Timeline for Invariants of exterior powers
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2017 at 4:35 | comment | added | Vanya | @Vit Tucek : Thank you for the solution. | |
| Dec 14, 2017 at 16:50 | comment | added | Vít Tuček | @AbdelmalekAbdesselam Sorry, I misunderstood your sentence "By the above it has at most that dimension." | |
| Dec 14, 2017 at 16:47 | comment | added | Abdelmalek Abdesselam | @VítTuček: I also did not get why you said "However" for $q=14$ when your computations confirm that my upper bound is in fact exact. Indeed, $14=13+1=9+5$ so there are two partitions with distinct parts congruent to 1 mod 4. | |
| Dec 14, 2017 at 16:46 | comment | added | Vít Tuček | @AbdelmalekAbdesselam It means the Cartan type. At least that's how they call it in Sage. So it's result for $O(7)$. | |
| Dec 14, 2017 at 16:24 | comment | added | Abdelmalek Abdesselam | @VítTuček: thanks for checking +1. Sorry I didn't get what $B_3$ means. | |
| Dec 14, 2017 at 16:20 | comment | added | Vít Tuček | Pattern $1,0,0,0,1$ for $q=1,2,3,4,5$ is confirmed on these few cases. However, for $B_3$ I got "Multiplicity of the trivial representation in the 14-th exterior power: 2" | |
| Dec 14, 2017 at 16:17 | comment | added | Abdelmalek Abdesselam | @VítTuček: Does your code confirm my predictions for $q=1,\ldots,5$ ? | |
| Dec 14, 2017 at 16:10 | history | edited | Vít Tuček | CC BY-SA 3.0 |
ad cocalc link
|
| Dec 14, 2017 at 15:57 | comment | added | Vít Tuček | @user49908 The representation of $O(V)$ on symmetric matrices is not irreducible. The symmetric form defining $O(V)$ is invariant! | |
| Dec 14, 2017 at 14:31 | history | edited | Vít Tuček | CC BY-SA 3.0 |
better wording
|
| Dec 14, 2017 at 14:12 | comment | added | Vanya | Thank you I was thinking along similar lines but not beyond the following: By schur's lemma it following that the given object is equal to the $\mathbb{C}$ times multiplicity of trivial representation in $\wedge^k \mathfrak{p}$. Then I thought, "IF" adjoint representation of $\mathbb{k}$ in $\mathfrak{p}$ is irreducible then we get zero for $k =1 $ and by duality for $k= n-1$. I could not proceed further. Thanks for the insight. | |
| Dec 14, 2017 at 13:18 | comment | added | Vít Tuček | Sign error. Happens to everybody. :) | |
| Dec 14, 2017 at 13:06 | comment | added | Abdelmalek Abdesselam | seems I misread the question the matrices are indeed symmetric instead of antisymmetric. | |
| Dec 14, 2017 at 13:02 | history | answered | Vít Tuček | CC BY-SA 3.0 |