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>>: Okay. So next speaker of this morning is Bianca Viray, who is going to talk about the
[inaudible] in genus-two.
>> Bianca Viray: I'd like to first thank the organizers for giving me the opportunity to speak. As
you see, I am going to be giving a board talk. So if it's too small, then just yell write larger, just
whenever you feel like it. Alternatively, you can sit closer. I promise I don't bite or will not spray
water guns or anything like that.
Okay. So I'm speaking on Igusa class polynomials embeddings of quartic CM fields and
arithmetic intersection theory. I'll give background on all of it so it hopefully won't be as
intimidating as it sounds.
This is all joint work with Helen Grudman, Jennifer Johnson Loung and Kristine Lauter Audriana
Solerno and Erica Whittenborn.
So the motivation for this problem is looking at the CM method for genus-two.
So first let me review the genus-one case. Francois Moran mentioned this briefly on day one but
I'm just going to review it. So step one is to compute the Hilbert class polynomial and as
mentioned there are three methods for doing this. There's the complex method.
This was first done by Atkin and Morah in '93. And then there's the Chinese remainder theorem
method which was done by Agashia Lauter and Vancutessen in '04. And then I guess, out of
order, but there's also the p-adic method, first studied by Covenia and Hinnuke [phonetic] in '02.
Okay. So first we compute this Hilbert class polynomial using your favorite method from here.
And then step two is to find a root of your Hilbert class polynomial mod P. And step 3 is then to
construct with an elliptic curve E with J invariant J 0. So I'm not going to go into details but this is
the outline. So what is the analog in the genus-two case.
Okay. So the J invariant gets replaced with the Igusa invariant. I-1, I-2, I-3. This isn't exactly
correct to completely determine every genus-two curve you actually need two Igusa invariants but
for the vast majority of cases -- for instance, if the first Igusa invariant is non-0 and characteristic
of the ground field is not 2 or 3, then these three values completely the isomorphism class. It's
not completely true, but it's true in enough cases that that's all we're going to work with. Then we
have the Hilbert class polynomial.
So this gives you genus-one curves which have complex multiplication by the ring of integers in a
quadratic imaginary field. So instead of looking at a quadratic imaginary field we look at the ring
of integers in K where K is is totally imaginary, quadratic extension.
That's probably small. Totally imaginary quadratic extension over a real quadratic field F.
So we call K a quartic CM field. And throughout my talk K will refer to a quartic CM field and F
will always be the quadratic sub field. Then the Hilbert class polynomial gets replaced with three
polynomials and you define it as you take the product over all genus-two curves and you have an
embedding of okay into the endomorphism ring of the Jacobean.
And then you take X minus the Igusa invariant at C. Okay. So we have three of these
polynomials. And then we want to do the same thing. Step one is compute HJ of OK for all J.
And again we have the exact analog. So there's a complex method, studied first by Shpalic
[phonetic] in '94 and Fonwonlin [phonetic] in '99 and Vang [phonetic] in '03. I should mention, this
is not a complete list of the people who have done this. There are many, many more people but
the list is much longer than on this side. So this one I only have the first people.
Okay. And then there's the CRT method, which was done by Isentrager [phonetic] and Lauter
in '05. And then the p-adic method, which is studied by Gaudry [phonetic], Hautman, Cole
[inaudible] and Vang. This was in '06.
Okay. So that's step one. And even though what I wrote on the board looks like an exact analog,
there's actually a major difference between the two cases.
So the Hilbert class polynomial is always -- has integer coefficients. So these methods, for
instance, the complex method you take the J invariants, compute them as complex numbers out
to a certain precision, multiply all the roots together and then recognize what the integer
coefficient should be. And as long as you have sufficient precision, this is possible.
But on the other hand the Igusa class polynomials, these have coefficients in Q. So if you drew
the same method, you need to have a bound on the size of the denominators in order to know
what precision you compute to. And in fact that's true for all of these methods. All of these
methods require as input not only the quartic CM field but also a bound on the denominators.
Okay. So for the CM method in genus-two what we really want, you want a bound on the
denominators of these Igusa class polynomials. And ideally you would like the bound to be very
sharp. I mean, if you think of how I described it, computing the precision, if you have a bound but
it's much too large then you're going to be computing lots of extraneous precision which can slow
down your algorithm quite a bit.
So we really -- it's not just that we want A bound but we'd like our bound to be as sharp as
possible. Okay. So what's been done so far -- so in 2003 Lauter conjectured that the primes
appearing in the denominators are bounded by the discriminant of your quartic CM field and that
they divide the discriminant of the field minus some integer squared.
Okay. So it turns out that this divisibility condition follows from work of Goran in '97 plus a few
additional observations. But the main bulk of the work was in this paper of Goran.
Okay. And then in 2007 Goran and Lauter proved a bound on the primes and then earlier this
year they proved a bound on the valuations. So the valuations of the primes, of the denominator.
Okay. So certainly while a bound is good, I mean this now turns these methods into something
which only require a quartic CM field as the input. You don't have to give anything else about the
denominators, because you can use these results. But unfortunately these bounds are not very
sharp. So we would still like to get something that's a little bit better.
Okay. So something I'll mention just as a note which we'll return to later, so both of these papers
heavily use what we call the embedding problem to prove the results. We'll return to this later.
So we've seen now two things out of the title, so that's good. So to continue and see how we can
maybe get a better handle on these denominators we need to look closer at what the definition of
these Igusa invariance is. All I told you before is they determine the isomorphism class but that
doesn't help you compute them.
Okay. So let me first mention that there are multiple ways of defining the Igusa invariants, some
of which may turn out for better for other methods. For what I'm going to say in this talk, the
definition I'll describe is the best. So this is only one of many other possible approaches, and I'm
not claiming this is the best in general. Just the most relevant for today.
So given a genus-two curve C, you can send it to an abelian surface. Namely by just taking its
Jacobean. And this map is injected. If you have two nonisomorphic genus-two curves, you send
them to your Jacobean, you'll get two nonisomorphic Jacobeans. So instead of thinking of the
Igusa invariants as functions on genus-two curves, we can think of these Igusa invariants as
functions on the modulo space of abelian surfaces. That's what we're going to do.
>>: [inaudible].
>> Bianca Viray: No. No particular field. Just a field. So think of Igusa invariants as functions
on the modelized space of principally polarized abelian varieties, abelian surfaces.
Okay. So I'm going to write out an expression for these Igusa invariants it's not going to make
much sense in the beginning but hopefully in a few seconds it will.
So we can write the first one in this form. The second one can be defined like this. Okay. So
they have all these kis and psis floating around. What they are is not so important. The remarks
that are important is that psi four psi six, ki ten and ki 12, so all of these functions that are
appearing have no poles on the modelized space of smooth abelian surfaces.
So that means if I take any smooth abelian surface and evaluate these functions on that surface, I
will always get a finite number. I will never get something that's infinity.
And the other remark is that ki 10 is the only function in the denominator. So let's tie this back
into what we were thinking about what we want to study is the denominator, what primes are
appearing and to what power they appear.
So if you have a prime appearing in the denominator, then if you look over FP, then that will be an
undefined function. You'll have a pole at that function.
So the primes that appear in the denominator correspond exactly to the primes such that there's
a genus-two curve with CM so that ki 10 has a 0 there. So none of these can be infinity. The
only way this function can be infinity is if ki 10 is 0. That's the only way this happens.
So that means -- so that means that the P valuation of the denominators is bounded above by the
number of 0s of ki 10 at the CM points over FP bar.
So you take a curve over FP bar with CM and if the function -- if ki 10 is 0 at that point of
whatever its order of managing is, you'll get that many primes appearing in your denominator. It's
a little bit annoying, but if you sort of unravel these definitions, this is exactly what it's saying. I
lied slightly because it's really bounded above by a suitable multiple. Because we have either K
10 to the 6 or K 10 cube but that factor is not so important. The upshot of this is that if you can
understand the 0s of ki 10 to what multiplicity they occur then you understand the denominators
of the Igusa class polynomials.
That was an annoying trip into modelized spaces and modulu interpretation and hopefully we'll
back into that and come back to something more concrete. Okay. So after doing this, the upshot
is you can get a statement like the following. So if you take one-sixth times the P valuation of the
denominator -Let me write bigger. This is maybe the main point. So if you can't see this then really you should
tell me. So 16 of the P valuation of the denominator of the constant term of H1 OK, this up to
cancellation -- so the only reason why I took H one here is because I want to know what factor to
multiply by up front. But you could choose H2 or H3 and then you would take ki one-fourth or
also one-fourth.
Okay. So using the modular interpretation, what I just said, so this value is equal to the number
of embeddings of OK into an endomorphism ring of a product of two elliptic curves. This is
counted up to isomorphism and with multiplicity.
So from going from here to here, so this by what I said is just basically telling you the number of
0s of ki 10. You can use the moduli interpretation of ki 10, the points in ki 10 where ki 10 is
0corresponds exactly to the abelian surfaces that are products of two elliptic curves. So your
abelian surface is just of the form E cross E prime. It has CM. So you want OK embedded into it.
The condition that OK embedded into it basically implies that these two elliptic curves are
supersingular. So this is just trying to back us out of thinking of a concrete interpretation of these
0s of ki 10. And then this is also equal to the P part of the arithmetic intersection number. So you
take DIV ki 10 dotted with the CM cycle and you want just the P part of that.
Okay. So now using this interpretation, this moduli interpretation, we get these three equivalent
things. And so maybe you're not so convinced. I told you I was trying to get to more concrete
things, but I have all these ugly words up here. I have this arithmetic section number. Looks like
maybe I just lied to you.
But I promise, I didn't. Okay.
Okay. So the point is that this ugly thing, there's a conjectural formula for, which removes all of
the arithmetic intersection theory and all of this. So in 2006, Grunya and Yang [phonetic] gave a
conjectural formula for this DIV ki 10, dot the CM cycle under the assumption that the discriminant
of the real quadratic field is 1 mod 4 and prime. Okay.
And actually Yang has proved this conjecture, assuming that D tilde, which is the norm of the
relative discriminant, it's prime, and two that OK is freely generated over OF and eta has a certain
form.
So I'm pretty sure this condition on the form of eta follows from this condition. But I haven't
worked out all the details. But anyway so the point is that this condition is not really that strong,
but this one is a very strong condition.
Okay. So let me now write their conjectural formula. So they actually gave a formula for a more
general case than this intersection number, but it gives you a formula for the intersection number
we care about.
Okay. So this P part -- this 2 ki 10 with the CM points at P, this is equal to you sum over an
integer M such that M is of the form D minus X squared over 4 for some X in Z.
You want M to be positive. And then you sum over prime ideals -- sorry. I need a little bit of
notation. So K tilde is the reflex field. And F tilde is the real sub field, real quadratic sub field of K
tilde.
You don't really need to have a good understanding of what the reflex field is. It's just some other
quartic CM field and there's a formula, an explicit way to get it from K. It's just some other field.
So you take primes in the ring of integers of OF, where they lie over your P and then you take
something of the form M plus M square root of D tilde over true D. You want this to be an inverse
of the relative discriminant ideal, and you want the absolute value of N to be less than M times
square root of D tilde.
And we have this BT of P. Okay. So this is 0 if P is split in K tilde. And we've got -- so we get
the P valuation of T plus 1 times this row value of T. Then we get the inertial degree these
otherwise.
Okay. And this row -- so row of an ideal A is equal to the number of integral ideals in the ring of
integers of K tilde such that the norm equals the one you started with.
Okay. So that was a long time to write. And it looks very ugly and involved, but if you look at it,
this is very easy to compute on a computer.
So if you -- as long as your D tilde and D are not so large, which we also have the case in
genus-one, these factors are very easy to compute. You're just looking at the splitting behavior of
an ideal in a quartic field and you're taking the P valuation. This is something that's easy to plug
in and easy to do, even though it's ugly to write down.
So if this formula -- so we have it proved in these two cases, if it was true in all cases, this would
be great. Because this 1-sixth of this, up to cancellation -- so this bound is very sharp. The only
time that it's not equal to the actual valuation of what's in the denominators is when you have 0s
of psi 6 ki 12 or psi 4 at the same points where you have 0s of ki 12. I don't know if this is a
theorem but certainly experimentally this rarely happens.
So we could use this conjecture, then this would give us a great bound to use for the genus-two
case.
>>: Course of two ->> Bianca Viray: No, it works for all of them. It's just a little bit.
You have to think of how you're combining the polynomial. You have to do something with the
symmetric functions and see what the bound of the denominator should be. I think it works out to
the same bound, but I haven't worked through all of the details.
>>: Even though you have sums.
>> Bianca Viray: You have sums but somehow you have functions like you really have poles
over an extension field and when you multiply them all together, it still gives you a bound, but just
the formula might not be as nice.
You just have to unravel the sums and think about what it means. But didn't want to tackle that
during my talk.
So we would like experimental evidence to verify this conjecture in cases where it's not proved.
So it's the one -- so if we only compare this formula with the valuation of the denominators, we
have this cancellation part which even though I said is is rare, it might happen and it might be
hiding some of the errors.
So ideally we'd also like additional ways to verify it. So what we're going to do is we're going to
compute all three of these things, except we don't know what the cancellation factor is.
We also don't know how to count multiplicity of embeddings, but the multiplicity is also something
that is not there so often. So we'll just count all three and see if that gives us enough evidence
for this conjecture. Okay?
So let me quickly explain how we go about counting these embeddings. And then I'll present the
evidence that we found.
Okay. So we're going to keep one of the assumptions, which I just erased. So we're going to
assume that OK is generated by just one element over OF. But we won't put any assumptions on
the form of OF. Just whatever you want.
Okay. So if we let omega equal D plus square root D over 2. So then 1 and omega are
generators for the ring of integers. And you have the trace of eta, let me choose alpha 0 and
alpha 1 so this is true. And beta 0 and beta 1 so that this is true.
And these are just the relative norms and traces down to the real quadratic field. Okay. Then an
embedding OK into an endomorphism ring, this is equivalent to giving elements lambda 1,
lambda 2, in the endomorphism ring, such that the following conditions hold.
We want lambda 1 plus lambda 1 joule. The joule represents the Rosadi revolution that equal
each other. We want lambda 1 squared minus D lambda 1 plus D squared minus D over 4 to
equal 0.
Lambda 2 plus lambda 2 dual plus alpha 0 plus alpha 1, lambda 1. Four that the product and five
that lambda 1 and lambda 2 commute. Probably isn't so hard for you to see why this is true.
Basically we're thinking of lambda 1 is the image of eta. If OK is of this form, then the image of
omega and image of eta and image of eta completely determine the embedding and these are
the necessary sufficient conditions that need to be satisfied.
So this seems like maybe a silly thing to do, because it's basically by definition. But the upshot is
that if you unravel these conditions, and it tells you that it's only a finite computation. There's only
finitely many possibilities for embeddings, and you just kind of go through them all and count
them.
The endomorphism ring of the product we can think of as matrices where you stick different
isogeny rings and endomorphism rings in the different places and then composition of functions
just corresponds to multiplying the matrix.
So if lambda 1 is equal to this ABCD and lambda 2 is XYZW, so then conditions 1 and 2 imply
that A is an integer, D is equal to the discriminant minus A. C is equal to B dual, and the degree
of B is equal to minus A squared plus DA minus D squared minus D over 4.
And this is positive. Okay. So this tells you that there's only finitely many possibilities for A and
B, because there's only so many integers where this will be a positive number and then you want
an isogeny of a fixed degree there's only finitely many of those.
So this restricts lambda 1 to just finitely many cases. And then three, four and five tell you that Z
is equal to alpha 1 B dual minus Y dual and that the degree of X, the degree of W and the degree
of YV dual are all bounded.
It gives you a bit more information than this. But this is enough to tell you that already this also
reduces lambda 2 to finitely many possibilities. So I'm not saying that this is the smartest way to
do it or the fastest way to do it, it is just a way that works and it's enough that you can compute it.
I think if you wanted to compute a lot of these or your discriminants were very large, then you
probably want to choose a correct basis, different omega and a different eta so that these are
short vectors and that sort of thing.
But if you just want to do it, then you can just pick whatever you want and hope it works. Okay.
So we take this. So then for a fixed E and E prime we have finitely many choices for what's in the
endomorphism ring, since our curves are supersingular, then we change this into a problem of
counting elements into orders and ideals in the quartarian algebra infinity. We have MAGMA
code that just goes through and does all of this. And then you count the number of bottom
morphisms [inaudible] out by the isomorphisms. It's kind of tedious but that's what a computer is
for.
That's how we count out the number of embeddings. And okay. So let me just explain a few
more things before I introduce the data.
So one thing is that even though we can't count the multiplicity of the embeddings, we can sort of
check if we're getting -- we can have a little bit of extra assurance, because the Bruinian Yang
formula also includes counting multiplicity. So the Bruinian Yang formula, this order of P to the T
plus 1, this counts the multiplicity.
So we can pull this out of the formula right as a separate factor and we can see at least if we
have the number correct and that all we're missing is the multiplicity. And the other thing is if
we're missing multiplicity and this up to cancellation, the formula is correct, it should always be
larger than the other two values.
If there's cancellation that we're not seeing, it will make the primes appearing in the denominator
smaller. If we're missing multiplicity, they will also reduce our number. So the Bruinian Yang
formula should only be overcounting things, never undercounting.
So I think that's all I need before I present the data. Does anyone have any questions before the
board disappears? Okay. Then could I get the slide, please. Okay. So the examples we used
were the 13 examples that Vonromblin [phonetic] computed. These are all of the genius 2 curves
with CM defined over Q. In this slide I have the CM field, what D tilde is. And then so this is
one-sixth of the constant term, and there's a little bit of fudging at 2 and 3, because these are
their extra powers. You have to normalize the little - if you want to know the exact details of how
we get the 2 and 3 powers you can ask me afterwards.
This is the case where the conjecture is proved so it better be all right. And you see we do get
agreement everywhere. The only place we're missing agreement is our number of embeddings,
we only get two embeddings at 3 where we're supposed to get three to the fourth in the
denominator but this is the multiplicity 2. This is the factor I pulled out.
So looks like everything is okay. We just missed that multiplicity 2. And you can see at least in
these five examples multiplicity doesn't appear that often. It only appeared at 1 prime. Only
once. Okay. So let's look at the next group. So these are ones where D tilde is no longer prime.
We have a square factor, but they're still all 1 mod 4. If you notice, we still you have all the
agreement there. So the reason why there's two values for the Igusa invariant is for this field
there are two curves with CM by that field and since these are all defined over Q, the Igusa
invariants are over Q. So this is actually giving you the Igusa invariant of each curve, not just the
product in the constant term.
But if you take the product then you see that we get, what you get by all the other cases. Okay.
So it's still looking good. And now we get to the times where D tilde is highly divisible by 2 and
you see kind of goes haywire.
So the ones where I have double stars, this one the discriminant of the real quadratic field is 8.
So this doesn't even fall under the conjecture. But it would be nice to have a formula in all cases
so we figured we'd run it anyway, see if we can see what happens and what discrepancy occurs.
So I'm going to divide these up into a few different cases.
So the first one is when D is 8, you'll see that there are just lots of extra primes. Many extra
powers. I mean, all the other ones the number of primes should at least, the primes that are
appearing should at least be agreeing. Maybe the powers are off. But the primes that are
appearing should always be right. If you can't -- I mean, we can't be missing a prime just
because we can't count multiplicity. We can find every embedding. So in this case Bruin yea and
Yang -- well, so in the proof Yang shows that you need to consider only need to consider integers
M and N such that M squared D tilde minus N squared over 4 G is a positive integer divisible by
P.
But if you go through his proof of why you only need to consider these, this really highly relies on
the fact that D is 1 mod 4 and prime. And so when we were looking through the data of what M
and N come up, it looks like you need an additional condition. So for these fields, our guess is
that you need the condition 8 M plus N is congruent to 0 mod 16. So this is for these fields what it
looked like numerically. If you impose that additional condition in the formula, then those all
disappear. So again it's not a proof, just something to think about.
Okay. So that looks much better. There's not these huge extra numbers, these ugly exponents.
Okay. So then the second condition, the second discrepancy I want to focus on is these 2s. So
you see most of the other differences or all of these 2s.
When we find an embedding, there is actually an embedding. So that means that there is a curve
that that 2 should actually be appearing in the denominator. So the reason why it's missing in the
denominators is probably because of cancellation. But we don't -- we haven't proved that. But I
mean we can miss things in the embedding but we can't count things that are not there.
So definitely all the 2s should be appearing. And then the fractional multiplicity part -- so I'm not
sure what's going wrong. But if you again go through Yang's proof, what Yang proves is that
under his assumptions all of the times you will get a non-zero row value this will happen for the
multiplicity, that the order of P at the integral prime you get M squared D tilde minus N squared
over 4 D plus 1, that this will be equal to the prime lying over it, the order of this element plus 1
and in particular that this order is always odd.
Okay. So this statement just falls apart when you're in these cases. I mean, if you try to work
through the formula and you see the M and N that are causing these three halves, you'll see that
this is not true. This is something like 16. So you're getting an even number. So you'll always
have a fractional multiplicity, and I don't understand the rest of it enough to say how to fix it. But
at least this is what's going well.
And then maybe you found the last mistake by looking at it. But you'll see in this last field from
the Igusa invariants from the embeddings we get a 23 to the fourth. But Bernie Yang [phonetic]
has a 23 squared. As I mentioned Bernie Yang should never undercount. If we're missing things
because of multiplicity or the because of this cancellation Bernie Yang should be larger.
So this is not -- this discrepancy is not arising because of multiplicity or cancellation. It's arising
because of something else. And this I cannot find in the proof. I mean, this discrepancy is very
mysterious to me and I don't see where in the proof the assumption that D tilde is prime is
causing you to get lesser, a lesser power for 23. Okay. So I'll leave you with that.
[applause]
>>: Thank you very much. Are there any questions?
>>: Yes. So these easy to compute and psis you had so I'm sure you know more about this than
I do but my experience there sort of modular form, some complex upper hand space.
>> Bianca Viray: Yes.
>>: So what does it mean for them to be [inaudible] B bar?
>> Bianca Viray: Well.
>>: Evaluating that [inaudible].
>> Bianca Viray: Okay. Okay. So Dave's question was in his experience these ki Is are modular
forms and they're evaluated and defined on some complex upper half space, and he wanted to
know how you evaluate them at FP points and Christine said if you evaluate them at CM points
then you get values which are in number fields so then you can think about them and reduce mod
P.
>>: [inaudible].
>>: Any questions? Thanks again.
[applause]