Timeline for Degenerate representation
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 18, 2022 at 18:16 | vote | accept | user56980 | ||
| Jul 18, 2022 at 18:15 | history | bounty ended | user56980 | ||
| Jul 18, 2022 at 18:04 | comment | added | user56980 | I can't edit the nice answer $\lambda_r^2=...e_i^{*2}$ instead of $e_i^*$ | |
| Jul 18, 2022 at 17:32 | comment | added | user56980 | Thank you. I reacted quickly, of course $q_{rs}$ is not linear! | |
| Jul 18, 2022 at 16:55 | comment | added | Jay Taylor | Relatedly $e_i^*e_j^*$ is also not zero. Let $v = e_i + e_j$ then $e_i^*(v) = e_j^*(v) = 1$ and so $(e_i^*e_j^*)(v) = e_i^*(v)e_j^*(v) = 1\cdot 1 = 1$. Recall we evaluate the product pointwise. In fact, in the example in the previous comment the quadratic form is $q = e_1^*e_2^*$. | |
| Jul 18, 2022 at 16:51 | comment | added | Jay Taylor | That doesn't imply $q_{rs} = 0$ because $q_{rs}$ is not linear. Take $n=2$ and let $\beta$ be the symmetric bilinear form with $\beta(e_1,e_2) = \beta(e_2,e_1) = 1$ and $\beta(e_1,e_1) = \beta(e_2,e_2) = 0$. We have a quadratic form given by $q(v) = \frac{1}{2}\beta(v,v)$ for all $v \in V$. Certainly $q(e_1) = q(e_2) = 0$ but $q(e_1+e_2) = \frac{1}{2}\beta(e_1+e_2,e_1+e_2) = 1 \neq 0$. So it is not enough to check this on a basis. | |
| Jul 18, 2022 at 16:38 | comment | added | user56980 | Many thanks Jay, but $q_{rs}$ is already identically zero: $q_{rs}(e_i)=0$ for all $i=1,\ldots,n$ also $e_i^*e_j^*=0$ | |
| Jul 16, 2022 at 15:43 | history | answered | Jay Taylor | CC BY-SA 4.0 |