Timeline for Why is every variety of bands determined by a single identity?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2015 at 1:34 | comment | added | Benjamin Steinberg | @Gerhard, yes I meant by a single finite semigroup | |
| Jun 7, 2015 at 0:23 | comment | added | Gerhard Paseman | I was wondering about your comment. Do you mean "generated by a single finite semigroup"? Otherwise one could take a certain (infinite) relatively free algebra as the generating algebra. Gerhard "Looking For Some Nontrivial Answers" Paseman, 2015.06.06 | |
| Jun 6, 2015 at 21:32 | comment | added | Benjamin Steinberg | No it is just to help fill in the picture. It is not true that every proper subvariety of a locally finite variety is finitely generated. | |
| Jun 6, 2015 at 16:47 | comment | added | John Baez | @Steinberg - is your remark related to how each variety of bands is defined by a single equation? I don't see how, but I can't help wondering. | |
| Jun 6, 2015 at 0:55 | comment | added | Benjamin Steinberg | Each proper subvariety of bands is generated by a single semigroup. | |
| Jun 5, 2015 at 23:15 | comment | added | John Baez | You write: "recall that an identity in $n$ variables implies (by substitution) several identities in $m$ variables for $m$ less than $n$." This reminds me of something else that Fennemore states: "There exist exactly $10(n- 1)$ distinct varieties which can be defined by sets of identities which imply an identity in $n$ variables, $n \gt 2$. | |
| Jun 5, 2015 at 23:12 | history | edited | The Masked Avenger | CC BY-SA 3.0 |
added 28 characters in body
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| Jun 5, 2015 at 23:05 | history | answered | The Masked Avenger | CC BY-SA 3.0 |