# What is the algebraic geometry version of the spheres?

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".

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There are motivic spheres, with cohomology of dimension $1$ in degrees $0$ and $2n$. The first motivic sphere is a scheme, the actual sphere $\mathbb P^1$. –  Will Sawin Dec 7 '12 at 0:15
I know you weren't interested in standard schemes, but projective spaces of dimension $n$ seem to fit your criteria: they are composed from gluing together two "disks" (although not along a lower dimensional sphere), they contain lower-dimensional projective spaces, and they are double covered topologically by actual spheres. –  Brian Rushton Dec 7 '12 at 3:03
@Chris: thanks for fixing the typos. –  Qfwfq Dec 7 '12 at 11:15
I'd be interested to see some precise ways of phrasing the sentiment that spheres are the simplest closed manifolds. –  Tom Leinster Dec 7 '12 at 12:18
@TL: for example if you $#$ (connected sum) a manifold with a sphere, you obtain something which is homeomorphic to the original manifold, so the sphere (of the appropriate dimension) is like a neutral element for $#$ - I don't know if there is an equivalent of this in algebraic geometry. I don't know anything of homotopy theory, but it seems the sphere spectrum is quite a fundamental object. Or maybe you're thinking of some categorical characterization in the style of mathoverflow.net/questions/83753/… ? –  Qfwfq Dec 7 '12 at 12:41

To expand on Tom Goodwillie's answer:

a precise definition of "motivic sphere" would be $$S^{p,q} = \big( \Delta^{p-q} / \partial \Delta^{p-q} \big) \wedge \big( \bigwedge^q \mathbb{G}_m\big)$$ which you can interpret as the $q$-fold smash product of the multiplicative group $\mathbb{G}_m$ smashed with a $(p-q)$-dimensional simplicial sphere. This smash product makes sense in a category of simplicial presheaves on smooth schemes (which is what you work with in A¹-homotopy theory).

The indices can be explained from looking at the motive $M(S^{p,q}) = \mathbb{Z}(q)[p]$ or at the realizations, where the complex realization gives $S^p$ and the real realization gives $S^{p-q}$. More about this can be read in the Morel-Voevodsky paper.

Now these are not algebraic varieties, by definition, so a good question is

For integers $p$ and $q$, does an algebraic variety $X$ exist which is A¹-weakly equivalent to $S^{p,q}$?

We'll also say "$X$ is a $S^{p,q}$". From looking at realizations you can already exclude $p,q$ negative. One positive example is (by definition) $\mathbb{G}_m$, which is a $S^{1,1}$.

As Tom Goodwillie explained, $\mathbb{A}^n / (\mathbb{A}^n \setminus 0)$ is a $S^{2n,n}$ (which some people shorten to "motivic $2n$-sphere") and $\mathbb{A}^n \setminus 0$ is a $S^{2n-1,n}$.

There is not much known beyond that, I suppose.

There are even more varieties that could count as algebraic versions of spheres. An interesting feature of algebraic geometry is that you can look at many fields of definition. Projective quadrics over the complex numbers all look the same (in each dimension), as they are isomorphic (given the same dimension) to the split quadrics $Q_{2n}^{split} = \{\sum_{i=0}^n x_iy_i = 0\}$ or $Q_{2n+1}^{split} = \{\sum_{i=0}^n x_iy_i = z^2\}$ as well as to the anisotropic quadrics $Q_{m} = \{\sum_{i=0}^m z_i^2\}$ (by change of basis). Over smaller, non-quadratically closed fields, these quadrics are no longer isomorphic.

Now look at the affine quadrics inside the projective quadrics (by removing a suitable hyperplane section). Conjecturally, the split ones are motivic spheres (at least they have the right motive), while the anisotropic ones aren't. You can consider these forms of motivic spheres.

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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

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Out of curiousity, what is the map $\mathbb{P}^1 \times \mathbb{P}^1 \to \mathbb{P}^2$? –  S. Carnahan Jun 16 '13 at 0:00
In homogeneous coordinates $((a:b),(c:d))\mapsto (ac:ad+bc:cd)$. More conceptually, multiply homogeneous linear polynomials $(aX+bY)(cX+dY)$. $P^n$ is the $n$-th symmetric power of $P^1$, by unique factorization of degree $n$ homogeneous polynomials. –  Tom Goodwillie Jun 16 '13 at 1:09

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.

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