List of open problems
of degree in
, . Let
max coefficient of
has a solution in
- For , is it true that
? Does this follow from Lang's conjectures? (Voloch)
- Instead, look at those with those with points locally everywhere
(Poonen). Is this a positive fraction, i.e. is
- As a special case, if you write down a plane cubic, how likely is it to have a rational point? (Voloch)
- For cubic surfaces, there are examples where the Hasse principle fails, but maybe for almost all values of a parameter in a family, the Hasse principle holds. We should have asympotically that the
? (Colliot-Thélène) For , we hope this will hold in general.
- The set of reducible such hypersurfaces are a very small fraction (usually codimension )--they affect this calculation very little (Poonen).
- Serre looked at ; here one has the Hasse principle. (Tschinkel) For , the proportion of everywhere locally solvable ones tends to zero; but we should exclude this case because of the codimension condition. (Heath-Brown) Instead, we should restrict to families such that the codimension of reducible fibers is at .
- Computational evidence is all over the place, so one must rephrase the question better to get some kind of answer. For example, for cubics, those with prime power discriminant and the general evidence are quite different. (Swinnerton-Dyer)
- How is this related to -torsion elements in
Update (4/21/04): [Poonen and Volloch have tackled this problem:]
If is smooth projective geometrically rationally connected defined over a number field , then the Brauer-Manin obstruction should be the only one to the Hasse principle. That is, is
As special cases, this should be the case for Fano varieties of dimension , e.g. (smooth) complete intersections of degree
- It would be worth doing a reasonably large search on diagonal quartic -folds. (Swinnerton-Dyer)
- It is an old problem that for nonsingular cubic forms in at least variables, the Hasse principle holds. For diagonal cubic forms over
, this is proved modulo finiteness of
- For the smooth intersection of two quadrics in
, the Hasse principle should hold? If it has a rational point, then it in fact satisfies weak approximation. The critical problem is in variables. (Colliot-Thélène)
define the smooth complete intersection of two quadrics
over a number field. For simplicity, assume
. (Sansuc, Swinnerton-Dyer, Colliot-Thélène)
- For ( variables), this can be provably done by the circle method. (Heath-Brown)
- For , there is the concrete problem: given two quadratic forms as above in variables over a -adic field, assume is smooth, so that
is separable. Does there exist
contains 3 hyperbolics (split off an extra in its decomposition). Solution to this problem would give a local-global principle for . (Colliot-Thélène)
- It is possible in odd characteristic after an odd degree field extension (Heath-Brown), unwritten.
- If in the pencil, there is one of rank , then it is possible; or other conditions with (e.g. two conjugate lines or contains a conic defined over the ground field).
Consider the hypersurface given by
. Take the height
and throw out the set where some , . Can one estimate the counting function? (Peyre)
Some people are working on this. What news?
, with height function
, and restrict to
. Excluding trivial solutions (on lines), can you prove a counting function which is
This is a singular cubic surface so we also expect an explicit constant. (Tschinkel)
- This has a -singular point. On top of the singularity, one gets a configuration of 4 lines all of which have self-intersection .
- Are numerics possible? (Voloch)
- This is a compactification of the affine plane (solve for ), but it is not equivariantly embedded. (Tschinkel)
- For the singular cubic surface
has established that the counting function has exact order of
Proceedings of the session in analytic number theory and Diophantine
equations, Bonner Math. Schriften 360 (2003).
- Progress: Using the universal torsor (as calculated by Hassett and
Tschinkel) Browning has established that the counting function has exact
order of magnitude
- Update (4/21/04) Tim Browning has shown
. He makes use of the universal torsor. See
- Takloo-Bighash: for
, have an effective
lower bound of , and should be able to get
by a similar method, but can't push it any further. Expected
upper bound in this case is again
Hassett: the universal torsor was in his lecture.
- Hassett: there are 13(ish) singular cubics (zero-dimensional in
moduli), the list is in a paper, reference available.
- Is there a surface over
which has a finite nonzero number of rational points
- Is there a nonsingular quartic surface with this property?
- Find a third rational point on
. (There are only 2 points with height
, and other non-public reasons to believe that there are only finitely many points.)
- Find a smooth quartic
would be the first place to try, but over any number field is OK. It seems as though this is possible for either all such families or no such family.
- There are surfaces with no rational lines with infinitely many rational points. (Peyre)
- Do you hope that the Brauer-Manin obstruction is the only one to weak approximation? (Harari)
- The rank of the Neron-Severi group over
is rank , but is only rank over
coming from the hyperplane section. Therefore it looks like a Kummer surface, the product of two CM elliptic curves. What are the elliptic curves? (Read the right paper of Shioda.) The CM is by
- It does not seem helpful to look over a finite extension. (Colliot-Thélène)
- Are there any heuristics looking modulo any primes? (Poonen) The zeta function does not say anything about solubility. (Swinnerton-Dyer)
, Elkies found another point, with smallest height on the order of (Colliot-Thélène); he uses a fibered pencil of elliptic curves, looks at values of the parameter for which there was a point everywhere locally and then looked for a global point. There is no such fibration in this case. Maybe one could go to an extension and then look for rational points (Villegas).
- What restrictions are necessary for such a surface to occur? (Poonen) The condition to ensure that
is a black-board full. (Swinnerton-Dyer)
- Is it possible for the given surface that
is finite? If so, there might be many rational points. (Harari) In the computations of Brauer-Manin obstructions, there are many with
, there are a lack of examples with `transcendental elements'. We expect
to be finite, proven in certain cases because of the Tate conjecture.
- Are there any known examples of quartic surfaces with
with infinitely many rational points? (Poonen) Maybe almost always they have infinitely many. (Swinnerton-Dyer)
- If there are infinitely many points on a quartic, will they be Zariski dense? Look in Mordell's book, perhaps. (Colliot-Thélène)
- Silverman has examples of surfaces in
with two noncommuting endomorphisms, so this gives infinitely many points, but this has Picard group rank 2. (Voloch)
The distribution of rational points on Enriques surfaces has not been
well studied. (Skorobogatov)
Let be an Enriques surface.
, is Zariski dense?
- Is there an which violates the Hasse principle?
- Is there an such that
- Is there an with
There is a
cover of the Enriques surface which is a
surface; is there some torsor over the surface for the torus
for which the total space is a torsor for under a nonabelian
group? For a bi-elliptic surface, is this possible? (Harari)
Let an Enriques surface over a number field ,
the double cover (K3 surface),
the Brauer-Manin set.
, lift it to an adelic point
on . Under what assumptions on will it be liftable
, or at least
Guess (reported by Harari): there should be some nonabelian torsor
for the group which is the semidirect
, where is Neron-Severi torus of .
should correspond to
About Pb/question 8 (which is closely related to Pb 7), David Harari
adds the following update (2004-09-25): ``Skorobogatov and myself have
recently proved that there exist Enriques surfaces with adelic
but not in the closure of the set of
rational points ("The Manin obstruction to weak approximation
is not the only one"). In particular, some adelic points of
are not liftable to
, see the
paper "Non-abelian descent and the arithmetic of Enriques surfaces"
Use the intermediate Jacobian (when it is an abelian variety) in arithmetic? To fix ideas, , look at algebraic -cycles modulo rational equivalence.
Compute this for a rigid Calabi-Yau -fold (over
, simply connected, , , , . In this case, the intermediate Jacobian
is of dimension . Determine given , i.e. give its -invariant.
- For two quadratic forms in variables, i.e.
, look at the Jacobian of the genus curve given by
. (Colliot-Thélène) Over
, you can prove the Weil conjecture for . Can you use this to prove something? You can also look at the variety of lines on , also a principal homogeneous space for an abelian variety; so over a finite field, this will have a rational point, so there will be a line over
. We can say something with contains a pair of skew conjugate lines or a conic defined over the ground field; how can you do these things such as finding a line over a quadratic field...?
- Explicit examples of rigid Calabi-Yau -folds? Take an elliptic curve with complex multiplication by
, take the kernel of the endomorphism , ;
has singular points, blowing up these points gives (Candela). Also the quintic hypersurface
with has nodes; the resolution has . There are more such examples. (Yui)
Are there results at the level of number fields arising from the techniques of rationally connected varieties? (Colliot-Thélène)
For example, recently Kollàr got nice results over local fields: if is a projective variety, we say
are -equivalent if you can link them by a chain of curves of genus zero over ; if is smooth with
, then is rationally connected if
is a single point. Kollar proved that if is a local field, and is rationally connected, then is finite. (Szobó)
Kollàr and Szabo proved that if is a number field, and is rationally connected, then is trivial for almost all .
If is a variety over a number field, and is rationally connected, then does there exist a field
such that for all
that consists of a point? (Ellenberg) Negative answer by a conic bundle over
Let be a smooth projective rationally connected variety with the cohomological dimension of . Does have a rational point? (Colliot-Thélène)
- There was a false proof for
- Yes if
. (Esnault) Yes if is the function field of a curve. (Harris, Graber, Starr)
- At least for surfaces, we hoped that universal torsors would be nice objects, e.g. they are birational to homogeneous spaces under a nice group, so they would be close to -rational if they had a -point. An example of ( a horrible field) a cubic surface with
. (Madore, Colliot-Thélène)
- Colliot-Thélène adds the following:
With hindsight, Problem 11, as phrased, had been settled by J. Ax in
Bull. Am. Math. Soc. 71 (1965) p. 717. Ax
produced a smooth hypersurface in 9-dimensional
projective space, of degree 5, over a field
of cohomological dimension 1, with no rational
point (that such hypersurfaces are rationally
connected was proven much later).
Ax'example has index 1, i.e. the g.c.d. of the degree of the finite
field extensions over which the hypersurface acquires a rational point is 1.
In Journal of the Inst. of Math. Jussieu (2004) 3 p. 1-16,
J.-L. Colliot-Thélène and D. Madore produce a
field of cohomological dimension 1
and a smooth cubic surface over that field which has index 3, thus settling
negatively a question of Kato and Kuzumaki (1986).
Describe all pairs of sets
, stable under
, where the elements of and are alternately placed around the unit circle. (Rodriguez-Villegas)
- There is a solution but it is much more complicated than the statement of the problem. The solution is used in the classification of algebraic hypergeometric functions.
- There is the infinite family
Find a smooth quintic hypersurface in
with Picard number 1 over
- This is used in constructing error correcting codes.
- If you compute the analytic rank (the zeta function), by Tate's theorem, the second Betti number is so compute the number of points up to something like
(Voloch). So testing them exhaustively would be very costly.
- Shioda has examples over
of Picard number , so they might be defined over
. (Raskind) This has been tried once. (Voloch)
Consider cubic hypersurfaces
- If soluble, give upper bound for smallest solution in terms of the .
- Look at , tabulate the size of the smallest solution and conjecture a particular growth rate in terms of .
has a solution
, how large is the smallest solution?
be a solution with
Swinnerton-Dyer had suggested: if
Wooley had suggested instead
Progress: Stoll, Stein:
computations up to
suggest an upper bound of
. Also, assuming Schinzel, finiteness of Sha, and one unproven lemma,
Stoll can produce
with least solution
. (Need the hypotheses to ensure that the examples
do have rational solutions.)
Shape of examples: if prime, look at
. Assume and are approximately of the same
, that is a cube mod but not mod
, and that 5 is a cube mod but not mod (plus some additional
- The growth rate should be like . If you do the corresponding thing with squares,
, the answer is . (Swinnerton-Dyer)
- Seems more like
- Is there a heuristic which suggests this? (Poonen) No. (Swinnerton-Dyer)
Characterize the rational numbers that can be written as
and fixed . (Poonen)
It is necessary that
for and . Is it sufficient?
- You might repeat this kind of problem with any rational function with no rational poles. (As in Waring's problem.) Something has already been done for a function with rational poles. (Poonen)
- You can phrase this problem a different kind of way: prove or disprove the Hasse principle for this equation for all
. (Swinnerton-Dyer) You can probably show for sufficiently large (and fixed) that there is a solution locally.
- There are applications to Diophantine definitions: this would show that inside
, the set of these rational numbers which are integral at half of the places, is Diophantine. (Poonen)
Solve the local-global solubility problem for finding lines on a cubic hypersurface. (Wooley)
- Find an example of a cubic hypersurface over
(in as many variables as possible) with no rational line.
- Find an example of a cubic hypersurface over
(in as many variables as possible) with no rational line.
- For (a), we must have at least variables, since there are cubic forms in variables with no point. If you have variables, then there is a rational line. (Wooley)
- For (b), for variables, there is a rational line. (Wooley)
- Also, find one with a rational point but no rational line. (Colliot-Thélène)
Draw a regular pentagon , construct the circle through the vertices , and consider the curve
. This is a quintic curve with double points, and therefore has geometric genus . This curve has five points at (given by the slopes of the lines). Compute the -torsion of . (McCallum)
- The points at are among the -torsion.
- The motivation is: these curves are principal homogeneous spaces, and are candidates for -torsion elements in
. If is the parameter on , then this is a twist of the universal elliptic curve of . Here we have explicit models. (McCallum)
- If you consider this as a pencil of elliptic curves, how does this relate to the talks at this conference? (Ellenberg)
- Does the pentagon have to be regular? (Voloch) There are various variations, such as replacing the circle with a star pentagon.
- Find a separable polynomial
such that the Jacobian of the hyperelliptic curve is isogeneous over
, an elliptic curve, with or and
- Related problem: Give an example of a map
, where acts diagonally, whose image does not lie in
for any subgroup
- The case and is possible, as is the case and . A consequence of this construction would be better bounds on the density of quadratic twists of of rank . (Silverberg)
- What if you ask this question over
? (Voloch) Can at least do and over
. (Poonen) Even over
, maybe it cannot be done for large . (Ellenberg)
- One might consider the curve over
, for composite. If , for example, it maps to ?
- Does it have to be hyperelliptic? (Rodriguez-Villegas) Yes, for applications. (Ellenberg)
- The random matrix heuristics suggest that there is a positive power, so there should be a curve there, and finding such a curve would give a proof of a density result. (Ellenberg)
pairwise isogeneous elliptic curves,
, and look at the set of principal polarizations on . Find a nonsplit principal polarization on ; then it comes from
, . Now you just need to show that is hyperelliptic.
Let be a variety over a number field and suppose that for every open Zariski dense
, the map
has a splitting (e.g. if is Zariski dense). In this case, if
, is there no Brauer-Manin obstruction to the Hasse principle for , i.e. is
- Is it possible that
follows from the splitting condition? (Poonen)
You might also ask the corresponding question for a local field. (McCallum)
- You might also ask this question for other obstructions. (Ellenberg)
. Write the system of two equations
; think of this as a one-parameter family of curves of genus one;
is empty and
- The secret reason for asking: then has a zero-cycle of degree . (Colliot-Thélène)
If you have a zero cycle of degree , then there is such a point.
- Even in a given number field of degree , looking at a random way, the evidence you will find is zero. (Colliot-Thélène)
- What about
modulo the action by three? (Ellenberg)
- Reduce the search by finding one elliptic curve with many rational points (fix , consider
), then search for . (Voloch) But the ratio is only dependent on . (Poonen)
- Do we expect
? (McCallum) Not necessarily. (Colliot-Thélène)
- Consider instead
; search now in the first curve for a cubic point, and search for . (Poonen) Put the first one in Weierstrass form, and generate the cubic fields.
be an algebra with
given by its multiplication table. Suppose you know that
; find an algorithm which gives an explicit isomorphism. (Stoll)
- The motivation comes from very explicit -descent on elliptic curves.
- The case of
reduces to finding rational points on conics. (Stoll)
- Is this an problem? (Voloch) In some sense, but one needs nine
-matrices, not one
-matrix; the equations are not all linear. (Stoll)
Let be a number field,
. Construct an algebraic set
for some such that the projection onto one of the coordinates is exactly the set of elements of with
for some (archimedean) absolute value of . (Shlapentokh)
This would imply results on Hilbert's tenth problem for rings of integers. It probably is hard because it is close to being equivalent.
be an elliptic curve over
, and suppose
. Is is finite?
- More generally,
gives a subset
; what others can you build?
It is possible to describe in a Diophantine way the set of points
; this is an infinite set. (Poonen)
Given an elliptic curve
, describe the set of primes such that
is dense in
. Is this set nonempty? Is there a modular interpretation of this problem? (Takloo-Bighash)
- You need surjectivity of the reduction map (Murty and Gupta have some results) and the surjectivity of the map on formal groups (Silverman). One suspects that for any such curve there exists a prime with this property.
- This is related to Brauer-type problems for surfaces.
- You should be able to collect data on this to see if there is a positive density. (McCallum)
Is there an algorithm to decide solubility of a system of linear equations
, together with equations of the form
- This is a problem in linear programming; add variables. (Voloch)
- A negative answer would have implied undecidability for Hilbert's tenth problem over
Describe the variety of curves of low degree on the Fermat variety
- Lines are known, but that is all. Hope that there are none or very few.
- The curves should be over
Given a plane curve of degree , the number of points on of
height at most is . There are curves which have the
number of rational points . Can you do better than ?
(Roger Heath-Brown, 2004-09-25) observes that actually is
easy, so the challenge should be to improve on this.
Let be an elliptic curve over
, , such that
be a cyclic
extension of degree of conductor , and let
number of distinct primes dividing . It is true that
? (Chantal David)
be a cyclic cubic extension. Then the -rank of the
class group of is
controlled by primes dividing discriminant of ; if there are
such primes, then divides the class number
(genus theory). Problem:
carry this over to a guaranteed contribution of 3-primary component
of the 3-Selmer group (over ) of an elliptic curve
Does divide the Selmer order?
with a rational 3-torsion point, and let
vary as above. Then there is a constant depending only on
such that divides the Selmer order.
(Clarification request: do the mean the algebraic part of the -function,
or the 3-part of the Selmer group, or the group obtained from a 3-descent?
That affects the value of .)
This has been checked computationally (by checking special values of
Other problems? Poonen: replace 3 by another prime , or replace
by another number field, etc.
If is a field such that all
-acyclic varieties over
have a point, is
topologically generated by one element?
Harmonic analysis and nonabelian torsors:
Takloo-Bighash: Do these give you methods to find rational points?
Hassett: Is there a harmonic analysis argument to count points
on the quintic del Pezzo surface? Wooley: no examples known where
you can combine harmonic analysis with torsors.
Given a family of hypersurfaces, show that ``almost all'' members
of the family satisfy the Hasse Principle in ``interesting''
Colliot-Thélène: maybe the opposite?
Poonen: the family was all hypersurfaces of degree in
with coefficients of height , over some fixed number field .
The problem was: find the proportion of these that satisfy the
Hasse Principle (originally, that have rational points, but we
can find the local points easily).
Wooley and Venkatesh (tentative):
is asymptotic to the product of local densities, at least for
. This is progress for large: for an
, one has HP (using current technology)
. Proof uses (of course) circle method.
de Jong: for cubics in
what do conclusions of this type say about 3-Selmer groups of elliptic curves?
Stoll: given local points, proportion that have rational points (i.e.,
satisfy HP) should be 0.
Mazur: can you put an exponent on that?
Stoll: wait for experimental
Instead of counting points of bounded height on a variety,
count points of bounded height in a Diophantine set. (I.e.,
counting points in a base the fibre above which has a rational point.)
rates of growth can you get? Example: conic bundle over an elliptic
curve. Heuristically, it appears rate of growth can be
whereas for varieties it always turns out to be
Replace Heath-Brown by Arakelov (et al).
. Analogue for
Weak approximation for complex function fields:
Weak approximation over a finite extension of
from Hassett's lecture.
Colliot-Thélène: Known for a connected linear algebraic group
over (reduce to reductive groups immediately; since field has
cohomological dimension 1, one has Borel subgroup, reduce to tori;
reduce to quasi-trivial tori, which are open subsets of affine space).
CT: , where is a subgroup of (not necessarily
normal). Same should go through if is connected (techniques of
Borovoi et al.).
CT: Question: Decide whether weak approximation holds for
where is a finite subgroup of
GL. This seems nontrivial. (de
Jong thinks he can do this; Graber is unsure.) CT says de Jong claims:
Let be an arbitrary smooth projective, geometrically and rationally
connected variety. Then weak approxmation holds. (de Jong: do this by
reducing to characteristic . Not in the dJ-H-S paper.)
CT: there is an Enriques surface over
that has a rational
point but does not satisfy weak approximation.
Yuri Tschinkel remarks that there is progress on Nr 35 due to
Colliot-Thelene/Gille and Madore.
To this Brendan Hasset adds that
Madore has announced a proof of weak approximation
in smooth fibers for a cubic surface over the function field
, where is a curve and is algebraically closed of
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