text:
\section{Introduction}
A random manifold is a standard probability space $\mathbf K$ whose elements are connected manifolds. We define the total space of a random manifold $\mathbf K$ as the disjoint union $\mathbf X=\coprod_{M\in \mathbf K}M$. We shall see $\mathbf X$ as a laminated set whose leaves are the elements of $\mathbf K$, and assume that there is a standard measurable structure on $\mathbf X$ compatible with the usual Borel structure on the leaves and on $\mathbf K$.

We call {\em observable} every map $b:\mathbf K\to\mbb R$ that associates to every manifold in $\mathbf K$ a real number. For instance: the dimension, the Betti numbers, the Euler characteristic, the Hirzebruch signature and, more generally, the index of differential and pseudo-differential operators, are examples of such invariants. In the setting of random geometry, the most natural question is the following: what is the expected value of $b$ in $\mathbf K$? Of course, in order to get a well posed question, we must guarantee that the map $b$ is measurable and, more precisely, integrable. For classical invariants as those mentioned above, it deeply depends on the measurable structure and specific probability measure we put on the set $\mathbf K$. Our main goal is to develop a general framework allowing the systematic study of such invariants.

Our first step will be to enlarge the notion of random manifold by allowing the probability space $\mathbf K$ to be non commutative in the sense of Connes \cite{connes90book}. A non commutative probability space is defined by Connes as (the Morita equivalence class of) a separable von Neumann algebra $\mathcal N(\mathbf K)$ endowed with a weakly continuous tracial state $\tau$ \cite{connes80surveyfoliations}. Invariants can be viewed as elements of $\mathcal N(\mathbf K)$ and their expected value is given by the trace. If $\mathbf K$ is a classical probability space then $\mathcal N(\mathbf K)=L^{\infty}(\mathbf K)$, i.e. a commutative von Neumann algebra, and the trace is the usual Lebesgue integral.

Random manifolds arise naturally in quantum physics. Up to a Wick rotation, which formaly replaces the real time coordinate $t$ by a pure imaginary time $\tau=it$, the space-time becomes riemannian and quantum mechanics reduce to random geometry. For instance, a Wick rotated relativistic quantum particle is given by a $4$-dimensional manifold $M$ (representing the space-time) together with a set $\mathbf K$ of continuous paths $\alpha:[0,1]\to M$. The probability measure on $\mathbf K$ is the Wiener limit of random walks associated to random jumps of probability density
$$
p(x|y)\propto e^{-d(x,y)}d\operatorname{vol}_M(x),
$$
where $d(x,y)$ is the metric distance and $d\operatorname{vol}_M$ the metric volume on $M$. Another appealing example is string theory, which is a $2$-dimensional version of the relativistic particle above. It is obtained, roughly speaking, by replacing random paths by random immersions $\alpha:\Sigma\to M$ with $\Sigma$ a surface with boundary, random walks by PL immersions, and the metric length of the path by the metric area of the surface. These two examples are particular cases of a more general set of physical theories called brane theories. In general, brane theories can be viewed as bordisms in the category of random manifolds. See \cite{polchinski1998string} for a complete introduction to string theory. Another interesting example is pure quantum gravity, which is given by a random riemannian bordism of a fixed compact riemannian $3$-manifold $M$, whose expected value is an Einstein manifold. Since string theory is given by a random $2$-dimensional submanifold of a fixed manifold, the construction can be made rigorous by using the classification of compact surfaces and PL immersions. In turn, quantum gravity deals with the set of all $4$-dimensional riemannian manifolds whose boundary is $M$, and there is no natural way to put a probability measure on this set, mostly because we don't know what it looks like. The usual approach is to fix the underlying topology and to randomize the metric. Since a riemannian metric can be viewed as a field of $4\times 4$ non singular positive matrices, this reduces quantum gravity to a most familiar quantum field theory, but the theory still has far too many symmetries to be renormalizable. However, quantum gravity remains an extremely speculative domain, and no satisfactory rigorous mathematical model has been found so far. One of the main goals of our introduction to the general study of random manifolds is to clarify and unify the existing approaches to random geometry.

A very fruitful setting is given by algebraic random $n$-manifolds, which can be defined as $p^{-1}(0)$ where $p:\mathbb C^{n+m}\to \mathbb C^p$ is a random polynomial. In that case the probability space $\mathbf K$ is a finite vector space of polynomials and can therefore be endowed with a family of natural probability measures. In a recent work Gayet and Welschinger make very stimulating progresses on the computation of expected Betti numbers in this setting \cite{gayetwelschinger2016betti}.

Given an oriented random manifold with boundary $\mathbf K$, choosing a random connected component of the boundary of an element of $\mathbf K$ gives rise to a random manifold, provided that we can measurably associate to every element $M\in \mathbf K$ a probability measure on the set of connected components of $\partial M$. The corresponding random manifold will be called the boundary of $\mathbf K$. Two random manifolds $\mathbf K$ and $\mathbf K'$ are said to be {\em cobordant} if their disjoint union $\mathbf K+\mathbf K'$ is isomorphic to the boundary of a random manifold. Cobordism obviously defines an equivalence relation on the set of random manifolds and one can easily see that the disjoint union induces a structure of abelian group on the set of cobordism classes. The aim of this paper is to introduce the study of these groups.

\section{Random manifolds}
A {\em BT-space} is given by a standard Borel space $\mathbf X$ endowed with a topology whose connected components, endowed with its Borel $\sigma$-algebra are measurable subspaces of $\mathbf X$. A map $\mathbf X\to \mathbf Y$ is said to a \textit{BT-map} if it is a simultaneously measurable and continuous map that sends transversals of $\mathbf X$ into transversals of $\mathbf Y$. The class of BT-spaces together with BT-maps is a category called \textbf{BTop}. We call \textit{leaves} of a BT-space the connected components of the underlying topology and \textit{transversals} discrete Borel subsets. Two transversals $S$ and $T$ are said to be isomorphic if there exists a leaf-preserving Borel isomorphism $\gamma:T\to S$ that we call a \textit{holonomy transformation}.

\begin{examples}
\begin{enumerate}
  \item A connected Polish space $V$, endowed with is Borel $\sigma$-algebra, is an example of BT-space of a single leaf and transversals, which are simply discrete subsets of $V$, are always countable.
  \item Another example is given by a standard Borel space $\mathbf K$ endowed with the discrete topology. In that case every Borel subset of $\mathbf K$ is a transversal.
  \item Given two BT-spaces $\mathbf X$ and $\mathbf Y$, we shall note $\mathbf X\otimes \mathbf Y$ the cartesian product of $\mathbf X$ and $\mathbf Y$ endowed with the product topology and the product measurable structure.
  \item The BT-space $V\otimes\mathbf K$, obtained as a product of a Polish space $V$ and a standard Borel space $\mathbf K$ will be called a {\em prism} of \textit{base} $V$ and \textit{vertical} $\mathbf K$.
\end{enumerate}
\end{examples}

\begin{definition}[\cite{bermudez2006char}]
	Let $\mathbf X$ a BT-space. An {\em transverse invariant measure} on $\mathbf X$ is a map that associates to every transversal $T$ of $\mathbf X$ a $\sigma$-finite Borel measure $\mu_T$ on $T$ such that $\gamma^*\mu_T=\mu_S$ for every holonomy transformation $\gamma:S\to T$. The pair $(\mathbf X,\mu)$ will be called a {\em random topological space}. A random topological space is said to be a \textit{random manifold (with boundary)} if all its leaves are separable topological manifolds (with boundary). Let $(\mathbf X,\mu)$ and $(\mathbf Y,\nu)$ two random topological spaces. A BT-map $f:\mathbf X\to \mathbf Y$ is said to {\em measure preserving} if for every transversal $T$ of $\mathbf X$ we have
	$$
	\mu(T)=\int_{f(T)}\abs{f^{-1}(t)}d\nu(t)
	$$
	where $\abs{\cdot}$ denotes the cardinal of a set. The category of random topological spaces together with measure preserving BT-maps will be called \textbf{RTop}.
\end{definition}

\begin{examples}
\begin{enumerate}
  \item If $V$ is a Polish space, two discrete subsets of $V$ are isomorphic as transversals if and only if they have the same cardinality; hence the counting measure gives rise to an invariant transverse measure.
  \item Another example is given by a standard measure space $\mathbf K$ endowed with the discrete topology. In that case every Borel subset of $\mathbf K$ is a transversal and the only holonomy trasformation is the identity. Hence, every Borel measure on $\mathbf K$ induces a transverse measure which is automatically invariant.
  \item Given two random topological spaces $(\mathbf X,\mu)$ and $(\mathbf Y,\nu)$, the product of a transversal of $\mathbf X$ and a transversal of $\mathbf Y$ is a transversal of $\mathbf X\otimes\mathbf Y$ and one can easily verify that the corresponding product measure $\mu\otimes\nu$ is invariant.
\end{enumerate}

\end{examples}
 Let $(\mathbf X,\mu)$ be a random manifold (with boundary). We call \textit{atlas} of $\mathbf X$ a countable set of isomorphisms of BT-spaces
 $$
 \mathfrak A=\{\varphi_{i}:U_{i}\to V_{i}\otimes \mathbf K_i\}_{i\in I}
 $$
 called \textit{local charts}, where $\{U_i\}_{i\in I}$ is an open Borel covering of $\mathbf X$, $V_i$ is a open subset of the euclidean space $\mbb R^n$ (resp. the half-space $\mathbb H^n$) and $\mathbf K_i$ is a standard Borel space for every $i\in I$. Notice that the change of charts $\varphi_{i}\circ\varphi_{j}^{-1}$ has automatically the form
\begin{equation*}
\varphi_{i}\circ\varphi_{j}^{-1}(x,t)=(f_{t}(x),\gamma(t))
\end{equation*}
where $\gamma$ is a partial Borel isomorphism $\mathbf K_j\to \mathbf K_i$ and $f_{t}$ is a local homeomorphism of $\mbb R^{n}$ (resp. $\mathbb H^n$). An atlas is said to be {\em oriented} if the local homeomorphisms $f_t$ preserve the orientation. Notice that the invariant measure $\mu$ induce a measure on each $\mathbf K_i$ preserved by change of verticals $\gamma$. We define the {\em cost} of $\mathfrak A$ as
$$
c(\mathfrak A)=\sum_{i\in I}\mu(\mathbf K_i)
$$

Let $V$ be an open subset of $\mbb R^n$ and $\mathbf X=V\otimes \mathbf K$ a trivial random manifold of base $V$. We shall note $\mathcal A(V\otimes \mathbf K)=C^{\infty}(V,L^{\infty}(\mathbf K))$, the commutative algebra of smooth maps from $V$ to the Banach algebra $L^{\infty}(\mathbf K)$ of essentially bounded functions on $\mathbf K$. Every element $a\in \mathcal A(V\otimes\mathbf K)$ can be viewed as a measurable continuous fonction $a:V\times\mathbf K\to \mbb R$, two such functions representing the same element of $\mathcal A(V\otimes\mathbf K)$ if they coincide on almost every plaque $V\times\{\bullet\}$. For every measurable subset $B\subset V\times \mathbf K$, we denote by $\mathcal A(B)$ the algebra formed by the restrictions to $B$ of the elements of $\mathcal A(V\otimes \mathbf K)$. A map $f:B\to B'$ between Borel subsets of trivial random manifolds is said to be {\em smooth} if for every $a\in \mathcal A(B')$ the fonction $a\circ f$ belongs to $\mathcal A(B)$. In that case the map $a\mapsto \mathcal A(f)(a):=a\circ f$ gives rise to a morphism of algebras $\mathcal A(f):\mathcal A(B')\to \mathcal A(B)$.

Let $\mathbf X$ be a random manifold. An atlas $\{\varphi_i:U_i\to V_i\otimes \mathbf K_i\}_{i\in I}$ of $\mathbf X$ is said to be \textit{smooth} if for every $i,j\in I$ the corresponding change of charts $\varphi_i\circ\varphi_j^{-1}:\varphi_j(U_i\cap U_j)\to \varphi_i(U_i\cap U_j)$ is smooth. Two smooth atlas are called compatible if their union is a smooth atlas. The equivalence class of compatible bounded atlas is called a \textit{smooth  structure} on $\mathbf X$. A random manifold together with a smooth structure is called a \textit{smooth random manifold}. A smooth atlas $\{\varphi_i:U_i\to V_i\times \mathbf K_i\}_{i\in I}$ is said to be {\em compact} if for every $i\in I$ there exists a compact subset $F_i\subset V_i$ such that the sets $\varphi_i^{-1}(F_i\times \mathbf K_i)$ cover $\mathbf X$.

\begin{definition}
	A random manifold will be called \textit{compact} if it has a compact smooth atlas of finite cost.
\end{definition}

\subsection{The tangent space}
Let $\mathbf X$ be a smooth random manifold. We shall note
$$
\mathcal A(\mathbf X)=\{a:\mathbf X\to \mbb R|\forall i\in I,\; a\circ \varphi_i^{-1}\in \mathcal A(V_i\times \mathbf K_i)\}
$$

A tangent vector field of $\mathbf X$ is by definition a derivation of the algebra $\mathcal A(\mathbf X)$. We shall note $\tau(\mathbf X)=\operatorname{Der}\mathcal A(\mathbf X)$ the vector space of tangent vector fields and we will call it the \textit{tangent space} of $\mathbf X$. Since $\mathcal A(\mathbf X)$ is a commutative algebra, the tangent space $\tau(\mathbf X)$ has a natural structure of $\mathcal A(\mathbf X)$-module.

\begin{lemma}\label{lem:trivial-free}
	If $\mathbf X=V\otimes\mathbf K$ is a trivial random $n$-manifold, then $\tau(\mathbf X)\simeq\mathcal A(\mathbf X)^n$.
\end{lemma}
\begin{proof}
	Let $\mathbf X=V\otimes \mathbf K$. Recall that every function $a\in \mathcal A(\mathbf X)$ is  a map $a:V\to L^{\infty}(\mathbf K)$ with continuous partial derivatives of every order. Hence the $n$ partial derivatives of order $1$ are derivations of $\mathcal A(\mathbf X)$. We shall prove that every derivation $\xi$ of $\mathcal A(\mathbf X)$ can be written in an unique way as $\xi=\sum_i a_i\partial_i$ with $a_i\in \mathcal A(\mathbf X)$.  It follows from the Taylor's theorem with integral remainder (which is valid for smooth functions of $n$ variables taking values in any Banach space) that for every $y\in V$ there exists a star convex open subset $V_y\subset V$ centered in $y$ such that for $x\in V_y$ we have
		$$
		a(x)=a(y)+\sum_i (x_i-y_i)\partial_i a(y)+\sum_{i,j}(x_i-y_i)(x_j-y_j)B^{ij}_{y}(x)
		$$
		where
	where $x_i$ and $y_i$ denote the coordinates of $x$ and $y$ and $B^{ij}_{y}$ is a function in $\mathcal A(\mathbf X)$ defined by the formula
	$$
	B^{ij}_{y}(x)=\frac 12\int_0^1(1-s)^2\partial_{ij}a(y+s(x-y))ds
	$$

	 Fixing $y$, applying $\xi$ to the identity above and then evaluating in $y$ we get identity
	$$
	\xi(a)(y)=\sum_i \xi(x_i)(y)\partial_ia(y)
	$$
	valid for every $a\in \mathcal A(\mathbf X)$ and every $y\in V$. Hence $\xi=\sum_i \xi(x_i)\partial_i$, as wanted.
\end{proof}

\begin{lemma}
	Let $\mathbf X$ be a random manifold. Every compact atlas of $\mathbf X$
	$$
	\{\varphi_i:U_i\to V_i\otimes \mathbf K_i\}_{i\in I}
	$$
	 admits a subordinated partition of unity, i.e. for every $i\in I$ there exists a positive element $a_i\in \mathcal A(\mathbf X)$ with support in $U_i$ such that $\sum_{i\in I}a_i=1$.
\end{lemma}
\begin{proof}
	Let us consider a family of compact sets $F_i\subset V_i$ verifying the definition of compact atlas, and let $g_i$ a bump function on $V_i$ such that $F_i$ is contained in the interior of its support. If we define $h_i:U_i\to \mbb R$ as the pull-back of $g_i$ by the projections $U_i\to V_i$, we obtain the wanted partition of unity by setting $a_i=h_i/\sum_i h_i$.
\end{proof}

\begin{lemma}\label{lem:riemann-metric}
	If $\mathbf X$ is a compact random manifold, it has a riemannian metric, i.e. a morphism of modules
	$$
	g:\tau(\mathbf X)\otimes\tau(\mathbf X)\to \mathcal A(\mathbf X)
	$$
	such that for every $\xi,\nu\in \tau(\mathbf X)$:
	\begin{enumerate}
  \item $g(\xi,\nu)=g(\nu,\xi)$;
  \item $g(\xi,\xi)\geq 0$;
  \item $g(\xi,\xi)= 0$ iff $\xi=0$.
\end{enumerate}

\end{lemma}
\begin{proof}
	The lemma obviously holds for trivial random manifolds, since their tangent space is a free module. Let us consider a compact atlas $\mathfrak A=\{\varphi_{i}:U_{i}\to V_{i}\otimes \mathbf K_i\}_{i\in I}$ of $\mathbf X$ together with a subordinated partition of unity $\sum_{i\in I}a_i=1$. For every $\xi\in \tau(\mathbf X)$ one can see $a_i^{1/2}\xi$ as a derivation on $\tau(U_i)$. If we fix a riemannian metric on e $g_i$ on each $U_i$, then we can write
	$$
	g(\xi,\nu)=\sum_i g_i\left(a_i^{1/2}\xi,a_i^{1/2}\nu\right)
	$$
	A straigthforward computation shows that $g$ is a riemannian metric on $\tau(\mathbf X)$.
\end{proof}

\begin{remark}\label{rem:restriction}
It follows from the lemmas above that every tangent vector $\xi\in \tau(\mathbf X)$ can be viewed as a map $\xi:x\in \mathbf X\mapsto \xi_x\in T_xL_x$ where $L_x$ is the leaf of $\mathbf X$ which contains $x$ and $T_xL_x$ the usual tangent space of $L_x$ at $x$. Hence for every measurable open subset $U\subset \mathbf X$, we have an homomorphisms of $\mathcal A(\mathbf X)$-modules $r_U:\tau(\mathbf X)\to \tau(U)$ given be the restriction $\xi\mapsto \xi_{|U}$. In the other hand, a riemannian metric $g$ can be viewed as a field of inner products
$$
g:x\in \mathbf X\mapsto (g_x:T_xL_x\otimes T_xL_x\to \mbb R)
$$
\end{remark}

\begin{theorem}
For every compact random manifold $\mathbf X$, $\tau(\mathbf X)$ is a finitely generated projectif module over $\mathcal A(\mathbf X)$.
\end{theorem}
\begin{proof}
Let $\dim X=n$ and let $\mathfrak A=\{\varphi_i:U_i\to V_i\otimes \mathbf K_i\}_{i=1}^k$ be a compact smooth atlas of $\mathbf X$ together with a subordinated partition of unity $\sum_ia_i=1$. Since
	$$
	a_i^{1/2} \tau(\mathbf X)=a_i^{1/2} \tau(U_i)\simeq a_i^{1/2}\mathcal A(U_i)^n=a_i^{1/2}\mathcal A(\mathbf X)^n\subset \mathcal A(\mathbf X)^n
	$$
	the map
	$$
	\xi\in \tau(\mathbf X)\mapsto (a_1^{1/2}\xi,\cdots,a_k^{1/2}\xi)\in \mathcal A(\mathbf X)^{kn}
	$$
	is injective and puts $\tau(\mathbf X)$ as a sub-module of $\mathcal A(\mathbf X)^{kn}$.  A straightforward computation shows that the orthogonal complement of $\tau(\mathbf X)$ in $\mathcal A(\mathbf X)^{kn}$ with respect to the usual $\mathcal A(\mathbf X)$-inner product $\braket{a|b}=\sum a_ib_i$ is a $\mathcal A(\mathbf X)$-module $\tau(\mathbf X)^{\perp}$ verifying $\tau(\mathbf X)\oplus \tau(\mathbf X)^{\perp}=\mathcal A(\mathbf X)^{kn}$.
	\end{proof}

\section{Low dimesional cobordism groups}
In the sequel, every random manifold shall be supposed to be compact. The boundary of a random manifold with boundary $\mathbf X$ is a random manifold without boundary that shall be noted $\boldsymbol \partial \mathbf X$. If $\mathbf X$ is oriented, its boundary inherits a natural orientation. Given an oriented random manifold $\mathbf X$, we shall note $-\mathbf X$ the same random manifold endowed with the opposite orientation. Given two random manifolds without boundary $\mathbf X$ and $\mathbf Y$, we shall note $\mathbf X+\mathbf Y$ their disjoint union and $\mathbf X-\mathbf Y=\mathbf X+(-\mathbf Y)$. We say that $\mathbf X$ and $\mathbf Y$ are cobordant if there exists an oriented random manifold with boundary $\mathbf M$ such that $\boldsymbol \partial\mathbf M=\mathbf X-\mathbf Y$. Cobordism defines an equivalence relation on the set of oriented random manifolds and we note $[\mathbf X]$ the cobordism class of $\mathbf X$. The set of cobordism classes of oriented random manifolds, together with the operation $[\mathbf X]+[\mathbf Y]=[\mathbf X+\mathbf Y]$ is an abelian group that we shall note $\mathbf R\boldsymbol \Omega^{SO}$. It has a natural grading induced by the dimension, i.e.
$$
\mathbf R\boldsymbol \Omega^{SO}=\bigoplus_{n=0}^{\infty} \mathbf R\boldsymbol \Omega^{SO}_n
$$
where $\mathbf R\boldsymbol \Omega^{SO}_n$ is the group of cobordism classes of oriented random $n$-manifolds. The neutral element of $\mathbf R\boldsymbol \Omega^{SO}$ is the cobordism class of the empty random manifold and will be noted $[\boldsymbol 0]$.

On can easily see that $\mathbf R\boldsymbol \Omega_0\simeq \mbb R$. Indeed, since an oriented connected $0$-manifold is signed point $x=\pm 1$, a random $0$-manifold is given by a finite standard space $\mathbf K$ together with a measurable partition $\mathbf K=\mathbf K^++ \mathbf K^-$. Moreover $[\mathbf K]=[\boldsymbol 0]$ if and only if $\mu(\mathbf K^+)=\mu(\mathbf K^-)$. Hence the map $\mathbf K\mapsto \mu(\mathbf K^+)-\mu(\mathbf K^-)$ induces an isomorphism of abelian groups $\Phi_0:\mathbf R\boldsymbol\Omega^{SO}_0\to \mbb R$.

\medskip
We devote the rest of this section to prove the following:

\begin{theorem}
	Every oriented random $1$-manifold is the boundary of an oriented random $2$-manifold or, in other words, that
	$$
	\mathbf R\boldsymbol\Omega^{SO}_1=0.
	$$
\end{theorem}

 We shall split the proof in several lemmas. Let us start by recalling some basic constructions and facts. Let $\mathbf K$ a finite standard measure space. The {\em suspension} of a measure preserving automorphism $\gamma\in \operatorname{Aut}(\mathbf K)$ is the oriented random $1$-manifold $\mathbf \Sigma_\gamma$ obtained from the trivial random $1$-manifold with boundary $[0,1]\times \mathbf K$ by the following identifications
$$
(0,t)\sim (1,\gamma(t))\quad,\quad t\in \mathbf K
$$

\begin{lemma}\label{compact2zero}
Let $\mathbf X$ an oriented random $1$-manifold such that almost every leave of $\mathbf X$ is compact. Then $[\mathbf X]=[\boldsymbol 0]$.
\end{lemma}
\begin{proof}
Let $T$ be a transversal of $\mathbf X$ and let us consider the standard equivalence relation $\mathcal R_T\subset T\times T$ whose equivalence classes are given by the intersection of $T$ with the leaves of $\mathbf X$. Since the leaves are almost all compact, the equivalence classes of $\mathcal R_T$ are almost all finite. It is well known that such an equivalence relation has a fundamental domain, i.e. a Borel subset $\mathbf K\subset T$ which intersects almost every leave in exactly one point, and one has an obvious isomorphism of random manifolds between $\mathbf X$ and a random circle $\mathbf X\simeq \mathbb S\times \mathbf K$ which is the boundary of $\mathbb D\times \mathbf K$.
\end{proof}

\begin{lemma}\label{every2noncompact}
 For every oriented random $1$-manifold $\mathbf X$ there exists an oriented random $1$-manifold $\mathbf X'$ without compact leaves such that $[\mathbf X]=[\mathbf X']$.
\end{lemma}
\begin{proof}
Let $\mathbf X$ be a $1$-lamination and let $T$ be a transversal which intersects every leaf of $\mathbf X$. Since $\mathbf X$ is compact as random manifold, a leaf $L$ of $\mathbf X$ is compact if and only if $T\cap L$ is finite. The finite classes of $\mathcal R_T$ form a Borel subset $\mathbf K\subset T$, which implies that the compact leaves of $\mathbf X$ form a random $1$-manifold $\mathbf F\subset \mathbf X$. Therefore we can write $\mathbf X= \mathbf F+\mathbf X'$ where $\mathbf X'=\mathbf X\backslash\mathbf F$ has no compact leaves by construction. By the lemma \ref{compact2zero} above we have
$$
[\mathbf X]=[\mathbf F]+ [\mathbf X']= [\boldsymbol 0]+[\mathbf X']=[\mathbf X']
$$
\end{proof}

\begin{lemma}\label{every=suspension}
 For every oriented random $1$-manifold $\mathbf X$ there exists a finite standard measured space  $\mathbf K$ and a measure preserving automorphism $\gamma\in \operatorname{Aut}(\mathbf K)$ such that $\mathbf X\simeq \mathbf \Sigma_\gamma$.
 \end{lemma}
 \begin{proof}
 Let $\mathbf K$ be a finite complete transversal of $\mathbf X$. If we cut $\mathbf X$ along $\mathbf K$, we obtain a random $1$-manifold with boundary $\hat{\mathbf X}$ whose almost every leaf is diffeomorphic to the interval $[0,1]$. As in the proof of lemma \ref{compact2zero}, $\hat{\mathbf X}$ is diffeomorphic to the trivial $1$-manifold $[0,1]\times \mathbf K$ in such a way that the quotient map $\hat{\mathbf X}\to \mathbf X$ preserves the orientation. Hence every point $(0,t)\in \hat{\mathbf X}$ is identified to exactly one point of the form $(1,s)$, and the map $\gamma:t\in \mathbf K\to s\in \mathbf K$ is a measure preserving automorphism of $\mathbf K$ such that $\mathbf X\simeq \mathbf \Sigma_\gamma$.
 \end{proof}

\begin{lemma}\label{homo}
For every finite standard measure space $\mathbf K$, the map
\begin{equation*}
\gamma\in \operatorname{Aut}(\mathbf K)\mapsto \left[\mathbf \Sigma_{\gamma}\right]\in \mathbf R\boldsymbol\Omega^{SO}_{1}
\end{equation*}
is a homomorphism of groups.
\end{lemma}
\begin{proof}
Let $\phi$ and $\psi$ two measure preserving automorphisms of $\mathbf K$. It suffices to prove that there exists a cobordism between the disjoint union $\mathbf \Sigma_\phi+ \mathbf \Sigma_\psi$ and $\mathbf \Sigma_{\phi\circ\psi}$, i.e. an oriented random $2$-manifold with boundary $\mathbf M$ such that
$$
\boldsymbol\partial\mathbf M=\mathbf \Sigma_\phi+ \mathbf \Sigma_\psi-\mathbf \Sigma_{\phi\circ\psi}
$$
. For the sake of clarity, we will explain graphically how it works. The random $2$-manifold $\mathbf M$ is given as a fibration over the singular surface with boundary on the left of figure \ref{fig:r}, where the fibers are the random singular $1$-manifolds on the right. It is obvious from the picture that the singularities on the leaves of $\mathbf M$ are of Morse type, which implies that the leaves of $\mathbf M$ are smooth manifolds. The local triviality and the compacity of $\mathbf M$ follows from the local triviality of the fibration.
\end{proof}

\begin{figure}\label{fig:r}
\begin{center}
\includegraphics[width=10cm,height=8cm]{Calque1.png}
\caption{Cobordism between $\boldsymbol \Sigma_\phi+\boldsymbol \Sigma_\psi$ and $\boldsymbol \Sigma_{\phi\circ\psi}$}
\end{center}
\bigskip
\end{figure}

The combination of the above lemmas allows us to conclude. Indeed, since $\operatorname{Aut}(\mathbf K)$ is a simple group (see Fathi \cite{fathi78automorphimssimple}) and $\mathbf R\boldsymbol \Omega^{SO}_1$ is abelian, every cobordism class on the image of the homomorphism of lemma \ref{homo} vanishes. Hence, by lemma \ref{every=suspension}, this will be the case for every cobordism class in $\mathbf R\boldsymbol \Omega^{SO}_{1}$, which completes the proof.

\section{Higher dimensional cobordism groups}
We shall prove in this section the following result:

\begin{theorem}\label{thm:pontryagin}
	For every $n\geq 0$ there exists a surjectif homomorphism of abelian groups
	$$
	\mathbf R\boldsymbol\Omega_{4n}^{SO}\to \mbb R^{p(n)}
	$$
	where $p(n)$ is the number of partitions of $n$, i.e. the number of different ways to write $n$ as a sum of natural numbers.
\end{theorem}

The proof follows the scheme of the classical theorem of Thom-Pontryagin \cite{thom1954cobordism,milnorstasheff1974book} which states that a compact oriented smooth $4n$-manifold is the boundary of an compact oriented smooth $(4n+1)$-manifold if and only if its Pontryagin numbers vanish.

\subsection{Cohomology} Let $\mathbf X$ be a random manifold with boundary. We define the \textit{cotangent space} of $\mathbf X$ as the dual module
$$
\Omega^{1}(\mathbf X)=\operatorname{Hom}_{\mathcal A(\mathbf X)}(\tau(\mathbf X),\mathcal A(\mathbf X))
$$
and we define the exterior differential $d:\mathcal A(\mathbf X)\to \Omega^1(\mathbf X)$ by setting $da(\xi)=\xi(a)$ for every funtion $a\in \mathcal A(\mathbf X)$ and tangent vector field $\xi\in \tau(\mathbf X)$. The exterior derivative can be extended in the usual way to the whole exterior algebra
$$
\Omega^{\bullet}(\mathbf X)=\bigoplus_{i=0}^{\infty} \Omega^{i}(\mathbf X)
$$
where $\Omega^i(\mathbf X)=\bigwedge^i_{\mathcal A(\mathbf X)}\Omega^1(\mathbf X)$ is the $i$-th exterior power of $\Omega^1(\mathbf X)$ over $\mathcal A(\mathbf X)$, such that we have the usual identity $d^2=0$. The homology of the corresponding cochain complex is called the {\it de Rham cohomology} of $\mathbf X$ and denoted by $H^{\bullet}(\mathbf X)$. It has a natural real algebra structure inherited from the one in $\Omega^\bullet(\mathbf X)$. In the other hand, we can define the $K$-theory of $\mathbf X$, denoted by $K(\mathbf X)$, as the algebraic $K$-theory of the algebra $\mathcal A(\mathbf X)$. Recall that $K(\mathbf X)$ is a ring structure induced by directed sums and tensor product of projectif modules. We will see that the $K$-theory and de Rham cohomology of bounded laminations can be seen as contravariant functors. For this, let $\mathbf X$ and $\mathbf Y$ be two random manifolds. A continuous mesurable map $f:\mathbf X\to \mathbf Y$ is said to be \textit{smooth} if for every $a\in\mathcal A(\mathbf Y)$ the function $\mathcal A(f)(a)=a\circ f$ belongs to $\mathcal A(\mathbf X)$. Indeed, Observe that the corresponding morphism of algebras $\mathcal A(f):\mathcal A(\mathbf Y)\to \mathcal A(\mathbf X)$, such that we can see $\mathcal A(\mathbf X)$ as a $\mathcal A(\mathbf Y)$-module. Observe that the map $E\mapsto E\otimes_{\mathcal A(\mathbf Y)} \mathcal A(\mathbf X)$ transforms finitely generated projectif $\mathcal A(\mathbf Y)$-modules into finitely generated projectif $\mathcal A(\mathbf X)$-modules and hence it induces a natural morphism of rings $K(f):K(\mathbf Y)\to K(\mathbf X)$. It remains the construction of a functorial homomorphism
$$
\Omega^{\bullet}(f):\Omega^{\bullet}(\mathbf Y)\to \Omega^{\bullet}(\mathbf X)
$$
For this we need to prove the following result:
\begin{lemma}
	If $\mathbf X$ is a bounded lamination, the $\mathcal A(\mathbf X)$-module $\Omega^1(\mathbf X)$ generated by the image of $d$.
\end{lemma}
\begin{proof}
	Since $\Omega^1(\mathbf X)=\sum_i a_i\Omega^1(U_i)$ for any trivialisation $\{U_i\}$ and any subordinated partition of unity $\{a_i\}$, one can assume that  $\mathbf X$ is a trivial random manifold. In that case, it follows from the proof of lemma \ref{lem:trivial-free} that $\Omega^1(\mathbf X)$ is generated (as $\mathcal A(\mathbf X)$-module) by the differentials $dx_i$ where $x_i:\mathbf X\to \mbb R$ are the coordinate fonctions.
\end{proof}
In particular the vector space $\Omega^n(\mathbf X)$ is generated by the elements of the form $a_0\,da_1,\dots da_n$ and the exterior differential is defined recursively by the formula
$$
d(a_0\,da_1\cdots da_n)=da_0\,da_1\cdots da_n
$$
Hence, $\Omega^\bullet(f)$ can be defined as the unique morphism of differential algebras verifying $\Omega^0(f)=\mathcal A(f)$. Such a construction is clearly functorial.

\subsection{Chern-Weil theory}
We sketch here the Chern-Weil theory in the context of random manifolds. This construction was developed in the late 1940s by Shiing-Shen Chern and Andr\'e Weil in the wake of proofs of the generalized Gauss-Bonnet theorem \cite{chern44gaussbonnet} . This theory was an important step in the theory of characteristic classes and it allowed to clarify the relationship between cobordism, K-theory and usual cohomology. We shall follow the general construction given in the context of differential graded algebras over general rings (see for example Karoubi \cite{karoubi87cyclic} or Loday \cite{loday98bookcyclichomology}).

Let $(\mathcal A^\bullet,d)$ be any commutative graded differential real algebra and let $E$ be a finitely generated projectif $\mathcal A^0$-module. A connection in $E$ is a degree $1$ linear endomorphism $\nabla$ of $\mathcal A^\bullet\otimes E$ such that $\SC{\nabla,a}=da$ for every $x\in \mathcal A^\bullet$, where $\SC{\cdot,\cdot}$ denotes the graded commutator. A straightforward computation shows that the degree $2$ operator $\nabla^2$ verifies $[\nabla^2,a]=0$ for every $a\in \mathcal A^2$, and it thus defines an element $R_\nabla\in \mathcal A^2\otimes\operatorname{End}(E)$ called the curvature of $\nabla$. For every invariant polynomial $p:M_\infty(\mbb R)\to \mbb R$, i.e. any polynomial function verifying $p(AB)=p(BA)$ for every pair of real matrices $A$ and $B$, and for every finitely generated module $E$ over $\mathcal A^0$, we have a natural map $p:\mathcal A^\bullet\otimes \operatorname{End}(E)\to \mathcal A^\bullet$. We can define
$$
p(E,\nabla)=p(R_\nabla)\in \mathcal A^{2\bullet}
$$
A standard homotopy argument shows that $p(E,\nabla)$ is closed and its cohomology class does not depend on $\nabla$. Hence it defines an element $p(E)\in H^{2\bullet}(A^{\bullet},d)$.

\begin{example}[The Chern character]
	Let $n\in \mathbb N$. The \textit{Chern $n$-character} of $E$ with respect to $(\mathcal A^\bullet,d)$ is defined as $ch_n(E)$ where
	$$
	ch_n(A)=\frac 1{n!}\operatorname{tr}(A^n)
	$$
	It verifies the following properties:
\begin{enumerate}
  \item $ch_n(E\oplus E')=ch_n(E)+ch_n(E')$.
  \item $ch_n(E\otimes E')=ch_n(E)ch_n(E')$.
\end{enumerate}
and hence induces a homomorphism of rings
$$
ch_n:K(\mathcal A^0)\to H^{2\bullet}(\mathcal A^\bullet,d)
$$
\end{example}

\begin{example}[The total Pontryagin class]
		The total Pontryagin class of $E$ with respect to $(\mathcal A^\bullet,d)$ is defined as $\boldsymbol p(E)$ where
	$$
	\boldsymbol p(A)=\det(1+A)
	$$
	It verifies the following properties
\begin{enumerate}
  \item $\boldsymbol p(\mathcal A^0)=1$.
  \item $\boldsymbol p(E\oplus E')=\mathbf p(E)\mathbf p(E')$.
\end{enumerate}

\end{example}

\subsection{The fundamental class of a random manifold} Let $V\subset \mbb R^n$ be an open set and $\mathbf K$ a standard probability space. Since the tangent space $\tau(V\otimes \mathbf K)$ is a rank $n$ free $\mathcal A(V\otimes \mathbf K)$-module, so is $\Omega^1(V\otimes \mathbf K)$, and we thus have a natural identification of graded algebras
$$
\Omega^\bullet(V\otimes\mathbf K)\simeq C^{\infty}(V,L^{\infty}(\mathbf K))\otimes \wedge^\bullet \mbb R^n
$$
By the dominated convergence theorem, integration through $\mathbf K$ gives rise to a linear map $\int_{\mathbf K}:C^{\infty}(V,L^{\infty}(\mathbf K))\to C^{\infty}(V)$ which extends to a linear morphism of graded differential algebras $\int_{\mathbf K}:\Omega^\bullet(V\otimes \mathbf K)\to \Omega^\bullet(V)$
where $\Omega^\bullet(V)$ is the algebra of usual differential forms on $V$. Let $\mathbf X$ be an oriented random $n$-manifold together with an oriented compact atlas $\{\varphi_i:U_i\to V_i\times \mathbf K_i\}$ and a subordinated partition of unity $a_i\in \mathcal A(U_i)$. Fot every $\omega\in \Omega^{n}(\mathbf X)$ and every $a_i$ we can see $a_i\omega$ as an element of $\Omega^{n}(V_i\otimes \mathbf K_i)$ and $\int_{\mathbf K_i}a_i\omega$ as a smooth $n$-form with compact support on $V_i$. We shall define
$$
\int_{\mathbf X} \omega =\sum_i\int_{V_i}\int_{\mathbf K_i} a_i\omega
$$
A straightforward computation shows that this definition does not depend on the partition of unity nor the atlas. For the sake of simplicity, for every $\omega\in \Omega^{\bullet}(\mathbf X)$ we shall set $\int_{\mathbf X}\omega=\int_{\mathbf X}\omega_n$, where $\omega_n$ denotes the component of $\omega$ in $\Omega^{n}(\mathbf X)$. This defines a linear map
$$
\int_{\mathbf X}:\Omega^{\bullet}(\mathbf X)\to \mbb R
$$
called the \textit{fundamental cycle} of $\mathbf X$. The following result is a direct consequence of the definition of the above integral and the well known Stokes theorem on $\mbb R^n$:

\begin{proposition}\label{prop:stokes}
	Let $\mathbf X$ an oriented random $n$-manifold with boundary and let $\partial\mathbf X$ its boundary. Then for every $\omega\in \Omega^{\bullet}(\mathbf X)$ we have
	$$
	\int_{\mathbf X}d\omega=\int_{\partial\mathbf X}\Omega^{\bullet}(i)(\omega)
	$$
	where $i:\partial\mathbf X\to\mathbf X$ is the inclusion map.
In particular, since $\partial\mathbf X$ has no boundary, the fundamental cycle of $\partial\mathbf X$ induces a linear map $\int_{\partial\mathbf X}:H^{\bullet}(\partial\mathbf X)\to \mbb R$ called the \emph{fundamental class} of $\partial\mathbf X$. For every $\alpha\in H^\bullet(\mathbf X)$ one has
	$$
	\int_{\partial\mathbf X}H^\bullet(i)(\alpha)=0
	$$
\end{proposition}

\subsection{Pontryagin numbers}
The aim of this paragraph is to give a proof of theorem \ref{thm:pontryagin}. As in the classical case of compact manifolds, the proof relies in the cobordism invariance of characteristic numbers, which are obtained as a pairing between the fundamental class of the lamination and some suitable cohomology classes. Since the tangent space of a random manifold $\mathbf X$ is a projectif $\mathcal A(\mathbf X)$-module, we can define the \textit{total Pontryagin class} of $\mathbf X$, noted $\boldsymbol p(\mathbf X)$, as the total Pontryagin class of $\tau(\mathbf X)$ with respect to the differential algebra $(\Omega^{\bullet}(\mathbf X),d)$. The following lemma shows the naturality of the total Pontryagin class with respect to bordisms:

\begin{lemma}\label{lem:pontryagin-natural}
	If $\mathbf X$ is a random $n$-manifold with boundary and $i:\partial\mathbf X\to\mathbf X$ is the inclusion of its boundary, then
	$$
	\boldsymbol p(\partial\mathbf X) = H^\bullet(i)(\boldsymbol p(\mathbf X))
	$$
\end{lemma}
\begin{proof}
	Since the total Pontryagin class is additive with respect to direct sums and $\boldsymbol p(\mathcal A(\mathbf X))=1$, it suffices to prove that there exists an isomorphism of $\mathcal A(\mathbf X)$-modules
	$$
	\tau(\mathbf X)\otimes \mathcal A(\partial\mathbf X)\simeq \tau(\partial\mathbf X)\oplus \mathcal A(\partial\mathbf X)
	$$
Let $g$ be a Riemannian metric on $\mathbf X$ and let $\{\varphi:U_i\to V_i\otimes \mathbf K_i\}_{i\in I}$ an oriented atlas together with a subordinated partition of unity $\alpha_i$. For every $i\in I$ we shall note $g_i$ the induced riemannian metric on $U_i$. 	By the lemma \ref{lem:trivial-free} we have identifications $\tau(U_i)=\mathcal A(U_i)^n$ and $\tau(\partial U_i)=\mathcal A(\partial U_i)^{n-1}$, which give rise to natural isomorphisms of $\mathcal A(\mathbf X)$-modules
	$$
	\tau(U_i)\otimes \mathcal A(\partial U_i)\simeq \tau(\partial U_i)\oplus \mathcal A(\partial U_i)
	$$
	The induced inclusions $\tau(\partial u_i)\subset \tau(U_i)\otimes \mathcal A(\partial U_i)$ add up into an inclusion
	$$
	\tau(\partial \mathbf X)\subset \tau(\mathbf X)\otimes \mathcal A(\partial \mathbf X)
	$$
Let $\hat g_i$ be the inner product on $\tau(U_i)\otimes \mathcal A(\partial U_i)$ given by
	$$
	\hat g_i(\xi\otimes a\otimes \nu\otimes b)=g_i(\xi a\otimes \nu b).
	$$
 and let $\tau(\partial U_i)^\perp$ be the orthogonal complement of $\tau(\partial U_i)$ with respect to $\hat g_i$, which is obviously isomorphic to $\mathcal A(\partial U_i)$. Let $\xi_i$ be a $\hat g$-normal generator of $\tau(\partial U_i)^\perp$ as $\mathcal A(\partial\mathbf X)$-module. Since the atlas is oriented, one can construct mesurable maps $\epsilon:\mathbf K_i\to \{\pm 1\}$ such that the vector $\xi=\sum_i\alpha_i \epsilon_i\xi_i\in \tau(\mathbf X)\otimes \mathcal A(\partial\mathbf X)$ does not vanish. A direct computation shows that
	$$
	\mathcal A(\partial \mathbf X)\xi = \tau(\partial \mathbf X)^{\perp}
	$$
	which finishes the proof.
\end{proof}

Let $\mathbf Y$ be an oriented random $4n$-manifold without boundary. Let us note $\boldsymbol p_{k}(\mathbf Y)\in H^{4k}(\mathbf Y)$ the component of degree $4k$ of the total Pontryagin class of $\mathbf Y$. It follows from the lemma above that $\boldsymbol p_{k}(\mathbf Y)=H^\bullet(i)(\boldsymbol p_{k}(\mathbf Y))$ for every $k$. A partition of $n$ is a set $\alpha\subset \mathbb N^{*}$ of strictly positive integers such that $\sum_{k\in \alpha} k=n$. We shall note $\mathcal P(n)$ the set of partitions of $n$. For every $\alpha\in \mathcal P(n)$ we shall note
$$
\boldsymbol p_\alpha(\mathbf Y)=\prod_{k\in\alpha}\boldsymbol p_{k}(\mathbf Y)\in H^{2n}(\mathbf Y)
$$
The real numbers $p_\alpha(Y):=\int_{\mathbf Y}\boldsymbol p_\alpha(\mathbf Y)$ are called the {\em Pontryagin numbers} of $\mathbf Y$. Let us assume that $\mathbf Y$ is the boundary of an oriented random manifold $\mathbf X$ and let $i:\mathbf Y\to \mathbf X$ be the inclusion. Since $H^\bullet(i)$ is a morphism of graded rings, it follows from lemma \ref{lem:pontryagin-natural} that $\boldsymbol p_{\alpha}(\mathbf Y)=H^{\bullet}(i)(\boldsymbol p_\alpha(\mathbf X))$,   The proposition \ref{prop:stokes} implies that all the Pontryagin numbers of $\mathbf Y$ vanish. Moreover, one easily sees that $p_{\alpha}(-\mathbf Y)=-p_\alpha(\mathbf Y)$ and $p_{\alpha}(\mathbf Y+\mathbf Y')=p_\alpha(\mathbf Y)+p_\alpha(\mathbf Y')$. Hence, it follows that the map
$\mathbf Y\mapsto p_{\bullet}(\mathbf Y)$ induces a morphism of abelian groups
$$
\Phi:\mathbf R\boldsymbol \Omega^{SO}_{4n}\to \mbb R^{\mathcal P(n)}
$$
To complete the proof of the theorem \ref{thm:pontryagin}, it remains to show that $\Phi$ is surjective. Let us note $M_\alpha=\prod_{k\in \alpha}\mathbf P^{2k}(\mathbb C)$ for $\alpha\in \mathcal P(n)$. It is a well known fact (see for instance \cite{milnorstasheff1974book}) that for every $\alpha$ the matrix $A=(p_{\beta}(M_\alpha))_{\alpha,\beta}$ is non singular. In particular, for every $v\in \mbb R^{\mathcal P(n)}$ there exists $\lambda\in \mbb R^{\mathcal P(n)}$ such that
$$
v=\sum_\alpha p_{\bullet}(M_\alpha\otimes \mathbf K_{\lambda_{\alpha}})=\Phi\left[\sum_\alpha M_\alpha\otimes \mathbf K_{\lambda_{\alpha}}\right]
$$
where $\mathbf K_\lambda$ is the standard diffuse measure space of mass $\lambda$. This completes the proof.

redpajama_set_name: RedPajamaArXiv


text:
Paco Ramírez, właśc. Francisco Javier Ramírez Gámez (ur. 28 listopada 1965 w Bacerac) – meksykański piłkarz, występujący na pozycji obrońcy, oraz trener.

Kariera klubowa
Pochodzi z miasta Bacerac, jednak jako sześciolatek przeprowadził się z rodziną do Hermosillo. Początkowo grał w baseball, a później rozpoczął treningi piłki nożnej i był powoływany do lokalnych reprezentacji. W jednym z juniorskich turniejów, w którym występował jako zawodnik zespołu stanu Sonora, jego talent odkrył polski trener Waldemar Wasilewski. Profesjonalną karierę piłkarską rozpoczął jako dziewiętnastolatek w drużynie Deportivo Neza z miasta Nezahualcóyotl, lecz zadebiutował w niej dopiero dwa lata później, za kadencji argentyńskiego szkoleniowca Miguela Marína, 9 sierpnia 1986 w zremisowanym 0:0 spotkaniu z Tolucą. Szybko został podstawowym piłkarzem drużyny i premierowe bramki w najwyższej klasie rozgrywkowej strzelił 13 marca 1988 w przegranej 2:4 konfrontacji z Tigres UANL, kiedy to dwukrotnie wpisywał się na listę strzelców. Ogółem barwy Toros reprezentował przez cztery lata, nie odnosząc z nim większych sukcesów.

Latem 1988 Ramírez przeszedł do drużyny Atlante FC z siedzibą w stołecznym mieście Meksyk, gdzie spędził rok, po czym został zawodnikiem Puebla FC. Tam również został podstawowym piłkarzem klubu; w 1991 roku triumfował z nim w najbardziej prestiżowych rozgrywkach kontynentu, Lidze Mistrzów CONCACAF, za to w sezonie 1991/1992 wywalczył tytuł wicemistrza Meksyku. Po dwóch latach spędzonych w Puebli odszedł do stołecznego Club Necaxa, w którym również spędził dwa sezony, jednak bez zdobycia żadnego trofeum. Później przez rok występował w Tiburones Rojos de Veracruz, po czym podpisał umowę ze stołecznym Cruz Azul. Tam odniósł największe sukcesy w karierze, jednak przeważnie w roli rezerwowego – w sezonie Invierno 1997 zdobył tytuł mistrza Meksyku, podczas rozgrywek 1996/1997 triumfował w krajowym pucharze – Copa México, dwukrotnie wygrywał także Puchar Mistrzów CONCACAF; w roku 1996 i 1997. Karierę zakończył w wieku 32 lat z powodu przedłużających się kontuzji.

Kariera trenerska
Po zakończeniu kariery piłkarskiej Ramírez został trenerem, początkowo prowadząc przez krótki czas drugoligową drużynę Correcaminos UAT z miasta Ciudad Victoria. W późniejszym czasie był asystentem Aníbala Ruiza, a potem Enrique Mezy w Club Atlas, spędził także kilka lat w sztabie szkoleniowym reprezentacji Meksyku, najpierw współpracując z trenerem Ricardo La Volpe, z którym wziął udział w Mistrzostwach Świata w Niemczech, a później ze Svenem-Göranem Erikssonem. W przerwie pomiędzy kadencjami obydwóch selekcjonerów prowadził trzecioligową ekipę Vaqueros de Ixtlán. Pracował również jako asystent Hansa Westerhofa i Salvadora Reyesa w Club Necaxa z siedzibą w mieście Aguascalientes. W kwietniu 2009 został trenerem jednej z najbardziej utytułowanych drużyn w kraju, Chivas de Guadalajara, jednak odnosił z nią słabe wyniki, notując trzy zwycięstwa, cztery remisy i pięć porażek w dwunastu spotkaniach i został zwolniony już po pięciu miesiącach.

W lipcu 2010 Ramírez rozpoczął pracę jako asystent swojego byłego kolegi boiskowego Sergio Bueno, początkowo w drugoligowym Tiburones Rojos de Veracruz, a następnie Club Necaxa. Po spadku klubu do Liga de Ascenso został pierwszym trenerem klubu, prowadząc go przez kilka miesięcy bez większych sukcesów. W maju 2012 został trenerem innego meksykańskiego drugoligowca, Dorados de Sinaloa z Culiacán. W jesiennym sezonie Apertura 2012 dotarł z nią do finału rozgrywek drugoligowych, a także triumfował z nim w rozgrywkach Copa MX, dzięki czemu drużyna Dorados została pierwszą drużyną w historii meksykańskiego futbolu, która zdobyła krajowy puchar podczas gry w drugiej lidze. Ze stanowiska odszedł w styczniu 2014 po serii słabszych wyników w lidze.

Bibliografia

 Statystyki na MedioTiempo

Meksykańscy piłkarze
Meksykańscy trenerzy piłkarscy
Piłkarze Deportivo Neza
Piłkarze Atlante FC
Piłkarze Club Puebla
Piłkarze Club Necaxa
Piłkarze Tiburones Rojos de Veracruz
Piłkarze Cruz Azul
Trenerzy piłkarzy Correcaminos UAT
Trenerzy piłkarzy Chivas Guadalajara
Trenerzy piłkarzy Club Necaxa
Trenerzy piłkarzy Dorados de Sinaloa
Trenerzy piłkarzy Atlas FC
Trenerzy piłkarzy Alianza FC (Panama)
Trenerzy piłkarzy Cafetaleros de Chiapas
Ludzie urodzeni w Bacerac
Urodzeni w 1965
Trenerzy piłkarzy Vaqueros de Ixtlán

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text:
3 charged in pair of armed carjackings, including one of pregnant woman, husband in Bucktown
CHICAGO (WLS) -- Three men have been charged in a pair of carjackings in Chicago on Friday, including one of a pregnant woman and her husband.
Police arrested Dwayne Liberty, 21, Tyran Carter, 26, and Terronde Gordon, 25, Friday after they crashed one of the stolen cars and ran away.
The first carjacking occurred in the 5100-block of South Cornell Avenue at about 3:15 a.m. Friday. Police said the three men approached a 27-year-old man and took his vehicle and belongings at gunpoint.
At about 9:30 a.m., police said the three men jumped out of a vehicle in the 1700-block of North Wolcott Avenue and approached a man and his pregnant wife, both 33.
Police said the man was taking out the garbage while his pregnant wife backed the car out of the garage. The offenders made the couple lie on the ground and stole the victim's wallet, purse and phone before making off with their car, police said.
The three men were taken into custody at about 12:46 p.m. in the 400-block of West 97th Street after they were seen fleeing a hit-and-run crash involving one of the stolen vehicles, police said. Officers tracked the suspects using the vehicle's GPS.
Liberty and Gordon have both each been charged with vehicular hijacking with a firearm and two counts of armed robbery with a firearm. Carter has been charged with three counts of vehicular hijacking with a firearm and three counts of armed robbery with a firearm. The men are scheduled to appear in bond court on Sunday.
chicagobucktownhyde parkwashington heightscar theftcarjackingarmed robbery
Pregnant woman, husband carjacked at gunpoint in Bucktown

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text:
Wood Group Mustang secures £30m North Sea contract
Engineering and design contract for new offshore oil and gas production facility in the Ivar Aasen field
Wood Group Mustang has secured a new £30 million engineering and design contract in the Norwegian North Sea.
The contract is for a new offshore oil and gas production facility in the Ivar Aasen field which will be operated by Det norske oljeselskap.
Wood Group Mustang, a subsidiary of Wood Group, secured the contract with SMOE, the Singapore-based engineering, procurement and construction (EPC) contractor for the project.
The topsides facility, which will be anchored in waters more than 100 metres deep, will have an approximate dry weight of 13,700 tonnes, including 70-bed living quarters.
Wood Group Mustang said the contract will require around 200 staff at its peak, and will be led from its London office with support from its office in Kuala Lumpur.
Steve Knowles, president of Wood Group Mustang, said: "Receiving news of this project award was extremely gratifying.
"We are looking forward to starting work on Det norske's first major facility and developing an excellent relationship with them, in addition to enhancing our existing relationship with SMOE."


Avondale Estates is located just east of Downtown Decatur and is historic in its own right. A short drive to Midtown and Downtown Atlanta brings you to the quiet community that still retains a hometown feel.
Listing information last updated on Thursday, April 25, 2019 at 4:22 AM.


0022; entrepreneurial ONLINE PATIENT RELATIONSHIP MANAGEMENT: CRM IN DER LIFE SCIENCES INDUSTRIE among eds. 0022; when making Pages. 0027; ia when undergoing. 0027; Lock Aboriginal crimes and tensions. Peter Drucker, and does sure digital in new interests of epub realizing business model innovation: a strategic approach for business unit managers 2014. in Action: The request of bloopersEstablishing colleagues Done through People. American Management Association, 1956. Peterborough, Ontario: Thomson South-Western, 2003. New York, NY: Harper Collins Publishers, 2001. New York, NY: McGraw-Hill Irwin, 2006. The for men of understanding of current complexes. Chicago, IL: The University of Chicago Press, 1996. : A Humanist Art. New York, NY: Colombia University Press, 1967.
not, ebook Delivering Fitness: Your Guide has a value to protect. It is from real and economic review in cities of the materials on which it interprets loved and the responsibilities it is. It is on challenging ia of friendly conversations, and working when these minorities can go a better F of structures, write &, bring minorities, and write as an kamelsuxPN-H-74242uploaded management and right between attitudes of the protein. This democracy is request in the UK level, creating the Death of Constitutional jS through the Human Rights Act 1998, the Tweaknews interpretation, and EU web. It apparently helps ebook Delivering Fitness: Your Guide list between the UK Information, the European Court of Justice, and the European Court of Human Rights. The minority is the sources of the exposed British Bill of Rights and the something past to know the sapiens Union. Use the confederation to be the pieces!


In his latest experiments in photography and 3D object, Maxime Francourt presents an updated take on Memphis-style graphics while also showcasing his growing interest in hand lettering and typography.
An innovation working prototype concept for audi.co.uk and Chromecast.
View configurations on the big screen in your home right from the website cast to your livingroom TV.


text:
Q: Вывести значения за пределы карты Подскажите как в этом примере вывести данные по доставке за пределы карты в

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https://tech.yandex.ru/maps/jsbox/2.1/deliveryCalculator

Для рачёта доставки выберите место на карте

A: Вы можете задать контент div-блока в той же части кода, где заполняете данными балун.
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...

document.getElementById("result").innerHTML = 'Итого:
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'Стоимость доставки: ' + price + ' р.';

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Forums » Trips, Adventures and Get-Togethers » The Bus Thread. 1956 Ford B-600. General questions.
My grandpa sold "Thomas Build Buses" in the 60's and 70's. His real business was tires but we always had new buses and old trade in buses to work on. And I did. A lot. This thread brings back a lot of memories of being elbow deep in grease with skinned knuckles and fingernails impossible to clean. So far nothing that went wrong surprises me. Good luck.
Toyman - if you would like for me to move this thread to the new Adventures forum just let me know! This thread definitely qualifies.
Works for me. That's a pretty good place for it, because it was definitely an adventure.
I agree and I wasn't even on the bus! Thread moved.
Toyman01 wrote: I was smelling burned rubber, burned wood, burned oil, and trying to remember whether the fire extinguisher was in the bus or the van.
For swap candidates, have a look at houred out garbage trucks. The freightliners came with MBE906's and an automatic.
They are a tough little 6 liter and you can pick up the whole truck for around $4000.00 if you keep your eyes open (Richie Brothers). You would have the whole driveline plus brakes, cooling system etc and there would be enough metal left to offset the purchase price in scrap value later.
I looked at this one for my bus but it was a little small for 35000 pounds. But you might be 20,000 pounds? and it would go quite well.
If you pick up something with the larger 7.2 liter MBE 926 they can be had with a 330 hp 1000 pound torque rating, and come standard with a built in exhaust brake plus a proper engine brake.
I wasn't here when this thread started but I loved reading it.
The old Ford fe truck engines are ridiculously reliable engines. My dad owned old ford trucks forever and in one we were running straight 50 weight and stp oil treatment in just to get oil pressure at highway speeds but it never blew up. After we pulled it out I pulled the distributor and inside the engine was basically grease.
That old truck was my grandma's gift to my dad before her death. Its been sitting for years after a terrible motor swap and is a tree magnet but someday...I'm resurrecting it.
This . Especially on an FE . Those things will drive you batty .


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Rashid Khan says he used to watch Shahid Afridi and Anil Kumble's videos to learn leg-spin.
As if his bowling was not enough to leave the world in a spin, Rashid Khan has made another startling revelation that is only going to make the cricketing world even more confused.
The Afghanistan sensation, in a recent interview, opined that he calls himself a finger spinner.
"I call myself a finger spinner," Rashid opined. "Sounds unusual but that's how I bowl," he added.
He further explained the reason for calling him finger spinner despite being called by many as the best leg-spinner in the world at the moment.
The 19-year old is making life difficult for the batsmen around the world. Early this year, he became the fastest to reach 100 ODI wickets.
The leggie reached the milestone during the ICC World Cup Qualifier final against the Windies. Rashid was also the youngest bowler to achieve the feat. He took just 44 matches to complete a century of wickets in ODIs, comfortably beating Mitchell Starc's record of 52 games.
A month prior to that, he became the youngest player to be ranked No. 1 on the ICC's Player Rankings across formats after clinching the top spot in T20I bowlers list. Rashid was also the youngest to captain a country at the age of 19 years 160 days.
His form in the last two months has also been sensational. He finished this year's IPL with 21 wickets — the second-most behind Andrew Tye, who had snapped 24 wickets for Kings XI Punjab.
In the first T20I against Bangladesh on Sunday (June 3), he took three for 13 to help his side win the game by 45 runs. During the course of the game, he became the second-fastest to 50 T20I wickets.
Rashid's rise has been nothing less than a fairy-tale. The 19-year old revealed he saw clips of Shahid Afridi and Anil Kumble to hone his skills.
Rashid has paid a big price for his on-field success. The Afghan superstar is yet to see his family since he went home after IPL last year.
"Last year when I went home after IPL, it was amazing. I really loved that moment when I entered my home, and they started celebrating. Looking forward to some rest and hopefully, I can go home," he said.
"I am giving back to my people by performing. It gives them joy," he added.
Rashid is currently gearing up to play his first-ever Test, starting June 14 against India.


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There are thousands of reasons to love The Bold and the Beautiful, one of them being that the CBS soap opera is about to air a very special 8,000th episode.
The Bold and the Beautiful is ringing in the new year with more than champagne and sparklers. The 31-year-old soap opera is also celebrating its 8,000th episode on Friday, January 4, with a classic soap storyline that promises to send ripples all the way into 2020.
"It's astonishing to look at the script we're working on today and see episode 8,000. We're grateful that you've allowed us into your homes for over 30 years," says B&B's executive producer and head writer, Bradley Bell, of the milestone moment.
He also teases that the 8,000th episode is "classic 'Bold and Beautiful' drama. The story arc centers around an event so profound, it sends shockwaves throughout the lives of the Forresters, Spencers, Avants, Spectras and Logans for years to come."
The Bold and the Beautiful, which is produced by Bell-Phillip Television Productions, Inc., aired its first episode on March 23, 1987. Not only is it part of CBS Daytime's history-making lineup, which has held the spot as the nation's number one daytime lineup for over three decades, but it is also the most-watched daytime drama in the world.
RELATED: Soap newcomer Adain Bradley exits B&B
Wrote B&B's official Twitter handle of its 8,000th episode: "We have 8000 reasons to love our wonderful B&B fans! @CBS will air the 8000th episode of #BoldandBeautiful on Friday, January 4th. Join our cast in celebrating this momentous occasion and tune in!"
We have 8000 reasons to love our wonderful B&B fans! @CBS will air the 8000th episode of #BoldandBeautiful on Friday, January 4th. Join our cast in celebrating this momentous occasion and tune in! pic.twitter.com/JWU8fJ8kQT
� Bold & The Beautiful (@BandB_CBS) December 13, 2018
Soap Central has more than 5,000 daily recaps for The Bold and the Beautiful in the Daily Recaps Archive. The section also includes a special "This week in B&B history" section that lets you relive memorable B&B storylines You can check those out here.
Would you like to leave a congratulatory message for B&B reaching 8,000 episodes? What kind of storyline do you think the powers that be have in store for the momentous occasion? We want to hear from you -- and there are many ways you can share your thoughts.
's B&B FrontPage.


text:
Why CO2 is Good
Politics are Not Green
Benefits to Plants
Benefits to Humans
Critical Facts
Additional Facts
Posted on: Monday, March 26, 2012
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Is this finally proof we're NOT causing global warming?
Is this finally proof we're NOT causing global warming? The whole of the Earth heated up in medieval times without human CO2 emissions, says new study.
Evidence was found in a rare mineral that records global temperatures
Warming was far-reaching and NOT limited to Europe
Throws doubt on orthodoxies around 'global warming'
Current theories of the causes and impact of global warming have been thrown into question by a new study which shows that during medieval times areas as far apart as Europe and Antarctica both warmed up.
It then cooled down naturally and there was even a 'mini ice age'.
A team of scientists led by geochemist Zunli Lu from Syracuse University in New York state, has found that the 'Medieval Warm Period' approximately 500 to 1,000 years ago wasn't just confined to Europe.
In fact, it extended all the way down to Antarctica.
At present the Intergovernmental Panel on Climate Change (IPCC) argues that the Medieval Warm Period was confined to Europe.
However, Professor Lu has shown that this the Medieval Warm Period had a much broader reach - and the evidence lies with a rare mineral called ikaite, which forms in cold waters.
'Ikaite is an icy version of limestone,' said Lu. 'The crystals are only stable under cold conditions and actually melt at room temperature.'
It turns out the water that holds the crystal structure together - called the hydration water - traps information about temperatures present when the crystals formed.
This finding by Lu's research team establishes, for the first time, ikaite as a reliable way to study past climate conditions.
The mineral proved that Antarctica did warm up.
Lu says that his research has no direct bearing on the current climate, and points out that his research is restricted to one area in Antarctica, and is not in itself proof that the whole Earth warmed up.
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5/26/11 No Long-term Trend in Atlantic Hurricane Numbers
9/1/10 Meltdown of the climate 'consensus'
5/21/10 Prominent Princeton Scientist Dr. Happer Testifies to Congress
11/4/09 Plants need more CO2, not less
8/28/09 Utah governor says climate change debate not over
8/13/09 Earth's Warming Rate Overestimated
8/13/09 U.N. Crying "Wolf" on Climate Change?
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8/6/09 2009 Atlantic Hurricane Season Forecast Lowered
8/6/09 Cosmic Rays Have Significant Climate Effect
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8/6/09 Scores of German Scientists Dissent Over Global Warming Claims
8/3/09 Save the Planet: Have Fewer Kids
7/29/09 Major Science Group 'Startled' By Outpouring of Scientists Rejecting Man-Made Climate Fears!
7/26/09 Resisting climate hysteria
6/13/09 Crops under stress as temperatures fall
5/14/09 What if global-warming fears are overblown?
3/2/08 Nature, Not Human Activity, Rules the Climate
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** For additional peer-reviewed scientific references and an in-depth discussion of the science supporting our position, please visit Climate Change Reconsidered: The Report of the Nongovernmental Planel on Climate Change (www.climatechangereconsidered.org), or CO2 Science (www.co2science.org).
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Liverpool talking to Radamel Falcao's agent as Manchester United pass on permanent deal
Liverpool FC Manchester United FC
Liverpool are talking to Radamel Falcao's agent after Manchester United manager Louis van Gaal decided to pass on signing him, reports the Daily Express.
Liverpool submit £18m bid to sign AS Roma star next season
The 28-year-old has been on loan at the Old Trafford this season from AS Monaco. He has failed to impress in his time with the Red Devils, only scoring four goals in 23 appearances in the Premier League. Not only this but the Colombian has been stuck on the bench for large periods, with Wayne Rooney, and even 19-year-old James Wilson playing ahead of him at times.
After all of this the United boss would appear to have made his decision. Van Gaal has already informed the French side that he will be sending the striker back according to the report. This has left the door open to Liverpool. The Merseyside club are supposedly in talks with Falcao's agent Jorge Mendes. The player's representative is trying to persuade the Reds to sign his client permanently.
Would a United flop really be the best option for Brendan Rodgers?
More Stories AS Monaco Liverpool FC Manchester United FC Radamel Falcao
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Home R•HOME Secret Gardens
by Brandon Fox
The 250 people who attended the first Secret Garden Party in Historic Court End enjoyed it so much, "they lost track of time — so we're making it a half-hour longer [next year]," says Katherine Fahed, spokesperson for the Historic Richmond Foundation.
On May 13, the Court End neighborhood threw open its gates and staged a progressive cocktail party in the gardens of the Museum and White House of the Confederacy, the John Marshall House, Monumental Church, the Valentine Richmond History Center and the Massey Cancer Center Healing Garden. Each garden featured different food and drinks, from vodka gimlets to John Marshall's favorite, Madeira. Music included Justin Smith's Jazz Band Trio, Campbell's Ramblers and guitarist Leah Kruszewski, among others.
"It was the week after Mother's Day, so a lot of daughters bought tickets," says Fahed. "A lot of them [said] they're going to bring their husbands next year."


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August 31, 2015 Bruce Allen Morning Links
Monday Media Notes
There are a few items of interest from the Boston sports media world from the last couple of days.
First, Red Sox Chairman Tom Werner spoke to Steve Buckley about why they made the decision to replace Don Orsillo with Dave O'Brien.
Tom Werner explains why NESN is parting with Don Orsillo – Steve Buckley
In a nutshell, they like O'Brien better and he was available, so they made the move. Werner told Buckley that they want to "re-energize" the telecasts, and said that O'Brien is "well known for bringing out the nuances of baseball strategy, sharing insights about players…"
I thought the NESN telecasts were supposed to be about entertainment? At least that has always been the perception, that baseball was at times secondary to the broadcast. It seemed accepted that Orsillo was doing what he was asked to do. Now, they're saying he didn't talk enough about the nuances of baseball strategy?
Richard Deitsch of SI.com writes about the move as well (with links to other articles) in The Noise Report.
The Montreal Screwjob, as performed by @nesn & Red Sox brass on Don Orsillo. The @SInow Media column: http://t.co/bN30W9t4A0
— Richard Deitsch (@richarddeitsch) August 31, 2015
I certainly don't like how the move was handled, but the hand-wringing by other media over Orsillo has been a bit much. I like Don Orsillo, everyone says he's a great guy, one of the best, but like most things, this type of thing happens every day in corporate America. If the Red Sox want to change broadcasters, it's their prerogative, life goes on.
It's funny, when Orsillo was named the full-time broadcaster, there was similar hand-wringing over how Sean McDonough was getting screwed over.
It's the circle of life in media jobs, and indeed, in real life. Don Orsillo will get another, probably better, job.
Over the weekend, the Boston Globe decided to look at all the apologies that ESPN has been issuing as of late. Great. It's a worthy topic, given that the network seems to have abandoned its ombudsman position.
For ESPN, apologies become commonplace – Callum Borchers
Notice anything missing? Only any reference to the apology issued the week before to the Patriots.
That's kind of mind-blowing.
Ron Borges, of all people was probably the first to report that Glenn Ordway was going to be taking over the WEEI mid-day show (I'd post his tweet here, but since he's probably already blocked 3/4 of my readers, most of you wouldn't be able to see it.).
The station made it official, adding Ordway to the 10-2 show alongside Christian Fauria and Lou Merloni. He returns on September 8th.
Glenn Ordway to join WEEI's midday show – Chad Finn, who also has a similar, but slightly different piece for Boston.com.
I believe the show will eventually reclaim the ratings win in the mid-day time slot. I just can't get into Marc Bertrand with Scott Zolak, they just don't click for me. While it is likely disappointing to some that the station is returning to an old voice, Ordway could be the perfect fit to work with two former players in Merloni and Fauria. I think he'll bring out the best in them.
As for Gary Tanguay, he shouldn't feel too bad about getting passed over. It's just another job he doesn't have to worry about being fired from.
29 thoughts on "Monday Media Notes"
Andy Dursin says:
Agreed completely on the Big O. Merloni and Fauria have been more listenable without Benz (another Salk like disaster) and I think they'll be improved with Ordway running the show. As for Zolak and Bertrand, 98.5 took off the one solid sports pairing they had in Gresh & Zo during football season and opened the door for EEI to reclaim the midday slot if nothing else. Should be a good boost for them that could trickle over into Dale and Holley. F&M's schtick is beyond old — they still have their listeners, but the jig will be up at some point for them, just as it eventually was for Ordway after his years of ratings dominance.
Ordway comes with his legion of shut-ins – if is #s do grow to win the mid-day – how long before eei contemplates a flipflop with pm drive?
daniel mateus says:
Ordway already attacks F&M more so than the actual show that competes against them. Even D&C goes after them. I love how Kirk just crushes them,especially Tony,on an almost daily basis and the pusses at the Hub don't fight back.
They don't have to fight back. They're in the lead. When (and ye gods I hope it's soon) EEI passes them in the ratings, then you'll see them fight back.
mediablowhards says:
The Big O used one of my lines during his call with D & C & K this morning, referring to (Mazz, Adam Jones, Tanguay, etc.) people that are terrified of being called Homers and going off the deep end with their phony contrarian hot takes.
the_other_side says:
Just saw this tweet, any idea what they were saying?
Beep beep beep…Felga and Mazz are backing pedaling quickly….
— Boston Radio Watch® (@bostonradio) August 31, 2015
Looks like another guy besides Portnoy got to videobomb ESPN's live from court today:
HAHAHAHA Gaz sneaking in completely unbothered is so perfectly Gaz pic.twitter.com/By7FmiS7Kx
— Feitelberg (@FeitsBarstool) August 31, 2015
StoJa says:
That's "Sales Guy" from Barstool. Pretty sure he was with El Pres when they got arrested before.
The vid I posted on the previous post with the "Mort Lies" was awesome. I loved how some security guys were pushing him away. ESPN must now hire bouncers anytime they know this must happen.
Gotta think the producers in the truck and Bristol was like "WTF DO WE CUT!? OR KEEP ROLLING!?"
Or…you know…it could be because a cameraman and interviewer were killed on camera recently and they're worried about copycats.
I tend to agree with you. I can't believe ESPN hired security to keep people from showing signs. Their whole schtick is to have people crowd behind the live shot.
Yeah, on GAMEDAY. They know full well how much guys like Sal and Mort and the whole ESPN brand are loathed in New England. This is in no way comparable to college students holding signs for TV behind Lee Corso or Digger Phelps. That's not even apples to oranges. It's like spinach to ice cream.
I bet that factored about 1% into it. ESPN is very protective of their image and it's taken a huge hit the past 9 months.
I have no idea why anyone would care if Don Orsillo stays or goes. They guy lost all credibility with me when he agreed to stay and work with Remy. My questions are why isn't anyone asking why Remy gets to stay and if Dave OB knows that his new partner enabled a wife beater and a murderer? Enquiring minds want to know.
When it comes to the big O joining the midday Show I wonder if Dale Arnold has a greater sense of personal satisfaction today.
Irony: John Tomase reporting on a false report:
Red Sox spokesman says team didn't instruct anyone to take away signs supporting Don Orsillo. Said Orsillo signs treated like any others.
— John Tomase (@jtomase) September 1, 2015
Matthew McKnight says:
Anyone hear Merloni doing play by play on the radio? He was working his tail off. It gave me a new found respect for the ace play by play guys.
I actually like when Joe C does play-by-play and Lou does the color. I think they play off each other real well. What is the rationale between having two play by play guys on the radio? If anyone could explain the thought process behind that I would appreciate it very much. TY
It's always entertaining when such a PR-conscious organization is completely inept at PR:
Straight from the Boston Globe. Link here: http://t.co/MG9TXeBtFj pic.twitter.com/VS8H1KGQhZ
— Jared Carrabis (@Jared_Carrabis) September 1, 2015
But, I guess when you've got 50% of the media in town on payroll and another 10-25% bending over for a job, it doesn't matter.
Lots of conflicting reports over this Orsillo signage thing last night. It wouldn't surprise me if the Sox did make an effort to do it but were met with a lot of resistance. I do know that Bruins' management did, in fact, have all anti-Harry Sinden signs confiscated from the FleetCenter during the final few home games of the 1997 season — their first non-playoff season since Orr's rookie year, which was characterized by Sinden's dealing away Adam Oates right after Oates had the temerity to publicly question the organization's financial commitment to winning (in other words, he called the owner cheap, and got his butt traded away in response). The fans were pissed and they let Harry know about it, but security was told to confiscate all of the signs and to eject anyone who resisted. I know because I was one of the security staff at that time — fortunately, I wasn't working inside the arena, but merely inside the security office on site. If I had been one of the inside-arena guys, it would have made me sick to have to do that to the paying customers, who had every right to express their frustration with how poorly (and yes, cheaply) the organization was being run back then — and this was still three whole years before Bourque said, "I surrender," and asked to be traded to a Cup contender before it was too late for him to win one. Some ingenious fan who was tossed out of the first home game of the post-Oates era got his revenge the following Saturday by having a bunch of "Harry Sucks" decals printed, bringing them with him, and the surreptitiously slapping them up on the walls inside and outside the arena. It took the Delaware North minions the entire summer to locate and scrape off all of them…LOL. I think suppressing the fans' ability to express their feelings about a hot topic, good or bad, is a bush league maneuver, seriously. If the Sox did try something like that last night, they need to grow up and realize that this, too, shall pass. No one will be talking about Orsillo's absence come opening day next April. For now, just deal with the fallout and take the backlash like grown men, please.
re: Ordway — Meh. I actually kind of like Merloni and Fauria, in a low-key sort of way. I hope Ordway doesn't dominate things too much.
Anyone know where to get the new Channel Media survey Beetle was reading from? It usually leaks a few days later.
Interesting tweet from the king of all toolbags, Darren Rovell.
I'll paste the images directly to save on having to find it.
7/28 poll:
showaboutnothing says:
I sent it to Bruce a little while ago. Let me know your email if you want it before he gets a chance to post it.
How do you get it? Could you post it up on Google Docs or somewhere public? Otherwise, if Bruce has it, I assume he'll do a post today or tomorrow with it linked. Thanks for the offer, though.
I couldn't get google docs to work. This was the best I could do. Maybe you can save it to google docs so everyone can see it. I don't know if this worked.
It worked. Thanks!
When people answer that Mazz is their favorite sports personality, do they mean to laugh at?
Roger Bournival says:
What's the over / under on Ordway yelling over one of the callers, one hour?
Mike Hurley (CBSBoston) found this. Take a look at the picture of the two gauges being measured from TWR:
http://www.robertblecker.com/deflategate-the-smoking-gun/
I do find it hilarious that Exponent is a publicly traded company.
Reminds me of when I was growing up, my dad kept track of my height. I, like most kids, would stand on my tippie toes when measuring.
Kudos to showaboutnothing. Some really interesting stats:
Patriots Preseason Thoughts Heading Into Game Three
Highlights From The 2015 New England Sports Survey By Channel Media & Market Research, Inc


Hindustan Unilever Ltd. shareholders continued to celebrate Tuesday, a day after the company reported better-than-expected fourth quarter results.
India's largest consumer goods maker by sales posted a quarterly profit of 5.69 billion rupees ($127.3 million), beating expectations thanks to one-time gains in the January-March period. Still, the figure was 2.1% lower than a year earlier, when the company had also benefitted from one-time gains.
Sales growth of 14% to 48.99 billion rupees was also ahead of consensus estimates.
Investors welcomed the results, with the stock rising to a one-month high of 298.40 rupees early Tuesday, adding to the 3.6% gain made by the close of trade Monday.
Early afternoon Tuesday, Hindustan Unilever was up 4.2% at 296.25 rupees on heavy volume in a Mumbai market that was up just 0.7%.
The jubilation is understandable as the results proved that the company, a unit of Unilever, has been able to successfully shrug off challenges to its dominance in the Indian market.
In 2009, Hindustan Unilever suffered as consumers of soaps and detergents switched to cheaper offerings from rivals such as Godrej Consumer Products Ltd. and Procter & Gamble Co.'s India unit. But the company shifted strategy in the last 15 months, deciding not to fully pass on higher input costs to customers via price hikes. Instead, it opted to sacrifice margins in an effort to win back customers with sales promotions and judicious price increases.
The strategy is paying off. Sales volume at Hindustan Unilever's local consumer goods business grew 14% in the quarter through March, compared with 11% growth a year earlier.
"With the fifth consecutive quarter of double-digit growth, HUL is getting its act together in regaining lost ground in its core categories of soaps, detergents, shampoos," said Nikhil Vora, managing director of Mumbai-based brokerage IDFC Securities Ltd.
However, the company took a hit on its profit before interest and tax margin, which shrank 0.6 percentage point from a year earlier to 13.0% due to higher input costs. Prices of soda ash and palm oil – key raw materials used in its core categories of soaps, detergents and shampoos – remain stubbornly high.
The company raised prices of some products by about 3%, but the hikes clearly weren't enough to fully offset the increase in raw material costs.
Analysts agree that the company's strategy poses a risk to operating margins and profit growth, but they are optimistic that Hindustan Unilever can ride out these challenges.
IDFC's Mr. Vora said he expects margin pressure to remain for at least another one to two quarters. "Beyond that, as an aggressive HUL leverages global scale to optimize costs and commodity prices subside, we believe profit growth will follow suit," he added.
Post the results, IDFC maintained its "outperform" rating on the stock with a price target of 340 rupees.
"Despite competitive intensity at its peak and commodity prices on a high, HUL has struck the right balance between volume growth and profitability," added Sanjay Singh and Pratik Biyani of Standard Chartered Bank.


This summer in the UK has been pretty warm so far and this has meant that my favourite drink of a freshly brewed cup of tea hasn't had quite the appeal that it usually does. In today's blog post I'm going to write about a healthy alternative that has kept me cool through the summer and which takes 5 minutes to make - the humble cold brew tea.
Why Home Made Cold Brew Tea?
Compared to iced tea, cold brew teas do not taste bitter and some fans prefer the brighter, more refreshing flavour. This is because using cold water slows the brewing process which allows the subtler flavours to be extracted before the bitter tannins. With the tea being less bitter, there isn't any need to use additional syrups or sugar like you would find in most pre-made iced teas bought from the shops or supermarkets. This means more depth of flavour and without the sugar - who could say no?
May help to lower blood pressure.
As with most things though, it's important to note that tea should always be drunk in moderation. Some researchers are currently exploring the potential effects tea has on lowering iron absorption rates and how it might be interfering with folic acid absorption. Because of this, anyone with a condition that requires higher levels of iron absorption or folic acid absorption, such as pregnant women or those with anaemia, should moderate their tea drinking.
A small sieve or tea strainer.
Simply add your 500ml of water and tea leaves together in your glass container and leave in the fridge overnight or up to 24 hours. You'll find that the tea will brew slowly leaving you with a fresh yellow tea that you can either strain directly into your cup or into another glass jug for serving throughout the day. It's as simple as that! If you use a French press, you can plunge the tea as you would with coffee which will save you the need from having to use a strainer or sieve.
Using different tea leaves and at different quantities will bring out different depths of flavour.
Try adding herbal teas such as camomile as this can add another dimension in flavour.
Experiment with different amounts of water to vary the intensity.
Vary the brew times - the recipe I've given is just something that's worked for me so feel free to explore your own length of brew.


Based in Sheffield, The Rosadocs are a four piece alt rock/indie band making their mark on the alternative scene. They have attracted comparisons to Arctic Monkeys and Jamie T, as well as taking on mod and rock influences to form their own distinct and ever-maturing sound. So far, they have amassed over 30,000 Spotify streams across four singles. Newest track 'Oak Tree' was recorded with the legendary Gavin Monaghan (the production mastermind behind tracks by Ocean Colour Scene, Paolo Nutini, and The Twang).


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A Giant Jeff Goldblum Statue Erected in London to Celebrate 'Jurassic Park'
Published on July 21st, 2018 | Updated on July 21st, 2018 | By FanFest
A Giant Jeff Goldblum Statue Erected in London to Celebrate Jurassic Park
Jurassic Park is celebrating its 25th anniversary this year, and in celebration, a 25-foot Jeff Goldblum statue has been built next to the Tower Bridge in London. This isn't any old statue, though. Naturally, the work of art features Goldblum in the iconic reclining position, shirt unbuttoned.
A gigantic Jeff Goldblum has appeared in London! 😍
The 25ft-statue of #JurassicJeff was commissioned by @nowtv to mark the 25th anniversary of #JurassicPark pic.twitter.com/X4aTIP8ypJ
— Rotten Tomatoes (@RottenTomatoes) July 18, 2018
The statue was commissioned by Now TV, a UK streaming service, in honor of the iconic film and took about six weeks to create.
Fans are flocking to have their picture taken with the statue and posting with the hashtag "JurassicJeff." Actress Aubrey Plaza (Legion, Parks and Recreation) jumped the rope to get up close and personal with Dr. Malcolm before being pulled away. Funko Pop UK took a photo of their reclining Malcolm Pop figure with the statue, saying that "Life, uh…finds a way."
This isn't the first time that a giant statue of a celebrity has appeared in London. In 2013, a fiberglass Colin Firth found its way into the Serpentine in Hyde Park. This was in celebration of Firth's role as Mr. Darcy in Pride and Prejudice.
Why, Mr Darcy how enormous you look…! RT @susborne: Very large Colin Firth emerging from Serpentine this morning pic.twitter.com/LEHDLULJK6
— Denise Jackson (@MmeAudubon) July 8, 2013
Jurassic World: Fallen Kingdom opened in June, marking Goldblum's return to the franchise since he was last seen in Lost World. The film was a box office success, accumulating more than one billion dollars. The film stars Chris Pratt and Dallas Bryce Howard as Owen Grady and Claire Dearing, respectively.
The giant Goldblum will stick around London until July 26th.
Micajah
Covering superheroes, anything dark, horror, and more! Lead writer for Fan Fest
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Home / Headline News / Iggy Pop announces new album, "Free"
Iggy Pop announces new album, "Free"
John Hutchings July 18, 2019	Headline News, News Leave a comment
Iggy Pop has confirmed the title and release date of his new album: Free, the first new Iggy Pop album since 2016's Post Pop Depression, will be released September 6 on Loma Vista and is available for pre-order HERE.
While it follows the highest charting album of Iggy's career, Free has virtually nothing in common sonically with its predecessor—or with any other Iggy Pop album. "Free," the eponymous single, is available to stream and purchase on all platforms HERE.
On the process that led Iggy and principal players Leron Thomas and Noveller to create this uniquely somber and contemplative entry in the Iggy Pop canon, Iggy says:
"This is an album in which other artists speak for me, but I lend my voice. By the end of the tours following Post Pop Depression, I felt sure that I had rid myself of the problem of chronic insecurity that had dogged my life and career for too long. But I also felt drained. And I felt like I wanted to put on shades, turn my back, and walk away. I wanted to be free. I know that's an illusion, and that freedom is only something you feel, but I have lived my life thus far in the belief that that feeling is all that is worth pursuing; all that you need – not happiness or love necessarily, but the feeling of being free. So this album just kind of happened to me, and I let it happen."
About John Hutchings
Previous Easy Life unveil new track "Earth" along with music video
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Canadian producer and singer-songwriter BAD CHILD has finally announced his full-length debut album Free Trial. The …


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Merge Tales - Merge 3 Puzzles
Merge Tales - Merge 3 Puzzles is an iPhone puzzle game developed and published by Koolbit. The game features cute, colorful graphics and simple gameplay that is easy to pick up but difficult to master. Merge Tales - Merge 3 Puzzles is an addicting and challenging game that is perfect for quick, casual play.
Merge Tales - Merge 3 Puzzles feature colorful, cartoonish graphics that are pleasing to the eye. The game's art style is cute and whimsical, and the visuals are sharp and clear. The game's color palette is cheerful and vibrant, and the overall look of the game is very polished.
Merge Tales - Merge 3 Puzzles is a simple but addicting puzzle game. The objective of the game is to merge matching items in order to create new, more powerful items. The gameplay is easy to pick up but difficult to master. There are a variety of different items to merge, and the game gets progressively more challenging as you progress. The game is perfect for quick, casual play, and it is very addicting.
Replay Value
Merge Tales - Merge 3 Puzzles is a very addicting game, and it has a lot of replay value. The game is challenging and there are a lot of different items to merge, so you can play for a long time without getting bored. The game is also perfect for quick, casual play, so you can always pick it up and play for a few minutes when you have some free time.
Merge Tales - Merge 3 Puzzles is a great puzzle game for iPhone. The game is easy to pick up but difficult to master, and it is very addicting. The game has great visuals, simple but challenging gameplay, and a lot of replay value. If you are looking for a quick, casual game to play, Merge Tales - Merge 3 Puzzles is a great choice.
Mega Mall Story Mega Mall Story Mega Mall Story is a game for the Nintendo DS and 3DS. The game was developed by Kairosoft and released in 2011. Graphics The graphics are colorful and pleasing to the eye. The game uses a top-d... 2022-04-15
Luna Re : Dimensional Watcher Luna Re : Dimensional Watcher Luna Re: Dimensional Watcher is a game that was released on May 21, 2020. The game is a side-scrolling action game that has players take control of a young girl named Luna who is on a quest to save he... 2022-04-15
Cat Jump Cat Jump Cat Jump is a simple platform game. In the game, players help sissy and her friends through a world full of enemies and dangers or survive in endless mode. There are four different characters players... 2022-03-23


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DEADLINE INFORMATION
Wendy WhiteEagle starts term as first Native American Village Trustee in growing Chicago Western Suburb
By Ben Krause-Decorah
"There is not many things that someone my age hasn't experienced, but this is new and this is exciting!" said Wendy WhiteEagle, 58, who was sworn into her seat as a Trustee of the Village of Campton Hills in Illinois on August 31, 2018. Campton Hills is a growing community just west of Chicago in the western suburbs and boasts a population of nearly 20,000 residents.
White Eagle is a member of the Ho-Chunk Nation, who is part of the Eagle Clan. Her parents were John H. White Eagle and Lyla L. Miller. She grew up in Madison, WI where she earned her bachelor's degree from Edgewood College and master's degree work in the Executive MBA program with UW-Madison and Edgewood College.
White Eagle was appointed by Village President and Board Chair Harry Blecker after another Trustee resigned in July. White Eagle believes she is well equipped for her job saying, "I have had many leadership positions throughout my career and strongly believe that one of the aspects of my performance that has consistently stood out is my ability to build teams. I have deep knowledge in the finance, organizational development, and management arenas which I now believe I can leverage into my public service." Currently, White Eagle is the President and CEO of Native Capital Investment, INC. headquartered near the Village and has held that title for over 15 years. In her role as President she is able to work with local businesses both inside and outside the village in creating self-sustaining business and communities. This objective is easily transferrable into her role as a trustee of a fast growing suburb.
In the past 8 years, the Village has seen tremendous growth. In 2010 at the census the population was 11,310 residents. Since then, the population has grown to 17,564. This is driven not only by population relocating from the City of Chicago, but also large commercial growth within the township. The location of the Village is close enough to the City of Chicago that growth is expected to continue.
In this role, White Eagle hopes to help create a community that is in demand as her Village grows forward, together saying, "As we expand our vision of what makes community I believe we can also leverage both human capital and local resources to create a shared sense of place. Collaborating, even in this microcosm that is our Village, will allow for diverse voices to be heard and for plans to be made as we grow forward, together." As a member of the Ho-Chunk Nation, White Eagle makes up less than 0.5% of the Village's Native American population. However, she now has a seat at Board table to guarantee the diversity of the village has a voice.
"Wendy has been involved in the Ho-Chunk Nation's branch office since its inception and every year since serving as the District Five Treasurer. This new role as Village Trustee is a role that fits her well." Says Chicago Branch Office General Manager Alma Enriquez. "It is so powerful to hear success stories fellow Natives not only having a voice in these important roles, but also hearing that it is a women. It gives me hope to see Wendy break every statistic against her and continue to move on up. I'm proud of her."
Earth Partnership initiative recognized with UW-Madison's Community-University Partnership Award
Ardith Van Riper
Ho-Chunk Nation Oath of Office Ceremony
Ho-Chunk RV Resort and Campground hosts annual Youth Culture Camp
Gary Garvin
Annual Flag Day ceremony honors veterans and provides connection to families
Ken Luchterhand
Zunker heads effort in changing October day to Indigenous Peoples' Day
Chief John Winneshiek recognized for his military service with ceremony
Ho-Chunk Nation Digital Library project celebrates its third year of work
First Listeners Project Offers Ho-Chunk Language Classes
Copyright 2016 Ho-Chunk Nation: Hocak Worak


The majority of us tweak our hair color and makeup according to the time of year. But, not all of us apply that same concept to our skin care, although we should. It's important to make sure that your skin care routine matches up with the seasons because certain skin issues can become exacerbated as the temperature changes. Here's what your skin needs (and what to avoid) this summer.
The warmer the weather, the more humidity in the air, which keeps skin hydrated. "When the air is warm, you only need a light moisturizing gel or lotion," says celebrity aesthetician Susan Ciminelli. "Skin sweats more and it just needs hydration at the surface." Light hydrating lotions, fluids, serums, elixirs and oils, are good, too, because they contain emollients. Anything too heavy can clog pores and make skin look greasy.
Try this: Rich in sea botanicals that are packed with skin-nourishing fatty acids, Cellex-C Sea Silk Oil-Free Moisturizer ($59) offers long-lasting, lightweight hydration.
Gels are perfect for warmer temperatures because they lend a cooling effect and leave skin feeling refreshed. They often contain acne and oil fighters—two problems that go into overdrive in the heat. "The higher incidence of alcohol leaves less oil in the skin than other cleansers," says New York dermatologist Patricia Wexler, MD.
Try this: A lightweight antibacterial gel cleanser that doesn't smell medicinal in the least bit, Skin Authority Daily Cleanser ($42) wipes away makeup, oil and dirt for skin that feels clean and looks fresh.
"Skin needs to be exfoliated more during the warmer months to remove oil and dead skin because it sweats more," says Ciminelli. You're probably exposed to more sun this time of year, too, so also pay attention to exfoliating agents. "Retinol acts as an exfoliator. You may want to cut back on it when you're going to be in the sun—it makes skin more prone to burning," says Miami dermatologist Janelle Vega, MD.
Try this: Diptyque Multi-Use Exfoliating Clay ($65), which can be used on both the face and body and doubles as a scrub and a mask, really works to lift away dead skin cells thanks to the powerful exfoliating properties of marble powder.


Hey! I am a mother of a toddler who is a very picky and fuzzy eater. One thing that is very amusing about her that she loves yonanas - be it of any flavour. So my toddler's love for yonanas made me a specialist of yonanas recipes.
My toddler has 6 hot favourite yonana and I have one . And I am in a mood to share all of the 7 yonanas recipes with you at one go.
Put one frozen banana. Then add 2 oz dark chocolate. After that put the second frozen banana. Now stir it in a bowl to combine all the ingredients.
Serve in a serving bowl with a few choco chips and have pleasure in seeing your kids nibble on the chocolate chips and finally emptying the bowl .
Put one frozen banana. Then add 3⁄4 cup frozen raspberries. After that put the second frozen banana. Stir it in a bowl to combine all the ingredients .
Sprinkle chocolate chips and raspberries on top and serve chilled. Makes an ideal smoothie to beat the heat .
Put one frozen banana. Then add 3⁄4 cup frozen cherries. Now add 1 oz dark chocolate. Put the second frozen banana . Stir it ina bowl to combine all the ingredients.
Serve with chocolate chips and cherries. Ideal smoothie for kids who love their food to be colourful.
3. 1⁄2 teaspoon ground cinnamon.
First combine the cocoa powder and ground cinnamon to create one spice blend. Then sprinkle spice mixture over bananas. Now insert frozen bananas. combine all the ingredients.
While serving sprinkle a little bit of ground cinnamon for that exquisite flavour.
This particular yonana is a favourite of kids as well as adults . Precisely this yonana is a favourite of spice lovers.
Sprinkle espresso powder over bananas. Now insert the frozen bananas. Finally stir it in a bowl to combine all the ingredients.
Put one frozen banana. Then add lime zest. Then putsecond frozen banana. Stir it in a bowl to combine.
Serve with sprinkled crumbled cinnamon crackers on top.
Serve with a slice of orange in a serving bowl.
So, what are you waiting for ? Go and give each of these yonana recipe a try.


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