text:
\section{Introduction}

 $L^2$-cohomology can be defined for any CW-complex with a proper and cocompact action by an infinite discrete group. Heuristically, it comes from restricting the cellular cochains to the subspace which are square summable. These cochains naturally form a Hilbert space, and this additional structure combined with the group action allows us to put a dimension on these cohomology groups, the \emph{$L^2$-Betti numbers}.

In this paper, we will study the $L^2$-cohomology of Coxeter groups acting on the Davis complex. In fact, in this setting there is a more general \emph{weighted } cohomology theory developed in \cite{Dymara} and \cite{DDJO}. The idea behind this is natural and studied outside the Coxeter group setting: one can modify the natural norm on chains by a weight function and consider the complex of associated square summable chains. One can think of these groups as follows: when the weight function is ``small'', it is easy to be square summable, and weighted $L^2$-cohomology behaves like ordinary cohomology. On the other hand, when the weight function is ``large'', it is difficult to be square summable, and  weighted $L^2$-cohomology behaves like cohomology with compact support.

However, the Coxeter group setting is unique in that one can define  \emph{weighted $L^2$-Betti numbers}. Weighted $L^2$-Betti numbers not only tie the theory of weighted $L^2$-cohomology to algebraic properties of the Coxeter group, but they also have interesting connections with other topics such as buildings, Hecke algebras, growth series, and the Hopf Conjecture concerning Euler characteristics of aspherical manifolds.

The results of this article can be summarized as follows.

\begin{itemize}
\item Using a spectral sequence argument, we show that for ranges of weights depending on certain subgroups of the Coxeter group, the weighted $L^2$-cohomology of the Davis complex is concentrated in low dimensions (Theorem \ref{thm:l2kskeleton}). Our result generalizes a theorem of Dymara \cite[Theorem 10.3]{Dymara}.

\item We use the Strong Atiyah Conjecture and the above argument to compute (in some cases) the weighted $L^2$-cohomology for a certain subcomplex of $\Sigma(W,S)$ called a ruin. This allows us to derive vanishing results for the weighted $L^2$-cohomology of many low-dimensional Coxeter groups (Theorems \ref{thm:AtiyahKn} and \ref{thm:Atiyahcomputation}).

\item We extend (to the weighted setting) a computation of Davis-Okun \cite[Theorem 3.1]{DO3} of the $L^2$-cohomology of right-angled Coxeter groups whose nerves are barycentric subdivisions of PL $n$-manifolds for $n = 6,8$. A consequence of this is that the Weighted Singer Conjecture is true for all even $n\leq 8$ whenever the Coxeter group is right-angled and the corresponding nerve is the barycentric subdivision of a PL-cellulation of a sphere.

\end{itemize}

\emph{Acknowledgements} We would like to thank Boris Okun for many helpful discussions.

\section{Coxeter Groups and the Davis Complex}

\subsection{Coxeter groups}
A \emph{Coxeter matrix} $M=(m_{st})$ on a set $S$ is an $S\times S$ symmetric matrix with entries in $\mathbb{N}\cup\{\infty\}$ such that
$$m_{st}=\begin{cases} 1 &\mbox{if } s=t \\
\geq 2 & \mbox{otherwise.} \end{cases}$$
One can associate to $M$ a presentation for a group $W$ as follows. Let $S$ be the set of generators and let $\mathcal{I}=\{(s,t)\in S\times S \mid m_{st}\neq\infty\}.$ The set of relations for $W$ is $$R=\{(st)^{m_{st}}\}_{(s,t)\in\mathcal{I}}.$$
The group defined by the presentation $\left<S,R\right>$ is a \emph{Coxeter group} and the pair $(W,S)$ is a \emph{Coxeter system}. If all off-diagonal entries of $M$ are either $2$ or $\infty$, then $W$ is \emph{right-angled}.

Given a subset $T\subset S$, define $W_T$ to be the subgroup of $W$ generated by the elements of $T$. Then $(W_T,T)$ is a Coxeter system. Subgroups of this form are \emph{special subgroups}. $W_T$ is a \emph{spherical subgroup} if $W_T$ is finite and, in this case, $T$ is a \emph{spherical subset}. We will let $\mathcal{S}$ denote the poset of spherical subsets (the partial order being inclusion).

Given $w\in W$, call an expression $w=s_1s_2\dotsm s_n$ \emph{reduced} if there exists no integer $k<n$ with $w=s_1's_2'\dotsm s_k'.$ We define the \emph{length} of $w$, denoted by $l(w)$, to be the integer $n$ so that $w=s_1s_2\dotsm s_n$ is a reduced expression for $w$.

If $T\subset S$ and $w\in W$, the coset $wW_T$ contains a unique element of shortest length. This element $w$ is said to be \emph{$(\emptyset, T)$-reduced}.

If $(W,S)$ is a Coxeter system, let $\mathbf{t} := (t_s)_{s \in S}$ denote an $S$-tuple of indeterminates, where $t_s = t_{s'}$ if $s$ and $s'$ are conjugate in $W$. If $s_1s_2 \dots s_n$ is a reduced expression for $w$, let $t_w$ be the monomial $t_w := t_{s_1}t_{s_2}\dots t_{s_n}$.

Note that $t_w$ is independent of choice of reduced expression due to Tits' solution to the word problem for Coxeter groups (see the discussion at the beginning of \cite[Chapter 17]{Davis}). The \emph{growth series} of $W$ is the power series in $\mathbf{t}$ defined by $$W(\mathbf{t})=\sum_{w\in W} t_w.$$

The \emph{region of convergence} $\mathcal{R}$ for $W(\mathbf{t})$ is defined to be $$\mathcal{R}:=\{\mathbf{t}\in (0,+\infty)^S \mid W(\mathbf{t}) \text{ converges}\}.$$

For each $T\subset S$, we denote the growth series of the special subgroup $W_T$ by $W_T(\mathbf{t})$, the respective region of convergence by $\mathcal{R}_T$, and define $\mathbf{t}^{-1}:=(t_s^{-1})_{s\in S}$.

\subsection{The basic construction} A \emph{mirror structure} over a set $S$ on a space $X$ is a family of subspaces $(X_s)_{s\in S}$ indexed by $S$. We say $X$ is a \emph{mirrored space over $S$}. Suppose that $(W,S)$ is a Coxeter system and that $X$ is a mirrored space over $S$. Put $S(x):=\{s\in S\mid x\in X_s\}$ and define an equivalence relation $\sim$ on $W\times X$ by $(w,x)\sim (w',y)$ if and only if $x=y$ and $w^{-1}w'\in W_{S(x)}$. Let $\mathcal{U}(W,X)$ denote the quotient space: $$\mathcal{U}(W,X)=(W\times X)/ \sim.$$
$\mathcal{U}(W,X)$ is the \emph{basic construction}. There is a natural $W$-action on $W\times X$ which respects the equivalence relation, and hence descends to an action on $\mathcal{U}(W,X)$.

\subsection{The Davis complex}\label{section:coxcell}

Recall that $\mathcal{S}$ the set of all spherical subsets of $S$. Let $K$ denote the geometric realization of the poset $\mathcal{S}$ and $L$ the geometric realization of the abstract simplicial complex $\mathcal{S}$. $K$ is the \textit{Davis chamber} and $L$ is called the \textit{nerve} of $(W,S)$. Note that $K$ is the cone on the barycentric subdivision of $L$ with cone point corresponding to $\emptyset$.

For each $s\in S$ let $$K_s:=|\mathcal{S}_{\geq\{s\}}|.$$
The family $(K_s)_{s\in S}$ is a mirror structure on $K$.

The \textit{Davis complex} $\Sigma(W,S)$ associated to the nerve $L$ is defined to be $\Sigma(W,S):=\mathcal{U}(W,K)$. Note that $\Sigma(W,S)$ is a simplicial complex. It is proved in \cite{Davis1} that $\Sigma(W,S)$ is contractible. Furthermore, if $L$ is a triangulation of an $(n-1)$-sphere, then $\Sigma(W,S)$ is an $n$-manifold.

The Davis complex admits a decomposition into \emph{Coxeter cells} as follows. For each $T\in\mathcal{S}$, let $v_T$ denote the corresponding barycenter in $K$. Let $c_T$ denote the union of simplices $c\subset\Sigma(W,S)$ such that $c\cap K_T=v_T$. The boundary of $c_T$ is cellulated by $wc_U$, where $w\in W_T$ and $U\subset T$. With its simplicial structure, the boundary $\partial c_T$ is the Coxeter complex corresponding to the Coxeter system $(W_T,T)$, which is a sphere since $W_T$ is finite. It follows that $c_T$ and its translates are cells, which are called the \emph{Coxeter cells of type $T$}. We denote $\Sigma(W,S)$ with this cellular decomposition by $\Sigma_{cc}(W,S)$. Note that $\Sigma_{cc}(W,S)$ is a regular CW-complex such that the poset of cells that can be identified with $W\mathcal{S}:=\{wW_U\mid w\in W, T\in\mathcal{S}\}$. The simplicial structure on $\Sigma(W,S)$ is the geometric realization of the poset $W\mathcal{S}$, hence $\Sigma(W,S)$ is the barycentric subdivision of $\Sigma_{cc}(W,S)$.

\section{Weighted $L^2$-cohomology}

In this section we give a brief introduction to weighted $L^2$-cohomology. Further details can be found in \cite{Davis}, \cite{DDJO}, \cite{Dymara} and \cite{eckmann}. We then compile some results about the weighted $L^2$-cohomology of the Davis complex $\Sigma(W,S)$

Let $(W,S)$ be a Coxeter system. For the remainder of this article, let $\mathbf{q}=(q_s)_{s\in S}$ denote an $S$-tuple of positive real numbers satisfying $q_s=q_{s'}$ whenever $s$ and $s'$ are conjugate in $W$. Set $\mathbf{q}^{-1}=(q^{-1}_s)_{s\in S}$. If $w=s_1\cdot\cdot\cdot s_n$ is a reduced expression for $w\in W$, we define $q_w:=q_{s_1}\cdot\cdot\cdot q_{s_n}.$

\subsection{Hecke-von Neumann algebras.} Let $\mathbb{R}W$ denote the group algebra of $W$, and let $\{e_w\}_{w\in W}$ denote the standard basis on $\mathbb{R}W$ (here $e_w$ denotes the characteristic function of $\{w\}$). Given a multiparameter $\mathbf{q}$ of positive real numbers as above, we deform the standard inner product on $\mathbb{R}W$ to an inner product $$\left<e_w,e_{w'}\right>_\mathbf{q}=\left\{
                \begin{array}{ll}
                  q_w, & \hbox{$w=w'$;} \\
                  0, & \hbox{otherwise.}
                \end{array}
              \right.
$$

Using the multiparameter $\mathbf{q}$, one can give $\mathbb{R}W$ the structure of a \textit{Hecke algebra}. We denote $\mathbb{R}W$ with this Hecke algebra structure by $\mathbb{R}_\mathbf{q} (W)$, and $L^2_\mathbf{q}(W)$ will denote the Hilbert space completion of $\mathbb{R}_\mathbf{q} (W)$ with respect to $\left<\hskip1mm,\hskip1mm\right>_\mathbf{q}$. There is a natural anti-involution on $\mathbb{R}_\mathbf{q} W$, which implies that there is an associated \textit{Hecke-von Neumann algebra} $\mathcal{N}_\mathbf{q}(W)$ acting on the right on $L^2_\mathbf{q}(W)$. It is the algebra of all bounded linear endomorphisms of $L^2_\mathbf{q}(W)$ which commute with the left $\mathbb{R}_\mathbf{q} (W)$-action.

Define the $\textit{von Neumann trace}$ of $\phi\in\mathcal{N}_\mathbf{q}(W)$ by $\textnormal{tr}_{\mathcal{N}_\mathbf{q}}(\phi):=\left<e_1\phi,e_1\right>_\mathbf{q}$. This extends to a trace on $(n\times n)$-matrices with coefficients in $\phi\in\mathcal{N}_\mathbf{q}(W)$ by taking the sum of the traces of diagonal elements.

Let $V\subseteq (L^2_\mathbf{q}(W))^n$ be invariant under the $W$-action. The orthogonal projection $p_V: (L^2_\mathbf{q}(W))^n\rightarrow (L^2_\mathbf{q}(W))^n$ is in $\mathcal{N}_\mathbf{q}(W)$. Therefore, we can attribute to $V$ a nonnegative real number called the \textit{von Neumann dimension}.

\subsection{Weighted $L^2$-cohomology.} Suppose $(W,S)$ is a Coxeter system and that $X$ is a mirrored finite $CW$-complex over $S$. Set $\mathcal{U}=\mathcal{U}(W,X)$. We first orient the cells of $X$ and equivariantly extend this orientation to $\mathcal{U}$ in such a way so that if $\sigma$ is a positively oriented cell of $X$, then $w\sigma$ is positively oriented for each $w\in W$.

We define a measure on the $w$-orbit of an $i$-cell $\sigma\in X$ by $$\mu_\mathbf{q}(w\sigma)=q_u,$$ where $u$ is $(\emptyset, S(\sigma))$-reduced, $S(\sigma):=\{s\in S|\sigma\subseteq X_s\}.$ This extends to a measure on the $i$-cells $\mathcal{U}^{(i)}$, which we also denote by $\mu_\mathbf{q}$.

There is a weighted inner product given by $$\left<f,g\right>_\mathbf{q}=\sum_\sigma f(\sigma)g(\sigma)\mu_\mathbf{q}(\sigma),$$ and we denote the induced norm by $||\text{ }||_\mathbf{q}$.

Define the \textit{$\mathbf{q}$-weighted $i$-dimensional $L^2$-(co)chains}, denoted by $L_\mathbf{q}^2C_i(\mathcal{U})$,  to be the Hilbert space of square-summable (with respect to the weighted inner product) $i$-(co)chains.

The boundary map $\partial_i: L_\mathbf{q}^2C_i(\mathcal{U})\rightarrow L_\mathbf{q}^2C_{i-1}(\mathcal{U})$ and coboundary map $\delta^i: L_\mathbf{q}^2C_i(\mathcal{U})\rightarrow L_\mathbf{q}^2C_{i+1}(\mathcal{U})$ are defined by the usual formulas, however there is one caveat: they are not adjoints with respect to this inner product whenever $\mathbf{q}\neq\mathbf{1}$. One remedies this issue by perturbing the boundary map $\partial_i$ to $\partial_i^\mathbf{q}$: $$\partial_i^\mathbf{q}(f)(\sigma^{i-1})=\sum_{\sigma^{i-1}\subset\alpha^{i}} [\sigma:\alpha]\mu_\mathbf{q}(\alpha)\mu_\mathbf{q}^{-1}(\sigma) f(\alpha).$$

A simple computation shows that $\partial_i^\mathbf{q}$ is the adjoint of $\delta$ with respect to the weighted inner product, hence $\left(L_\mathbf{q}^2C_\ast(\mathcal{U}),\partial_i^\mathbf{q}\right)$ is also a chain complex. The \textit{reduced $\mathbf{q}$-weighted $L^2$-cohomology} is defined to be $$L_\mathbf{q}^2H_i(\mathcal{U})=\ker \partial_i^\mathbf{q}/\overline{\im \partial_{i+1}^\mathbf{q}},$$ $$L_\mathbf{q}^2H^i(\mathcal{U})=\ker\delta^i/\overline{\im\delta^{i-1}}.$$

\begin{lemma} We recall some properties of weighted $L^2$-cohomology \cite{DDJO}.

\begin{itemize}

\item (Hodge decomposition): $L_\mathbf{q}^2H^i(\mathcal{U})= (\ker \delta^i \cap \partial_i^{\mathbf{q}}) \oplus \overline{\im \delta^{i-1}} \oplus \overline{\im \partial^{\mathbf{q}}_{i+1}}$

\item $L_\mathbf{q}^2H^i(\mathcal{U}) \cong L_\mathbf{q}^2H_i(\mathcal{U})$

\item (Poincar\'{e} Duality): $L_\mathbf{q}^2H_i(\mathcal{U})\cong L_{\mathbf{q}^{-1}}^2H_{n-i}(\mathcal{U},\partial\mathcal{U}).$

\item (Invariance of cellulation)\label{prop:l2coxcell} If $\Sigma_{cc}(W,S)$ denotes $\Sigma(W,S)$ with the Coxeter cellulation, then $L^2_\mathbf{q} H_\ast(\Sigma(W,S))\cong L^2_\mathbf{q} H_\ast(\Sigma_{cc}(W,S))$.

\end{itemize}

\end{lemma}

The \textit{$i$-th $L_\mathbf{q}^2$-Betti number of $\mathcal{U}$}, denoted by $L_\mathbf{q}^2b_i(\mathcal{U})$, is the von Neumann dimension of $L^2_\mathbf{q} H_i(\mathcal{U})$. We define the \textit{weighted Euler characteristic of $\mathcal{U}$}: $$\chi_\mathbf{q}(\mathcal{U})=\sum (-1)^iL_\mathbf{q}^2b_i(\mathcal{U}).$$

\begin{lemma}\label{lemma:propertiesbetti} We record some further properties of weighted $L^2$-Betti numbers

\begin{itemize}
\item (Weighted Atiyah Formula) \cite[Corollary 3.4]{Dymara}: $\chi_\mathbf{q}(\Sigma(W,S)) = \frac{1}{W(\mathbf{q})}$

\item (Weighted Kunneth Formula): $ L_\mathbf{q}^2b_k(\mathcal{U} \times \mathcal{U}') = \sum_{i + j = k}L_\mathbf{q}^2b_i(\mathcal{U})L_\mathbf{q}^2b_k(\mathcal{U}')$

\item ($0$-dimensional homology) \cite[Theorem 10.3]{Dymara} \label{prop:betti0}
$L_\mathbf{q}^2b_0(\Sigma(W,S))\neq 0$ if and only if $\mathbf{q}\in\mathcal{R}$. Moreover, when $\mathbf{q}\in\mathcal{R}$, $L_\mathbf{q}^2b_\ast(\Sigma(W,S))$ is concentrated in dimension $0$.

\item (Top homology) \cite[Lemma 4.8]{wm2} \label{Pushing0}
Suppose that $\Sigma(W,S)$ is $n$-dimensional and that $L^2_\mathbf{1}b_n(\Sigma(W,S))=0$. Then $L^2_\mathbf{q} b_n(\Sigma(W,S))=0$ for $\mathbf{q}\leq\mathbf{1}$.

\end{itemize}

\end{lemma}

\subsection{An alternate definition of $L_\mathbf{q}^2$--Betti numbers}\label{sec:altbetti} As discussed in \cite[$\S$6]{DO2}, L\"{u}ck defined an algebraic version of $L_\mathbf{q}^2$-Betti numbers \cite{Luckbook}. The main point is that there is an equivalence of categories between the category of Hilbert $\mathcal{N}_\mathbf{q}$-modules and projective $\mathcal{N}_\mathbf{q}$-modules. Hence one can define $\dim_{\mathcal{N}_\mathbf{q}} M$ for a finitely generated projective $\mathcal{N}_\mathbf{q}$-module $M$ which agrees with the dimension of the corresponding Hilbert $\mathcal{N}_\mathbf{q}$-module. So, $\dim_{\mathcal{N}_\mathbf{q}} M$ for an arbitrary $\mathcal{N}_\mathbf{q}$-module is then defined to be the dimension of its projective part.

As in \cite{Luckbook}, define the cohomology groups $H_W^\ast(\mathcal{U}(W,X),\mathcal{N}_\mathbf{q}(W))$ to be the cohomology of the complex $$C_W^\ast(\mathcal{U},\mathcal{N}_\mathbf{q}(W)):=\text{Hom}_W(C_\ast(\mathcal{U}),\mathcal{N}_\mathbf{q}(W)).$$

It follows that $$L_\mathbf{q}^2b_i(\mathcal{U})=\dim_{\mathcal{N}_\mathbf{q}}H_W^i(\mathcal{U},\mathcal{N}_\mathbf{q}(W)).$$ The advantage of this definition is that we do not need to take closures of images (which allows us e.g. to use spectral sequences).

\subsection{Conjectures in $L^2_\mathbf{q}$-cohomology.}

The following conjecture strengthens a classical conjecture of Singer, which predicts that the $L^2$-cohomology of a closed, aspherical $n$-manifold is concentrated in the middle dimension.

\begin{wsinger}[Weighted Singer Conjecture]
\label{conj:weightedSinger}
Suppose that $L$ is a triangulation of $S^{n-1}$, so that $\Sigma(W,S)$ is a contractible $n$-manifold. Then $$L_\mathbf{q}^2H_i(\Sigma(W,S))=0 \text{ for } i>\frac{n}{2} \text{ and } \mathbf{q}\leq \mathbf{1}.$$
\end{wsinger}

By weighted Poincar\'{e} duality, this is equivalent to the conjecture that if $\mathbf{q}\geq \mathbf{1}$ and $i<\frac{n}{2}$, then $L_\mathbf{q}^2H_i(\Sigma(W,S))$ vanishes. Conjecture \ref{conj:weightedSinger} is true by \cite[Theorem 2.1]{Dymara} for $n\leq 2$, and in \cite{DDJO}, it was proved for the case where $W$ is right-angled and $n\leq 4$, and furthermore it was shown that Conjecture \ref{conj:weightedSinger} for $n$ odd implies Conjecture \ref{conj:weightedSinger} for $n$ even (also under the assumption that $W$ is right-angled). Recently, the first author proved the conjecture under the assumption that $L$ is a flag triangulation and $n\leq 4$ \cite{wm}.

If $\mathbf{q}=\mathbf{1}$, these are the ordinary $L^2$-cohomology groups, and one recovers the original Singer Conjecture. The Singer Conjecture holds by a result of Lott and L\"{u}ck \cite{LottLuck}, in conjunction with the validity of the Geometrization Conjecture for $3$-manifolds \cite{p02,p03}, for $n=3$. It was proved by Davis-Okun \cite{do} for the case where $W$ is right-angled and $n\leq 4$. It was later proved for the case where $W$ is an even Coxeter group and $n=4$ by Schroeder \cite{Schroeder} under the restriction that $L$ is a flag complex. Recently Okun--Schreve \cite[Theorem 4.9]{os14} gave a proof of the Singer Conjecture for the case $n=4$, so now the Singer Conjecture is known in full generality for Coxeter groups in dimensions $n\leq 4$.

Recall that a subcomplex $A$ of $L$ is \emph{full} if the vertices of a simplex of $L$ lie in $A$, then the entire simplex lies in $A$. In \cite{DDJO}, the following key strengthening of the Weighted Singer Conjecture was proved in low dimensions:

\begin{theorem}[{\cite[Theorem 16.13]{DDJO}}]\label{RAGeneralizedSinger}
Suppose that $W_L$ is right-angled, $L$ is a triangulation of $S^{n-1}$ with $n\leq 4$ and that $A$ is any full subcomplex. Furthermore, suppose that $\mathbf{q}\leq\mathbf{1}$. Then $L^2_\mathbf{q} b_i(\Sigma(W_A))=0$ for $i>\frac{n}{2}$.
\end{theorem}

\section{Ruins}\label{sec:ruins}
\subsection{Some Hilbert $\mathcal{N}_\mathbf{q}(W)$-submodules of $L^2_\mathbf{q}(W)$}\label{sec:hts}
We begin by considering the following self-adjoint idempotents in $\mathcal{N}_\mathbf{q}(W)$:

\begin{lemma}[{\cite[Lemma 19.2.6]{Davis}}]
Given a subset $T\subset S$ and and $\mathbf{q}\in\mathcal{R}^{-1}_T$, there is an idempotent $h_T\in\mathcal{N}_\mathbf{q}(W)$ defined by
$$h_T:=\frac{1}{W_T(\mathbf{q}^{-1})}\sum_{w\in W_T} \varepsilon_w q_w^{-1} e_w,$$
where $\varepsilon_w=(-1)^{l(w)}.$
\end{lemma}

Therefore, the map defined by $x\rightarrow h_T x$ is an orthogonal projection (whenever $h_T$ is defined), and we denote the image by $H_T$. Note that by \cite[Lemma 19.2.13]{Davis}, $$H_T=\bigcap_{s\in T} H_s.$$

Using these submodules, we define a chain complex as follows. For a spherical subset of cardinality $k$, $T\in\mathcal{S}^{(k)}$, let $$C_i(H_T):=\bigoplus_{U\in (\mathcal{S}_{\geq T})^{(i+k)}} H _U.$$

Fix some ordering of $\{s\in S-T\mid T\cup\{s\}\in\mathcal{S}\}$. Whenever $U\subset V$, we have an inclusion $i_V^U: H_V\hookrightarrow H_U$. Therefore,  the boundary map $\partial : C_{i+1}(H_T)\rightarrow C_i(H_T)$ corresponds to a matrix $(\partial_{UV})$, where $\partial_{UV}=0$ unless $U\subset V$, and is equal to $(-1)^j i_V^U$ if $U$ is obtained by deleting the $j^{\text{th}}$ element of $V$. This turns $C_\ast(H_T)$ into a chain complex of Hilbert $\mathcal{N}_\mathbf{q}(W)$-modules. Similarly, whenever $U\subset V$ we have the projection $p_V^U: H_U\rightarrow H_V$. Thus we have a coboundary map where the matrix entries consist of projections, and we get a cochain complex $C^\ast(H_T)$ of Hilbert $\mathcal{N}_\mathbf{q}(W)$-modules.

\subsection{Ruins} For $U\subset S$, set $\mathcal{S}(U):=\{T\in\mathcal{S}\mid T\subset U\}$. Define $\Sigma(U)$ to be the subcomplex of $\Sigma_{cc}(W,S)$ consisting of all (closed) Coxeter cells of type $T$ with $T\in\mathcal{S}(U)$. Given $T\in\mathcal{S}(U)$, we define the following subcomplexes of $\Sigma(U)$:

\begin{align*}
\Omega(U,T): & \hskip2mm \text{the union of closed cells of type }T', \text{ with }T'\in\mathcal{S}(U)_{\geq T},\\
\partial\Omega(U,T): & \hskip2mm \text{the cells of }\Omega(U,T)\text{ of type } T'', \text{ with } T''\not\in\mathcal{S}(U)_{\geq T}.
\end{align*}

The pair $(\Omega(U,T),\partial\Omega(U,T))$ is the \emph{$(U,T)$-ruin}. Note that if $T=\emptyset$, then $\Omega(U,T)=\Sigma(U)$ and $\partial\Omega(U,T)=\emptyset$. Ruins can also be expressed in terms of the basic construction. Define $K(U,T):=\Omega(U,T)\cap K$ and $\partial K(U,T):=\partial\Omega(U,T)\cap K$, where $K$ is the Davis chamber. Then $K(U,T)$ and $\partial K(U,T)$ have an induced mirror structure, and it follows that

$$\Omega(U,T)=\mathcal{U}(W,K(U,T)),\hskip2mm \partial\Omega(U,T)=\mathcal{U}(W,\partial K(U,T)).$$

The $(S,T)$-ruin has a chain complex that looks like this:

\begin{prop}[{\cite[Lemma 20.6.21]{Davis}}]
\label{prop:ruinchaincpx}
For $T\in\mathcal{S}^{(k)}$, the chain complexes $C_\ast(H_T)$ and $L_\mathbf{q}^2C_{\ast+k}(\Omega(S,T),\partial\Omega(S,T))$ of $\mathcal{N}_\mathbf{q}(W)$-modules are isomorphic. In particular, $$L_\mathbf{q}^2C_{m}(\Omega(S,T),\partial\Omega(S,T))=0 \text{ for } m<k.$$
\end{prop}

For brevity, we write $(\Omega(U,T),\partial)$. For $s\in T$, set $U'=U-s$ and $T'=T-s$. As in \cite[Proof of Theorem 8.3]{DDJO}, we have the following weak exact sequence:
\begin{center}
\begin{tikzpicture}[baseline= (a).base]\label{eq:ruinsequence}\node[scale=1] (a) at (0,0){
\begin{tikzcd}[column sep=small]
 L_\mathbf{q}^2H_{\ast}(\Omega(U',T'),\partial) \arrow{r} & L_\mathbf{q}^2H_{\ast}(\Omega(U,T'),\partial) \arrow{r} & L_\mathbf{q}^2H_{\ast}(\Omega(U,T),\partial)
\end{tikzcd}
};
\end{tikzpicture}
\end{center}
For the special case when $U=S$ and $T=\{s\}$ the above sequence becomes:

\begin{center}
\begin{tikzpicture}[baseline= (a).base]\node[scale=1] (a) at (0,0){
\begin{tikzcd}[column sep=small]\label{eq:ruinsequence}
  L_\mathbf{q}^2H_{\ast}(\Sigma(S-s)) \arrow{r} & L_\mathbf{q}^2H_{\ast}(\Sigma(S))\arrow{r} & L_\mathbf{q}^2H_{\ast}(\Omega(S,s),\partial)
\end{tikzcd}
};
\end{tikzpicture}
\end{center}

\subsection{$L^2_\mathbf{q}$-(co)homology of ruins}
Given a Coxeter system $(W,S)$, for $T\in\mathcal{S}$ and $T\subseteq V\subseteq S$, define $$\St(T,V):=\bigcup_{\substack{U\subseteq V \\ U\cup T\in\mathcal{S}}} U,$$ and $$Lk(T,V):=\St(T,V)\setminus T.$$ If $V=S$, then we write $\St(T)$ and $Lk(T)$ instead of $\St(T,S)$ and $Lk(T,S)$. If $T=\emptyset$, we make the convention that $S(T,U)=U$.

\begin{theorem}[Compare {\cite[Theorem 20.6.22]{Davis}}]
\label{thm:homologykruin}
Suppose that $T\in\mathcal{S}^{(k)}$ and that $\mathbf{q}\in \mathcal{R}_{\St(T)}$. Then $L^2_\mathbf{q} H_\ast(\Omega(S,T),\partial\Omega(S,T))$ is concentrated in dimension $k$.

\end{theorem}
\begin{proof}
 We first make an observation about ruins. We note that for every $V\subseteq S$, $\Omega(V,T)=\Omega(\St(T,V),T)$, the point being that $\Omega(V,T)$ consists of Coxeter cells corresponding to spherical subsets of $V$ containing $T$, and $\St(T,V)$ is the union of all such subsets. In particular, $\Omega(U,T)=\Omega(\St(T,U),T))$, and hence $\Omega(U,T)$ is a subcomplex of $\Sigma(\St(T,U))$.

The proof is by induction on $k$. We will show that for $U\subset S$ and $T\in\mathcal{S}(U)^{(k)}$, $L^2_\mathbf{q} H_\ast(\Omega(U,T),\partial)$ is concentrated in dimension $k$. For the base case $k=0$, note that $\St(\emptyset,U)=U$, $\Omega(U,\emptyset)=\Sigma(U)$, and $\partial\Omega(U,\emptyset)=\emptyset$. Hence, for $k=0$, the theorem asserts that for $\mathbf{q}\in\mathcal{R}_U$, $L^2_\mathbf{q} H_\ast(\Sigma(U))$ is concentrated in dimension $0$, which is Lemma \ref{prop:betti0}.

Suppose the theorem is true for $k-1$ and let $T\in\mathcal{S}(U)^{(k)}$. Let $s\in T$, $V=T-s$ and consider the long exact sequence:

$$\longrightarrow L_\mathbf{q}^2H_\ast(\Omega(\St(T,U)-s,V),\partial)\longrightarrow L_\mathbf{q}^2H_\ast(\Omega(\St(T,U),V),\partial) \longrightarrow\dotsm$$
$$
\dotsm\longrightarrow L_\mathbf{q}^2H_\ast(\Omega(\St(T,U),T),\partial)\longrightarrow
$$

Note that $$\Omega(\St(T,U),V)=\Omega(\St(V,\St(T,U)),V)$$ and $$\Omega(\St(T,U)-s,V)=\Omega(\St(V,\St(T,U)-s),V).$$

Since $\St(V,\St(T,U))\subseteq \St(T,U)$ and $\St(V,\St(T,U)-s)\subseteq \St(T,U)$, it follows that $\mathcal{R}_{\St(T,U)}\subseteq\mathcal{R}_{\St(V,\St(T,U))}$ and $\mathcal{R}_{\St(T,U)}\subseteq\mathcal{R}_{\St(V,\St(T,U)-s)}$. Since $\mathbf{q}\in \mathcal{R}_{\St(T,U)}$, it follows by induction that the left-hand term and the middle term of the exact sequence are both concentrated in dimension $k-1$. Since $L_\mathbf{q}^2H_\ast(\Omega(\St(T,U),T),\partial)$ vanishes for $\ast<k$ (Proposition \ref{prop:ruinchaincpx}), it follows that $L_\mathbf{q}^2H_\ast(\Omega(\St(T,U),T),\partial)=L_\mathbf{q}^2H_\ast(\Omega(U,T),\partial)$ is concentrated in dimension $k$.

\end{proof}

\begin{remark}\label{remark:concentrationsubruins}
Suppose that $T\in\mathcal{S}$ and that $\mathbf{q}\in \mathcal{R}_{\St(T)}$. Then, for $U\in \mathcal{S}_{\geq T}$, Theorem \ref{thm:homologykruin} implies that $L^2_\mathbf{q} H_\ast(\Omega(S,U),\partial)$ is concentrated in dimension $|U|$. This is because, if $T\subset U$, then $\St(U)\subset St(T)$. Therefore $\mathbf{q}\in \mathcal{R}_{\St(T)}\subseteq\mathcal{R}_{\St(U)}$.

\end{remark}

\section{A spectral sequence} In this section, we define a spectral sequence following the line laid down in \cite[$\S$ 2]{DO2}.

A \textit{poset of coefficients} is a contravariant functor $\mathcal{A}$ from a poset $\mathcal{P}$ to the category of abelian groups. In other words, it is a collection $\{\mathcal{A}\}_{a\in\mathcal{P}}$ of abelian groups together with homomorphisms $\phi_{ba}: \mathcal{A}_a\rightarrow\mathcal{A}_b$, defined whenever $a>b$, such that $\phi_{ca}=\phi_{cb}\phi_{ba}$, whenever $a>b>c$. The functor $\mathcal{A}$ gives us a system of coefficients on the cell complex Flag($\mathcal{P}$): it associates to the simplex $\sigma$ the abelian group $\mathcal{A}_{\min(\sigma)}$. Hence, we get a cochain complex $$C^j(\text{Flag}(\mathcal{P});\mathcal{A}):=\bigoplus_{\sigma\in \text{Flag}(\mathcal{P})^{(j)}}\mathcal{A}_{\min(\sigma)},$$

where $\text{Flag}(\mathcal{P})^{(j)}$ denotes the set of $j$-simplices in Flag($\mathcal{P}$).

Let $Y$ be a CW complex. A \textit{poset of spaces} in $Y$ over $\mathcal{P}$ is a cover $\mathcal{V}=\{Y_a\}_{a\in\mathcal{P}}$ of $Y$ by subcomplexes indexed by $\mathcal{P}$ so that if $N(\mathcal{V})$ denotes the nerve of the cover, then:

\begin{enumerate}[(i)]
  \item $a<b \Longrightarrow Y_a\subset Y_b$,
  \item the vertex set Vert($\sigma$) of each simplex in $N(\mathcal{V})$ has the greatest lower bound $\wedge\sigma$ in $\mathcal{P}$, and
  \item $\mathcal{V}$ is closed under taking finite nonempty intersections, i.e., for any simplex $\sigma$ of $N(\mathcal{V})$, $$\bigcap_{a\in\sigma} Y_a=Y_{\wedge\sigma}.$$
\end{enumerate}

Note that any cover leads to a poset of spaces by taking all nonempty intersections as elements of the new cover and removing duplicates. The resulting poset is the set of all nonempty intersections, ordered by inclusion.

The following lemmas appearing in \cite{DO2} define a spectral sequence associated to a poset of spaces, and give conditions for the sequence to degenerate.

\begin{lemma}[{\cite[Lemma 2.1]{DO2}}]
\label{lemma:spectralseq}
Suppose $\mathcal{V}=\{Y_a\}_{a\in\mathcal{P}}$ is a poset of spaces for $Y$ over $\mathcal{P}$. There is a Mayer-Vietoris type spectral sequence converging to $H^*(Y)$ with $E_1$-term: $$E_1^{i,j}=C^i(\text{Flag}(\mathcal{P}); \mathcal{H}^j(\mathcal{V})),$$ and $E_2$-term: $$E_2^{i,j}=H^i(\text{Flag}(\mathcal{P}); \mathcal{H}^j(\mathcal{V})),$$ where the coefficient system $\mathcal{H}^j(\mathcal{V})$ is given by $\mathcal{H}^j(\mathcal{V})(\sigma)=H^j(Y_{\min(\sigma)})$.
\end{lemma}

\begin{lemma}[{\cite[Lemma 2.2]{DO2}}]
\label{lemma:spectralseqdeg}
Suppose that $\mathcal{V}:=\{Y_a\}_{a\in\mathcal{P}}$ is a poset of spaces for $Y$ over $\mathcal{P}$. If for every $a\in\mathcal{P}$, the induced homomorphism $H^\ast(Y_a)\rightarrow H^\ast(Y_{<a})$ is the zero map, then the spectral sequence degenerates at $E_2$ and $$H^\ast(Y)=\bigoplus_{a\in\mathcal{P}}H^i(\text{Flag}(\mathcal{P}_{\geq a}),\text{Flag}(\mathcal{P}_{>a}),H^j(Y_a)).$$
\end{lemma}

\section{$L_\mathbf{q}^2$-(co)homology of $(\Sigma,\Sigma^{(k-1)})$}

To simplify notation, write $\Sigma$ for $\Sigma_{cc}(W,S)$ and let $\Sigma^{(k-1)}$ denote the $(k-1)$-skeleton of $\Sigma_{cc}(W,S)$. For the proofs in this section, we will also write $H^\ast_\mathbf{q}(\mathcal{U})$ for $H_W^\ast(\mathcal{U},\mathcal{N}_\mathbf{q}(W))$. Refer to Section \ref{sec:altbetti} for the definition of $H_W^\ast(\mathcal{U},\mathcal{N}_\mathbf{q}(W))$.

The following lemma is standard.
\begin{lemma}
\label{lemma:homologyskeltonsimplex}
Let $\Delta$ denote the standard $n$-simplex and let $\Delta^{(k)}$ be its $k$-skeleton, $k<n$. Then the reduced homology $\tilde{H}_\ast(\Delta^{(k)})$ (with coefficients in $\mathbb{R}$) is concentrated in dimension $k$.
\end{lemma}

\begin{theorem}
\label{thm:l2kskeleton}
Let $k\geq 1$. Suppose that for every $T\in\mathcal{S}^{(k)}$, $\mathbf{q}\in\mathcal{R}_{St(T)}$, and let $\Sigma^{(k-1)}$ denote the $(k-1)$-skeleton of $\Sigma$. Then $L_\mathbf{q}^2 H_\ast(\Sigma,\Sigma^{(k-1)})$ is concentrated in dimension $k$.

\end{theorem}
\begin{proof}
We will show that $H_\mathbf{q}^\ast(\Sigma,\Sigma^{(k-1)})$ is concentrated in dimension $k$.

Consider the relative cochain complex $L^2_\mathbf{q} C^\ast(\Sigma,\Sigma^{(k-1)})$. We have that
$$L^2_\mathbf{q} C^i(\Sigma,\Sigma^{(k-1)})=\left\{                                                      \begin{array}{ll}
                                                                    0, & \hbox{$i\leq k-1$;} \\
                                                                    \bigoplus_{U\in\mathcal{S}^{(i)}} H_U, & \hbox{$i>k-1$.}
                                                                  \end{array}
                                                                \right.$$
Set $C_{-i}(\Sigma,\Sigma^{(k-1)})=L^2_\mathbf{q} C^i(\Sigma,\Sigma^{(k-1)})$. Then $C_\ast(\Sigma,\Sigma^{(k-1)})$ is a chain complex. For every $T\in\mathcal{S}$ let $\tilde{\Omega}_T$ denote the complex with reindexed chain complex $C_{-i}(\tilde{\Omega}_T)=L^2_\mathbf{q} C^i(\Omega(S,T),\partial\Omega(S,T))$. In this way, we have made the cochain complex of every $(S,T)$-ruin a subcomplex of the re-indexed relative cochain complex $C_\ast(\Sigma,\Sigma^{(k-1)})$.

Let $\mathcal{P}$ be the poset $\mathcal{S}^{(\geq k)}$ with the order reversed. By the above re-indexing of cochain complexes, $\{\tilde{\Omega}_T\}_{T\in\mathcal{P}}$ is a poset of spaces over the complex $Y$ with chain complex $C_\ast(\Sigma,\Sigma^{(k-1)})$, and hence we have the spectral sequence of Lemma \ref{lemma:spectralseq}.

We first establish the condition of Lemma \ref{lemma:spectralseqdeg}. So, we claim that for every $U\in\mathcal{P}$, the induced map $H_\mathbf{q}^\ast(\tilde{\Omega}_U)\rightarrow H_\mathbf{q}^\ast(\tilde{\Omega}_{<U})$ is the zero map. To prove the claim we will show that $H_\mathbf{q}^{-|U|}(\tilde{\Omega}_{<U})=0$, as this implies $H_\mathbf{q}^\ast(\tilde{\Omega}_U)\rightarrow H_\mathbf{q}^\ast(\tilde{\Omega}_{<U})$ is the zero map since $H_\mathbf{q}^\ast(\tilde{\Omega}_U)$ is concentrated in dimension $-|U|$ (see Remark \ref{remark:concentrationsubruins}). Recall that $\tilde{\Omega}_{<U}=\bigcup_{T\in\mathcal{P}_{<U}}\tilde{\Omega}_{T}$. The proof is by induction. For the base case, note that for every spherical $V$ properly containing $U$, the induced map $H_\mathbf{q}^\ast(\tilde{\Omega}_U)\rightarrow H_\mathbf{q}^\ast(\tilde{\Omega}_{V})$ is the zero map. This is because of Theorem \ref{thm:homologykruin}, which states that $H_\mathbf{q}^\ast(\tilde{\Omega}_U)$ and $H_\mathbf{q}^\ast(\tilde{\Omega}_{V})$ are concentrated in dimension $-|U|$ and $-|V|$, respectively, and $-|V|<-|U|$. Let $\mathcal{C}$ be a subcollection of elements of $\mathcal{P}_{<U}$ and let $B=\bigcup_{T\in\mathcal{C}}\tilde{\Omega}_{T}$. We wish to show that $H_\mathbf{q}^{-|U|}(B)=0$. Write $B=A\cup \tilde{\Omega}_V$, where $V\in\mathcal{C}$ and $A=\bigcup_{\substack{T\in\mathcal{C}\\ T\neq V}}\tilde{\Omega}_T$. Then we have the Mayer-Vietoris sequence:

$$\begin{tikzcd}[column sep=small]
 H_\mathbf{q}^{-|U|-1}(A\cap \tilde{\Omega}_V) \arrow{r} &  H_\mathbf{q}^{-|U|}(B) \arrow{r} & H_\mathbf{q}^{-|U|}(A)\oplus H_\mathbf{q}^{-|U|}(\tilde{\Omega}_V)
\end{tikzcd}$$

By induction, $H_\mathbf{q}^\ast(A)$ vanishes for $\ast\geq -|U|$, and by Theorem \ref{thm:homologykruin}, $H_\mathbf{q}^\ast(\tilde{\Omega}_V)$ is concentrated in $-|V|<-|U|$. We now claim $H_\mathbf{q}^\ast(A\cap \tilde{\Omega}_V)$ vanishes for $\ast\geq -|U|-1$, as this implies $H_\mathbf{q}^{-|U|}(B)=0$. We observe that

$$A\cap\tilde{\Omega}_V=\bigcup_{\substack{T\in\mathcal{C}\\ T\neq V}} \tilde{\Omega}_{T}\cap\tilde{\Omega}_V=\bigcup_{\substack{T\in\mathcal{C}\\ T\cup V\in\mathcal{S}}} \tilde{\Omega}_{T\cup V}.$$

The last equality follows from the fact that $H_T\cap H_V=H_{T\cup V}$ whenever $T\cup V$ is spherical (see Section \ref{sec:hts}). Thus $A\cap \tilde{\Omega}_V$ is the union of elements of the form $\tilde{\Omega}_{T\cup V}$, where $|T\cup V|\geq |V|+1> |U|+1$. By Theorem \ref{thm:homologykruin}, $H_\mathbf{q}^\ast(\tilde{\Omega}_{T\cup V})$ vanishes for $\ast\geq -|U|-1$, so by induction $H_\mathbf{q}^{-|U|-1}(A\cap \tilde{\Omega}_V)=0$.

We have established the condition in Lemma \ref{lemma:spectralseqdeg}, and hence
\begin{equation}\label{eq:totalhom}H_\mathbf{q}^{-n}(Y)=\bigoplus_{U\in\mathcal{P}}H^{|U|-n}(\text{Flag}(\mathcal{P}_{\geq U}),\text{Flag}(\mathcal{P}_{>U}); H_\mathbf{q}^{-|U|}(\tilde{\Omega}_U))\end{equation}

The strategy of the proof is as follows. By Theorem \ref{thm:homologykruin}, for every $U\in\mathcal{P}$, $H_\mathbf{q}^\ast(\tilde{\Omega}_U)$ is concentrated in dimension $-|U|$. So, we are done if we show that for every $U\in\mathcal{P}$, $H^\ast(\text{Flag}(\mathcal{P}_{\geq U}),\text{Flag}(\mathcal{P}_{>U}))$ is concentrated in dimension $|U|-k$. This implies $E^{i,j}_2=0$ unless $i+j=-k$, and by (\ref{eq:totalhom}), $H_\mathbf{q}^\ast(Y)$ is concentrated in dimension $-k$. Re-indexing our complexes, it follows that the cohomology of the complex $L^2_\mathbf{q} C^\ast(\Sigma,\Sigma^{(k-1)})$ is concentrated in dimension $k$.

We claim that for $U\in\mathcal{P}$ with $m=|U|$, $H^\ast(\text{Flag}(\mathcal{P}_{\geq U}),\text{Flag}(\mathcal{P}_{>U}))$ is concentrated in dimension $m-k$.

Since the geometric realization of $\text{Flag}(\mathcal{P}_{\geq U})$ is contractible, by the long exact sequence for the pair it suffices to show that the reduced cohomology $\tilde{H}^\ast(\text{Flag}(\mathcal{P}_{>U}))$ is concentrated in dimension $m-k-1$. Note that if $m=k$, then we are done since $\text{Flag}(\mathcal{P}_{> U})=\emptyset$. Also, note that for the special case where $m-k=1$, the map $H^0(\text{Flag}(\mathcal{P}_{\geq U}))\rightarrow H^0(\text{Flag}(\mathcal{P}_{>U}))$ in the long exact sequence for the pair $(\text{Flag}(\mathcal{P}_{\geq U}),\text{Flag}(\mathcal{P}_{> U}))$ is injective ($\text{Flag}(\mathcal{P}_{\geq U})$ is the cone on $\text{Flag}(\mathcal{P}_{>U})$), so showing $\tilde{H}^\ast(\text{Flag}(\mathcal{P}_{>U}))$ is concentrated in dimension $m-k-1=0$ does in fact suffice.

Consider the poset $\mathcal{S}^{op}$, which is the poset $\mathcal{S}$ of spherical subsets with order reversed. Note that $\text{Flag}(\mathcal{S}^{op}_{> U})\cong \text{Flag}(\mathcal{S}_{< U})$. The geometric realization of $\text{Flag}(\mathcal{S}_{< U})$ is $b\Delta$, where $b\Delta$ is the barycentric subdivision of the $(m-1)$-dimensional simplex $\Delta$. This is because $\mathcal{S}_{< U}$ is the poset of proper subsets of $U$. Note that $\text{Flag}(\mathcal{P}_{>U})$ is a subcomplex of $\text{Flag}(\mathcal{S}^{op}_{> U})$, and more precisely, the geometric realization of $\text{Flag}(\mathcal{P}_{> U})$ is the subcomplex of barycentric subdivision of $\partial\Delta$ (recall that $k\geq 1$) obtained by removing barycenters corresponding to spherical subsets of cardinality less than $k$.

\begin{figure}[H]
\setlength{\tabcolsep}{15pt}
\begin{tabular}{c c c}
\input{ruinsproof1} & \input{ruinsproof2} & \input{ruinsproof3}\\
$\text{Flag}(\mathcal{S}^{op}_{> U})$ & $\text{Flag}(\mathcal{P}_{> U})$, $k=1 $ &$\text{Flag}(\mathcal{P}_{> U})$, $k=2 $
\end{tabular}
\caption{:\hskip2mm Geometric realizations when $U=\{s,t,u\}$}
\end{figure}

These barycenters correspond to faces of $\partial\Delta$ of dimension less than or equal to $k-2$. Hence $$\tilde{H}^\ast(\text{Flag}(\mathcal{P}_{> U}))\cong \tilde{H}^\ast(\partial\Delta-\Delta^{(k-2)}),$$ where $\Delta^{(k-2)}$ denotes the $(k-2)$-skeleton of $\Delta$. Note that if $k=1$, then $\Delta^{(k-2)}=\emptyset$, and we are done as the reduced homology $\tilde{H}^\ast(\partial\Delta)$ is concentrated in dimension $m-2$. So, suppose $k>1$. By Alexander Duality, $$\tilde{H}_\ast(\Delta^{(k-2)})\cong \tilde{H}^{m-\ast-3}(\partial\Delta-\Delta^{(k-2)}).$$ By Lemma \ref{lemma:homologyskeltonsimplex}, the reduced homology $\tilde{H}_\ast(\Delta^{(k-2)})$ is concentrated in dimension $k-2$.
 It follows that $\tilde{H}^\ast(\partial\Delta-\Delta^{(k-2)})$ is concentrated in dimension $m-k-1$, and therefore the same holds for $\tilde{H}^\ast(\text{Flag}(\mathcal{P}_{>U}))$.

\end{proof}

\begin{remark}\label{remark:ruinconcentration}
In fact, the proof of Theorem \ref{thm:l2kskeleton} implies the following statement: if for every $U\in\mathcal{S}^{\geq k}$ we have that $L^2_\mathbf{q} H_\ast(\Omega(S,U),\partial)$ is concentrated in dimension $|U|$, then $L_\mathbf{q}^2 H_\ast(\Sigma,\Sigma^{(k-1)})$ is concentrated in dimension $k$. Thus, if one can show the aforementioned concentration of ruins, then Corollary \ref{cor:vanishingstars} implies that $L_\mathbf{q}^2 H_n(\Sigma)=0$ for $n>k$.
\end{remark}

\begin{corollary}
\label{cor:vanishingstars}
Suppose that for every $T\in\mathcal{S}^{(k)}$, $\mathbf{q}\in\mathcal{R}_{St(T)}$. Then $$L_\mathbf{q}^2 H_n(\Sigma)=0 \text{ for } n>k.$$

\end{corollary}
\begin{proof}
For the case where $k=0$, this is just Proposition \ref{prop:betti0}, so suppose $k\geq 1$. By Theorem \ref{thm:l2kskeleton}, $L_\mathbf{q}^2 H_\ast(\Sigma,\Sigma^{(k-1)})$ is concentrated in dimension $k$. Therefore the long exact sequence for the pair $(\Sigma,\Sigma^{(k-1)})$ implies the assertion.
\end{proof}

We shall need the following corollary in Section $7$.

\begin{corollary}\label{cor:ruins2dim}
Suppose that $\Sigma(W,S)$ is two-dimensional and that for every $t\in S, L_\mathbf{1}^2H_\ast(\Omega(S,t),\partial)$ is concentrated in dimension $1$. Then $$L_\mathbf{q}^2 H_2 (\Sigma(W,S))=0 \text{ for } \mathbf{q}\leq\mathbf{1}.$$
\end{corollary}
\begin{proof} Let $T\in\mathcal{S}^{(2)}$. As $\Sigma_L$ is two-dimensional, Proposition \ref{prop:ruinchaincpx} implies that $L_\mathbf{1}^2H_\ast(\Omega(S,T),\partial)$ is concentrated in dimension $2$. This and the assumption imply that all ruins are concentrated in the correct dimension, and hence $L_\mathbf{1}^2 H_2 (\Sigma(W,S))=0$ (see Remark \ref{remark:ruinconcentration}). The statement for $\mathbf{q}\leq\mathbf{1}$ follows from Lemma \ref{lemma:propertiesbetti}.
\end{proof}

\begin{example}

Suppose that $L$ is a flag triangulation of a closed orientable surface, and let $(W,S)$ be the associated right-angled Coxeter group. Following \cite[Example 17.4.3]{Davis}, we have that

$$\frac{1}{W(t)}=1-(f_0-3)t+(f_0+3-3\chi(L))t^2-(\chi(L)-1)t^3,$$

where $\chi(L)$ is the Euler characteristic of $L$ and $f_0$ is the number of vertices of $L$. The radius of convergence $\rho$ of $W(t)$ is the smallest modulus of a root of the above polynomial.

The link of every vertex of $L$ is a $k$-gon, with $k\geq 4$. If $W_{Lk(v)}$ denotes the special subgroup corresponding the the link of a vertex $v$, and $Lk(v)$ has $k$ vertices, then by an easy computation we have that

$$\rho_{Lk(v)}=\frac{(k-2)-\sqrt{k^2-4k}}{2}.$$

If $v_0$ is the vertex of $L$ whose link has the maximal number of vertices, then Corollary \ref{cor:vanishingstars} implies that $L_q^2 H_\ast(\Sigma(W,S))=0$ is concentrated in dimension $1$ whenever $\rho<q<\rho_{Lk(v_0)}$.

The main point is that $\rho$ is explicitly computable. For example, it is still open if $L_q^2 b_\ast(\Sigma(W,S))$ is concentrated in degree $2$ for $q=1$ (see \cite[Conjecture 11.5.1]{do}), but Corollary \ref{cor:vanishingstars} lets us conclude that $L_q^2 b_2(\Sigma(W,S))=0$ for $q<\rho_{Lk(v_0)}$.

\end{example}

\section{Further computations}

\subsection{The Strong Atiyah Conjecture and weighted $L^2$-Betti numbers} In this section, we will assume that the nerve $L$ of the Coxeter system $(W,S)$ is a graph (so that $\Sigma(W,S)$ is two-dimensional). Furthermore, we need only consider the Davis complex with the Coxeter cellulation. For convenience, we will label the edge $[s_is_j]$ of the nerve with $m_{ij}$. This allows us to recover the Coxeter system by reading the presentation off of the nerve.

\begin{Aconj} Let $G$ be a discrete group. For any proper cocompact $G$-CW complex $Y$
	\[
		\lcm G \cdot L^2_\mathbf{1}b_k(Y;G) \in \mathbb{Z},
	\]
	where $\lcm G$ denotes the least common multiple of the orders of finite subgroups of $G$.
\end{Aconj}

In \cite{ks}, the second author proved the Strong Atiyah Conjecture for groups that are virtually cocompact special. Restricting to the Coxeter group setting, we have the following:

\begin{theorem}[{\cite[Corollary 6.5]{ks}}]\label{thm:atiyah} Let $W$ be a Coxeter group without Euclidean triangle special subgroups. If $W$ is a finite index subgroup of $W'$, then the Strong Atiyah Conjecture holds for $W'$.
\end{theorem}

We will exploit Theorem \ref{thm:atiyah} to make computations of $L^2$-Betti numbers for many examples of Coxeter groups. For these examples, the strategy will be to show that the $L^2$-cohomology of every $(S,t)$-ruin is concentrated in dimension $1$. We can then apply Corollary \ref{cor:ruins2dim} to compute the $L^2$-cohomology of the Davis complex.

For every $t\in S$, the $(S,t)$-ruin has the property that $\Omega(S,t)=\Omega(St(t),t)$. Therefore, we can compute the $L^2$-Betti numbers $L_\mathbf{1}^2b_\ast(\Omega(S,t),\partial)$ using the subgroup $W_{St(t)}$.

Suppose the labeled $\St(t)$ has an automorphism group $G$ that is vertex transitive and edge transitive. Furthermore, suppose $G$ does not flip edges, so that we can think of the edge stabilizer as a subgroup of the vertex stabilizer.

Since $G$ acts by labeled automorphism on $\St(t)$, it acts on $W_{St(t)}$. We can form the semidirect product $W_{St(t)} \rtimes G$. Obviously, the semidirect product also acts on the Davis complex $\Sigma(W_{\St(t)})$. Furthermore, the following lemma of Davis tells us that the stabilizers of Coxeter cells under this action are only the obvious subgroups.

\begin{lemma}[{\cite[Proposition 9.1.9]{Davis}}]\label{prop:stabilizers}
For $U\in\mathcal{S}(St(t))$ and $w\in W_{St(t)}$, let $wc_U$ denote the Coxeter cell of type $U$, and let $G_U$ denote the stabilizer of the simplex $U$ in $\St(t)$ under the $G$-action. The isotropy subgroup of $W_{St(t)} \rtimes G$ fixing $wc_U$ is the subgroup $wW_Uw^{-1} \rtimes G_U$.
\end{lemma}

Let $G_t$ denote the stabilizer of the vertex $t$ in $\St(t)$ under the $G$-action. Then the group $W_{St(t)} \rtimes G_t$ acts on the $(St(t),t)$-ruin. By multiplicativity of $L^2$-Betti numbers we have the following.

\begin{lemma}\label{lemma:multiplicative}
 $$L_\mathbf{1}^2b_\ast(\Omega(St(t),t),\partial, W_{St(t)} \rtimes G_t) = |G_t|L^2_\mathbf{1}b_\ast(\Omega(St(t),t),\partial,W_{St(t)}).$$
\end{lemma}

The following simple lemma will allow us to use the Strong Atiyah Conjecture to make computations.

\begin{lemma}\label{lcm12}
Suppose that $X$ is a $G$-complex with associated $L^2$-chain complex $$0 \rightarrow L^2C_n(X) \rightarrow L^2C_{n-1}(X) \rightarrow \dots \rightarrow L^2C_1(X) \rightarrow L^2C_0(X) \rightarrow 0.$$ If $\partial_i: L^2C_i(X) \rightarrow L^2C_{i-1}(X)$ or $\partial_i: L^2C_i(X) \rightarrow L^2C_{i-1}(X)$ is nonzero, and $\dim_G(L^2C_i(X)) = \frac{1}{\lcm(G)}$, and $G$ satisfies the Strong Atiyah Conjecture, then $L^2H_i(X,G) = 0$.
\end{lemma}

\begin{proof}
Since $G$ satisfies the Strong Atiyah Conjecture, we know that if $L^2b_i(X,G) < \frac{1}{\lcm(G)}$, then $L^2b_i(X,G) = 0$. Since one of the boundary maps is nonzero, $L^2H_i(X,G)$ is strictly smaller than $L^2C_i(X)$.

By assumption, $\dim_G(L^2C_i(X)) = \frac{1}{\lcm(G)}$, and this implies $L^2H_i(X,G) = 0$.
\end{proof}

\begin{lemma}\label{vanishing}
Suppose that $\St(t)$ is labeled with $m$'s and let $\deg(t)$ denote the degree of the vertex $t$ in $\St(t)$. Furthermore, suppose that $G_t$ acts transitively on the edges emanating from $t$. If $m$ is a multiple of $\deg(t)$, then $L_\mathbf{1}^2H_\ast(\Omega(S,t),\partial)$ is concentrated in dimension $1$.
\end{lemma}
\begin{proof}
By Proposition \ref{prop:ruinchaincpx}, we have that $$\dim_{W_{St(t)}} L_\mathbf{1}^2 C_2(\Omega(S,t),\partial,W_{St(t)})=\frac{\deg(t)}{2m}.$$

If $\frac{\deg(t)}{2m} = \frac{|G_t|}{\lcm(W_{St(t)} \rtimes G_t)}$, then $L_\mathbf{1}^2b_2(\Omega(St(t),t),\partial, W_{St(t)} \rtimes G_t)=0$ by Lemma \ref{lcm12}, and hence Lemma \ref{lemma:multiplicative} implies $$L_\mathbf{1}^2b_2(\Omega(S,t),\partial)=L^2_\mathbf{1}b_2(\Omega(St(t),t),\partial,W_{St(t)})=0.$$

Let $E_T$ denote the stabilizer of the submodule $H_T$ of the chain complex $L_\mathbf{1}^2 C_2(\Omega(St(t),t),\partial,W_{St(t)})$ under the $W_{St(t)} \rtimes G_t$-action. Since $G_t$ acts transitively on the edges adjacent to $t$, it acts transitively on the submodules $H_T$, and hence $|G_t|/|E_{T}|=\deg(t)$. If $m$ is a multiple of $\deg(t)$, then $$\lcm (2|G_t|,2m|E_T|)=2|G_t|\frac{m}{\deg(t)}.$$

It follows that $\lcm(W_{St(t)} \rtimes G_t)=2|G_t|\frac{m}{\deg(t)}$ and this completes the proof.

\end{proof}

\begin{theorem}\label{thm:AtiyahKn}
Let $K_n$ denote the complete graph on $n$-vertices. Suppose that the nerve $L=K_n$ is labeled by uniform labels $m\geq n-1$. Then $$L_\mathbf{q}^2 H_2 (\Sigma(W,S))=0 \text{ for } \mathbf{q}\leq\mathbf{1}.$$
\end{theorem}
\begin{proof}
 It suffices to consider the case where $m = n-1$; if $m > n-1$ then the group $W_{K_n}$ is a Coxeter subgroup of $W_{K_{m+1}}$, and vanishing for the larger group implies vanishing of the subgroup in the top dimension. For every $t\in S$, the alternating group $A_{n-1}$ acts on $St(t)$ and the action is transitive on the edges adjacent to $t$. Therefore, by Lemma \ref{vanishing}, we obtain that the $L^2$-cohomology of every $(S,t)$-ruin is concentrated in dimension $1$. The theorem now follows by Corollary \ref{cor:ruins2dim}.

\end{proof}

\begin{theorem}\label{thm:Atiyahcomputation}
Suppose the labeled nerve $L$ with uniform labels $m$ is a graph with no $3$-cycles. If $m$ is a multiple of $\lcm_{s\in S}\deg(s)$, then $$L_\mathbf{q}^2 H_2 (\Sigma(W,S))=0 \text{ for } \mathbf{q}\leq\mathbf{1}.$$
\end{theorem}
\begin{proof}
If $L$ has no $3$-cycles, then for every $t\in S$, $Lk(t)$ is a full subcomplex of $L$. Since $Lk(t)$ consists of $\deg(t)$ points, it follows that the alternating group $A_{\deg(t)}$ acts on the $(St(t),t)$-ruin, and moreover, the action is transitive on the edges of $St(t)$. Again, by Lemma \ref{vanishing}, we obtain that every $(S,t)$-ruin has $L^2$-cohomology concentrated in dimension $1$, and the theorem follows by Corollary \ref{cor:ruins2dim}.

\end{proof}

\subsection{Weighted $L^2$-homology of barycentric subdivisions}
In this section, all Coxeter groups are assumed to be right-angled. Unless otherwise stated, the standing assumption in this section is that $\mathbf{q}\leq\mathbf{1}$. For simplicity, we will use the following notation: if $A$ is a full subcomplex of the nerve $L$, then $$b_i^\mathbf{q}(L):=L^2_\mathbf{q} b_i(\Sigma(W,S))\text{ and } b_i^\mathbf{q}(A):=L^2_\mathbf{q} b_i(W_L\Sigma_A).$$

Let $L$ be a regular CW-complex and let $bL$ denote its barycentric subdivision. Since $bL$ is a flag complex, there is a natural right-angled Coxeter group associated to $bL$ denoted by $W_{bL}$. The vertices of $bL$ are naturally graded by ``dimension'': each vertex $v$ is the barycenter of a unique cell $\sigma_v$. The dimension of $v$ is the dimension of $\sigma_v$. For a $k$-cell $\sigma^k$, we denote its boundary complex by $\partial\sigma^k$. We begin with the following set of lemmas.

\begin{lemma}[{\cite[Lemma 2.1]{DO3}}]\label{Links}
Let $bL$ be the barycentric subdivision of a regular CW-complex $L$, and let $v$ be a vertex of dimension $k$. Then $$\Link_{bL}(v)=b\partial\sigma_v^k\ast b\Link_L \sigma_v^k.$$
\end{lemma}

\begin{remark}\label{LinksRemark}
Suppose that $L$ is a PL-cellulation of an $(n-1)$-manifold, then $\Link_{bL}(v)=S^{k-1}\ast S^{n-2-k},$ where $S^{k-1}$ and $S^{n-2-k}$ are barycentric subdivisions of PL-cellulations of spheres of corresponding dimensions.
\end{remark}

\begin{lemma}[{\cite[Lemma 3.2]{DO3}}]\label{AffectingLinks}
Let $bL$ be the barycentric subdivision of a regular CW-complex $L$, and let $v$ be a vertex of dimension $k$. The removal of $v$ from $bL$ does not change the first factor in the decomposition of Lemma \ref{Links} of links for vertices of dimension less or equal than $k$, and it does not change the second factor for vertices of dimension greater or equal than $k$.
\end{lemma}

\begin{lemma}\label{HomologyAffectedLink}
Suppose that $A$ is a full subcomplex of a triangulation of $S^m$, where $m\leq 3$. If Conjecture \ref{conj:weightedSinger} holds for $S^p$, then $b_i^\mathbf{q}(S^p\ast A)=0$ for $i>\frac{p+m}{2}+1$.
\end{lemma}
\begin{proof}
By Theorem \ref{RAGeneralizedSinger}, $b_i^\mathbf{q}(A)=0$ for all $i>\frac{m+1}{2}$. Since Conjecture \ref{conj:weightedSinger} holds for $S^p$, it follows that $b_i^\mathbf{q}(S^p)=0$ for $i>\frac{p+1}{2}$. The assertion follows by the weighted K\"{u}nneth formula.
\end{proof}

\begin{lemma}\label{RemovingLinks}
Suppose the nerve $L$ is $(n-1)$-dimensional and that $v$ is a $k$-vertex of $bL$. Let $bL-v$ denote the complex obtained by removing the vertex $v$ from $bL$. If $\Link_{bL}(v)$ satisfies the hypothesis of Lemma \ref{HomologyAffectedLink}, then $$b_i^\mathbf{q}(bL)=b_i^\mathbf{q}(bL-v)\text{ for } i>\frac{n}{2}.$$
\end{lemma}
\begin{proof}
Set $\Lk(v)=\Link_{bL}(v)$ and let $C\Lk(v)$ denote the right-angled cone on $\Lk(v)$. Consider the Mayer-Vietoris sequence:
 $$\begin{tikzcd}[column sep=small]
  L^2_\mathbf{q}H_\ast(\Sigma_{\Lk(v)}) \arrow{r} & L^2_\mathbf{q}H_\ast(\Sigma_{C\Lk(v)}) \oplus L^2_\mathbf{q}H_\ast(\Sigma_{bL-v}) \arrow{r} & L^2_\mathbf{q}H_\ast(\Sigma_{bL})
\end{tikzcd}$$

By assumption, $L^2_\mathbf{q}H_\ast(\Sigma_{\Lk(v)})=0$ for $\ast>\frac{n-1}{2}$, and hence by the weighted K\"{u}nneth formula, $L^2_\mathbf{q}H_\ast(\Sigma_{C\Lk(v)})=0$ for $\ast>\frac{n-1}{2}$. The lemma follows from the weak exactness of the Mayer-Vietoris sequence.
\end{proof}

In \cite{DO3}, Davis and Okun computed the ordinary $L^2$-Betti numbers of the right-angled Coxeter group $W_{bL}$ where $L$ is a PL-cellulation of an $(n-1)$-manifold for $n = 6,8$: they show that they always vanish above the middle dimension. We prove the analogous statements for weighted $L^2$-cohomology.

\begin{theorem}\label{WeightedManifold}
Suppose that $L$ is a PL-cellulation of an $(n-1)$-manifold with $n=6,8$. Then $b_i^\mathbf{q}(bL)=0$ for $i>\frac{n}{2}$.
\end{theorem}

\begin{proof}
We first consider the case where $n=6$. By Lemma \ref{Links}, vertices of dimension $3$ and $1$ have respective links $S^2\ast S^{1}$ and $S^0\ast S^{3}$, and the link of a vertex of dimension $4$ is $S^3\ast S^{0}$ (all spheres involved are barycentric subdivisions of PL-cellulations of spheres). By the validity of Conjecture \ref{conj:weightedSinger} for $n\leq 4$, these links satisfy the hypothesis of Lemma \ref{HomologyAffectedLink}. We obtain a $2$-complex $T$ by removing vertices of $L$ of dimensions $4$, $3$, $1$ (in that order). Note that Lemma \ref{AffectingLinks} implies that all of the first factors of the join decompositions of the links are not affected. By Lemma \ref{RemovingLinks}, $b_i^\mathbf{q}(bL)=b_i^\mathbf{q}(T)$ for $i>3$. Since $T$ is $2$-dimensional, the theorem follows.

For the case $n=8$, we invoke a similar procedure to obtain a $4$-complex $T$. We first begin by removing vertices of dimension $6$ (with respective links in $bL$ equal to $S^5\ast S^{0}$), then vertices dimension $4$ (with respective links in $bL$ equal to $S^3\ast S^{2}$), and finally of dimension $3$ (with respective links in $bL$ equal to $S^2\ast S^{3}$). By the previous computation for the case $n=6$ and the validity of Conjecture \ref{conj:weightedSinger} for $n\leq 4$, we have that, at each stage of removing a vertex, the remaining affected links satisfy Lemma \ref{HomologyAffectedLink}. Hence, Lemma \ref{RemovingLinks} implies that $b_i^\mathbf{q}(bL)=b_i^\mathbf{q}(T)$ for $i>4$.

 \cite[Theorem 3.1]{DO3} implies that $b_5^\mathbf{1}(T)=b_5^\mathbf{1}(bL)=0$. Since $\Sigma_{T}$ is $5$-dimensional, Lemma \ref{Pushing0} implies that $b_5^\mathbf{q}(T)=0$ for $\mathbf{q}\leq\mathbf{1}$, thus completing the proof of the assertion.
\end{proof}

\begin{remark}
In fact, the statement of Theorem \ref{WeightedManifold} holds when $L$ is assumed to be a PL-cell complex that is a generalized homology manifold. The point is that, given a $k$-vertex, its link  $\Link_{bL}(v)=S^{k-1}\ast GHS^{n-2-k},$ where $S^{k-1}$ is a $(k-1)$-sphere and $GHS^{n-2-k}$ is a generalized homology $(n-2-k)$-sphere. Moreover, Theorem \ref{RAGeneralizedSinger} and Conjecture \ref{conj:weightedSinger} (for $n\leq 4$) remain true when $L$ is assumed to be a generalized homology sphere. \end{remark}

In fact, if we assume $L$ to be a PL-cellulation of a sphere, we get the following corollary.

\begin{corollary}
Suppose that $L$ is a PL-cellulation of $S^{n-1}$. Then Conjecture \ref{conj:weightedSinger} is true for $W_{bL}$ for all even $n\leq 8$.
\end{corollary}

redpajama_set_name: RedPajamaArXiv


I have an easy technique for you today – one that you are going use over and over because it is so easy! I chose to showcase this technique on the Grown with Love bundle because it's the perfect bundle for this! When you view my video you'll see why I chose this technique for this week. . . . but first let me share the card I was creating when I got the inspiration from my class attendees.
Love this card – and you can make it as it will be part of my Card Buffet in September. Who wouldn't be happy when they received this card! Yep, it's pretty cool! And the best part is that the flower pot and the flowers are cut out with dies! Now, check out the flowers and the greeting – they are all colored with markers!
Here's another Grown with Love card – and this one too is colored with markers. It is one of the easiest techniques ever!
And here's the one I'll show you in my video. Love it! So perfect for fall – and it is so simple!
And there you have it! I hope you enjoy the Tuesday series! Remember to shop my online store for papercrafting needs!


text:
UNHINGED: CNN Analyst Says Free Speech Is 'Nonsense'
CNN Analyst Robert Baer says free speech is "nonsense."
Baer, a former CIA case officer, appeared on "CNN Newsroom" with host Boris Sanchez to discuss Elon Musk's new pro-free speech policies on Twitter.
TheBlaze reported on Saturday that Baer bizarrely accused Musk of promoting free speech online to benefit Russia in its war against Ukraine.
"Well, Boris, I can tell you one thing, Putin will be all over Twitter if there's no regulations on this," he said of reinstated censored accounts. "Fake accounts, spoofed accounts, the rest of it, this is a great opportunity for him."
Baer then went on to suggest that Musk was personally colluding with Russian intelligence officials against the United States.
"And so when he's talking about the popular voice, Musk, he's really talking about Russian intelligence," he said. "The Russians are waiting for something like this."
He also claimed that Musk was trying to interfere with U.S. support of Ukraine.
"They need a propaganda campaign against the United States and against our support for Ukraine," he continued. "And they're going to be all over Twitter. I guarantee this. Supporting the far right, plans, demands to stop arming Ukraine. You just wait."
Baer's comments became even more unhinged as claimed that free speech as "nonsense."
"[Y]ou know, this freedom of speech is just nonsense because you can't go in a movie theater and yell 'fire.' It's against the law," Baer continued, incorrectly labeling legally protected speech illegal.
When Sanchez asked Baer what advice he would give to Musk for the future, he said, "I'd say this Libertarian nonsense is destructive to American national security. And he has got to reinstitute the same restrictions that were on Twitter before he bought it. There's no other choice."
Conservatives are under attack. Contact your representatives and demand that Big Tech be held to account to mirror the First Amendment while providing transparency, clarity on "hate speech" and equal footing for conservatives. If you have been censored, contact us at the Media Research Center contact form, and help us hold Big Tech accountable.
Walmart raises minimum wage as retail labor market remains tight
There have been 275 attacks on Catholic churches since May 2020: Report
Trump defends Mike Pence over classified documents found at his Indiana home: 'Leave him alone!!!'
Sopranos Star John Ventimiglia's Daughter Dead at 25
California gunman found dead in van after killing 10 in Chinese New Year mass shooting

redpajama_set_name: RedPajamaCommonCrawl


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Please help get the word about this event so that we might recover security camera video or cell phone captures.
If you have a video or still image please send to me @ drtanuki@gmail.com ; Thank you!
I saw this fireball while awaiting sunrise at Mesa Arch, Canyonlands National Park outside Moab, UT. The streak went quickly orange - white - green before fading out.


text:
package grn;

import java.io.FileInputStream;
import java.io.FileOutputStream;
import java.io.ObjectInputStream;
import java.io.ObjectOutputStream;
import java.io.Serializable;
import java.util.Random;
import java.util.Vector;

public class GRNModel implements Serializable {
	private static final long serialVersionUID = 1L;

	double beta;
	double delta;
	int currentStep;
	double maxEnhance;
	double maxInhibit;
	double enhanceMatching[][];
	double inhibitMatching[][];

	public Vector<GRNProtein> proteins;

	public GRNModel(Vector<GRNProtein> p, double b, double d) {
		proteins=p;
		beta=b;
		delta=d;

		maxEnhance=0;
		maxInhibit=0;

		updateSignatures();
	}

	public GRNModel() {
		proteins=new Vector<GRNProtein>();
		beta=1;
		delta=1;
	}

	public GRNModel(GRNModel grn) {
		proteins=new Vector<GRNProtein>();
		for (int i=0; i<grn.proteins.size(); i++) {
			proteins.add(new GRNProtein(grn.proteins.get(i)));
		}

		beta=grn.beta;
		delta=grn.delta;

		maxEnhance=grn.maxEnhance;
		maxInhibit=grn.maxInhibit;

		enhanceMatching=new double[grn.enhanceMatching.length][grn.enhanceMatching[0].length];
		for (int i=0; i<grn.enhanceMatching.length; i++) {
			for (int j=0; j<grn.enhanceMatching[i].length; j++) {
				enhanceMatching[i][j]=grn.enhanceMatching[i][j];
			}
		}
		inhibitMatching=new double[grn.inhibitMatching.length][grn.inhibitMatching[0].length];
		for (int i=0; i<grn.inhibitMatching.length; i++) {
			for (int j=0; j<grn.inhibitMatching[i].length; j++) {
				inhibitMatching[i][j]=grn.inhibitMatching[i][j];
			}
		}

		currentStep=grn.currentStep;
	}

	public static GRNModel loadFromFile(String fileName) throws Exception {
		FileInputStream fis = new FileInputStream(fileName);
		ObjectInputStream ois = new ObjectInputStream(fis);

		GRNModel res = (GRNModel)ois.readObject();

		fis.close();
		ois.close();

		return res;
	}

	public Vector<GRNProtein> getProteins() {
		return proteins;
	}

	//		inline unsigned int getNbProteins() {return nbProteins;};
	public void evolve(int nbSteps) {
		double enhance, inhibit;
		int i,j,k;
		double sumConcentration;
		//		for (i=0; i<proteins.size(); i++) {
		//			cerr << proteins[i].concentration << "\t";
		//		}
		//		cerr << proteins.size() << endl;

		for (int step=0; step<nbSteps; step++) {
			Vector<GRNProtein> nextProteins=new Vector<GRNProtein>();

			// Calculating next step protein concentrations
			for (j=0; j<proteins.size(); j++) {
				// For an input protein, the concentration is not calculated
				if (proteins.get(j).type==GRNProtein.INPUT_PROTEIN) {
					nextProteins.add(proteins.get(j));
				} else {
					enhance=0;
					inhibit=0;
					// Calculating enhancing and inhibiting factor for the current protein
					for (k=0; k<proteins.size(); k++) {
						if (proteins.get(k).type!=GRNProtein.OUTPUT_PROTEIN) {
							enhance+=proteins.get(k).concentration*enhanceMatching[j][k];
							inhibit+=proteins.get(k).concentration*inhibitMatching[j][k];
						}
					}
					// if (j=5) cout << enhance << "   " << inhibit << endl;
					// Calculating the next concentration of current protein
					nextProteins.add(new GRNProtein(proteins.get(j).id, proteins.get(j).type, Math.max(0.0,proteins.get(j).concentration+delta/proteins.size()*(enhance-inhibit)), proteins.get(j).enhancer, proteins.get(j).inhibiter));
				}
			}

			//		for (i=0; i<nextProteins.size(); i++) {
			//			cout << nextProteins[i].id << "=" << nextProteins[i].concentration << "\t";
			//		}
			//		cout << endl;
			// Reajusting proteins concentration so that their sum stay equal to 1
			sumConcentration = 0;
			for (i=0; i<proteins.size(); i++) {
				if (proteins.get(i).type!=GRNProtein.INPUT_PROTEIN) {
					sumConcentration+=nextProteins.get(i).concentration;
				}
			}
			//cout << sumConcentration << "\t" << nbDiv << "\t" << (sumConcentration-1)/nbDiv << "\t";
			if (sumConcentration!=0) {
				for (i=0; i<proteins.size(); i++) {
					if (proteins.get(i).type!=GRNProtein.INPUT_PROTEIN) {
						nextProteins.get(i).concentration/=sumConcentration;
					}
				}
			}

			// Finalizing the step by switching the vector
			proteins=nextProteins;

			currentStep++;
		}

	}

	public void duplicateProteins(int nbDup, double mutProb, Random rng) {
		//srand(time(NULL))
		int nbProteins=proteins.size();
		for (int dup=0; dup<nbDup; dup++) {
			for (int i=0; i<nbProteins; i++) {
/*				proteins.add(new GRNProtein(
						Math.random()>mutProb?proteins.get(i).id:(int)(Math.random()*GRNProtein.IDSIZE),
								proteins.get(i).type,
								proteins.get(i).concentration,
								Math.random()>mutProb?proteins.get(i).enhancer:(int)(Math.random()*GRNProtein.IDSIZE),
										Math.random()>mutProb?proteins.get(i).inhibiter:(int)(Math.random()*GRNProtein.IDSIZE)));*/
				proteins.add(new GRNProtein(
						rng.nextDouble()>mutProb?proteins.get(i).id:(int)(rng.nextDouble()*GRNProtein.IDSIZE),
								proteins.get(i).type,
								proteins.get(i).concentration,
								rng.nextDouble()>mutProb?proteins.get(i).enhancer:(int)(rng.nextDouble()*GRNProtein.IDSIZE),
										rng.nextDouble()>mutProb?proteins.get(i).inhibiter:(int)(rng.nextDouble()*GRNProtein.IDSIZE)));

			}
		}
		updateSignatures();

	}

	private void updateSignatures() {
		// calculating signatures
		enhanceMatching = new double[proteins.size()][proteins.size()];
		inhibitMatching = new double[proteins.size()][proteins.size()];
		for (int j=0; j<proteins.size(); j++) {
			for (int k=0; k<proteins.size(); k++) {
				enhanceMatching[j][k] = GRNProtein.IDSIZE-Math.abs(proteins.get(j).enhancer-proteins.get(k).id);
				maxEnhance=Math.max(maxEnhance, enhanceMatching[j][k]);
				inhibitMatching[j][k] = GRNProtein.IDSIZE-Math.abs(proteins.get(j).inhibiter-proteins.get(k).id);
				maxInhibit=Math.max(maxInhibit, inhibitMatching[j][k]);
			}
		}
		for (int j=0; j<proteins.size(); j++) {
			for (int k=0; k<proteins.size(); k++) {
				enhanceMatching[j][k]=Math.exp(beta*(double)enhanceMatching[j][k]-(double)maxEnhance);
				inhibitMatching[j][k]=Math.exp(beta*(double)inhibitMatching[j][k]-(double)maxInhibit);
			}
		}
	}

	public void writeToFile(String fileName) throws Exception {
		FileOutputStream fos = new FileOutputStream(fileName);
		ObjectOutputStream oos = new ObjectOutputStream(fos);

		oos.writeObject(this);

		oos.close();
		fos.close();
	}

	public GRNModel copy() {
		GRNModel grn = new GRNModel();
		for (int i=0; i<this.proteins.size(); i++) {
			grn.proteins.add(new GRNProtein(this.proteins.get(i)));
		}

		grn.beta=this.beta;
		grn.delta=this.delta;

		grn.maxEnhance=this.maxEnhance;
		grn.maxInhibit=this.maxInhibit;

		grn.enhanceMatching=new double[this.enhanceMatching.length][this.enhanceMatching[0].length];
		for (int i=0; i<this.enhanceMatching.length; i++) {
			for (int j=0; j<this.enhanceMatching[i].length; j++) {
				grn.enhanceMatching[i][j]=this.enhanceMatching[i][j];
			}
		}
		grn.inhibitMatching=new double[this.inhibitMatching.length][this.inhibitMatching[0].length];
		for (int i=0; i<this.inhibitMatching.length; i++) {
			for (int j=0; j<this.inhibitMatching[i].length; j++) {
				grn.inhibitMatching[i][j]=this.inhibitMatching[i][j];
			}
		}

		grn.currentStep=this.currentStep;

		return grn;

	}

	public void reset() {
		for (GRNProtein p : proteins) {
			p.concentration=1.0/(double)proteins.size();
		}
	}

	public String toString() {
		String res="";
		for (GRNProtein p : proteins) {
			res+=p.toString();
		}
		return res;
	}

	public long size() {
		return proteins.size();
	}

	public double distanceTo(GRNModel g, boolean compareDynamicsCoeff, double protCoef, double enhCoef, double inhCoef, double betaMax, double betaMin, double deltaMax, double deltaMin ) {
		if (g==null) {
			System.err.println("(GRNModel.distanceTo) Genome is null!");
		}
		double distance=0;
		GRNModel gs, gl;
		if (size()<g.size()) {
			gs=this;
			gl=g;
		} else {
			gs=g;
			gl=this;
		}

		// comparing all types
		for (GRNProtein gi1 : gl.proteins) {
			double minD = Double.MAX_VALUE;
			for (GRNProtein gi2 : gs.proteins) {
				double d = gi1.distanceTo( gi2, protCoef, enhCoef, inhCoef);
				if( gi1.type == gi2.type && d < minD )
					minD = d;
			}
			distance+= minD;
		}

		if (compareDynamicsCoeff) {
			// take beta and delta to the distance calculation
			distance+=Math.abs(beta-g.beta)/(betaMax-betaMin);
			distance+=Math.abs(delta-g.delta)/(deltaMax-deltaMin);
			return ((double)distance)/(gl.size()+2.0);
		} else {
			return ((double)distance)/gl.size();
		}
	}

	static public void main(String args[]) {
		/* For compiling a JAR that outputs the distance between 2 genomes */
		try {
			GRNModel g1 = GRNModel.loadFromFile( args[0] );
			GRNModel g2 = GRNModel.loadFromFile( args[1] );

			boolean compDynamicsCoef = Boolean.parseBoolean( args[2] );
			double protCoef = Double.parseDouble( args[3] );
			double enhCoef  = Double.parseDouble( args[4] );
			double inhCoef  = Double.parseDouble( args[5] );
			double betaMax  = Double.parseDouble( args[6] );
			double betaMin  = Double.parseDouble( args[7] );
			double deltaMax  = Double.parseDouble( args[8] );
			double deltaMin  = Double.parseDouble( args[9] );

			System.out.println( g1.distanceTo( g2, compDynamicsCoef, protCoef, enhCoef, inhCoef, betaMax, betaMin, deltaMax, deltaMin ) );
		} catch (Exception e) {
			e.printStackTrace();
		}
	}

}

redpajama_set_name: RedPajamaGithub


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To become better at your craft, you need to practice. Whether you are writing emails, love letters, blog posts, magazine articles, short stories, or a novel, it's important to write regularly. Pay attention and put some effort into each email. Make each blog post count.
The best writers read. A ton. However, you don't have to read Shakespeare, Hemingway, or Austen to become a great writer. Reading what you might consider bad writing is just as important as reading what you might consider great writing. You'll learn from other writers mistakes and quickly figure out what NOT to do.
It's also crucial to read a variety of materials. Read books, magazine articles, blog posts, brochures, and short stories. Read chick lit, memoirs, poetry, and children's books. Each genre will teach you something different. Each author will show you their own personal style of writing.
Whether you are a copywriter, ghost writer, or journalist, it's still important to write what you love. When you write what you love, you write with passion.
So if you write magazine articles for a living but really love poetry, write poetry. If you love gardening, write about gardening. Find what you love and find the time to write about it. You'll quickly see how much more you put into your work. You'll also see how great your writing really is.
Unless you work for a company or in an office, you likely write from home. Alone. Writers are often solitary creatures. You may do your best work alone. However, it's important to connect with others to grow as a writer.
You can connect with writers all over the world via Twitter, Google+, or other social networking platforms. It's a great way to see what others are doing, discuss all aspects of writing, and maybe even find a mentor or develop a friendship.
If you write for a living and it's your bread and butter, you might not ever consider taking a break. Again, writers are solitary creatures and often write from home. This might mean writing all day, all night, or all weekend. TAKE A BREAK.
Working on a novel, short story, or article? Step away from it for an hour, day, or even a weekend. Being too close to something for too long is never a good idea. When you come back to your work, with fresh eyes, you'll find all your mistakes.
If you regularly practice the five tips above, you'll find that your writing will improve quickly and dramatically.
Do you have any special tricks you use to improve your writing? Share them with us in the comments section below.


Did you know that the tea bag was invented inadvertently?
In 1908 The New York tea importer Thomas Sullivan was sending tasting tea samples to his customers. Because he could not ship loose tea, and the shipping was too expensive in heavy tin cans, he packed the samples in small mousseline bags. However, the customers did not unwrappe the tea and dipped the whole bag into the water. They found the invention of the tea bag pretty handy - even though if it was not foreseen by Sullivan.
According to one legend Bodhidharma founder of the Zen school of Buddhism, after meditating for nine years, he accidentally fell asleep. He woke up in such disgust at his weakness, he cut off his eyelids and they fell to the ground and took root, growing into tea bushes.
In one popular Chinese legend around 2737 BC, Shennung, the legendary Emperor of China, inventor of Chinese medicine, was drinking a bowl of boiling water. The wind blew and a few leaves from a nearby tree fell into his water and began to change its color. The ever inquisitive and curious monarch took a sip of the brew and was pleasantly surprised by its flavor and its restorative properties.
Farhad Pirouz, managing director and C.E.O. of dph tea bag machine company in Germany, explains that "the greatest value addition in tea is the packaging segment." During a presentation at the 5th Global Dubai Tea Forum last April Pirouz said that tea bags account for 12% of the volume of world production but "their sales value is much higher".
In France tea exports averaged $17.33 per kilo in 2010. India's tea exports averaged $3.31 and China $3,92 per kilo.
"In order to meet growing demand, where growth in population and a decrease in production is forecast, is it wiser ... (a) to invest in increasing production of tea or (b) invest in improving the infrastructure of our value chain?" asks Pirouz.
We at dph have decided to represent machines which are "the machines of the future". Based on this vision, we have decided in favour of HST and its range of teabag packing machines to be the right machine range for the market, for our customer and friends.
At the very end, our customers want to produce "state of the art" teabags for your markets, using production lines that are cost effective, simple to run / maintain, serving you a long time into the future.
It's time to "Decide For Future" !


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Insane in the Ukraine
Posted by historymaniacmegan on September 24, 2019
In mid-September 2019, according to The New York Times, an unidentified internal Trump administration whistleblower filed a complaint about "multiple acts" by a shitty excuse for a president Donald Trump. The whistleblower in question is part of the US intelligence community and filed this complaint back in August, which was passed to their inspector general. That inspector general determined it credible and a matter of "urgent concern" – legal standard normally requiring notifying congressional oversight committees. He then concluded the complaint, "relates to one of the most significant and important of the DNI's responsibilities to the American people." However, Trump's acting national intelligence director stepped in to block key congressional committee chairs from receiving the whistleblower complaint's details, which remain murky. An act some legal analysts claim is breaking the law.
Now despite the murky details, the whistleblower's complaint reportedly involves a broader set of events than a single phone call. But not surprisingly, the Trump administration is trying to prevent further info from coming to light. For some time, it's been rumored Donald Trump tried pressuring Ukraine's government into launching an investigation of former Vice President and current Democratic presidential frontrunner Joe Biden, possibly by withholding military aid to the country unless they complied. On August 28, 2019, Politico reported that the Trump administration was, "slow-walking $250 million in military assistance to Ukraine." According to the site, Trump had personally asked his national security team to review the program, supposedly to ensure the money was being spent on American interests, writing, "The funds for Ukraine can't be spent while they're under review and the money expires at the Sept. 30 end of the fiscal year." Now it's not confirmed if the whistleblower complaint has anything to with this Ukranian debacle, but both cases seem closely related.
Naturally given Donald Trump's affinity for Russian President Vladmir Putin and Russia's war with Ukraine, critics instantly accused him of supporting Putin's policies again. On September 5, 2019, Washington Post editorial claimed they've been told that Trump was trying to force the Ukranian government to investigate Joe Biden. They write:
"Some suspect Mr. Trump is once again catering to Mr. Putin, who is dedicated to undermining Ukrainian democracy and independence. But we're reliably told that the president has a second and more venal agenda: He is attempting to force Mr. Zelensky to intervene in the 2020 U.S. presidential election by launching an investigation of the leading Democratic candidate, Joe Biden. Mr. Trump is not just soliciting Ukraine's help with his presidential campaign; he is using U.S. military aid the country desperately needs in an attempt to extort it."
During a September 2 press conference in Warsaw, Associated Press' Jill Colvin asked Vice President Mike Pence, "Can you assure Ukraine that the hold-up of that money has absolutely nothing to do with efforts, including by Rudy Giuliani, to try to dig up dirt on the Biden family?" Pence conspicuously didn't make that kind of assurance. Instead, he replied, "as President Trump had me make clear, we have great concerns about issues of corruption." However, the notion that the Trump administration has any great concern about corruption issues is basically akin to Pig Pen having any concern about personal hygiene. Because we all know that Trump and his cronies engage in corruption on a regular basis that the swamp he's promised to drain has now become a reeking cesspit of hazardous waste. Hell, the only time the Trump administration shows any concern about corruption is when it pertains to someone they don't like because it makes them look bad. So naturally, they're looking for dirt.
On Friday, September 20, 2019, The Wall Street Journal reported that, during a July phone call with Ukraine President Volodymyr Zelensky, Donald Trump pressured him "about eight times" to work with his sell out lawyer Rudy Giuliani on an investigation into Biden's son, Hunter. That Thursday, Giuliani tweeted that if Trump told Ukraine to "investigate corruption that affects US" he'd just be "doing his job," and complaining that "the Biden Family… bilked millions from Ukraine." He even later confirmed that he himself has been trying to get Ukraine to investigate Biden. Strange Trump didn't call the Ukrainian government to investigate his own campaign manager Paul Manafort back in 2016, because he actually bilked millions from the Ukraine and is serving prison time for it. However, if Trump did this as president, it would be a shockingly corrupt use of his foreign policy powers. Since he's basically demanding a foreign country intervene in the 2020 election by digging up dirt on a potential opponent, or have its security put at risk.
The idea that Donald Trump's team would try getting the Ukranian government to investigate Joe Biden's family isn't just theoretical. Even Trump lawyer Rudy Giuliani has openly admitted he's been doing just that. As he told the New York Times in May, "We're not meddling in an election, we're meddling in an investigation, which we have a right to do." Ukraine-related corruption has already played an outsized role in Trump scandals. Paul Manafort's prosecution for financial and lobbying crimes related to his work for a former Ukranian regime was a major part of the Mueller probe. And during the summer of 2016 back when Manafort was Trump's campaign chair, he was plagued by reports that the Ukranian government was looking into his payments. So Donald Trump's team apparently has the idea to try and cook up a similar scandal involving Joe Biden.
The details relate to Joe Biden's ne'er-do-well son Hunter who joined a Ukrainian natural gas company Burisma's board in 2014. Now the company's owner was under investigation for corruption and money laundering. Two years later, Ukraine's prosecutor general Viktor Shokin was fired, after pressure from Vice President Biden and other Western officials along with many Ukrainian officials and citizens. Biden just happened to have the loudest voice. Shokin has reportedly claimed he was pushed out because he was investigating Burisma's payments to Hunter Biden. However, the New York Times writes, "there is no credible evidence that Biden sought Shokin's removal in order to protect Hunter." Instead, the rationale was said he wasn't doing enough to investigate the corruption. Now, in an effort to cause political problems in Biden's 2020 campaign, Giuliani has been pushing the new Ukrainian government to open an investigation into the Biden matter, as well as whether there was any foul play in the earlier Ukrainian Manafort investigation. Giuliani confirmed he was doing all this to the Times back in May. The effort continued through August. But Giuliani was cagey in Trump's personal role in the scheme. He told the Times in May that Trump supports his endeavors and "he basically knows what I'm doing, sure, as his lawyer." In August, he told the Times he was just acting as a private citizen. Despite that State Department officials were involved in Giuliani's communications with Ukrainian officials for some reason.
Now that Donald Trump has all but openly admitted that he pushed Ukranian President Volodymyr Zelensky to investigate former Vice President Joe Biden and his son Hunter, Congress must impeach him. Impeaching Trump over Robert Mueller's findings in the Russia investigation would've been an attempt to address past offenses. Impeaching Trump over these calls would be an attempt to halt what surely resembles an ongoing attempt to hijack American foreign policy in service of his reelection. Democrats are obligated to stop this before it gets any further. Sure, impeachment is virtually guaranteed to fail in the Republican-controlled Senate so there's no real chance of actually removing Trump from office. Public opinion about the Russian scandal became more set along partisan lines as time went on, making it unlikely that drawing attention to it would galvanize the public against Trump in 2020. Since that would risk distracting Democrats on which Trump is genuinely unpopular like on healthcare and climate change and jeopardize the House Democratic majority with marginal gain.
But the new Ukraine scandal challenges this logic. There is now an obvious and immediate pragmatic upside to impeachment: stopping an ongoing abuse of presidential power that could undermine the 2020 election's integrity. Thanks to an intelligence community whistleblower, investigative journalists, and Donald Trump's own public statements, Trump seems to have repeatedly attempted to convince the Ukranian government to open an investigation into Hunter Biden's Ukraine business dealings and Joe Biden's alleged involvement in protecting his son from prosecutorial attention. But there's no evidence of illegal conduct by either Biden in the Ukraine dealings. Hunter's partnership with a corrupt Ukranian oligarch was arguably unethical. But there's no reason to believe his dad was involved in it. Still, even if either Biden was implicated in anything illegal, Trump's actions would still be as impeachable. Because he's trying to get a foreign power to investigate a potential political opponent on the pretense of turning Biden's fake Ukraine scandal into "her emails" 2.0. Thus, he actively working to weaponize the presidency to boost his political fortunes.
Hell, it may be even worse. Donald Trump himself has linked the Biden issue to US to Ukraine aide. On Sunday, he told reporters, he "had every right" to push Ukraine about Joe Biden because "we don't want a country that we're giving massive aid to be corrupting our system." If Trump threatened to condition aid to Ukraine on its Biden investigation, then he's been nakedly twisting US foreign policy to suit his own ends. This is a grotesque and seemingly ongoing abuse of power with potential implications for an election's integrity next year. Whereas the Russia investigation an attempt to find out exactly what happened in a prior election, the Ukraine scandal reflects Trump's contemporary and future-looking behavior. Given that the goal is no longer retrospective accountability, this dramatically changes the logic of impeachment. Since it's now about stopping his current behavior. The hope would be that impeachment would bring so much attention and scrutiny to Trump's Ukraine push that he can't get away with undermining another election.
Any impeachment proceeding would be the story in American politics, sucking up media attention and congressional investigative resources. A House majority vote to impeach would lead to a trial in the Senate, attracting more scrutiny even if Majority Leader Mitch McConnell refuses to take the proceedings seriously. The aim would be to prevent Trump from making some kind of shady, behind-the-scenes agreement with some Ukrainian authorities and make him think twice about any other similar scheme for using his powers for electoral gain. Such level of attention seems like the best available tool for preventing Donald Trump from continuing his efforts to undermine the 2020 election. Moreover, such high levels of press coverage and partisan furor would also make it harder to imagine the Ukrainian government would make any corrupt deal with Trump. Democratic posturing would serve as a counterweight to Trump's pressure on Ukraine, signaling the country's leadership that any cooperation with Trump's inappropriate demands could seriously fray relations with the US in the next administration. Under this logic, it doesn't matter if impeachment will invariably fail in the Senate. Just shining a light on Trump's misbehavior will limit his freedom to act. Because if you have a president actively trying to abuse his power in order to invite foreign meddling in the next presidential election, you need to do what you can to stop him. Impeachment is the biggest and most powerful tool in the Democrats' inventory. Because impeaching Trump is about signaling that his conduct is unacceptable as well trying to impose accountability on him and setting a standard for future ones.
Should impeachment be used not only to signal disapproval but actually work to head off an ongoing threat to American democracy, then the normative power of the proceedings might be reestablished. They won't just be futile raging at American politics' debased nature under Donald Trump, but an effective means of actually changing these politics for the better. For Trump's impeachment to actually serve as a means of accountability to show future officeholders that misbehavior carries costs, there needs to be actual bite to them. Otherwise, they really risk sending the opposite intended signal that nothing really matters and that the president can do whatever he wants as long as at least 34 senators support him. But if impeachment can plausibly constrains Donald Trump, preventing him from engaging in abuse of power for political gain, then the Trump administration's lesson would be that actions carry consequences, that Congress' ultimate constitutional power can still be used to rein in a president even in a political environment seemingly defined by extreme partisanship. Furthermore, impeachment sends the strongest and most high-profile signal possible that Trump's actions are unacceptable, both now and to future presidents.
Nonetheless, Donald Trump's behavior in this Ukraine situation should worry anyone who cares about the health of American democracy. If this isn't impeachable behavior, then I don't know what is. Could impeachment potentially rein in Trump? I'm not sure since Trump never learns from his misconduct. But it will limit him on what he can get away with. Will a formal impeachment inquiry hurt the House Democrats' chances to retain the House? Who knows. But seeing how the Ukraine scandal drove a painful reality home of an emboldened Trump appearing to meddle in an upcoming US election again, right before our eyes, Congress must impeach.
This entry was posted in Society and tagged crime, Donald Trump, Elections, government, Law, politics, Ukraine, United States. Bookmark the permalink.
The Wonderful World of Scarecrows (Fifth Edition)
The Wonderful World of Vintage Postcards (Eighth Edition)


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Those that have been long time readers of IFTLE and its predecessor PFTLE know that I sometimes look at 3D IC through the eyes of someone that was part of the bumping/WLP technology explosion that occurred in the late 1990s/early 2000s. Further, if you know this history you know that while I give most of the credit for the development of most if not all of that early low cost bump/WLP technology to start ups FCT and Unitive in the USA, it was the Taiwanese who saw the power of this technology, licensed it from the aforementioned startups and put the capacity in place to make this the key technology that it is today. While fan-in WLP has evolved into fan-out WLP and copper pillar technology it has in some sense has become synonymous with "Advanced Packaging".
The Semicon Taiwan 3D Technology Forum was chaired by Ho-Ming Tong, General Manager and CTO for ASE. Speakers included representatives from Yole, Nokia, Qualcomm, UMC, SIliconware, Verigy, Applied, ITRI, IME and Sematech. The next two blogs will cover significant material from this meeting.
Dr. Tong, who some say coined the term "2.5D" for the use of silicon or glass interposers with TSV, indicated that this technology "..is ready to move to the next stage" Tong expects commercialization of 2.5D chip technology to take place in two years.
Tong notes that 2.5D IC should not be regarded as a transitional integration technology. 2.5D will enable packaging of chips in the 32-22 nm nodes where the fragile mechanical stability of the low-K dielectrics used in these products will require their bonding to an intermediate silicon interposer before final placement in a standard package.
Tong believes that commercialized products made using 2.5D IC and 3D IC technologies, including smartphones and computing applications, will hit the market in the next five years. This last comment made headlines at several news outlets [ link], which Quoted Dr. Tong as saying " Despite tremendous progress in recent years, 3D IC with through silicon vias (TSVs) still presents significant challenges in cost, design, manufacturing, test and supply chain readiness and the technology is still three to five years away from mass production".
After checking with Dr. Tongs colleagues in ASE I have been assured that these comments were meant to indicate that widespread use in products is still 3-5 years out but that ASE stands by the recent roadmap they shared at Semicon West this summer [ see IFTLE 9 "3D In and Around the Moscone Part 1" ] which indicates that 2.5 D readiness is imminent and full 3D for wide IO DRAM / logic bonding would be coming in the 2011 timeframe.
Shan-Chieh Chien, VP of ATD at UMC called 3D stacking with TSV the "big elephant" technology for foundries. UMC, which recently announced an alliance with Elpida and Powertech [ see IFTLE 8 "3D Infrastructure Announcements and Rumors" ] commented that logic + wide IO DRAM stacking will occur in 2011-2012 consistent with the comments of his other Asian colleagues. UMC also sees a significant future for silicon interposers indicating that UMC will use their Cu dual damascene BEOL process for passive device and fine pitch RDL on these interposers.
A new IC Insights image sensor market report forecast that CMOS image sensors sales will rise 34% in 2010 to $5.2 billion from ~ $3.9 billion in 2009. Between 2009 and 2014, CMOS image sensor sales are projected to increase at a 17% compound average growth rate (CAGR), reaching $8.3 billion by 2014. CMOS image sensors dominate portable systems applications, such as camera phones, webcams in notebook computers, and other embedded cameras in handheld products, but higher-speed CMOS imagers are also being aimed at automotive systems, medical equipment, and wireless video security networks.


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Adriana González Carrillo (Mexico, née le ) est une femme politique mexicaine, membre du Partido Acción Nacional, a occupé les fonctions de députée fédérale et de sénatrice.

Lien externe
  Page officielle de Adriana González Carrillo

Femme politique mexicaine
Naissance en mars 1975

redpajama_set_name: RedPajamaWikipedia