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
Tag: representations
The
previous part of this sequence was [quiver representations][1]. When $A$
is a formally smooth algebra, we have an infinite family of smooth
affine varieties $\mathbf{rep}_n~A$, the varieties of finite dimensional
representations. On $\mathbf{rep}_n~A$ there is a basechange action of
$GL_n$ and we are really interested in _isomorphism classes_ of
representations, that is, orbits under this action. Mind you, an orbit
space does not always exist due to the erxistence of non-closed orbits
so one often has to restrict to suitable representations of $A$ for
which it _is_ possible to construct an orbit-space. But first, let us
give a motivating example to illustrate the fact that many interesting
classification problems can be translated into the setting of this
non-commutative algebraic geometry. Let $X$ be a smooth projective
curve of genus $g$ (that is, a Riemann surface with $g$ holes). A
classical object of study is $M = M_X^{ss}(0,n)$ the _moduli space
of semi-stable vectorbundles on $X$ of rank $n$ and degree $0$_. This
space has an open subset (corresponding to the _stable_ vectorbundles)
which classify the isomorphism classes of unitary simple representations
$\pi_1(X) = \frac{\langle x_1,\ldots,x_g,y_1,\ldots,y_g
\rangle}{([x_1,y_1] \ldots [x_g,y_g])} \rightarrow U_n(\mathbb{C})$ of the
fundamental group of $X$. Let $Y$ be an affine open subset of the
projective curve $X$, then we have the formally smooth algebra $A =
\begin{bmatrix} \mathbb{C} & 0 \\ \mathbb{C}[Y] & \mathbb{C}[Y] \end{bmatrix}$ As $A$ has two
orthogonal idempotents, its representation varieties decompose into
connected components according to dimension vectors $\mathbf{rep}_m~A
= \bigsqcup_{p+q=m} \mathbf{rep}_{(p,q)}~A$ all of which are smooth
varieties. As mentioned before it is not possible to construct a
variety classifying the orbits in one of these components, but there are
two methods to approximate the orbit space. The first one is the
_algebraic quotient variety_ of which the coordinate ring is the ring of
invariant functions. In this case one merely recovers for this quotient
$\mathbf{rep}_{(p,q)}~A // GL_{p+q} = S^q(Y)$ the symmetric product
of $Y$. A better approximation is the _moduli space of semi-stable
representations_ which is an algebraic quotient of the open subset of
all representations having no subrepresentation of dimension vector
$(u,v)$ such that $-uq+vp < 0$ (that is, cover this open set by
$GL_{p+q}$ stable affine opens and construct for each the algebraic
quotient and glue them together). Denote this moduli space by
$M_{(p,q)}(A,\theta)$. It is an unpublished result of Aidan Schofield
that the moduli spaces of semi-stable vectorbundles are birational
equivalent to specific ones of these moduli spaces
$M_X^{ss}(0,n)~\sim^{bir}~M_{(n,gn)}(A,\theta)$ Rather than studying
the moduli spaces of semi-stable vectorbundles $M^{ss}_X(0,n)$ on the
curve $X$ one at a time for each rank $n$, non-commutative algebraic
geometry allows us (via the translation to the formally smooth algebra
$A$) to obtain common features on all these moduli spaces and hence to
study $\bigsqcup_n~M^{ss}_X(0,n)$ the moduli space of all
semi-stable bundles on $X$ of degree zero (but of varying ranks).
There exists a procedure to associate to any formally smooth algebra $A$
a quiver $Q_A$ (playing roughly the role of the tangent space to the
manifold determined by $A$). If we do this for the algebra described
above we find the quiver $\xymatrix{\vtx{} \ar[rr] & & \vtx{}
\ar@(ur,dr)}$ and hence the representation theory of this quiver plays
an important role in studying the geometric properties of the moduli
spaces $M^{ss}_X(0,n)$, for instance it allows to determine the smooth
loci of these varieties. Move on the the [next part.
[1]: http://www.neverendingbooks.org/index.php/quiver-representations.html
The previous post can be found [here][1].
Pierre Gabriel invented a lot of new notation (see his book [Representations of finite dimensional algebras][2] for a rather extreme case) and is responsible for calling a directed graph a quiver. For example,
$\xymatrix{\vtx{} \ar@/^/[rr] & & \vtx{} \ar@(u,ur) \ar@(d,dr) \ar@/^/[ll]} $
is a quiver. Note than it is allowed to have multiple arrows between vertices, as well as loops in vertices. For us it will be important that a quiver $Q $ depicts how to compute in a certain non-commutative algebra : the path algebra $\mathbb{C} Q $. If the quiver has $k $ vertices and $l $ arrows (including loops) then the path algebra $\mathbb{C} Q $ is a subalgebra of the full $k \times k $ matrix-algebra over the free algebra in $l $ non-commuting variables
$\mathbb{C} Q \subset M_k(\mathbb{C} \langle x_1,\ldots,x_l \rangle) $
Under this map, a vertex $v_i $ is mapped to the basis $i $-th diagonal matrix-idempotent and an arrow
$\xymatrix{\vtx{v_i} \ar[rr]^{x_a} & & \vtx{v_j}} $
is mapped to the matrix having all its entries zero except the $(j,i) $-entry which is equal to $x_a $. That is, in our main example
$\xymatrix{\vtx{e} \ar@/^/[rr]^a & & \vtx{f} \ar@(u,ur)^x \ar@(d,dr)_y \ar@/^/[ll]^b} $
the corresponding path algebra is the subalgebra of $M_2(\mathbb{C} \langle a,b,x,y \rangle) $ generated by the matrices
$e \mapsto \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} $ $ f \mapsto \begin{bmatrix} 0 & 0 \\ 0 & 1 \end{bmatrix} $
$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 name \’path algebra\’ comes from the fact that the subspace of $\mathbb{C} Q $ at the $(j,i) $-place is the vectorspace spanned by all paths in the quiver starting at vertex $v_i $ and ending in vertex $v_j $. For an easier and concrete example of a path algebra. consider the quiver
$\xymatrix{\vtx{e} \ar[rr]^a & & \vtx{f} \ar@(ur,dr)^x} $
and verify that in this case, the path algebra is just
$\mathbb{C} Q = \begin{bmatrix} \mathbb{C} & 0 \\ \mathbb{C}[x]a & \mathbb{C}[x] \end{bmatrix} $
Observe that we write and read paths in the quiver from right to left. The reason for this strange convention is that later we will be interested in left-modules rather than right-modules. Right-minder people can go for the more natural left to right convention for writing paths.
Why are path algebras of quivers of interest in non-commutative geometry? Well, to begin they are examples of _formally smooth algebras_ (some say _quasi-free algebras_, I just call them _qurves_). These algebras were introduced and studied by Joachim Cuntz and Daniel Quillen and they are precisely the algebras allowing a good theory of non-commutative differential forms.
So you should think of formally smooth algebras as being non-commutative manifolds and under this analogy path algebras of quivers correspond to _affine spaces_. That is, one expects path algebras of quivers to turn up in two instances : (1) given a non-commutative manifold (aka formally smooth algebra) it must be \’embedded\’ in some non-commutative affine space (aka path algebra of a quiver) and (2) given a non-commutative manifold, the \’tangent spaces\’ should be determined by path algebras of quivers.
The first fact is easy enough to prove, every affine $\mathbb{C} $-algebra is an epimorphic image of a free algebra in say $l $ generators, which is just the path algebra of the _bouquet quiver_ having $l $ loops
$\xymatrix{\vtx{} \ar@(dl,l)^{x_1} \ar@(l,ul)^{x_2} \ar@(ur,r)^{x_i} \ar@(r,dr)^{x_l}} $
The second statement requires more work. For a first attempt to clarify this you can consult my preprint [Qurves and quivers][3] but I\’ll come back to this in another post. For now, just take my word for it : if formally smooth algebras are the non-commutative analogon of manifolds then path algebras of quivers are the non-commutative version of affine spaces!
[1]: http://www.neverendingbooks.org/index.php?p=71
[2]: http://www.booxtra.de/verteiler.asp?site=artikel.asp&wea=1070000&sh=homehome&artikelnummer=000000689724
[3]: http://www.arxiv.org/abs/math.RA/0406618