Timeline for Isometric immersion of $\mathbb H^2$ into $\mathbb R^\infty$ built by Bieberbach
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 2, 2023 at 22:05 | comment | added | Zaragosa | You are right now I am editing it so that everything is understood, except that due to a great abuse of notation I put $\mathbb R^\infty$. | |
| Jan 2, 2023 at 22:04 | history | edited | Zaragosa | CC BY-SA 4.0 |
added 7 characters in body
|
| Jan 2, 2023 at 20:09 | comment | added | Terry Tao | Technically, this is an isometric embedding into $\ell^2$ rather than ${\mathbb R}^\infty$ (the $\ell^2$ metric is not well-defined on the latter space). | |
| Jan 2, 2023 at 18:00 | review | Close votes | |||
| Jan 7, 2023 at 3:09 | |||||
| Jan 2, 2023 at 16:35 | vote | accept | Zaragosa | ||
| Jan 2, 2023 at 16:34 | answer | added | Zaragosa | timeline score: 5 | |
| Jan 2, 2023 at 16:33 | history | edited | Zaragosa | CC BY-SA 4.0 |
deleted 641 characters in body
|
| Oct 28, 2021 at 14:19 | history | edited | Zaragosa | CC BY-SA 4.0 |
added 124 characters in body
|
| Oct 28, 2021 at 1:07 | comment | added | Zaragosa | @MoisheKohan Thank you, actually, the metric of the Poincaré disk is $g_{D}=4\frac{dx^2+dy^2}{(1-(x^2+y^2))^2}$, but multiplying everything by a constant can fix this fact. A query I know that if the pullback $f^*g_{\mathbb R^2}=M g_{\mathbb H^2}$ then the image is a surface of curvature $-1/M$, it is easy to prove it by the Gauss' formula but in the $n-$dimensional case, for example, $f^*g_{\mathbb R^2}=M g_{\mathbb H^2}$ could be proved. | |
| Oct 28, 2021 at 0:52 | history | edited | Zaragosa | CC BY-SA 4.0 |
deleted 2 characters in body
|
| Oct 28, 2021 at 0:45 | comment | added | Moishe Kohan | Apart from the typo in the last line of your computation (should be just $(1-x^2-y^2)^{2}$), this is the standard formula for the hyperbolic metric on the unit disk. | |
| Oct 27, 2021 at 20:55 | history | edited | Zaragosa | CC BY-SA 4.0 |
deleted 16 characters in body
|
| Oct 27, 2021 at 19:58 | history | edited | Zaragosa | CC BY-SA 4.0 |
added 14 characters in body
|
| Oct 27, 2021 at 19:50 | history | edited | Zaragosa | CC BY-SA 4.0 |
added 669 characters in body
|
| Oct 27, 2021 at 19:06 | comment | added | Zaragosa | I'm going to write how I have attacked it, I don't like it but maybe someone else has a more pleasant solution. | |
| Oct 27, 2021 at 19:04 | comment | added | Zaragosa | @MoisheKohan I understand, I'm sorry. | |
| Oct 27, 2021 at 18:14 | comment | added | Moishe Kohan | Cross-posted at MSE: math.stackexchange.com/questions/4288146/…. As a general rule, you should avoid such simultaneous cross-posting in order to avoid duplication of efforts. Post your question on one site (say, MSE, wait a week or so and if there is no satisfactory answer, make a note of cross-posting and post on the other site (say, MO). | |
| Oct 27, 2021 at 18:11 | comment | added | Moishe Kohan | From the context, I think, ${\mathbb R}^\infty$ means $\ell_2$ and $dx_i$ denotes the linear functional on $\ell_2$ defined by $dx_i((x_1, x_2,....))=x_i$. Your notation $f^*g_{\mathbb R^\infty}=\displaystyle\sum_{m=1}^\infty dx_m^2$ is horrifying. What you mean is that you define the standard inner product on $\ell^2$, treat it as a Riemannian metric $g$ and take its pull-back by $f$. | |
| Oct 27, 2021 at 12:48 | history | edited | Ben McKay | CC BY-SA 4.0 |
edited title
|
| Oct 27, 2021 at 7:26 | comment | added | YCor | What do you denote by $\mathbf{R}^\infty$? I understand it consists of certain sequences, but with which condition and which norm/distance, and I don't know what you denote by $dx_m^2$. | |
| S Oct 26, 2021 at 23:07 | history | suggested | Steven Stadnicki | CC BY-SA 4.0 |
Changed the title to match the text
|
| Oct 26, 2021 at 23:04 | review | Suggested edits | |||
| S Oct 26, 2021 at 23:07 | |||||
| Oct 26, 2021 at 22:54 | history | asked | Zaragosa | CC BY-SA 4.0 |