cotangent bundles

By lieven

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}
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 \tilde{Q}~:~\xymatrix{\vtx{} \ar@/^/[rr]^a && \vtx{}
\ar@(u,ur)^x \ar@(d,dr)_{x^_} \ar@/^/[ll]^{a^_}} 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)}
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
\xymatrix{\vtx{} \ar@/^/[rr]^{a+da} \ar@/_/[rr]_{a-da} & & \vtx{}
\ar@(u,ur)^{x+dx} \ar@(d,dr)_{x-dx}} and the labeling of the arrows may help you to work through some sections of the Cuntz-Quillen paper…

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2 Responses to “cotangent bundles”

  1. differential forms | neverendingbooks Says:

    [...] 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 [...]

  2. moduli spaces | neverendingbooks Says:

    [...] 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 [...]

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