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Another questions about unidetermined monads.

EDIT: Here a note with a few more details on what they are: link

Let $T : C^o \to C$ be such a monad, so that the multiplication $\mu_A$ is determined by the unit from the equation $T\eta_A = \mu_A$.

Let moreover $T$ be lax idempotent, so that $T$ satisfies the following property:

let $a : TA\to A$ be an algebra; then $a\dashv \eta_A$ is an adjunction

(so that for example $a$ is uniquely determined up to iso).

Apply $T$ to this adjunction, and you will get an adjunction $Ta\dashv \mu_A$.

In the special case where $T$ is the presheaf construction $T$ not only reverses 1-cells, but also 2-cells; a similar definition applies.

For such a $T$, $\eta_A$ is the Yoneda embedding of $A$ and $T\eta$ the functor $[(TA)^o,Set]\to TA$ that restricts the functor $\lambda F.\zeta(F)$ to representables, giving a presheaf $\lambda a.\zeta(A(-,a))$.

But then its left adjoint $Ta$ must be determined by the universal property of $\text{Lan}_{\eta_A}$, and it turns out that if $F\in TA$ its yoneda extension acts as follows $$ \begin{align*} \text{Lan}_{\eta_A}F(G) &\cong \int^{a :A}Fa\times TA^o(\eta_A(a),G)\\ &\cong \int^{a:A}Fa\times TA(G,A(-,a))\\ &\cong \int^{a:A}Fa\times \mathcal O(G)(a) \end{align*} $$ where $\cal O$ is the Isbell functor.

In other words, $\text{Lan}_{\eta_A}F(-)\cong F\boxtimes \mathcal O(-)$, the functor tensor product of F, "twisted" by $\cal O$.

Even more suggestively, $\mu_A=T\eta_A$ has a right adjoint as well. It is the right extension along $\eta_A$, and a similar computation shows that

$\text{Ran}_\eta F$ is equal to the functor hom $\{\text{Spec}(F), G\}$ where again the Spec functor comes from Isbell duality.

I find the notation quite serendipitous[¹], I would be inclined to call the functor $G\mapsto F\boxtimes \mathcal O(G)$ the "extension of scalars" of ${\cal O}(G)$ by $F$, and the functor $G\mapsto \{\text{Spec}(F), G\}$ the functor that takes the "$F$-points of $G$".

What precisely am I after, here? Is it (part of) the reason why they're called Spec and $\cal O$?

[¹] A "serendipity" is the "fortuitous discovery of a pleasant truth, unexpectedly found while we were searching something else." I guess this definition applies here, where I was looking for something completely different, and I ended up suggesting a justification for the algebro-geometric notation for Isbell duality.

fosco
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    I still am not sure how to understand the notion of unidetermined contramonad on its own. Or, at least I'm getting confused by what the associativity axiom would be. For that matter, I'm not clear on how to understand "naturality" of $\eta$, although I might be able to make something up. – Todd Trimble Oct 22 '18 at 19:08
  • Things become weird, I agree; the point is that "being induced by the unit" only makes sense if $\mu = T(A\to TA) : TTA\to TA$. Sure, you can assume $T$ has an adjoint, but I dont think it is the right track to follow. (I guess $\eta$ and $\mu$ become dinatural, or even lax dinatural, right?) – fosco Oct 22 '18 at 19:18
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    I don't know what "dinatural" would mean in this context. Maybe I'm being dense. Can you write some equations down? [Also, I don't understand why you think it's not the right track to follow (I can't seem to parse your meanings without some assumption like that). But I'd be happy to pursue this later, if you can answer me by writing down some equations.] – Todd Trimble Oct 22 '18 at 19:23
  • Sorry, I begin to be confused and tired at the end of my day :-) What I mean is that whatever the property that makes the presheaf construction be such that $P(f)$ has adjoints on both sides, I'm unsure this is the right thing to ask to a generic contramonad: the $P$'s coming from more generic Yoneda structures can yield non-cocomplete presheaf-objects. – fosco Oct 22 '18 at 19:41
  • And... sure I can add a little more explanation and diagrams – fosco Oct 22 '18 at 19:41
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    Please excuse me, this (and the linked) stuff is very interesting but I am forced to say - this is one of the most unclear-what-you're-asking kind of questions that I've ever seen – მამუკა ჯიბლაძე Oct 22 '18 at 19:45
  • When one studies a topic for a long time, they always end up forgetting what is obvious and what instead needs a detailed explanation because it asks too much a leap to the reader. Please, tell me how can I be clearer. The question though, I felt it's pretty clear: "what is this thing? Why is such an equation true?" – fosco Oct 22 '18 at 20:02
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    I think one thing I might find helpful (esp. regarding "not the right track"), besides the request for more precision surrounding the notion of unidetermined contramonad, are more examples besides the ones of free cocompletion type. [Perhaps that would be more appropriate at the other thread, the one you linked to here.] I hope it's clear that I'm rooting you on! – Todd Trimble Oct 22 '18 at 21:52
  • I meant phrases of the question ending with question marks (the two last ones) - they are very different from those in your comment. The latter are also unclear for me, but much less so. Moreover I am very far from understanding what does serendipity have to do with all this. – მამუკა ჯიბლაძე Oct 22 '18 at 22:49
  • I am sorry I must reschedule the edit I planned until tomorrow; Im' too tired to rely on my writing skills. I can at least answer the last question: "serendipity" is the fortuitous discovery of a pleasant truth, unexpectedly found while we were searching something else. I wouldn't define otherwise an unexpected notational congruence between algebraic geometry and Isbell duality. – fosco Oct 22 '18 at 23:09
  • But has this notation been used anytime before the nlab entry? And there, I believe, it is intentionally hinting at algebro-geometric setup as one of leading examples, no? – მამუკა ჯიბლაძე Oct 23 '18 at 07:51
  • Let me also mention the MO question Theme of Isbell duality as it is not (yet?) in the Related column on this page. – მამუკა ჯიბლაძე Oct 23 '18 at 07:53
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    I've edited the thread with an external link to a pdf containing (1) the definition of unidetermined contramonad; (2) the definition of lax idempotent I'm working with. – fosco Oct 23 '18 at 10:57

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