Timeline for Semigroups related to iterated orthogonal complement
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Jan 10, 2023 at 19:09 | comment | added | Benjamin Steinberg | I think the issue here is that your r,l are order-reversing and order-reversing galois connections are a bit messy under composition. | |
| Jan 10, 2023 at 18:31 | comment | added | Benjamin Steinberg | This is somehow related to residuation theory by Blyth and otehrs, but because your Galois connection reverses order, I am a bit confused. You might look at the paper Regular monoids generated by two galois connections by Martin Skorsky, Semigroup Forum volume 39, pages 263–293 (1989). He looks at something similar to what you do but not exactly the same. He gets a regular semigroup out of this and you might look at his section 2 for ideas. | |
| Jan 10, 2023 at 17:02 | history | edited | user494312 | CC BY-SA 4.0 |
corrections due to comments by @BenjaminSteinberg
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| Jan 10, 2023 at 16:56 | comment | added | user494312 | @BenjaminSteinberg: Oh, you are right. It is not a semi-group. And what I wrote using $S^{-1}$ is nonsense, and is unfixable. I meant to use take orthogonals with respect to $R^{-1}$, but do not see now how it might helps. Let me edit the question as you suggest. | |
| Jan 10, 2023 at 14:31 | comment | added | Benjamin Steinberg | Also, I don't believe what you define is actually a semigroup if you leave $S$ fixed. Your equivalence relation, as I understand it, is that the free monoid on l and r act on the right P(V) by your definition above. But then, the equivalence relation is only a right congruence and not a congruence. Just because $S^u=S^v$, I don't see any reason why $S^wu=S^wv$. I think you would get a congruence if you define u equivalent to v if T^u=T^v for all subsets T of V. | |
| Jan 10, 2023 at 14:28 | comment | added | Benjamin Steinberg | But $S$ is a set not a relation | |
| Jan 10, 2023 at 12:36 | comment | added | user494312 | The inverse of $S$: $S^{-1}:=\{ (y,x)\in V\times V: (x,y)\in S\}$. That is, $(x,y)\in S$ iff $(y,x)\in S^{-1}$. | |
| Jan 10, 2023 at 12:09 | comment | added | Benjamin Steinberg | What does S^{-1} mean? | |
| Jan 10, 2023 at 9:28 | comment | added | user494312 | @BenjaminSteinberg: Let $u^T$ denote the result of replacing each letter $l$ by $r$, and each $r$ by $l$, i.e. $l^T:=r$, $r^T:=l$, and $(xy)^T:=x^Ty^T$. Let $S^{-1}:=\{(y,x): (x,y)\in S\}$. Then $u=_{S,R}u u^T u$, as follows from the following. $S^r = ((S^{-1})^l)^{-1}$, and, more generally, $S^u=((S^{-1})^{u^T})^{-1}$. This implies that $l=_{S^u,V} lrl$ means that $ul=_{S,V} ul u^T r ul = (ul) (ul)^T (ul)$. | |
| Jan 10, 2023 at 9:27 | comment | added | user494312 | @BenjaminSteinberg: Yes, I think so, though it requires a little argument. Here is a (corrected) argument. | |
| Jan 10, 2023 at 1:45 | comment | added | Benjamin Steinberg | Are you sure the semigroup is regular? It isn't enough for the generators to be regular. | |
| Jan 9, 2023 at 17:45 | history | edited | user494312 | CC BY-SA 4.0 |
added 36 characters in body; edited title
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| Jan 9, 2023 at 17:45 | history | edited | YCor | CC BY-SA 4.0 |
formatting
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| Jan 9, 2023 at 17:43 | history | asked | user494312 | CC BY-SA 4.0 |