The n-Category Café

http://www.penguin.cz/~utx/amr
The problem is proprietary codecs which require a license before you compile them on VLC which is open source.
Because you are academics (not commercial) these will probably be free. I dual boot, and couldn’t play the video on Quicktime Pro on Windows. So I watch the video on VLC
on one smaller window and play the audio on another smaller window, and sync them manually (start video first).
As long as you can play the video on VLC in Linux, then JB and crew should be able to redistribute the audio separately in mp3 format and you could sync them as I did. It seems to me that the app which created the audio portion should also have a ‘save as’ feature for the non-proprietary .mp3 format or there is likely a converter to mp3 from the original audio format. This solution assumes there is a problem in recreating the video download files in another format, redoing all of the released file.

Both Jacques’s link and

http://ncatlab.org/jamesdolan/published/Algebraic+Geometry

take me to the appropriate pages without any login.

Mike wrote:

Why do people say “doctrine” rather than 2-monad?

The main reason is outlined in my reply to James below: the word ‘doctrine’ brings its own intuitions, which are valuable in themselves, in part because they haven’t been completely formalized yet. There are things that perhaps should be doctrines that aren’t 2-monads.

The vague idea of a ‘doctrine’ is that it’s a kind of ‘language’ in which you can write theories. Some doctrines are more powerful than others. In a powerful doctrine you can say fancy things; in a weak one you can only say very simple things.

But then something funny happens: if we use category theory to make the concepts of ‘theory’ and ‘doctrine’ precise, we realize that a ‘doctrine’ is like a categorified version of a ‘theory’.

Why? Because more technically, a theory $T$ is a category equipped with extra stuff! A ‘model’ of this theory is a functor $F: T \to Set$, where $F$ preserves this extra stuff. The idea is that this functor picks out a set and equips it with extra structure.

So, a theory is a category equipped with extra stuff… but what sort of extra stuff? Well, that’s what your doctrine tells you!

So, we see:

A ‘theory’ is something you use to describe sets equipped with extra structure (or more generally objects in categories with extra structure).

A ‘doctrine’ is something you use to describe categories equipped with extra stuff (or more generally objects in 2-categories equipped with extra stuff).

So a doctrine is like a categorified theory.

Now, I’m sure you know this all very well, Mike — I just felt like explaining it to everyone else. But the point is that there are examples of this pattern that don’t arise from 2-monads — at least, not as easily as you might want. (See below.)

Why do people say “doctrine” rather than 2-monad? Anyone who hangs out around here should be able to guess what a “2-monad” is even if they’ve never heard of one before, but “doctrine” conveys no intuition (to me) until someone tells you that it means “2-monad.”

“anyone who hangs out around here” is not a good description of my intended audience; there are fashions of speaking and thinking around here that strike me as unhelpfully complicated.

since you seem to suggest that pre-formal intuition can be useful, i’ll give some of my own pre-formal intuition about doctrines here:

first of all, forget 2-monads; they’re a red herring here.

a doctrine is a 2-category in which it’s appropriate to think of the objects as “theories” and to think of a hom-category hom(x,y) as “the category of interpretations of theory x into theory y”
or as “the category of models of the theory x in the environment y”.

in other words, as james borger put it, the concept of doctrine is all about “logic-inspired words like ‘theory’ and ‘model’”.

the important point is that all you need to proceed with the applications to algebraic geometry is:

doctrine = 2-category

theory of a doctrine = object of a 2-category

model of a theory of a doctrine = morphism out of an object of a 2-category

further discussion of how to sharpen the definition of “doctrine” (or of some substitute word) is of course possible but arguably futile unless preceded by an understanding of the intended examples (the main ones being the doctrine of “dimensional theories”, the doctrine of “multi-dimensional theories”, and the doctrine of “coherently locally ringed toposes”) and more especially of the relevant morphisms between them, reflecting the way in which these doctrines fit into a “hierarchy” or “ladder” of “increasing expressiveness”.

(of the 3 doctrines listed above, the first and third have stable precise definitions; the definition (and name) of the second doctrine is more provisional, but an example of the kind of category that qualifies as a theory of the second doctrine is the category of coherent sheaves over a scheme.)

there’s no pseudomonad for cartesian closed categories

There is; it just lives on the 2-category $Cat_g$ of categories, functors, and natural isomorphisms rather than $Cat$.

Also, there must be a version of Beck’s theorem in this context.

One version can be found in

• Le Creurer, I. J. and Marmolejo, F. and Vitale, E. M. Beck’s theorem for pseudo-monads. JPAA 173 (2002), 293–313.

Another is in the appendix of

• Hermida, Claudio. Descent on 2-fibrations and strongly 2-regular 2-categories. Appl. Categ. Structures 12 (2004), 427–459.

Note, though, that most definitions of “pseudomonad” require the functor part to be a strict 2-functor. Thus, this is not the “completely weak” version of a 2-monad which would consist of a weak 2-functor on a weak 2-category, etc.

By the way, when I say there’s no pseudomonad for CCC’s, I mean there’s no pseudomonad on Cat whose (pseudo)algebras are CCC’s. There is, I believe, a pseudomonad on

$$Cat_g = [categories, functors, natural isomorphisms]$$

whose algebras are CCC’s.

(This has been heavily on my mind lately since I’m working on this topic with Mike Stay.)

So, I’m not saying 2-monads (or better, pseudomonads) are ‘no good’ as a formalization of the intuitive concept of ‘doctrine’. I’m really just trying to say that there’s an intuitive concept there which is a bit tricky to formalize and thus deserves a name different from the various proposed formalizations.

Math is full of words like this: ‘space’, ‘structure’, ‘system’, ‘geometry’, ‘theory’….

So, I don’t say ‘doctrine’ when I know I mean ‘2-monad’ or ‘pseudomonad’. I say it when I want to be a bit vague. And I think James Dolan does that too.

And, just to make things even more intimidating, Marmolejo says ‘Gray-category’ where I say ‘3-category’, and ‘Gray’ where I say ‘2Cat’.

Well, I don’t think Francisco is setting out to intimidate people! :-) He’s just being precise: $Gray$ just means 2Cat but with the Gray tensor product as monoidal product (the “right” monoidal product for our purposes here). And a $Gray$-category means a category “enriched in Gray” – a precise way of saying what people in these parts usually mean by a 3-category (as opposed to strict 3-category).

You know this of course, but let’s not give the idea that Marmolejo is being unwarrantedly jargon-y! He’s fine.

lawvere discusses beck’s concept here

more specifically, on page 22 at the above link

I don’t get much out of Lawvere’s comments at that link, except what John already said. And that still looks to me as though it is well-described by a 2-monad, or at least a finitary 2-monad (aka 2-Lawvere-theory). Or equivalently the category of algebras for such a 2-monad, which of course is equivalent data to the 2-monad if you remember its forgetful functor.

I would like to discuss what you are actually trying to explain, instead of (or in addition to) arguing about terminology, because it sounds really nifty! But unfortunately I don’t have time to watch 6+ hours of video at the moment, and won’t for at least another week.

thinking of a doctrine as a 2-monad is definitely on the wrong track here (for reasons that i may get around to explaining).

i’ll try to get around to it right now:

the problem is that there are important examples of morphisms between doctrines that are not examples of the “natural” (in an informal sense) kind of morphism between 2-monads.

we can see what the natural kind of morphism between 2-monads is by de-categorifying and looking at the natural kind of morphism between monads (that live on a single category s). monads on s are monoids in the monoidal category of endo-functors of s, so arguably the natural morphisms between the monads should be the monoid homomorphisms, and in fact this corresponds to the intuitive concept of “interpretation between algebraic theories” when you think of a monad as an algebraic theory. given such an interpretation f : t1 -> t2 of algebraic theories, we get both a contravariantly induced forgetful functor from the category of t2-algebras to the category of t1-algebras and, if the base category s has appropriate (co-)completeness properties, a left adjoint to the forgetful functor.

but consider the case where s is the category of sets and t1 is the algebraic theory of abelian groups and t2 is the algebraic theory of commutative rings and where we have a “forgetful functor” from the category of t2-algebras to the category of t1-algebras, with a left adjoint going the other way, that _doesn’t_ come from an “interpretation of algebraic theories” in the sense that we’ve been talking about so far. namely, consider the forgetful functor that assigns to a commutative ring its multiplicative group; its left adjoint is the “group ring” functor. thus if you want to count this adjunction as an “interpretation of algebraic theories” then you’ll have to modify your concept of “algebraic theory” and/or your concept of “interpretation of algebraic theories”.

a categorified version of this adjunction is very important in algebraic geometry. a dimensional theory is analogous to an abelian group (because dimensions have inverse dimensions and the tensor product of dimensions is symmetric), and there’s another kind of theory, perhaps we should call it a “multi-dimensional theory” for now, that’s analogous to a commutative ring.

(the category of coherent sheaves over a variety is an example of such a multi-dimensional theory. coherent sheaves are slightly generalized vector bundles, and vector bundles are multi-dimensional objects, in contrast to line bundles which are one-dimensional objects (aka “dimensions”). a category of one-dimensional objects is closed under tensor product and inverse and thus like a categorified abelian group, whereas a category of multi-dimensional objects is closed under tensor product and direct sum and thus like a categorified commutative ring.)

so there’s a forgetful 2-functor from the doctrine of multi-dimensional theories to the doctrine of dimensional theories, assigning to a multi-dimensional theory its subcategory of one-dimensional objects (thus for example assigning to the multi-dimensional theory of coherent sheaves over a variety the dimensional theory of line bundles over the variety). furthermore this forgetful 2-functor has a left adjoint, which is a sort of categorified group ring construction. and this adjunction is (for reasons that hopefully i’ll get around to explaining at some point) of great importance in algebraic geometry. but in analogy to the de-categorified case, we can’t count this adjunction as an “interpretation of doctrines” unless we modify our concept of “doctrine” and/or our concept of “interpretation of doctrines”. that (together with a few similar examples) is my motivation for not being satisfied with defining “doctrine” as “2-monad”.

i should mention that one reason that lawvere’s suggestion to define “doctrine” to mean (approximately) “2-monad” was a somewhat reasonable suggestion is that it does lead to the natural kind of morphism between doctrines being a special kind of adjunction. it’s important (for reasons that i hope to explain) that interpretations of doctrines should be adjunctions, but there are some important adjunctions that don’t qualify as the natural kind of morphism between 2-monads, and that motivates my dissatisfaction with defining “doctrine” to mean “2-monad”.

Yes! Your guess is right, though for full credit you need to tell me how you’re making $K[x,y]$ into a graded commutative algebra.

Here’s the idea:

We want to get a graded commutative algebra, say $A$, from a theory of physics with certain ‘dimensions’, certain ‘quantities’ having these dimensions, and certain ‘relations’ between these quantities.

Note: when I say ‘graded commutative algebra’ I mean this in a very general sense. We don’t need the grades to be integers; they can live in any abelian group $G$. In practice, this will often be a free abelian group.

Here’s how it works:

• The ‘dimensions’ are the generators of the abelian group $G$ which serves as the grading group for our algebra.
• Each ‘quantity’ is a generator of our algebra $A$. The ‘dimension’ of that quantity says what grade this generator lives in.
• Each ‘relation’ between quantities gives us an equation that holds in our algebra $A$. We are only allowed to equate quantities that live in the same grade!

You’ll notice this is like a ‘generators and relations’ description of an algebra, except now there are three levels instead of two: dimensions, quantities and relations. There’s a profound reason for that — it has to do with categorification.

But never mind. Let’s do the example.

I said “we have the theory of two particles each of which has a mass… and that’s all!”

So, I’m telling you one dimension, mass. And this generates a free abelian group $G$, namely $\mathbb{Z}$. Its elements are

$$..., mass^{-2}, mass^{-1}, 1, mass, mass^2, ...$$

This should make you think of dimensional analysis in physics. And that’s the whole point.

Next, I’m saying there are two quantities with dimensions of mass: the masses of our two particles, say $m_1$ and $m_2$.

So, our graded commutative algebra has two generators, $m_1$ and $m_2$, both of which live in grade 1.

And then I said “that’s all!” — so there are no relations.

So, the answer is

$$G = \mathbb{Z}$$

$$A = K[m_1, m_2]$$

where both $m_1$ and $m_1$ live in grade 1.

And as you know, this graded commutative algebra is related to the projective line. We should say much more about the physical meaning of that process: the process of getting a space from our graded commutative algebra. But this is enough for now.

I can give it a try. We have two dimensions, mass and velocity, so our grading group ought to be just the free group on these two generators, i.e. $G=\mathbb{Z}mass \oplus\mathbb{Z}velocity$. The quantities are the masses $m_1, m_2$ and velocities $v_1,v_2$ of the two particles. Hence $$A=k[m_1,m_2,v_1,v_2],$$ with the grading specified by $$\vert m_1\vert=(1,0)=\vert m_2\vert,\quad \vert v_1\vert=(0,1)=\vert v_2\vert.$$ In other words, $A$ is the ring $k[m_1,m_2]\otimes k[v_1,v_2]$ with its usual bigraded structure. Geometrically, $A$ corresponds to $\mathbb{P}^1\times\mathbb{P}^1$. There is a natural inclusion $$k[m_1,m_2]\rightarrow A, f\mapsto f\otimes 1$$ which geometrically is the projection of $\mathbb{P}^1\times\mathbb{P}^1$ onto its first factor.

A nice subvariety of $\mathbb{P}^1\times\mathbb{P}^1$ is the curve of bidegree $(1,2)$ cut out by the polynomial $m_1v_1^2-m_2v_2^2$.

Excellent!

So, you’re saying that that our new theory (two particles with mass and velocity) is some sort of ‘product’ of two copies of our old theory (two particles with mass).

But technically, I guess you’re saying it’s a kind of coproduct, because you’re hinting that there are two ways to include old theory into our new one.

But when we convert theories into spaces, things get turned around. We see that the space for our old theory is the projective line, and the space for our new theory is a product of two copies of the projective line. That’s because this process of converting theories into spaces is contravariant.

And we should probably talk more about that process right now.

Given a theory $T$ — or to be precise, a dimensional theory — we start by forming a category $hom(T,K)$ of ‘models’ of this theory in some rather simple environment $K$. Then we form the set of isomorphism classes of models — usually called the ‘moduli space’ of models. And this is almost the space we want.

But what’s $K$? $K$ is the boring theory where all quantities are dimensionless: the theory of dimensionless numbers! As a graded commutative algebra, its grading group is the trivial group, so there’s only one grade. And, the quantities in this grade are just the numbers in our favorite commutative ring $k$ — the one we fixed right at the start of this game.

In other words, $K$ is just a glamorized version of the ring $k$.

Okay, let’s see if everyone understands this stuff. Let me pose two more puzzles.

For starters, what does a morphism

$$f : T \to K$$

amount to when $T$ is the theory of two particles with mass? We call such a morphism a model of our theory $T$. But what does it amount to, concretely?

(This isn’t hard. Remember that we can think of $T$ and $K$ as commutative graded algebras. So, $f$ is just a homomorphism of commutative graded algebras. So, anyone who knows what all these words means can figure out what it amounts to.)

More interestingly, when do two models

$$f, g: T \to K$$

count as isomorphic? For this we need to understand morphisms between models. And for this, we need to realize that we can also think of $T$ and $K$ as categories of some sort. Namely, ‘dimensional categories’, as defined in the paper.

From this viewpoint, the models $f, g : T \to K$ are actually functors of some sort… and a morphism between models is a natural transformation of some sort.

So, if you’re a category theorist, you can use this to figure out when two models of $T$ in $K$ are isomorphic.

There must also be a way to do this thinking of $T$ and $K$ as graded commutative algebras, but that might be harder, or less fun.

But there’s also a way to do this quite intuitively, thinking of $T$ and $K$ as physical theories — with $K$ being a physical theory of a very degenerate sort, suitable only for mathematicians: the theory of nothing at all except numbers!

You see, a model of $T$ in $K$ is a way of attaching numerical values to physical quantities. But two such recipes should count as isomorphic if they differ only by ‘change of units’. If you’ve ever thought about dimensional analysis, you can guess when this happens without knowing any category theory!

Of course the fun part is how the formal category-theoretic approach to the puzzle matches the intuitive physical approach.

john has posed a couple of puzzles of the general form: consider such and such a dimensional theory; what graded commutative algebra does it correspond to?

i have another such puzzle, for people who haven’t seen it before. but while the theories that john has described have been physics-motivated, the theory in this next puzzle is of a different flavor, designed to illustrate a different aspect of the doctrine of dimensional theories.

consider the dimensional theory of “a pair (d,e) of lines equipped with a lie algebra structure on the direct sum d+e”. what graded commutative algebra does it correspond to?

(your description of the graded commutative algebra should be fairly explicit, as usual; for example by explicit generators and relations. extra credit for stuff like giving the dimension of the moduli space of models of the theory, or for generalizing to the case of an n-tuple of lines where n>2, for example n=3.)

Yeah, soon after I posted that John said something about other doctrines and coherent sheaves. It will be interesting to read!

Regarding the permutation group analogy, I don’t think we really disagree. I just meant to say that a projective embedding of a variety is an auxiliary piece of data, much like giving an action of a group on some set. These are both useful tools for studying the original variety / group, but the variety / group exists prior to the embedding / action. (Of course, groups do have natural permutation representations, the regular representations, whereas varieties essentially never have a distinguished projective embedding. So the analogy is pretty poor, but I couldn’t think of anything better.)

I wouldn’t take the “disease” comment as very indicative of algebraic geometers’ views. In modern terminology, giving a specific projective embedding (modulo automorphisms of the ambient projective space) is the same as equipping the variety X with a very ample line bundle (the pull-back of the line bundle O(1), whose global sections are linear homogeneous polynomials, on the projective space). Such a line bundle is often referred to as a polarization of X.

(I think that “very ample” is chosen just to indicate that such a line bundle has a “very ample” supply of global sections, while the “polarization” terminology, which may seem a bit opaque, is used for historical reasons: it dates back to Riemann, as far as I know.)

Algebraic geometers frequently study varieties equipped with polarizations. (For example, polarizations often have to be included if one wants to get reasonable moduli spaces.)

The thing that sparked my question was the definition:

Definition. An object x in a symmetric monoidal category K is called a line object if it has an inverse object with respect to tensor product, and if the canonical “switching” morphism x⊗x→x⊗x is the identity morphism.

I’m not sure what it is about this definition that warrants the word “line”. I’m guessing that a better name for this would also serve as a better name for “line bundle”. I know the terminology will never change, but a better descriptive name would probably help me overcome some mental barriers.

oh. no, i don’t think that we need a better descriptive name here; we probably just need a better explanation of what about the definition warrants the word “line”. (i may have completely neglected to give such an explanation.)

the intuition behind it is pretty rudimentary: multi-dimensional objects have no tensor inverses because the multiplicativity of tensor product with respect to dimensionality would require their inverses to be fractional-dimensional; thus having a tensor inverse is the hallmark of a one-dimensional object (aka a “line”, aka a “dimension”).

the other clause in the definition of “line object”, requiring that its “self-braiding” be trivial, seems less crucial to me, so it should be fun to experiment with omitting that clause.

Eric wrote:

I’m not sure what it is about this definition that warrants the word “line”. I’m guessing that a better name for this would also serve as a better name for “line bundle”. I know the terminology will never change…

… because Jim invented it and he’s incredibly stubborn?

but a better descriptive name would probably help me overcome some mental barriers.

Feel free to call it a ‘line bundle’ if you like. As Jim mentioned in his reply to your first question, there’s a certain sophisticated crowd who ruthlessly downplays the distinction between an ‘$X$’ and an ‘$X$ bundle’. The reason is that both of these can be thought of as an ‘$X$ object’, thanks to internalization. And then after a while you get sick of saying ‘$X$ object’ and just say ‘$X$’!

For example, I used to use ‘torsor’ as the name for a fiber of a principal bundle, and I got really confused when that sophisticated crowd used ‘torsor’ to mean a principal bundle. I would have been happy if they said ‘torsor bundle’. But then I grew up: I learned that a torsor bundle really is a ‘torsor object’ in some category of bundles, and then I got sick of saying ‘torsor object’ and acquiesced to simply saying ‘torsor’.

But it’s not good to pretend you’re sophisticated before you actually are. So, you should feel perfectly fine about calling the objects in a dimensional category ‘line bundle objects’ or ‘line bundles’ or ‘line objects’ or ‘lines’ — whatever aids your intuition most.

When I first saw this definition it looked like rubbish, since the switch map in a symmetric monoidal category is almost never the identity. In fact, that’s basically what prevents every symmetric monoidal category from being equivalent to a strictly symmetric one. But then when I looked again at the example of line bundles it made sense: the switch map for the unit object (by which I mean the identity object of the monoidal structure) is always the identity, and a line bundle is basically a “twisted” version of the unit object (the ground ring). So it seems to me that a “line object” in general can be thought of as a “twisted form of the unit object.”

presumably this is related somehow to the purely homotopy-theoretic theory of k-line bundles for k a topological (commutative) field, which is controlled by cohomology with coefficients in the strict topological abelian group given by the multiplicative group of k, and for that reason exhibits strict symmetry of the tensor product.

Let’s have a go. Perhaps the second of these:

• Dimensional analysis by Percy Williams Bridgman
• Dimensional analysis and theory of models by Henry Louis Langhaar

I see Hassler Whitney wrote something on the subject:

• The mathematics of physical quantities. Part I. Mathematical models for measurement, Amer. Math. Monthly 75 (1968), 115-138.
• The mathematics of physical quantities. Part II. Quantity structures and dimensional analysis, Amer. Math. Monthly 75 (1968), 227-256.

Evan wrote:

I believe the moduli space for $T′$ will be the image of the Segré embedding of the two copies of the moduli space for $T$.

Right! That’s exactly the buzzword I was alluding to: ‘Segré embedding’. And even better, the obvious ‘inclusion’ morphism of dimensional theories

$$T' \to T + T$$

induces a map of models

$$hom(T + T, K) \to hom(T', K)$$

which gives the Segré embedding

$$\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^3$$

when we take isomorphism classes of models and discard the so-called ‘irrelevant’ ones.

And I’m so bad at algebraic geometry that this really helps me feel I understand the Segré embedding!

As James mentioned…

Just to be clear, we’ve got an algebraic geometer named James contributing to this thread, and also a guy who calls himself james dolan, who is the one you’re talking about. I’m calling the former James and the latter Jim.

If Jim Stasheff joins in, we’re doomed.

Gee, this puzzle was supposed to be pretty easy. Maybe people are scared because they don’t see how take a dimensional theory, throw in an isomorphism, and get a new dimensional theory? Maybe you’re scared because you’re not sure what the grading group of this theory is? Maybe you can’t guess what its commutative graded algebra is?

well i can tell you why i’m a bit unsure as to exactly what answers to your questions you’re looking for. my reasons might be a bit different from some other people’s reasons, at least on the surface, but they’re probably secretly the same- you also seem to be hinting at the same reasons yourself.

for me, the confusion is this:

we’re talking here about graded commutative algebras, but we’ve alluded to some sort of theorem that says that they’re equivalent in some sense to “dimensional categories”. but dimensional categories, being categories, can be distinguished either up to strict isomorphism (often a bad idea), or up to a coarser relation given by some kind of “equivalence of categories” (often a good idea; perhaps here it should be some sort of “symmetric monoidal equivalence” because of dimensional categories being a kind of symmetric monoidal categories.) so this makes me wonder whether there’s also some sort of “coarse equivalence of graded commutative algebras” that we should be concerned with. or more precisely, it makes me try to remember what i concluded about this issue the last time that i thought about it. because it seems like i might need to resolve this issue in order to give sensible answers to your questions.

anyway, i made no attempt to avoid a blatant category-theory bias in the answer that i just gave. maybe someone else should try to give more of an intuitive “dimensional analysis” take on the situation. it has something to do with: when do two dimensions count as “equivalent”, and how do you deal with that? do you just get rid of one of them as being “redundant”, and if so then how do you know which one is the redundant one?? and stuff like that…

Maybe people are scared…

…or too busy.

Let’s have a stab while I have a moment. You’d think that having the unit of mass would be as though there were a third (massive) particle, kept in the vaults of a scientific institute to provide an absolute standard.

Assuming $k$ is a field, the space of models would then be

$$(k \backslash \{0\} \times k \times k) / (k \backslash \{0\}) \cong k^2.$$

Would we be less inclined to drop the two massless particles model?

I’ve thought about this myself, but I’m not too convinced. In ordinary algebraic geometry, say where we complete affine space to projective space, there is a certain homogeneity where every point in projective space looks like every other point, whereas in the completion you’re proposing, the archimedean places “at infinity” are qualitatively different from the nonarchimedean ones (the latter are ultrametric, for example).

I mean, maybe you’re right and the analogy is a really good one, but I don’t see that as yet. Could you elaborate?

Jim Dolan wrote had written:

but perhaps i can perform an experiment. […] i recommended working out the example of the coherent sheaves over projective n-space as the free symmetric monoidal finitely cocomplete algebroid on a 1-dimensional object linearly embedded in standard linear [n+1]-space. maybe if you could work out this example and post a description of it here then i could get a sense of how the ideas that i’m trying to explain are subsumed by some larger body of ideas that you’re referring to.

If I understand you correctly, you would like me to tell you how I would suggest to view this situation. In that respect, here is an attempt:

it seems to me that the concept that you are getting at is closely related to that of a perfect stack.

A stack is called perfect if its compact quasicoherent sheaves are precisely the dualizable quasicoherent sheaves and if the whole category of quasicoherent sheaves is generated from these compact/dualizable objects.

(In particular affine, projective and generally quasi-projective schemes (and $\infty$-schemes) are perfect stacks ($\infty$-stacks).)

Could that be?