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.

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    $\begingroup$ 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. $\endgroup$ – Todd Trimble Oct 22 '18 at 19:08
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    $\begingroup$ 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.] $\endgroup$ – Todd Trimble Oct 22 '18 at 19:23
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    $\begingroup$ 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 $\endgroup$ – მამუკა ჯიბლაძე Oct 22 '18 at 19:45
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    $\begingroup$ 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! $\endgroup$ – Todd Trimble Oct 22 '18 at 21:52
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    $\begingroup$ 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. $\endgroup$ – Fosco Loregian Oct 23 '18 at 10:57

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