Timeline for Is there a projection from $G$ to the Levi subgroup of a Parabolic subgroup?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| May 21, 2018 at 20:14 | comment | added | Mikhail Borovoi | In the case $G={\rm SL}_{n,\mathbb{C}}$, $L\subsetneqq G$, there is no nontrivial homomorphism $G\to L$ (for example, because $G$ is a simple group and the dimension of $L$ is smaller than the dimension of $G$). | |
| May 21, 2018 at 20:03 | comment | added | Jianrong Li | @Mikhail, thank you very much. I mean a homomorphism of algebraic groups. | |
| May 21, 2018 at 16:19 | comment | added | Mikhail Borovoi | What do you mean by a projection? A homomorphism of algebraic groups? A morphism of varieties? Or what? | |
| May 20, 2018 at 15:25 | comment | added | Jianrong Li | @dhy, thank you very much. The only property I need is the image of the projection $G \to L_J$ is $L_J$. Does such projection exist? | |
| May 20, 2018 at 14:40 | review | Close votes | |||
| May 20, 2018 at 22:46 | |||||
| May 20, 2018 at 13:56 | comment | added | dhy | I'm not sure what properties you want your projection to have, but I doubt there will be such a projection w/ any reasonable properties. For example, taking $G$ to be $SL_2(\mathbb{C})$ and $P=B$, you don't have any projection $G\rightarrow T$ that sends $T$ to itself (here I'm identifying $L$ and $T$ with diagonal matrices.) To see this, note that the map $\pi_1(T)\rightarrow\pi_1(G)$ is not injective. | |
| May 20, 2018 at 13:43 | history | asked | Jianrong Li | CC BY-SA 4.0 |