Duplicate:
Exercise from Stein with partial differential operator

Let $S$ be a linear partial differential operator in $\mathbb{R}^{n}$, $n \geq 2$ and consider the vector space of $f \in C^{\infty}(\mathbb{R}^{n})$ such that $Sf = 0$. Why is this vector space not finite dimensional?

REPLY [2 votes]: A function can be constant in one direction and non-constant in another.
For instance, any smooth function of $y$ like $e^y$ or $\sin(y)$ can be thought of as a function on $\mathbb{R}^2$ which is smooth and annihilated by $\frac{\partial}{\partial x}$, and that's an infinite dimensional vector space of two-variable smooth functions.


text:
\begin{document}

\begin{center}
{\Large \textbf{On the infinitude of Prime $k$-tuples}} \vskip 4em

{ J. LaChapelle}\\
\vskip 2em
\end{center}

\begin{abstract}
 Starting with Zhang's theorem  on the infinitude of prime doubles \cite{Z}, we give an inductive argument that there exists an infinite number of prime $k$-tuples for at least one admissible set $\mathcal{H}_k=\{h_1,\ldots,h_k\}$ for each $k$.
\end{abstract}

\section{Introduction}

The eventual lesson learned from counting single primes was ``If you want the asymptotics of primes, you have to look at $\zeta(s)|_{\Re(s)=1}$". Indeed, in the single prime case it required a thorough understanding of $\zeta(s)$ to finally nail down the prime number theorem (PNT). Even the Selberg-Erd\"{o}s `elementary proof' of the PNT conceals $\zeta(s)$ lurking in the background \cite{HA}. Consequently, it seems unlikely that the Hardy-Littlewood $k$-tuple conjecture \cite{HL} can be settled without possessing the $k$-tuple analog of zeta.

Nevertheless, number theorists have made impressive gains in the quest to count prime $k$-tuples --- especially recently. As important as these recent advances are, the situation for counting prime $k$-tuples is rather like that for single primes prior to Riemann, Hadamard, and de la Vall\'{e}e Poussin: Without a $k$-tuple zeta function to exploit, the focus has been on showing the infinitude of prime $k$-tuples.

Perhaps the most germane are Zhang's theorem \cite{Z} and the Maynard-Tao theorem \cite{M},\cite{T} paraphrased by Granville \cite{GR};
\begin{theorem}
There exists an integer $k$ such that the following is true: If
$x+a_1,\ldots,x+a_k$ is an admissible set of forms then there are infinitely many integers
$n$ such that at least two of $n+a_1,\ldots,n+a_k$ are prime numbers.
\end{theorem}
\begin{theorem}
For any given integer $m\geq2$, there exists an integer $k$
such that the following is true: If $x+a_1,\ldots,x+a_k$ is an admissible set of forms then there are infinitely many integers $n$ such that at least $m$ of $n+a_1,\ldots,n+a_k$ are prime numbers.
\end{theorem}

Both of these theorems imply a corollary, again given by Granville \cite{GR};
\begin{corollary}
There is an integer $h$; $0<h\leq B$ such that there are infinitely many
pairs of primes $p,\,p+h$. (for some finite bound $B$)
\end{corollary}
To justify this corollary, choose a sufficiently large finite interval $(0,B]$ and string it together to cover the positive integers in the obvious way. Each interval contains at least two primes out of a finite number of combinations that could occur. Since there are an infinitude of intervals, at least one of those combinations must be represented an infinite number of times. Moreover, such a combination is necessarily admissible because $h$ will be even.

Unfortunately, the same reasoning doesn't work for $k$-tuples with $k>2$. The problem is, one can deduce an infinite number of at least one particular combination but there is no guarantee it will be an \emph{admissible} $k$-tuple. This is disappointing because belief in the $k$-tuple conjecture is strong, so one strongly expects an infinitude of admissible prime $k$-tuples.

Of course Euclid (and later several others) figured out a way to get the total number of primes without using $\zeta(s)$. And Zhang first did it for (certain)  prime doubles. On the other hand, Euler found a way to utilize $\zeta(s)$ to deduce the infinitude of single primes. Can one generalize Euler to the prime double case and thereby get a handle on a $k$-tuple zeta function? Unfortunately, it is well-known the sum over prime-double reciprocals does not diverge. So a straightforward generalization is thwarted from the start.

But maybe a straightforward generalization is not the best approach. Let's briefly re-interpret Euler's method for guidance.
First recall that, if $\mathcal{P}$ is the set of all primes, then
\begin{equation*}
\log(\zeta(s))\asymp\sum_{p\in\mathcal{P}}p^{-s}\;.
\end{equation*}
We aim to show that $\lim_{s\rightarrow1^+}\log(\zeta(s))=\infty$, but suppose we do not know Euler's product representation of $\zeta(s)$. We can still conclude the result by a simple argument.

Assume the contrary. Then $\sum\Lambda(n)/\log(n)n^s$ would converge uniformly at $s=1$, and we would have
\begin{equation*}
\lim_{s\rightarrow1^+}\sum_{n=1}^\infty\frac{\Lambda(n)}{\log(n)n^s}=\lim_{N\rightarrow\infty}\sum_{n\leq N}\frac{\Lambda(n)}{\log(n)n}\;.
\end{equation*}
Hence, partial summation would yield
\begin{equation*}
\lim_{s\rightarrow1^+}\log(\zeta(s))=\log(\log(N_{\mathrm{max}}))-\epsilon+O\left(\frac{1}{\log(N_{\mathrm{max}})}\right)
\end{equation*}
where $\epsilon$ is an end-point contribution and
\begin{equation*}
N_{\mathrm{max}}:=\lim_{N\rightarrow\infty}\sum_{n=1}^N\Lambda(n)=\lim_{N\rightarrow\infty}\sum_{p^k\leq N}\log(p).
\end{equation*}
But from Euclid we know the right-most sum must diverge, and so we have a contradiction.
This $\log(\log(N))$ behavior for $\sum_{p\leq N}1/p$ is well-known, and it shows the reciprocal of primes `just barely' diverges with $N$.

The point of reviewing Euler's method is to give a preview of our plan for prime doubles. We don't yet have a representation of the prime double zeta function $\zeta_{(2)}(s)$. So we will adapt the above argument to the prime-double case and utilize Zhang's result to infer the divergence of a certain pertinent sum for at least one admissible $\mathcal{H}_2=\{0,h\}$. It turns out that the pertinent sum to consider is
\begin{equation*}
\log'(\zeta_{(2)}(s)):=\sum_{n=1}^\infty\frac{\Lambda(n)\Lambda(n+h)}{\log(n)\log(n+h)}\frac{\log(n(n+h)^{1/2})}{(n(n+h))^{s/2}}
\asymp\sum_{p\in\mathcal{P}}\frac{\log(p(p+h))}{(p(p+h))^{s/2}}\;.
\end{equation*}
This sum happens to exhibit the $\log(\log(N))$ behavior.

Of course one could just guess this sum. But it is more satisfying and reassuring to see that it comes from the definition
\begin{equation*}
\log(\zeta_{(2)}(s)):=\sum_{n=1}^\infty\frac{\Lambda(n)\Lambda(n+h)}{\log(n)\log(n+h)}\frac{1}{(n(n+h))^{s/2}}\;,
\end{equation*}
which in turn comes from explicit formulae relating exact and average summatory functions for prime doubles \cite{LA1},\cite{LA2}.

The prime-double case is then extended to $k$-tuples by induction. The reasoning relies crucially on Zhang's theorem: Given an admissible prime $k$-tuple $(p+h_1,\ldots,p+h_k)$, Zhang's theorem implies there exists at least one $h$ such that the amended $(k+1)$-tuple $(p+h_1,\ldots,p+h_k, p+h_k+h)$ is also prime (but not necessarily admissible). Of course this is consistent with the Maynard-Tao theorem.

The final step is to show there must be at least one \emph{admissible} prime $(k+1)$-tuple by this construction. Here we rely on a lemma establishing the fortunate circumstance that $\log'(\zeta_{(2)}(s))\asymp\log'(\zeta_{[2i]}(s))$ where the left side is associated with admissible $h$ and the right side represents an equivalence class of prime doubles $[2i]$ determined by $(2i)^{l'}\sim 2i$ for all $l'\in\mathbb{N}_+$ with $2i$ such that $(2i)^l=h$ for some $i,l\in\mathbb{N}_+$.

The end game and our main result is the following theorem:
\vskip 1em
\noindent\textbf{Theorem} \,\emph{Assume $\mathcal{H}_k=\{0, h_2,\ldots, h_k\}$ is admissible. Then}
\begin{equation}
\lim_{s\rightarrow 1^+}\frac{\sum_{\mathfrak{p}_k\in\mathfrak{P}_k}\frac{\log^{k-1}(p_{(k)})}{p^s_{(k)}}}
{(-1)^{k-1}\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)}=1
\end{equation}
\emph{and}
\begin{equation}\label{divergence}
\lim_{s\rightarrow1^+}(-1)^{k-1}\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)=\infty
\end{equation}
\emph{for at least one admissible $\mathcal{H}_k$.}

\noindent Here $\mathfrak{P}_k$ is the set of admissible prime $k$-tuples and $p_{(k)}:=(p(p+h_2)\cdots(p+h_k))^{1/k}$.

To briefly recap, possessing exact and average summatory functions for admissible prime $k$-tuples allows us to infer certain relevant objects $\log(\zeta_{(k)}(s))$. Although $\zeta_{(k)}(s)$ remains elusive, $\log(\zeta_{(k)}(s))$ together with Euler's method and Zhang's theorem allow us to deduce an infinitude of at least one admissible prime $k$-tuple for all $k$. History suggests we will need to understand $\zeta_{(k)}(s)$ to go further.

\section{Definitions and lemmas}
This section will introduce some notation/definitions and establish some useful lemmas.
\begin{itemize}
  \item $\mathfrak{P}_k$ is the set of admissible prime $k$-tuples.
  \item $\mathfrak{P}_k\ni\mathfrak{p}_k:=(p,p+h_2,\ldots,p+h_k)$ where $p$ is prime and $\mathcal{H}_k=\{0, h_2,\ldots, h_k\}$ is admissible.
  \item $n_{(k)}:=(n(n+h_2)(n+h_k))^{1/k}$ for integer $n$.\footnote{The subscript $(k)$ is supposed to indicate the level $k$ and implicitly the admissible $\mathcal{H}_k$. Sometimes we will make the dependence on $\mathcal{H}_k$ explicit by writing for example $p_{(h_2)}$ for a particular prime double.}
  \item $\mu_{(k)}(n):=(-1)^k\mu(n)\cdots\mu(n+h_k)\,$.
  \item $\lambda_{(k)}(n):=\Lambda(n)\cdots\Lambda(n+h_k)/\log(n)\cdots\log(n+h_k)\,$.
  \item $\Lambda_{(k)}(n):=\lambda_{(k)}(n)\log^{k}(n_{(k)})\,$.
  \item $\log\left(\zeta_{(k)}(s)\right):=\sum_{n=1}^\infty\lambda_{(k)}(n)/n_{(k)}^s\,,\;\;\;\;\Re{(s)}>1\,$.
\end{itemize}

\noindent Now some lemmas.
\begin{lemma}\label{lemma 1}
\begin{equation}\label{bound}
\log\left(\zeta_{(2)}(s)\right)\asymp\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\frac{1}{p_{(2)}^{s}} =:\log\left(\mathfrak{z}_{(2)}(s)\right)\;.
\end{equation}
\end{lemma}
\emph{proof}:
By definition,
\begin{eqnarray}
\log\left(\zeta_{(2)}(s)\right)
&=&\sum_{\mathfrak{p}^\omega_2\in\mathfrak{P}_2}\frac{1}{\omega p^{\omega s/2}}\frac{1}{\omega' (p^\omega+2i)^{\omega's/2}}\;,\;\;\;\;\;\;\Re(s)>1\notag\\
&=&\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\frac{1}{p^{s/2}}\frac{1}{ (p+2i)^{s/2}}+\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\sum_{\omega=2}^\infty\sum_{\omega'=2}^\infty\frac{1}{\omega p^{\omega s/2}}\frac{1}{\omega' (p^\omega+2i)^{\omega's/2}}\notag\\\notag\\
&=:&\log\left(\mathfrak{z}_{(2)}(s)\right)+S(s)
\end{eqnarray}
where $\mathfrak{p}^\Bold{\omega}_2:=(p^\omega,(p^\omega+2i)^{\omega'})$ is a prime-power double.
But, for $s=\sigma+it$ with $\sigma, t\in\R$ and $\sigma>1$,
\begin{eqnarray}
\left|S(s)\right|&\leq&\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\sum_{\omega=2}^\infty\sum_{\omega'=2}^\infty\left|\frac{1}{\omega p^{\omega s/2}}\frac{1}{\omega' (p^\omega+2i)^{\omega's/2}}\right|\notag\\
&<&\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\sum_{\omega=2}^\infty\sum_{\omega'=2}^\infty\left|\frac{1}{ p^{\omega s/2}}\frac{1}{ (p^\omega+2i)^{\omega's/2}}\right|\notag\\
&<&\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\sum_{\omega=2}^\infty\sum_{\omega'=2}^\infty\left|\frac{1}{ p^{\omega s/2}}\frac{1}{ (p+2i)^{\omega's/2}}\right|\notag\\
&\leq&\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\frac{1}{\left(p^\sigma-p^{\sigma/2}\right)}
\frac{1}{\left((p+2i)^\sigma-(p+2i)^{\sigma/2}\right)}\notag\\
&<&\sum_{n=2}^\infty\frac{1}{\left(n^\sigma-n^{\sigma/2}\right)}\frac{1}{\left((n+2i)^\sigma-(n+2i)^{\sigma/2}\right)}\notag\\
&<&\sum_{n=2}^\infty\frac{1}{n^{1/2}(n^{1/2}-1)}\frac{1}{(n+2i)^{1/2}((n+2i)^{1/2}-1)}\notag\\
&<&\sum_{n=2}^\infty\frac{1}{(n-1)^{3/2}}=\zeta(3/2)\;.
\end{eqnarray}$\qed$

The following two lemmas utilize a particularly useful interpretation of sums of the form $\sum \mu_{(2)}(n)\Lambda_{(2)}(n)f(n)$. Since the pre-factor localizes onto prime doubles \cite{LA1}, it is advantageous to view the sum as a double sum over the coprime $2$-lattice.

\begin{lemma}\label{lemma 2}
Denote the coprime $2$-lattice by $\{(n_1,n_2)\in\mathbb{N}_+^2 \;|\; \mathrm{gcd}(n_1,n_2)=1\}$, and let $2i\leq2j=(2i)^l$ with fixed $l\in\mathbb{N}_+$. Then
\begin{equation}
\sum_{n=1}^{\infty}\frac{\mu_{(2i)}(n)\lambda_{(2i)}(n)}{n^{s/2}(n+2i)^{s/2}}\log(n(n+2i))
\geq\sum_{n=1}^{\infty}\frac{\mu_{(2j)}(n)\lambda_{(2j)}(n)}{n^{s/2}(n+2j)^{s/2}}\log(n(n+2j))-R(2j,2i;\infty)
\end{equation}
where $R(2j,2i;\infty):=\lim_{N\rightarrow\infty}\sum_{n=N+1-(2j-2i)}^{N}\frac{\mu_{(2j)}(n)\lambda_{(2j)}(n)}{n^{s/2}(n+2j)^{s/2}}\log(n(n+2j))$.
\end{lemma}
\emph{proof}: For $N>1+(2j-2i)$,
\begin{eqnarray}
\sum_{n=1}^{N}\frac{\mu_{(2i)}(n)\lambda_{(2i)}(n)}{n^{s/2}(n+2i)^{s/2}}\log(n(n+2i))\notag\\
&&\hspace{-2.0in}=\sum_{n_1\leq N}\sum_{n_2\leq N+2i}
\frac{\mu(n_1)\lambda_{(1)}(n_1)\mu(n_2)\lambda_{(1)}(n_2)}{n_1^{s/2}n_2^{s/2}}\log(n_1n_2)
\,\delta(n_2\,,\, n_1+2i)\notag\\
&&\hspace{-2.0in}=\sum_{n_1=1}^{N}\,\frac{\mu(n_1)\lambda(n_1)}{n_1^{s/2}}
\left(\sum_{n_2\leq N+2i}
\frac{\mu(n_2)\lambda(n_2)}{n_2^{s/2}}\,\log(n_1n_2)\,\delta(n_2\,,\, n_1+2i)\right)\notag\\
&&\hspace{-2.0in}\geq\sum_{n_1=1+(2j-2i)}^{N}\,\frac{\mu(n_1)\lambda(n_1)}{n_1^{s/2}}
\left(\sum_{n_2\leq N+2i}
\frac{\mu(n_2)\lambda(n_2)}{n_2^{s/2}}\,\log(n_1n_2)\,\delta(n_2\,,\, n_1+2i)\right)\notag\\
&&\hspace{-2.0in}=\sum_{n'_1=1}^{ N-(2j-2i)}\frac{\mu(n'_1)\lambda(n'_1)}{{n'}_1^{s/2}}
\left(\sum_{n_2\leq N+2i}\frac{\mu(n_2)\lambda(n_2)}{n_2^{s/2}}\,\log(n'_1n_2)
\,\delta(n_2\,,\, n'_1+2j)\right)\notag\\
&&\hspace{-2.0in}=\sum_{n=1}^{N-(2j-2i)}\frac{\mu_{(2j)}(n)\lambda_{(2j)}(n)}{n^{s/2}(n+2j)^{s/2}}\log(n(n+2j))\;.
\end{eqnarray}
The delta function in the second line restricts the double sum to the appropriate ray $\mathfrak{r}_{(2i)}$ in the coprime $2$-lattice. The fourth line is a simple truncation of the outer sum. The fifth line holds since $n_1\mapsto n'_1=n_1+2i-2j$ bijectively maps lattice intersections of $\mathfrak{r}_{(2i)}$
to lattice intersections of $\mathfrak{r}_{(2j)}$ precisely because $2j=(2i)^l$ which ensures that $\mathrm{gcd}(n'_1,n'_1+2j)=1$ and the inner sum gets evaluated on a congruent set of coprimes. The result follows as $N\rightarrow\infty$ since both series converge for $\Re{(s)}>1$.
$\qed$

By the same token,
\begin{lemma}\label{lemma 3}
Let $2j\leq2m=(2j)^l$ with fixed $l\in\mathbb{N}_+$, then
\begin{equation}
\sum_{n=1}^{\infty}\frac{\mu_{(2m)}(n)\lambda_{(2m)}(n)}{n^{s/2}(n+2m)^{s/2}}\log(n(n+2m))
\geq\sum_{n=1}^{\infty}\frac{\mu_{(2j)}(n)\lambda_{(2j)}(n)}{n^{s/2}(n+2j)^{s/2}}\log(n(n+2j))-R(2m,2j;1)
\end{equation}
where $R(2m,2j;1):=\sum_{n=1}^{(2m-2j)}\frac{\mu_{(2j)}(n)\lambda_{(2j)}(n)}{n^{s/2}(n+2j)^{s/2}}\log(n(n+2j))$.
\end{lemma}
\emph{proof}:
The proof follows the same argument as the preceding lemma.
\begin{eqnarray}
\sum_{n=1}^{N}\frac{\mu_{(2m)}(n)\lambda_{(2m)}(n)}{n^{s/2}(n+2m)^{s/2}}\log(n(n+2m))\notag\\
&&\hspace{-2.0in}=\sum_{n'_1=1+(2m-2j)}^{N+(2m-2j)}\,\frac{\mu(n'_1)\lambda(n'_1)}{{n'_1}^{s/2}}
\left(\sum_{n_2\leq N+2m}
\frac{\mu(n_2)\lambda(n_2)}{n_2^{s/2}}\,\log(n'_1n_2)\,\delta(n_2\,,\, n'_1+2j)\right)\notag\\
&&\hspace{-2.0in}\geq\sum_{n'_1=1+(2m-2j)}^{N}\,\frac{\mu(n'_1)\lambda(n'_1)}{{n'_1}^{s/2}}
\left(\sum_{n_2\leq N+2m}
\frac{\mu(n_2)\lambda(n_2)}{n_2^{s/2}}\,\log(n'_1n_2)\,\delta(n_2\,,\, n'_1+2j)\right)\notag\\
&&\hspace{-2.0in}=\sum_{n=1}^{N}\frac{\mu_{(2j)}(n)\lambda_{(2j)}(n)}{n^{s/2}(n+2j)^{s/2}}\log(n(n+2j))
-R(2m,2j;1)\;.
\end{eqnarray}
Again, the crucial fact is that $\mathrm{gcd}(n_1,n_1+2m)=1$  and the map $n_1\mapsto n'_1=n_1+2m-2j$ ensures that $\mathrm{gcd}(n'_1,n'_1+2j)=1$.
$\qed$

Therefore choosing $2i\leq2j=(2i)^l$ or $(2j)^l=2i\geq 2j$,
\begin{eqnarray}
\log'\left(\mathfrak{z}_{(2i)}(s)\right)
&=&\sum_{n=1}^\infty\frac{\mu_{(2i)}(n)\lambda_{(2i)}(n)}{n^{s/2}(n+2i)^{s/2}}\log(n_{(2i)})\notag\\
&\geq&\sum_{n=1}^\infty\frac{\mu_{(2j)}(n)\lambda_{(2j)}(n)}{n^{s/2}(n+2j)^{s/2}}\log(n_{(2j)})\pm |2j-2i|\notag\\
&=&\log'\left(\mathfrak{z}_{(2j)}(s)\right)\pm|2j-2i|
\end{eqnarray}
follows from Lemma \ref{lemma 2} and Lemma \ref{lemma 3} because the remainder in both cases is certainly bounded by $|2j-2i|$.
So if $2i$ is square-free and $2j=(2i)^l$, then
\begin{equation}
\log'\left(\mathfrak{z}_{[2i]}(s)\right)\asymp\log'\left(\mathfrak{z}_{(2j)}(s)\right),\;\;\;\;\Re{(s)}>1
\end{equation}
where the equivalence class includes all integer powers of $2i$.

Together with Lemma \ref{lemma 1} conclude
\begin{corollary}\label{equivalence}
\begin{equation}
\log'\left(\zeta_{[2i])}(s)\right)\asymp\log'\left(\zeta_{(2j)}(s)\right),\;\;\;\;\Re{(s)}>1\;.
\end{equation}
\end{corollary}

\section{Euler's lead}
With these preliminaries, we can follow Euler's method for prime doubles. Recall the definition
\begin{definition}
\begin{equation}
\log\left(\zeta_{(2)}(s)\right):=\sum_{n=1}^\infty\frac{\lambda_{(2)}(n)}{n_{(2)}^s}\,,\;\;\;\;\Re{(s)}>1\;.
\end{equation}
\end{definition}
We have
\begin{proposition}\label{proposition}
Let $2j=h$ where $h$ comes from Zhang's theorem. Then
\begin{equation}
\lim_{s\rightarrow1^+}\log'\left(\zeta_{(2j)}(s)\right)=\infty\;.
\end{equation}
\end{proposition}
\emph{Proof}:
Assume the contrary. Then $\lim_{s\rightarrow1^+}\log'\left(\zeta_{2j}(s)\right)$ is bounded and so converges. Consequently, for every $\varepsilon>0$ there exists an $M$ such that $m>M$ implies
\begin{equation}
\left|\sum_{n=m+1}^{m+l}\frac{\lambda_{(2j)}(n)}{(n+2j)}\log(n_{(2j)})\right|<\varepsilon
\end{equation}
for each $l\in\mathbb{N}_+$. It follows that $\log'\left(\zeta_{2j}(s)\right)$ converges uniformly for $\Re{(s)}\geq 1$.

Now, the PNT implies $\lambda_{(1)}(n+2j)=\lambda(n+2j)=O(1/\log(n+2j))$. Moreover, $\log(n_{(2)})/\log(n)= O(1)$. So by uniform convergence, the PNT, and partial summation we get
\begin{eqnarray}
\lim_{s\rightarrow1^+}\log'\left(\zeta_{(2j)}(s)\right)
&>&-\lim_{N\rightarrow\infty}\sum_{n=1}^N\frac{\lambda_{(2j)}(n)}{(n+2j)}\log(n_{(2j)})\notag\\
&\sim&-\lim_{N\rightarrow\infty}
\sum_{n=1}^N\frac{w_{(2j)}(n)\Lambda(n)}{(n+2j)\log(n+2j)}\notag\\
&=&\log(\log(N_{\mathrm{max}}^{(j)}))-\epsilon_{2j}+O\left(\frac{1}{\log(N_{\mathrm{max}}^{(j)})}\right)
\end{eqnarray}
where the weight  $w_{(2j)}(n)=1$ if $\Lambda(n)\neq0\wedge\Lambda(n+2j)\neq0$ and $w_{(2j)}(n)=0$ otherwise, $N_{\mathrm{max}}^{(j)}:=\lim_{N\rightarrow\infty}\sum_{n=1}^{N}w_{(2j)}(n)\Lambda(n)$, and the constant $\epsilon_{2j}$ is an inconsequential end-point contribution. But the PNT and Zhang's theorem imply $N_{\mathrm{max}}^{(j)}=\infty$, and we arrive at a contradiction.
$\QED$

Therefore, for at least one $2j$, Proposition \ref{proposition} and Corollary \ref{equivalence} imply
\begin{corollary}
\begin{equation}
\lim_{s\rightarrow1^+}\log'\left(\mathfrak{z}_{[2i]}(s)\right)
=\sum_{\mathfrak{p}_2\in\mathfrak{P}_2}\frac{\log(p_{[2i]})}{p_{[2i]}}
=\infty
\end{equation}
where $[2i]$ is the equivalence class determined by $(2i)^{l'}\sim 2i$ for all $l'\in\mathbb{N}_+$ with $2i$ such that $(2i)^l=2j$ for some $i,l\in\mathbb{N}_+$.
\end{corollary}
The sum must include an infinite number of terms, and so there are infinitely many $\mathfrak{p}_{2}\in\mathfrak{P}_2$ for each admissible $\mathcal{H}_{2}=\{0,[2i]\}$.

\section{Proof of Theorem}
We restate the theorem for easy reference.
\begin{theorem}\label{theorem}
Assume $\mathcal{H}_k$ is admissible. Then
\begin{equation}
\lim_{s\rightarrow 1^+}\frac{\sum_{\mathfrak{p}_k\in\mathfrak{P}_k}\frac{\log^{k-1}(p_{(k)})}{p^s_{(k)}}}
{(-1)^{k-1}\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)}=1
\end{equation}
and
\begin{equation}\label{divergence}
\lim_{s\rightarrow 1^+}(-1)^{k-1}\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)=\infty
\end{equation}
for at least one admissible $\mathcal{H}_k$.
\end{theorem}

\emph{Proof}: It is straightforward to show that (\ref{bound}) generalizes to $\log\left(\zeta_{(k)}(s)\right)\asymp\sum_{\mathfrak{p}_k\in\mathfrak{P}_k}p_{(k)}^{-s}$
with $\left|S(s)\right|<\zeta(\frac{k+1}{k})$. So
\begin{equation}
\sum_{\mathfrak{p}_k\in\mathfrak{P}_k}\frac{\log^{k-1}(p_{(k)})}{p_{(k)}^{s}}
\asymp(-1)^{k-1}\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)\;.
\end{equation}

Now let $w_{(k+1)}(n)=1$ encode the condition that $\Lambda_{(k)}(n)\neq0\wedge\Lambda(n+h_{k+1})\neq0$ and $w_{(k+1)}(n)=0$ otherwise. Evidently $w_{(k+1)}(n)= w_{(k)}(n)w_{(1)}(n+h_{k+1})$. In particular
 \begin{equation}
w_{(k+1)}(n)\Lambda(n+h_{k+1})=w_{(k)}(n)\Lambda(n+h_{k+1})\;.
\end{equation}
Also note that
\begin{eqnarray}
\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)
&=&(-1)^{k-1}\sum_{n=1}^\infty\frac{\lambda_{(k)}(n)}{n^s_{(k)}}\log^{k-1}(n_{(k)})\notag\\
&>&(-1)^{k-1}\sum_{n=1}^\infty\frac{\lambda_{(k)}(n)}{(n+h_{k})^s}\log^{k-1}(n_{(k)})\notag\\
&=&(-1)^{k-1}\sum_{n=1}^\infty \frac{w_{(k)}(n)\lambda_{(k)}(n)}{\log^{1-k}(n_{(k)})}\,\frac{1}{(n+h_{k})^s}\;.
\end{eqnarray}
So assuming $w_{(k)}(n)\lambda_{(k)}(n)=O(\lambda(n+h_k)\log^{1-k}(n_{(k)}))$ with $\mathcal{H}_k$ admissible is tantamount to assuming (\ref{divergence}) which in turn implies infinitely many $\mathfrak{p}_k\in\mathfrak{P}_k$.

Consequently, adopting the assumption that $w_{(k)}(n)\lambda_{(k)}(n)=O(\lambda(n+h_k)\log^{1-k}(n_{(k)}))$ for at least one admissible $\mathcal{H}_k$ --- which is tautological for $k=1$, get
\begin{eqnarray}\label{condition}
(-1)^k\log^{(k)'}\left(\zeta_{(k+1)}(s)\right)
&=&\sum_{n=1}^\infty\frac{\lambda_{(k+1)}(n)}{n^s_{(k+1)}}
\log^k(n_{(k+1)})\notag\\
&>&\sum_{n=1}^\infty\frac{\lambda_{(k+1)}(n)}{(n+h_{k+1})^s}
\log^k(n_{(k+1)})\notag\\
&>&\sum_{n=1}^\infty\frac{\lambda_{(k+1)}(n)}{(n+h_{k+1})^s}
\left(\frac{k}{k+1}\right)^k\log^k(n_{(k)})\notag\\
&\sim&\left(\frac{k}{k+1}\right)^k\sum_{n=1}^\infty
\frac{w_{(2)}(n+h_k)\Lambda(n+h_k)}{\log(n+h_{k+1})(n+h_{k+1})^s}\notag\\
\end{eqnarray}
where the binomial expansion was used in the third line (keeping only the first term), and the last line uses the assumption as well as $\lambda(n+h_{k+1})=O(1/\log (n+h_{k+1}))$ implied by the PNT and $\log(n_{(k)})/\log(n+h_k)=O(1)$.

As in the prime double case, assume $(-1)^k\log^{(k)'}\left(\zeta_{(k+1)}(s)\right)$ is bounded at $s=1$. Partial summation yields
\begin{equation}
\lim_{s\rightarrow 1^+}(-1)^k\log^{(k)'}\left(\zeta_{(k+1)}(s)\right)
\sim\log(\log(\mathrm{max}\,{ x}))-\epsilon_{h_{k+1}}\;.
\end{equation}
But again we have a contradiction, because $\mathrm{max}\,{ x}=\lim_{N\rightarrow\infty}\sum_{n=1+h_k}^{N+h_k}w_{(2)}(n)
\Lambda(n)=\infty$ follows from Zhang's work for at least one $2j_{k+1}$ with $0<2j_{k+1}\leq h_{k+1}-h_k$ provided $h_{k+1}-h_k$ is chosen large enough. Hence, $\lim_{s\rightarrow 1^+}(-1)^k\log^{(k)'}\left(\zeta_{(k+1)}(s)\right)=\infty$ for the associated equivalence class $[2i_{k+1}]$ by previous arguments.

If it happens that $2i_{k+1}\equiv0$ (mod $k+1$), then $h_k+[2i_{k+1}]$ and $h_k$ belong to the same residue class  (mod $k+1$) so $\mathcal{H}_{k+1}=\{0,\ldots,h_k,h_k+[2i_{k+1}]\}$ is automatically admissible. Conversely, if $2i_{k+1}\equiv\!\!\!\!\not \;\;0$ (mod $k+1$), then $\mathcal{H}_{k+1}$ can be rendered admissible for a suitable choice of representative in $[2i_{k+1}]$. For example, if $2i_{k+1}\equiv\!\!\!\!\not \;\;0$ (mod $k+1$), then $h_{k+1}=h_k+(2i_{k+1})$ and $h_{k+1}=h_k+(2i_{k+1})^2$ belong to different residue classes  (mod $k+1$). Consequently at least one of them will yield an admissible $\mathcal{H}_{k+1}$, because at least one case will not occupy the complete set of residue classes modulo primes.

Finally, Zhang's result guarantees the induction assumption is true at $k=2$ for at least one $\mathcal{H}_2=\{0,2j_2\}$. It follows that $\lim_{s\rightarrow 1^+}(-1)^k\log^{(k-1)'}\left(\zeta_{(k)}(s)\right)=\infty$ for at least one admissible $\mathcal{H}_{k}=\{0,[2i_2],[2i_2]+[2i_3],\ldots,\sum_{\hat{k}\leq k}[2i_{\hat{k}}]\}$  for all $k$ by induction.
$\QED$

\begin{corollary}
Given an admissible $\mathcal{H}_k$ satisfying Theorem \ref{theorem}, if some $2j_{\hat{k}}=2^a$ for integer $a$, then there are infinitely many twin primes.
\end{corollary}

\emph{Proof}: The $(0,\ldots,\hat{k},\ldots)$ components of $\mathfrak{p}_k$ determine an admissible $\mathcal{H}_2=\{0,2^a\}$ and its associated prime doubles $\mathfrak{p}_2$ of which there are necessarily infinitely many. Then previous arguments imply the claim. Of course similar statements for other types of prime doubles can be made for any component of $\mathfrak{p}_k$ with $2j_{\hat{k}}=(2i_{\hat{k}})^a$.
$\QED$

One would like to extend this theorem to \emph{all} admissible $\mathcal{H}_k$. But such an extension would require that $\mathrm{max}\,{ x}\rightarrow\infty$ for \emph{all} prime doubles. In this case, the assumption used for the induction argument could be made for all admissible $\mathcal{H}_k$ and verified at $k=2$.

\begin{corollary}
If $\zeta_{(2)}(s)$ is meromorphic on $\C$ with a single simple pole at $s=1$, then Theorem \ref{theorem} holds for all admissible $\mathcal{H}_k$.
\end{corollary}

\section{Acknowledgments}
I thank A. Granville for offering some helpful advice.


TITLE: Combinatorial problem - multisets
QUESTION [0 upvotes]: As I am solving some basic combinatorial problems today, I found out this problem: How many different 5-digit numbers can be formed from digits 2, 2, 7, 7, 9? Can someone guide me to a solution for this problem? It's probably something related to multisets. I'm beginner in this area. Thanks!

REPLY [3 votes]: So, you have five spots to put $2,2,7,7,9$ into. First, you pick two spots out of the five for the $2$'s, for which there are $\binom{5}{2}$ ways. Now, there are three spots remaining, and I want to pick two spots out of the three for the $7$'s. For this there are $\binom{3}{2}$ ways. (Note that if you were to pick one spot for the $9$, $\binom{3}{1} = \binom{3}{2}$. Hence, there are
$$\binom{5}{2}\binom{3}{2} = \frac{5!}{2!2!1!} = 30$$
ways of doing this.

REPLY [1 votes]: It is just asking you indirectly that
How many arrangements can be made with the given numbers.
If the numbers would have been different than it would have simply been 5! without any repetition, But since 2 and 7 are repeated twice it can be arranged in 5!/2!2!.


TITLE: Group containing both $5\mathbb{Z}$ and $7\mathbb{Z}$
QUESTION [2 upvotes]: Prove that there doesnot exist any proper subgroup of $(\mathbb{Z},+)$ containing both $5\mathbb{Z}$ and $7\mathbb{Z}$
 Since $5$ and $7$ are coprime hence there cannot exist any proper subgroup containing multiples of both $5$ and $7$ if it contains then it must be  the group $\mathbb{Z}$ itself
Is m above reason correct please help?

REPLY [0 votes]: The key is that 5 and 7 are indeed coprime, but the numbers themselves are irrelevant. In fact, you might see it clearer if you turn to a more general situation:
Claim. If $m,n \in \mathbb{Z}$ are two coprime integers, then the only subgroup of $(\mathbb{Z},+)$ containing $m\mathbb{Z}$ and $n\mathbb{Z}$ is precisely $\mathbb{Z}$.
Proof. Let $H \subseteq \mathbb{Z}$ be a subgrup of $(\mathbb{Z},+)$ containing $m\mathbb{Z}$ and $n\mathbb{Z}$. Clearly $\mathbb{Z}=1\mathbb{Z}$, so it suffices to show that $1 \in \mathbb{Z}$.
Since $m$ and $n$ are coprime, by Bézout's identity we have that
$$mx+ny=1$$
for some $x,y \in \mathbb{Z}$.
Besides, $m,n \in H$ and $H$ is closed under addition, so it follows that $1 \in H$ and we are done.


TITLE: Show that $\mathfrak{sl}_n$ is the Lie algebra of the algebraic group $SL_n$
QUESTION [2 upvotes]: I am currently reading the book "Linear Algebraic Group" by Springer, more precisely in chapter 4 where Lie algebras of linear algebraic groups are introduced. I would like to prove that the Lie algebra of $SL_n$ is given by the algebra $\mathfrak{sl}_n$ of $n\times n$ matrices with trace zero using the formalism developped in the book.
Let $k$ be an algebraically closed field and consider the linear algebraic group $G=GL_n$, whose affine algebra is given by $k[G]=k[T_{i,j},\Delta^{-1}]_{1\leq i,j \leq n}$ where $\Delta=\det(T_{i,j})$. Inside $G$, we have the subgroup $H = SL_n$ which is Zariski closed in $G$, hence it is again a linear algebraic group whose affine algebra is $k[H]=k[T_{i,j}]/(\Delta - 1)$.
I know already that the Lie algebra $L(G)$ of $G$ can be identified with the algebra $\mathfrak{gl}_n$ of $n\times n$ matrices with usual Lie brackets. That is because the $k$-derivations of $k[G]$ commuting with all left translations are exactly the $D_X$ for $X=(x_{i,j})\in \mathfrak{gl}_n$, defined by
$$D_XT_{i,j}:=-\sum_{h=1}^nT_{ih}x_{hj}$$
To compute the Lie algebra $L(H)$ of $SL_n$, the book recommends to proceed in the following manner. I let $J=(\Delta-1)$ be the ideal of $k[G]$ defining $H$, and I consider $\mathcal D_{G,H}$ the set of all the $k$-derivations of $k[G]$ preserving $J$. Then, there is a natural isomorphism
$$\mathcal D_{G,H}\cap L(G) \cong L(H)$$
Whence, to prove that $L(H)$ identifies with $\mathfrak{sl}_n$, I need to show the following:

We have $D_X(\Delta - 1) \in (\Delta - 1)$ if and only if $\operatorname{Trace}(X)=0$.

By argueing on dimension, it is actually enough to just prove that $D_X(\Delta - 1) \in (\Delta - 1)$ for $X = E_{i,j}$ or $E_{i,i}-E_{j,j}$ for $i\not = j$. I thought that it would be easy enough, but well... The computations are terrible and I can't reach the desired conclusion (even considering the case $n=2$).
Is there any "clever" way to tackle this problem ? Also, is there maybe another way to characterize the Lie algebra that would make the argument more direct ?

REPLY [1 votes]: I believe I have found another way of argueing that $L(SL_n)$ is identified with $\mathfrak{sl}_n$ by computing the tangent space $T_{I_n}SL_n$ rather than by considering left-invariant derivations. Recall that $G = GL_n$ and $H = SL_n$.
By definition of the module of differentials $\Omega_H:=\Omega_{k[H]/k}$, I know that there are isomorphisms of $k$-vector spaces$$T_{I_n}H \cong \operatorname{Hom}_{k[H]}(\Omega_H,k_{I_n})\cong \operatorname{Hom}_k(\Omega_H(I_n),k)$$
where $k_{I_n}$ is $k$ seen as a $k[H]$-module via the identification $k[H]/M_{I_n}\cong k$, $M_{I_n}$ being the maximal ideal in $k[H]$ of regular functions vanishing at $I_n$ ; and then $\Omega_H(I_n) = \Omega_H/M_{I_n}\Omega_H$ (these notations are the ones used in Springer's book).
Now, because $k[H] = k[T_{i,j}]/(\Delta - 1)$, by denoting $t_{i,j}$ the image of $T_{i,j}$ in $k[H]$, I know that $\Omega_H$ is generated by the $dt_{i,j}$. If I consider the canonical basis $(e_{i,j})$ of $k[H]^{n^2}$ and define the map $k[H]^{n^2}\rightarrow \Omega_H$ sending $e_{i,j}$ to $dt_{i,j}$, then the kernel is the submodule generated by the element $\sum_{i,j=1}^n \frac{\partial \Delta}{\partial T_{i,j}}(t)e_{i,j}$ (the term "$-1$" disappears in the derivation). Thus I have a presentation of $\Omega_H$ as the quotient of the $n^2$-dimensional free module over $k[H]$ with the above kernel.
Now, I would like to understand the quotient $\Omega_H(I_n)$. This amounts to tensoring $\Omega_H$ with $k[H]/M_{I_n}\cong k$ over $k[H]$, the isomorphism with $k$ being given by evaluating in $I_n$. Thus $$\Omega_H(I_n)\cong k^{n^2}/(\sum_{i,j=1}^n  \frac{\partial \Delta}{\partial T_{i,j}}(I_n)e_{i,j})$$
But now, if $X$ is the matrix whose $(i,j)-$coefficient is $T_{i,j}$, then $\frac{\partial \Delta}{\partial T_{i,j}}$ is its $(i,j)-$cofactor. Evaluating it at $I_n$ (that is, taking $T_{i,j}=\delta_{i,j}$), these partial derivatives are $1$ when $i=j$ and $0$ otherwise. Thus, $\Omega_H(I_n)\cong k^{n^2}/(\sum_{i}^n  e_{i,i})$.
I need now to compute the dual of this vector space. That is the set of all linear form on $k^{n^2}\cong M_n(k)$ vanishing on the identity. Because every linear form of $M_n(k)$ has the form $M\mapsto \operatorname{Trace}(XM)$ for some $X\in M_n(k)$, I see now that $T_{I_n}H$ can naturally be identified with the space $\mathfrak sl_n$. It would remain to check that the Lie algebra structure of $T_{I_n}H$ coincide with that of $\mathfrak sl_n$. That is something I am still unsure, as the Lie algebra structure on the tangent space comes from that on the derivations of $k[H]$, so it's a little abstract.


TITLE: Confusion about meaning of 1, 1', and 2 in finding $S_3$ irreps, and how they correspond to the representation of the group elements
QUESTION [1 upvotes]: I am working through the paper Non-Abelian Discrete Symmetries in Particle Physics, which can be found here. I am confused about the following highlighted passage found on pg. 8:

I understand this follows from the fact that there are two one dimensional irreps and one two dimensional irreps. My main question is asking if my following interpretation of this is correct:
The above statement is saying that the representation of each group $g\in G$ element can be written as the direct sum
\begin{align}
D(g)=\mathbf{1}_{g}\oplus\mathbf{1'}_g\oplus \mathbf{2}_g.
\end{align}
I don't think this is correct, but I'm a bit stuck at the moment. Could somebody please clarify how I go about finding $\mathbf{1}, \mathbf{1'}$, and $\mathbf{2}$, and how exactly that corresponds to the representation of the group elements?

REPLY [2 votes]: Some context might be helpful:
In the paper you refer to, $m_n$ denotes the number of $n$-dimensional irreps of $S_3$. The passage you are reading to is trying to count how many irreps there are of each dimension.
The paper quotes two well-known facts:

The order of the group equals $m_1 + 4m_2 + 9m_3 + 16m_4 + \dots $
The number of conjugacy classes equals $m_1 + m_2 + m_3 + m_4 + \dots$

Since the order of $S_3$ is $6$, and the number of conjugacy classes is $3$, the only possibility is that
$$ m_1 = 2, \ \  \ m_2 = 1, \ \  \ m_3 = m_4 = \dots = 0.$$
The irreps of $S_3$ are:

The one-dimensional "identity" irrep $\mathbf{1}$, where every element is represented by $1$.
The one-dimensioanl "alternating" irrep $\mathbf{1}'$, where every even element is represented by $1$ and every odd element is represented by $-1$.
The two-dimensional irrep $\mathbf{2}$, obtained by viewing elements of $S_3$ as $2\times 2$ matrices describing symmetries of a triangle.

Your statement that "the representation of each group element $g \in G$ can be written as the direct sum $\mathbf{1}_g \oplus \mathbf{1}'_g \oplus \mathbf{2}_g$" doesn't make a great deal of sense to me, I'm afraid! There is no such thing as "the representation". There are many possible representations of $S_3$: any direct sum of $\mathbf{1}$, $\mathbf{1}'$ and $\mathbf{2}$ will do.


text:
\DOC GEN_BETA_CONV

\TYPE {GEN_BETA_CONV : term -> thm}

\SYNOPSIS
Beta-reduces general beta-redexes (e.g. paired ones).

\KEYWORDS
conversion.

\DESCRIBE
The conversion {GEN_BETA_CONV} will perform beta-reduction of simple
beta-redexes in the manner of {BETA_CONV}, or of generalized beta-redexes such
as paired redexes.

\FAILURE
{GEN_BETA_CONV tm} fails if {tm} is neither a simple nor a tupled beta-redex.

\EXAMPLE
The following examples show the action of {GEN_BETA_CONV} on tupled redexes and
others:
{
  # GEN_BETA_CONV `(\x. x + 1) 2`;;
  val it : thm = |- (\x. x + 1) 2 = 2 + 1

  # GEN_BETA_CONV `(\(x,y,z). x + y + z) (1,2,3)`;;
  val it : thm = |- (\(x,y,z). x + y + z) (1,2,3) = 1 + 2 + 3

  # GEN_BETA_CONV `(\[a;b;c]. b) [1;2;3]`;;
  val it : thm = |- (\[a; b; c]. b) [1; 2; 3] = 2
}
However, it will fail if there is a mismatch between the varstruct and the
argument, or if it is unable to make sense of the generalized abstraction:
{
  # GEN_BETA_CONV `(\(SUC n). n) 3`;;
  Exception: Failure "term_pmatch".

  # GEN_BETA_CONV `(\(x,y,z). x + y + z) (1,x)`;;
  Exception: Failure "dest_comb: not a combination".
}

\SEEALSO
BETA_CONV, MATCH_CONV.

\ENDDOC

subset_name: curated


TITLE: Higher Ext vanish iff module is free
QUESTION [1 upvotes]: Let $\Lambda$ be a commutative ring with $1$. Let $M$ be a $\Lambda$-module. Is it true that $\mathrm{Ext}^i_\Lambda(M,\Lambda)=0$ for all $i\ge1$ if and only if $M$ is free over $\Lambda$?
The 'if' part of the question is clear: compute the Ext groups by a projective resolution of $M$. Since $M$ is free, we may choose the trivial resolution, which makes all Ext groups vanish.
The question is, does the converse hold?

REPLY [2 votes]: The converse is false; if $M$ is any projective module, then all higher Ext groups vanish, for exactly the same reason (compute the Ext groups by the trivial projective resolution). In general, there are plenty of examples of projective modules that are not free. For example, take a non-principal ideal of a Dedekind domain.