Timeline for Jacobson-Morozov theorem
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 7, 2015 at 5:00 | comment | added | Anupam Singh | I must say that I did the computations for root generators. | |
| Jul 6, 2015 at 11:32 | comment | added | Anupam Singh | Some basic computation for classical groups in the "defining" representation shows following: If in the image of $I+te_{1,2}$ the term $t^2$ appears then it has kernel else it does not have kernel. I hope my computations are right. Thus in B_l case we have maps from PSL_2 else all embedding of root generators are via SL_2. | |
| Jul 5, 2015 at 19:38 | comment | added | YCor | clear? every theorem is a clear consequence of any stronger theorem :) Ben's argument shows that it follows just from the classification of $\mathfrak{sl}_2$-representations | |
| Jul 5, 2015 at 10:48 | comment | added | David Stewart | @YCor. The independence of the unipotent element is in fact clear, since by Kostant's work, all $\mathfrak{sl}_2$ subalgebras containing a given nilpotent element $e$ are conjugate by an element of the centraliser $G_e$ of $e$. (This works over $\mathbb{C}$ and fields of sufficiently large characteristic.) | |
| Jul 3, 2015 at 21:14 | comment | added | YCor | Possibly this criterion is not very practical; at least it shows hat for a unipotent element, if we perform the Jacobson-Morozov theorem, the kernel of the resulting homomorphism from $SL_2$ (which is trivial, $\pm 1$, or all of $SL_2$) only depends on the unipotent element and not on the choice of homomorphism. | |
| Jul 3, 2015 at 21:00 | comment | added | Jim Humphreys | @YCor: Thanks for the comment. I've done a bit of editing but will also need to think over the strategy Ben uses when $\phi$ involves more than one summand. It seems awkward to have to consult tables like those of Lawther (or to make up one's own for classical types), but maybe there is no alternative. | |
| Jul 3, 2015 at 20:56 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
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| Jul 3, 2015 at 18:34 | comment | added | YCor | Ben is right. If you have a rep of $SL_2$ which is a sum of $n_i$-dimensional irreducibles $V_{n_i}$, $1\le i\le k$, then the unipotent of $SL_2$ acts with Jordan blocks of size $n_i$, $1\le i\le k$. If all $n_i$ are odd then each $SL_2\to GL(V_{n_i})$ has $-1$ in the kernel, so the whole rep factors through $PGL_2$, while otherwise (i.e. if at least one block has even size) one of this maps is faithful so the whole rep is faithful (i.e. doesn't factor). The point is that although $SL_2$ has irreducibles in all dimensions, it does not have faithful irreducibles in all dimensions. | |
| Jul 3, 2015 at 18:21 | history | answered | Jim Humphreys | CC BY-SA 3.0 |