Why is this not an algebraic space? - MathOverflow

Why is this not an algebraic space?

2009-12-15T21:06:01Z

This question is related to the following question which I've just seen which was posted by Anton. His question is whether a algebraic space which is a group object is necessarily a group scheme, and the answer appears to be YES. Now my naive idea of what an algebraic space is is that it is the quotient of a scheme by an etale equivalence relation, but I seem to be confusing myself. I'm hoping someone will help lead me out of my confusion.

Let me first consider an analogous topological situation, where the answer is no. We can consider the category of smooth spaces, by which I mean the category of sheaves on the site of smooth manifolds which are quotients of manifolds by smooth equivalence relations (with discrete fibers).

Here is an example: If we have a discrete group G acting (smoothly) on a space X, we can form the equivalence relation $R \subset X \times X$, where R consists of all pairs of points $(x,y) \in X \times X$ where $y= gx$ for some $g \in G$. If R is a manifold, then the sheaf $[X/R]$ (which is defined as a coequalizer of sheaves) is a smooth space.

Here is an example of a group object in smooth spaces: We start with the commutative Lie group $S^1 = U(1)$. Now pick an irrational number $r \in \mathbb{R}$ which we think of as the point $w = e^{2 \pi i r}$. We let $\mathbb{Z}$ act on $S^1$ by "rotation by r" i.e.

$\mathbb{Z} \times S^1 \to S^1$

$ (n, z) \mapsto w^n z$

This gives us an equivalence relation $R = \mathbb{Z} \times S^1 \rightrightarrows S^1$, where one map is the action and the other projection. The fibers are discrete and the quotient sheaf is thus a smooth space, which is not a manifold. However the groupoid $R \rightrightarrows S^1$ has extra structure. It is a group object in groupoids, and this gives the quotient sheaf a group structure.

The group structure on the objects $S^1$ and morphisms $R$ are just given by the obvious group structures. Incidentally, this group object in groupoids serves as a sort of model for the "quantum torus", 0604.5405.

Now what happens when we try to copy this example in the setting of algebraic spaces and schemes?

Let's make it easy and work over the complex numbers. An analog of $S^1$ is the group scheme $\mathbb{G}_m / \mathbb{C} = spec \mathbb{C}[t,t^{-1}]$.

Any discrete group gives rise to a group scheme over $spec \mathbb{C}$ by viewing the set $G$ as the scheme

$\sqcup_G spec \mathbb{C}$

So for example we can view the integers $\mathbb{Z}$ as a group scheme. This (commutative) group scheme should have the property that a homomorphism from it to any other group scheme is the same as specifying a single $spec \mathbb{C}$-point of the target (commutative) group scheme.

A $spec \mathbb{C}$ point of $spec \mathbb{C}[t,t^{-1}]$ is specified by an invertible element of $\mathbb{C}$. Let's fix one, namely the one given by the element $w \in S^1 \subset \mathbb{C}^\times$. So this gives rise to a homomorphism $\mathbb{Z} \to \mathbb{G}_m$ and hence to an action of $\mathbb{Z}$ on $\mathbb{G}_m$.

Naively, the same construction seems to work to produce a group object in algebraic spaces which is not a scheme. So my question is: where does this break down?

There are a few possibilities I thought of, but haven't been able to check:

Does $R = \mathbb{Z} \times \mathbb{G}_m$ fail to be an equivalence relation for some technical reason?
Do the maps $R \rightrightarrows \mathbb{G}_m$ fail to be etale?
Is there something else that I am missing?

Answer by Jonathan Wise for Why is this not an algebraic space?

2009-12-15T21:29:46Z

If $X$ denotes the quotient sheaf $\mathbf{G}_m / \mathbf{Z}$ then the inclusion

$\mathbf{G}_m \times_X \mathbf{G}_m \rightarrow \mathbf{G}_m \times \mathbf{G}_m$

can be identified with the action map

$\mathbf{G}_m \times \mathbf{Z} \rightarrow \mathbf{G}_m \times \mathbf{G}_m$.

Since this map is not quasi-compact, $X$ is not quasi-separated, so it is not an algebraic space by Knutson's definition.

Answer by James Borger for Why is this not an algebraic space?

2009-12-15T22:15:19Z

I agree with Jonathan Wise's answer, but what about possibly non-quasi-separated algebraic spaces, i.e. according to Raynaud-Gruson or To\"en-Vaqui\'e?

It seems that the answer is amazingly... Yes, that quotient is an algebraic space! Actually, you've already given the proof: R is a scheme and an etale equivalence relation. It's a rather crazy algebraic space in that it doesn't have an open subscheme. The theorem in Knutson that says open subschemes always exist uses quasi-separatedness, as it apparently must. So it seems you've discovered a group object in the category of algebraic spaces which is not quasi-separated or a scheme. Well done! I never would have guessed such a thing would exist, much less be so simple. (An even slightly simpler example would be the additive group modulo Z, at least in characteristic 0.)

Answer by Anton Geraschenko for Why is this not an algebraic space?

2009-12-24T17:52:07Z

Though Knutson requires algebraic spaces to be quasi-separated, I don't think it's a reasonable requirement. After all, not all schemes are quasi-separated, and we certainly want all schemes to be algebraic spaces.

Perhaps $\mathbb{G}_m/\mathbb{Z}$ feels like a strange algebraic space because it is not locally separated (i.e. its diagonal is not an immersion). You can see this because the morphism $R\to \mathbb{G}_m\times \mathbb{G}_m$ is not an immersion (this map is a base change of the diagonal, and immersions are stable under base change). The other classic example of a non-locally-separated (but quasi-separated) algebraic space is the line with a doubled tangent direction (Example 1 on page 9 of Knutson's book).