Timeline for Is every monomorphism of commutative Hopf algebras (over a field) injective?
Current License: CC BY-SA 2.5
10 events
| when toggle format | what | by | license | comment | |
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| Feb 9, 2010 at 21:23 | comment | added | Nicolas Schmidt | Ah, now I see what you mean. No, I think the argument is perfectly valid. | |
| Feb 7, 2010 at 17:34 | comment | added | Pavel Etingof | I'd like to add that even if only $f$ is given, you can determine the decomposition of $V$ into irreducible $sl(2)$-modules. Namely, the irreducible summands correspond to Jordan blocks of $f$. In particular, $V$ is irreducible iff $f$ is a regular nilpotent matrix, i.e. $f^{dim(V)-1}$ is nonzero. | |
| Feb 7, 2010 at 17:26 | comment | added | Pavel Etingof | I don't think I used that the representation of $sl(2)$ is irreducible. I claimed that if V is any representation of $sl(2)$ and we are given the operators $h$ and $f$ on this representation, then we can uniquely reconstruct the action of $e$. Namely, 1) if $v$ is a nonzero vector with $hv=mv$ and $f^{m+1}v=0$, then $ev=0$ (this is true for any finite dimensional $sl(2)$-module); 2) if $u=f^nv$, where $v$ is as in (1), then $ef^nv$ is computed in the usual way using $[e,f]=h$. Do you disagree with this? | |
| Feb 7, 2010 at 13:37 | comment | added | Nicolas Schmidt | Your argument basically depends on the fact, that an irreducible lie representation of sl(2) on a finite dimensional vector space $V$ is completely determined by the endomorphisms representing e and h: Take $v$ any eigenvector of $h$ with $ev = 0$, then we can construct canonically a basis of $V$, where the action of $f$ is a priori known. However splitting a given representation into irreducible ones, or even determining whether it is irreducible seems to depend on the whole action of sl(2). | |
| Feb 6, 2010 at 23:20 | comment | added | Pavel Etingof | Can you please explain what may be a problem with this argument and which technical assumptions do you mean? | |
| Feb 6, 2010 at 21:50 | vote | accept | Nicolas Schmidt | ||
| Feb 6, 2010 at 21:50 | comment | added | Nicolas Schmidt | I'm not sure if this argument is 100% correct, but it seems to work under some mild technical assumptions. Either way I am convinced now that the answer is "no". As I have understood the article of Chirvasitu mentioned below, the reason for this should be the lack of faithful coflatness. | |
| Feb 5, 2010 at 12:19 | history | edited | Pavel Etingof | CC BY-SA 2.5 |
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| Feb 5, 2010 at 5:56 | history | edited | Pavel Etingof | CC BY-SA 2.5 |
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| Feb 5, 2010 at 5:47 | history | answered | Pavel Etingof | CC BY-SA 2.5 |