Timeline for Riemannian submersions from complex hyperbolic space into the hyperbolic space
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Aug 10, 2017 at 15:32 | comment | added | Marcos Petrúcio Cavalcante | @RobertBryant: Thank you very much!! | |
| Aug 10, 2017 at 15:28 | vote | accept | Marcos Petrúcio Cavalcante | ||
| Aug 9, 2017 at 12:14 | comment | added | Igor Belegradek | Thank you, this helped me understand the situation better. I was not familiar with this way of thinking about $\mathbb{CH}^n$. | |
| Aug 9, 2017 at 8:55 | comment | added | Robert Bryant | @IgorBelegradek: In the new version, which provides an example for all $n$ (and simplifies the $n=2$ example), the kernel of $\rho_n$ is simply an abelian group of dimension $n$. (The structure equations for the kernel are got by setting all of the $\beta_i$ to zero in the structure equations for $G_n$ itself. This leaves $\mathrm{d}\alpha_i=0$ for $1\le i\le n$, which are the structure equations of an abelian group.) The fibers of $\pi_n$ are copies of $\mathbb{R}^n$ endowed with a flat, complete metric (but, of course, they are not totally geodesic in $\mathbb{CH}^n$). $G_n$ has no center. | |
| Aug 9, 2017 at 1:27 | comment | added | Igor Belegradek | While I understand the idea of your construction, I still do not see how to answer the question in my first comment, i.e., what is the kernel of $\rho$? | |
| Aug 9, 2017 at 0:55 | comment | added | Robert Bryant | @IgorBelegradek: I hope that the revised and generalized answer is helpful. Let me know if you still have questions. | |
| Aug 8, 2017 at 13:48 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Fixed an error in the formula for O'Neill's T
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| Aug 8, 2017 at 13:22 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Simplified the example and extended it to all n and computed the A and T tensors
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| Aug 6, 2017 at 11:37 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Added an explicit description of the Riemannian submersion
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| Aug 6, 2017 at 2:44 | comment | added | Igor Belegradek | I gather $G$ and $\rho(G)$ must be the stabilizers of points at infinity in $\mathrm{Isom}(\mathbb{CH}^2))$ and $\mathrm{Isom}(\mathbb{H}^2)$, respectively. I think the former is a semidirect product of a Heisenberg group with $\mathbb R$, where $\mathbb R$ permutes horospheres. Now I have trouble visualising the kernel of $\rho$. The center of the Heisenber group must lie in the kernel, while the above mentioned copy of $\mathbb R$ should not. I do not think such $\rho $ exist. What am I missing? Could you describe $\rho$? | |
| Aug 5, 2017 at 13:04 | history | edited | Robert Bryant | CC BY-SA 3.0 |
Clarified the relation between the complex hyperbolic line and the real hyperbolic 2-plane.
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| Aug 4, 2017 at 22:37 | history | answered | Robert Bryant | CC BY-SA 3.0 |