Timeline for Jordan algebra and quaternionic projective space
Current License: CC BY-SA 3.0
11 events
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| May 22, 2012 at 19:13 | comment | added | Vít Tuček | The subtlety lies in the fact that not every matrix of rank one is a projector onto a one-dimensional subspace. One has to fix its eigenvalue to one in order to obtain a genuine projection. For rank one matrices it is equivalent to have trace equal to one and to have the only nonzero eigenvalue equal to one. Your statement that the quaternionic projective space is space of rank one elements is incorrect. Rather one should say that it is the (real) projectivization of such elements, i.e. you should consider two rank one elements equal if they differ by a nonzero real multiple. | |
| May 19, 2012 at 2:59 | vote | accept | Mirjana | ||
| May 19, 2012 at 2:50 | comment | added | Mirjana | One more question. I can understand why is the quaternionic projective space the space of elemenents of $J_{n}(H)$ of rank 1, because the quaternionic projective space is the space of 1-dimensional subspaces of $H^{n}$. But, why the space of elements of $J_{n}(H)$ of trace 1? | |
| May 17, 2012 at 11:53 | comment | added | Vít Tuček | Yes. I believe that $Sp(n)$ is the group of isometries of the quaternionic projective space and it is also the group of automorphisms of $J_n(\mathbb{H})$. But you should take my statements about quaternionic matters only as an slightly educated guess, since what I really know only concerns the octonionic case. | |
| May 17, 2012 at 9:33 | comment | added | Mirjana | Is the condition $\overline{A}^{T}=A$ here because these matrices preserve $H$-inner product? Is there any connection between $Sp(n)$ and and $J_{n}(H)$? | |
| May 16, 2012 at 12:08 | comment | added | Vít Tuček | You are right, the conditions $\overline{A}^t=A$ and being of rank one and trace one are sufficient to define the projective space and its metric. The Jordan multiplication lurks in the background, because - at least in the octonionic case - the group of isometries is the group of automorphisms of the Jordan algebra. | |
| May 16, 2012 at 12:05 | comment | added | Vít Tuček | I do not know of a nice direct way to connect Jordan algebra structure and the geometry of the projective space. In the octonionic case, the incidence relation is given by the cross product which can be expressed in terms of Jordan product and traces. (See arxiv.org/abs/0902.0431) | |
| May 16, 2012 at 12:00 | comment | added | Alexander Chervov | @robot Still I am not clear about the role of Jordan algebra structure. If you will omit this word and just write matrices A^t=A , norm Tr(A^2)... nothing changes.. Any way thank you for yours answer, it useful (at least for me). | |
| May 16, 2012 at 11:46 | history | edited | Vít Tuček | CC BY-SA 3.0 |
expanded ; added 272 characters in body
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| May 16, 2012 at 10:28 | comment | added | Alexander Chervov | How does the structure of the Jordan algebra comes into play ? Is the space of rank 1 closed with respect to Jordan's product ? I would be very thankful for further comments... For the long time I keep some papers on Jordan algs nearby me but always does not see key how to open them :) | |
| May 16, 2012 at 9:00 | history | answered | Vít Tuček | CC BY-SA 3.0 |