Timeline for Putting algebraic curves in $\mathbb{R}^3$
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 25, 2017 at 10:31 | answer | added | Neil Strickland | timeline score: 1 | |
| Jan 14, 2017 at 18:33 | answer | added | Sam Nead | timeline score: 1 | |
| Jan 14, 2017 at 18:06 | answer | added | Robert Bryant | timeline score: 4 | |
| Jan 14, 2017 at 17:09 | comment | added | Robert Bryant | @Holonomia: I can't answer this for David. He'll have to do that. I just pointed out that this construction doesn't answer David's actual question. | |
| Jan 14, 2017 at 15:23 | comment | added | Holonomia | @Robert Bryant: David's picture is going to be inside some "box". So if the problem is the presence of boundary curves I think they do not disappear with technical assumtion of the immersion being closed. Thus, I am putting more stress on David's aim to draw pretty pictures than on the precise technical request. I'm misunderstanding David's motivation? | |
| Jan 14, 2017 at 15:03 | comment | added | Robert Bryant | @Holonomia: Whether David will be interested in this particular construction or not, I can't say. I'll let him answer that. I'm just pointing out that this construction won't satisfy his request that the immersion be closed. Instead, each of the $n$ missing points on the line at infinity will show up as circles that are boundaries of the surface in $\mathbb{R}^3$ that results from your construction. | |
| Jan 14, 2017 at 12:52 | comment | added | Holonomia | So you think that David is not interested into radial + stereographic projection of the Fermat's curves, isn't it? | |
| Jan 14, 2017 at 12:37 | comment | added | Robert Bryant | @Holonomia: Yes, that's an example of a curve for which the $S^3$ projection doesn't have pinch points (or even self-intersections). However, such pinch points will exist for most algebraic curves. Also, contrary to David's requirement, the image of your curve in $\mathbb{R}^3$ won't be closed. The image of the radial projection into $S^3\subset\mathbb{C}^2$ will have $n$ great circles in its closure that don't belong to the image (corresponding to the $n$ points on the line at infinity on the original curve), and these will survive as non-closed points after you stereographically project. | |
| Jan 14, 2017 at 11:24 | comment | added | Holonomia | @Robert Bryant: I was aware of that. Actually, I had in mind pictures of Fermat's curves $x^n + y^n - 1 = F(x,y) $. If I made not mistakes there are not pinch points when projecting radially to $S^3$. Pinch points must satisfy the system $F(x,y)=0 , F_x \cdot x + F_y \cdot y = 0$. Thus if David wants to drawn "some" pretty pictures perhaps this can be interesting for him. | |
| Jan 14, 2017 at 11:05 | comment | added | Robert Bryant | @Holonomia: For most algebraic curves $X\subset\mathbb{C}^2$, the union of the (complex) tangent lines to $X$ is all of $\mathbb{C}^2$, so radially projecting $X$ onto a $3$-sphere in $\mathbb{R}^4=\mathbb{C}^2$ will introduce some 'pinch' points in the image (generically speaking, the same number as the dual degree of the curve minus the number of points on the line at infinity). In general, to do what David wants to do, you'd at least have to choose a smooth nonvanishing section of the normal bundle of $X$ in $\mathbb{C}^2$, which is possible (since $X$ is not compact) but not easy. | |
| Jan 14, 2017 at 5:50 | comment | added | მამუკა ჯიბლაძე | Would not the "usual" way satisfy you? I mean, putting Riemann surfaces in ${\mathbb C}\times{\mathbb R}$ by, say, $(f(x+iy),|g(x+iy)|)$ where $(f,g)$ is some (local) parametrization of $X$? | |
| Jan 13, 2017 at 20:51 | comment | added | Holonomia | Are you interested in the case that $\phi$ is the defined just in $\mathbb{C}^2 \setminus \{ 0 \}$ ? For example, taking first a radial projection from $\mathbb{R}^4$ to $S^3 \subset \mathbb{R}^4$ then stereographic projection to some tangente space. In any case, I do not know if the pictures you get doing so are going to be pretty. | |
| Jan 13, 2017 at 18:22 | comment | added | Margaret Friedland | Not quite what you are asking, but related questions were studied in the context of mininal surfaces. One result says that under certain assumptions on a region $D \subset \mathbb{R}^n, \ n \geq 3$ every bordered Riemann surface $M$ admits a continuous map to $\overline{D}$ which is is a conformal full complete proper minimal immersion on $M^\circ$, in MR3407187 Alarc\'on, A.Drinovec Drnov\v sek, B.; Forstneri\v c, F; L\'opez, F. J. Every bordered Riemann surface is a complete conformal minimal surface bounded by Jordan curves. Proc. Lond. Math. Soc. (3) 111 (2015), no. 4, 851–886 | |
| Jan 13, 2017 at 17:33 | comment | added | David E Speyer | Mikhalkin arxiv.org/abs/math/0010087 shows that, if $X$ is a Harnack curve, then $(z_1, z_2) \mapsto (\mathrm{Re}(z_1), \mathrm{Re}(z_2))$ covers a closed region in $\mathbb{R}^2$, mapping $2 \to 1$ onto the interior and $1 \to 1$ onto the boundary. The preimage of the boundary is $X \cap \mathbb{R}^2$, and this disconnects $X$. So I just need to write down a function which has different signs on the two components of $X \setminus \mathbb{R}^2$ and take that as the third coordinate of the map. Abstract arguments show such a function exists, but I haven't found an explicit formula. | |
| Jan 13, 2017 at 17:29 | comment | added | David E Speyer | @T.Amdeberhan I'm not sure what a serious test would mean. Here are minor things I've thought: Lines are easy. Conics are fine -- I can change coordinates to $z_1 z_2=1$ and then use $(z_1, z_2) \mapsto (\mathrm{Re}(z_1), \mathrm{Im}(z_1), |z_2|^2)$. (continued) | |
| Jan 13, 2017 at 17:29 | comment | added | Mohan | Absolutely. Just can't think of some intrinsic nice maps other than projections. | |
| Jan 13, 2017 at 17:27 | comment | added | David E Speyer | @Mohan This argument shows linear maps can't work. I agree with that, and have checked that the secant map does fill up in concrete cases. It says nothing about more interesting smooth maps. | |
| Jan 13, 2017 at 17:26 | comment | added | Mohan | I would suspect not, since the real dimension of $X$ is two, the secant variety fills up (it will be four dimensional in general) the Euclidean space, giving you no room to project to get an injective map. Of course, this does not mean it is impossible, but seems unlikely. | |
| Jan 13, 2017 at 17:21 | comment | added | T. Amdeberhan | Have you already tested some (nontrivial) examples? | |
| Jan 13, 2017 at 17:15 | history | asked | David E Speyer | CC BY-SA 3.0 |