Timeline for A local isometric immersion from $\mathbb H^{n}$ into $\mathbb R^{2n-1}$
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| S Oct 6, 2021 at 6:00 | history | bounty ended | CommunityBot | ||
| S Oct 6, 2021 at 6:00 | history | notice removed | CommunityBot | ||
| Oct 4, 2021 at 20:32 | comment | added | Christos | @MattF. Choose non-zero real numbers $a_i,\; 1\leq i\leq n-1$, such that $\sum_i a_i^2=1$. The immersion $$f\colon D:=\{(x_1,\dots,x_n)\in\mathbb R^n\;|\; x_n<0\}\to \mathbb R^{2n-1},\; (x_1,\dots,x_n)\mapsto (y_1,\dots,y_{2n-1})$$ defined by $$y_{2i-1}=a_i e^{x_n}\cos(x_i/a_i), y_{2i}=a_i e^{x_n}\sin(x_i/a_i),\; 1\leq i\leq n-1,\; y_{2n-1}=\int_0^{x_n} \sqrt{1-e^{2u}} du$$ induces on $D$ a non complete metric of constant negative curvature. | |
| Oct 4, 2021 at 15:32 | comment | added | user44143 | The integral in the question is badly stated, since it has $z_n$ both as a limit and as the argument of integration, and since it is not obvious what lower limit makes it converge. I'd prefer giving the result of the integral, using $a\sqrt{n-1}$ as the lower limit: $$ - \sqrt {u} + \log (\frac {1 + \sqrt {u}} {\sqrt {1 - u}})\ \text {with}\ u = 1 - \frac {(n - 1) a^2} {x_n^2} $$ | |
| Oct 3, 2021 at 21:44 | vote | accept | Zaragosa | ||
| Oct 3, 2021 at 15:13 | answer | added | Christos | timeline score: 4 | |
| Oct 2, 2021 at 20:44 | comment | added | Zaragosa | @RobertBryant Thank you very much for your answer, you can do the detail I ask you please. This problem has driven me crazy for many days now, I am making several errors in the calculation of the sectional curvature by means of the Christoffel symbols. I understand what you say but when I do the detail of the (not) completeness and the rest I get a little confused. If I got to the direct calculation you mention, at first I thought that that might be enough for the sectional curvature to be $-1/a^2$ but I think that's not the case. | |
| Oct 2, 2021 at 9:39 | comment | added | Robert Bryant | I'm not sure what you are asking. By direct calculation, $$\phi^*({dx_1}^2 + \cdots + {dx_{2n-1}}^2) = (a/z_n)^2\,\bigl({dz_1}^2 + \cdots + {dz_{n}}^2\bigr)$$ when $(z_n/a)^2>(n{-}1)$, so this answers Q1 and Q2 once you know that the formula on the RHS is a metric of sectional curvature $K=-1/a^2$. As for Q3, the metric on the RHS is defined for $z_n>0$ and is complete on that half-space, as it is homogeneous (translation in the first $n{-}1$ coordinates and scaling in $z_i$). Clearly, it is not complete on the half-space $z_n>a\sqrt{n{-}1}$, so the image of $\phi$ cannot be complete either. | |
| S Sep 28, 2021 at 4:10 | history | bounty started | Zaragosa | ||
| S Sep 28, 2021 at 4:10 | history | notice added | Zaragosa | Authoritative reference needed | |
| Sep 28, 2021 at 0:22 | history | edited | Zaragosa | CC BY-SA 4.0 |
added 39 characters in body
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| Sep 25, 2021 at 6:52 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor grammar fixes
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| Sep 24, 2021 at 23:55 | history | asked | Zaragosa | CC BY-SA 4.0 |