I am looking for a reference regarding the maximal proper connected algebraic subgroups of $PGL_3$ and $PGL_2 \times PGL_2$ respectively when the base field is any algebraically closed field (of characteristic $\neq 2$).

For instance in the case $G=PGL_2 \times PGL_2$, if $H$ is a maximal proper connected algebraic subgroup of $G$, then I believe that $H$ is as follows:

  • $PGL_2 \times B$, where $B$ is a Borel subgroup of $PGL_2$;
  • $B \times PGL_2$; or
  • conjugate to the diagonal embedding of $PGL_2$ in $G$.

I know a quite simple proof of this result over a field of characteristic zero but it uses the correspondence between Lie subalgebras and algebraic subgroups.

Thank you in advance for your help!

  • 1
    $\begingroup$ First you have max. parabolics. Putting this aside you can assume your groups is reductive. Moreover, it has no center, otherwise it is contained in a parablic. So it is semi-simple of rank 1 or 2. In both cases the rank 2 case must be the full group. The rank 1 case must be the irreducible rep of $\text{SL}_2$ in the case of $\text{PGL}_3$ and the graph of an automorphism in the product case. $\endgroup$
    – Uri Bader
    Jul 19 '17 at 6:23
  • $\begingroup$ In the case of $PGL_3$ you also have a maximal conjugacy class for $PGL_2$, and again to be sure that you have only one conjugacy class for $SL_2$ (contained in a parabolic conjugacy class, so it doesn't yield maximal connected algebraic subrgoups of $PGL_3$) and one conjugacy class for $PGL_2$ (which yields maximal connected algebraic subrgoups of $PGL_3$), it seems that one needs Lie algebra arguments or at least some classifications results (like the paper of Seitz pointed out by Jim). $\endgroup$
    – sabrebooth
    Jul 19 '17 at 8:18

Gary Seitz (and various collaborators over the years) have worked out lots of concrete information about maximal closed subgroups of classical groups and exceptional algebraic groups over an algebraically closed field of prime characteristic. Much of this is used in the study of maximal subgroups of corresponding finite groups of Lie type. Anyway, a basic 1987 paper on classical algebraic groups by Seitz is here.

I don't have all this information at my fingertips, but I suspect you can already draw a complete answer to your question from this substantial paper. He imitates some of the methods used by Dynkin in characteristic 0, though his results are more subtle.

[ADDED] Probably you don't need to get into the rather elaborate inductive framework of the long paper by Seitz, since you are dealing with semisimple groups of rank 2. (Also, it doesn't seem to matter whether you allow characteristic 2 here or what the isogeny type of your groups may be.)

The older 1971 paper of Borel-Tits here already implies that any maximal proper closed (say connected) subgroup of a (connected) reductive group $G$ must be either parabolic or reductive. This is written up in Theorem 30.4(a) in my 1975 Springer text on linear algebraic groups. Parabolic subgroups are easy to describe (up to conjugacy), even if the root system of $G$ isn't irreducible. So for example in your direct product of two copies of a rank 1 group, proper parabolics involve the direct product of a Borel subgroup $B$ in one factor and the entire group in the other factor. To get a maximal proper reductive subgroup, you can use as indicated a diagonal embedding.

[P.S.] I've never thought systematically about the classification of proper maximal reductive subgroups in a semisimple group $G$ which isn't simple (as an algebraic group), assuming the simple case is already known. Is there a systematic method for general $G$ to work out the list of maximal (closed, connected) reductive subgroups? I don't recall any literature which treats this directly.


Here is another answer in case $G = \mathrm{PGL}_2 \times \mathrm{PGL}_2$, just using elementary calculations. So let $H \subset G$ be a connected maximal algebraic subgroup. For $i = 1, 2$, denote by $p_i \colon G \to \mathrm{PGL}_2$ the projection to the $i$-th factor.

If $p_1(H) \neq \mathrm{PGL}_2$, then $p_1(H)$ is a maximal (connected) subgroup of $\mathrm{PGL}_2$ and thus $H = B \times \mathrm{PGL}_2$ for some Borel subgroup $B$ of $\mathrm{PGL}_2$. Analogously, if $p_2(H) \neq \mathrm{PGL}_2$, then $H = \mathrm{PGL}_2 \times B$.

Now, assume $p_1(H) = p_2(H) = \mathrm{PGL}_2$. Let $N \subseteq H$ be a normal subgroup of $H$. Since $\mathrm{PGL}_2$ is simple, $p_i(N) = 1$ for $i =1, 2$ and hence $N$ lies in the kernel of $p_1$ and of $p_2$. Thus $N = 1$ and therefore $H$ is simple. Since $H$ is a subgroup of $\mathrm{PGL}_2 \times \mathrm{PGL}_2$, it must be isomorphic to $\mathrm{PGL}_2$. Thus, $p_i \colon H \to \mathrm{PGL}_2$ is an isomorphism for $i=1, 2$ and hence H is conjugate to the diagonal embedding (here one uses the fact that all automorphisms of $\mathrm{PGL}_2$ are inner).

[ADDED] There is a more direct argument in case $p_1(H) = p_2(H) = \mathrm{PGL}_2$: the kernel of $p_1 |_H \colon H \to \mathrm{PGL}_2$ is the normal subgroup $H \cap \{ 1 \} \times \mathrm{PGL}_2$ of $H$. Since $p_2(H) = \mathrm{PGL}_2$, one can easily see that $H \cap \{ 1 \} \times \mathrm{PGL}_2$ is normal in $\{ 1 \} \times \mathrm{PGL}_2$. Since $H \subsetneq \mathrm{PGL}_2 \times \mathrm{PGL}_2$, this implies that $H \cap \{ 1 \} \times \mathrm{PGL}_2$ is trivial and thus $p_1 |_H$ is an isomorphism. Analogously one sees that $p_2 |_H$ is an isomorphism. Hence $H$ is conjugate to the diagonal embedding (here one uses the fact that all automorphisms of $\mathrm{PGL}_2$ are inner).

[ERRATUM] The direct argument above only shows that $p_1 |_H$ and $p_2 |_H$ are bijective.

  • 1
    $\begingroup$ Thank you for your answer, but that's precisely to justify the statement "Since $H$ is a subgroup of $PGL_2 \times PGL_2$, it must be isomorphic to $PGL_2$." that I'm a bit annoyed. It seems elementary but I can't find a short argument without using the classification of semisimple Lie algebras (and of course this can't be used in positive characteristic). Also, even if we know that $H\cong PGL_2$, it is not direct that there are exactly three conjugacy classes for $PGL_2$ in $G=PGL_2 \times PGL_2$ (the way I see it requires to consider conugacy classes of $sl_2$-triples in $Lie(G)$). $\endgroup$
    – sabrebooth
    Jul 18 '17 at 20:26
  • $\begingroup$ First of all, we know that the simple group $H$ has rank $\leq 2$, as it is a subgroup of $\mathrm{PGL}_2 \times \mathrm{PGL}_2$. However, if the rank of $H$ would be $2$, then it cannot be of dimension $\leq 5$ (yes, here we use the classification of simple root systems (which is available in any characteristic)). Thus the rank of $H$ is one and since $H$ is simple it must be isomorphic to $\mathrm{PGL}_2$. I'm sorry, but the second statement in your comment I don't understand. $\endgroup$
    – Anonymous
    Jul 18 '17 at 21:40
  • $\begingroup$ I added a more direct argument in case $p_1(H) = p_2(H) = \mathrm{PGL}_2$ in my answer. $\endgroup$
    – Anonymous
    Jul 19 '17 at 5:16
  • $\begingroup$ I agree that you get $H \cong PGL_2$, but then it is not clear that, up to conjugacy in $PGL_2 \times PGL_2$, you only have one possibility to embed $PGL_2$ (ie, the diagonal embedding). A priori there could be other ways to embed $PGL_2$ in $PGL_2 \times PGL_2$ such that it surjects on both factors (in characteristic zero it is easy to check via $sl_2$-triples general theory but in positive characteristic it seems to me that one needs another argument). $\endgroup$
    – sabrebooth
    Jul 19 '17 at 8:25
  • $\begingroup$ $\{ (x, f(x)) : x \in \operatorname{PGL}_2 \}$ for any surjective $f: \operatorname{PGL}_2 \rightarrow \operatorname{PGL}_2$ would be maximal, right? So in positive characteristic would you get infinitely many conjugacy classes since $f$ could be a power of the Frobenius map? Maybe at the end of your answer $p_1$ is an isomorphism of groups, but not an isomorphism of algebraic groups. $\endgroup$
    – spin
    Jul 19 '17 at 8:52

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