text:
\section{Introduction and preliminaries}
The goal of this paper is to introduce and study the notion of ends for
algebras. First let us recall the definition of ends of groups (see
e.g. Freudenthal (1945), Epstein (1962)).
Let $\Gamma$ be a group, then we call $P\subset \Gamma$ an almost invariant
set if for any $g\in \Gamma$, $Pg$ and $P$ differ only by finitely
many elements. That is $\{(Pg\cup P)\backslash P\}$ is
a finite set. Obviously, if $P$ is almost invariant, then $\Gamma\backslash P$
is almost invariant as well. If two almost invariant sets differ only by finitely many elements
then we call them equivalent, $P\simeq Q$. The equivalence classes of
almost invariant sets are the endsets.
We say that the group $\Gamma$ has $k$ ends if it can be partitioned into the
disjoint union of $k$ infinite almost invariant sets, but it can not be
partitioned into the disjoint union of $k+1$ infinite almost invariant sets.
If for any $k\geq 1$, the group has such partitions, then $\Gamma$ has
infinitely many ends. By the theorem of Hopf (1944), a group may have $0,1,2$ or
$\infty$ many ends. Finite groups have zero ends, the group ${\mathbb Z}^2$ has
one end, the group ${\mathbb Z}$ has two ends and the free group of two generators has infinitely
many ends.

\noindent
Now let $K$ be a commutative field and $\mbox{$\cal A$}$ be a unital $K$-algebra. We say that
a $K$-linear subspace $V\subset \mbox{$\cal A$}$ is an {\bf almost invariant subspace} if for any $x\in\mbox{$\cal A$}$,
$(Vx+V)/V$ is finite dimensional.
\begin{definition}
\label{d2}
The algebra $\mbox{$\cal A$}$ has $k$ ends if $\mbox{$\cal A$}=V_1\oplus V_2\oplus\dots\oplus V_k$,
where $\{V_i\}_{i=1}^k$ are infinite dimensional
almost invariant subspaces, but
$\mbox{$\cal A$}$ can not be written as the direct sum of $k+1$ infinite dimensional
almost invariant subspaces. We say that $\mbox{$\cal A$}$ has infinitely many ends, if
for any $k\geq 1$, $\mbox{$\cal A$}$ has such direct sum decomposition. We apply the
following convention; finite dimensional algebras have zero ends.
\end{definition}
Let us see some examples.
\begin{example}
$\mbox{$\cal A$}=K[x]$, the polynomial ring of one variable.
\end{example}
Let $S\subset \mbox{$\cal A$}$ be an infinite dimensional almost invariant subspace.
Let $n>0$, such that $Sx+S\subset S+\mbox{span}_K\,\{1,x,x^2,\dots,x^n\}$. Suppose
that $p_m\in S$ is a polynomial of degree $m>n$. Then $p_mx\in
 S+\mbox{span}_K\,\{1,x,x^2,\dots,x^n\}$. That is $S$ contains a polynomial of degree $m+1$.
Inductively, $S$ contains a polynomial of degree $k$ for any $k\geq m$.
Hence $K[x]=S+\mbox{span}_K\,\{1,x,x^2,\dots,x^{m-1}\}$. Therefore $K[x]$ has only one end.
\begin{example}
$\mbox{$\cal A$}=K[x,x^{-1}]=K{\mathbb Z}$.
\end{example}
If $S\subset \mbox{$\cal A$}$ is an infinite dimensional almost invariant subspace,
then our previous example shows that $K[x]\cap S$ is either finite or
cofinite dimensional, similarly,  $K[x^{-1}]\cap S$ is either finite or
cofinite dimensional. Hence $K[x,x^{-1}]$ has two ends.
\begin{example}
The $\ast$-algebra $\Sigma_n(K)=\oplus_{1\leq i \leq n} K[x_i]$
\end{example}
One can easily prove that if $\mbox{$\cal A$}$ and $\mbox{$\cal B$}$ are algebras, having respectively
$k$ and $l$
ends then $\mbox{$\cal A$}\oplus\mbox{$\cal B$}$ has $k+l$ ends.  Thus $\Sigma_n(K)$ has exactly $n$ ends.
\begin{example}
The algebra $\mbox{$\cal A$}=K[x,y]/J$, where the ideal $J$ is generated by the monomials
of $x$ and $y$ containing at least $2$ $y$'s.
\end{example}
Note that $S_k={\mbox span}\,\{x^k,x^ky, x^kyx, x^kyx^2,\dots\}$ is an almost
invariant subspace. Then
$$\mbox{$\cal A$}=S_1\oplus S_2\oplus\dots S_n\oplus \{K\oplus S_{n+1}\oplus
S_{n+2}\dots\}$$
is a decomposition of $\mbox{$\cal A$}$ into the direct sum of $n+1$ infinite dimensional
almost invariant subspaces. Hence $\mbox{$\cal A$}$ has infinitely many ends.

\noindent

We shall define endclasses of algebras, similarly to the endsets of groups, and
identify the endclasses with the idempotents of a certain natural extension
of our algebras. Thus one-endedness of a certain algebra
is equivalent to the fact that this extension has no non-trivial idempotents.
Houghton (1972) proved that if a group is infinite then the number of ends
is exactly the dimension of the first cohomology group $H^1(\Gamma,{\mathbb C}\Gamma)$
plus one.
 Using a Fredholm-module construction
we associate to the endclasses of $\mbox{$\cal A$}$ elements of the first cyclic cohomology
group $HC^1(\mbox{$\cal A$})$. We also  prove that algebras of Gelfand-Kirillov
dimension $1$ have only finitely many ends.
Finally, we study the case of group algebras.
\section{The Fredholm extension}
Let $K$ be a field and $K^\infty$ be the countable dimensional vector space
over $K$. Consider the endomorphism ring $\mbox{End}_K(K^\infty)$.
Let $I\subset \mbox{End}_K(K^\infty)$ be the ideal of endomorphisms of finite rank.
If $T\in I$ then let $\widetilde{T}$ be the restriction of $T$ onto the
range of $T$. Then $\widetilde{T}$ is a finite dimensional linear
transformation hence $Trace({\widetilde{T}})$ is well-defined.
If $T'$ is the restriction of $T$ onto a finite dimensional
space containing the range of $T$, then of course $Trace(\widetilde{T})=
Trace (T')$. We shall denote $ Trace(\widetilde{T})$ by $\mbox{Tr}\,(T)$.
Obviously, $\mbox{Tr}\,: I\rightarrow K$ is a linear functional.
\begin{lemma}
\label{l4}
If $A\in\mbox{End}_K(K^\infty)$ and $B\in I$, then $\mbox{Tr}\,(AB)=\mbox{Tr}\,(BA)$.
\end{lemma}
\smallskip\noindent{\it Proof.}  Let $\{e_i\}_{i\geq 1}$ be a basis of $K^\infty$ such that
$e_1,e_2,\dots, e_n$ span the range of $B$ and the span of
$\{e_1,e_2,\dots,e_m\}$ contains the range of $AB$, where $m>n$.
Let $Be_i=\sum^\infty_{j=1} b_{ij} e_j$, $Ae_i=\sum^\infty_{j=1} a_{ij} e_j$.
Then $b_{ij}=0$ if $j>n$ and $a_{ij}=0$, if $j>m$ and $i\leq n$.
Hence $a_{ij} b_{ji}=0$, if either $i>m$ or $j>m$. Thus
$$
\mbox{Tr}\,(AB)=\sum^m_{i=1}\sum^\infty_{j=1} a_{ij}b_{ji}=
\sum^m_{i=1}\sum^m_{j=1} a_{ij}b_{ji}=
\sum^m_{i=1}\sum^\infty_{j=1} a_{ij}b_{ji}=\mbox{Tr}\,(BA)\quad\qed
$$
Now let us suppose that $\mbox{$\cal A$}$ is a countable dimensional unital $K$-algebra.
For $x\in \mbox{$\cal A$}$, denote by $l_x$ (resp. $r_x$) the left (resp. right)
multiplications by $x$.
Obviously, if $T\in\mbox{End}_K(\mbox{$\cal A$})$ and $T$ commutes with $r_x$ for any $x\in\mbox{$\cal A$}$,
then $T= l_{T(1)}$. We can identify the set $\{l_x\}_{x\in\mbox{$\cal A$}}$ with $\mbox{$\cal A$}$
itself and the set $\{r_x\}_{x\in\mbox{$\cal A$}}$ with the opposite algebra $\mbox{$\cal A$}^o$.
Let
$$
\widetilde{\mbox{$\cal A$}}=\{T\in \mbox{End}_K(\mbox{$\cal A$})\,\mid\, Tr_x-r_xT\in I,\,\mbox{if}\,
x\in\mbox{$\cal A$}\}\, .
$$

Obviously $\mbox{$\cal A$}\subset\widetilde{A}, I\subset\widetilde{A}$ and
$\widetilde{A}$ is a unital subalgebra of $\mbox{End}_K(\mbox{$\cal A$})$.
\begin{definition}
The Fredholm extension of $\mbox{$\cal A$}$ is defined as $\mbox{Fred}\,(\mbox{$\cal A$})=\widetilde{\mbox{$\cal A$}}/I$.
\end{definition}
Note that
if for any $0\neq x\in\mbox{$\cal A$}$, $l_x$ is of infinite dimensional range then
$\pi:\mbox{$\cal A$}\rightarrow\mbox{Fred}\,(\mbox{$\cal A$})$ is injective. If $\mbox{$\cal A$}$ is the polynomial ring $K[x]$,
then $\mbox{Fred}\,(\mbox{$\cal A$})$ is just the Toeplitz-algebra.

\begin{proposition}
\label{p6}
Suppose that $x$ is a Fredholm element,that is both $\mbox{Ker}\, (l_x)$ and
$\mbox{Coker}\, (l_x)$ is finite dimensional, then $[x]$ is invertible in
$\mbox{Fred}\,(\mbox{$\cal A$})$.
\end{proposition}
\smallskip\noindent{\it Proof.}
First of all note that if $W\subseteq\mbox{$\cal A$}$ is a finite codimensional subspace
and $T\in\mbox{End}_K(\mbox{$\cal A$})$, then there exists
a finite codimensional subspace $W_T\subset W$ such that $T(W_T)\subseteq W$.
Indeed, consider
the linear map $\pi_W\circ T: W\rightarrow \mbox{$\cal A$}/W\,.$
Then $W_T=\mbox{Ker}\,(\pi_W\circ T)$ is clearly finite codimensional
and $T(W_T)\subseteq W$.
Now let $W^x\subset\mbox{$\cal A$}$ be a subspace, $\mbox{Ker}\, (l_x)\oplus W^x=\mbox{$\cal A$}$.
Then $ l_x\mid_{W^x}$ is injective. Consider
the finite codimensional subspace $l_x(W^x)$.
All elements in $l_x(W^x)$ can be uniquely written as
 $xw$,
where $w\in W^x$.
Let $\mbox{$\cal A$}=l_x(W^x)\oplus B$, where $B$ is finite dimensional.
Define $T\in\mbox{End}_K(\mbox{$\cal A$})$ to be $0$ on $B$ and $T(xw)=w$ on
$l_x(W^x)$.
Then $T\in\widetilde{\mbox{$\cal A$}}$. Indeed, let $s\in\mbox{$\cal A$}$ and let
$W^x_{r_s}\subset W^x$ be
a finite codimensional space as above, that is
 $r_s(W^x_{r_s})\subset W^x$. If
$w\in W^x_{r_s}$, then
$$Tr_s(xw)=T(xws)=ws\quad\mbox{and}\quad r_sT(xw)=ws\,.$$
Hence, $[T]$ is the inverse of $[x]$ in $\mbox{Fred}\,{\mbox{$\cal A$}}$\qed

Let $\mbox{$\cal A$}=V\oplus W$ be a decomposition of $\mbox{$\cal A$}$ into the direct sum of
two almost invariant subspaces, where $V$ is infinite dimensional
. A decomposition
$\mbox{$\cal A$}=V'\oplus W'$ is {\bf equivalent} to the decomposition above, if
$V$ and $V'$ respectively $W$ and $W'$ differ only by a finite dimensional
subspace. That is $(V+V')/V$, $(V+V')/V'$, $(W+W')/W$ and $(W+W')/W'$ are
all finite dimensional. We call the equivalence classes of such decompositions
{\bf the endclasses of the algebra $\mbox{$\cal A$}$}. Note that $\mbox{$\cal A$}=V\oplus W$ and
$\mbox{$\cal A$}= W\oplus V$ are determining different endclasses. Of course, if $V$ is an
almost invariant subspace it might not have a complementary almost invariant
subspace at all. The following example shows that in some pathological
situations an almost invariant subspace can represent more than one endclasses.
\begin{example}
$\mbox{$\cal A$}=\mbox{End}_K(K^\infty)$.
\end{example}
Let $\{e_i\}_{i=1}^\infty$ be a basis of $K^\infty$.
Let $A=\mbox{span}_K\,\{e_1, e_3, e_5, \dots\}, B=\mbox{span}_K\,\{e_2, e_4, e_6, \dots\},
C=\mbox{span}_K\,\{e_1+e_2, e_3+e_4, e_5+e_6,\dots\}$. For $X\subset K^\infty$,
let $$N_X=\{T\in \mbox{End}_K{K^\infty}\,\mid\, T(v)=0\,\mbox{for any $v\in
X$}\}\,.$$
Then clearly, $N_A, N_B$ and $N_C$ are pairwise inequivalent almost
invariant subspaces, and $\mbox{End}_K{K^\infty}=N_A\oplus N_B= N_A\oplus N_C=
N_B\oplus N_C\,.$

\begin{theorem}
The nonzero idempotents of $\mbox{Fred}\,(\mbox{$\cal A$})$ are in bijective correspondence
with the endclasses of $\mbox{$\cal A$}$.
\end{theorem}
\smallskip\noindent{\it Proof.}
Let $\mbox{$\cal A$}=V\oplus W$ be a decomposition of our algebra into the direct
sum of almost invariant subspaces. Let $\pi_V$ (resp. $\pi_W$) be the
projection
onto $V$ (resp. $W$). Then $\pi_V$ is an idempotent in $\mbox{$\cal A$}$. Indeed
if $x\in \mbox{$\cal A$}$, then by the observation in Proposition \ref{p6}, there exist
$R^V_x\subseteq V$, $R^W_x\subseteq W$ finite codimensional subspaces
such that $r_xR^V_x\subseteq V$ and $r_xR^W_x\subseteq W$.
Hence if $a\in R^V_x$, then $r_x\pi_V(a)=\pi_V(ax)=ax$,
$r_x\pi_W(a)=\pi_W(ax)=0$, thus $[\pi_V,r_x]\in I$.
If $A=V'\oplus W'$ an equivalent decomposition then clearly $[\pi_V]=
[\pi_{V'}]$ in $\mbox{Fred}\,(\mbox{$\cal A$})$.
Now, if $P\in \mbox{Fred}\,(\mbox{$\cal A$})$ is a non-zero idempotent, then let
$\widetilde{P}\in\widetilde{\mbox{$\cal A$}}$ be a representative of $P$ and
$\widetilde{Q}\in\widetilde{\mbox{$\cal A$}}$ be a representative of $1-P$.
It is easy to check that $\mbox{Ran}\,(\widetilde{P})$ and $\mbox{Ran}\,(\widetilde{Q})$
are almost invariant subspaces, such that $\mbox{Ran}\,(\widetilde{P})+
\mbox{Ran}\,(\widetilde{Q})$ is finite codimensional in $\mbox{$\cal A$}$ and
$\mbox{Ran}\,(\widetilde{P})\cap \mbox{Ran}\,(\widetilde{Q})$ is finite dimensional.
Hence there exist almost invariant subspaces $V,W$, such that
$\mbox{$\cal A$}=V\oplus W$ and $V$ is equivalent to $\mbox{Ran}\,(\widetilde{P})$,
$W$ is equivalent to $\mbox{Ran}\,(\widetilde{P})$, where $V$ is infinite dimensional.
Then $[\pi_V]=P$, $[\pi_W]= 1-P$, thus the Theorem follows. \qed

\section{The noncommutative geometry of the ends}

In this section we apply the Fredholm module construction of Connes (1994).
First
recall the notion of a graded differential algebra of length $1$.
Let $\mbox{$\cal A$}$ be a unital $K$-algebra and $\Omega$ be an $\mbox{$\cal A$}$-bimodule with
a $K$-linear map $\int:\Omega\rightarrow K$ and a $K$-homomorphism
$d:\mbox{$\cal A$}\rightarrow \Omega$ satisfying the following conditions
\begin{itemize}
\item $d(ab)=a\cdot db+da\cdot b$, if $a,b\in\mbox{$\cal A$}$.
\item $\int\, a\omega=\int\, \omega a$, if $a\in\mbox{$\cal A$}$,\,$\omega\in \Omega$.
\item $\int\,da=0$, if $a\in\mbox{$\cal A$}$.
\end{itemize}
Then $\tau:\mbox{$\cal A$}\otimes_K\mbox{$\cal A$}\rightarrow K$, $\tau(a,b)=\int a\cdot db$ defines a
cyclic cohomology of $\mbox{$\cal A$}$, $\tau\in HC^1(\mbox{$\cal A$})$.
Indeed, one can easily check that
$$\tau(a,b)=-\tau(b,a)$$
and
$$\tau(ab,c)-\tau(a,bc)+\tau(ca,b)=0\,.$$
Also, $\tau$ defines a character $\phi_\tau:GL_1(\mbox{$\cal A$})\rightarrow K$, by
 $\phi_\tau(a)=\tau(a^{-1},a)$.

\noindent
The classical example is of course the algebra of smooth complex valued
functions on the unit circle. Then,
$$\tau(f,g)=\frac{1}{2\pi i}\int f(z) dg(z)\quad.$$
 $\phi_\tau(f)=\frac{1}{2\pi i}\int \frac{f'(z)}{f(z)} dz$
is just the winding number of $f$.

\noindent
Now let $\mbox{$\cal A$}$ be a finitely generated $K$-algebra and let $\mbox{$\cal A$}=V\oplus W$ be
a decomposition of $\mbox{$\cal A$}$ into the direct sum of two infinite dimensional
almost invariant subspaces. Let $F$ be defined the following way,
$F(v)=v$, if $v\in V$, $F(w)=w$, if $w\in W$. That is $F\in\widetilde{\mbox{$\cal A$}}$
is an endomorphism such that $F^2=1$.
Then $(\mbox{$\cal A$},I,F)$ is a Fredholm-module, where $I\in\mbox{End}_K(\mbox{$\cal A$})$ is
the trace ideal. The coboundary operator $d:\mbox{$\cal A$}\rightarrow I$ is given by
$dx=[F,r_x]$, $\int:I\rightarrow K$ is defined as $\int \omega=\mbox{Tr}\,(\omega)$.

\noindent
Clearly, $d(xy)=dx\cdot y+x\cdot dy$ and $\int a\omega=\int \omega a$.
One should only check that
$\int da=0$, if $a\in\mbox{$\cal A$}$.
Pick a basis $\{e_i\}^\infty_{i=1}$ for $V$ and
a basis $\{f_i\}^\infty_{i=1}$ for $W$ such that the span
of $\{e_1,e_2,\dots,e_n,f_1,f_2,\dots, f_n\}$ contains the range
of $[F,r_a]$. Let $e_ia=v_i+w_i$, $f_ia=v'_i+w'_i$, where
$v_i,v'_i\in V$ and $w_i,w'_i\in W$. Then $[F,r_a]e_i=-2w_i$ and
$[F,r_a]f_i=2v_i$, hence $\mbox{Tr}\, [F,r_a]$ must be zero.
Therefore we associated a first cyclic cohomology class in $HC^1(\mbox{$\cal A$})$ to
the decomposition. Now if $\mbox{$\cal A$}=V'\oplus W'$ is an equivalent decomposition
then $F-F'\in I$. Thus,

$$\int a (Fb-bF)-\int a(F'b-bF')=\int (a(F-F')b-ab(F-F'))=\int
(ab-ba)(F-F')\,.$$
This shows that $\tau-\tau'$ is a cyclic coboundary. Hence we proved the
following theorem.
\begin{theorem}
If $\{A,I,F\}$ is a triple as above associated to a decomposition $\mbox{$\cal A$}=V\oplus
W$, then $p(a,b)=\mbox{Tr}\, (a[F,b])$ is a cyclic
cocycle. The cyclic cohomology element represented by $p$ in $HC^1(\mbox{$\cal A$})$
depends only on the endclass associated to the decomposition.
\end{theorem}

\noindent
Finally, let us show that at least in some cases one can obtain
non-trivial cyclic cohomologies.
Let $K[x,x^{-1}]=\mbox{$\cal A$}=V\oplus W$, where $V=\mbox{span}_K\,\{1,x,x^2,\dots\}$,
$W=\{x^{-1},x^{-2},\dots\}$.
Let us calculate $\phi_\tau(x)=\mbox{Tr}\,(x^{-1}[F,r_x])$.
Clearly $(Fr_x-r_xF) (x^k)=0$ if $k\neq -1$. On the other hand
$(Fr_x-r_xF) (x^{-1})=2$ thus $\mbox{Tr}\,(x^{-1}[F,r_x])=2.$ That is the
associated cyclic cohomology is non-trivial.
\section{Algebras of linear growth}
In this section $\mbox{$\cal A$}$ shall be a finitely generated $K$-algebra. Let $t_1,t_2,\dots t_n$
be a generator system of $\mbox{$\cal A$}$ containing the unit. Let $S_1$ be the vector
space spanned by $\{t_1,t_2,\dots t_n\}$. Set $S_2=S_1S_1, S_3=S_2S_1$,
inductively $S_{n+1}=S_nS_1$. Then $S_1\subset S_2\subset\dots$ are finite
dimensional vector spaces and $\cup^\infty_{n=1}S_n=\mbox{$\cal A$}$. Now we briefly
review some basic notions on the Gelfand-Kirillov dimension (see Krause and
Lenagan (2000)).
The growth function of $\mbox{$\cal A$}$, relative to the generator system is the function
given by $f(n):={\rm dim} (S_n)$, and the Gelfand-Kirillov dimension is
$$GK{\rm dim}(\mbox{$\cal A$})=\limsup\frac{\log(f(n))}{\log n}\,.$$
Note that the Gelfand-Kirillov dimension does not depend on the choice
of the generators. If $GK{\rm dim}(\mbox{$\cal A$})=1$, then by Bergman's theorem, $\mbox{$\cal A$}$ in
fact has linear growth, that is $f(n)\leq cn$, for some real number $c>0$.
Bergman also showed that the linear growth of an algebra
 is equivalent to the existence of a constant $K>0$, such that
$f(n+1)-f(n)<K$.

\noindent
By definition, if $V\subseteq A$ is an almost invariant subspace then there
exists $k_V>0$ such that $VS_1\subseteq V+S_{k_V}$.
\begin{lemma}
\label{luj1}
Let $\mbox{$\cal A$}=V\oplus W$ be a direct sum decomposition into almost invariant
subspaces, then for some $l>0$,
$S_n\subseteq (V\cap S_{n+l})\oplus (W\cap S_{n+l})$, holds for any $n>0$.
\end{lemma}
\smallskip\noindent{\it Proof.}
Let $k>0$ be an integer such that $VS_1\subseteq V+S_k$,
$WS_1\subseteq W+S_k$ and let $j>k$ be an integer such that
$S_k\subseteq (V\cap S_j)\oplus (W\cap S_j)$. If $z\in S_k$,
$z=v+w, v\in V\cap S_j, w\in W\cap S_j$, then $zt_i=vt_i+wt_i$, where
$vt_i\in (V\cap S_{j+1})+S_k$ and $t_iw\in (W\cap S_{j+1})+S_k$.
Hence $S_{k+1}\subseteq (V\cap S_{j+1})\oplus (V\cap S_{j+1})$. Inductively,
$S_{k+n}\subseteq (V\cap S_{j+n})\oplus (V\cap S_{j+n})$. \qed

\noindent
We call a subspace $V\subset \mbox{$\cal A$}$ {\bf sparse} if for infinitely many integer
$m$, $V\cap S_m=V\cap S_{m+1}$.
\begin{lemma}
\label{luj2}
If $\mbox{$\cal A$}$ is a finitely generated algebra and $V\subset\mbox{$\cal A$}$ is a sparse almost
invariant subspace, then $V$ is finite dimensional. \end{lemma}
\smallskip\noindent{\it Proof.}
Let $\mbox{$\cal A$}=V\oplus W$ as above , $m>j$ and $V\cap S_m=V\cap S_{m+1}$. Suppose
that $z\in S_k$, $z= v+ w$, $v\in V\cap S_m$, $w\in W\cap S_{m+1}$.
Then $zt_i\in (V\cap S_{m+1})\oplus (W\cap S_{m+2})=(V\cap S_m)\oplus
(W\cap S_{m+2}).$ Inductively, $S_n\subseteq (V\cap S_m)\oplus
(W\cap S_{m+n})$. Hence, $(V\cap S_m)\oplus W=\mbox{$\cal A$}$, thus $V\subset
S_m$.\qed

\begin{theorem}
If $\mbox{$\cal A$}$ is a finitely generated algebra of Gelfand-Kirillov dimension $1$, then
$\mbox{$\cal A$}$ has only finitely many ends. Also if the growth function satisfies
$f(n)\leq Cn$, then $\mbox{$\cal A$}$ has at most $C$ ends.
\end{theorem}
\smallskip\noindent{\it Proof.}
Suppose that for some $C>0$, ${\rm dim}_K S_n\leq C\,n$ and
 $\mbox{$\cal A$}=V_1\oplus V_2\oplus \dots V_k$ and $\{V_i\}_{i=1}^k$ are infinite
dimensional almost invariant subspaces. Then none of the $V_i$'s are
sparse. Hence, for some $m>0$, $V_i\cap(S_{m+1}\backslash S_m)\neq \emptyset$,
for all $1\leq i \leq k$. Thus ${\rm dim}_K\,S_n\geq k(n-m)$, therefore
$k\leq C$. \qed

\noindent
Groups of linear growth must have two ends, however it follows from Example 3.
that the number of ends of algebras of linear growth may reach any positive
integer. Groups of non-exponential growth have only finitely many ends,
nevertheless the algebra of Example 4. in the Introduction, has quadratic growth
and infinitely many ends.

\begin{question}
If $GK{\rm dim}(\mbox{$\cal A$})$ is finite and $\mbox{$\cal A$}$ is finitely presented, is it possible
that $\mbox{$\cal A$}$ has infinitely many ends ?
\end{question}

\section{The ends of group algebras}

Let $\Gamma$ be a group and $K$ is a commutative field.
If $V\subset\Gamma$ is an almost invariant subset then $KV\subset K\Gamma$
is an almost invariant subspace. Therefore if $\Gamma$ has at least $k$
ends, then $K\Gamma$ has at least $k$ ends as well. On the other hand let us
suppose that $\Gamma$ is a one-ended group, then $\Gamma\times C_2$ is
one-ended as well, where $C_2$ is the cyclic group of two elements. Then
$\frac{1+a} {2} $ is an idempotent in ${\mathbb C}(\Gamma\times C_2)$, where $a$ is the
generator of $C_2$. In general, complex group algebras of infinite groups with torsion
always has more than one ends. Of course finite groups have zero ends and
its group algebras are finite dimensional, so they have zero ends as well.

\noindent
According to the Idempotent Conjecture, if $\Gamma$ is torsion-free, then
${\mathbb C}\Gamma$ has no non-trivial idempotent. The following question can be viewed
as a version of the Idempotent Conjecture.
\begin{question}
If  $\Gamma$ is torsion-free and has only one end, is it true that ${\mathbb C}\Gamma$
has one end as well ?
\end{question}
Note that the answer is yes if and only if $\mbox{Fred}\,({\mathbb C}\Gamma)$ has no
non-trivial idempotent.
In the Abelian case we can answer the Question.

\begin{theorem}
If $\Gamma\simeq{\mathbb Z}^n$, $n\geq 2$, then the group algebra $K\Gamma$
has only one end.
\end{theorem}
We introduce some notations. If $P\in K{\mathbb Z}^n$ and $P=\sum_{\underline{v}\in {\mathbb Z}^n}k_{\underline{v}}
\underline{v}$, $k_{\underline{v}}\in K$, where $[\underline{v}]$ is the representative of the element
$\underline{v}\in {\mathbb Z}^n$ in $K{\mathbb Z}^n$, thus $[\underline{v}][\underline{w}]=[\underline{v}+
\underline{w}]$. Then let
$$T(P)=\{\underline{v}\in {\mathbb Z}^n\,\mid\,k_{\underline{v}}\neq 0\}\quad.$$
If $\underline{v}=(k_1,k_2,\dots,k_n)\in {\mathbb Z}^n$, then $s(\underline{v})=\sup_{1\leq j \leq n} k_j$,
$i(\underline{v})=\inf_{1\leq j \leq n} k_j$. If $P\in K{\mathbb Z}^n$, then
$s(P)=\sup_{\underline{v}\in T(P)} s(\underline{v}),$ \, $i(P)=\inf_{\underline{v}\in T(P)} i(\underline{v}).$
If $m\leq t$ are integers, then
$$A_{m,t}=\{(k_1,k_2,\dots,k_n)\in {\mathbb Z}^n\,\mid\, m\leq k_j\leq t,\,
\mbox{if}\, 1\leq j \leq n\}\quad,\,A_m=\cup^\infty_{t=m} A_{m,t}\,.$$
The leading term of $P\in {\mathbb Z}^n$, $l(P)$ is the greatest element of $T(P)$ in the
lexicographic ordering.
Furthermore, let $x_1=[(1,0,\dots,0)], x_2=[(0,1,0\dots,0)],\dots,
x_n=[(0,0,\dots,0,1)]\in K{\mathbb Z}^n$.
\begin{lemma}
Let $V\subset K{\mathbb Z}^n$ be an infinite dimensional almost invariant subspace.
Then if $d\in{\mathbb N}$ is large enough, there exists $Q\in V$ such that
$l(Q)=(d,d,\dots,d)$.
\end{lemma}
\smallskip\noindent{\it Proof.}
By definition, if $V$ is an infinite dimensional almost invariant
subspace of $K{\mathbb Z}^n$, then there exists a finite set
$\{\underline{b}_1, \underline{b}_2,\dots,\underline{b}_s\}$ such
that for any $R\in V, Rx_i= R'+L$, where
$R'\in V, L\in\mbox{span}_K\, \{\underline{b}_1,
\underline{b}_2,\dots,\underline{b}_s\}$.
Let $i(b)=\inf_{1\leq j \leq s} i(\underline{b}_j),
s(b)=\sup_{1\leq j \leq s} s(\underline{b}_j).$ so the symmetric
difference of $T(R')$ and $T(R)+(0,0,\dots,1,\dots,0)$ is in the hypercube
$A_{i(b),s(b)}.$ Naively speaking we have control on the terms only
outside this hypercube.
Start with an element $P_1\in V$ with a term $\underline{v}^1\notin A_{i(b),s(b)}\,.$
Then there exists $P_2\in V$ with a term $\underline{v}^2\notin A_{i(b),s(b)}\,$ and
recursively we can obtain $P_l\in V$ with a term $\underline{v}^l$, such that
$\underline{v}^l>
\underline{b}_j$ in the lexicographic ordering, for any $1\leq j \leq s$.
Then repeating the process, we can obtain $Q\in V$, $l(Q)=(d,d,\dots,d)$,
$s(b)<d$.
\qed

\noindent
Now  fix an integer
$d>0$ such that for some $Q\in V$, $l(Q)=(d,d,\dots,d)$ and
$s(b)<d$. We also fix an integer
$m\leq 0$ such that $m<i(b)$ and
$m\leq\inf_{\underline{v}\in T(Q)} i(\underline{v})$, for the same $Q\in V$.
Let $r=s(Q)-d$. Then $Qx_1=Q'+L'$, where $Q'\in V$ and $L'\in
\mbox{span}_K\, \{\underline{b}_1,
\underline{b}_2,\dots,\underline{b}_s\}\,.$
 Obviously, $l(Q')=(d+1,d,\dots,d)$
and $s(Q')-(d+1)=r$. By induction, we can obtain the following lemma.
\begin{lemma}
For any $\underline{v}\in{\mathbb Z}^n$ such that $i(\underline{v})\geq d$, there exists $Q\in V$
such that $l(Q)=\underline{v}$, $T(Q)\subset A_m$ and $s(Q)-s(\underline{v})=r$.
\end{lemma}
\begin{lemma}
$$\lim_{t\rightarrow\infty}
\frac{{\rm dim}_K(V\cap \mbox{span}_K\,(A_{m,t}))}{|A_{m,t}|}=1$$
\end{lemma}
\smallskip\noindent{\it Proof.}
First note that if $P_1,P_2,\dots,P_l\subset K{\mathbb Z}^n$ have different
leading terms then they form an independent system over $K$. By the previous
lemma, if $t$ is large enough then for any $\underline{v}$, $d\leq i(\underline{v})\leq t-r$,
there exists $Q\in V$ such that $l(Q)=\underline{v}$ and $T(Q)\subset A_{m,t}$.
Hence,
$$\frac{{\rm dim}_K(V\cap \mbox{span}_K\,(A_{m,t}))}{|A_{m,t}|}\geq
\frac{|A_d\cap A_{m,t-r}|}{|A_{m,t}|}\,.$$
However, $\lim_{t\rightarrow\infty} \frac{|A_d\cap A_{m,t-r}|}{|A_{m,t}|}=1$\qed

\noindent
Now the Theorem follows. Indeed, by the previous lemma, any two infinite
dimensional almost invariant subspaces have non-zero intersections. \qed

\section{Some questions on domains}

Among the intensively studied ``geometric'' properties of groups, one
has the growth, amenability and ends (see Gromov (1993)).
As mentioned in the course of this paper, these concepts are
closely related. The three concepts have been introduced in the
context of algebras (See Elek (2003) and
Samet (2000) for discussions on the amenability of algebras).
Groups of polynomial growth has
integer degree of growth. Algebras of polynomial growth may have
any real number greater or equal two as Gelfand-Kirillov dimension
(Krause-Lenagan (2000)).
 It is an old problem, whether all
 domains of polynomial growth have integer Gelfand-Kirillov
dimension.

\noindent
By Hopf Theorem, a group may have zero, one, two or infinitely many ends.
All the domains we can check has zero, one, two or
infinitely many ends. So we can ask a similar question.
\begin{question}
Is it true, that a domain may have only zero, one, two or infinitely many
\\ ends ?
\end{question}

\noindent
Finally, it is well known, that amenable groups have finitely many
ends. The algebra in Example 3. has quadratic growth and so it is amenable.
On the other hand, it has infinitely many ends. Again, we can ask about
the similarity of domains to groups.

\begin{question}
Is it true, that any amenable domain has only finitely many ends ?
\end{question}

\vskip0.2in
\noindent
{\large{\bf References}}
\vskip0.1in \noindent
Connes, A. (1994). {\it Noncommutative Geometry.}
Academic Press Inc: San Diego.
\vskip0.1in \noindent
Elek, G. (2003). {\it The amenability of affine algebras.}
J. Algebra 264:469-478.
\vskip0.1in \noindent
Epstein, D. (1962). {\it Ends}.
In: {\it Topology of 3-manifolds and related results},
Ford M. Ed., Prentice Hall, pp. 110-117.
\vskip0.1in \noindent
Freudenthal, H. (1945). {\it \"Uber die Enden diskreter R\"aume und Gruppen}
Comment. Math. Helv. 17:1-38.
\vskip0.1in \noindent
Gromov, M. (1993). {\it Asymptotic invariants of infinite groups.}
in: {\it Geometric Group Theory}, Graham A. Niblo and Martin A. Roller,
Eds., Vol. 2, London Mathematical Society Lecture Note Series 182,
Cambridge Univ. Press, Cambridge, UK, pp. 1-295.
\vskip0.1in \noindent
Hopf, H. (1944). {\it Enden offener R\"aume und unendliche diskontinuierliche
Gruppen.} Comment. Math. Helv. 16:81-100.
\vskip0.1in \noindent
Houghton, C. (1972). {\it Ends of groups and the associated first cohomology
group.} J. London Math. Soc. 6:81-92.
\vskip0.1in \noindent
Krause, G., Lenagan, T.(2000) {\it Growth of algebras and the Gelfand-Kirillov
dimension} (Revised edition), Graduate Studies in Mathematics, 22,
American Mathematical Society: Providence, RI.
\vskip0.1in \noindent
Samet-Vaillant, A. Y. (2000) {\it $C^*$-algebras, Gelfand-Kirillov dimension
and Folner sets.} J. Funct. Anal. 171:346-365.

\end{document}

redpajama_set_name: RedPajamaArXiv


Ronald Davis, author of The Gift of Dyslexia , has created a list of thirty-seven common traits and behaviors (symptoms) of people who have dyslexia.
If a person possesses nine or ten of these traits or symptoms of dyslexia, the Dyslexia Correction ®Program may be helpful in correcting the problems associated with dyslexia.
Appears bright, highly intelligent, and articulate, but unable to read, write, or spell at grade level.
Isn't "behind enough" or "bad enough" to receive extra help in the school setting.
Feels dumb; has poor self esteem; hides or covers up weaknesses with ingenious compensatory strategies; easily frustrated and emotional about school reading or testing.
Talented in art, drama, music, sports, mechanics, story telling, sales, business, designing, building, or engineering.
Seems to "zone out" or daydream often; gets "lost" easily or loses track of time.
Learns best through hands on experience, demonstrations, experimentation, observation, and visual aids.
Complains of dizziness, headaches, or stomach aches while reading.
Reading or writing shows repetitions, additions, transpositions, omissions, substitutions, and reversals in letters, numbers, and/or words.
Complains of feeling or seeing non existent movement while reading, writing, or copying.
Extremely keen-sighted and observant, or lacks depth perception and peripheral vision.
long words, or transposes phrases, words, and syllables when speaking.
Excellent long term memory for experiences, locations, and faces.
Poor memory for sequences, facts, and information that has not been experienced.
Thinks primarily with images and feeling, not sounds or words (little internal dialogue.
Can be class clown, troublemaker, or too quiet.
Had unusually early or late developmental stages (talking, crawling, walking, tying shoes, etc.).


« Sukkot "Booths" also The birth of Yeshua?
If You Don't Want the Truth Don't Read This!
We've heard it said that you 'can't see the forest for the trees'. That may well be the case for us here in Israel. We are surrounded by bloodthirsty Islamic enemies, who live to kill Jews and Christians. We may well be blinded by the trees while the forest is on fire all around us. Survival and self preservation is a very strong motivation, so strong at times that little matters outside of your immediate situation. The problem with that is when you do take your head out of the hole and look around you see a world on fire.
I woke this morning with a clear vision of things I hadn't associated with before because my calling is to rescue European Jews – as many as possible. The Holy Spirit (Ruach Hakodesh) begin to show me how the coming USD collapse, Ebola and the Fema camps around the US are all tied together and why, or how, Obama can go about his business or golfing with that evil smile on his face. He knows what is about to happen and knows there is nothing he can do about it. In fact he is part of the problem.
So let me put it in a nutshell and not elaborate on each aspect of it. Ebola is being deliberately introduced into the US and it is far more dangerous than is being reported. In fact it can't be stopped once it starts spreading. The Fema Camps should now be called 'Ebola Camps', and in order to fill them it will take martial law and it is right around the corner. Sure, I know how insane this sounds – but you can either believe it now or wait until it is too late for you, and be forced into one of these camps with some very sick people. It will begin with anyone who has a cold, flu or any fever. Or if they even think you have been exposed to Ebola. You will be put into the camps with others who have Ebola and the pandemic will accelerate at a very rapid pace.
The Camps will be easy to fill with millions because of the collapse of the dollar and the riots will be out of control across the US. Most of the folks arrested and thrown into these camps will not be sick – at least not at first – but will have no protection against the ones who are sick.
And for those who do not believe there are Fema Camps all over the US – new and sitting empty but staffed – just run a search on Google. Most of this letter looks like I am talking about the US only, but in fact it is the entire world that is on fire, but when the dollar falls and America goes completely crazy trying to find food and water, the rest of the world will follow suit.
Sure, we are in serious trouble here in Israel, but what's new in that? The God's eyes are on Israel He loves Israel above all the nations and will bless those who bless Israel. I am 100% certain Israel will survive but I cannot say that for the US or any other country. For those God has joined to this calling to save Jewish lives I feel a deep love and compassion – and want to say or do whatever I can to wake you up and help you survive.
Obama knows he has to finish what he has been called to do, and that is destroying the USA before he leaves office. He knows how much he is hated by so many but he also knows he has very little time before the dollar collapses. He is putting many things into place around the US and the Fema Camps are just one of them.
I began this by saying Israel can't see the forest for the trees. Well, I have just explained your trees and forest, but if you allow even that to take your eyes off Yeshua and His calling on your life, you lose. And that loss is for eternity.
God called this ministry to save Jewish lives. We will do just that. How many depends on the obedience of the believers He has joined together with this ministry – and their love for Israel and God's Word.
I am well aware that many will read this and say goodbye – but before you go please think about your family and loved ones. Do what you can, while you can, to prepare for them. Take a moment and read this before you go. I-Tim. 5:8 But if any provide not for his own, and specially for those of his own house, he hath denied the faith, and is worse than an infidel.
If God called you to this ministry we need to hear from you for time is so very important.


Former Manchester United captain Gary Neville believes Manchester City have the best player in the Premier League - but it's not Sergio Aguero.
If you're familiar with Neville's work on Sky Sports, you could probably guess that he would have named David Silva.
Neville can often be heard eulogising over the Spain international when working, either as a co-commentator or pundit, because of the expert way in which the 29-year-old is able to control matches.
Neville was the special guest at a charity event last night and revealed his thoughts to fellow Sky Sports colleague Guillem Balague live on stage.
"The best player in the Premier League is David Silva," Neville was quoted as saying by Football Beyond Borders, the charity Thursday night's event was in aid of.
"He's wearing the wrong colour shirt but it's him," the current England coach added, according to Full Time Devils, who attended the event.
Alexis, Hazard and Aguero better than Silva?
While Silva is certainly one of the most technically-gifted players in the Premier League, is he really the best?
There aren't too many times when football fans disagree with Neville, the best pundit on television, but this could be one of them.
Arsenal fans would probably argue that Alexis Sanchez is a better player while Chelsea fans would probably say that Eden Hazard - the PFA's current Premier League Player of the Year - is also superior.
Is Gary Neville right to call David Silva the best player in the Premier League? Have your say by leaving a comment in the box below!


Home / Celebrities / Brock Lesnar: Steroids or Natural?
Does Terry Crews Take Steroids?
WWE Superstar and former American football player Brock Lesnar is super buff. Does the wrestler juice or is he all natural? Let's find out.
No matter where you support him, the man-mountain that is Brock Lesnar delivers. During his football run with the Minnesota Vikings, he achieved a range of impressive scores. The first of these would be him bench pressing 225 pounds for 30 reps. He followed this up by running a 40-yard dash in 4.65 seconds. And finally, he had the vertical jump that reached 35 inches. All of these are impressive results for this six feet three tall, 286 pounds massive beast.
Even during his Vikings days, Lesnar was a force of nature. Someone you'd not want to irritate or try to challenge. But it's his presence at the WWE that confirmed these words as fact. As a wrestler, Brock Lesnar managed to beat both John Cena and The Rock. That earned him the championship at 25, making him, at the time, one of the youngest champions in history. Naturally, it was only uphill from there for Lesnar. When he started in MMA, he beat Min Soo Kim. Later matches would see him face Frank Mir and Randy Couture. Defeating these gentlemen earned him the title of UFC heavyweight champion.
No matter how you slice it, Brock Lesnar is a massive force to be reckoned. However, his physique and track record got him in hot water, as people kept on wondering if Lesnar took steroids. Is his impressive frame natural, or did the big giant get some substance help? Join us as we try to find out.
Now, this in and of itself is not proof of anything, but it's a great start. When asked about the topic in 2016, Brock Lesnar responded with 'I'm white, and I'm big. Deal with it.' This statement he followed up with 'What do you want me to say?' And really, what can an athlete say in this situation? Admitting to using an enhancing substance means the death of your sports career. So, at a bare minimum, you can understand Lesnar's lack of concrete response.
However, this is also a telling sign. Lesnar never outright said 'I'm not using enhancements' or 'I've never used steroids before.' The words 'what do you want me to say' are just as telling. These are words that come out of the mouth of someone under a lot of pressure. Or, potentially, someone hiding something important. In other words, him replying with 'what do you want me to say' is halfway to an admission itself.
But as we stated in the first sentence of this section, this is hardly concrete evidence. For that, we have to dig a bit more deeply.
One cursory glance at the man tells you that his muscles don't LOOK natural. Brock Lesnar is stacked, yes, but he looks hugely bloated. It's as if his body is trying to retain a lot of water, which is the case with most steroid users. His gut matches the guts of athletes on Human Growth Hormones, or HGH. Furthermore, there are his cobra-like trapezii (or 'traps' – no, not the anime kind, the kind us Bulk Hulks flex).
Another, more evident sign of steroid use, is Lesnar's skin tone. While he might be 'white and jacked,' as he puts it, he usually looks pink-to-reddish. This effect is the result of high bodily temperature. The man might be too hot to handle. Using steroids can spike up your blood pressure and body heat, making you both red in appearance and far sweatier than usual. Lesnar's sweat level is also pretty obvious from his photos alone.
But again, these might not be evidence in and of themselves. So let's dig even further and see the timeline of Brock Lesnar's growth.
Their transformation is probably the best way to judge someone's physique and spot potential steroid use. Brock Lesnar is a Minnesotan child of a farmer. If any of you readers have happened to ever work on a farm, you know very well what that means for your body.
Farm work is intense, requiring a lot of work with your whole body. Naturally, you would be physically fitter than your non-farmer peers, and this is true for Lesnar. One look at the picture of a 15-year-old Brock says it all – that's a robust and healthy adolescent boy.
But just one year later, at 16, Brock Lesnar is 80lbs heavier. Eighty pounds! In a year! That cannot be achieved by farm work alone. Or intense training, for that matter. No, this is probably a combination of steroids and a fierce increase in food intake.
Someone with a naturally bulkier build would be more significant than Lesnar at 15, but would also grow into a bigger man gradually. Lesnar, on the other hand, gained an insane amount of weight in a short period. Besides, him juicing as a teenager would make sense, as most teenagers set on having that perfect body start at around the age of 15.
However, there is more concrete, more damning evidence of Brock Lesnar roiding it up.
In the mid-summer of 2016, Brock Lesnar fought and won against Mark Hunt. However, he failed no less than two drug tests at the time, one before and one after the fight with Hunt. The substances he used were anti-estrogens clomiphene and hydroxy-clomiphene. You would typically use these after a steroid cycle, mainly to get your testosterone production back to normal. On a side-note, Lesnar wound up paying a $250.000 fine for this, followed by a two-year-long suspension from wrestling.
Fifteen years before this, however, Lesnar was arrested for allegedly possessing steroids. Four months after this he was let go because what he had were growth hormones, which don't count as steroids. That might not seem like evidence that he used steroids, but think about it – if he's not opposed to using growth hormones, what's stopping him from using actual steroids?
Both direct and circumstantial evidence point to Brock Lesnar using steroids, and that he has done so since at least early high school. Well, imagine my shock, as Paul Joseph Watson of Infowars tends to say. What makes us confident of this verdict, though? Well, four particular elements. First off, he has apparent visual symptoms of a steroid user, with his gut, skin colour, and face. Next, when you look at his timeline, all signs point to him using enhancement substances as a teenager. Furthermore, the man himself never denied taking steroids in the past. Lastly, Lesnar got caught and fined for using anti-estrogens.
Yeah, there's no dodging this – Brock Lesnar probably used steroids.
As stated earlier, Brock Lesnar has what seems to be HGH gut. Naturally, this means he used growth hormones, which got him into police custody. Other telltale signs of HGH use are the head size, as well as nose growth – both likely with Lesnar. HGH promotes tissue and bone growth, so using it will give you a more massive noggin.
His massive traps you can explain away with a combined intake of HGH and insulin. Judging by the white mountain's HGH history, this is possibly the case.
But his massive gains probably didn't come about with HGH and insulin alone. For this, Lesnar had to have used bulking steroids, and several fit that bill. Those are Dianabol, Deca Durabolin, and Anadrol. Both Dianabol and Deca Durabolin would give Lesnar the thick, full-looking musculature he has. But Lesnar appears to be retaining a lot of water. With that in mind, Anadrol would be fit, as it is known to build mass quickly and keep the excess water.
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text:
7 Ways to "Cheat" Writing Inspiration
by Jeff Vasishta
Waiting for inspiration to strike could take a lifetime. Here are 7 ways to jumpstart your writing without having to wait around.
Songwriters are known for using personal trauma as the catalyst for hit tunes. The Police classic "King Of Pain" was written by front man Sting about his long and messy divorce from his first wife. Eric Clapton's "Tears In Heaven" was written about the death of his toddler son. But structuring a three or four minute song around a moment of grief is a lot easier that crafting three hundred page novel or even a short story. That's not to be said that it can't be done. It often has been. Author, Akhil Sharma's gut wrenching Family Life, which was basically the thinly fictionalized story of his own family's tragedy, when a freak accident left his brother severely brain-damaged, and the other lost and virtually orphaned in a strange land, is one recent novel that springs to mind. But it took Sharma 12 often painful, soul-searching years to write. Most people, Sharma included, don't want to take that long to write a novel, and wallowing in anguish in order to create is not recommended on an ongoing basis.
If there's something in your personal life that inspires you to write, that's wonderful and you should go for it. But what about the times when inspiration doesn't strike? Are you going sit around and wait? The good news is that you don't have to. There are several ways to "cheat" inspiration—a kind of artificial insemination which can still give birth to a wonderful story and keep you writing in the process. We'll take a look at seven of the most popular ways, which many of the authors I've interviewed over the years have told me they use.
1. Local Newspapers
Notice I've mentioned local newspapers here as opposed to national ones. That's because the kind of intimate, folksy stories that often make the best fiction can be found in smaller papers. I'm not saying you should completely hijack a story and use it as your own but rather use it as a springboard for your imagination to take over. You might read about an immigrant family surviving extreme hardship to send their youngest son/daughter to an Ivy League school on a scholarship. What about if one of the family members had to do something which was a little shady to get the money for private tutoring or to pay the fees? A father or older sibling worked for a criminal, the mother found out something incriminating about an employer and extorted money? Now, as their child stands on the precipice of a great career which will transform all their lives, payback comes along and threatens to ruin it all?
Just out of interest when writing this article I went online to my local newspaper and read the following headline: "'It was a bad scene' judge says of severed hand home invasion". I'm already thinking of possible scenarios! The great thing about a local newspaper is that it is filled with details that can be further researched and ones you will probably be familiar with. That's where the authenticity and credibility comes in—filling in those fine shadings with information only you and residents of your local town could know about—the school math teacher who used to run a New York modeling agency, the Indian restaurant owner who's sister was raped and killed by an abusive husband she was trying to leave.
If you're a budding Tom Clancy or John Le Carré national newspapers may provide inspiration. You would already have the kind of background that could read a story about an arms deal or diplomats or spies being expelled and use your imagination to create something brilliant. But that's not most of us. Rather, I think we all know the kind of small town characters that appeared in Elizabeth Strout's brilliant new novel, Anything Is Possible. What she was able to do was insert the darkness, pain and secrecy under the surface, of seemingly everyday characters, which made her novel so compelling.
2. Family & Friends
I once listened to a lecture during my MFA which argued that most stories in fiction can be traced back to famous fairy tales, which themselves can be traced back to stories thousands of years old. There's a lot of validity in it. Revenge, lust, betrayal, escape etc. are all themes which can be found in fairy tales or Greek mythology or even many religious texts. Basically, when it comes to stories, to use a cliché, there's nothing new under the sun. What is new, and what makes everyone's story unique, are the characters and intimacy of the world they inhabit.
Where better to find that intimacy than your own family?
There are countless books that have been inspired by family events. I already mentioned Akhil Sharma's which was specifically about his own family. I'll also mention Ann Patchett's engrossing Commonwealth as a recent novel which was inspired by her growing up in a blended family. Your extended family is also a great resource. They may know about your ancestors and may even own artifacts which can help in your research. Next Thanksgiving or family cook-out, get chatting to your relatives and ask them what's going on in their lives—just don't let them see you taking notes!
3. Famous Pop Culture Events From The Past
When I was first sent a galley of Marlon James' 2015 Booker Prize winning novel, A Brief History of Seven Killings, I read the accompanying press release which stated that it was centered around the attempted assassination of Bob Marley in 1976. I was immediately interested without even reading a word of the text. There was something magical about '70's Jamaica when Marley was at his peak. The era was sexy, dangerous, and filled with great music. When I read more about it and saw that there were cameos from Mick Jagger, a CIA operative, drug lords, a Rolling Stone Magazine reporter, and the action switched from Kingston to New York, I couldn't wait to get started. It turned out I wasn't alone in wanting to write about it. Almost every magazine and newspaper featured James and his novel. He'd intrigued us all and in doing so gave the novel a head of steam that few other books that year could match.
The bad news is that Marley is now taken. The good news is that there's no shortage of pop culture events that ring the nostalgia bells in thousands of readers. They can incorporate film, TV, news, and sporting events—anything that helps frame the time period of your novel and inspires both you and thus your readers to delve deeper into that world.
Jane Anne Phillips wove fiction in her 2013 novel, Quiet Dell which was centered around the actual brutal slaying of a family in the Chicago suburbs in the 1950's. Jeffrey Renard Allen's Song Of The Shank: A Novel (2014), was a fictionalized tale about the real life, largely unknown blind piano prodigy and autistic savant Thomas Wiggins (Blind Tom). He was born into slavery in 1849 and went on the play at the White House and in concert halls in Europe. Ryan Gattis' All Involved (2015) which depicted Latino gang rivalry was set around the 1992 LA Riots.
4. Reading Book Summaries And Synopses
I've often attended writing workshops when agents have asked authors to describe their novel in three sentences or less. Sometimes I'll browse through websites and see how some of the most successful novels can be broken down to its core elements, whether in a few sentences, or in one page synopsis. This is not only a great exercise for writers see if their work can stand up to being stripped down but can be incredibly inspiring for other writers who are trying to create compelling plots.
5. Getting Out of Your Comfort Zone
For many years I was a music journalist and record labels would pay for me to fly around the world to interview their artists. I know, tough life! I used to love those trips, not simply for the travel, fancy hotel and free food but because I was always filled with ideas when it came to writing the story. It made a huge difference compared to the customary phone interview or sterile record label conference room setting. I would often find myself incorporating snippets of conversations with a taxi driver, descriptions of the city we were in, gossipy details from backstage at the show. When you are out of your normal surroundings your antennae are raised and you're naturally more inspired to create. You don't have to fly half way around the world, just take a trip to somewhere you've never been and you see what a difference it makes.
If you're struggling to come up with fully formed characters, why not borrow some? Fan fiction, where authors write their own spin offs/prequels/sequels of already popular novels or films, is probably the easiest way to get writing. You already know the characters and what their personalities are like. What a fan fiction writer has to do is create a new scenario for them and see how they'll react based on what you already know. Theoretically, you have a potential audience hun-gry to take a look. Don't think that fan fiction is just for amateurs. Fifty Shades of Grey, one of the best selling novels of all time, started out that way as an homage to the Twilight vampire novels. It first surfaced on a small website with no agent attached and then proceeded to hit the stratosphere.
There are dozens of websites that publish fan fiction. One of the best known is wattpad.com, which carries user written fiction about absolutely everything you can think of—celebrities, video game characters, YouTubers. Its latest trend is fiction in the form of text messages known as chat stories. Sure, it's not Nabokov but it's fiction, easy to write and can be oddly addictive. Stories that have appeared on the site, which is free to use, have since gone on to get major book deals become TV series and major movies.
7. The Day Job
Most people want to become writers so they can quit their day job but the day job is where a writer might receive their greatest inspiration. Anytime a disparate group of people are brought together, placed under stress, and forced to react with one another, you are bound to get something you can use in your writing. Even the most banal and boring jobs can produce great drama. Case in point, "The Office". Ricky Gervais used his experience of working mind numbing office jobs to create some of the most memorable television ever made because so many people could relate. We all know what it's like to work for an annoying, egomaniac of a boss, so how much fun must it be to satirize them?
A year ago I was given a writing assignment that required me to work as an in-house writer for a real estate publication. At first I was unsure about the corporate nature of the job but it was decent money to put words on a page and most writers will tell you that can be a rarity. What a wellspring of inspiration it turned out to be. I sat next to a good-looking Chinese/American millennial lady going through relationship drama. Across the room was an aging hipster who seemed to spend as much time on his Tinder account as his work computer and my supervisor was a cool, gay Latina who used to play soccer for Venezuela. There was also a single Russian mother who was an avid kick boxer and blasted thrash metal on her headphones. I'm sure if they were all writers they'd have a suitably apt description for me too. Spacey Anglo-Indian writer who became morbidly depressed when his soccer team lost! So now you know, the next time an agent reads your manuscript and succinctly tells you not to quit your day job, it might be the best advice you've ever received.
If variety is the spice of life, Jeff Vasishta's writing career is a vindaloo. As a music journalist with Rolling Stone, Yahoo, and Billboard, he's interviewed Prince, Beyonce, and Quincy Jones. Man Booker Prize winner Marlon James, along with Richard Price and Ann Patchett have been quizzed by him for Interview Magazine and he also writes about gentrification for amNEWYORK.
Do you have what it takes to be a writer? Show us a sample of your work here.
Tiffany Dickinson
Great article and good ideas. I particularly appreciate the idea of testing whether your book can "stand up to being stripped down." Also, the local newspaper can be a gold mine. In our small city in the Pacific Northwest, I am often amazed at the background stories and undercurrents that pop up on the page. Thanks.

redpajama_set_name: RedPajamaCommonCrawl


There's still plenty of autumn left, but you can already feel the holiday season approaching.
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More specifically: data they collected during previous years.
Looking at past campaigns to see how well they performed is not only the easiest but also the most crucial step in your marketing strategy. Thoroughly analyzing it is vital for better executing your campaigns in 2017.
In fact, that sort of data might be the biggest advantage established businesses have against new ones.
Kevin George is the head of marketing at EmailMonks, an email design and coding company that specializes in crafting beautiful email templates, PSD-to-HTML email conversion, and free master email templates.


Updated 05/16/12 – 8:43 p.m.
CHICAGO (CBS) — A Cook County judge has released the video of convicted killer William Balfour being questioned by police in the murders of Jennifer Hudson's family members.
Balfour was convicted of first-degree murder last week in the slayings of Hudson's mother, brother and nephew.
CBS 2's Dana Kozlov reports the video of Balfour's interview with police was a key piece of evidence for the prosecution.
The interview with Balfour took place the night Hudson's family was killed on Oct. 24, 2008. Balfour allegedly shot and killed 57-year-old Darnell Donerson, 29-year-old Jason Hudson, and 7-year-old Julian King, because he was jealous his then-estranged wife Julia Hudson was seeing another man.
Police took Balfour into custody just hours after the murders, when Julian King was still missing.
Chicago police detectives didn't waste time, taking Balfour's picture and collecting his clothes as possible evidence, right after taking him into an interrogation room.
"Why am I here," Balfour asked detectives.
"You're here because your mother-in-law – your ex-mother-in-law and ex-brother-in-law were found dead," a detective told him.
"What that got to do with me being here?" Balfour asked.
"That's what we're here to find out," a detective said.
Balfour was still married to Julia Hudson at the time, although they were estranged. Police brought Balfour in for questioning four hours after the bodies of Hudson's mother and brother were found, but while Julian King was still missing.
"What I'm interested in is this kid. Where's the kid?" a detective asked Balfour.
"Ok. Here we go with this. I don't know where he is. I don't," Balfour said.
Balfour was initially pretty calm. He didn't ask for a lawyer and, early on, suggested Jason Hudson's drug dealing was the cause of the murders.
Detectives continued questioning Balfour and, after about an hour, began zeroing in on some apparent gaps in Balfour's story.
"She also says she saw you with a .45 … with a gun, a .45," one detective said.
"A couple months back, and she told you to put it up under the bed, and she never saw it again after you left," the other detective told Balfour.
"She ain't never seen me with no gun," Balfour said.
Balfour allegedly stole that gun from Jason Hudson and used it to kill all three victims.
Balfour was charged with the crimes more than a month later, but never left custody. He was kept in jail on a probation violation until he was charged, and was then ordered held without bail until his trial last month.
He's been convicted of first-degree murder in all three deaths and faces a mandatory life sentence.