Timeline for Are the finite dimensional von Neumann algebras, singly generated?
Current License: CC BY-SA 3.0
18 events
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| Dec 22, 2014 at 7:20 | comment | added | Sebastien Palcoux | @DavePenneys; if I'm not mistaken, Vijay Kodiyalam and Srikanth Tupurani do not allow the involution operation $A \mapsto A^*$ that's why it's generated by a single $2k$-box and not a $k$-box. Their paper seems to have being fixed because in its last version here it's $min(2k,k+4)$ instead of $k+1$. Anyway I don't understand why they don't allow the involution, it's not a tangle operation, but a planar algebra is a planar *-algebra. | |
| Nov 24, 2013 at 0:13 | comment | added | Julien | @SébastienPalcoux Yeah, that's what I had overlooked... | |
| Nov 24, 2013 at 0:11 | comment | added | Sebastien Palcoux | @julien : oh ok, because a von Neumann algebra is equal to its bicommutant as a subalgebra of a $B(H)$, not as a subalgebra of a $\oplus_i B(H_i)$. | |
| Nov 24, 2013 at 0:00 | comment | added | Sebastien Palcoux | @julien : I have an obvious question, perhaps it's related to what you overlooked : why $(\mathbb{C}I)'' = \oplus_i \mathbb{C}1_{n_i} (\neq \mathbb{C}I$) ? | |
| Nov 23, 2013 at 23:41 | vote | accept | Sebastien Palcoux | ||
| Nov 23, 2013 at 23:33 | comment | added | Sebastien Palcoux | @AndréHenriques : yes of course. Thanks. | |
| Nov 23, 2013 at 23:25 | comment | added | Julien | @DavePenneys Right... thanks... I for sure overlooked something... | |
| Nov 23, 2013 at 23:24 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Nov 23, 2013 at 23:19 | comment | added | André Henriques | @Sébastien: $A-aI$ is not an element in the $i$th summand of your algebra (because $I$ is not an element of that summand). That's why you need to use $A^2-aA$ instead. | |
| Nov 23, 2013 at 23:10 | comment | added | Dave Penneys | @julien: if each of the $n_i=1$, then the element is zero. | |
| Nov 23, 2013 at 22:40 | comment | added | Julien | I might have overlooked something, but why can't you just take $\lambda_1=\ldots=\lambda_k=0$? | |
| Nov 23, 2013 at 22:08 | comment | added | Sebastien Palcoux | Why do you take $N:=A^2-aA$ instead of $A-aI$ ? | |
| Nov 23, 2013 at 21:51 | comment | added | Sebastien Palcoux | You're right, but my deep reason is that I'm looking for a translation into the planar algebra framework of the property of being cyclic for a subfactor (i.e. distributive lattice of intermediate). Note that the cyclic "group subfactors" are exactly the "cyclic group" subfactors. Bisch and Jones wrote papers on singly generated planar algebras of small dimension. Singly generated subfactor planar algebras are not cyclic in general, so I look for some restriction on the depth of the generator. | |
| Nov 23, 2013 at 21:40 | comment | added | André Henriques | Why is it interesting to know that something is singly generated? Often (e.g. in group theory), having few generators means that the defining relations become horrible, and it is better to have more generators and to keep the set of relations under control. | |
| Nov 23, 2013 at 21:36 | comment | added | Sebastien Palcoux | Application (after Dave Penneys): let $\mathcal{P}$ be a finite index, depth $k$ subfactor planar algebra. Then $\mathcal{P}_k$ generates $\mathcal{P}$, but $\mathcal{P}_k$ is a finite dimensional von Neumann algebra, so through your answer, $\mathcal{P}$ is generated by a single $k$-box. But look at this paper (suggested by Dave), they prove that it is generated by a single $(k+1)$-box. So your argument is better, isn't it ? | |
| Nov 23, 2013 at 21:35 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Nov 23, 2013 at 21:08 | history | edited | André Henriques | CC BY-SA 3.0 |
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| Nov 23, 2013 at 20:42 | history | answered | André Henriques | CC BY-SA 3.0 |