Is there a name/reference for the following object? We have a vector space $V$ over some field with two associative bilinear operations $\circ,*:V \times V \to V$ which satisfy the interchange law, i.e., $$(a*b)\circ(c*d) = (a\circ c)*(b\circ d) $$ for all $a,b,c,d$ in $V$. My first instinct was to search for "bialgebra" but of course that is an algebra and a coalgebra structure rather than two algebra structures...
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asked Nov 23, 2022 at 21:24
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3$\begingroup$ If we ignore the requirement that V is a vector space and $\circ,*$ are associative and bilinear, then this question gives some examples of such structures: mathoverflow.net/q/154550/22277 $\endgroup$– Joseph Van NameCommented Nov 23, 2022 at 21:42
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$\begingroup$ There's a related notion in Definition 1 of my paper arxiv.org/pdf/0912.5307.pdf $\endgroup$– André HenriquesCommented Nov 29, 2022 at 11:06
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$\begingroup$ Thanks, @AndréHenriques! Serves me right for asking here before walking over to ask you or Chris :) In my case there is no unital structure to the two algebras, so I am saved from Eckmann Hilton through other means... $\endgroup$– Vidit NandaCommented Nov 29, 2022 at 13:14
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1 Answer
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If the operations have units ($a * 1 = a$ etc), then that is simply called a commutative algebra: https://en.wikipedia.org/wiki/Eckmann%E2%80%93Hilton_argument Indeed in that case, $a*b = a \circ b = b * a = b \circ a$.
Otherwise, I don't know if there is an established name. This is an algebra over the Boardman-Vogt tensor product of operads $\mathrm{Ass} \otimes_{BV} \mathrm{Ass}$, where $\mathrm{Ass}$ is the operad encoding associative algebras. This is not an uninteresting structure, and it exhibits nontrivial hidden commutativity properties. Here is a summary taken from Boardman--Vogt tensor products of absolutely free operads (Bremner-Dotsenko)
The quoted results are from:
- J. Kock: Note on commutativity in double semigroups and two-fold monoidal categories. Journal of Homotopy and Related Structures 2 (2007) no. 2, 217–228.
- M. Bremner, S. Madariaga: Permutation of elements in double semigroups. Semigroup Forum 92 (2016), no. 2, 335–360.
If I had to choose a name, I would call them "double algebras", but I do not believe this is standard. "Double semigroup" certainly is standard terminology, but there is not requirement of linearity.
answered Nov 23, 2022 at 22:07
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5$\begingroup$ I will confess that I had to read Prop 1.6 about 5 times before I realized that the two sides of the equality were actually different! For others suffering similarly, note that $x_6$ and $x_7$ are swapped. (And yes, I was indeed aware of Eckmann-Hilton but none of my products are unital). I will wait a bit to see if anyone else has seen the structure and given it a catchy name, and will happily accept this answer otherwise :) $\endgroup$ Commented Nov 24, 2022 at 0:42