Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $K$ be a field and $G:=SL_2(K)$, then $G$ is a $K-$split reductive group (to use some big words). These groups are classified by a based root datum $(X,D,X',D')$. Let $G'$ be group associated to $(X',D',X,D)$, the so called dual group. Is it correct that $G'=PGL_2(K)$?

I"m wondering how $PSL_2(K)$ fits into this picture. I'm aware of the fact that if C is algebraically closed, then $PSL_2(C) \cong PGL_2(C)$ as abstract groups; can this be made into an isomorphism of algebraic groups, i.e. is $PSL$ a $K-$form of $PGL$?

share|improve this question

3 Answers 3

up vote 37 down vote accepted

Yes, the dual of $SL_2$ is $PGL_2$.

But you're not going down the right track with $PSL_2$. The problem with $PSL_2$ is that it's not a variety at all! You can quotient out the variety $SL_2$ by the subgroup $\pm1$ but the quotient is the variety $PGL_2$ (recall that quotients in the category of sheaves (for these are really fppf sheaves) don't have to be surjective on global sections, so the statement that there's a surjection $SL_2\to PGL_2$ does not imply that the induced map $SL_2(\mathbf{Q})\to PGL_2(\mathbf{Q})$ is a surjection).

The problem with $PSL_2$ is that it is a functor from, say, rings to groups, but it's not a representable one, so in particular it's not an algebraic group. If you like, you can imagine $PSL_2$ as a presheaf quotient, and $PGL_2$ as the associated (representable) sheaf.


Edit: I tried to think of a way to make this observation more enlightening. Let's for example try and build an affine variety over $\mathbf{Q}$ representing the $PSL_2$ functor. Well we can certainly build an affine variety over $\mathbf{Q}$ representing the $SL_2$ functor: it's just $A:=\mathbf{Q}[a,b,c,d]/(ad-bc-1)$. Now let's see what happens if we try and quotient out by the group $\pm1$. The quotient is affine, and is represented by the invariants of the action, that is, the subring of $A$ consisting of polynomials in $a$, $b$, $c$, $d$ with the property that every monomial mentioned in the polynomial has total degree even. Now here's the problem: I can see a $\mathbf{Q}$-point of this subring (that is, a map from this subring to $\mathbf{Q}$) that corresponds to the matrix $(s,0;0,1/s)$ for $s=\sqrt{p}$, $p$ a prime number! It's the point that sends $a^2$ to $p$, $ab$ to $0$, and so on and so on, and finally $d^2$ goes to $1/p$ and $ad$ goes to $1$, $bc$ goes to $0$, so $ad-bc=1$. The map from the subring to $\mathbf{Q}$ doesn't extend to a map from the whole ring to $\mathbf{Q}$, so our putative construction has failed because the $\mathbf{Q}$-points of this ring are a group that canonically but strictly contains $PSL_2(\mathbf{Q})$ (this subgroup being the $\mathbf{Q}$-points that extend to $\mathbf{Q}$-points of $A$).

share|improve this answer
    
Thank you for your answer! Maybe part of my confusion was due to the fact that the construction you gave in your edit is denoted $PSL_2$ in Springer's book on algebraic groups, p. 24, while as you clearly point out, this should rather be denoted $PGL_2$. –  Guntram Feb 23 '10 at 12:52
2  
Here's a proof that the PSL_2 functor isn't representable, using basically the same trick. Let me stick to varieties over Q, but the same trick will work over any base. If it were representable by some variety $V$, then the non-trivial map Q(sqrt(p))-->Q(sqrt(p)) induces a map V(Q(sqrt(p))-->V(Q(sqrt(p)) and the fixed points will be precisely V(Q). However the element (s,0;0,1/s) in PSL_2(Q(sqrt(p)) above, with s=sqrt(p), is fixed under the involution but is not in the image of PSL_2(Q). Note that this trick works even over C: just base extend to C(t) and use sqrt(t). –  Kevin Buzzard Feb 23 '10 at 13:05

As Kevin says, the "right" definition of ${\rm{PSL}}_n$ is as representing the quotient sheaf ${\rm{SL}}_n/\mu_n$, just as one defines ${\rm{PSO}}(q) = {\rm{SO}}(q)/Z_{{\rm{SO}}(q)}$ (with $Z_G$ denoting the scheme-theoretic center of a smooth group $G$). So really, there is no difference between ${\rm{PSL}}_n$ and ${\rm{PGL}}_n$ when defined correctly, and likewise ${\rm{PSO}}(q) = {\rm{PGO}}(q)$. Personally, I avoid the notation ${\rm{PSL}}_n$ like the plague, since it creates too much confusion.

Lest this seem like a flippant answer, let me point out that for a general ring $R$ with nontrivial Picard group, it is likewise not true that ${\rm{PGL}}_n(R)$ is the "naive" thing either! For example, if $R$ is a Dedekind domain whose Picard group has nontrivial 2-torsion then ${\rm{PGL}} _2(R)$ is generally bigger than ${\rm{GL}} _2(R)/R^{\times}$. And this is not a quirk with algebraic geometry. The same thing happens with Lie groups: if a smooth manifold $M$ has nontrivial 2-torsion line bundles it can and does happen that there are $C^{\infty}$ maps $f:M \rightarrow {\rm{PGL}} _2(\mathbf{R})$ which do not arise from a map to ${\rm{GL}} _2(\mathbf{R})$ (concretely, pulling back the quotient map ${\rm{GL}} _2(\mathbf{R}) \rightarrow {\rm{PGL}} _2( \mathbf{R})$ along $f$ yields a line bundle on $M$ that may be non-trivial).

And the "weirdness" of it all (based on experience over an algebraically closed field) is also seen by the fact that the concrete definition of the group scheme ${\rm{PGL}}_n$ is as a basic affine open in the projective space of $n \times n$ matrices, and we know that "points" of projective spaces are a subtle thing (compared with the case of geometric points) when the source has nontrivial line bundles (whether a scheme or manifold).

Since one cannot get by with field-valued points when doing representability arguments, the same "problem" which one sees for the naive viewpoint on ${\rm{PSL}} _n$ is also relevant when doing proofs for ${\rm{PGL}} _n$. The difference is that for the latter one has to work more "globally" to see the surprise because the degree-1 Zariski and fppf cohomologies for $\mathbf{G} _m$ coincide (so for local rings nothing funny happens, as they have vanishing higher Zariski sheaf cohomology) whereas for the former there is already a funny thing happening for local rings and even fields (which can have nontrivial degree-1 fppf cohomology for $\mu_n$). In more concrete terms, it is equivalent to define the functor ${\rm{PGL}} _n$ as a quotient sheaf for either the Zariski or fppf topologies, whereas for ${\rm{PSL}} _n$ one has to sheafify for the fppf topology (etale ok when $n$ is a unit on the base), and in either case the naive functor on rings (inspired by the case of algebraically closed fields) is not even a Zariski sheaf and hence beyond local rings something has to be done to get the right functor (e.g., one that is representable).

Many books on linear algebraic groups use a version of algebraic geometry that is not well-suited to the subtleties of quotient considerations (e.g., Borel uses Serre's clever method "quotient by $p$-Lie algebra" to handle quotients by infinitesimal groups without saying "infinitesimal group", and some of his quotient arguments would be much shorter if he could have used flatness systematically). In particular, Springer's book has some serious errors when the ground field is not algebraically closed (in the later parts, where he discusses $F$-reductive groups and related things). For example, he uses the incorrect argument that surjectivity of an $F$-map between smooth $F$-varieties can be checked on $F_s$-points, which is not true when $F$ is not perfect (remove an inseparable point from the affine line and consider the inclusion into the line). So be careful in that part of his book. (Some statements are false, not just proofs.)

share|improve this answer
1  
Just to clarify: for anyone who thinks "pah, who thinks about non-perfect fields anyway", the answer is absolutely that the theory of connected reductive groups over fields like k(T) and k((T)), k finite, are an integral part of the Langlands programme, and neither of these fields are perfect. To Brian: I've found Platonov-Rapinchuk quite readable. It took me a long time to find a book that wasn't SGA3 but which did reductive groups over fields in a way I understood. –  Kevin Buzzard Feb 23 '10 at 16:55
1  
Kevin, if you recommend Platonov-Rapinchuk to your students, just warn them to assume everything is affine in the discussion of adelic points there. Their exposition on adelic points claims to avoid affineness hypotheses, but what they define is actually not well-defined in such generality (so doesn't work for G/B and so forth). –  BCnrd Feb 23 '10 at 17:05
3  
Kevin, one other thing: I just checked out P-R from the library, and I am reminded that they assume from the outset that their ground field is characteristic 0! Really a shame. :) –  BCnrd Feb 24 '10 at 1:06
    
Heh, well, I guess I'm letting slip where my interests lie :-) Whyever you are learning this stuff? I'm trying to teach myself automorphic forms over number fields. –  Kevin Buzzard Feb 24 '10 at 20:31
5  
Kevin, you mean "over global fields", right? :) Anyway, I taught myself about algebraic groups to better understand the automorphic stuff. But this PSL_n business doesn't involve anything serious about algebraic groups, just general quotient nonsense from SGA3 which is relevant for fppf/etale sheaves in general. I learned the serious parts in a better way during my recent work on pseudo-reductive groups (which, to my amusement, seems to be a handy reference for over 50% of what I respond to on this website). Care to ask a question about smooth unipotent groups over imperfect fields? :) –  BCnrd Feb 24 '10 at 21:24

Another way of saying essentially the same thing as Kevin has is the following: when do you call a map of algebraic groups a quotient? Considering just the maps $x \mapsto x^n$ from the multiplicative $\mathbb G_m$ to itself. The quotient group functor isn't representable, because if it was, then we would have $$R^\times/(R^\times)^n \to S^\times/(S^\times)^n$$
corresponding to $Hom(A,R) \to Hom(A,S)$ and this latter is clearly injective if $R \to S$ is, while the former isn't (just take $R,S$ fields and $S$ algebraically closed say, e.g. $\mathbb R$, and $\mathbb C$ with $n=2$).

So to stay in the land of algebraic group functors, you probably want to declare $x \mapsto x^n$ a quotient map. One solution to this is to say $G\to H$ is onto if the associated pullback on functions is an injection (as it certainly is for the power maps). Then you can translate this into the context of group functors by showing that this is the same as requiring that the group functors are "surjective" after a finitely generated faithfully flat extension, i.e. given $h \in H(R)$ the quotient, I can find such an extension $S$ of $R$ and a $g \in G(S)$ mapping to the image of $h$ in $H(S)$.

share|improve this answer
1  
Right! So in particular the map from $SL_n$ to $PGL_n$ is surjective, because you can take the $n$'th root you need to pull back to $SL_n$, not over the base, but over a finite flat extension (namely, the one generated by the $n$th root). –  Kevin Buzzard Feb 23 '10 at 14:14

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.