Timeline for About cluster variables obtained by (sequentially) mutating at exchangeable variables from an initial cluster
Current License: CC BY-SA 4.0
11 events
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| Dec 29, 2021 at 21:25 | answer | added | F. C. | timeline score: 1 | |
| Aug 17, 2021 at 8:20 | comment | added | amator2357 | Thank you both for your comments, you have been very helpful. | |
| Aug 17, 2021 at 6:58 | comment | added | Jan Grabowski | Thanks for explaining @amator2357; because you'd not named the variables in your fixed ex, I misunderstood. I agree with @SamHopkins: this will almost never happen. In particular if you can find a cluster with no variables from ex such that no mutations of any of these are in ex. If you're working on surfaces, this might not be too hard, since you just have to construct a triangulation that looks "transverse" to your initial one (i.e. has its edges as different as possible). For general CAs, I'd conjecture your equality only holds for a small number of cases in very small rank. | |
| Aug 17, 2021 at 4:00 | comment | added | Sam Hopkins | As I said before, there are clearly only finitely many sequences of the kind you have in mind, which means for infinite type cluster algebras there's no chance of reaching every cluster variable. But even for finite type (like Type A which you mentioned), I don't think you'll be able to get to every cluster variable in that few mutations. Indeed the diameter of the associahedron apparently grows like $2n$ (see arxiv.org/abs/1207.6296), which is more than the number of flips you're allowing. | |
| Aug 16, 2021 at 22:05 | history | edited | amator2357 | CC BY-SA 4.0 |
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| Aug 16, 2021 at 21:58 | history | edited | amator2357 | CC BY-SA 4.0 |
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| Aug 16, 2021 at 21:54 | comment | added | amator2357 | Hi @JanGrabowski, thank you for your comment. Seems like I am failing in explaining myself properly. Say we have an initial seed arising from a triangulation (of a marked surface). I can flip a diagonal $a$ to get a diagonal $c$, say, then flip some diagonal $b$. Then I could flip $c$, but I don't want to allow for that as $c$ wasn't in the initial cluster. On the other hand, flipping $a$ then $b$ is fine. Does that make more sense? | |
| Aug 16, 2021 at 21:41 | comment | added | Jan Grabowski | The usual definition of a cluster variable is something obtained under iterated mutation (by arbitrarily long but finite sequences of mutations) from the initial seed, so your equality is necessarily true by that definition. There are other characterisations in certain situations (and in particular some things to think/worry about in the infinite rank case) so if your definition of cluster variable isn't this, it would be helpful to know what you are using. | |
| Aug 16, 2021 at 21:22 | comment | added | amator2357 | Hi @SamHopkins, thanks for the comment. I am asking if we can get every cluster variable by mutating at variables from the initial seed only. So say if $ex=\{x_1,\dots,x_k\}$ then, for instance, we can do $\mu_{x_k}\circ \cdots \circ \mu_{x_1}$ or just $\mu_{x_1} \circ \mu_{x_2}$, etc . Yeah, that's too much to ask, as you pointed out. Maybe it is true for finite type guys. | |
| Aug 16, 2021 at 21:10 | comment | added | Sam Hopkins | Are you asking if we get every cluster variable by mutating at most once at each vertex? That doesn't seem possibly true: there are usually infinitely many cluster variables. Maybe I'm confused about what you are asking though... | |
| Aug 16, 2021 at 20:55 | history | asked | amator2357 | CC BY-SA 4.0 |