Timeline for degenerating surface
Current License: CC BY-SA 3.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 4, 2011 at 1:00 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
added 9 characters in body; Post Made Community Wiki
|
| Nov 23, 2011 at 14:01 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
added 17 characters in body
|
| Nov 23, 2011 at 0:16 | comment | added | Vitali Kapovitch | @Robert Paul edited the question as it originally did seem to ask if the curvature is always bounded above - I got caught on that too (see my original answer to 2) above). But he made clear what he wants in the comments. You should claim the bounty as you did answer the main question and I don't think it can be split anyway. | |
| Nov 22, 2011 at 23:57 | comment | added | Robert Bryant | @Vitali: Thanks. Actually, I didn't realize that this is what Paul was asking in 1). I thought that he was asking whether you could prove an upper bound for the curvature, not whether there was some example with an upper bound. Now that I read over the question again, I realize that your interpretation makes more sense than mine. I did know that the curvature was nonpositive everywhere, since, after all, these immersions are minimal surfaces. I guess I should claim the reward, or should we split it. | |
| Nov 22, 2011 at 22:32 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
edited body
|
| Nov 22, 2011 at 22:23 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
added 274 characters in body
|
| Nov 22, 2011 at 22:19 | comment | added | Vitali Kapovitch | @Robert Bryant This is a really nice example!! It was pretty clear to me that a $C^0$ approximation like this near 0 is possible but I wasn't sure if it can still happen with $C^2$ convergence. Note that your example also answers the original question by the OP because $ds_t^2$ has nonpositive sectional curvature for all $t>0$. | |
| Nov 22, 2011 at 22:15 | comment | added | Paul | Thank you Vitali and Robert for your proof of the fact an immersion is possible if and only if $k$ is odd. | |
| Nov 22, 2011 at 21:25 | comment | added | Robert Bryant | The case $k=2m+1$ does occur. Consider the (complex) $1$-parameter family of maps $u_t:\mathbb{C}\to\mathbb{R}^3$ given by $$ u_t(z) = \bigl(Re(z^{2m+1}-(2m{+}1)t^2z),\ Im(z^{2m+1}+(2m{+}1)t^2z),\ \tfrac{4m+2}{m+1} Re(t z^{m+1})\ \bigr). $$ These smooth maps converge smoothly to $u_0$ as $t\to0$ and $u_t$ induces the metric $$ ds_t^2 = (2m{+}1)^2\bigl(|z|^{2m}+|t|^2\bigr)^2 |dz|^2. $$ Thus, $u_t$ is an immersion for $t\not=0$, while $u_0(z) = \bigl(Re(z^{2m+1}),\ Im(z^{2m+1}),\ 0\ \bigr)$. | |
| Nov 22, 2011 at 17:54 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
added 3 characters in body
|
| Nov 22, 2011 at 17:52 | comment | added | Vitali Kapovitch | got you. then please disregard my original answer about 2). Note however that my edit above shows that when $k$ is even you can not at all extend the immersion $z\to z^k$ near the boundary $S^1$ to an immersion $D^2\to \mathbb R^3$. this does take care of both 1) and 2) in that case. I don't really want to think about possible geometric restrictions when $k$ is odd until I'm certain it's actually possible. | |
| Nov 22, 2011 at 17:44 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
added 77 characters in body
|
| Nov 22, 2011 at 17:28 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
edited body
|
| Nov 22, 2011 at 17:22 | history | edited | Vitali Kapovitch | CC BY-SA 3.0 |
added 1761 characters in body; deleted 5 characters in body
|
| Nov 21, 2011 at 14:37 | history | answered | Vitali Kapovitch | CC BY-SA 3.0 |