Why is this not an algebraic space? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T04:41:19Z http://mathoverflow.net/feeds/question/9043 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/9043/why-is-this-not-an-algebraic-space Why is this not an algebraic space? Chris Schommer-Pries 2009-12-15T21:06:01Z 2009-12-24T17:52:07Z <p>This question is related to the following <a href="http://mathoverflow.net/questions/8918/is-an-algebraic-space-group-always-a-scheme" rel="nofollow">question</a> 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 <em>is</em> 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.</p> <p>Let me first consider an analogous topological situation, where the answer is no. We can consider the category of <em>smooth spaces</em>, 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). </p> <p>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.</p> <p>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.</p> <p>$\mathbb{Z} \times S^1 \to S^1$</p> <p>$(n, z) \mapsto w^n z$</p> <p>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.</p> <p>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", <a href="http://front.math.ucdavis.edu/0604.5405" rel="nofollow">0604.5405</a>.</p> <p>Now what happens when we try to copy this example in the setting of algebraic spaces and schemes?</p> <p>Let's make it easy and work over the complex numbers. An analog of $S^1$ is the group scheme $\mathbb{G}_m / \mathbb{C} = spec \mathbb{C}[t,t^{-1}]$. </p> <p>Any discrete group gives rise to a group scheme over $spec \mathbb{C}$ by viewing the set $G$ as the scheme</p> <p>$\sqcup_G spec \mathbb{C}$</p> <p>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.</p> <p>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 $\mathbb{G}_m$.</p> <p>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?</p> <p>There are a few possibilities I thought of, but haven't been able to check:</p> <ol> <li>Does $R = \mathbb{Z} \times \mathbb{G}_m$ fail to be an equivalence relation for some technical reason?</li> <li>Do the maps $R \rightrightarrows \mathbb{G}_m$ fail to be etale?</li> <li>Is there something else that I am missing?</li> </ol> http://mathoverflow.net/questions/9043/why-is-this-not-an-algebraic-space/9044#9044 Answer by Jonathan Wise for Why is this not an algebraic space? Jonathan Wise 2009-12-15T21:29:46Z 2009-12-15T21:29:46Z <p>If $X$ denotes the quotient sheaf $\mathbf{G}_m / \mathbf{Z}$ then the inclusion</p> <p>$\mathbf{G}_m \times_X \mathbf{G}_m \rightarrow \mathbf{G}_m \times \mathbf{G}_m$</p> <p>can be identified with the action map</p> <p>$\mathbf{G}_m \times \mathbf{Z} \rightarrow \mathbf{G}_m \times \mathbf{G}_m$. </p> <p>Since this map is not quasi-compact, $X$ is not quasi-separated, so it is not an algebraic space by Knutson's definition.</p> http://mathoverflow.net/questions/9043/why-is-this-not-an-algebraic-space/9048#9048 Answer by James Borger for Why is this not an algebraic space? James Borger 2009-12-15T22:15:19Z 2009-12-16T00:30:58Z <p>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?</p> <p>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.)</p> http://mathoverflow.net/questions/9043/why-is-this-not-an-algebraic-space/9689#9689 Answer by Anton Geraschenko for Why is this not an algebraic space? Anton Geraschenko 2009-12-24T17:52:07Z 2009-12-24T17:52:07Z <p>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.</p> <p>Perhaps $\mathbb{G}_m/\mathbb{Z}$ feels like a strange algebraic space because it is <strong>not locally separated</strong> (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).</p>