Timeline for Explicit construction of a (the?) dual symmetric space
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 19, 2022 at 13:49 | comment | added | Callum | @S.T. Having done a little more digging I am more convinced that there is a problem. According to "Embeddings from noncompact symmetric spaces to their noncompact duals" by Chen, Huang and Leung there is exactly one non-compact symmetric space in such a family (which is simply connected) but a finite number of compact ones (with one being the universal cover of the others). So the correct statement is that for a compact symmetric space there is a unique up to isometry non-compact dual, while for a non-compact space the compact dual is only unique up to covering and isometry. | |
| Sep 18, 2022 at 13:38 | comment | added | Callum | My only concern is that if we fix a simply connected symmetric space and its dual that the (not simply connected) symmetric spaces which have those as covers might not pair up neatly. This may be an unfounded concern but I don't see a way to prove that easily. Most sources I see simplify either to the linear case or the simply connected case so I haven't found a clear result in the more general picture | |
| Sep 18, 2022 at 11:46 | comment | added | S.T. | Moreover, in the linear case, we can indeed just consider the usual complexification of linear Lie groups as just viewing them as complex matrices. However, I would like to see something that works in the general case. Why do you think there may be symmetric spaces without duals? | |
| Sep 18, 2022 at 11:45 | comment | added | S.T. | Of course for an orthogonal symmetric Lie algebra $(\mathfrak{g}, \Theta)$, we can take its dual $(\mathfrak{g}^*, \Theta^*)$, take the unique simply-connected (real) Lie group $G^*$ with this Lie algebra, and then take $K^*$ to be the unique connected Lie subgroup with Lie algebra equal to $\ker(\Theta^* - 1)$. But why can $K$ be identified with $K^*$ (as isomorphic maybe?). They have the same Lie algebra, but why are they "basically the same stabilizer group"? | |
| Sep 16, 2022 at 12:32 | comment | added | Callum | If the definition uses the explicit construction of the object then it is unique up to the choices that the definition requires in that construction. In this case, that is the choice of orthogonal symmetric Lie algebra $(\mathfrak{g},\Theta)$. So this construction is unique up to a choice of Lie algebra: $\mathfrak{g}$ (all of which are isomorphic) and a matching involution on $\mathfrak{g}$: $\Theta$ (which are all conjugate under the action of the group). We also need a choice of $G^\mathbb{C}$ and so on but e.g. $K, G^*$ are defined by our other choices by the subgroups-subalgebras theorem | |
| Sep 16, 2022 at 12:04 | comment | added | LSpice | Why does constructing something make it unique? | |
| Sep 16, 2022 at 8:43 | history | answered | Callum | CC BY-SA 4.0 |