\documentclass{amsart}
\usepackage{color}
\usepackage{amssymb}
\usepackage{amsfonts}
\usepackage{enumerate}
\usepackage{eucal}
%\usepackage[all]{xy}

\newcommand{\Fam}[1]{\widetilde{\mathcal{#1}}}
\newcommand{\Comp}[1]{\mathcal{#1}}
\newcommand{\jamfam}[2]{\Fam{#1}\!\rtimes\Fam{#2}}
\newcommand{\onto}{\twoheadrightarrow}

\DeclareMathOperator{\ext}{ext}
\DeclareMathOperator{\Ext}{Ext}
\DeclareMathOperator{\Hom}{Hom}
\DeclareMathOperator{\End}{End}


\newtheorem{theorem}{Theorem}
\newtheorem{theoremdef}[theorem]{Theorem-Definiton}
\newtheorem{definition}[theorem]{Definition}

\setlength{\parskip}{1ex plus 0.2ex minus 0.1ex}

\begin{document}

$\Fam{F}$, $\Comp{G}$

\section{Notational Conventions}

Objects of an Abelian category are denoted by capital latin letters:
$F$.

Components of parameterizing spaces are written $\Comp{F}$.

Families carried by such are written $\Fam{F}$.

\section{A Lemma}

Throughout let \(\Lambda\) be a $k$-algebra.  Denote by \(M(\Lambda,
d)\) the moduli space of \(d\) dimensional right \(\Lambda\) modules.

\begin{theoremdef}
Let \(\Comp{F}\) and \(\Comp{G}\) be components of \(M(\Lambda, f)\)
and \(M(\Lambda, g)\) respectively.  Further, suppose that
\(\ext(\Fam{F}, \Fam{G}) = 0\).  Then there exists a unique component
of \(M(\Lambda, f + g)\), \(\Comp{F\!\rtimes\! G}\) such that the
general member of \(\jamfam{F}{G}\) of extensions of objects of
type \(\Fam{F}\) by objects of type \(\Fam{G}\) and the objects of
type \(\Fam{F}\) that occur as subobjects and the objects of type
\(\Fam{G}\) that occur as factors are general.
\end{theoremdef}
\begin{proof}
Let \(\Comp{U} = \{(p,q)\in\Comp{F\times G} : \ext(G_q, F_p)\text{ is
  minimal}\}\).  Then one can define a vector bundle \(\Comp{E}\) over
\(\Comp{U}\) by \(\Comp{E}_{(p,q)} = \Ext(G_q, F_p)\).

\(\Comp{E}\) has an obvious map to \(M(\Lambda, f+g)\), and as
\(\Comp{E}\) is the total space of a vector bundle hence irreducible,
its image \(\overline{\Comp{E}}\) must lie within a component.

But by [Bill \& Jan] and the hypothesis that \(\ext(\Fam{F}, \Fam{G})
= 0\) we also know that \(\overline{\Comp{E}}\) is open.  So the
closure of \(\overline{\Comp{E}}\) in \(M(\Lambda, f+g)\) must be the
desired component.

For another component \(\Comp{C}\) satisfying the definition of
\(\Comp{F\!\rtimes\! G}\), let \(\Comp{C'}\) be an open subset of
points \(p\) such that \(C_p\) is an extension of an object ...
\end{proof}

\begin{definition}
Given an object \(F\) of some Abelian category, a
\emph{P-decomposition} is an ordered direct sum \(F =
\oplus_{i=1}^nF_i\) such that
\begin{enumerate}
\item \(i\neq j \implies \Ext(F_i, F_j) = 0\)
\item \(i<j \implies \Hom(F_i, F_j) = 0\)
\end{enumerate}
\end{definition}

\begin{definition}
We say a type \(\Fam{F}\) \emph{has a P-decomposition} if the general
object \(F\) has a P-decomposition \(\oplus_{i=1}^nF_i\) such that
each \(F_i\) is a member of a type \(\Fam{F}_i\) and that the general
object of each type occurs in the P-decomposition of some object of
\(\Fam{F}\).
\end{definition}

This somewhat generalizes the canonical decomposition of a dimension
vector for a quiver, in the sense that the canonical decomposition of
a dimension vector is a P-decomposition with the additional constraint
that \(\End{\Fam{F}_i} = k\). XXX nearly.

(notice that for any family \(\Fam{F} = \Fam{F}\)
is a P-decomposition).  

\begin{definition}
A type \(\Fam{F}\) is a \emph{quiver type} if ...
\end{definition}

\begin{definition}
A type \(\Fam{F}\) \emph{has Q-decomposition} if it has a
P-decomposition \(\Fam{F} = \oplus_{i=1}^n\Fam{F}_i\) where each
\(\Fam{F}_i\) is a quiver type.
\end{definition}

\begin{theorem}
Suppose \(\Fam{F}\) and \(\Fam{G}\) are families with
Q-decompositions, that \(\ext(\Fam{F}, \Fam{G}) = 0\) and that the
general member of \(\jamfam{F}{G}\) contains a unique object of type
\(\Fam{F}\).  Then \(\jamfam{F}{G}\) has a Q-decomposition.
\end{theorem}

\begin{proof}
Let the length of the Q-decompositions for \(\Fam{F}\) and \(\Fam{G}\)
be \(e\) and \(f\) respectively and let \(e = \ext(\Fam{G}, \Fam{F})\).
The proof goes by induction on \((e, f+g)\) (with lexicographical
ordering).

If \(e = 0\) the result is trivial (the rigidity requirements imply
that \(\hom(\Fam{F}, \Fam{G}) = 0\), so concatenating the
decompositions of \(\Fam{F}\) and \(\Fam{G}\) suffices.).

So, assume not.  We wish to reduce to families \(\Fam{F'}\) and
\(\Fam{G'}\) such that \(\ext(\Fam{G'}, \Fam{F'}) < \ext(\Fam{G},
\Fam{F})\).

If \(f+g=2\), i.e. \(\Fam{F}\) and \(\Fam{G}\) have Q-decompositions
of length 1, i.e. if they are both families of quiver type, then
theorem 6.1 of Aidan's ``Quivers'' paper essentially says that
\(\jamfam{F}{G}\) is of quiver type.

Assuming \(f>1\) and that the result is known for all ``smaller''
cases, we note that for the general member \(F\) of \(\Fam{F}\) and
the general member \(R\) of \(\jamfam{F}{G}\) we have short exact
sequences
\begin{align}
0 \to \oplus_{i=1}^{f-1}F_i & \to F \to F_f \to 0 \\
0 \to F \to R \to G \to 0
\end{align}
Write \(\oplus_{i=1}^{f-1}F_i\) as \(F'\).

Let \(E\) be the quotient of \(R\) by \(\oplus_{i=1}^{f-1}F_i\).
\(E\) has a Q-decomposition by induction.  Write this as
\(E=\oplus_{i=1}^e\).  We have another ses:
\begin{equation}
0 \to F_f \to E \to G \to 0
\end{equation}
Applying the functor \(\Hom(\boldsymbol-,F')\) yields the long exact
sequence:
\begin{equation}
\begin{split}
0 & \to \Hom(G, F') \to \Hom(E, F') \to \Hom(F_f, F') \\
  & \to \Ext^1(G, F') \to \Ext^1(E, F') \to \Ext^1(F_f, F') \to \cdots
\end{split}
\end{equation}
from which we deduce that \(\Ext^1(G, F')\) surjects onto \(\Ext^1(E,
F')\).  As \(\Ext^1(G, F')\) is plainly a summand of \(\Ext^1(G,F)\),
we have surjections
\begin{equation}
\Ext^1(G,F) \onto \Ext^1(G, F') \onto \Ext^1(E, F')
\end{equation}
and it suffices to show that at least one of these is not an
isomorphism.


\end{proof}


\end{document}