Skip to main content
16 events
when toggle format what by license comment
Oct 2, 2015 at 7:54 comment added Najib Idrissi Just to be clear, what is called an "operad" in the paper you linked is the standard version of operad. However, they also consider "Cartesian operads" = "operads of type $\mathbb{F}^{op}$", and this is a different beast, which is apparently equivalent to a Lawvere theory. If I were you, I would drop the word "operad" altogether from the question and say "Lawvere theory" from the get-go...
Oct 2, 2015 at 4:31 vote accept user40276
Oct 2, 2015 at 2:10 comment added Qiaochu Yuan @Samuele: in that question this identity was encoded by showing that it was equivalent (given suitable assumptions) to other identities which don't involve multiple occurrences of a given variable. As far as I know, there's no way to do this for the inverse identity $a a^{-1} = a^{-1} a = 1$. For flexible algebras you're looking for an operad in vector spaces so there are tricks you can play with multilinearity, but here the best we can hope for is an operad in sets.
Oct 1, 2015 at 21:16 comment added Samuele Giraudo @QiaochuYuan: it is not true that operads cannot encode relations with multiple occurrences of a given variable. For instance flexible algebras, that are nonassociative algebras wherein product satisfies $(xy)x = x(yx)$, can be encoded by a symmetric operad (see mathoverflow.net/questions/128547/…? for instance).
Oct 1, 2015 at 20:16 answer added Todd Trimble timeline score: 9
Oct 1, 2015 at 19:19 comment added Qiaochu Yuan ...that the machinery of resolutions of operads (in the ordinary sense) continues to apply.
Oct 1, 2015 at 19:17 comment added Qiaochu Yuan @user40276: that's a pretty nonstandard use of the term "operad" as far as I can tell. In any case, for going up to groupoids there's an additional problem even for writing down a Lawvere theory. First, it has to be two-sorted, since groupoids have both an object of objects and an object of morphisms. Second, and much more importantly, composition isn't a morphism from some product to some other product; instead it involves a pullback. So you need something more sophisticated than a Lawvere theory, namely (I think?) a finite limit sketch. And at this point I see no reason to expect...
Oct 1, 2015 at 19:13 history edited user40276 CC BY-SA 3.0
added 290 characters in body
Oct 1, 2015 at 18:52 comment added Todd Trimble (Operads in Qiaochu's last comment referring to the default notion of permutative operad = monoid with respect to plethystic monoidal structure on the category $Set^\mathbb{P}$ where $\mathbb{P}$ is the category of finite permutations. One could stretch the meaning of operad so that a Lawvere theory is a "cartesian operad". But then that stretch should be made clear in the question!)
Oct 1, 2015 at 18:50 comment added user40276 @QiaochuYuan Maybe my knowledge about operads is too poor, but ins't a Lawvere theory the same thing as an operad where the monoidal structure is the cartesian one (for instance, see page 38 here cs.ox.ac.uk/people/bob.coecke/andrei.pdf).
Oct 1, 2015 at 18:07 comment added Qiaochu Yuan @user40276: the problem is that the defining property of inverses, namely that $a^{-1} a = a a^{-1} = 1$, requires that the same variable appear twice in an equation. Operads can't encode these sorts of equational laws, although Lawvere theories can.
Oct 1, 2015 at 18:03 comment added user40276 @GijsHeuts Why not? When there's the pullback $\mathcal{G}_{1} \times_{s, t} \mathcal{G}_1$, it's possible to define a map (called inversion) $i: \mathcal{G}_1 \rightarrow \mathcal{G}_1$ satisfying some diagrams commutativity
Oct 1, 2015 at 17:31 comment added Gijs Heuts Even in Cartesian monoidal categories, groups and groupoids cannot be encoded as algebras over operads. There is no way to encode the existence of inverses.
Oct 1, 2015 at 17:22 comment added user40276 @QiaochuYuan There is the notion of internal category in a monoidal category ncatlab.org/nlab/show/internal+category+in+a+monoidal+category . However in my case, I'm interested when the pullback exists (not necessarily all pullbacks though)
Oct 1, 2015 at 17:17 comment added Qiaochu Yuan What do you even mean by an internal groupoid in a monoidal category that isn't cartesian?
Oct 1, 2015 at 17:10 history asked user40276 CC BY-SA 3.0