Timeline for Topology of the algebra $\mathbb{C}\{A\}$ for a LCA group $A$
Current License: CC BY-SA 3.0
11 events
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| May 29, 2017 at 13:38 | comment | added | YCor | when I read contradictory statements, I try to fix them. So I understood $\mathbf{C}\{\mathcal{A}\}$ to be the subalgebra generated by $A$ and I asked about the meaning of the last sentence. I apologize not to have guessed the right meaning. Now it sounds clear. | |
| May 29, 2017 at 13:20 | comment | added | Bedovlat | If it were an additive subgroup, the multiplicative structure of $\mathcal{A}$ would be irrelevant, and it would be hard to make sense of $\mathbb{C}\{A\}$ as a subalgebra of $\mathcal{A}$. This should have stopped the reader already in the first paragraph. Anyway, I have amended the statement. Hope it's clear now. | |
| May 29, 2017 at 13:11 | history | edited | Bedovlat | CC BY-SA 3.0 |
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| May 29, 2017 at 13:10 | comment | added | YCor | Yes you mean $A\subset\mathcal{A}^\times$ a multiplicative subgroup, I indeed understood an additive subgroup. | |
| May 29, 2017 at 13:07 | comment | added | Bedovlat | Ok, you might have confused an additive subgroup with a multiplicative subgroup. If that is the case, I can understand. I will amend the statement. | |
| May 29, 2017 at 13:01 | comment | added | Bedovlat | First of all, $A\subset\mathcal{A}$ is a subGROUP, so $0\in A$ is excluded. Of course, it may happen that $A$ is not a linearly independent set. That is exactly why I have written that sentence. So, again, what is abusrd here? | |
| May 29, 2017 at 12:22 | comment | added | YCor | If you have a linear space $\mathcal{A}$ and a subset $Z$, to say that "element of $Z$ are linearly independent in $\mathcal{A}$ has some standard sense", namely there is no nontrivial linear combination between them. If $0\in Z$, or if $Z$ contains some vector $v$ as well as $2v$, this is not the case. | |
| May 29, 2017 at 12:09 | comment | added | Bedovlat | What exactly does sound to you like absurd? | |
| May 29, 2017 at 11:15 | comment | added | YCor | What do you mean by "elements of $A$ are linearly independent in $\mathcal{A}$"? it sounds absurd. | |
| May 29, 2017 at 0:59 | history | edited | Bedovlat | CC BY-SA 3.0 |
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| May 28, 2017 at 23:07 | history | asked | Bedovlat | CC BY-SA 3.0 |