Timeline for Isomorphism between Spin(3,2) and Sp(4, R)
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| May 7, 2023 at 21:15 | comment | added | paul garrett | @SeewooLee, yes, using quadratic extensions of the base field, and/or the unique quaternion division algebra of $p$-adic local fields, gives similar results. But/and instead of "signature", a slightly different classification is needed... | |
| May 7, 2023 at 18:00 | comment | added | Seewoo Lee | Hi, I hope you find this comment (for your 10 years ago answer). Are the exceptional isogenies also hold over nonarchimedean fields? It seems like most of the arguments of proofs does not actively use archimedeaness of $\mathbb{R}$ and $\mathbb{C}$, so we may replace $\mathbb{C}/\mathbb{R}$ with quadratic extensions of local fields $E/F$ - am I right? | |
| Jul 23, 2013 at 12:25 | comment | added | paul garrett | @FrancoisZiegler Aha! Thanks! I suspected something of that sort, since upon extending scalars it would have to be ok. Yes, a further section is called-for! :) | |
| Jul 23, 2013 at 4:30 | comment | added | Francois Ziegler | @paulgarrett: $SU(3,1)$ that you ask about maps to $SO^*(6)$ (a.k.a. $SO(3,\mathbb H)$), according to Helgason (§X.6.4) or Besse (Einstein Manifolds, p. 201). You may need to add a third section "Over $\mathbb H$"! | |
| Jul 17, 2013 at 1:10 | history | edited | paul garrett | CC BY-SA 3.0 |
added 2 characters in body
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| Jul 16, 2013 at 16:44 | history | edited | paul garrett | CC BY-SA 3.0 |
added 591 characters in body
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| Jul 16, 2013 at 10:57 | comment | added | The User | Great, that seems to be very practical! | |
| Jul 15, 2013 at 23:51 | comment | added | paul garrett | @YvesCornulier, thx for filling in! :) | |
| Jul 14, 2013 at 21:57 | comment | added | YCor | For completeness: Paul lets $Sp_4(R)$ act on the 16-dimensional matrix algebra $M_4(R)$ by conjugation. The usual bilinear form $tr(xy)$ is invariant. The Lie algebra $RI_4+sp_4(R)$ is an invariant subspace on which this form is nondegenerate ($sp_4(R)$ being the 10-dimensional Lie algebra), so its orthogonal is also nondegenerate, it is thus (16-(10+1))-dimensional, i.e. 5-dimensional. Explicit coordinates show the signature is (3,2). This defines a 2-to-1 map from $Sp_4(R)$ onto $SO(3,2)$. | |
| Jul 14, 2013 at 20:37 | history | answered | paul garrett | CC BY-SA 3.0 |