Timeline for Coalgebras(or quantum groups) which admit a linear operator satisfying certain functional equation
Current License: CC BY-SA 4.0
31 events
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| Apr 26, 2019 at 8:49 | comment | added | Ali Taghavi | @KonstantinosKanakoglou Thanks for your suggestion. I read your comment very late(just now). i am sorry about that. | |
| Feb 16, 2019 at 1:16 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Feb 2, 2019 at 2:02 | comment | added | Konstantinos Kanakoglou | @ user44191: Nice! and maybe it would be even better, if all these notes and comments, including the OP and the answer, were recorded in the form of a new question here (towards particular or general solutions of the OP functional equation) or maybe in a post in nLab. | |
| Feb 1, 2019 at 3:00 | comment | added | user44191 | As a note, the example you have is in fact dual to a case of what @მამუკაჯიბლაძე wrote in the comments to the main post; the dual algebra to the group coalgebra is commutative, and the action you've chosen comes from multiplication by the idempotent $\mathbf{1}_S$ (which is central because the algebra is commutative). | |
| Jan 31, 2019 at 22:10 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 31, 2019 at 21:24 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 31, 2019 at 21:15 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 31, 2019 at 15:34 | comment | added | user44191 | I've made a chat:chat.stackexchange.com/rooms/89079/… | |
| Jan 31, 2019 at 1:12 | comment | added | Konstantinos Kanakoglou | I am not sure why $T^{-1}(0)$ should be a coideal. Am i missing something simple ? | |
| Jan 30, 2019 at 5:07 | comment | added | user44191 | And the statement that $T^*$ acts on $T^*(1)$ on $T^*(C^*)$ should be equivalent to a statement on the quotient coalgebra $C/T^{-1}$. | |
| Jan 30, 2019 at 2:41 | comment | added | user44191 | There is no direct connection between $T(C)$ and $T^*(C^*)$; instead, $T^{-1}(0)$ is the subspace of $C$ that "identifies" $T^*(C^*)$, the subalgebra in the dual of $C$, and so is its natural counterpart. The right side of the resulting equation is then clear: $(\Delta \circ T^2)(T^{-1}(0)) = 0$, so $((T \otimes T) \circ \Delta)(T^{-1}(0)) = 0$. I think that should imply that $T^{-1}(0)$ is a coideal. | |
| Jan 29, 2019 at 23:24 | comment | added | Konstantinos Kanakoglou | @user44191, my initial thought when i read your dual comment, was that $T(C)$ or $T^2(C)$ should be picked and that they would be coideals rather than subcoalgebras, in order to "read" the argument in the coalgebra language. However your proposal on picking $T^{-1}(0)$ seems interesting as well. | |
| Jan 29, 2019 at 22:13 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 29, 2019 at 22:10 | comment | added | Konstantinos Kanakoglou | In any case, this (i mean your dual statement on algebras) seems to be an interesting remark. Maybe it would be useful to discuss this and its possible translation, in a separate question (since this is not what the OP is asking about). | |
| Jan 29, 2019 at 22:07 | comment | added | Konstantinos Kanakoglou | @user44191, thank you for your feedback. You are probably right in your observation that $T(C)$ need not be a subcoalgebra in general. So the first point of the "translation" is probably incorrect. I will remove it and try to come back after i will give it some more thought. | |
| Jan 29, 2019 at 21:23 | comment | added | user44191 | I think instead of $T(C)$, $T^{-1}(0)$ is likely to be more useful. | |
| Jan 29, 2019 at 21:02 | comment | added | user44191 | I think your translation is incorrect; removing the $T$ (analogous to where I essentially said $T(xy) \in C$), the inclusion goes in the other direction, that is, $(T \otimes T) \circ \Delta(C) = \Delta(T(T(C)) \subseteq \Delta(T(C))$. | |
| Jan 29, 2019 at 20:25 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 29, 2019 at 18:52 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 29, 2019 at 5:57 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 29, 2019 at 5:45 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 29, 2019 at 5:12 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 29, 2019 at 4:58 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 28, 2019 at 23:48 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 28, 2019 at 23:42 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 28, 2019 at 2:42 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 27, 2019 at 5:55 | comment | added | Ali Taghavi | Thank you very much for your answer. | |
| Jan 26, 2019 at 18:23 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 25, 2019 at 23:05 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 25, 2019 at 22:58 | history | edited | Konstantinos Kanakoglou | CC BY-SA 4.0 |
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| Jan 25, 2019 at 22:53 | history | answered | Konstantinos Kanakoglou | CC BY-SA 4.0 |