non-commutative geometry
- Brauer’s forgotten group
- connected component coalgebra
- Galois and the Brauer group
- a noncommutative Grothendieck topology
- noncommutative geometry
- noncommutative geometry 2
- projects in noncommutative geometry
- points and lines
- more noncommutative manifolds
- the necklace Lie bialgebra
- the one quiver for GL(2,Z)
- representation spaces
- quiver representations
- moduli spaces
- cotangent bundles
- differential forms
- curvatures
- Brauer-Severi varieties
- smooth Brauer-Severis
- hyper-resolutions
- a cosmic Galois group
- double Poisson algebras
- A for aggregates
- B for bricks
- necklaces (again)
- seen this quiver?
- why nag? (2)
- why nag? (3)
- sexing up curves
- the Klein stack
- Alain Connes on everything
- noncommutative topology (1)
- a noncommutative topology 2
- noncommutative topology (3)
- noncommutative topology (4)
- non-geometry
- non-(commutative) geometry
- noncommutative Fourier transform
- noncommutative bookmarks
- noncommutative geometry : a medieval science?
The
previous post in this sequence was
moduli spaces. Why did we spend
time explaining the connection of the quiver
![Q~:~\xymatrix{\vtx{} \ar[rr]^a & & \vtx{} \ar@(ur,dr)^x} Q~:~\xymatrix{\vtx{} \ar[rr]^a & & \vtx{} \ar@(ur,dr)^x}](/latexrender/pictures/20b1cf17f357ac30f0b2ce1d2f53919a.gif)
to moduli spaces of vectorbundles on curves and moduli spaces of linear
control systems? At the start I said we would concentrate on its double
quiver
Clearly,
this already gives away the answer : if the path algebra $C Q$
determines a (non-commutative) manifold $M$, then the path algebra $C
\tilde{Q}$ determines the cotangent bundle of $M$. Recall that for a
commutative manifold $M$, the cotangent bundle is the vectorbundle
having at the point $p \in M$ as fiber the linear dual $(Tp M)^$ of
the tangent space. So, why do we claim that $C \tilde{Q}$
corresponds to the cotangent bundle of $C Q$? Fix a dimension vector
$\alpha = (m,n)$ then the representation space
$\mathbf{rep}{\alpha}~Q = M{n \times m}(C) \oplus Mn(C)$ is just
an affine space so in its point the tangent space is the representation
space itself. To define its linear dual use the non-degeneracy of the
_trace pairings $M{n \times m}(C) \times M{m \times n}(C)
\rightarrow C~:~(A,B) \mapsto tr(AB)$ $Mn(C) \times Mn(C)
\rightarrow C~:~(C,D) \mapsto tr(CD)$ and therefore the linear dual
$\mathbf{rep}_{\alpha}~Q^ = M{m \times n}(C) \oplus Mn(C)$ which is
the representation space $\mathbf{rep}{\alpha}~Q^s$ of the quiver
![Q^s~:~\xymatrix{\vtx{} & & \vtx{} \ar[ll] \ar@(ur,dr)} Q^s~:~\xymatrix{\vtx{} & & \vtx{} \ar[ll] \ar@(ur,dr)}](/latexrender/pictures/c03cbdd2112bd021e2df3e5e9876406a.gif)
and therefore we have that the cotangent bundle to the representation
space $\mathbf{rep}{\alpha}~Q$ $T^* \mathbf{rep}{\alpha}~Q =
\mathbf{rep}{\alpha}~\tilde{Q}$ Important for us will be that any
cotangent bundle has a natural symplectic structure. For a good
introduction to this see the
course notes “Symplectic geometry and
quivers” by
Geert Van de Weyer. As a consequence $C \tilde{Q}$
can be viewed as a non-commutative symplectic manifold with the
symplectic structure determined by the non-commutative 2-form
$\omega = da^* da + dx^* dx$ but before we can define all this we
will have to recall some facts on non-commutative differential forms.
Maybe
next time. For the impatient : have a look at the paper by
Victor Ginzburg Non-commutative Symplectic Geometry, Quiver varieties,
and Operads or my paper with Raf Bocklandt Necklace Lie algebras
and noncommutative symplectic geometry. Now that we have a
cotangent bundle of $C Q$ is there also a tangent bundle and does it
again correspond to a new quiver? Well yes, here it is
and the labeling of the
arrows may help you to work through some sections of the Cuntz-Quillen
paper…
art, arxiv, Cuntz, differential, geometry, Ginzburg, moduli, necklace, non-commutative, noncommutative, Quillen, quivers
2 comments
Posted in geometry
Written on Thu, 09 September 2004 at 11:40 am
Tags: art, arxiv, Cuntz, differential, geometry, Ginzburg, moduli, necklace, non-commutative, noncommutative, Quillen, quivers
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January 13th, 2008 at 11:47 am
[...] previous post in this sequence was (co)tangent bundles. Let $A$ be a $V$-algebra where $V = C times hdots times C$ is the subalgebra generated by a [...]
January 29th, 2008 at 8:12 pm
[...] Grassmannian. But why do we stress this particular quiver so much? This will be partly explained next time. I Love Social BookmarkingSubscribeDiggdel.icio.usMa.gnoliaStumbleUponTechnorati Previous in series [...]