Timeline for Map of the Klein quartic from $CP^2$ to $R^3$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Oct 2, 2022 at 2:35 | comment | added | David E Speyer | My new current strategy is to take the $4$ dimensional representation $V \oplus \epsilon$ of $S_4$, where $V$ is the $O(3)$ symmetries of the tetrahedron and $\epsilon$ is the sign character. This gives a map $S_4 \to SO(4)$ and thus an action of $S_4$ on the sphere $S^3$. Topologicially, I think I can visualize embedding a genus $3$ surface into $S^3$ so that the $S_4$ action restricts to it correctly. So now I am looking for functions which equivariantly map $Q$ to this $S^3$. (Then I'll use stereographic projection to go to $\mathbb{R}^3$.) | |
| Oct 2, 2022 at 2:31 | comment | added | David E Speyer | To see that the second one can't work, notice that any $4$-cycle $\sigma$ in $S_4$ acts by a $90^{\circ}$ rotation about a line. So $\sigma$ and $\sigma^2$ have the same fixed points in $\mathbb{R}^3$ (namely, that rotational axis). But $\sigma$ fixes no points of $Q$ and $\sigma^2$ fixes four of them (see Elkies paper). | |
| Oct 2, 2022 at 2:30 | comment | added | David E Speyer | To see that the first one can't work, note that the action of $S_4$ on $Q$ is orientation preserving (since it is holomorphic). But, also, any linear action would preserve the direction of the outward normal to $Q$. So $S_4$ must act in an orientation preserving way on $\mathbb{R}^3$, and must land in $SO(3)$. | |
| Oct 2, 2022 at 2:28 | comment | added | David E Speyer | Every once in a while I come back to this question and think about it. Here is today's observation: There is no $S_4$-equivariant embedding of $Q$ into $\mathbb{R}^3$, with $S_4$ acting linearly on $\mathbb{R}^3$. Proof: There are two faithful representations of $S_4$ on $\mathbb{R}^3$ -- subgroup of $O(3)$ preserving the tetrahedron, and the subgroup of $SO(3)$ preserving the cube. | |
| Oct 26, 2021 at 17:55 | comment | added | David E Speyer | Elkies library.msri.org/books/Book35/files/elkies.pdf construct what he calls the $S_4$-model of the Klein quartic: Coordinates in which the equation of the quartic is $x^4+y^4+z^4+3 \tfrac{-1+\sqrt{-7}}{2} (x^2 y^2+x^2 z^2 + y^2 z^2)$ and the $S_4$ action is by signed permutation matrices (modulo $\pm \mathrm{Id}$.) In these coordinates, $V_3$ is spanned by $(\mathrm{Re}(x \bar{y}), \mathrm{Re}(y \bar{z}), \mathrm{Re}(z \bar{x}))$ and $W_3$ is spanned by $(\mathrm{Im}(x \bar{y}), \mathrm{Im}(y \bar{z}), \mathrm{Im}(z \bar{x}))$. | |
| Oct 26, 2021 at 13:17 | comment | added | David E Speyer | To be more vague, let $V$ be the span of $(x,y,z)$ and let $\bar{V}$ be the span of $(\bar{x}, \bar{y}, \bar{z})$. If we restrict $V \otimes \bar{V}$ to the copy of $S_4$ contained in $PSL_2(\mathbb{F}_7)$, it decomposes as $1 \oplus V_3 \oplus W_3 \oplus X_2$. Here $1$ is the trivial rep spanned by $x \bar{x} + y \bar{y} + z \bar{z}$, the action on $V_3$ is by the rotations and reflections of a tetrahedron and the action on $W_3$ is by the rotations of a cube. Taking a basis for one of the three dimensional reps and dividing it by $x \bar{x} + y \bar{y} + z \bar{z}$ might work. | |
| Nov 15, 2020 at 19:57 | history | edited | David Lehavi | CC BY-SA 4.0 |
fixed broken link
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| Aug 29, 2015 at 13:59 | history | edited | Henry Segerman | CC BY-SA 3.0 |
Matching notation with our new usage outside of Math
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| Mar 21, 2015 at 19:08 | comment | added | Henry Segerman | Will Sawin: I think polynomials invariant by A_4 would map twelve points of the surface to one - we could perhaps use equivariant polynomials? | |
| Mar 17, 2015 at 18:00 | comment | added | Will Sawin | The obvious functions to try are polynomials in $x, \overline{x}, y, \overline{y}, z, \overline{z}$ divided by powers of $\sqrt{ x \overline{x} + y \overline{y} + z \overline{z}}$. You could try the first few polynomials that are invariant by $A_4$ and see if they give you an embedding. | |
| Mar 17, 2015 at 1:18 | history | edited | Sam Nead | CC BY-SA 3.0 |
edited to make it shorter, and changed the symbol from \chi to \mathcal{X}
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| Mar 17, 2015 at 0:30 | comment | added | Henry Segerman | Yes, thank you Robert! It is a great resource. | |
| Mar 16, 2015 at 23:55 | comment | added | Robert Bryant | Yes, I knew that the Eight-fold way was not an algebraic model, but I thought that you would find it useful or at least interesting if you didn't know it already. | |
| Mar 16, 2015 at 21:05 | comment | added | Henry Segerman | We don't have the physical book, but it is online: (library.msri.org/books/Book35/contents.html). We have looked at Elkies' article somewhat closely, but couldn't work out a suitable map. In trying to figure out how Helaman Ferguson made the sculpture we also looked at his article with Claire Ferguson. In this he says that he: "carved it in a qualitative free form process known as direct carving, paying attention to the combinatorics and topology but not rigid or measured geometry." So the Eight-fold way" is also a topological model. | |
| Mar 16, 2015 at 20:51 | comment | added | Robert Bryant | If you can get your hands on it, you should have a look at "The Eight-fold Way: The Beauty of Klein's Quartic Curve", edited by Silvio Levy, which has some very beautiful pictures of the Helaman Ferguson sculpture, as well as solid discussions of the mathematics. | |
| Mar 16, 2015 at 20:02 | history | asked | Henry Segerman | CC BY-SA 3.0 |