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Nov 21, 2022 at 2:40 vote accept Arshak Aivazian
Nov 20, 2022 at 7:35 comment added მამუკა ჯიბლაძე @SimonHenry Yes, for general $X$ the operadic case is simpler. Just to note, still there is a very similar formula here too: $(\coprod_nX^n\times T(n))/\sim$, where $(X^f(\xi),\omega)\sim(\xi,T(f)(\omega)))$ for $f:m\to n$, $\xi\in X^n$, $\omega\in T(m)$. So the only complication is that one has to take also those $f$ which are not bijections.
Nov 20, 2022 at 5:02 comment added Simon Henry @მამუკაჯიბლაძე : for a general set of generator $X$, the formula for the free algebra for a Lawvere theory is $Colim_{n \to X} T(n)$ with the colimits over all finite set with a map to $X$, for a symmetric operad it is $\sum X^n \times O(n) / \Sigma_n$. Basically, in the case of operads you can clearly separate the contribution of the various power of X, while in the case of Lawvere theory they are mixed. It might not strike everybody as such, but these two formula behave very differently because of this, and I agree that this is one of the key difference between the two.
Nov 19, 2022 at 23:31 history edited Arshak Aivazian CC BY-SA 4.0
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Nov 19, 2022 at 13:41 comment added მამუკა ჯიბლაძე @RuneHaugseng Sorry, I don't understand. Is not the free $n$-generated algebra over a Lawvere theory $T$ just $\hom_T(x^n,x)$?
Nov 17, 2022 at 8:50 comment added Rune Haugseng One reason operads are nicer than general Lawvere theories is that there is a simple formula for their free algebras, which does not exist for arbitrary Lawvere theories.
Nov 16, 2022 at 18:57 answer added Simon Henry timeline score: 22
Nov 16, 2022 at 17:47 history edited Arshak Aivazian CC BY-SA 4.0
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Nov 16, 2022 at 17:41 history edited Arshak Aivazian CC BY-SA 4.0
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Nov 16, 2022 at 17:02 history edited Arshak Aivazian CC BY-SA 4.0
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Nov 16, 2022 at 16:59 comment added Arshak Aivazian What "unwanted behavior" means (in the examples that can be given) I leave to the community to decide. The description of algebraic theories defined by nonsymmetric operads over $\rm{Set}$ is known to me, yes.
Nov 16, 2022 at 6:40 history became hot network question
Nov 16, 2022 at 3:53 comment added მამუკა ჯიბლაძე Very good question (except I don't quite understand what could "undesirable behavior" mean in this context). I am not ready for a full-fledged answer, but at least in standard set-theoretic context, varieties of algebras that can be captured by operads are those definable by identities avoiding repeated uses of the same variable in a term. For example, $(xy)z=x(yz)$ is "operadable" but $x^7=x^4$ is not.
Nov 15, 2022 at 23:30 history edited David White CC BY-SA 4.0
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Nov 15, 2022 at 23:29 answer added David White timeline score: 2
Nov 15, 2022 at 23:21 history edited Arshak Aivazian CC BY-SA 4.0
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Nov 15, 2022 at 23:15 history edited Arshak Aivazian CC BY-SA 4.0
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Nov 15, 2022 at 23:05 history edited Arshak Aivazian CC BY-SA 4.0
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Nov 15, 2022 at 22:45 history edited Arshak Aivazian CC BY-SA 4.0
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Nov 15, 2022 at 22:40 history asked Arshak Aivazian CC BY-SA 4.0