Question

In topology the spheres $S^n$ are the "simplest" closed manifolds, and they are like "Dirac's delta at $n$" for (reduced) cohomology groups. Furthermore they are boundaries of the simplest compact manifolds-with-boundary, i.e. the disks $D^{n+1}$, which are contractible. And $S^{n}$ is obtained by glueing two copies of $D^{n}$ along their boundary $S^{n-1}$. My question is:

Are there some objects of algebraic geometric nature that somehow reproduce the same pattern, or that are considerable as the equivalent of spheres from topology?

More generally, are there "homology spheres" for some homology theory like -say- Chow groups? What about an "algebraic Poincaré conjecture"?

If they do exist, I don't expect them to be standard varieties or schemes, otherwise they probably would have made their appearence "classically".

I imagine some experts in homotopy theory may find this question naive or trivial; well, I apologise if it is so...

Answer 1

To expand on Will Sawin's comment in a vague sort of way (which is the best I can do):

The cofiber of the pair $(\mathbb P^n,\mathbb P^{n-1})$ (which doesn't exist as a scheme but does exist if you widen your scope to a suitable category of (pre)sheaves) can be called a $2n$-sphere, the motivic $2n$-sphere.

Also, the smash product of the motivic $2m$-sphere with the motivic $2n$-sphere is the motivic $2(m+n)$-sphere. For example, you can make a map of pairs $$(\mathbb P^1\times \mathbb P^1,\mathbb P^1\vee \mathbb P^1) \to (\mathbb P^2,\mathbb P^1)$$ inducing an isomorphism of cofibers.

In some sense, based on considering maps $X\times \mathbb A^1\to Y$ as homotopies, $\mathbb P^{n-1}$ is a deformation retract of $\mathbb P^n-\ast$, the complement of a point in projective $n$-space. So the homotopy cofiber of the pair $(\mathbb P^n,\mathbb P^n-\ast)$ is, up to homotopy, the same thing.

If you believe in excision, then the homotopy cofiber of the pair $(\mathbb A^n,\mathbb A^n-\ast)$is another model for the same sphere. And since $\mathbb A^n$ is contractible, that makes $\mathbb A^n-\ast$ look a lot like a $(2n-1)$-sphere, with the motivic $2n$-sphere being in some sense its suspension.

But the motivic $2(n+1)$-sphere is not in the same sense the double suspension of the motivic $2n$-sphere; the cohomology is wrong

Answer 2

Probably not, at least treating the question naively, since algebraic geometry focuses on smooth projective varieties, and these have nonzero cohomology (for basically any cohomology theory) in degrees $0, 2, \ldots 2\cdot \dim$: the class of the hyperplane section in $H^2$ has nonzero top cup product with itself.