Timeline for Abelian variety with CM defined over real numbers
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| S Aug 24, 2021 at 3:58 | vote | accept | Sophie | ||
| S Aug 24, 2021 at 3:58 | vote | accept | Sophie | ||
| S Aug 24, 2021 at 3:58 | |||||
| Aug 24, 2021 at 3:57 | vote | accept | Sophie | ||
| S Aug 24, 2021 at 3:58 | |||||
| Aug 24, 2021 at 3:56 | comment | added | Sophie | Thanks so much for your comments and explanations. I am actually mainly interested in the case of elliptic curves and I was hoping to find an explicit elliptic curve but by the answers this is unfortunately not possible. | |
| Aug 18, 2021 at 4:14 | answer | added | Kapil | timeline score: 5 | |
| Aug 17, 2021 at 21:29 | history | became hot network question | |||
| Aug 17, 2021 at 19:43 | comment | added | alpoge | Alternatively (lemme know if I’m saying something stupid in this argument since I’m a bit surprised about it) if you take the connected component of the identity in the real points of A you get (S^1)^k with k at most n, and K acts on the singular cohomology of that with rational coefficients, providing an embedding (since 1 goes to the identity) K \to M_k(\Q), so a primitive element of K satisfies a degree k polynomial, contradiction. | |
| Aug 17, 2021 at 17:56 | comment | added | alpoge | Nope, K would then preserve Lie(A/\R), but the action of K on Lie(A/\C) is the direct sum of characters in the CM type of A, and now let \chi be one such character and use Minkowski to produce an element x of \o_K with \chi(x) having huge imaginary part and very small other embeddings —- thus if \chi appears then so must its conjugate (otherwise the trace of the action of x on Lie(A/\C) can’t be real), contradiction. | |
| Aug 17, 2021 at 16:57 | answer | added | Will Sawin | timeline score: 10 | |
| Aug 17, 2021 at 16:20 | comment | added | abx | So in fact the answer is negative for $n=1$, since $\operatorname{End}_{\mathbb{R}} (A)\otimes \mathbb{Q}$ must act faithfully on the 1-dimensional $\mathbb{R}$-vector space $H^0(A,\Omega ^1_A)$. | |
| Aug 17, 2021 at 16:19 | comment | added | abx | @Laurent Moret-Bailly: Oops, you are right of course! I overlooked the "$\mathbb{R}$" in $\operatorname{End}_{\mathbb{R}} $. | |
| Aug 17, 2021 at 15:18 | comment | added | Laurent Moret-Bailly | @abx: Wait, in this case $\mathrm{End}_\mathbb{R}(E)=\mathbb{Z}$, right? To get the full CM for $E$ you need the ground field to contain $i$. | |
| Aug 17, 2021 at 15:11 | comment | added | abx | Take $A=E^n$, where $E$ is an elliptic curve with CM defined over $\mathbb{Q}$ — e.g. $\ y^2=x^4-1$. Then $\operatorname{End}(A)\otimes \mathbb{Q}=M_n(L) $, with $L=\operatorname{End}(E)\otimes \mathbb{Q}$. Any extension of $L$ of degree $n$ injects into $M_n(L)$. | |
| Aug 17, 2021 at 13:29 | history | asked | Sophie | CC BY-SA 4.0 |