Timeline for Sub-representations of the affine group
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 30, 2010 at 16:00 | comment | added | Bugs Bunny | I think these are all subrepresentations. To prove this, use complete reducibility as a representation of the multiplicative group. This will tell you that any subrepresentation (of affine group) is spanned by some of $X^t$ and all it remains is to invert some funny determinant, which proves that the span of $(X+\alpha)^m$ is what I said it is. Judging by you comment, you have done it already. | |
| Jun 30, 2010 at 14:13 | vote | accept | Klim Efremenko | ||
| Jun 30, 2010 at 14:13 | comment | added | Klim Efremenko | Yes, all sub-representations looks like this. Proof: Let $a=a_0+a_1p+a_2p^2+a_k p^k$ then ${a \choose b}$ non zero mod p only if $b_i< a_i$. | |
| Jun 30, 2010 at 13:33 | comment | added | Klim Efremenko | Does it characterize all sub-representations? Let $c_1, \ldots c_k$ be integer numbers. Consider the set $S=\{t=a_0+a_1p+a_2p^2 \ldots a_k p^k | a_i<c_i\}$. Then span of $X^t$ for $t\in S$ will be sub-representation. Does it all sub-representations? If yes, how can you prove it? | |
| Jun 30, 2010 at 13:17 | vote | accept | Klim Efremenko | ||
| Jun 30, 2010 at 13:22 | |||||
| Jun 30, 2010 at 12:48 | vote | accept | Klim Efremenko | ||
| Jun 30, 2010 at 13:17 | |||||
| Jun 30, 2010 at 11:44 | history | answered | Bugs Bunny | CC BY-SA 2.5 |