Jan Grabowski
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Registered User
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Apr 8 |
awarded | ● Commentator |
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Apr 8 |
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Quantum-Jimbo Algebras: Why Such Fuss About Roots of Unity? This isn't a precise technical answer but the root of unity or not divide is (in ways that can be made precise) analogous to the characteristic zero versus characteristic p divide. Away from the root of unity case, most things are nice; in the root of unity case they go just as badly wrong as Lie algebras in characteristic p. (NB. The interesting shift here is that the characteristic of the actual field underlying $U_{q}(\mathfrak{g})$ isn't in the foreground: the split in behaviour described above happens even for $U_{q}(\mathfrak{g})$ as an algebra over the complex numbers.) |
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Feb 24 |
awarded | ● Yearling |
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Jan 11 |
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Fuss-Catalan algebras and non-commutative Hilbert schemes Catalan numbers count among other things triangulations of polygons (another example from my own interests, in cluster algebras). The OEIS page oeis.org/A000108 lists lots more! So $m=2$ is likely to be speculation and I'd concur that $m=3$ is the place to start. |
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Jan 11 |
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Fuss-Catalan algebras and non-commutative Hilbert schemes Nice question! If it had just been the ordinary Catalan numbers, I'd be much less surprised but I've seen very few instances of the Fuss-Catalan numbers in my time (which many moons ago included some work on the Fuss-Catalan algebras...). |
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Jan 10 |
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Tensor product of quivers I know it might be tricky to do but is there any chance of a picture..? The only other quiver product I've come across in my travels is the natural quiver version of the square or Cartesian product of graphs - see en.wikipedia.org/wiki/Cartesian_product_of_graphs for example. I don't think what you've got is that - I think the units are different (for the Cartesian product, I think the unit should be the null graph/quiver on one vertex.) |
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Dec 28 |
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Isomorphisms of quantum planes Thanks, I saw this already. (By one of those coincidences, it appeared not long after I originally asked the question.) But thanks for posting the link for future readers of the question. |
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Dec 13 |
awarded | ● Critic |
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Nov 27 |
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Intersecting Family of Triangulations Have you thought about using a cluster algebra mutation argument? Triangulations having all but one diagonal in common correspond to adjacent clusters in a cluster algebra of type $A$, so maybe this helps? |
