quiver representations

In what
way is a formally smooth algebra a _machine_ producing families of
manifolds? Consider the special case of the path algebra $\mathbb{C}Q$ of a
quiver and recall that an $n$-dimensional representation is an algebra
map $\mathbb{C}Q{\to }^{\varphi }{M}_n(\mathbb{C})$ or, equivalently, an
$n$-dimensional left $\mathbb{C}Q$-module ${\mathbb{C}}_{\varphi }^n$ with the action
determined by the rule $a.v=\varphi (a)v\mathrm{\forall }v\in {\mathbb{C}}_{\varphi }^n,\mathrm{\forall }a\in \mathbb{C}Q$ If the $e_{i}1\le i\le k$ are the idempotents
in $\mathbb{C}Q$ corresponding to the vertices (see this [post][1]) then we get
a direct sum decomposition ${\mathbb{C}}_{\varphi }^n=\varphi (e_1){\mathbb{C}}_{\varphi }^n\oplus \dots \oplus \varphi (e_k){\mathbb{C}}_{\varphi }^n$ and so every $n$-dimensional
representation does determine a _dimension vector_ $\alpha =(a_1,\dots ,a_k)\text{with}{a}_i=di{m}_{\mathbb{C}}{V}_i=di{m}_{\mathbb{C}}\varphi (e_i){\mathbb{C}}_{\varphi }^n$ with $|\alpha |=\sum _i{a}_i=n$. Further,
for every arrow we have (because $e_j.a.e_i=a$ that $\varphi (a)$
defines a linear map $\varphi (a):V_i\to V_j$ (that was the
whole point of writing paths in the quiver from right to left so that a
representation is determined by its _vertex spaces_ $V_i$ and as many
linear maps between them as there are arrows). Fixing vectorspace bases
in the vertex-spaces one observes that the space of all
$\alpha$-dimensional representations of the quiver is just an affine
space ${\mathbf{rep}}_{\alpha }Q={\oplus }_a{M}_{a_j\times a_i}(\mathbb{C})$ and
base-change in the vertex-spaces does determine the action of the
_base-change group_ $GL(\alpha )=G{L}_{a_1}\times \dots \times G{L}_{a_k}$ on this space. Finally, as all this started out with fixing
a bases in ${\mathbb{C}}_{\varphi }^n$ we get the affine variety of all
$n$-dimensional representations by bringing in the base-change
$G{L}_n$-action (by conjugation) and have ${\mathbf{rep}}_n\mathbb{C}Q=\underset{|\alpha |=n}{⨆}G{L}_n{\times }^{GL(\alpha )}{\mathbf{rep}}_{\alpha }Q$ and in this decomposition the connected
components are no longer just affine spaces with a groupaction but
essentially equal to them as there is a natural one-to-one
correspondence between $G{L}_n$-orbits in the fiber-bundle $G{L}_n{\times }^{GL(\alpha )}{\mathbf{rep}}_{\alpha }Q$ and $GL(\alpha )$-orbits in the
affine space ${\mathbf{rep}}_{\alpha }Q$. In our main example
an $n$-dimensional representation
determines vertex-spaces $V=\varphi (e){\mathbb{C}}_{\varphi }^n$ and $W=\varphi (f){\mathbb{C}}_{\varphi }^n$ of dimensions $p,q\text{with}p+q=n$. The arrows
determine linear maps between these spaces and if we fix a set of bases in these two
vertex-spaces, we can represent these maps by matrices
which can be considered as block
$n\times n$ matrices $a\mapsto \left[\begin{array}{cc}0& 0\\ A& 0\end{array}\right]b\mapsto \left[\begin{array}{cc}0& B\\ 0& 0\end{array}\right]$
$x\mapsto \left[\begin{array}{cc}0& 0\\ 0& X\end{array}\right]y\mapsto \left[\begin{array}{cc}0& 0\\ 0& Y\end{array}\right]$ The basechange group
$GL(\alpha )=G{L}_p\times G{L}_q$ is the diagonal subgroup of $G{L}_n$
$GL(\alpha )=\left[\begin{array}{cc}G{L}_p& 0\\ 0& G{L}_q\end{array}\right]$ and
acts on the representation space ${\mathbf{rep}}_{\alpha }Q=M_{q\times p}(\mathbb{C})\oplus M_{p\times q}(\mathbb{C})\oplus M_q(\mathbb{C})\oplus M_q(\mathbb{C})$
(embedded as block-matrices in $M_n(\mathbb{C}{)}^{\oplus 4}$ as above) by
simultaneous conjugation. More generally, if $A$ is a formally smooth
algebra, then all its representation varieties ${\mathbf{rep}}_{n}A$ are
affine smooth varieties equipped with a $G{L}_n$-action. This follows more
or less immediately from the definition and [Grothendieck][2]\’s
characterization of commutative regular algebras. For the record, an
algebra $A$ is said to be _formally smooth_ if for every algebra map $A\to B/I$ with $I$ a nilpotent ideal of $B$ there exists a lift
$A\to B$. The path algebra of a quiver is formally smooth
because for every map $\varphi :\mathbb{C}Q\to B/I$ the images of the
vertex-idempotents form an orthogonal set of idempotents which is known
to lift modulo nilpotent ideals and call this lift $\psi$. But then also
every arrow lifts as we can send it to an arbitrary element of
$\psi (e_j){\pi }^{-1}(\varphi (a))\psi (e_i)$. In case $A$ is commutative and
$B$ is allowed to run over all commutative algebras, then by
Grothendieck\’s criterium $A$ is a commutative regular algebra. This
also clarifies why so few commutative regular algebras are formally
smooth : being formally smooth is a vastly more restrictive property as
the lifting property extends to all algebras $B$ and whenever the
dimension of the commutative variety is at least two one can think of
maps from its coordinate ring to the commutative quotient of a
non-commutative ring by a nilpotent ideal which do not lift (for an
example, see for example [this preprint][3]). The aim of
non-commutative algebraic geometry is to study _families_ of manifolds
${\mathbf{rep}}_{n}A$ associated to the formally-smooth algebra $A$. [1]:
https://lievenlb.local/wp-trackback.php/10 [2]:
http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/Grothendieck.
html [3]: http://www.arxiv.org/abs/math.AG/9904171