>> Kristin Lauter: Okay. So today, we're very pleased to have Tonghai Yang
visiting us from University of Wisconsin. Tonghai is a great friend and
collaborator for a long time, ever since we were post-docs together at University
of Michigan. He was a pose dock at university of Michigan and at the Institute for
Advanced Study in Princeton. He was the winner of the AMS Centennial
fellowship at Harvard in 1999, along with a number of other awards and
incredibly prestigious and complicated articles and books that he has published
on the subject of arithmetic geometry. So we're very pleased to have him visiting
us here for the first time.
He'll be speaking to us on some work that's very close to my own interests and
relate to genus two cryptography. But I should also mention he's also the
founder of a very successful charity organization called Hometown Education
Foundation, which he'll also tell you about.
>> Dr. Tonghai Yang: Okay.
>> Kristin Lauter: Thank you.
>> Dr. Tonghai Yang: Thank you very much. Thank you. Thanks Kristin very
much for the invitation. It's great to be here. It's great to see Microsoft in person
instead of on computer. And so this is just I want to give a one minute or two
minute brief introduction to this charity called Hometown Education Foundation. I
established in 2004 to have the poor kids in my home country, in China, a rural
area, to have them just finish their very best elementary through high school
education. Because they have to pay for their books and some other fees.
Unlike the US, if you are poor, you don't pay anything, you're totally free, they
have to pay. And some of the kids don't have money to pay, and they in danger
of dropping out of school.
And really pretty minimal, actually. In elementary school is about 50 dollars a
year and middle school about 80 dollars a year and high school much more
expensive. But you know, it's very hard to get to high school, too, it's not totally -you need to take up competition to get to the high school. I mean 60 percent of
the rural area, about 90 or 80 to 90 of the city people go to high school, but only
about 60 rural people get to high school kids. So I hope you can help us to buy
just donations. And we don't take any like over half. All the money you donated
we give it to the students. I pay all overhead here. My colleagues there pay all
overhead there. And you know what they have out here using goodshop.com
they just dial web page or this online story there. Once you go there, just
indicate you support Hometown Education Foundation and just verify it, and just
from there you shop as usual. You don't do anything else. It's free. And every
time you do they will pay like one percent of proceeds to us as a donation from
them. So it's sort of very nice like typical like Amazon.com, like travelocity.com
or, you know, or there's or the other companies, eBay out there. And also
Microsoft very nice. I have this mention of a [inaudible] and you donate one
dollar, they give another one dollar, so two dollars very good for the kids.
So it's a lot -- well, these now here -- my mouse stuck. Make this bigger. So I'll
talk about the arithmetic in the section and the motivation is conjecture of Kristin
years ago. So she was very nice explain it to me. We try to work it out. Some
got stuck in the middle.
>>: I'm sorry. Could you [inaudible].
>> Dr. Tonghai Yang: Say it again.
>>: Could you [inaudible].
>> Dr. Tonghai Yang: Okay. Sure. I forgot about that. Not a very ->>: [inaudible].
>> Dr. Tonghai Yang: Okay. Is this good enough. Or still bigger? How do I get
it bigger? Control L?
>>: [inaudible].
>> Dr. Tonghai Yang: Okay. Good. Okay. So the best idea I'm trying to start
with cryptography to prove [inaudible] I have no idea what I'm talking about it, but
I heard that the best idea for the discreet logarithmic cryptography is if you give
an opinion group you can see equation NP equals Q, then if you know P and you
also know N, it's sort of easy to figure out your addition. So very simple to get a
Q. So that's a good part.
The even better part is it's hard to find N if you know P and Q. So somebody
cannot break it even they know P and Q. Okay. So that's the idea. So what you
do in that case, A or B, are the typical depends on the biggest prime factor of the
order of G. And oppression on G because oppression complicated because it
would be more expensive. So practically this important too.
So one thing to do for mathematician maybe to find a good opinion group such
as that. Order is a big prime or a big prime time a couple very small primes so
we can save the storage and this money. Say there's these oppression on G is
hard to be both hard, mean we are more secure and one's easy and more
efficient depends against a very important [inaudible].
So here is what you want to do. And so how you get that group? So public
group I think probably the first public key are cryptographies started with these.
They have a lawsuit between this one and RSA. G equal to the -- yes?
>>: [inaudible] RSA?
>>: No, he didn't say.
>>: [inaudible].
>> Dr. Tonghai Yang: This is before. As I think they have a patent lawsuit. One
of my former teacher was expert on -- witness. And this is FP star, and the other
one would be more newer one these curve E and looking at these rational points
and these have been in group, hopefully be a prime order would be [inaudible]
would be very nice. If you taking the curve it's a one dimensional project space
with group structure you can also think, well, what about that opinion? So the
mathematician don't prevent that to do that. So A is opinion variety over FP. So
the problem with -- on the general opinion where I have lots of [inaudible]
produce lots of them. The problem is so how to describe the U curve very
simple, just Y squared equals FX and degree three is written down so [inaudible]
in geometry just go write it down.
But in general variety cannot write them down hard dimension. Second, it's hard
to determine or even estimate conventional points. How many are there?
There's not [inaudible] very hard. Depending on the [inaudible]. So how do you
solve this problem? Around '90s and, you know, 2000s, and this person Spelnik
[phonetic] and also [inaudible] in the middle '90s and also Harry Cohn and Kristin,
Microsoft, and then when around 2000 found you can do that using Jacobi of CM
just two curves to overcome this difficulties.
Now I take just two curves if Jacobi is have complex multiplication I will define it.
so have more structure. If you have a general opinion where it's very hard, but if
you know it's Jacobi it's so much easier. But if we want to be even better, you
want to be extra symmetry, so it's structure there, that would be easier to control.
That special idea is if you don't know the sym, don't worry about it, just extra
structures.
So let me try to explain what's a sym opinion is. I started with a -- started with a
real quadratic field, QH over square of D, D is positive. Then I take another
extension for F. By this time I will be totally imagine the quadratic stage of F, so
we unit called a quadratic degree four over Q, sym number field. What it really
means is you take a [inaudible] here is negative, and his [inaudible] also
negative. So got a total negative. So you take this, then they become 20
imaginary. Okay? It is biquadratic we could take Q equal to negative integer,
say P equals zero. Otherwise we go non-biquadratic. The OK will be [inaudible]
of integers over K and sym will be variety by OK [inaudible] variety with an OK
action. Meaning there are lots of amorphisms of A. So that's called symmetry
[inaudible] there where you have a OK, there's a degree four over Z.
And embedded. So there's a lot of action there, so nice. This gives you lots of
restriction on A, make it easier to understand. So sometime we call sym type
because K have a four embed into [inaudible] degree four because sym type if
sigma 1, sigma 2 are embedded such that we get to F that's still different. You
don't want to take an embedding and take [inaudible] a conjugation. They don't
give you good thing. So we want the same type. From these you can produce -how you produce sym variety, I just tell you one way to produce over C? How do
you do that? You take A to be an idea of K or fractional idea of K such as this
guy say a free OF module.
You don't need that, but for my purpose this is a good one. Free OF module.
Then the complex [inaudible] you take the C square, you embed a K to C square
because that's why I have sigma 1, sigma 2. That's the same [inaudible]. So A
would be a [inaudible] in there. Okay? And these will be OK action. So what
you do, you take sigma 1A axis first coordinate, sigma 2A axis second
coordinates. This is the one here. So sigma 1, you XL here, sigma XL second
one, these will be -- you can check this is a combinatorials have OF actually is a
sym variety is by OK. And with the principle position I don't want to explain this,
this is extra condition you have to do to make these work. Okay?
So what about a facts about how this can be [inaudible]. So have some facts in
general. We have a variety. Surfaces actually are determined by three equals
[inaudible] J1, J2, J3. Very similar how you determine a curve. We determine a
curve over complex number field or any algebre [inaudible] J1 [inaudible] here
are the three of them. Because this is what come [inaudible] okay? And with JI
they equal the infinity, I'll call it J, taking in a curve is J will never be infinity, but if
I take the -- this variety could be infinity, whether the infinity is A is [inaudible] to
product of two equal curves. [inaudible] okay?
So the fact that two is if they are not infinity then you actually this opinion is
surface can be written as a Jacobi -- that's a bad J, and it's different from that J.
This is Jacobi of a 2-2 curve. And actually even more than that, so [inaudible]
taught me how to -- given C -- given these JIs, you can produce actually model
for C. [inaudible] purpose. You can really write down what's J -- the see the
equation is about. That's the [inaudible]. You can not do that in general opinion
writing. It's very hard to do this here. Okay. So for a CM variety, this [inaudible]
this guy equals the infinity even only if K is biquadratic. So then from now on we
just assume it's not a biquadratic because that is not a very [inaudible] curve.
Okay. [Inaudible] so what's good about [inaudible] surfaces. You can take any
sym surface. What's good about this is they are very concrete. You can sort of
compute.
First thing is you tell this guy here take the J. Then we can [inaudible] really
more like complex metaphors over C. Actually not depend over C, that depend
over number series. That's what we have with number series. So that algebraic
numbers and these C can explicit, we can explain that in terms of JI over A, then
you found, well, they all in number field, then you know what will happen. So A is
defined over number field. This is sort of a very down to earth proof that
[inaudible] if I [inaudible] number field because that's genuine number field.
Okay?
So that's cool. You can start with somewhere. Second, you take a P to be split
completing K, then you first see that this device denominator you see this is
algebra number then you take the norm of the guy, become an integer -- no, a
rational number. If P does not divide the number of the rational number, then
you take A to be the reduction of A module prime above P, then you can tell you
this [inaudible] and [inaudible] because the JIs are not -- because you see if the
denominator means -- well that means that mod P is not infinity. And mod P is
not infinity, then it's also Jacobi and Jacobi over C, mod P. And turns out C
minus P is all -- just two curve instead of singularity don't have a singularity.
That's cool. Not only that, so you gave out very nice ones. This is what I really
care about it from [inaudible] point of view. [inaudible] for number field but you
[inaudible] for FP, FP. That's why I was [inaudible] split completely the arrow
maybe -- the P also split the complete arrow, and this will be a [inaudible] surface
defined over FP. Okay? They ask the question what's this guy here, and using
conversant multiplication theory, you can really give a good handle of that.
So this guy give you a two good things you need. I've been in service. You have
explicit construction, write down equation. Second, you can bound all your
integer from the computer number you hope it will be a big prime, just compute
them and see what it is. So sure we can do it.
So this is a very -- this is the reason why we use these two CM variety,
[inaudible] service. Okay? So the present question will be first of all, which
prime that divides the norm? And second, how you burn these denominator?
The first one always very interesting, right? Very interesting. That's -- so the
[inaudible] [inaudible] very interesting question because you know where you
want to do. Second and also important and here explained to me I don't know
why he would care about that is [inaudible] it seem important when you try to
compute them because this algebraic number the JIs actually is a [inaudible]
function when it's a complex function is a for expansion. You want
approximations because we're ready to compute them. We know algebraic
numbers are sort of using [inaudible] and compute them, but if the integer is easy
but if it's a rational number you don't know what the number [inaudible] that's why
I have [inaudible] numbers. You need to solve the two problems.
So that's why [inaudible]. And here's a question. She has where inside of idea
about what P should be? Assume that T, TK is a discriminate of your number
field, TK is already read T squared and [inaudible]. T is not the integer. And
then if no P is divided denominator for the same abelian surface [inaudible] then
P have to divides this integer. Okay. For some integer N, for some integer N
such as that the [inaudible] side is a part of integer. And here this is that also
have the part of integer 2. Or I should have say it's T squared minus M, M
should be smaller than D. So square of D. So this is not a big condition. The
thing about is really only a [inaudible] many condition in there. N is burned, M is
burned. Okay. So if you don't start with the condition every [inaudible] you got
are good. Okay? And in particular, NOP has at least smarter than T times
[inaudible] divided by 40 -- 64. You take A equals zero, M equals zero. Okay?
So this -- we take a -- I don't want to take it, but if you take a P, the bigger than
this number, then everything's good. Okay? Everything's good. So that's her
conjecture. And she also made a progress on this with her co-author Corine is
saying P, if T divides these, then P have to be T square a little bit bigger than the
[inaudible] square and trace that square. [inaudible] A plus B squared of T, you
take trace, and if K equals this here. Okay? So that's a very nice progress, you
know. It's not too big here. And this detail [inaudible] bigger than trace T.
So now I'm going to say today [inaudible] to prove her conjecture and also bound
as denominator when T and T [inaudible] are the primes. That's what I'm going
to do. I'm going to show you how to do that. So on a set of notations and
conditions K would be non-biquadratic, [inaudible] sym number field. And as
explained -- and we also assume D is [inaudible] primes count one out of four,
okay? That's just the [inaudible]. And also want to be OK to be a free OF
module. Make my life easier.
This is technical. This is sort of important somehow. If it's not, you have different
-- become more complicated. And to give a condition of this we actually needed
something called a refresh field take the data and data prime. Prime means the
conjugate. This is another quadratic sym number field. And she has a real
quadratic field is detailed in the data here. Really this equal to [inaudible] data
prime up to a square.
So that's another reason why I needed K to be non-biquadratic, otherwise this
would be just quadratic, this would be Q, too small. Too small. Okay. We can
deal with that more another different story.
>>: [inaudible].
>> Dr. Tonghai Yang: No. When this [inaudible] they're the same. [inaudible]
they're the same. In general we don't need to be -- could be the same and could
be different. If they're the same, it's a second extension, if it's different, then it's
not even [inaudible] you can do. Okay.
So I want to define some numbers. These numbers come from arithmetic. It
comes from the K under F third [inaudible]. This is sort of like a string first
[inaudible] then really T is not too bad because every time we've told you it's a
curve so only have one quadratic field showed up. They just have a field and
everything there.
Once they started with the same opinion variety, they always have the same
type. The type will tell you, you really -- [inaudible] not defined over K, it's
defined some extension of the reflex field of that type. So it's really how to deal
with the reflex field. [Inaudible] a choice, okay? So what you would do is you
put the sum over all the primes, only one or two. F third provides above P.
Secondly, you look at all these numbers, there's [inaudible] because absolute
value of N is less than M squared off T third. So only [inaudible] M is fixed.
Okay? M is fixed. And this TK [inaudible] is relative different -- relative
discriminate, the inverse. Okay. So this is some number over define. Next size
is this.
BP is either zero. It would be very easy if the P split in this case delta and only
do anything to zero. If not split, it means [inaudible] or inert. It's always very
simple. Just look at this prime T, [inaudible] order plus one and log P just means
his norm, okay, norm. Then take a look. Okay. And this part is the counting
function. It's counting function. Okay. Let me tell you what we've got here. So
first let me realize it's here. It's T times T [inaudible] F prime, this would be in OF
is an integral idea of OF, okay?
And you can divide -- what about divided by P? That means this guy also have
to be a multiple of P, otherwise you'll get zero. So this rule of A is just counting
from [inaudible] theory A is in F [inaudible] it can harmony integral idea of K delta
such as norm equals your idea. This is a local condition actually, the conjugate
rather local product. So but actually his count, you know, you get a computer
program, you [inaudible] computed that showed up. Give you answer, integers
and not a P, K third. So PN, PM or P really is a positive or a zero integer,
[inaudible] integer, real integer, okay. So I tell you. So PM is equal to N up.
That's also [inaudible] because for almost all P it's got to be zero.
So here is a theorem I want to explain how this conjecture. And above
conditions and that A [inaudible] with OK action defined above. Okay? Then
take the norm of my J [inaudible] equals [inaudible] take a norm and then MI and
NI, then you can really tell what's NY equal to? NY is a factor of this. Okay?
This is [inaudible] because this guy just log P. So explain and take the log, really
K third really sum P to the B -- I wrote it bad. I should be writing more nicer. Can
I write here? Do I have anything? Okay. So really just the P of -- P of this guy P
power in time this, multiply that too. So this is a factor. Very nice would run.
The WK I explain in a minute. N to N3 would be just instead of six would be four,
okay? And WK is a member of a [inaudible] unity in K. Typical is two, though in
one special case it's 10. It's K equal to Q, Q over zeta 5. So really this just really
just two or 10. That's all. Okay?
So this is really just integers that tell you here. And this is a slight refinement of
Kristin's conjecture because we not only give you which prime or to tell you what
the bound is. So the boundaries are really boundaries is a factor because equal
to. Okay? And why this was two here is because you can look at this guy here,
DN of these is because if you look at -- let me back -- oh, no, back. Let's see if I
can put it there. Put it in wrong place. So if I look at the P of these, [inaudible]
this way and Z is only thing that happened is these -- Z square minus M squared
T delta divided by 4T would be a multiple of P because of this here. That's what
P was used for here. Okay? So P divides this here. That's why the condition is
about. Okay?
So that's the conjecture. How do I prove it? I will give you a rough idea how to
prove it using arithmetic geometry. So you really don't think that you want, the
point of this arithmetic geometry, the modular space is you don't think opinion
surfaces or curves individually. You put them together, think it's a part of a
bigger group. That's called modulized space. Let's take X and want to be
modulized space of curves, okay? That we know the known really if you look at
his C points really as the [inaudible] to Z. That's when N is -- this is the same
thing as P1. So what you take, you can take a tau, the complex number, I don't
have [inaudible] just produce a curve as you do this C [inaudible] you take the J
tau. That's how the J, call [inaudible] curve. This also thing with J because I
write it this way here is a modular function. Okay? That's why you really should
think.
Similarly, you can look at our opinion services, a modular space over opinion
services that's very complicated, but if you add another thing called a principle
prization, then it's a very good modular space. What it looks like? It was very
good. A lot of -- this X, write it wrong way is this is Z equal [inaudible] and you
ask P for X on there, and this case is a theorem of [inaudible] earlier is [inaudible]
P3.
So the 3-fold is much more complicated than these, put it's still P3 feel much
better if you [inaudible].
So how do you do that? What's this mean? You take each tau is two by two
matrix [inaudible]. You just do like this. This is Z two, tau is not a matrices here,
so this is C squared. Then this have [inaudible] surfaces. And what you do, how
you identify them with P3, you take J1, J2, J3, that should be A3 if they're not
zero, but sometimes infinity that's why you think it's the P3. Okay?
So what's H2? I want to define it quickly. H2 just a symmetric two by two
matrices, because you measure the part is a part of a definite matrices, that's all.
If you think that you are [inaudible] how you think symmetrical matrices one by
one symmetrical matrices and you measure it by [inaudible] it's exact same.
Okay. How do you X? SP for X and on these you write ABCD. But A is 2
[inaudible] everybody is two by two matrices. That T4. AX how you add 8 tau
plus B, then you C tau plus D I cannot write device because I don't -- in order is
important in matrices okay? So that's what the [inaudible] is about. So this is Z
three-fourth.
So what does [inaudible] looks like if you see in terms of tau? If you are -- this is
the modular function really, SP for invariant is called a function. You can write a
fraction of two stuff, okay. Even though the denominator is very core is called a
key 10, very good name, key 10 to the sixth power and key 10 to the fourth
power and key 10 to the first power. That's why the 6 and 4, 4 showed up in the
different, the N1, N2, N3. F2, F3, F4. F1, F2, F3, what's that? Key 10 is a very
famous polymorphic [inaudible] formula with 10. They call it key 10. It's given by
Igusa called, Igusa key 10. I don't know he like it or not.
The F1, F2, F3 are just a modular form, polymorphic modular forms of weight.
This will be 60, this will be 40, this would be 40. They're all very explicit, can
compute. Okay? But they are for expansions. You cannot just plug in the exact
number, you have to an approximation. Okay?
Another good thing about this is key 10 and FIs have integral coefficient.
Coefficient are integral. But this tells you, this tells you actually these divisors -what divisors, divisor means the zeros and the [inaudible] no zeros, just
[inaudible] zeros. The zeros are divisors on the arithmetic. 4 forward is not just
[inaudible]. So it's very nice, it's really easy -- you think this divisor over Z then
this will be 44 because over C is three-fourths over Z add one, because Z is
[inaudible] is a curve. Okay?
So this got divisors. Very nice. And so what's make the theorem work? Two
things. First is a key 10 is a very beautiful thing. Not only because Igusa found
it, also he found that his divisor is nice. His zeros you knew where the zeros are.
The zeros are two key base of G1. The GMs are so called Mth surface X2.
[inaudible] G1 is a very explain -- very easy to explain. What's that guy equals
to? Really just over CM opinion services who cannot be written as Jacobians of
[inaudible] 2 curve. They are product of two equal curves. So it's very explain.
Let's make this so nice, make the work, see it works. Okay? So but on the other
hand, the problem of making the theorem imperfect because it's a service said
NI, a factor of this guy, right? I didn't say I hope I want to say the guys are equal
to. That would be much better theorem. And I cannot do that because I don't
know anything about a divisor of FI. If I knew then I might have a good chance.
Especially if the divisor you can describe nice in terms of Hamburg surfaces. If
we could do that, it would be much nicer theorem. Okay.
So that's the idea. So why this show up?
>>: [inaudible].
>> Dr. Tonghai Yang: Oh, because I -- if we know the CMs my proof will be
really give exact formula. Okay? I will show you when I give you a see how the
-- so it's hard to find, generally it's very hard to find a zero to polymorphic form.
Even for the [inaudible] we take a modular form. K you know how to do it. Zero
is in the [inaudible]. And I say here of the [inaudible] you know that, [inaudible]
that's pretty much you know. But the others are very hard, okay?
So that will give you a rough idea how to prove these. I look at a CM of X I
define to be modular space EX 2 over sym, I've been in services of OK action
and principle priorization. [inaudible] take all these sym in variety I constructed
where C really just stay there and you add them up. Actually twice of them
because I didn't tell you what the sym type is. Okay? So really just that up.
They use a simple idea for arithmetic geometry called [inaudible] theory on
arithmetic in the section theory. You can see that it would take this one, take J1,
okay. You really take -- instead of trying to [inaudible] integer, I just take an
absolute value, positive or negative, and take a log of them, give them a one
additive, and this is just a number you do the theories. It's one over -- which this
is from here to here didn't do anything, just tell you CM of K is double of all the
opinion where we defined it, put them together. That's a two become one.
That's all. Okay? The J of A, the different As. Here's that A run all of them.
The normal [inaudible] tell you to all the As together, all the conjugated
[inaudible] here. Okay. So you compare these and use [inaudible] theory really
very simple, just efficient, actually, is just a computing the arithmetic in the
section between the divisor of J1 and CM of K. Is it on in X2? The [inaudible]
X2. This is dimension one because the only fundamental points over C. Z is a
dimension 1, this is dimension 1. And Z's divisor in C there was a divisor Z is this
dimension 3 because you have one dimension [inaudible], just say one equation,
so [inaudible]. Means this. Let's think of you have -- let's think of a more
complex one. You have a metaphor, say 40 dimension measure metaphor.
Okay? You have a three dimension metaphor floating around, three dimension
floating around there. You are another curve. If they -- what do you mean by in
the section? Which is very intuitive in nature means so you have a curve, you
have a three dimension over what you said in terms of the points in general. If
they are right, okay? Just kind of how many points there. That's all.
So arithmetic [inaudible] is the same. It's the same. You have over Z, you have
a metaphor of four dimensional the third dimension over C. You take three
dimensional there, take one dimension here, that don't intersect over C because
over C one is two dimensional, another one is zero dimensional over three
dimensional [inaudible].
So only [inaudible] each P. For each P the [inaudible] because the points might
become in the -- in your metaphor divisor. So for each P you have [inaudible]
numbers. But a definite P you should [inaudible] a different [inaudible] okay? It
just make sense. So you can add weight log P. You count how many points are
there and you just log P. Okay? Yeah?
>>: C over K, so you said that's modulized subspace of X2. Is that a subvariety
or a post variety?
>> Dr. Tonghai Yang: No, it's just you take -- you take -- so you just count. So
what does 2 XTs come to -- you just take all the CM services, right, and these
you take the same [inaudible] with complex multiplication.
>>: [inaudible].
>> Dr. Tonghai Yang: That's just [inaudible] kind of all this -- oh, X2 is ->>: No, I mean, the C over K, is it a subvariety of X2?
>> Dr. Tonghai Yang: Yes, subvarieties. No, it's just a bunch of -- over C is a
bunch of points. So it's the closest team of dimension zero over C is.
>>: Oh, I see.
>> Dr. Tonghai Yang: Dimension close to -- close to variety of dimension one
over C. P also over [inaudible]. So this is really some AP log P. And it was a
[inaudible] a little bit of a section over C so very easy just -- said kind of
[inaudible] the over on FP [inaudible] more complicated, you have to do the
[inaudible] up. But that's just a -- roughly speaking [inaudible] so AP not P.
Some of them.
>>: Well, that first equal sign there ->> Dr. Tonghai Yang: Yes.
>>: No, sorry.
>> Dr. Tonghai Yang: This one?
>>: Yes. So that's like the content here, the hard part is ->> Dr. Tonghai Yang: No, that the not hard. That's very easy.
>>: So how do you, when you define the [inaudible] in your section theory don't
you define it in terms of --
>> Dr. Tonghai Yang: Yes [inaudible] so this is the function because J1 is
rational function so in the section with this guy over the zero. So the reason I -this infinite part of this in the section, this is the final part of this section. And
because the JIs are rational functions on the modular space, right, so here's in
the section where it's zero. That's why the infinite part equal the finite part.
That's exactly what they did with the [inaudible]. There can be a final part, there
can be an infinite part. They check them.
They reason for them to check is a much bigger theorem. So [inaudible] right, so
it would be happy. So far we don't need it, we know it's zero. So the infinite part
you out a finite part like this. Okay? So I just compute this part. So what's this
guy? We just write a J1 equal F1 divided by key 10 to the sixth power so you
really [inaudible] equal to divide F1 minus, because denominator, 60 -- 6 times
divisor of key 10 times 6. Okay. And this part I don't know what [inaudible] this
part there, and minus the key 10 divisor is G1. Two G1's is 12 of this guy here.
Okay. That's what I really want.
So to compute these really how to compute this part and this part, okay? This
part. That's really the [inaudible] try to compute this part and compute this part.
So what I can tell you is all they are getting compute is this part. The G1 part.
And actually I can come GM. GM you come to this equal, too. Equal to this guy
here. Equal to the summation of N [inaudible] zero and this, and DN minus N
square over 4. So here the G should be M should be and [inaudible] shall not be
a square. Otherwise I have been in trouble. Okay? This I define it before.
[inaudible] so that's all I can compute. Now, that's the main theorem I used to
prove it, okay?
On the other hand, this part is party or zero. Because counting the section
numbers. Okay. This F1 and the sym K that don't intersect at the -- over C. So
really this is just really just the effective numbers. So really just really [inaudible]
or zero. Okay?
So let's see what you get now. Look at back see here. This equals this, right?
This is party or zero, and this is we know how to compute this and that's equal to
this guy here. That's the reason eventually knows that this AP is not a P, so this
you know actually this guy is a divisor. That's the reason. That's the proof of
that. That's the simple proof of this. That's the idea. Okay? Any questions so
far?
So now let me -- that's how you prove this logic conjecture, repeat all the same.
F is the same [inaudible] with OK action and above and this [inaudible] then this
will be R factor this guy here as far as showed up here. And N towards 3 is 4,
this 4 has come from that key to the sixth power to the first power. This
[inaudible] K2 for 2 or 10, not very important. This is the formula. You can
compute actually. So give your bound. Not [inaudible] but really give you a
factorization that is a factor of this guy here. Okay?
So how do you prove that? So now I want to do another, you know, 10 to 20
minutes. I want to tell you how to prove that theorem. I want to tell you in the
section from a purely geometry. You take seen there and you compute what's in
the section, okay? That turns out to be a very hard [inaudible] in general. See
the X2 is a huge -- it's a three dimensional metaphor over C. There you go over
Z, you been on four dimensional. These are harder theories. The device is a
three dimensional theme and you get another one where you want to -- say it's
very hard. So another [inaudible] over -- it will occur much easier because
original only just one dimensional, you put two dimensional very simple. I want to
cut those [inaudible] you cannot do it by -- there's something in between. There's
a one dimensional, three dimensional or from two dimensional, four dimensional,
there's something called modular surfaces.
So this is where you take a modular space of [inaudible] with an OF action plus
some priorization. OF have a real quadratic field. And you cut it, become
smaller than X2. So this actually is I can tell them what looks like. You tell here's
a C points really looks like H squared, two copies of H. And [inaudible] 2 of OF.
That's because it got X. This is dimension two metaphor surfaces, called
surfaces. Okay. Hilbert modular surface. Because the two dimensional.
Here you can see I cannot say not equal to P2. Actually it's not. More generally,
in general F this has got a general tab, very complicated. This come to one
dimensional. That's nice. And how you [inaudible] take gamma [inaudible] just
X to gamma of Z1 and a gamma prime over Z2. Prime gamma conjugate. So
that's where [inaudible] equal curves.
Here you look at three things. You do X1, X and X2. Meet in the middle. So
easier. [inaudible] X1 to X you take E, you get 10 to be OF, become two
dimensional surfaces, certainly have OF action here. So that's the core thing to
do. And his image is called his first his divisors of X. So very nice here. And
you also have a -- from X to X2. So X, what's X is his point R, a [inaudible]
services with real multiplication, right? So you can know the real multiplication.
So forget about [inaudible] you get eight. And that's where be your point DX2.
And, these, [inaudible] also very nice is the T Hamburg services actually. It's
very nice. So how do you -- so now I want to do is a trace net of Cs from the X2
to X, the [inaudible] so let me toll you even more about the [inaudible] divisors.
These things are all the [inaudible] beautiful things that started with his [inaudible]
in 1970s. They try to prove some conjecture was there and you treat all this
divisor here. This is also part of a small varieties, they help divide it. So take
[inaudible] space V is [inaudible] F such as you should transpose equal to
[inaudible] conjugate. We compute with AB, number, number, prime. Okay. It
would take the correct form and it will be AB this is the one-one, this will be
signature will be two-two. Okay. And if you care about his -- this, if you use this
[inaudible] space you do small writing Hilbert modular surface. And take a lattice
like this, AB and AB is this number is different inverse, then you can define some
curves inside this varieties.
But you do take any A in this [inaudible] and determine [inaudible] you just
compute these quadratic equation. And this will be a curve in copies of H2, H
and a small h. Why is that? I just -- this is something very explicit. You write
[inaudible] like this. Okay? Then if you do it again, Z2 really determined by Z1.
So that's why the one cup of H but twist it. Twist it. The twist is important.
Okay? So what's [inaudible] divisors are very simple, just take all the As
[inaudible] typical to M over T, FA, and take all of them, you take SO to OF action
there, that's where we give you our divisor. What's [inaudible] is then symbol
[inaudible] to one and in general is just a final union of curve in X otherwise. And
they are small curves or modular curves. Some small and some modular curves.
Okay? And T1, whatever decide before, this is really just diagonal map here.
We rather see one carefully in terms of over C, really just Z, the ZZ. It's that cool.
Okay. And [inaudible] the prime splitting F, then the TQ maybe looks like a twist
embedding of the open modular curve Y0Q. The modular curve embed to the
small curve, to the Hilbert modular surface.
In general is more complex, sometimes small not just modular curve, okay? So
we put back the [inaudible] how you do this. That's called TM. Some call it
closure of TM in here. Oh, this is wrong. This is X. This X. X is a [inaudible]
surface. Then here the beautiful thing whatever can do. You tell the Hamburg
surface [inaudible] understand because if you care about [inaudible] really put
this guy [inaudible] really equal to this. They're his [inaudible] curves. That's all
the user understands here. And this is dimension one over a C, dimension two
over a Z. One dimension newer than here. Make your life a lot better. Okay?
And CM is if you have an okay [inaudible] suddenly have OF again so you can
watch the thing is a sym cycle X because K is bigger than F. Okay. So then you
compute them. This one I want to compute my theorem. Then it become
[inaudible] X2 in the section you can switch it to X in the section here, TM. Here
the M will be several orders conditions are here. It's kind of M really just really be
TM minus N square over 4. So here is some [inaudible] okay? So, this should
be wrong M. This [inaudible].
>>: [inaudible].
>> Dr. Tonghai Yang: No, no. This is N, and this should be N. Some kind of
[inaudible] like this. Okay? T of -- this M should be equal to TM minus M square
over four. And some [inaudible] some old [inaudible] so this is how [inaudible].
So the main theorem I want to compute is the TM, for the TM I know take some
[inaudible]. So TM will equal this. If it's [inaudible] the OK that 3 OF module and
what [inaudible] anyway, then the section on these Hilbert modular surface is
equal to half of, the DM I defined before. That's where if you haven't used this
theorem then you can get back to that theorem. So really what we're working on
the Hilbert instead of the other one. Okay?
So how you do this? So even that is sort of hard actually. For general M is very
hard to do because the M, the TM is hard to describe. Some number smart
curve, some of them modular curve, small curve is a very complicated thing to
do, it's very hard to describe. So it's hard to compute these in general. And what
I did is used some trick. We come to the sum of what I can, others I use
modularity. So first the one that can be a weaker version of this. Okay? Other
version, you take the same idea. Only instead of take M, you take the prime
splitting F. That's why you have a modular curve. You have a [inaudible] curves.
Then I can compute this guy equal to. I cannot do it actually, even that I cannot
do it. I can only figure equal to half of PQ. PQ is what this one is supposed to
be, right? But I don't know what happen with Q, the Q place. So I don't care, just
as I know it's different by log Q by some virtual numbers.
Later on I just argue from general theorem this CQ had to be zero. And so really
very weak version. First have to be split. You will split, I don't care where the
place had a Q. Okay? That's what I proved. This is [inaudible] computation
really seriously to intersections here. The others is not. Others used the version,
then what I do is use the same value form general formula I proved by Bruinier
and myself and then use this arithmetic intersection theory I just plan before, use
more general, not just degree zero, the very general machine have to use zen.
Zen plus co-modularity of [inaudible]. This guy will take these TZ and TM, put
them together, actually is a modular [inaudible] right in this here. And this better
tell you what -- tell you this. And these also same as X2. These are a real nice
[inaudible] when I talk to Kristin this morning [inaudible] without is already known
by his [inaudible] okay? He's much more general. You look at also the
[inaudible] same as H2 in this case. H2 [inaudible] is easy okay? There they tell
you this TMs, you don't need the word TMs. They are really a linear combination
of [inaudible] TMs. And you can take that [inaudible] to be the split [inaudible]
talk. So that's what you don't need [inaudible] you just need the TQs then you
can just take the other linear computation of them.
The machine of this guys, they're linear. So you don't need to deal with
everything. That's the trick. Okay? So you knew [inaudible] group that's done
by a Bruinier [inaudible]. Okay? And they divide the general case. So what's
[inaudible] talk with the second part? It's like this. You have this guy here. You
don't need to be free. It just take any modular form. If the integral coefficients
and [inaudible] if you take all the [inaudible] together, you take [inaudible] should
be equal to one. And such that his divisor is known, if we know his divisor, then
you are happy. If his divisor equals linear combination of TMs. Then their value,
their value just you take the product and take the log, the [inaudible] here I
explain in a minute equal two parts. One part is what have we got here? Is he
here? And the other part is very transcendental, okay? This is the part I wanted.
And for this -- this is where I want to prove my original one okay? So first ->>: [inaudible].
>> Dr. Tonghai Yang: I will explain these. So first is a [inaudible] metric, really
just mean. You take absolute value. But that's not really defined if this function
form. So what you do, you take Y1, Y2, and take the K over two. This 16 pi
square just a normalization seems nice over Z. Okay? The beta is the very
weird C here by the very transcendental can see. [inaudible] function. 4 pis
transcendental. And this guys supposed to be transcendental, too. This guy,
take the [inaudible] key of over F, that's the [inaudible] to the quadratic extension
and technically it's a function you complete it. And that's it. That's a log
[inaudible] of this. Okay?
This actually is another theorem. This part would be equal to we proved in -once we proved this, this is always equal to the height of your opinion surface.
The height tells you how complicated this guy. This actually is called. Now I've
been in Chowla-Selberg formula. Chowla-Selberg formula is a very beautiful
formula in 1950s when rephrased as curves of height of [inaudible] equal to
[inaudible]. Okay? And this is a part of a conjecture [inaudible].
So that's what I was down there before. So that's really see -- that theorem plus
that before we can prove the theorem. I don't want to get detailed. Okay. So
this, I want to tell you some -- couple minutes to tell you other applications. So
one application used to -- just two curves, we talk about that, the J [inaudible]
[inaudible] geometry language. Okay? Like this. If a K quadratic field as above
if you have a C is a just two curves or number field, such as his Jacobi is not the
same but okay. That's exactly what we do, right? And this kind of always give
you [inaudible]. And if had a good reduction everywhere that J that already have
good reduction everywhere and that's always true. We can always do that.
Same already where always make it good reduction everywhere. However, take
error to be a prime. If C had the bad reduction in some prime above L, even
though J -- Jacobi have everywhere, C is [inaudible]. How I determine it, by the
number of key 10. The [inaudible]. That's exactly what you really say here. If
you have paradise in a place then must be set of a condition for some this guy
here this now zero. Okay? That's because the [inaudible] means P divides the
norm of the other guy and that's a fact of this, so this can not be zero. Okay.
That's the idea.
Conversely if this is not zero for some these are here, then they already exist just
two curves several condition. Now I realize you're talking about this here. It's
actually for every curve we always have a better reduction over subprime
because this [inaudible] that's all the prime and subprime, okay. It's here. And
the idea is see here the [inaudible] if and only if the J is in Jacobi of the
[inaudible] in the G1 and the P is really [inaudible] section points. That's the B1,
B can be about. Okay. And that's the idea. So another application just measure
briefly is now I've been in generalization of celebrated [inaudible] formula. Okay.
I don't thing I want to give you proof of the [inaudible] version I want to say
[inaudible] I don't think it's time is over. I don't want -- if you want to know, I can
explain to you later. But I just stop here. Thanks a lot.
[applause].
>> Kristin Lauter: Okay. So let's thank Tonghai again and we're going to go
have lunch in the cafeteria.
[applause].
>> Dr. Tonghai Yang: Thanks.