>> Kristin Lauter: Okay. So today we're very pleased to have Nils Bruin visiting us
from Simon Fraser University. And he'll speak on genus 2 curves with (4,4)-split
Jacobians, and which is joint with his student Kevin Doerksen and which one the best
poster prize at ANTS last year. Thanks, Nils.
>> Nils Bruin: Thank you, Kristen. So, yeah, first of all, I of course would like to thank
Kristen and Reinier Broker for inviting me and giving me this opportunity to talk about
this subject.
It's -- okay. Let's just first start to look at the setting a little bit, and then we can discuss
what we'll actually be doing in the talk.
Okay, so we'll be looking at -- well, K will be field characteristic different from 2. C in
effect will later on just -- oh, yeah -- no, different from 2 will be enough actually.
We'll consider a curve C of genus 2 over K and its Jacobian, and we say that a Jacobian J
is split over K, or in general in abelian variety is split if it's isogenous to a product of
elliptic curves. So if it admits a finite morphism that preserves the group structure to a
product of elliptic curves, so for abelian surfaces there's actually a nice result if you work
through it.
There's only very limited ways in which you can get split abelian surfaces, and this is the
construction if you actually have a Jacobian of a genus 2 curve that is split, then in fact
the curve C will have to admit morphisms to those elliptic curves themselves, each of
degree N.
Well, a priori, though, those maps are not well defined because you can change these
factors themselves by isogenies again. But you can actually pick them out in such a way
that the morphisms are of degree N, each -- your -- the isogeny from E1 to E2 times E2 to
the Jacobian can be expressed really quite easily in terms of pull backs along your
morphism.
And in fact your Jacobian is of a very specific form. You will have an isomorphism from
the n-torsion on the one curve to the n-torsion on the other curve. You will get a -- that
allows you to embed a -- well, basically the n-torsion diagonally into the product, and
that group will actually be exactly the kernel of this isogeny over here.
So if J is a Jacobian, then over C it's principally polarized, and then you see that your
delta N has to be maximally isotropic with respect to Weil-pairing, and that really means
that the N here is -- well, it's an isomorphism of group schemes, but in fact it would have
to be an NT isometry with respect to the Weil-pairing. So it should flip the sign on the
Weil-pairing.
So you can make these things using two elliptic curves with NT isometric n-torsion.
Okay. So the goal we set ourselves today is to try and find universal genus 2 curves with
a split (n,n) Jacobian, so sort of a model from which you can obtain any genus 2 curve
with an N and split Jacobian, it will not be a universal model in the strict sense of the
word because the moduli-space that we're looking at is not fine here.
But just as for Weierstrass models -- I mean, a Weierstrass model itself is not really a
universal model for an elliptic curve. But, nonetheless, you can get any elliptic curve
from specializing the parameters in a Weierstrass model. So in that sense, we will try and
create something that really looks a little bit like a universal model.
Okay. So the approach for N2, 3, and 5 is to really use those maps phi 1 and phi 2 that
you get from the genus 2 curve to the elliptic factors of the Jacobian. And Frey and Kani
have a very nice result where they show that those morphisms actually factor through the
hyper -- they can be made so that they -- so that you can get a cover from a P1 to a P1
that behaves nicely with respect to the hyperelliptic evolution.
So if you work it out, it turns out you can find a diagram like this. And now you can try
and describe your curve's C by trying to see, okay, well, if you have an elliptic cover of a
P1 here and you have a cover of P1s over here of degree N, it would have to have the
same degree as that map such that the fiber product of these two curves is actually of
genus 2. That's actually hard.
If you take a fiber product like this in general, if this N is big, then the genus of this curve
would be very high. So you have to have a lot of canceling ramification between these
morphisms, and that is what allows you to get a lot of conditions on what these pi 1s can
look like in terms of the morphism that you have over there.
And that gives you some equations that you can actually just solve it and get equations
for what these things can look like, and then you just take the fiber product to get your
genus 2 curve.
For NS4, in fact, it turns out there is some partial work already going back till 1887. This
was pointed out to me by Everett Howe, who would know the history of these things very
well. So, yeah, he actually sent me an e-mail. He saw the preprint and he said, oh, you
should look at this paper.
I have to say I haven't done so in detail yet, but it really looks like he says, okay, well, if
you have this hyperelliptic integral with certain symmetries to it, then you can do this
degree 4 transformation and express it in elliptic integrals, which ultimately boils down to
something like this. And in his settings, the degree 4 really corresponds to the 4-by-4
splitness.
Okay. So for historical point of view, this is very interesting. Of course, as much of this
work, it's over C. And we would be interested in doing this over algebraically non-closed
fields, and as we will see actually that is where a couple of the interesting things happen.
Okay. So our approach for (4,4)-split Jacobians will be the following. As you saw, what
you need to create an (n,n)-splits Jacobian is to take two elliptic curves with isomorphic
n-torsion, really NT isometric if you take the -- that pairing into account.
And if the 4-torsion of E1 and E2 is isomorphic, then that's certainly the case for the
2-torsion as well. So your (4,4)-isogeny will actually factor through a (2,2)-isogeny. So
if we understand (2,2)-split Jacobians, that should get us a long way towards
understanding the (4,4)-split ones as well.
So we will really try and use this diagram over here. So the steps that we will be using
here is first we want to give a description of the (2,2)-split Jacobians. That's completely
classical. It's easy. But, in addition, we want that this J2, which will have a (2,2)-isogeny
to this product here admits a second (2,2)-isogeny to this J4 here.
And if you work it out, they have to have trivially intersecting kernels. And that turns out
to be an important bit of information, because there's other configurations where that
doesn't happen. And those would not correspond to (4,4)-split Jacobians.
So in order to check that the kernels trivially intersect, you can just look at what the
isogeny [inaudible] that's the push forward along these phi 1 and phi 2 of your divisors.
The kernel of that will really just be the image of the 2-torsion over there because the
composition of the 2 really is multiplication by 2.
And that really does intersect -- oh, I guess -- that's completely misspelled on this slide.
This should be delta 4 modulo delta 2, which would be the kernel of this thing over here.
Sorry. So this is wrong in the slide. Of course, these two things do not intersect trivially
at all. They're identical.
Okay. So once we figured out what the condition is for that, then the next bit is to
actually identify J4, so that means that we actually have to take the image under a
(2,2)-isogeny. (2,2)-isogenies are also called Richelot isogenies for genus 2 Jacobians.
They're well understood, at least geometrically. And it turns out that in our case actually
to do this over an algebraically non -- non-algebraically closed field, we have to do a
little bit of extra work here. Because people have only considered Richelot isogenies
with very specific Galois structures on the kernel. There's still a little bit to do there, and
that was -- perhaps that was one of the things I found most satisfying out of this project,
that finally I saw how that worked out. It turns out to have a very easy answer.
Okay. So in a way this project is really just a little bit of housekeeping in the arithmetic
of genus 2 curves. It's just plugging together some results and completing things here
and there where they are not quite complete yet.
Okay. So the (2,2)-split Jacobians, I will describe them here in a very straightforward
way. They also turn out to fit in a diagram like this. So (2,2)-splits Jacobian has to -- the
genus 2 curve will have to be a double cover of two distinct elliptic curves, which it has
to be a bielliptic genus 2 curve, which really means that you have to have an extra
involution or you have to have a bunch of extra involutions or your C2 [inaudible] as this
thing is Galois here, so there's an involution over C2 over E1, which means that you get
an extra involution over here which factors down to the quotient by the hyperelliptic
involution, which gives you a further P1 down here.
So you see that those -- that these bielliptic genus 2 curves have to be V4 covers of a
projective line, and two of the factors that you get in there will be elliptic curves. That's
exactly what you're looking at. So if you do that just in general, if you prescribe yourself
one of those elliptic curves, we can without loss of generality, at least if we assume that
our characteristic is not 2 or 3, you can write this elliptic curve in this form over here.
Just a short Weierstrass form.
And the other elliptic curve here, if you work it out, has to be ramified at three of the
points where your E1 is ramified as well in order for this thing to be of genus 2, so that
means that this E2 will probably not be ramified over say infinity, but it has to be
ramified over another point, A here, and A can still allow yourself a twist.
And from that you can just compute the fiber product of these things to get your equation
for C2, so it's really just taking the cubic polynomial here and substitute something
quadratic inside.
Okay. So that gives us a good description of all -- so this is an example of what I'll
consider to be a universal model for a (2,2) -- for a genus 2 curve with a (2,2)-split
Jacobian. Here you have a bunch of parameters, U, B, and C, and you get additionally an
A and a D that you can specialize to values to actually get genus 2 curves with a
(2,2)-split Jacobian. And any such curve can actually be obtained from specializing these
particular models.
But as you see, there are many distinct values that you can plug in here that give you
isomorphic curves. So it's not really a universal family.
Okay. So the next step is to understand our (2,2)-isogenies a little bit better, Richelot
isogenies. So just a quick recap. I think most of the people in this audience already
know this, so I can skip over this pretty quickly.
If you write your elliptic curve as Y squared equals a sextic, then you can just write down
your 2-torsion in a Jacobian as differences of the ramification points of your C over the
canonical model. So these are the branch points of C over the X line, and clearly those
points -- well, if you take a divisor of this form and multiply it by 2, then you get the
divisor of a rational function.
So these are clearly two torsion points, and now you just have to work out that this
[inaudible].
And in addition we can actually see the Weil-pairing from this representation really quite
easily, too, if you take two of those two torsion points, you simply look at how big the
overlap is between their support and if the overlap at -- well, if they overlap at two points
that you're looking at the same point, so that should pair trivially; if they're overlapping at
exactly one point, then you're using three points in your support in total, so then it's, oh,
those points pair nontrivially. And if they have disjoint support, then they pair trivially
again. That turns out to be exactly the Weil-pairing on your genus 2 Jacobian.
>> Kristin Lauter: So, sorry, but aren't there more 2-torsion ones than just the ones like P
minus -- oh, sorry, that's not [inaudible] so it's NE, I and J?
>> Nils Bruin: Yeah, there's -- it's two 2s [inaudible] yeah, so it's six choose 2, so that's
exactly 515 plus 16, so that's -- so the -- yeah, okay. So yeah. Okay. It's not -- right. So
if you take I equal to J, then you get the origin and that's the only point that has multiple
representations in this [inaudible].
So, yeah, no, because this is genus 2, this is really all. If you go to a higher genus, higher
elliptic curves, yes, then you have to consider larger linear combinations of the
Weierstrass points as well.
Okay. So this gives us a good understanding of the 2-torsion. That's good. Because now
we can start looking at Richelot isogenies.
So if you have an isogeny between Jacobians or between a Jacobian and a product of
elliptic curves, so if you have an isogeny between abelian varieties that over an
algebraically closed field would be principally polarized, then you get the result that it's -in order for these things to actually be compatible with the Weil-pairing, the kernel of
your isogeny has to be maximally isotropic; that is to say, the kernel has to be such that
the Weil-pairing restricts to trivial pairing on that kernel and the kernel will actually be of
maximal size to still allow that property.
And for NS2, this means that, well, your kernel is going to be of order 4. And it has to
consist of points that pair-wise trivially, and that means that the distinct points have to
be -- has -- have to have disjoint support.
So in order to be able to write down a subgroup of the Jacobian like this, essentially what
you have to do is pair up the roots in three groups, three groups of two. So what it of
course points to is to write down a quadratic splitting of your polynomial -- your sextic
polynomial F in the following way.
Well, there's a leading coefficient to take care of. So it will be good if the leading
coefficient is a cube, because then you can just equally distribute the N coefficient over
the three quadratic factors.
But changing your polynomial out because the equation of your curve C is actually just Y
squared equals F of X, so changing the leading coefficient by a square doesn't hurt you at
all, so you can easily assume that the leading coefficient is indeed a cube by just
multiplying by the leading coefficient squared. And now you can just nicely distribute
that over the three quadratic factors that you want to use in your F.
Okay. So your isogeny is going to be defined over the base field. Well, a necessary
condition for that is that at least a kernel has to be Galois stable. So you should make
sure that your quadratic splitting that you're writing down here is Galois stable. And that
turns out to be enough.
Okay. So recapping, we are considering a genus 2 curve, Y squared F of X. And its
Jacobian will admit a Richelot isogeny over the base field if and only if we can group the
roots in three groups of two in a Galois stable way. So what can the Galois group -- so
the Galois action on the roots of course factors through a 6, of a degree 6 polynomial, so
there's some permutation action on the roots.
So what can the Galois group still do? Well, you can swap any of these so that if you see
two for each of the vectors, and you can swap -- you can permute each of the tuples in an
arbitrary way. So that's just S3 acting semidirectly or -- on the C2 cube. So you get a
wreath product of C2 and S3. That's the normal thing that you get when you consider
these things.
So this is a nice subgroup of S6. And that turns out to be the one-point stabilizer in this
set of quadratic splittings under S6.
So once you've restricted your action to this group to this one-point stabilizer, then the
quadratic splittings still have two orbits. You'll have the orbit containing quadratic
splittings that share a pair with these, and you have an orbit of quadratic splittings that do
not share a pair with these.
So you get two orbits and that means that if you want to look at two-point stabilizers, you
get two conjugacy classes of those, you get a -- if you work it out, you get a C2 times C -V4, so that's just by stabilizing one pair and allowing the different ways of grouping these
things in other pairs.
And there is a copy of S3 in there, which I'll denote as S3 twiddle. It's a, group that
[inaudible] on the 6 roots, it's actually the image of the normal S3 inside of S6 under the
outer automorphism of S6.
Okay. So those are two subgroups that we can consider. And really in order to have
multiple Richelot isogenies, you should have that Galois X via one of those subgroups on
your roots. So if your elliptic curve -- if your genus 2 curve admits multiple Richelot
isogenies, this polynomial F is actually going to have a not too big splitting field. This is
a group of order eight, and this is a group of order six. That's a much smaller splitting
field than you'd normally expect for a sextic.
Okay. So you can write down a little Galois theory diagram for this. On the bottom you
have your S6, so there you're putting no level structure on your genus 2 curve at all. If
you want to put full-level structure -- full 2-level structure on your Jacobian, then you
would have to, well, basically allow for no action at all on your roots.
You have to be able to distinguish all the 2-torsion points individually. You're not
allowed to permute them at all anymore. So in terms of moduli-spaces, this level would
of course bounce to the moduli-space of genus 2 Jacobians, genus 2 curves. And all the
way up there we would have our moduli-space of genus 2 curves with full two-level
structure.
And in between we have this level where we are fixing one Richelot isogeny, so that's the
analog of X nought 2 for elliptic curves, you're fixing one 2-isogeny on your elliptic
curve. Here a Richelot isogeny on your genus 2 Jacobian, and in between here we're
finding two distinct ways of putting a little bit finer but not full two-level structure on
your genus 2 Jacobian corresponding to prescribing different Richelot isogenies.
And you can tell them apart by what their kernels look like, because that's how we found
our orbits originally here. So if you go and look, you find that, okay, it's our curve here,
the original one, this group -- that's the stabilizer of just -- well, I mean, any conjugate of
this would be a stabilizer of something conjugate to that.
But if we just fix a notation, it would say fix this quadratic splitting, C2 times V4, also
stabilizes a splitting like this, 1, 2, 3, 5, 4, 6, so now you have swapped two things here.
So these would correspond to -- so if you find this is a Galois group for your polynomial
F, you would find that there's two Richelot isogenies at least -- in fact, there are three.
There's a third permutation here which would also be stable under this group. But they
would all share this vector here. So those kernels wouldn't be disjoint, or they wouldn't
intersect trivially.
And for your S3 twiddle here, there you find that there's also another quadratic splitting
that gets stabilized of this form, and here you indeed see there's no sharing of pairs here.
The kernels of the two isogenies here would in fact intersect trivially, so for our purpose
really what we're interested in is this S3 twiddle.
So now we know that our curve C2 has to be -- has to be (2,2)-splits to start with. Well,
we already determined what the model for that was. We could write those things as a
fiber product of a Weierstrass form and a double cover of the projective line ramified at
infinity in some other point A. And solving that, we can write down the model for your
curve C2 in this form.
And now if this -- if its Jacobian admits two Richelot isogenies with trivially intersecting
kernels, then this FX should really split over an S3 twiddle extension. So you know what
the Galois action is on this polynomial F that we have here. Well, we really obtained our
polynomial here from a cubic polynomial to start with, which -- so if this big polynomial
F here splits completely, then this small polynomial F is going to split as well. And its
Galois group would normally be S3. So it's -- if you work it out, it's actually the same
splitting field. So really what you need is that your big F here has the same splitting field
as your cubic polynomial small F.
So that's a very straightforward condition. So now you get your resolvents out and you
just write out what it means for your Galois groups to be like that. Then you can actually
find that you should choose your As and Ds of a specific form. Well, your Ds are only
defined up to squares, of course, because it doesn't matter if you change D by its square,
it's just doing a twist, and it turns out you find these relations here if we find -- if we put
A and D in these forms here, depending on the parameter S, then we will actually have
the right kind of splitting over polynomial F of X.
And it's completely non-coincidental that the polynomial A that you get here, if you look
at this, if you look at how you multiply points by two on Weierstrass models, this is the X
coordinate. So if you take a point S comma something and multiply by two on the
elliptic curve with that short Weierstrass model, then you get that. That's completely
expected. And as you see here, the D has some very recognizable components in it as
well. You get the discriminant of your cubic in there again.
Okay. So that means that we're basically there. We now have a model of our C2 because
we should have this extra Richelot isogeny, which we can ensure by having our As and
Ds of this form. And by writing our C2 like that in the first place, we've already made
sure that we have a (2,2)-split Jacobian. So now we know that our J4 will actually be the
image of the other Richelot isogeny on this thing.
Okay. So that's what we have here now. So given -- yes.
>> Kristin Lauter: [inaudible] notation [inaudible].
>> Nils Bruin: Yeah.
>> Kristin Lauter: [inaudible] quotients delta 2 and delta [inaudible].
>> Nils Bruin: Yeah, yeah, exactly. Yeah. So I can go quickly back to this diagram.
Here. Okay. So we have our elliptic curve, E1, that's sort of the data that we started
with. We now want to create the genus 2 curves that are (4,4)-isogeny to E1 times -well, whatever comes next to that. And it turns out that the things that come next to that
are the E2s of this form, you just fill in your As and Ds that you got from your
(2,2)-splitness, and now by taking you're a and D in the special way, you're making sure
that you can actually go this extra step further to actually get something that's (4,4)-split.
So we already see now as well that this E2 here has to have isomorphic 4-torsion to our
given curve E1, so we're actually recovering from this work that Rubin and Silverberg
already did where they actually wrote down explicit models of elliptic curves with
prescribed 2-, 3-, 4-, or 5-torsion.
So in this case we're recovering the result for prescribed 4-torsion, but we are actually
getting NT isometric 4-torsion here. So we twist by something. But it turns out, yeah,
okay, that's actually some extra result in order to flip the Weil-pairing on the 4-torsion of
an elliptic curve, it turns out that you have to twist by the discriminant of your elliptic
curve. That's unique to 4.
In general, you can't really just flip the Weil-pairing and stay geometrically identical.
You get different models. But okay.
Okay. So really we're just looking at our C2 here and we now have to find the image of
the second Richelot isogeny that we have on the Jacobian of that curve. So for that we
need a little bit of extra work of Richelot isogenies, because it turns out people always
took the easy case there.
Well, geometrically they've really just -- they've only looked at the geometric case.
That's essentially what I should say.
So if you have your elliptic curve C2 with a -- or your genus 2 curve with a given
quadratic splitting, we can write down a couple of quantities. We take this determinant D
here that you get by taking the 3-by-3 matrix that you get by putting the coefficients of
your quadratic factors in a matrix.
So that gives you this delta. And you define three other quadratic factors now by taking
appropriate determinants of your Fs and derivatives thereof. If you do the computation
over here, a priori you think, well, these are degree 1 polynomial, these are degree 2
polynomials, so if you take this determinant, you're going to get something of degree 3.
But the leading term actually cancels, so you get something off degree 2. That's a very
nice thing that happens.
I mean, just look at the leading coefficients here. Only putting in two things. Okay. And
we're creating these other effects by just permuting our IJK cyclically, so that gives us
three things. And now we can write down another genus 2 curve by just using these
quadratic factors instead. And we are allowing ourselves a twist in front here because,
well, who knows. So it's completely classical and that's really where Bolza already
started in his work too.
It's completely classical, that this is how you write down the genus 2 curve of which the
Jacobian is Richelot isogenous to what you started with in the geometric case.
So if your F1, F2, and F3 were defined over the base field to start with, then we don't
need this delta here. We can just take a one here. And then so the Jacobians of these
things are indeed 2 to isogenous.
So if we disregard the problems that we get from our base field, then we've written down
the right thing here. Well, what can the base field introduce that is still a twist. So in
general -- in general a genus 2 curve only admits quadratic twists. So that's what we're
getting here.
So we have to figure out what delta is. Because that's not given. The nice thing about
this whole construction here, and you see that from the fact that you just permute these
things cyclicly, is that whatever Galois does in FIs it will do exactly the same thing on the
GIs.
So the curve I'm writing down here, even if the FIs are not individually defined over K
but only Galois stable as a set over K, the equation I'm writing down here is still the finite
over the ground field. So that's the observation that's important here. We have the -- we
have a descent of the curve, we just have to figure out which twists to put in front.
>> Kristen Lauter: [inaudible] some notations, so the delta you mean this to be an
integer, you're talking about like the symmetric ->> Nils Bruin: Yeah, yeah. So ->> Kristin Lauter: [inaudible]
>> Nils Bruin: Oh, no, no, no. No, no. No. This is just -- I was already using a
subscript, so, no, this notation is just an ornament. It's putting delta somewhere on our C.
For elliptic curves it gets used as a twist. No, it's not a symmetric product.
>> Kristin Lauter: And so this delta in general, you're saying it's ->> Nils Bruin: It's an element of the field.
>> Kristin Lauter: [inaudible] integer?
>> Nils Bruin: It's an element of your field K. Its value is only important up to squares.
Okay? Sorry about -- you're right, especially in a talk where you use genus 2 curves.
And Jacobians, people might thing it's this symmetric product. No, no, no. It is just a
curve. I did -- perhaps I should have written C4 comma delta. That might have removed
the confusion, but at the expense of putting to [inaudible].
Okay. So which twists in general? Okay. There's actually a nice result, and that is
basically how you prove Richelot -- that Richelot isogenies in general actually work.
Isogenies are given by correspondences between curves. So you don't actually get a map
from one curve to the other curve, but you get a correspondence from one curve to the
other.
So that's to say if you take the [inaudible] product of one curve with the other one,
normally a function would give you a graph in this product. Here a correspondence, well,
you can talk about its graph, but it won't give this nice 1 to some -- or many to 1 thing
that you get from a map. It's a 2-2-2 correspondence in this case in fact.
And in this case in fact you can be completely explicit with the curve that you write
down, so if we have our C and C twiddle here obtained from the Richelot isogeny on the
previous page, then you can consider the curve gamma delta in this product given by the
following equations over here.
So this is giving us some biquadratic relation on the Xs. And these relations give us a
way of obtaining the Y coordinates from that. I mean, this is biquadratic [inaudible]
gives us the 2-by-2 correspondence bit and then given the Xs this is bilinear in the Y. So
if you prescribe 1Y, you get the others.
Okay. So if we take our delta to be 1 here again, then this is just the classical Richelot
correspondence. This is exactly the thing that proves if you work it out that -- proves that
you get this Richelot isogeny. So the properties of this gamma inside this product here,
it's class in the Picard group of the surface actually prescribes what kind of -- if it defines
an isogeny or not.
Okay. But in our case, we only -- we can just use that geometrically that all works out.
We only have to figure out what twist to put in.
So in this case, as you see, so if Galois acts trivially on our G1, G2 and G3, then nothing
happens here. And if we then just take our delta to be 1, then this all works out as well.
So if our F1 and F2 are in fact Galois conjugate with respect to the quadratic extension
that you get by adjoining the square root of delta -- in other words, if you have F1 and F2
quadratic quadratically conjugate and you just take delta to be the discriminant of that
quadratic extension, then if you go and figure out what happens here, if -- well, if F1 and
F2 can be exchanged by Galois, then -- well, it will do exactly the same thing on the Gs.
So this equation remains unchanged. That's already Galois stable. And these two
equations, in fact, if you go and look at it, Galois would interchange these two and it
would interchange the signs on these. So that means that as an ideal this thing would in
fact again be Galois stable. So that's the right choice to take.
So this solves the case when your F3 is over the base field and your F1 and F2 are
quadratically conjugate. Well, that's really the only case you have to consider. Because
we are looking at quadratic twists, making cubic extensions. It's not going to make
anything a square that wasn't a square before. So you just go to the -- if F3 isn't over the
base field already, just go to the cubic extension over which it does become over the base
field.
Then you use this result. You get the right value of your delta. And since the only thing
that's important for your delta is its value up to squares, that doesn't get -- the cubic
extension doesn't interfere with that. So then you're just done.
And ultimately this really just means that you should take delta to be the discriminant of
your big polynomial F. So the delta is not trivial, even though you were writing down the
model here already for delta equal 1, you can just write down this curve.
That's actually not the right choice for your Richelot isogeny. This should be the
discriminant of your big polynomial F. So I thought one -- I was very happy when I
found that. And of course in the end, in hindsight, if you look at it, of course that's the
only thing it really can be. But okay.
So that gives us all the ingredients because now we can just write down our curve, F of X
here, with those parameters. And now we just have to write down the Richelot isogenies
for this thing. The fact that its Galois group is so restricted already gave us that there are
other Richelot isogenies available.
So what we now can just do is write down this polynomial here. We only have -- well,
we have three parameters now. We have our BC and we have our S. Because the As and
Ds, that's just functions in S. And B and C. So just three unknowns. And now going to
this splitting field of F is putting a degree 6 extension on top of that. That's something
that you can actually still compute.
And even with these three unknowns in there, that's something that you can just about do.
And, you know, that F actually splits over that [inaudible] so you can just write down its
six roots over that formal splitting field of the entire thing. And now you can just
multiply those roots together in all the possible ways to get quadratic splitting because
there's only 15 possibilities, and there is going to be a couple of them that turn out to be
already defined over the base field that are actually Galois stable.
So you just do that explicitly, and you find which Richelot isogenies turn out to be
defined over the base field already. The descent happens for free basically. And that
means that you can just get your model of C4, because we know how to choose our delta
in the Richelot isogeny as well.
So you can do that and get your model. So it fits on one page if you reduce your -- if you
reduce your type a little bit. So the -- oh, yeah. I'm sorry about the delta here is actually
the discriminant of E1. So it's a minus 4 times B cubed minus 27 times C squared.
Okay. So it's a model. So you know that any (4,4)-split Jacobian can be obtained by
specializing your Bs, Cs, and S, Ss in this equation, and in fact you know what all the
other data is in terms of those parameters as well. You just fill in the formulas that we
got in our other slides.
So if you have a Jacobian that you suspect of being (4,4)-split, I would certainly not
suggest trying to see if you can write it in -- if you can solve for your Bs, Cs, and Ss of
that, because in that case it's much easier to just recognize the Richelot isogenies on that
thing to start with. And once you look at its images under the Richelot isogenies that
your variety actually admits, you can check for those, whether they're (2,2)-split. So for
that it's not so important.
Although, there's something else that you can do, and that's sort of a popular pastime for
people, and that is to try and find equations of Humbert surfaces, the locus of inside the
moduli-space of genus 2, Jacobians of Jacobians with larger endomorphism ring, and in
this case you'd get Humbert surfaces of square discriminant, where the endomorphism
ring really just is a [inaudible].
And since you have a model here, you can generate plenty of points. You just specialize
your parameters and compute Igusa invariants. And, well, if you can generate points on
your surface, and you know your surface is an algebraic surface, it has to satisfy some
equation of limited degree, it can just interplay it. So from there it's really just linear
algebra to try and find an equation of your surface.
But there are other techniques that are essentially -- I mean, essentially boil down to this
trick but work for arbitrary Humbert surfaces whereas this only works for (4,4)-split
Jacobians.
So there's something else that you can do, though, and that's always interesting when you
have a family, and that is to look at what happens for the different bad values. Because
there's plenty of values here where you specialize your S to a value, for instance, and you
might end up getting something that's actually not describing a genus 2 curve. So how
does this interact with, well, the boundary of at least this model and how does that
actually interact with the boundaries in your moduli-space.
So one thing that you can do here, for instance, is move your point S to infinity. And if
we go back to our formulas here, if we move S to infinity, well, it's not clear what
happens to this parameter. We'll just ignore that for now. But geometrically at least it
means that you're moving A to infinity.
Now, A was the branch, because our elliptic curve E2 was branched over the same three
finite points that E1 was branched over, and then it was also branched over to point A.
Now, moving that point to infinity, that's the point where E1 was already ramified. So
this is taking a limit where E2 actually goes to E1, at least geometrically.
So that's a valid construction. So that there you'd get E1 times E1. And, well, it's
possible that you get (4,4)-split Jacobians that look like that. So if you do that in this
family, if you do it in this formula, then things degenerate, because the degrees don't
match up. But you can fix that. You can just make little repair metrications so that the
thing does become regular for S goes to infinity.
And if you do that and let S go to infinity you get this model. So indeed this curve C
admits an involution. If you look carefully at the equation, you can send X to delta
divided by 4X. That means that the Jacobian of this curve geometrically at least is
(2,2)-split.
But what happened here is we went to the NT isometric family. And if you take E1 and
you want to write down an elliptic curve with the same J invariant but now with NT
isometric 4-torsion, it's not going to be E1 itself; it's going to be the twist of E1 by its
discriminate. So that gives you the E2 here.
So really we find that our intermediate object that we got here showed that the thing that
was originally a (2,2)-split Jacobian is really the Weil restriction with respect to the
quadratic extension where you join the discriminant of E1 -- oh, there's again a typo
here -- the Weil restriction of E1.
So those are also examples of abelian varieties, Weil restrictions of elliptic curves. In
geometric stories you never see them because, well, they would just be isomorphic to a
product of elliptic curve with itself. But if you're working over non-algebraically closed
fields, then the value restriction of elliptic curve with respect to a quadratic extension is
actually something different.
So what you get here is indeed that you can make (4,4)-split Jacobians by taking elliptic
curve and taking its direct product with the twist of the elliptic curve by its discriminate,
so it won't work if the discriminant is not a square. The intermediate thing, your J2, will
just be this Weil restriction. And the (4,4)-split Jacobian that is geometrically (2,2)-split
will be the Jacobian of this curve C over here. So you do get a nice boundary case in this
particular situation.
>> Kristin Lauter: Are you saying that you can glue it to its twist but you can't glue it to
itself?
>> Nils Bruin: You can glue it to itself, but the problem is that the Weil-pairing will not
restrict to trivial pairing on the diagonal embedding in that case. So if you mod out by
that diagonal, which is really the only subgroup that you can get ahold of, you would not
get a principally polarized abelian variety.
Yeah. So it's -- that's this special thing about -- so originally I didn't know that twisting
by the discriminate flips the Weil-pairing for 4-torsion. So I originally did this
computation and found this, and then I said, hey, and then once you know that you want
to prove that, it's easy to prove that quadratic twisting by the discriminant flips the
Weil-pairing on the 4-torsion. It's straightforward proven.
Then if you read a little bit further, you find that it was already known as well. So it's
one of those things that actually a nice crosscheck if you did your computations correctly.
If you get something unexpected or that I wasn't expecting at least, that actually matches
up with theory. It helps a little bit.
>> Kristin Lauter: So you can glue it to its twist as long as delta is not a square, right?
Or did you ->> Nils Bruin: So a consequence of this is if you take an elliptic curve that has squared
discriminant, then the 4-torsion will actually also be NT isometric to itself.
>> Kristin Lauter: Oh, yeah. Okay. So then [inaudible].
>> Nils Bruin: Yeah. So there you don't see the difference anymore. Okay. Yeah. I
think I used up about 50 minutes, so --
>> Kristin Lauter: Great.
[applause]
>> Kristin Lauter: Any more questions?
>>: So what's the smallest height over [inaudible] a curve with [inaudible] in terms of
the equation?
>> Nils Bruin: I haven't looked at that at all. I mean, usually isogenies don't -- I mean,
once you start with the data, you can just take an elliptic curve and, okay, yes, so -- so
you're really asking if you take a curve with prescribed 4-torsion, what is going to be the
height of other elliptic curves with NT isometric 4-torsion. Because if -- the isogenies
really are not going to change any heights.
But, yeah, so, yeah, if you want to get a small example, I guess you could just try and fill
in some small numbers in here. So you want me to sort of give you an estimate of the
degrees that happen in here, right? So this D here that happens to occur in cubes in this
denominator here, so you could simplify that a little bit, that's of degree 4 NS. And well
it's up to degree 2 and 3 in B and C.
So it's something [inaudible] where you have to assign the weights to B, C and S. I don't
immediately see which things -- it goes up to degree 11 in S here, so filling in the values
for S is going to make things go up quite quickly. Do you add a green 9 here, and in your
Bs and Cs here, I see squares 7th power. So, yeah, I haven't looked at that particular
question.
>>: Also in this family, do you expect there to be a rational point with about the same
frequency as you would for general [inaudible]?
>> Nils Bruin: I haven't thought about that, so my default answer would be yes in those
questions because I don't immediately see how it would influence it. I mean, the
frequency of rational [inaudible] I mean, asymptotically zero, so, yeah. I don't
immediately see how the Jacobian would interfere too much with having rational points
on it or not.
Okay? Well, thank you for your attention.