Timeline for Primitive elements of a tensor product of bialgebras
Current License: CC BY-SA 3.0
35 events
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| Aug 21, 2015 at 11:19 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
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| Aug 15, 2015 at 16:05 | comment | added | Duchamp Gérard H. E. | @darijgrinberg Thanks for your edits ! | |
| Aug 15, 2015 at 13:51 | history | edited | darij grinberg | CC BY-SA 3.0 |
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| Aug 15, 2015 at 10:53 | history | edited | darij grinberg | CC BY-SA 3.0 |
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| Aug 15, 2015 at 8:26 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
summability and substitution has been thouroughly tested since (I withdrawed the doubt !)
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| Aug 29, 2013 at 2:37 | comment | added | darij grinberg | I think the counterexample is not a counterexample. See my email for details. | |
| Mar 5, 2013 at 6:22 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
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| Feb 27, 2013 at 18:22 | comment | added | Duchamp Gérard H. E. | "What is a transcendence basis of a noncommutative algebra? (as in Lemma 8))" -> in fact, when ϕ is commutative, the deformation $(\mathbb{Q}<Y>,stuffle_\phi)$ is so. I will make this more clear. | |
| Feb 26, 2013 at 5:42 | comment | added | Duchamp Gérard H. E. | Cannot one edit the comments ? I have a few typos that I would like to fix ! | |
| Feb 25, 2013 at 12:04 | comment | added | Duchamp Gérard H. E. | I think that you can rather safely add to you list of on mathoverflow.net/questions/61954 "k is a $Q$-algebra" as one can apply the symmetrisors above to the case of enveloping algebras. | |
| Feb 25, 2013 at 11:54 | comment | added | Duchamp Gérard H. E. | The (counter) example I was speaking about is the following with $A=k[x]$ ($k$ is a field of characteristic zero). Let $Y$ be an alphabet and $A<Y>$ be the usual free algebra (the space of non-commutative polynomials over $Y$) and $\epsilon$, the "constant term" linear form. Let $conc$ be the concatenation and $\Delta$ the unshuffling. Then $(A<Y>,conc,1,\Delta,\epsilon)$ is a Hopf algebras (it is the enveloping algebra of the Lie polynomials). Let $A_+<Y>=ker(\epsilon)$ and $J_N=x^N.A_+<Y>$ then, for $N\geq 1$, $J_N$ is a Hopf ideal and $Prim(A<Y>/(J_N))$ is never free (no basis). | |
| Feb 25, 2013 at 7:06 | comment | added | Duchamp Gérard H. E. | Dear Darij, Thank you for your remarks and comments. "I didn't see the condition that be commutative... is it a standing assumption" No, I'll fix this. I think (and hope !) that the section about the CQMM is the most stable (I had to put this paper a bit in a hurry because this section was an answer about a question with the algebraists). I go across the references you give but I suspect that $Prim(B)$ be linearly free everytime. I have the counterexample where $Prim(B)$ is NOT linearly free (it is a bit artficial). I check it and put it quickly. Thanks again | |
| Feb 24, 2013 at 23:19 | comment | added | darij grinberg | (As far as I can judge from the first few pages of Higgins' paper, it's by far the most readable of these four sources.) | |
| Feb 24, 2013 at 23:18 | comment | added | darij grinberg | Finally, there is a proof by P. M. Cohn; see the discussion on mathoverflow.net/questions/61954 . Unfortunately it is very indirect and (IMHO) hard to understand. | |
| Feb 24, 2013 at 23:17 | comment | added | darij grinberg | Also, Higgins gave another proof (see Theorem 7 (ii) in his "Baer Invariants and the Birkhoff-Witt theorem", sciencedirect.com/science/article/pii/0021869369900866 ). I cannot vow for that one because I still haven't read the paper blush. | |
| Feb 24, 2013 at 23:15 | comment | added | darij grinberg | About [1]: The proof given in [1] only uses that the ground ring is a commutative $\mathbb Q$-algebra (field or not). An alternative proof is given in Emanuela Petracci's thesis iecn.u-nancy.fr/~petracci/tesi.pdf chapter 2. (I have checked both of the proofs, though I had to insert a lot of steps in the one given in [1], and replace a certain trick I don't understand by an induction in Petracci's thesis. It seems nobody has the time to write detailed proofs these days.) | |
| Feb 24, 2013 at 23:13 | comment | added | darij grinberg | About the $\phi$-stuffle: I didn't see the condition that $\phi$ be commutative... is it a standing assumption? | |
| Feb 24, 2013 at 14:06 | comment | added | Duchamp Gérard H. E. | In p45 of [1], P. E. and al. assume that the ring of scalars is a field of characteristic 0. But, although I have, for the time being, no "natural" counterexample, I cannot restrict to fields because, "my" scalars will be either rings of analytic functions (where inversion creates poles) or arithmetic functions. So, I was obliged to redo the CQMM theorem without supposing any basis. I thank you for the reference [1] that I'll add in the further versions of my paper. [1] Pavel Etingof et al. Quantum Fields and Strings: A Course for Mathematicians – Duchamp Gérard H. E. 7 hours ago | |
| Feb 24, 2013 at 6:11 | comment | added | Duchamp Gérard H. E. | In particular for the stuffle product. | |
| Feb 24, 2013 at 0:10 | comment | added | darij grinberg | (One question about your paper, which I hope to find time to delve into during the next weeks: What is a transcendence basis of a noncommutative algebra? (as in Lemma 8)) | |
| Feb 24, 2013 at 0:09 | comment | added | darij grinberg | I'm glad that Cartier-Milnor-Moore over $\mathbb Q$-algebras has been written up; it has been floating around in the folklore for way too long. But one remark: Over a commutative $\mathbb Q$-algebra, every Lie algebra satisfies PBW. See, for example, the proof cited in mathoverflow.net/questions/66683 . | |
| Feb 23, 2013 at 21:48 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
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| Feb 23, 2013 at 19:00 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
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| Feb 23, 2013 at 8:32 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
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| Feb 23, 2013 at 8:30 | comment | added | Duchamp Gérard H. E. | (Also, the statement "in fact the image of [...]") is slightly wrong.) I correct it ! sorry | |
| Feb 23, 2013 at 8:29 | comment | added | Duchamp Gérard H. E. | Yes it is this paper. As we aim at coefficients which are rings of function spaces, we had to revisit the theorem of Cartier-Milnor-Moore without PBW (because it is not assumed that Prim(B) has a linear basis). It turns out that only the fact that the ring contains the rationals is needed. | |
| Feb 23, 2013 at 5:41 | comment | added | darij grinberg | (Also, the statement "in fact the image of [...]") is slightly wrong.) | |
| Feb 23, 2013 at 5:38 | vote | accept | darij grinberg | ||
| Feb 23, 2013 at 5:38 | comment | added | darij grinberg | Ah!! Very nice proof. (You forgot an $I_1$ in $(\log_{*_1}\otimes e_2)\circ (I_1\otimes e_2)=(I_1\otimes e_2)\circ \log_{*_{12}}(I_1\otimes I_2)$.) The paper you are talking about, is it hal.archives-ouvertes.fr/hal-00793118 ? Because the link doesn't work for me. | |
| Feb 23, 2013 at 4:21 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
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| Feb 23, 2013 at 4:10 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
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| Feb 22, 2013 at 22:43 | history | edited | Duchamp Gérard H. E. | CC BY-SA 3.0 |
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| Feb 22, 2013 at 18:40 | comment | added | Duchamp Gérard H. E. | The idea is to restrict to the algebra $H$ generated by the primitive elements of $A\otimes B$ (I do not now is this has a name in general). This algebra is automativally cocommutative. For details, you can have a look at my recent paper hal.archives-ouvertes.fr/… In order to know the degree of generality, the point is to examine in detail the "domains of convergence" of the elements of the aforementioned computation. For example, $\log_{∗_1}(I_1)\otimes e_2$ projects onto $Prim(A_1)\otimes k$, from $H$. | |
| Feb 22, 2013 at 18:11 | comment | added | darij grinberg | This is a nice argument handling a multitude of cases; thanks a lot. I am not sure about its generality, though. How do we know that $I_1\otimes e_2$ is well-defined as an endomorphism of the subalgebra of $A\otimes B$ generated by the primitives? How do we know that $\log_{*_1}\left(I_1\right) \otimes e_2$ projects onto $Prim\left(A_1\right)\otimes k$ (without having $A_1$ cocommutative to begin with)? | |
| Feb 22, 2013 at 6:16 | history | answered | Duchamp Gérard H. E. | CC BY-SA 3.0 |