Can one compare monads arising from homotopy equivalent adjunctions?

Question

Suppose we have a Quillen adjunction $L\colon {\mathcal C} \leftrightarrow {\mathcal D}: R$. For convenience let us assume that all objects of $\mathcal C$ are cofibrant and all objects of $\mathcal D$ are fibrant, so that $L$ and $R$ are homotopy functors. Let $(L', R')$ be another Quillen adjunction. Let's suppose that we have a natural transformation $\alpha_L\colon L \to L'$. It induces a natural transformation $\alpha_R\colon R' \to R$. Let us assume that $\alpha_L$ is a weak equivalence. If I am not mistaken, it follows that $\alpha_R$ is a weak equivalence.

Let $T$ and $T'$ be the monads $RL$ and $R'L'$ respectively. As far as I see, in general there is no map of monads between $T$ and $T'$.

Question : Are the homotopy categories of $T$ algebras and $T'$-algebras equivalent?

Remark : The adjunction $(L, R)$ induces a Quillen adjunction between $T$-algebras and $\mathcal D$. This adjunction is sometimes (quite often?) a Quillen equivalence. In this case one may say that $T$ satisfies homotopy descent. It seems that if $T$ satisfies homotopy descent, then so does $T'$, and in this case Ho$(T-\mathrm{alg})$ and Ho$(T'-\mathrm{alg})$ are equivalent, since both are equivalent to Ho$(\mathcal D)$. I want to know if there is a direct way to compare algebras over $T$ and $T'$, without using descent.

Answer 1

Hi Greg, I'll give an alternative, elementary, starting point towards an answer. Dylan, there is a question about using the bar construction that seems maybe to be missing from your answer, and I don't understand your statements about dealing with multiplicative structure or with uniqueness. Also, if you will forgive a little teasing, if your favorite tool is a sledgehammer, then every nail looks like a rock.

I'll use the two-sided bar construction from "The geometry of iterated loop spaces", [12] on my web page; see [112] for a modernized discussion and some category theory relevant to the question. For a $T$-algebra $A$, define a $T'$-algebra $F(A)$ by $F(A) = R'B(L,T,A)$, where $T$ acts on the right of $L$ via $\epsilon R$.
This is a $T'$-algebra since $R'D$ is a $T'$-algebra for any $D\in \mathcal D$. Similarly, For a $T'$-algebra $A'$, define a $T$-algebra $G(A')$ by $G(A') = RB(L',T',A')$. Of course, $F$ and $G$ are functors.

I assume that we have natural weak equivalences $$\gamma'\colon B(R'L, T, A) \to R'B(L,T,A)$$ and $$ \gamma\colon B(RL', T', A') \to RB(L',T',A').$$ We expect the second to come from a weak equivalence of the general form $|RX|\to R|X|$ for a simplicial object $X\in \mathcal C$, and similarly for the first. With $R = \Omega$ and $\mathcal C$ the category of based spaces, proving that there is such a weak equivalence $\gamma$ was the hardest thing technically in [12]. It is something like this that I find missing in your sketch, Dylan, but if we can see directly that $B(R'L, T, A)$ is a $T'$-algebra (say by commuting $T'$ past $B$) and similarly for $B(RL',T',A')$, then perhaps that is not needed here.

Since $T = RL$, we have the chain of natural weak equivalences (ignoring algebra structure) $$ A \leftarrow B(T,T,A) \leftarrow B(R'L,T,A) \to F(A) $$ where the middle arrow is induced by $\alpha\colon R'\to R$, which I'll assume given (rather than taking a conjugate). Similarly, we have the chain of weak equivalences $$ A' \leftarrow B(T',T',A') \rightarrow B(RL',T',A') \to G(A'). $$ Therefore $F(G(A'))\simeq A'$ and $G(F(A))\simeq A$. It remains to prove that these equivalences (or related ones) are equivalences of $T'$-algebras and $T$-algebras. I haven't had time to try.

Answer 2

Unless I'm mistaken (very possible), the answer is "yes".

Here's the general idea: Every algebra may be obtained as the geometric realization of its bar complex. So we may reduce to the case of a free algebra, i.e. we need only construct some sort of natural weak equivalence $TX \rightarrow T'X$ preserving the multiplication structure. I don't know how to make precise this choice in a model category setting without using an ugly zig-zag and convoluted argument (but that is more likely due to ignorance than anything else.)

If we replace all model categories in sight with their underlying $\infty$-category (I'm ignoring some set-theoretic issues here, or assuming that the categories are simplicially enriched), then the above becomes an actual proof: The Quillen adjunction induces an adjunction of $\infty$-categories, and in this case a natural isomorphism of functors admits an $\infty$-categorical inverse. So we can define the map $$ T \rightarrow T' $$ as the composition $$\alpha_R^{-1}L' \circ R\alpha_L: RL \rightarrow RL' \rightarrow R'L'$$

(Here is where, in the model category case, I would have a zig-zag, and would have to do some sort of work... I don't actually know how the argument would go off the top of my head.)

This map respects the multiplication (the only proof I can think of uses that adjoint pairs are unique up to unique isomorphism, where "unique" in this setting means "parameterized by a trivial Kan complex". But it's possible this is more obvious than I think.)

It is a natural isomorphism, and so we're good.

I don't think we can get away with any other type of proof: you need some sort of "multiplication preserving zig-zag of weak equivalences of functors" argument to prove this at all, because if you provide some natural zig-zag of weak equivalences between $T$-algebras and $T'$-algebras then, in particular, you need to do it for free $T$-algebras and $T'$-algebras, and multiplication preserving weak-equivalences like this can only come in the form of some zig-zag of natural transformations of $T$ with $T'$ since that's where the algebra structure on free algebras comes from.