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 $\mathbb{C} Q \rightarrow^{\phi} M_n(\mathbb{C})$ or, equivalently, an

$n$-dimensional left $\mathbb{C} Q$-module $\mathbb{C}^n_{\phi}$ with the action

determined by the rule $a.v = \phi(a) v~\forall v \in \mathbb{C}^n_{\phi},

\forall a \in \mathbb{C} Q$ If the $e_i~1 \leq i \leq k$ are the idempotents

in $\mathbb{C} Q$ corresponding to the vertices (see this [post][1]) then we get

a direct sum decomposition $\mathbb{C}^n_{\phi} = \phi(e_1)\mathbb{C}^n_{\phi} \oplus

\ldots \oplus \phi(e_k)\mathbb{C}^n_{\phi}$ and so every $n$-dimensional

representation does determine a _dimension vector_ $\alpha =

(a_1,\ldots,a_k)~\text{with}~a_i = dim_{\mathbb{C}} V_i = dim_{\mathbb{C}}

\phi(e_i)\mathbb{C}^n_{\phi}$ with $ | \alpha | = \sum_i a_i = n$. Further,

for every arrow $\xymatrix{\vtx{e_i} \ar[rr]^a & &

\vtx{e_j}} $ we have (because $e_j.a.e_i = a$ that $\phi(a)$

defines a linear map $\phi(a)~:~V_i \rightarrow V_j$ (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) = GL_{a_1} \times \ldots \times

GL_{a_k}$ on this space. Finally, as all this started out with fixing

a bases in $\mathbb{C}^n_{\phi}$ we get the affine variety of all

$n$-dimensional representations by bringing in the base-change

$GL_n$-action (by conjugation) and have $\mathbf{rep}_n~\mathbb{C} Q =

\bigsqcup_{| \alpha | = n} GL_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 $GL_n$-orbits in the fiber-bundle $GL_n

\times^{GL(\alpha)} \mathbf{rep}_{\alpha}~Q$ and $GL(\alpha)$-orbits in the

affine space $\mathbf{rep}_{\alpha}~Q$. In our main example

$\xymatrix{\vtx{e} \ar@/^/[rr]^a & & \vtx{f} \ar@(u,ur)^x

\ar@(d,dr)_y \ar@/^/[ll]^b} $ an $n$-dimensional representation

determines vertex-spaces $V = \phi(e) \mathbb{C}^n_{\phi}$ and $W = \phi(f)

\mathbb{C}^n_{\phi}$ of dimensions $p,q~\text{with}~p+q = n$. The arrows

determine linear maps between these spaces $\xymatrix{V

\ar@/^/[rr]^{\phi(a)} & & W \ar@(u,ur)^{\phi(x)} \ar@(d,dr)_{\phi(y)}

\ar@/^/[ll]^{\phi(b)}} $ and if we fix a set of bases in these two

vertex-spaces, we can represent these maps by matrices

$\xymatrix{\mathbb{C}^p \ar@/^/[rr]^{A} & & \mathbb{C}^q \ar@(u,ur)^{X}

\ar@(d,dr)_{Y} \ar@/^/[ll]^{B}} $ which can be considered as block

$n \times n$ matrices $a \mapsto \begin{bmatrix} 0 & 0 \\ A & 0

\end{bmatrix}~b \mapsto \begin{bmatrix} 0 & B \\ 0 & 0 \end{bmatrix}$

$x \mapsto \begin{bmatrix} 0 & 0 \\ 0 & X \end{bmatrix}~y \mapsto

\begin{bmatrix} 0 & 0 \\ 0 & Y \end{bmatrix}$ The basechange group

$GL(\alpha) = GL_p \times GL_q$ is the diagonal subgroup of $GL_n$

$GL(\alpha) = \begin{bmatrix} GL_p & 0 \\ 0 & GL_q \end{bmatrix}$ 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 $GL_n$-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

\rightarrow B/I$ with $I$ a nilpotent ideal of $B$ there exists a lift

$A \rightarrow B$. The path algebra of a quiver is formally smooth

because for every map $\phi~:~\mathbb{C} Q \rightarrow B/I$ 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

$\psi(e_j)\pi^{-1}(\phi(a))\psi(e_i)$. 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]:

http://www.matrix.ua.ac.be/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