Timeline for Finite dimensional spherical representation of $SO(n,1)(\mathbb{R})$
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 27, 2011 at 8:43 | vote | accept | Ludo Marquis | ||
| Aug 7, 2011 at 21:18 | answer | added | Alain Valette | timeline score: 6 | |
| Aug 7, 2011 at 20:44 | comment | added | Alain Valette | "Unknown control sequence" seems to refer to a TeX error... | |
| Sep 3, 2010 at 5:14 | comment | added | Victor Protsak | No, the adjoint representation is not spherical for $n\geq 3.$ You need to take the defining representation of $SO(n,1)$ on $\mathbb{R}^{n+1},$ which is self-adjoint and spherical (the last basis vector is $K$-invariant), and its symmetric powers (which can be realized in the homogeneous polynomials of degree $d$) are spherical and, moreover, they exhaust all finite-dimensional spherical representations of $SO(n,1).$ I have no clue what you mean by "the Lie Algebra Unknown control sequence". | |
| Aug 27, 2010 at 16:03 | comment | added | Ludo Marquis | Sorry. "pherical" mean "spherical". Let me sum up (just to be sure), i take the Lie Algebra Unknown control sequence $\mathfrak{so}(n,1)$, i take the adjoint representation which is spherical, and the symmetric power of the adjoint representation gives me all the spherical representation of SO(n1). | |
| Aug 26, 2010 at 17:16 | comment | added | Jim Humphreys | I've edited the header following Victor's remark, since there is no likely mathematical term "pherical". | |
| Aug 26, 2010 at 17:15 | history | edited | Jim Humphreys | CC BY-SA 2.5 |
edited title
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| Aug 26, 2010 at 17:12 | comment | added | Victor Protsak | My guess is that "pherical" should have been "spherical", which means "contains a K-invariant vector", where $K$ is a maximal compact subgroup. In that case, the answer is given by the symmetric powers of the defining representation. | |
| Aug 26, 2010 at 16:03 | comment | added | José Figueroa-O'Farrill | I am not sure what you mean by "explicit", but I think that they are all given by tensors of the ($n+1$)-dimensional vector representation. The representation ring of the Lie algebra $\mathfrak{so}(n,1)$ is generated by the vector and spinor representations, but the spinorial representations are not representations of $SO(n,1)$. | |
| Aug 26, 2010 at 16:00 | comment | added | José Figueroa-O'Farrill | What does "pherical" mean? | |
| Aug 26, 2010 at 14:18 | history | asked | Ludo Marquis | CC BY-SA 2.5 |