cotangent bundles

The previous post in this sequence was moduli spaces. Why did we spend time explaining the connection of the quiver

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

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…

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