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Stream: learning: questions

Topic: Operads vs morphisms with product domains

Bernd Losert (Feb 13 2024 at 00:24):

I just started learning about operads a little bit. I like the idea of using operads to build complicated systems out of smaller systems. However, there is a question on my mind that I have not seen addressed anywhere in the stuff I have read: It seems to me that every morphism in an operad is just a morphism whose domain is a product (or a tensored product), so why bother with operads if I can basically do the same with morphisms that have product domains?

Mike Shulman (Feb 13 2024 at 01:21):

Are you using "operad" to mean "colored operad", a.k.a. multicategory?

Bernd Losert (Feb 13 2024 at 01:26):

I think colored operad, but does it matter?

Mike Shulman (Feb 13 2024 at 01:34):

Well, operads have multiple uses. Uncolored operads, and "small" colored operads, are used as presentations of algebraic theories. In this case there is almost never an actual "tensor product object" available: the idea is to represent directly the "$n$-ary operations" in the theory for all $n$.

On the other hand, in "large" colored operads, which are often called multicategories, like the multicategory of sets, it is much more common for there to be an actual representing tensor product object so that the $n$-ary morphisms are bijective to the ordinary morphisms out of the $n$-ary tensor product. In this case the advantages of the multicategorical approach are different. E.g. an algebra for a "small" operad in a "large" one is just an operad morphism from one to the other, and a multicategory structure is often "prior" to the tensor product and can be used to characterize it by a universal property and thereby prove easily that it is associative and so on.

(I put "small" and "large" in quotes to indicate that I don't mean the formal set-theoretic size.)

Bernd Losert (Feb 13 2024 at 01:44):

Hmm.. so it seems to me that if you are in a setting where you have a tensor product object available, then there is not much to be gained from working with operads other than some conveniences.

Kevin Arlin (Feb 13 2024 at 02:19):

It’s true that many people just use monoidal categories, and if you don’t mind muddying your conceptual world up that most often works fine. But the very first interesting tensor product one learns, that of vector spaces, is always introduced in terms of its multicategorical universal property and has no ordinary universal property. This illustrates that there’s an inevitable conceptual need for operads/multicategories. Whether you find such an argument compelling is up to you; the more hard-nosed are likely to prefer to motivate operads via Mike’s smaller examples, such as the $A_n$- and $E_n$-operads that are used regularly by topologists and algebraists for purely pragmatic reasons.

Josselin Poiret (Feb 13 2024 at 09:14):

also to note is that the definition of an operad/multicategory is unbiased compared to that of a monoidal category, although the latter could also be unbiased as well ([[biased definition]])

Bernd Losert (Feb 13 2024 at 12:01):

But the very first interesting tensor product one learns, that of vector spaces, is always introduced in terms of its multicategorical universal property and has no ordinary universal property.

Interesting. Do you have a reference comparing the two?

Patrick Nicodemus (Feb 13 2024 at 12:23):

I don't think there is exactly a rigorous distinction between a "multicategorical universal property" and an "ordinary universal property", although I don't disagree with Josselin, I think what they're saying is correct if understood informally. (This is not to criticize Josselin, my point is to answer your question, that I don't think there is a reference explaining this distinction because it doesn't formally exist)
Formally speaking, for the sake of this conversation, a "universal property" in a category $C$ is a covariant functor
$P: C\to Set$. An object $c\in C$ "has the universal property" if it is equipped with a natural isomorphism $y(c)\cong P$. Equivalently, by Yoneda, an object has the universal property if it is equipped with a distinguished element $\iota \in P(c)$ with the property that for all objects $d$ in $C$, the function $f \mapsto P(f)(\iota)$ determines a one to one correspondence $Hom(c,d)\cong P(d)$.

Patrick Nicodemus (Feb 13 2024 at 12:24):

In this case, for two modules $M,N$, the "universal property of the tensor product" refers to the covariant functor $\operatorname{Bilin}(M, N; -)$, which sends a module $X$ to the set of bilinear maps from $M, N$ to $X$. This is inherently a notion of the multicategory of vector spaces.

Patrick Nicodemus (Feb 13 2024 at 12:25):

The next simplest way to describe an isomorphic presheaf (i.e., to describe the same universal property) is to define $Q(X)$ as the set of linear maps from $M$ to the $R$-module $Hom_R(N,X)$ of morphisms, which can again be checked to be functorial in $X$. Conceptually this is arguably more complicated as it requires us to already understand the concept that the category of $R$-modules is enriched over itself. It also has the downside that it is not immediately obvious that this is a symmetric notion in $M,N$, i.e. the presentation itself is obviously asymmetric in the roles of $M$ and $N$.

Patrick Nicodemus (Feb 13 2024 at 12:35):

One example (that may be motivating?) of a multicategory which is not a monoidal category, is as follows. Let $C$ be a large but locally small category. We define a multicategory whose objects are profunctors on $C$ taking values in small sets. There is a standard notion of profunctor composition that would allow us to make $Prof(C)$ into a monoidal category except that the composition of profunctors valued in small sets may no longer be valued in small sets. Therefore you cannot talk about $P_1\otimes \dots P_n$. Nevertheless the universal property of $P_1\otimes\dots \otimes P_n$ is still sensible, so that we can talk about multi-natural transformations $P_1,\dots, P_n \Rightarrow Q$.

Patrick Nicodemus (Feb 13 2024 at 12:35):

(One can resolve this difficulty by simply passing to a larger universe but I find this conceptually inelegant, as it is essentially just kicking size issues down the road until they rear their head somewhere else. I don't have a good example off the top of my head of a formal situation where it would be fatal to do this, but there are some obvious infelicities, for example most categories do not have large colimits so you can no longer talk about weighted colimits with weights in an arbitrary presheaf when you allow presheaves to be valued in large sets.)

Bernd Losert (Feb 13 2024 at 12:38):

Thanks for the substantial reply. I will have to meditate on this for a bit.

Patrick Nicodemus (Feb 13 2024 at 12:42):

More simply, and Tom Leinster points this out in his book on "Higher operads" - if $C$ is a monoidal category, then any subcategory $M$ of $C$ which is not closed under the monoidal product is a multicategory which is not a monoidal category. So if you ever need to restrict yourself to a subcategory where you cannot actually talk about the monoidal product because it doesn't live in your category, you would have to use multicategories.

As a contrived example, take, say, the category of vector spaces less than dimension $10$.

Patrick Nicodemus (Feb 13 2024 at 12:44):

I guess my example is a special case of this as well. We do have a monoidal category of profunctors valued in large sets, but we care about profunctors valued in small sets because these are the ones which are important in the theory of weighted colimits, so I exclude profunctors with large sets from my category.

James Deikun (Feb 13 2024 at 12:47):

Thinking in terms of multicategories also adds conceptual clarity to the primary importance of lax (rather than oplax or pseudo) monoidal functors. Lax monoidal functors are just the functors that preserve the multicategory structure of multicategories that happen to be monoidal, rather than preserving the structure of a monoidal category itself.

Mike Shulman (Feb 13 2024 at 17:28):

I think probably what Kevin meant by "has no ordinary universal property" is that there is (at least apparently) no purely categorical construction or characterization that can be applied to the bare category of vector spaces to produce the tensor product on it. Specifying the functor $\mathrm{Bilin}(M,N;-)$ (which is almost tantamount to enhancing Vect to a multicategory) or the internal-hom $\mathrm{Hom}_R(M,N)$ is giving extra structure to the category Vect, in terms of which we can then characterize the tensor product, but it's unclear whether there is any way to characterize it purely by reference to the category Vect. In other words, there does not appear to be a predicate depending on an abstract category $C$ and three objects of $C$ such that when $C=\rm Vect$ the predicate isolates precisely the pairs $(M,N,M\otimes N)$.

John Baez (Feb 13 2024 at 21:06):

A bunch of us say the tensor product on Vect doesn't arise just from the category Vect (it would be nice to make this precise in various ways and prove it) but from the adjunction between Vect and Set. This adjunction makes it possible to define bilinear (and multilinear) maps between vector spaces.

Kevin Arlin (Feb 14 2024 at 22:47):

That's an interesting point because it might tempt you to think you could define "bilinear" maps $(A,B)\to C$ for $A,B,C$ in (say) any category $T$-Alg algebraic over Set as functions $A\times B\to C$ such that for every point of the underlying sets of $A$ or of $B$, the induced maps $A\to C$ lie in $T$-Alg. But I guess this won't really work when $T$ isn't a commutative theory. Maybe what happens with this is you get to define a "tensor product" that collapses into the center of the theory $T$? I vaguely seem to remember that's what happens for groups.