Timeline for Automorphisms of Generic Abelian Varieties
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 4, 2014 at 18:52 | answer | added | jacob | timeline score: 4 | |
| Apr 25, 2013 at 20:43 | comment | added | JHM | @ulrich: i was concerned that the product polarization needs to be nondegenerate. If the elliptic curve $E$ is totally isotropic (w.r.t. the polarization of the ambient ppav in which it lies), then the $\mathbb{Z}/2$ automorphism of $E$ may not be an automorphism of the ambient ppav since it may not preserve the polarization. For instance, w.r.t. standard symplectic basis $e_1, e_2, f_1, f_2$, the order two element $diag(1,1,-1,-1)$ is not symplectic. Otherwise I had (stupidly) not considered that split lattices have (barely) nontrivial automorphism groups. | |
| Apr 25, 2013 at 4:46 | comment | added | naf | @J. Martel: There is no problem with the polarisation at all; one just takes the product polarisation! | |
| Apr 24, 2013 at 23:05 | comment | added | JHM | @Lennart: so ulrich has a point. If a euclidean lattice $\Lambda$ splits orthogonally into two lattices $\Lambda' \oplus \Lambda''$, then we at least get a nontrivial copy of $\mathbb{Z}/2 \times \mathbb{Z}/2$ in the usual automorphism group. But there's an issue of whether or not this $\mathbb{Z}/2 \times \mathbb{Z}/2$ subgroup preserves the polarization, i.e. is the subgroup symplectic? This requires that $\Lambda', \Lambda''$ also be orthogonal w.r.t. the symplectic form and each be unimodular and symplectic. | |
| Apr 24, 2013 at 20:12 | comment | added | Lennart Meier | @Jeremy: I mean always automorphisms respecting the group structure; for elliptic curves this is the same as automorphisms fixing the zero point. @J. Martel: If the polarization is also the product, then the automorphism group of the product should be at least the product of the automorphism groups, shouldnt it? | |
| Apr 24, 2013 at 19:29 | comment | added | JHM | @ulrich: splitting a ppav $X$ as a product with an elliptic curve $E$ does not guarantee the automorphism group is nontrivial. Moreover (and possibly irrelevant) i think it is possible for a fixed point set to intersect both the period locus and complement nontrivially. I.e. the fixed point sets corresponding to a split ppav could `leak' into the period locus. | |
| Apr 24, 2013 at 11:07 | comment | added | Jérémy Blanc | I have a stupid question: what do you mean by automorphisms?? Because for me, an elliptic curve over an algebraically closed always have an infinite group of automorphisms (take translations). So is your question about the quotient of the automorphism group by the group of translations? Or you mean the automorphisms which preserve the group structure of the variety? Sorry if I am pointing out something that was clear for everybody anyway. | |
| Apr 24, 2013 at 5:54 | comment | added | naf | I would guess that for your first question if $g>1$ then the ppavs that are products of an elliptic curve and a ppav of dimension $g-1$ would give a maximal codimensional component of the locus of ppavs with extra automorphisms (and this is probably the unique component of this dimension if $g>2$). | |
| Apr 24, 2013 at 2:14 | comment | added | JHM | i believe i've said something false. Namely, a `very' $\mathbb{R}$-reducible finite subgroup does not guarantee a large fixed point set. In the symplectic case, if the form $\omega$ is anisotropic on the irreducible subspaces, then the centralizer will be a compact subgroup of $Sp_{2g} \mathbb{R}$. E.g. the lattice formed by taking $g$ $\omega$-orthogonal and $\langle, \rangle$-orthogonal copies of the unimodular hexagonal lattice supported on $g$ symplectic 2-planes in $\mathbb{R}^{2g}, \omega$ has automorphism group containing a direct product $\mathbb{Z}/6 ^g$ with fixed point set a point. | |
| Apr 23, 2013 at 22:07 | comment | added | JHM | Moreover if $H$ is a finite subgroup of $Sp_{2g}\mathbb{R}$ acting irreducibly on $(\mathbb{R}^{2g}, \omega)$ then the corresponding fixed point set in $\mathbb{h}_g$ will be a point. We'll find that finite subgroups which are very $\mathbb{R}$-reducible yield larger dimensional fixed point sets. | |
| Apr 23, 2013 at 21:50 | comment | added | JHM | For instance, these finite subgroups (all of which are cyclic) are computed in Connelly and Kozniewski's "Finiteness properties of classifying spaces for $\Gamma$-actions", wherein they refer to a paper by K.Ueno "On fibre spaces of normally polarized abelian varieties of dimension 2", I, J. Fac. Sci. Univ. Tokyo 18 (1971) 37-95. I was unable to locate a copy of this Ueno paper. | |
| Apr 23, 2013 at 21:45 | comment | added | JHM | The integral symplectic group $Sp_{2g}\mathbb{Z}$ acts biholomorphically on the Siegel upper half space $\mathfrak{h}_g$. So for $H$ a finite subgroup of $Sp_{2g}\mathbb{Z}$ the fixed point sets $\mathfrak{h}_g^H$ are totally geodesic contractible holomorphic subvarieties, hence all have even dimension. Explicitly we need to determine the dimension of the centralizer $Z(H)$ of $H$ in $Sp_{2g}\mathbb{R}$. For genus $g=2$, we can exhibit explicit finite subgroups whose fixed point sets are 4-dimensional subspaces of the 6-dimensional $\mathfrak{h}_2$. | |
| Apr 23, 2013 at 21:00 | history | asked | Lennart Meier | CC BY-SA 3.0 |