Timeline for Some special complex tori
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Nov 17, 2013 at 22:23 | history | edited | JHM | CC BY-SA 3.0 |
corrected minor error
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| Nov 16, 2013 at 18:53 | history | edited | JHM | CC BY-SA 3.0 |
clarified and corrected.
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| Nov 16, 2013 at 17:16 | history | edited | JHM | CC BY-SA 3.0 |
expanded
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| Nov 16, 2013 at 16:35 | comment | added | Will Sawin | If you take an elliptic curve with lattice $ \mathbb Z[\sqrt{-D}]$, then a Hermitian form whose imaginary part is integral must have the norm $N z \overline{w} / \sqrt{D}$ for some integer $N$. A Hermitian form whose real part is intetgral must have the form $M z \overline{w}$ for some integer $M$. These are inconsistent. If the principal polarization is not equal to the form with integral real part, I don't see how your argument works. | |
| Nov 16, 2013 at 13:39 | comment | added | JHM |
@WillSawin, it seems i have misread complex torus' as meaning principally polarized' complex torus, i.e. having $j^2=-1$. However, I don't understand your first point that the ppav's i'm describing are isogenous to direct sums of elliptic curves with $Q(i)$ CM. Elliptic curves with CM (besides the `square' elliptic curve at $\tau=i$) are not isogenous to $Q(i)$ CM curves, i.e. do not have the standard square lattice as sublattice.
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| Nov 16, 2013 at 6:25 | comment | added | abx | Thanks to both of you, but I insist that I want only the real part of $H$ to be integral on the lattice. I expect that most of these complex tori are non algebraic. | |
| Nov 16, 2013 at 5:07 | comment | added | Will Sawin | Another way to say this is that you have a rational matrix $j$ which is the complex structure up to a scalar, so $j^2=-D$ for some $D$, and so you make the lattice tensored with $\mathbb Q$ a vector space over $\mathbb Q(j)=\mathbb Q(\sqrt{-D})$. Then all endomorphisms of this vector space will give endomorphisms of the abelian variety, up to multiplication by an integer to clear the denominator. However this relies on the form having both real and imaginary parts integral, rather than just real. | |
| Nov 16, 2013 at 5:05 | comment | added | Will Sawin | You have to be a bit careful. If you want both the real and imaginary part to literally be integers, not only is it a CM variety, it is isogenous to a direct sum of elliptic curves with CM by $\mathbb Q(i)$! This is because not only do you know that MT is a torus, you know "which" torus it is. If you just demand they be integers after scalling, you get a direct sum of elliptic curves with CM by $\mathbb Q(\sqrt{-D})$ for some $D$ depending on the scaling factor. This is because MT is a one-dimensional torus. | |
| Nov 16, 2013 at 3:56 | history | answered | JHM | CC BY-SA 3.0 |