Comparing algebraic group orbits over big and small algebraically closed fields - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:09:34Z http://mathoverflow.net/feeds/question/49885 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/49885/comparing-algebraic-group-orbits-over-big-and-small-algebraically-closed-fields Comparing algebraic group orbits over big and small algebraically closed fields Jim Humphreys 2010-12-19T15:46:34Z 2013-02-23T21:17:45Z <p>For an affine algebraic group <code>$G$</code> it's often convenient (and harmless) to work concretely over an algebraically closed field of definition <code>$k$</code> while identifying <code>$G$</code> with its group of rational points over <code>$k$</code>. Once in a while, however, results obtained over <code>$k$</code> need to be compared with results over a bigger algebraically closed field <code>$K$</code>. </p> <p>For example, in his 1976 <em>Inventiones</em> <a href="http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=GDZPPN002092506" rel="nofollow">paper</a> completing the proof that a semisimple algebraic group always has a finite number of unipotent conjugacy classes, Lusztig observed that in characteristic <code>$p&gt;0$</code> it suffices to assume that <code>$k$</code> is an algebraic closure of the prime field. This in turn allowed him to apply indirectly the Deligne-Lusztig construction of characters for subgroups of <code>$G$</code> over finite subfields of <code>$k$</code>. Here the number of unipotent classes is moreover the same for any algebraically closed field. To justify his reduction, he cites a "simple argument" shown to him by Deligne (which he later told me he viewed in retrospect as "obvious"). </p> <p>Independently, a formal statement of the principle was written down and proved as Proposition 1.1 in a 1997 <em>Journal of Algebra</em> paper: MR1474171 (98j:20058), Guralnick, Robert M. (1-SCA); Liebeck, MartinW. (4-LNDIC); Macpherson, Dugald (4-LEED); Seitz, Gary M. (1-OR), Modules for algebraic groups with finitely many orbits on subspaces. J. Algebra 196 (1997), no. 1, 211–250. Here the proof is fairly elementary, but requires more than a journal page to write down and involves an induction step left to the reader. </p> <blockquote> <p>Is there a short and transparent proof (perhaps from the scheme viewpoint) that finiteness of the number of orbits of a semisimple group acting on an affine variety is the same when an algebraically closed field of definition is extended to another such field, while the number of orbits is unchanged?</p> </blockquote> <p>It would also be interesting to know of other situations in which such a comparison occurs. (Historical remark: In the older version of foundations for algebraic geometry developed by Weil and others it was standard procedure to work over a "universal domain" having infinite transcendence degree over its prime field, to permit for instance the use of "generic points". Then it was usually troublesome to descend to a countable field.) </p> <p>ADDED: To be more precise about "the number of orbits is unchanged", implicit in Lusztig's work and explicit in the 1997 paper cited is the natural requirement on such a bijection that each orbit over <code>$K$</code> should contain a point over <code>$k$</code>. (It's hard to visualize a proof that gives a bijection without this refinement.) On the other hand, it's unclear to me whether special assumptions on <code>$G$</code> such as "reductive" and "connected" are essential for the proof of a general comparison principle.</p> http://mathoverflow.net/questions/49885/comparing-algebraic-group-orbits-over-big-and-small-algebraically-closed-fields/49890#49890 Answer by Torsten Ekedahl for Comparing algebraic group orbits over big and small algebraically closed fields Torsten Ekedahl 2010-12-19T17:12:16Z 2010-12-21T05:18:47Z <p>I think this will work. There are a finite number of orbits of the action of $G$ on $X$ precisely when there is an open orbit and a finite number of orbits on the complement of the orbit. Hence, it is enough to show that if there is an open orbit of a point over the smaller field precisely when there is an open orbit over the larger field. One direction is clear so it is enough to show that there is an open orbit over the larger field, there is one over the smaller field. However, consider the closed subscheme <code>$S:=\{(g,x)\in G\times X | gx=x\}$</code> and the projection $S\to X$ on the second variable. A point of $X$ has an open orbit precisely when the fibre has the smallest possible dimension (when $X$ is irreducible which we may assume). However, there is an open subset (defined over the smaller field) of $X$ with fibres of minimal dimension.</p> <p><b>Addendum</b>: As for the bijection between the orbits this is proved the same way, we have an open orbit which gives one orbit over each field and then we use Noetherian induction.</p> http://mathoverflow.net/questions/49885/comparing-algebraic-group-orbits-over-big-and-small-algebraically-closed-fields/49899#49899 Answer by BCnrd for Comparing algebraic group orbits over big and small algebraically closed fields BCnrd 2010-12-19T18:58:58Z 2010-12-19T21:12:44Z <p>Since you ask about other situations where this sort of thing occurs, let me describe a general principle (applied to the context of the original question) which is widely applied in EGA and elsewhere, often called "spreading out and clever specialization". It is not as efficient as Torsten's argument, and may look quite long at first glance, but hopefully by the end you'll see that it conveys a quite simple idea that is broadly useful for this kind of issue. In particular, it doesn't use the finer information about local closedness of orbits (simple as that may be), so it is applicable to contexts way beyond the one of group actions. </p> <p>I will work with quite weak hypotheses to emphasize the general applicability and flexibility of the basic idea. Then you'll also see that in a discussion between two experts, this would all be disposed of in a few sentences (so the length of what follows may create the wrong impression about the complexity).</p> <p>Setup: Suppose a finite type group scheme $G$ over an algebraically closed field $k$ acts on a finite type $k$-scheme $X$ (assume $G$ and $X$ are affine if you wish), and $K$ is a nonzero $k$-algebra (perhaps not an algebraically closed field). I claim $j:X(k)/G(k) \rightarrow X(K)/G(K)$ is injective (so the source is finite whenever the target is finite), and (more interestingly) that it is also surjective if $X(k)/G(k)$ is finite and $K$ is an algebraically closed field. That will answer the original equivalence question.</p> <p>First let's do injectivity (which will be easy, and so correspondingly not so interesting). Since $K$ exhausted by finite type $k$-subalgebras $K_i$ (definitely <em>not</em> fields in general), we have $X(K)= \varinjlim X(K_i)$ and $G(K)= \varinjlim G(K_i)$ (as $X$ and $G$ are finite type, or alternatively it is clear in the affine case). Thus, $X(K)/G(K) = \varinjlim X(K_i)/G(K_i)$, so it enough to treat the $K_i$ in place of $K$. So we can assume $K$ is finitely generated as a $k$-algebra. [This is a powerful idea, even when the original $K$ is a field.] By the Nullstellensatz there is a $k$-algebra map $s:K \rightarrow k$ (quotient by any maximal ideal) with $k \rightarrow K$ as section; this is the "specialization" trick. It defines a map of sets $X(K)/G(K) \rightarrow X(k)/G(k)$ with the original map $j$ as a section (as $A \rightsquigarrow X(A)/G(A)$ is a functor on $k$-algebras $A$), so $j$ is injective. </p> <p>That was a more or less a formal kind of silliness (despite neat use of the Nullstellensatz), so now we come to the interesting part: assuming $K$ is an algebraically closed field and $X(K)/G(K)$ has at least $n$ points then so does $X(k)/G(k)$ (so surjectivity follows when $X(k)/G(k)$ is finite). The basic principle is this:</p> <blockquote> <p>whatever finite amount of stuff happens over an algebraically closed extension of an algebraically closed field already happens over the ground field via well-chosen specialization. (Kind of like those ads about Las Vegas.)</p> </blockquote> <p>Say $x_1,\dots,x_n$ in $X(K)$ lie in distinct orbits. Exhausting $K$ by finitely generated $k$-subalgebras $K_i$ as above, we can find a big enough $K_i$, call it $A$, so that $x_1,\dots,x_n \in X(A)$. We want to show that for a "sufficiently generic" specialization map $A \rightarrow k$, their images in $X(k)/G(k)$ remains distinct. Here, the valuable geometric intuition (which makes sense even within classical algebraic geometry, since $k$ is algebraically closed and $S :=$ Spec($A$) is basically a classical irreducible variety, as $A$ is a domain of finite type over $k$) is that the $x_i \in X(A)$ are sections to the projection $X \times S \rightarrow S$ such that on the <em>geometric</em> generic fiber over $S$ (i.e., pullbacks along $A \rightarrow K$) they are in pairwise distinct $G$-orbits, and we want to claim that under specialization over some dense open in $S$ they remain in pairwise distinct $G$-orbits. In other words, we aim to "verify" an instance of the</p> <blockquote> <p>Principle of the Geometric Generic Fiber: for a finite collection of finite type schemes over an irreducible noetherian scheme $S$, and any "finite information" structure involving them (maps among them, coherent sheaves on them, etc.), any reasonable property of this structure that holds over a geometric generic point of $S$ also holds on fibers over the geometric points supported in some dense open in $S$. </p> </blockquote> <p>[In practice it isn't always obvious that certain properties are "finite information", such as flatness or surjectivity of $S$-maps, but EGA IV$_3$ lays out the whole story on this principle.]</p> <p>Since any intersection of finitely many non-empty opens in the irreducible $S$ contains a $k$-point (Nullstellensatz once again), it suffices to prove a more general fact for a pair of points $x, x' \in X(A)$ (to then be applied to each of the <em>finitely many</em> pairs $x_i, x_{i'} \in X(A)$ with $i \ne i'$): I claim that if their images in $X(K)$ (the "geometric generic fiber") are in distinct $G(K)$-orbits, then there's a dense open $U$ in $S$ such that for any $u \in U$ (e.g., a $k$-point!) the specializations $x(u), x'(u) \in X(k(u))$ have disjoint orbits under the action of $G_{k(u)}$ on $X_{k(u)}$ (in the sense of their orbit subvarieties over $k(u)$, or geometric points thereof, which comes to the same thing). This will clearly do the job.</p> <p>OK, now comes the step where we make an actual group-theoretic construction (akin to what Torsten did more efficiently) to produce the required open: we view $G \times S$ as an $S$-group acting on the $S$-scheme $X \times S$ and form a "transporter scheme". That is, consider the closed subscheme <code>$$T_{x,x'} = \{g \in G \times S\,|\,g(x) = x'\} \subset G \times S$$</code> over $S$. In more precise terms, writing $G_S$ and $X_S$ as shorthand for $G \times S$ and $X \times S$ to save space, we have the action map $G_S \rightarrow X_S$ over $S$ defined functorially by $g \mapsto g.x$, and $T_{x,x'}$ is the preimage of the closed subscheme of the target given by the (closed immersion) section $x':S \rightarrow X_S$ over $S$. For example, if $s \in S(F)$ for a field $F/k$ then the $s$-fiber of $T_{x,x'}$ is the closed subscheme of $G_F$ defined by the condition "$g.x(s) = x'(s)$" for the points $x(s), x'(s) \in X(F)$. In effect, $T_{x,x'}$ is just the relative version of this latter classical transporter construction as we vary across the pairs $(x(s),x'(s))$ for $s$ wandering in $S$.</p> <p>Finally we have assembled enough to finish. Consider the structural morphism $q:T_{x,x'} \rightarrow S$. This is a map between finite type $k$-schemes. What is its fiber (i.e., pullback) over a point $s:{\rm{Spec}}(F) \rightarrow S$ (such as a $k$-point, or more importantly the "geometric generic point" ${\rm{Spec}}(K) \rightarrow S$)? Well, we just saw what this is: it is the "classical" transporter for $x(s), x'(s) \in X(F)$ inside of $G_F$. So the fiber of $q$ over a physical point $s \in S$ is <em>empty</em> precisely when the corresponding transporter (a finite type $k(s)$-scheme) is empty, which is to say that $x(s), x'(s) \in X(k(s))$ have disjoint orbits under $G_{k(s)}$ acting on $X_{k(s)}$ (i.e., in distinct orbits under $G(\overline{k(s)})$ acting on $X(\overline{k(s)})$, not merely under $G(k(s))$ acting on $X(k(s))$, since emptiness of a finite type $k(s)$-scheme amounts to the absence of $\overline{k(s)}$-points and not merely of $k(s)$-points). Excellent, so if the image of $q:T_{x,x'} \rightarrow S$ misses a dense open $U$, that open will do the job (i.e., for all $u \in U$, the points $x(u), x'(u) \in X(k(u))$ lie in distinct $G_{k(u)}$-orbits in $X_{k(u)}$). Aha, but by (the scheme version of!) Chevalley we know that the image of $q$ is a <em>constructible</em> set even at the level of schemes, so if it misses the generic point then it misses a dense open as desired. So we are reduced to proving that $q$ has empty fiber over the generic point of $S$. But that in turn is exactly the original hypothesis that on the <em>geometric</em> generic fiber over ${\rm{Spec}}(K) \rightarrow S$ our points $x, x' \in X(K)$ lie in distinct orbits under the $G(K)$-action. Voila. QED</p> <p>Now you can see the one serious ingredient that uses schemes (going beyond classical algebraic geometry) in an essential way: the validity of Chevalley's theorem on constructible images in the scheme framework, and the ability to apply it in conjunction with the literal generic point (and geometric points over that). Hopefully you can see that (together with specialization) this is a broadly useful technique for propogating results from an algebraically closed extension of an algebraically closed field back down to the ground field (such as surjectivity on points valued in an algebraically closed field). And that once one realizes this idea, it is sort of simple in the end. In effect, the Principle of the Geometric Generic Fiber above (which is made precise in EGA IV$_3$) is the scheme-theoretic replacement for Weil's "universal domain" concept.</p> http://mathoverflow.net/questions/49885/comparing-algebraic-group-orbits-over-big-and-small-algebraically-closed-fields/49933#49933 Answer by Thomas Scanlon for Comparing algebraic group orbits over big and small algebraically closed fields Thomas Scanlon 2010-12-20T06:24:39Z 2010-12-20T06:24:39Z <p>I think that the simplest explanation has nothing to do with schemes: the first-order theory of algebraically closed fields of a fixed characteristic is complete. The assertion that the group of $k$-points $G(k)$ of the algebraic group $G$ has $n$ orbits on the set of $k$-points $X(k)$ of the algebraic variety $X$ may be naturally expressed as a first-order logical formula and hence is true in one algebraically closed field of characteristic $p \geq 0$ just in case it is true in every such algebraically closed field.</p> http://mathoverflow.net/questions/49885/comparing-algebraic-group-orbits-over-big-and-small-algebraically-closed-fields/50012#50012 Answer by George McNinch for Comparing algebraic group orbits over big and small algebraically closed fields George McNinch 2010-12-20T23:22:59Z 2010-12-20T23:22:59Z <p>In section 1 of the following, I wrote down a proof of the result in question:</p> <blockquote> <p>McNinch, George "On the centralizer of the sum of commuting nilpotent elements." J. Pure Appl. Algebra 206 (2006), no. 1-2, 123–140. <a href="http://front.math.ucdavis.edu/0412.5283" rel="nofollow">[arXiv version]</a></p> </blockquote> <p>The argument I gave is a less powerful application of some of the tools used in the nice answer given by BCnrd -- it is really just an application of Chevalley's Theorem (I used the form found in Springer's "Linear Algebraic Groups" which is good enough for varieties but not for application to schemes in general). And (again in contrast to BCnrd's answer) my argument used the fact that orbits are locally closed.</p> <p>I am glad to have read what is probably the "right" level of generality for this argument found in BCnrd's answer.</p> <p>If $k \subset K$ is an extension of algebraically closed fields, Prop. 4 of loc. cit. show that each $G_{/K}$ orbit has a point rational over $k$ (when $G= G_{/k}$ has finitely many orbits). This gives the bijection between orbits over $k$ and over $K$ -- which of course is already a consequence of BCnrd's answer -- as in Scanlon's answer.</p> http://mathoverflow.net/questions/49885/comparing-algebraic-group-orbits-over-big-and-small-algebraically-closed-fields/122666#122666 Answer by anon for Comparing algebraic group orbits over big and small algebraically closed fields anon 2013-02-22T22:03:59Z 2013-02-23T21:17:45Z <p>The key point is to show the following: let $f:V\rightarrow W$ be a regular map of varieties over an algebraically closed field $k$; if $V(k)\rightarrow W(k)$ is surjective, then $V(K)\rightarrow W(K)$ is surjective for every algebraically closed field $K$ containing $k$.</p> <p>We prove this by induction on the dimension of $W$. We may suppose that $W$ is an irreducible closed subvariety of some affine space. Let $P\in W(K)$ be not in the image. We know that $tr.deg.k(P)\leq dimW$. If equality holds, then $P$ and its conjugates under $Aut(K/k)$ are Zariski dense in $W_{K}$, contradicting the fact that $f$ is (obviously) dominant. Hence $P\in Z(K)$ for some proper closed irreducible subvariety $Z$ of $W$. Now apply induction to $f^{-1}(Z)\rightarrow Z$ to get a contradiction.</p> <p>Edited: I add the rest of the argument. No hypotheses on $G$ are needed.</p> <p>We prove: Let $G\times V\rightarrow V$ be an action of the group variety $G$ on the variety $V$, and let $K$ be an algebraically closed field containing $k$. Then $G$ has finitely many orbits on $V$ if and only if $G_{K}$ has finitely many orbits on $V_{K}$, in which case the numbers of orbits are the same and each $K$-orbit has a $k$-point.</p> <p>For the proof, we first remark that if $v_{1},v_{2}\in V(k)$ lie in distinct $G$-orbits, then they lie in distinct $G_{K}$-orbits. To see this, let $Z$ be the inverse image of $v_{2}$ under the regular map $g\mapsto gv_{1}\colon G\rightarrow V$. Then $Z(k)$ is empty if and only if $Z$ is the empty variety if and only if $Z(K)$ is empty.</p> <p>Suppose that $G$ has only finitely many orbits on $V$, and let $v_{1} ,\ldots,v_{m}\in V(k)$ represent the different orbits. The regular map $(g,v_{i})\mapsto gv_{i}\colon G(k)\times{v_{1},\ldots,v_{m}}\rightarrow V(k)$ is surjective, and hence remains surjective with $K$ for $k$. Together with the first remark, this shows that $v_{1},\ldots,v_{m}$ represent the different orbits of $G_{K}$ on $V_{K}$.</p> <p>Finally, suppose that $G_{K}$ has only finitely many orbits on $V_{K}$. Then the first remark shows that $G$ has only finitely many orbits on $V$, and the previous argument applies.</p>