Question

Let $\mathbf{C}$ be a cofibrantly generated model category (assume for simplicity that all objects are fibrant) and $\mathbf{C}^{\mathrm{T}}$ the category of $\mathrm{T}$-algebras with the induced model structure (same weak equivalences and fibrations as in the underlying model category $\mathbf{C}$). By definition, the adjuction $\mathrm{T}:\mathbf{C}\rightleftharpoons\mathbf{C}^{\mathrm{T}}: \mathrm{U}$ is monadic. How about the homotopical version, i.e, $\mathbb{L}\mathrm{T}:Ho\mathbf{C}\rightleftharpoons Ho(\mathbf{C}^{\mathrm{T}}): \mathbb{R}\mathrm{U}$

is there any result about the "homotopical" monadicity theorem, which compares $Ho(\mathbf{C}^{\mathrm{T}})$ and $Ho(\mathbf{C})^{\mathbb{L}\mathrm{T}}$.

Answer 1

$Ho(C^T)$ is almost never monadic over $Ho(C)$. The objects of $Ho(C^T)$ are $T$-algebras in $C$, which means in particular that their $T$-algebra structure commutes strictly, whereas the algebras for the induced monad on $Ho(C)$ will only have algebra structure commuting up to (non-specified, non-coherent) isomorphism.

For instance, if $T$ is the monad derived from an $E_\infty$-operad, then $T$-algebras are $E_\infty$-spaces, whereas $Ho(T)$-algebras are "$H_\infty$-spaces". These have an obstruction theory specifying when they can be $E_\infty$-ized.