Timeline for (Non trivial) coidempotents(Co-$K$-theory)
Current License: CC BY-SA 4.0
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| S Apr 7, 2020 at 4:04 | history | bounty ended | CommunityBot | ||
| S Apr 7, 2020 at 4:04 | history | notice removed | CommunityBot | ||
| Mar 30, 2020 at 3:35 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Mar 30, 2020 at 2:36 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| Mar 30, 2020 at 2:26 | history | edited | Ali Taghavi | CC BY-SA 4.0 |
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| S Mar 30, 2020 at 2:16 | history | bounty started | Ali Taghavi | ||
| S Mar 30, 2020 at 2:16 | history | notice added | Ali Taghavi | Draw attention | |
| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 17, 2014 at 3:55 | comment | added | Mariano Suárez-Álvarez | Moreover, as soon as a group G is non-trivial is should be obvious to you that $\mathbb CG$ has non-trivial coidempotents! | |
| Apr 17, 2014 at 3:51 | comment | added | Mariano Suárez-Álvarez | The isomorphism type of that coalgebra depends only on the cardinal of the group and not on the group structure. | |
| Apr 17, 2014 at 3:43 | comment | added | Ali Taghavi | @MarianoSuárez-Alvarez I consider the same structure as you mentioned :$\Delta(g)=g\otimes g$. could you please explain why you excpect that the answer is no(what is the role of cardinal)? | |
| Apr 17, 2014 at 3:08 | comment | added | Mariano Suárez-Álvarez | How do you consider $\mathbb CG$ as a coalgebra? Its standard coalgebra structure is the one in which all elements of G are grouplikes and depends only on the cardinal of G, so it should be more or less expected that the answer to your last question is no... | |
| Apr 17, 2014 at 2:00 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 17, 2014 at 0:48 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 16, 2014 at 7:51 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 16, 2014 at 7:42 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 16, 2014 at 6:57 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 16, 2014 at 5:20 | comment | added | Mariano Suárez-Álvarez | over itself, so there are no non-trivial coidempotents. | |
| Apr 16, 2014 at 5:13 | comment | added | Mariano Suárez-Álvarez | Your definition of idempotent in an algebra is simply that of an idempotent in the endomorphism álgebra of its right regular module; your definition of a coidempotent is, similarly, that of an idempotent in the álgebra of endomorphisms of the regular right comodule. In particular, coidempotents correspond to direct summands of the regular right comodule. In the case of comatrix coalgebras, the right regular comodule is a direct sum of n simple comodules —just as in the álgebra case); $\mathbb C[x]$ with the coalgebra structure which makes x primitive, is indecomposable as a comodule | |
| Apr 16, 2014 at 5:11 | comment | added | Mariano Suárez-Álvarez | The álgebra of matrices with entries in an algebra is just the tensor product of that álgebra with $M_n(\mathbb C)$; it is natural to define the comatrix coalgebra with entries in a coalgebra C as the tensor product of C and the comatrix coalgebra (which is just the coalgebra dual to the álgebra $M_n(\mathbb C)$) | |
| Apr 16, 2014 at 5:07 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 16, 2014 at 5:02 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 15, 2014 at 21:43 | comment | added | Ali Taghavi | Now define $\tilde{\Delta_{n}}$ with $\tilde {\Delta_{n}}(\delta{ij}(c))$= the above product of $\Delta_{C}(c)$ by $\Delta_{n}(\delta_{ij})$. Is this $\tilde{\Delta_{n}}$ a coproduct on $M_{n}(C)$? | |
| Apr 15, 2014 at 21:35 | comment | added | Ali Taghavi | Let $(C,\Delta_{C})$ be a coalgebra and $\Delta_{n}$ be the standard coproduct structure on $M_{n}(\mathbb{C}$. Let $\delta_{ij}$ be the element of $M{n}(\mathbb{C})$ with $1$ in i-j place and zero other wise. for $c\in C$ let $\delta_{ij}(c)$ be the matric in $M_{n}(C)$ with only one non zero element $c$ in i-j place. Define a special product as follows; the product of $x\otimes y \in C\otimes C$ and $\delta_{ij}\otimes \delta_{jk} \in M_{n}(\mathbb{C}) \otimes M_{n}(\mathbb{C}$ is $\delta_{ij}(x) \otimes \delta_{jk}(y)$, as an element in $M_{n}(C) \otimes M_{n}(C)$. | |
| Apr 15, 2014 at 8:33 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 14, 2014 at 18:53 | history | edited | Ali Taghavi | CC BY-SA 3.0 |
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| Apr 14, 2014 at 18:44 | history | asked | Ali Taghavi | CC BY-SA 3.0 |