Timeline for Connected algebraic subgroup of $PGL_3$ and $PGL_2 \times PGL_2$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 19, 2017 at 10:51 | comment | added | Anonymous | Yes, my direct arument only shows that $p_1 |_H$ and $p_2 |_H$ are bijective (which is enough in characteristic zero). | |
| Jul 19, 2017 at 10:50 | history | edited | Anonymous | CC BY-SA 3.0 |
Added an explanation
|
| Jul 19, 2017 at 8:52 | comment | added | spin | $\{ (x, f(x)) : x \in \operatorname{PGL}_2 \}$ 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. | |
| Jul 19, 2017 at 8:25 | comment | added | sabrebooth | 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). | |
| Jul 19, 2017 at 5:16 | comment | added | Anonymous | I added a more direct argument in case $p_1(H) = p_2(H) = \mathrm{PGL}_2$ in my answer. | |
| Jul 19, 2017 at 5:15 | history | edited | Anonymous | CC BY-SA 3.0 |
Added a more direct argument in case $p_1(H) = p_2(H) = \mathrm{PGL}_2$
|
| Jul 18, 2017 at 21:40 | comment | added | Anonymous | 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. | |
| Jul 18, 2017 at 20:26 | comment | added | sabrebooth | 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 $Lie(G)$). | |
| Jul 18, 2017 at 12:08 | history | answered | Anonymous | CC BY-SA 3.0 |