Question 1: Are there some publications or preprints that provide $\Bbb A^1$-fibrant replacements of certain classes of (smooth) schemes?

Of course, smooth schemes that are $\Bbb A^1$-fibrant are $\Bbb A^1$-fibrant replacements for themselves. By proposition 3.19 of Morel-Voevodsky's '$\Bbb A^1$-homotopy theory of schemes' and the properties of the Nisnevich topology, one sees that a smooth scheme $X$ over a field $k$ is an $\Bbb A^1$-fibrant object if and only if it is $\Bbb A^1$-rigid, i.e. for which the canonical map $$ Hom_{Sm/k}(U,X)\rightarrow Hom_{Sm/k}(U\times \Bbb A^1,X) $$ is a bijection for every smooth scheme $U$ over $k$.

Example 2.1.10 and the argument after lemma 2.1.11 in Asok-Morel's Smooth varieties up to $\Bbb A^1$-homotopy and algebraic $h$-cobordisms provide examples of such schemes:

  • $0$-dimensional $k$-schemes,
  • abelian $k$-integral schemes,
  • smooth complex integral schemes that can be realised as quotients of bounded Hermitian symmetric domains by actions of discrete groups,
  • any open subscheme of $\mathbb G_m$,
  • closed integral (smooth) subschemes of $\Bbb A^1$-rigid schemes, and
  • product of $\Bbb A^1$-rigid schemes.

Question 1': What are other known classes of $\Bbb A^1$-rigid (smooth) schemes? Are there any known restrictions on $\Bbb A^1$-rigid schemes?

For both questions, I would like to have as many examples as there exist.

PS Schemes are taken to be separated and of finite type over the base field.

  • $\begingroup$ I don't get question 1: the formula $(L_{nis}Sing^{\mathbb{A}^1})^{\omega}$ always works for any simplicial presheaf. For question 2: $\mathbb{A}^1$-rigid schemes have no higher homotopy groups; if $X$ is a monoid scheme then $\Omega BX$ is $\mathbb{A}^1$-local (a "topos"-like property). For a more exotic property: the zero-th $S^1$-slice of the Eilenberg-Maclane $S^1$-spectrum on the free abelian group of an $\mathbb{A}^1$-rigid scheme is not a homotopy invariant presheaf with transfers (see the "An Example" section of Levine's "Slices and Transfers"). $\endgroup$ – Elden Elmanto Jul 20 '16 at 21:59
  • $\begingroup$ Oh I guess a countable class of examples not included above are all smooth projective curves of genus $\geq 1$. $\endgroup$ – Elden Elmanto Jul 20 '16 at 22:01
  • $\begingroup$ @elden-elmanto: I am asking if there exists a simplified (not necessary functorial) formula for smooth schemes, as it is the case for $\mathbb{A}^n$ and $\mathbb{A}^1$-rigid schemes. So, is it known what other classes of smooth schemes that admit a relatively simple to express $\mathbb{A}^1$-fibrant replacements, not requiring infinite iterations. I need is compute the hom-sets $[X,Y]_{\mathbb{A}^1}\cong [X,L_{\mathbb{A}^1} Y]$ for some smooth schemes $X,Y$ over a field. I do not know how to make use of the general formulas given in Hirschhorn's book or in Morel-Voevodsky's paper here. $\endgroup$ – user24453 Jul 20 '16 at 23:13
  • $\begingroup$ There's a general criterion for when you can replace the infinite iteration with just doing it once/twice - see the papers of Asok-Hoyois-Wendt. It works for "(pre)stacky" things as opposed to schemes, but then if you can present your schemes as a fiber of maps between these guys you get the same stoppage of iteration - so, for example, certain homogeneous spaces. Perhaps another class of results of this flavor involves proving that certain schemes are cellular so you can replace just using spheres; the relevant names are Dugger, Isaksen, Wendt and Voelkel. $\endgroup$ – Elden Elmanto Jul 20 '16 at 23:25
  • $\begingroup$ Heuristically I think it's pretty impossible to get a simple expression: the localization functor is a left adjoint hence calculated as a colimit. You can ask: "what kind?" as a measure for simplicity. Well you can't do finite coproducts since that clearly doesn't do anything so you are left with sifted ones. You can't just do filtered: spheres are compact so you wont have any interesting homotopy sheaves. You are left with geometric realization: so you need to build a simplicial object resolving your scheme - I cant think of anything easier than Sing! $\endgroup$ – Elden Elmanto Jul 20 '16 at 23:56

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