Timeline for Linear symmetric spaces are spaces with ''orthogonal complements''?
Current License: CC BY-SA 3.0
8 events
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| Apr 3, 2012 at 23:59 | comment | added | Misha | @jmart: Cartan involutions are the antipodal maps! 2-nd Satake's paper actually goes through the classification of symmetric spaces which admit rational Cartan involutions; of course, you would have to read the first paper too. If you are in the US, you should be able to get the 1st paper through the inter-library loan. You can also ask somebody more knowledgable in algebraic groups over number fields on MO, like Mikhail Borovoi or Jim Humphries, for instance. I am a geometer and know geometry of buildings and symmetric spaces, but my knowledge of algebraic groups is quite basic. | |
| Apr 2, 2012 at 18:08 | vote | accept | JHM | ||
| Mar 31, 2012 at 4:08 | comment | added | Misha | @jmart: See the update in my answer. | |
| Mar 31, 2012 at 4:08 | history | edited | Misha | CC BY-SA 3.0 |
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| Mar 31, 2012 at 3:58 | history | edited | Misha | CC BY-SA 3.0 |
update
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| Mar 30, 2012 at 23:56 | comment | added | JHM | Moreover, in the Borel-Serre compactifications of such symmetric spaces (really, of the associated locally symmetric space $U_g \backslash Sp_{2g} / Sp_{2g}(\mathbb{Z})$) there is a large role played by flags of totally isotropic subspaces. I wanted to know to which extent non-linearity is the cause for this non-canonicity. NOTE: In above comment, I should have said "the complement $W^o$ affords us an embedding of $GL(W)$ into $SL(V)". | |
| Mar 30, 2012 at 23:52 | comment | added | JHM | In a euclidean vector space $V$, for given subspace $W$ we can identify a canonical (coordinate free) complement $W^o$. This complement affords us an embedding, say of $GL(W) \times GL(W^o) \to SL(V)$. Given a nondegenerate symplectic vector space $(V, \omega)$ we have no obvious (as far as I can tell) means of assigning canonically (ie. coordinate free-ly) an analagous complement for a given totally isotropic subspace $W$. That is, there is no coordinate free means of embedding $GL(W)$ into $Sp(V)$. Now supposedly $U_g(\mathbb{C}) \backslash Sp_{2g}(\mathbb{R})$ is non-linear symmetric space. | |
| Mar 30, 2012 at 13:22 | history | answered | Misha | CC BY-SA 3.0 |