# (really) basic intuition for $\mathbb A^1$-homotopy theory

Apologies in advance if this question is inappropriate for MO.

I'm trying to read here and there about $\mathbb A^1$-homotopy theory in algebraic geometry. I understand some abstract machinery is needed, in particular sheaves over sites, some simplicial stuff, and some model categories. The wiki article outlines the setup and formal definition.

From what I understand, this theory is supposed to fit in with geometric intuition from topological spaces somehow. Where can I find basic examples illustrating the geometric picture? I'm looking for truly basic stuff to build on, e.g how to identify if a scheme is contractible (if that makes sense), do geometric vector bundles interact well with $\mathbb A^1$-homotopy, what's the picture of two morphisms being homotopic, etc.

If these questions are way off (or completely meaningless) and the similarity of $\mathbb A^1$-homotopy is and the usual notion in $\mathsf{Top}$ is purely formal in nature, what is the right way to think about this theory geometrically?

• Maybe this paper is going to be helpful (it was for me): arxiv.org/abs/math/0007070. Expecially the case where he works out the corresponding theory for smooth manifolds. Morally $\mathbb{A}^1$-homotopy theory is the universal homotopy theory on algebraic varieties where $\mathbb{A}^1=*$. – Denis Nardin Jan 10 '16 at 14:11
• @DenisNardin Dugger's papers are invaluable, but It seems the focus is on explaining the formalism which is used to define the whole story. My problem is I have no geometric intuition for what contractible or even homotopic "looks like", so I can't "guess" anything. That's what I'm looking for. – Arrow Jan 10 '16 at 21:10
• If this is the case, I might not suggest starting at $\mathbb{A}^1$-homotopy theory but rather with classical homotopy theory. – Sean Tilson Jan 11 '16 at 15:54
• @SeanTilson why? I have intuition for classical homotopy with the unit interval, but that doesn't help me tell when a scheme is $\mathbb A_1$-contractible. – Arrow Jan 11 '16 at 17:50
• @Arrow I agree that classical homotopy intuition will not give you $\mathbb{A}^1$-homotopy intuition. I think that motivic homotopy theory is hard enough that you may not get a satisfactory answer. Topologies in AG are significantly different from working over $\mathbb{R}$ with the metric topology that I think the only hope you would have is to dive in to AG itself. Then again, I shouldn't be someone answering this question (which is why I am not). It seems like you want an answer like "well, when things look like ___, then you know that ___" and I am not convinced that such an answer exists. – Sean Tilson Jan 12 '16 at 10:37

Let me try to answer the specific subquestion "What are (intuitively) maps between schemes in the $\mathbb{A}^1$-homotopy category and what does it mean for maps to be homotopic?"

The general-machinery answer: computation of maps $f:X\to Y$ in the homotopy category requires a fibrant replacement $\tilde{Y}$ of $Y$, a cofibrant replacement $\tilde{X}$ of $X$ and then working out the homotopy relation on $\operatorname{Map}(\tilde{X},\tilde{Y})$. Assume we are working in a model structure where representables are cofibrant, so we only need to deal with the fibrant replacement. Morel and Voevodsky show that fibrant replacement is given by iterating infinitely often the fibrant replacement for simplicial sheaves and the explicit $\mathbb{A}^1$-singular resolution functor. We'll try to unwind this, in the special case where we only have one step instead of infinitely many, i.e., we want to compute maps $f:X\to \operatorname{Sing}_\bullet^{\mathbb{A}^1}(Y)$ (in the homotopy category of sheaves, not $\mathbb{A}^1$-localized). The Verdier hypercovering theorem theorem tells us how to do this, see e.g. the exposition in Dugger-Hollander-Isaksen: Hypercovers and simplicial presheaves. Any map has a representative of the form $$X\leftarrow \mathfrak{U}_\bullet\to\operatorname{Sing}_\bullet^{\mathbb{A}^1}(Y)$$ where the first map is a Nisnevich hypercovering and the second map a morphism of simplicial sheaves. Now unwind this: looking at simplicial degree $0$, we see that locally our morphism has algebraic representatives. (Ok, not strictly true because a Nisnevich open is not a subset, but this is building in implicit functions.) However, these local algebraic representatives do not glue to a globally defined map $X\to Y$ because they do not need to agree -- in degree 1, we see that the restrictions of two local algebraic representatives $f_1:U_1\to Y$ and $f_2:U_2\to Y$ to the fiber product $U_1\times_X U_2$ are only locally naively $\mathbb{A}^1$-homotopic. More precisely, there is a Nisnevich covering of $U_1\times_X U_2$, and locally on this covering, there will be naive $\mathbb{A}^1$-homotopies $H:V_1\times\mathbb{A}^1\to Y$. The higher simplicial degrees add further locally defined coherences between the restrictions of the algebraic representatives to multiple intersections.

Well, rather terrible. And this is only maps from $X$ to the sheaf-fibrant replacement of a single application of the singular resolution. Usually, infinitely many such applications are required so that, in a way, there is no geometric intuition to be had here. But on an intuitive level, I guess one can say that

maps in the $\mathbb{A}^1$-homotopy category are locally algebraic, and the local algebraic representatives are locally $\mathbb{A}^1$-homotopic in a compatible way.

A similar description applies to maps being homotopic. Generally, you will want to avoid making the above really precise because it'll be a hell of a time to construct maps in the homotopy category like this. Nevertheless, it's possible to do some computations: you can see quite instructively how to work with such locally algebraic maps up to locally-defined $\mathbb{A}^1$-homotopies in the papers of Balwe-Hogdai-Sawant and Balwe-Sawant. (watch out for ghost homotopies)

• C. Balwe, A. Hogadi and A. Sawant. $\mathbb{A}^1$-connected components of schemes. Adv. Math. 282 (2015), 335--361.

A nicer, more geometric situation: If $X$ is smooth affine and $Y$ is an isotropic reductive group or a suitable homogeneous space, then it is in fact possible to prove that all maps $X\to Y$ in the $\mathbb{A}^1$-homotopy category have algebraic representatives, and homotopies are really chains of naive $\mathbb{A}^1$-homotopies $X\times\mathbb{A}^1\to Y$. Technically, the statement is that the singular resolution $\mathcal{F}=\operatorname{Sing}_\bullet^{\mathbb{A}^1}(Y)$ is almost fibrant, i.e., the fibrant replacement $\mathcal{F}\to R\mathcal{F}$ induces weak equivalences on sections over all smooth affine schemes. (A consequence of this is that the classical theory of homotopy classification of principal bundles works in these cases, answering the question on geometric vector bundles.) There is quite a bit of machinery that goes into the proof; the first result of this type (for general linear groups and Grassmannians) was proved by Morel (see the Lecture Notes on $\mathbb{A^1}$-algebraic toplogy) and there has been a recent extension in work of Aravind Asok, Marc Hoyois and myself (can be found on the arXiv here).

Comparison to classical topology: There is a different line of thinking to get the intuition started which you may have found already. The classical homotopy theory can be recovered by starting with the category of smooth manifolds (with Grothendieck topology given by open coverings) and then taking the Bousfield localization at $\mathbb{R}$ of the model category of simplicial sheaves on the site of manifolds. The simple reason why the result is the classical homotopy category is that every manifold has a good hypercovering, i.e., one where all spaces involved are balls in $\mathbb{R}^n$. All the homotopy information is then contained in the combinatorics of the nerve of the covering, which is just a simplicial set. In algebraic geometry, schemes look rather different even locally, so there is no way to have the structure of a scheme encoded in a single simplicial set. (Hence the need to work with a simplicial sheaf which provides a simplicial set for each scheme.)

Of course, you can compute maps in this homotopy category by the Verdier hypercovering theorem. Why aren't people doing this? Well, continuous maps are very flexible. Assume we have topological spaces $X$ and $Y$, a covering $U_1\cup U_2=X$ and maps $f_i:U_i\to Y$ which are homotopic over the intersection $U_1\cap U_2$. We can extend this homotopy on $U_1\cap U_2$ to a continuous homotopy on $U_1$, and then subsequently change to the situation where we can actually glue the maps. So the generality of locally defined maps, locally homotopic etc., is not needed for continuous maps of topological spaces. Again, this kind of flexibility is impossible to get in algebraic geometry.

I guess, you can say generally that it is fairly difficult to relate the algebraic geometry constructions to statements about the $\mathbb{A}^1$-homotopy category. A lot more difficult than in classical topology. Nevertheless, sometimes it is possible to translate classical topological statements (like the theory of the Euler class) into $\mathbb{A}^1$-homotopy, and then it produces interesting algebraic statements. Let me stop here before it turns into some lecture notes, I hope this gives a bit intuition what maps in $\mathbb{A}^1$-homotopy theory look like.

• Thanks for taking the time to write this answer! I learned a lot. – Arrow Jan 13 '16 at 13:11