User jan grabowski - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T01:33:36Z http://mathoverflow.net/feeds/user/13215 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/105727/isomorphisms-of-quantum-planes Isomorphisms of quantum planes Jan Grabowski 2012-08-28T14:32:27Z 2012-12-27T20:49:42Z <p>Let $k$ be a field and $q\in k^{*}$. The quantum plane $k_{q}[x,y]$ is the algebra <code>$k\langle x,y\rangle/\langle xy=qyx \rangle$</code> (i.e. the quotient of the free non-commutative $k$-algebra on two variables $x$ and $y$ modulo the ideal given). </p> <blockquote> <p>Question: For $q,r\in k^{*}$ and $q\neq r$, when is $k_{q}[x,y]$ isomorphic (as an algebra) to $k_{r}[x',y']$?</p> </blockquote> <p>I fully expect this is known but after (what I think is) fairly comprehensive literature searching, including a large proportion of the best-known quantum groups texts, I have been unable to find an answer. A reference would be appreciated just as much as a proof.</p> <p>Some comments:</p> <ul> <li>I know the (algebra) automorphism group: by work of Alev-Chamarie this is $(k^{*})^2$ unless $q=-1$ (when it is a semi-direct product of the torus with the group of order two generated by the map that interchanges the two variables). Hence I don't need to worry about $q=r$.</li> <li>I want algebra isomorphisms but information on Hopf algebra maps would be nice too (NB. the Hopf automorphisms for the usual Hopf structure are also those just described)</li> <li>if $q$ has finite order $N$ in $k^{*}$ and $r$ is of infinite order then the corresponding quantum planes are not isomorphic, as in the first case the centre is non-trivial (generated by $x^{N}$ and $y^{N}$) but in the second the centre is just $k$</li> <li>if $q$ has order $M$ and $r$ has order $N\neq M$, then the quotients by the centres are both finite-dimensional but of different dimension, hence the quantum planes are not isomorphic</li> <li>I would be happy to know the answer just for $k=\mathbb{C}$ </li> </ul> http://mathoverflow.net/questions/108797/what-do-cluster-algebras-tell-us-about-grassmannians/108956#108956 Answer by Jan Grabowski for What do cluster algebras tell us about Grassmannians? Jan Grabowski 2012-10-05T20:19:01Z 2012-10-05T20:19:01Z <p>I'm afraid that as far as I know, the answer is no. That is, the cluster structure hasn't (yet) told us anything new. There are two reasons why we might have expected that, though. </p> <p>Firstly, the Grassmannians have been studied over a long period of time in many different settings, in which much of the time this same phenomenon has happened - Grassmannians provide the nice, small examples, that we understand "completely". So asking the cluster structure to prove something new is quite a big ask!</p> <p>Secondly, cluster algebras (and their quantum analogues) haven't been around very long in mathematical terms. There are very few results of the form "Let $X$ be a cluster algebra. Then $X$ is a $Y$.", where $Y$ is a ring or algebra property.</p> <p>Some results this direction include recent work of Phillip Lampe, but these are the only ones that spring to mind. Others may know of more, of course.</p> <p>I have also given a number of cluster algebra talks, mostly about quantum Grassmannians, and my answer has been to point out that precisely because Grassmannians turn up everywhere, they are ideal for looking at as a first set of examples. Just as a group theorist might say "and this is what it looks like for Abelian groups", or $p$-groups, say. It's natural when studying projective varieties to turn to Grassmannians as the first non-trivial examples.</p> http://mathoverflow.net/questions/107621/status-of-a-conjectural-definition-of-h-nakajima/107662#107662 Answer by Jan Grabowski for Status of a conjectural definition of H. Nakajima Jan Grabowski 2012-09-20T08:56:33Z 2012-09-20T08:56:33Z <p>The following paper might help get you up to date, by working through its references:</p> <p><em>Yoshiyuki Kimura, Fan Qin. Graded quiver varieties, quantum cluster algebras and dual canonical basis.</em> </p> <p><a href="http://arxiv.org/abs/1205.2066" rel="nofollow">http://arxiv.org/abs/1205.2066</a></p> <p>It contains some material on $q,t$-characters and twisted versions of these, in relation to some problems in the theory of quantum cluster algebras. </p> <p>In particular it references</p> <p><em>David Hernandez. Algebraic approach to q,t-characters. Advances in Mathematics 187 (2004), no. 1, 1–52.</em></p> <p>which I imagine (though I've not read it) should be pretty close to what you want.</p> http://mathoverflow.net/questions/686/handling-arxiv-feeds-to-avoid-duplicates/106794#106794 Answer by Jan Grabowski for Handling arXiv feeds to avoid duplicates Jan Grabowski 2012-09-10T09:13:40Z 2012-09-10T09:13:40Z <p>I also use Google Reader in (nearly) the way you suggest - but I have simply subscribed to the "all new maths items" feed. It means that I see about 150 new papers in the feed each day but personally I don't find this to be a massive problem (except when I've been away for a week or so). I flick Reader into "list view" and just scroll down the titles, expanding to see the abstract when I think something might be relevant. The majority of authors choose their title well enough to make this work!</p> <p>It doesn't take so long to scan this list and I know that on a regular basis I see things outside of the subject categories I'd normally focus on. So I do pick up combinatorics, physics and algebraic geometry papers that I'd otherwise not find if I just stuck to math.QA and math.RT. Starring is nice and quick for holding on to things I might come back to later and I maybe download one or two a day too.</p> <p>I appreciate that this might not be the answer you were looking for but, much as I love tech, sometimes it's quicker and easier to make the organic bit of the process do a bit more work.</p> http://mathoverflow.net/questions/105221/quantum-groups-not-via-presentations/105228#105228 Answer by Jan Grabowski for quantum groups... not via presentations Jan Grabowski 2012-08-22T11:46:05Z 2012-08-22T11:46:05Z <p>Some possible partial answers might be:</p> <ul> <li>one could follow Lusztig and do away with the Lie algebra completely, just starting from a root datum. Then do some geometry...</li> <li>Majid's reinterpretation of Lusztig's construction, as exposited in his "A Quantum Groups Primer", is a (good, IMHO) attempt to explain where the formulae come from. The definition of the positive part of $U_{q}(\mathfrak{g})$ as natural (braided) object acted on by "the Cartan part" explains most of the formulae. Taking the Drinfeld double "explains" the cross-relations.</li> <li>another explanation for the formulae, especially the quantized Serre relations, is given via the Ringel-Hall categorical approach (Hall algebras); one should also mention Green at this point. This has been extended of late to double Hall algebras, trying to construct the whole quantum group and not just the positive part, but this is still essentially done via the Drinfeld double construction, just at a categorical level.</li> <li>a quite non-standard route would be to go from $\mathfrak{g}$ to the algebraic group $G$, construct the quantized coordinate ring - where you might find the formulae more to your taste and/or better motivated along Grothendieck/Manin lines (see the quantum groups book of Brown-Goodearl for example) - then dualize to get $U_{q}(\mathfrak{g})$. I say non-standard because most people want to go the other way, to figure out what the quantized coordinate ring should be.</li> </ul> <p>Apologies for the dearth of precise references: I can try to provide some if any of the above is helpful. Kassel's book specifically covers the FRT construction, by the way.</p> <p>Edit: The full Hopf structure, as opposed to just the algebra structure, is - again IMHO - essentially canonically determined. The coproduct is pretty canonical, by comparison with the natural one for $U(\mathfrak{g})$. There's almost no choice for the counit and the antipode formula is (if I remember correctly) forced from the requirement that various maps are algebra morphisms. Alternatively, the Hopf structure can be seen to drop out of the Drinfeld double construction; that might not be a helpful thing to say, of course.</p> http://mathoverflow.net/questions/65355/which-cluster-algebras-have-been-categorified/65447#65447 Answer by Jan Grabowski for Which cluster algebras have been categorified? Jan Grabowski 2011-05-19T15:19:43Z 2011-05-19T15:19:43Z <p>There are several different questions in here and what follows are only partial answers to some of these, mostly consisting of pointers to pieces of the literature.</p> <ul> <li>"In what other instances have cluster categories been constructed?"</li> </ul> <p>For surveys on cluster categories, I would recommend:</p> <p>Cluster algebras, quiver representations and triangulated categories, Bernhard Keller, <a href="http://arxiv.org/abs/0807.1960" rel="nofollow">arXiv:0807.1960</a></p> <p>Categorification of acyclic cluster algebras: an introduction, Bernhard Keller, <a href="http://arxiv.org/abs/0801.3103" rel="nofollow">arXiv:0801.3103</a></p> <p>Cluster categories, Idun Reiten, <a href="http://arxiv.org/abs/1012.4949" rel="nofollow">arXiv:1012.4949</a> (given as an ICM 2010 lecture)</p> <ul> <li>Cluster algebras with coefficients</li> </ul> <p>These don't fall inside the usual cluster category setting, as the coefficients (frozen variables) are expected to correspond to projective objects and typically cluster category theory works at the stable level, where these are not present. (I am not an expert on 2-Calabi-Yau categories, etc., so if someone else wishes to tidy up this claim a little, that would be helpful.) So one suggestion has been that one ought to work in a Frobenius category whose stable category would give the usual cluster category. This corresponds to the theorems that say that for each cluster algebra type, you can have different cluster algebras with different numbers of coefficients present but they all have the same cluster combinatorics.</p> <p>Work on getting such categories is in its early stages but being actively pursued by a number of researchers. One very notable success to date is the work of Geiss-Leclerc-Schroer on cluster algebra structures on coordinate rings associated to partial flag varieties:</p> <p>Partial flag varieties and preprojective algebras, Christof Geiss, Bernard Leclerc, Jan Schröer, <a href="http://arxiv.org/abs/math/0609138" rel="nofollow">arXiv:math/0609138</a></p> <p>Preprojective algebras and cluster algebras, Christof Geiss, Bernard Leclerc, Jan Schröer, <a href="http://arxiv.org/abs/0804.3168" rel="nofollow">arXiv:0804.3168</a> (a survey article covering the previously-listed paper and some earlier ones)</p> <p>Kac-Moody groups and cluster algebras, Christof Geiss, Bernard Leclerc, Jan Schröer, <a href="http://arxiv.org/abs/1001.3545" rel="nofollow">arXiv:1001.3545</a> (generalization of the above)</p> <p>Cluster algebras and representation theory, Bernard Leclerc, <a href="http://arxiv.org/abs/1009.4552" rel="nofollow">arXiv:1009.4552</a> (another more recent survey, from the 2010 ICM)</p> <ul> <li>Surface and tame cluster algebras</li> </ul> <p>For things to do with more general surface cluster algebras, I suggest you start with this survey:</p> <p>Geometric construction of cluster algebras and cluster categories, Karin Baur, <a href="http://arxiv.org/abs/0804.4065" rel="nofollow">arXiv:0804.4065</a></p> <p>I don't know enough about tame cluster algebras to say anything about these, other than noting that for example the work of Geiss-Leclerc-Schroer mentioned above includes many cluster algebras that are not of finite type.</p> <ul> <li>Other comments</li> </ul> <p>I would also add that there is work going on on quantum versions of these and infinite versions (see for example a paper by Holm and Jorgensen, <a href="http://arxiv.org/abs/0902.4125" rel="nofollow">arXiv:0902.4125</a>).</p> <p>Also, one can look at your question from a different angle, namely to ask which algebras occuring "in nature" admit (possibly quantum) cluster algebra structures; any good search engine will show you that this is the area I work in... Then the question becomes (a) are these categorified just as algebras and (b) are they categorified in a way that is compatible with the cluster structure? For example, Geiss-Leclerc-Schroer says "yes" to both for a large class of important examples.</p> http://mathoverflow.net/questions/56518/infinite-grassmannians-and-their-coordinate-rings Infinite Grassmannians and their coordinate rings Jan Grabowski 2011-02-24T13:05:26Z 2011-03-02T14:17:58Z <p>I'm currently thinking about some combinatorics associated to an infinite analogue of the coordinate rings of the Grassmannians $Gr(2,n)$. The combinatorics should be thought of as relating to Plucker coordinates $\Delta^{ij}$ but with $i &lt; j$ arbitrary integers, rather than restricted to ${1,\ldots ,n}$. So I've been trying to find the right infinite Grassmannian to have this coordinate ring (or, if the regular functions are a bit more complicated in the infinite case, to at least have these Plucker coordinates in there). I've looked (briefly) into:</p> <p>(a) Kac's construction of infinite Grassmannians ([Kac, Infinite dimensional Lie algebras, 3rd ed.], Exercise 14.32, p.339)</p> <p>(b) taking the union of the finite Grassmannians to get a classifying space for $O(n)$ or $U(n)$</p> <p>(c) infinite Grassmannians coming from Hilbert spaces ([Pressley and Segal, Loop groups])</p> <p>but none of these seem to quite describe what I want. Some of these are working with $\mathbb{N}$-dimensional space rather than $\mathbb{Z}$-dimensional space (i.e. something more like $\mathbb{C}[t]$ than $\mathbb{C}[t,t^{-1}]$), usually from a colimit of the finite ones, and I can't see how to alter the definition and be sure of keeping the theorems. And with the others that do work with something like $\mathbb{C}[t,t^{-1}]$, I can't see a description that corresponds to planes in that space (and I definitely need just the planes).</p> <p>I feel sure this is well-known so does anybody know a reference for both the construction I want and also enough information about its coordinate ring?</p> <p>Edit: Having thought about this a little more, I want to formulate the question more specifically as:</p> <blockquote> <p>Let $V=\mathbb{C}[t,t^{-1}]$ and define $Gr(2,V)$ to be the set of 2-dimensional subspaces of $V$. Does the finite-dimensional machinery of the Plucker embedding work in this setting and give Plucker coordinates in $\mathbb{C}[Gr(2,V)]$ of the form $\Delta^{ij}$ for integers $i &lt; j$?</p> </blockquote> http://mathoverflow.net/questions/132134/temperley-lieb-algebras-for-other-weyl-groups/132136#132136 Comment by Jan Grabowski Jan Grabowski 2013-05-28T21:02:10Z 2013-05-28T21:02:10Z Catalan numbers appear <i>everywhere</i>! :-) More seriously, I've thought a little in the first direction you propose in the last paragraph but haven't seen a cluster algebra connection so far. The TL relations are not obviously cluster algebra-like, but part of the thing about cluster algebras is that often one needs to look at more elements than in the defining presentation (usually the game is to find small generating sets for one's favourite algebra but this is absolutely not the case for cluster algebras, where one wants &quot;simple&quot; relations and have to live with needing large generating sets). http://mathoverflow.net/questions/126847/quantum-jimbo-algebras-why-such-fuss-about-roots-of-unity Comment by Jan Grabowski Jan Grabowski 2013-04-08T15:32:21Z 2013-04-08T15:32:21Z This isn't a precise technical answer but the root of unity or not divide is (in ways that <i>can</i> 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.) http://mathoverflow.net/questions/118643/fuss-catalan-algebras-and-non-commutative-hilbert-schemes Comment by Jan Grabowski Jan Grabowski 2013-01-11T17:15:34Z 2013-01-11T17:15:34Z Catalan numbers count among other things triangulations of polygons (another example from my own interests, in cluster algebras). The OEIS page <a href="http://oeis.org/A000108" rel="nofollow">oeis.org/A000108</a> lists lots more! So $m=2$ is likely to be speculation and I'd concur that $m=3$ is the place to start. http://mathoverflow.net/questions/118643/fuss-catalan-algebras-and-non-commutative-hilbert-schemes Comment by Jan Grabowski Jan Grabowski 2013-01-11T16:41:08Z 2013-01-11T16:41:08Z 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...). http://mathoverflow.net/questions/118539/tensor-product-of-quivers Comment by Jan Grabowski Jan Grabowski 2013-01-10T15:42:16Z 2013-01-10T15:42:16Z 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 <a href="http://en.wikipedia.org/wiki/Cartesian_product_of_graphs" rel="nofollow">en.wikipedia.org/wiki/Cartesian_product_of_graphs</a> for example. I don't think what you've got is that - I think the units are different (for the Cartesian product, I <i>think</i> the unit should be the null graph/quiver on one vertex.) http://mathoverflow.net/questions/105727/isomorphisms-of-quantum-planes/117357#117357 Comment by Jan Grabowski Jan Grabowski 2012-12-28T09:20:52Z 2012-12-28T09:20:52Z 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. http://mathoverflow.net/questions/114646/intersecting-family-of-triangulations Comment by Jan Grabowski Jan Grabowski 2012-11-27T11:09:55Z 2012-11-27T11:09:55Z 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? http://mathoverflow.net/questions/105727/isomorphisms-of-quantum-planes/105738#105738 Comment by Jan Grabowski Jan Grabowski 2012-08-29T15:51:47Z 2012-08-29T15:51:47Z @Mariano Thanks - it's on my desk, waiting to be read (busy with other duties today...) http://mathoverflow.net/questions/105727/isomorphisms-of-quantum-planes Comment by Jan Grabowski Jan Grabowski 2012-08-29T11:43:27Z 2012-08-29T11:43:27Z @B. Bischof: this paper of Awami, van den Bergh and Oystaeyen seems to be difficult to get hold of. Do you know a location where I might get it? (I can probably get it on an inter-library loan but would be grateful for faster access.) http://mathoverflow.net/questions/105727/isomorphisms-of-quantum-planes/105738#105738 Comment by Jan Grabowski Jan Grabowski 2012-08-29T11:32:01Z 2012-08-29T11:32:01Z Thank you both for this sequence of comments - they are very helpful. A reference dealing with the root of unity case would be very nice, if one exists. If not, would you mind expanding your last sentence a little, @eithil? http://mathoverflow.net/questions/56518/infinite-grassmannians-and-their-coordinate-rings/56519#56519 Comment by Jan Grabowski Jan Grabowski 2011-03-02T11:48:51Z 2011-03-02T11:48:51Z @Neil: thanks for this link. Perhaps it's that I'm not very expert in this but I can't quite see what I want in your notes. Would you be able to give me a pointer to the exact part I should look at it more detail? Or even give a brief outline sketch of the idea here? Thanks!