5
$\begingroup$

OK, so I asked a similar question before; $B{\mathbb G}_m$ is a simplicial presheaf over number field $k$. I see that there is some $A^1$-homotopy equivalence between the sheaf represented by ind-scheme ${\mathbb P}^\infty$ and that by $B{\mathbb G}_m$. As in the paper $A^1$-homotopy theory [MV], the construction is through Nisnevich site. I would like to know if some description of this map exists that stays within the etale site. MV gives a general construction good for any algebraic group; but I feel like something explicit can be said for the simplest case, namely ${\mathbb G}_m={\mbox{Spec}}\ k[t,t^{-1}]$.

$\endgroup$

2 Answers 2

6
$\begingroup$

This is more like an extended comment to David Roberts' answer. There (at least) two ways to see that $\mathbb{A}^\infty\setminus\{0\}$ is $\mathbb{A}^1$-contractible.

  1. Using Morel's unstable connectivity theorem which can be found in his book "$\mathbb{A}^1$-algebraic topology over a field", in the chapter on $\mathbb{A}^1$-homotopy and $\mathbb{A}^1$-homology sheaves. The result holds for the $\mathbb{A}^1$-homotopy theory over an infinite perfect field, and states that any $n$-connected simplicial presheaf is also $\mathbb{A}^1$-$n$-connected. It is claimed in Example 3.2.20 of [MV] and explained in Examples 2.11 of Dugger-Isaksen "Motivic cell structures" that there is an $\mathbb{A}^1$-weak equivalence $\mathbb{A}^n\setminus\{0\}\cong S^{n-1}\wedge\mathbb{G}_m^{\wedge n}$. In particular, $\mathbb{A}^n\setminus\{0\}$ is simplicially $(n-2)$-connected, and by Morel's unstable connectivity theorem also $\mathbb{A}^1$-$(n-2)$-connected. As mentioned in S. Carnahan's comment, taking the colimit of $\mathbb{A}^n\setminus\{0\}$ via the obvious inclusions therefore shows that $\mathbb{A}^\infty\setminus\{0\}$ is $\mathbb{A}^1$-contractible.

  2. The alternative is the classical topological argument that $S^\infty$ is contractible (see the MO-discussion How do you show that $S^{\infty}$ is contractible?), made algebraic: the shift-by-1 map is $$ S:\mathbb{A}^\infty\setminus\{0\}\to\mathbb{A}^\infty\setminus\{0\}: (x_1,x_2,\dots)\mapsto (0,x_1,x_2,\dots) $$ which is homotopic to the identity via $$ f_T:\mathbb{A}^\infty\setminus\{0\}\times\mathbb{A}^1\to \mathbb{A}^\infty\setminus\{0\}: (x_1,x_2,\dots)\mapsto (1-T)(x_1,x_2,\dots)+TS(x_1,x_2,\dots). $$ This works since the straight line through $(x_1,x_2,\dots)$ and $(0,x_1,\dots)$ does not go through $0$. Then the image of the shift map $S$ can be contracted to a point via $$ g_T:\mathbb{A}^\infty\setminus\{0\}\times\mathbb{A}^1\to \mathbb{A}^\infty\setminus\{0\}: (x_1,x_2,\dots)=(1-T)(0,x_1,x_2,\dots)+T(1,0,0,\dots). $$ This is just the same argument as in topology. In topology, the above maps must be renormalized to induce maps on $S^\infty$, but that is not necessary here. As the above maps are polynomial, they induce even naive $\mathbb{A}^1$-homotopies from the identity to the constant map, so $\mathbb{A}^\infty\setminus\{0\}$ is $\mathbb{A}^1$-contractible.

The explicit map is then given as in David Robert's answer. Formulated slightly differently, take $\mathbb{A}^\infty\setminus\{0\}\to\mathbb{P}^\infty$ to be the universal $\mathbb{G}_m$-bundle, and take the standard covering $\mathcal{U}$ of $\mathbb{P}^\infty$ by the $\mathbb{A}^\infty$ with coordinates $x_i\neq 0$, $i\in\mathbb{N}$. The universal $\mathbb{G}_m$-bundle induces a map $\check{C}(\mathcal{U})\to B\mathbb{G}_m$. Here, $\check{C}(\mathcal{U})$ is the Cech nerve of the cover $\mathcal{U}$ which is simplicially equivalent to the action groupoid. The map itself is the $\mathbb{G}_m$-cocycle for the universal bundle in degree $1$, and can be extended to a map of the whole Cech nerve since the cocycle condition is satisfied. The diagram $\mathbb{P}^\infty\leftarrow\check{C}(\mathcal{U})\to B\mathbb{G}_m$ gives the required weak equivalence in $\mathbb{A}^1$-homotopy category. Note that the map itself is already defined in the simplicial model category because $\check{C}(\mathcal{U})\to \mathbb{P}^\infty$ is a simplicial weak equivalence.

$\endgroup$
1
  • $\begingroup$ Thanks, Matthias! That algebraisation of the topological proof is really nice, and immediately graspable without any deep A^1 homotopy theory. $\endgroup$
    – David Roberts
    Jun 30, 2014 at 10:57
3
$\begingroup$

OK, here's a first attempt, with a small gap.

The ind-scheme $\mathbb{P}^\infty$ is the quotient of the ind-scheme $\mathbb{A}^\infty \setminus\lbrace 0\rbrace = \lim_{n\to \infty}\mathbb{A}^n\setminus\lbrace0\rbrace$ by the free action of $\mathbb{G}_m$, so here is a weak equivalence of simplicial presheaves $N(\mathbb{G}_m \curvearrowright\mathbb{A}^\infty \setminus\lbrace 0\rbrace) \to \mathbb{P}^\infty$, where the source is the nerve of the action groupoid and the target is the constant simplicial presheaf. I say 'weak equivalence', but more properly this is I guess a weak equivalence in the local projective model structure on simplicial presheaves, where 'local means whatever topology the $\mathbb{G}_m$ bundle over $\mathbb{P}^\infty$ is trivial in (I'm guessing étale, or possibly even Zariski). Next, there is a map of simplicial sheaves $N(\mathbb{G}_m \curvearrowright\mathbb{A}^\infty \setminus\lbrace 0\rbrace) \to B\mathbb{G}_m$, which I conjecture is a weak equivalence in the $\mathbb{A}^1$-homotopy sense. This should be true as long as $\mathbb{A}^\infty \setminus\lbrace 0\rbrace \to \ast$ is a $\mathbb{A}^1$-weak equivalence. This last statement is the one that I feel is a gap, in that I have no clue how to prove this, only argue by analogy with the differential geometric or topological case.

The span $$ \mathbb{P}^\infty \leftarrow N(\mathbb{G}_m \curvearrowright\mathbb{A}^\infty \setminus\lbrace 0\rbrace) \to B\mathbb{G}_m, $$ assuming that one can fill in the gap, should represent an $\mathbb{A}^1$-weak equivalence, and my guess, based on the assumption that the $\mathbb{G}_m$ bundle over $\mathbb{P}^\infty$ trivialises in the étale (resp. Zariski) topology, is that this construction uses the étale (resp. Zariski) site.

$\endgroup$
2
  • 1
    $\begingroup$ For your gap, could you show that the closed immersion $\mathbb{A}^n \setminus \{0\} \to \mathbb{A}^{n+1} \setminus \{ 0 \}$ is something like a suspension, and take a colimit? $\endgroup$
    – S. Carnahan
    Jun 12, 2014 at 23:45
  • $\begingroup$ @S.Carnahan, maybe, but I don't know the A^1 homotopy machinery so as to make that rigorous. $\endgroup$
    – David Roberts
    Jun 13, 2014 at 0:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.