Timeline for Free monoids on posets
Current License: CC BY-SA 4.0
26 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 21, 2021 at 22:16 | comment | added | Benjamin Steinberg | OK. So if S is not totally ordered we can't first make the order total. | |
| Apr 21, 2021 at 18:44 | comment | added | Jeff Strom | @BenjaminSteinberg I definitely want the order on $S$ to remain unchanged. | |
| Apr 21, 2021 at 16:06 | comment | added | Benjamin Steinberg | Does the order on F(S) have to restrict to the original order on S or can it extend the order? | |
| Apr 21, 2021 at 16:04 | comment | added | Benjamin Steinberg | I don't know a description of that space. It might be massive | |
| Apr 21, 2021 at 15:58 | comment | added | Jeff Strom | @BenjaminSteinberg Sure! If you make it an answer, I'll certainly upvote it. | |
| Apr 21, 2021 at 15:53 | comment | added | Benjamin Steinberg | The compatible partial orders on the free monoid form a compact totally disconnected space and the space of such orders that agree with the original order on S is a closed subspace. If S is finite it is even clopen. Do you want a description of that space? | |
| Apr 21, 2021 at 15:48 | comment | added | Benjamin Steinberg | This looks a bit like term orderings except that they often require total orders. | |
| Apr 21, 2021 at 15:23 | comment | added | Jeff Strom | @SamHopkins Yes, that's right. | |
| Apr 21, 2021 at 15:22 | history | edited | Jeff Strom | CC BY-SA 4.0 |
added 280 characters in body
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| Apr 21, 2021 at 15:19 | comment | added | Sam Hopkins | I see, thanks. So as Benjamin Steinberg notes you are not really talking about a single construction. | |
| Apr 21, 2021 at 15:19 | comment | added | Jeff Strom | @SamHopkins I said if and did not mean only if -- clarified now. | |
| Apr 21, 2021 at 15:18 | history | edited | Jeff Strom | CC BY-SA 4.0 |
more clarification on inequality
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| Apr 21, 2021 at 15:15 | comment | added | Benjamin Steinberg | I thought you were defining the order that way. Then your question is ambiguous since there are many orderings extending the ordering on S. | |
| Apr 21, 2021 at 15:10 | comment | added | Jeff Strom | @BenjaminSteinberg I don't think so. For example, say $S = \{ x,y,z\}$ with $x< z$ and $y< z$. Then in the extended partial order, it can happen that $xy< z$ or not. | |
| Apr 21, 2021 at 15:01 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Apr 21, 2021 at 15:01 | comment | added | Benjamin Steinberg | This seems to just be the disjoint union of the product orders on the sets $S^n$ and then observing that concatenation obviously respects this. | |
| Apr 21, 2021 at 14:56 | comment | added | Benjamin Steinberg | Do you mean $s_i\leq t_i$? | |
| Apr 21, 2021 at 14:55 | comment | added | Sam Hopkins | It seems that the poset you defined is just the disjoint union of $S^n$ over all $n\geq 0$; how much more could there be to say about it? | |
| Apr 21, 2021 at 14:53 | comment | added | Jeff Strom | @BenjaminSteinberg It's like I've found a new gadget, and I'm looking for the instruction book. I don't specifically have in mind that left adjoint, but I'd be happy to read about it. | |
| Apr 21, 2021 at 14:52 | comment | added | Jeff Strom | @SamHopkins No, it is genuinely free. | |
| Apr 21, 2021 at 14:51 | comment | added | Jeff Strom | @YCor Clarified! | |
| Apr 21, 2021 at 14:51 | history | edited | Jeff Strom | CC BY-SA 4.0 |
clarified order preservation
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| Apr 21, 2021 at 14:31 | comment | added | Benjamin Steinberg | Are you looking for a left adjoint to the forgetful functor from pomonoids to posets? | |
| Apr 21, 2021 at 14:12 | comment | added | Sam Hopkins | Do you mean a quotient of the free monoid? | |
| Apr 21, 2021 at 14:11 | comment | added | YCor | What do you mean by "multiplication respects the partial order"? the order is a priori on $S$, not on $F(S)$. | |
| Apr 21, 2021 at 14:10 | history | asked | Jeff Strom | CC BY-SA 4.0 |