Timeline for Elliptic curves and connected components
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2015 at 14:18 | comment | added | Will Sawin | and this is essentially the same as the continuity argument that Steven sketched, which is also valid. | |
| Jan 24, 2015 at 12:26 | vote | accept | user21956 | ||
| Jan 24, 2015 at 8:13 | comment | added | Laurent Moret-Bailly | In Joe's argument, all you need to know is that $E(\mathbb{R})$ is a topological group; the connected component of the identity is then a subgroup. | |
| Jan 24, 2015 at 1:21 | comment | added | Steven Stadnicki | @JoeSilverman, David: Those are both excellent explanations - it's great to get both the group-theoretic and geometric versions; thank you! | |
| Jan 24, 2015 at 1:05 | comment | added | David Lampert | @StevenStadnicki A line that intersects the finite component must intersect it twice, so (by the geometric interpretation of the group law) if P and Q are on the infinite (identity) component then so are P+Q, -P. For E,P above (we're told that) P generates all of E(Q). | |
| Jan 23, 2015 at 23:09 | comment | added | Joe Silverman | @StevenStadnicki Not sure if this will satisfy you, but $E(\mathbb{R})$ is a compact real Lie group, so it's real-analytically isomorphic to $S^1\times\Phi$, where~$S^1$ is the circle group and $\Phi$ is a finite group. (For elliptic curves, of course, $\Phi$ has to have order 1 or 2.) Anyway, the connected component of $E(\mathbb{R})$ corresponds to the group $S^1$ in this identification. | |
| Jan 23, 2015 at 22:39 | comment | added | Steven Stadnicki | For my own understanding, why does the latter statement hold? At least at first glance, it's not clear to me that addition 'respects' components; is it just a continuity argument ($P+0$ is, so $P+(1/n)P$ is, so...)? | |
| Jan 23, 2015 at 21:33 | history | answered | Michael Stoll | CC BY-SA 3.0 |