Base change HMF #1975
Comments
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I agree that we need precise definitions of both 'is_base_change' and 'is_CM'. A page such as the one above has blank knowls for both terms (they link to nonexistent mf.cm and mf.base_change). the cm should consistent with the existing knowl mf.elliptic.cm_form. So a very easy first step would be for someone to add text to those knowls: easy in the technical sense, of course getting the definition right might not be so easy. |
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I have not properly investigated, but at a glance it seems that the form could be a base-change from the subfield Q(sqrt(-5)), while you may have been thinking about base-change from Q only. |
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The Galois closure of the base field is this field http://beta.lmfdb.org/NumberField/8.8.700945700625.1 which does not have Q(sqrt(-5)) as subfield, but Q(sqrt(5)) (the Galois closure is totally real too). I think I checked the forms over Q(sqrt(5)) in the right level range (level norm 1 over the quartic field, means that the norm of the level for the form over Q(sqrt(5)) should only be divisible by 3,5,61 which are the factors in the discriminant) and none was giving the right base change, but I can be wrong. |
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Above p=11 tere are two primes of degree 1 (norm 11) both with eigenvalue 0, and one of degree 2 (norm 121) with eigenvalue +22. But base-changing locally from GF(11) to GF(11^2) would take eigenvalue a_11 to (a_11)^2-2*11 = -22 in this case. That would be true whether the b.c. was coming from Q or the subfield Q(sqrt(5)), in which 11 splits. Secondly, the 4 primes of norm 19 do not have the same eigenvalue except up to sign. I also looked at p=29 and p=31 which split as for 11; the eigenvalues for norm 29 are both 0 but that for norm 29^2 is 0+229=58; the eigenvalues for norm 31 are -8 and that for norm 31^2 is 2=(-8)^2-231. This suggests to me that the form is a twist of a base-change, possibly of a form from Q. But none of the related elliptic curves has rational j-invariant. |
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p=2 is interessting: unramified and inert, with eigenvalue -7, so the associated Frobenius polynomial is X^2+7X+16 with roots in Q(sqrt(-15)) which are not fourth powers (there are no elements of norm 2 in that field since 2 splits into nonprincipal primes). That means that our form cannot be a base change from Q of a form with rational eigenvalues. It could be b.c. from Q ofa form with eigenvalue sqrt(5) or sqrt(3) if my calculations are correct. And it could be a b.c. from Q(sqrt(5)) since (1+sqrt(-15))/2 and its conjugate (or their negatives) satisfy X^2-X+4 (or X^2+X+4), a polynomial whose squares satisfy X^2+7X+16. So over Q(sqrt(5)) we would be looking for a form with eigenvalue +1 or -1 at the prime above 2. If b.c. from Q (even up to twist) what would the level be? Presumably only divisible by the ramified primes 3, 5, 61. |
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@sanni85 Yes, I meant Q(sqrt(5)), not -5. It is a subfield of the field considered here, the Galois closure is irrelevant. |
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Here's a possible definition of base change to account for the base field being non-Galois. For simplicity assume that we want to define "base change from Q" and that we are talking about weight 2 (over Q, with parallel weight 2 over K). Ignore primes dividing the level or ramified in K/Q. For all other primes p, the form over Q will have Hecke eigenvalue a_p and char.poly. of Frobenius in the associated Galois repn X^2-a_p*X+p with roots alpha, beta (say). Then for every prime P of K above p, with degree f (so norm p^f) the Frobenius poly. will be X^2-(alpha^f+beta^f)*X+p^f, so the eigenvalue will be a_P=alpha^f+beta^f. Now if we have several P|p with different f's we can test whether they are consistent with this recipe (details omitted for now). Example: [K:Q]=3 non-Galois with p splitting as P_1*P_2 where P_j has degree j. Then a_P_1=a_p while a_P_2=a_p^2-2p. However I don't know a theorem which says, in the situation of the example that if this holds for all primes (together with equal a_P's when p splits), that the form actually is a base-change. I do know of such a theorem in the case where K/Q is cyclic (Galois) only. |
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Hi all! Sorry for any confusion caused--I have on my list to add knowls for these HMF properties, but you can probably imagine that is low priority. I think it would be worse than the empty set to have something wrong or slapdash. When we write "no" in CM or base change, we mean we did a computation and provably it is not. When we write "yes", we mean "looks likely", but it is not proven. I was told not to write that as "yes?" for consistency. Steve and I briefly explain the algorithms employed here: Up to isogeny, we find a CM elliptic curve (http://beta.lmfdb.org/EllipticCurve/4.4.13725.1/1.1/a/8) attached to the HMF and its j-invariant belongs to F_0. So the HMF over F is the twist of the base change of an HMF over F_0. So I guess we'd need to look for an HMF of parallel weight 2 and level equal to the conductor of the abelian extension F/F_0 with character either trivial or the quadratic character of F/F_0. (@JohnCremona, you have some experience with this situation, right? You take a classical modular form with quadratic character and base change it to the corresponding quadratic field. Stuff happens. I don't think it's so very different here with an HMF, it just starts life over F_0 = QQ(sqrt(5)) instead of over QQ.) I don't see a good candidate for trivial character (someone could double check this), so we probably need an HMF with character. If it's very important to see this explicitly, I can run some hacky code to compute the space with character to see what turns up. In any case, I think this issue should be closed: I would be sad if our notion of base change did not capture this kind of situation. Two final comments, not really relevant to this example. First, if there's a base change, it needn't come from an elliptic curve over F_0. As @JohnCremona explains for trivial character, to get the base change coefficients from F_0 to F, start with a form over F_0 with character chi and coefficients a_{pp_0}; then for pp above pp_0 not dividing the level NN_0, Second, a giant red flag to wave: we don't claim anything about the character of the forms in the LMFDB, and when the narrow class number is not 1, the forms presented can have nontrivial character coming from this class group. I don't think that's in play here, again because an elliptic curve was found, but be warned. |
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First a comment on the "red flag" -- up to now I have assumed that the forms had trivial character, but I understand exactly what you are saying -- and it means that if tere are some forms with nontrivial character, there will not be an elliptic curve, so I should not be looking for one! Perhaps this could explain some of the curves I cannot find. Secondly, I really think that as a matter of urgency we define what we mean by base-change in the LMFDB as it is at present, using the definition you @jvoight helpfully wrote out for us, since that is not obvious and neither I nor @sanni85 guessed it quite right. Writing that knowl will only take a few minutes! |
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Thanks to @JohnCremona and @jvoight for the explanations and the computations. When I think about base change I think about Galois representations and L-function properties (now you see why I was considering the Galois closure of the field). I did not think about forms with characters, and hence my confusion (I was just looking for examples of elliptic curves with everywhere good reduction which were not base change nor Q-curves and I clicked on the HMF, seeing that the form was a base change surprised me). |
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@sanni85 even if you think in terms of Galois representations, the Galois closure has nothing to do with the description of base-change. If L/K is an arbitrary extension of number fields, then you can restrict a representation of G_K to G_L. That is the base-change of the Galois representation, and it does not need the extension to be Galois. I have looked for a potential elliptic curve over Q(sqrt 5) with the correct ramification properties and a few Frobenius traces using Steve Donnelly's search function, and did not find any. This suggests that the HNF over Q(sqrt 5) that we are looking for may have a nontrivial character, even though there is no guarantee that the search was complete. |
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Is this correct or too naive?
In terms of L-functions (and using "F" for L-functions because
you already used "L"!) F_2 is a base change just means that
F_2 = F_1 x zeta_{L/K}
where F_1 is an L-function and zeta_{L/K} is the zeta-function of the
extension L/K, and x is Rankin-Selberg convolution?
…On Wed, 1 Mar 2017, AurelPage wrote:
@sanni85 even if you think in terms of Galois representations, the Galois closure has nothing to
do with the description of base-change. If L/K is an arbitrary extension of number fields, then
you can restrict a representation of G_K to G_L. That is the base-change of the Galois
representation, and it does not need the extension to be Galois.
I have looked for a potential elliptic curve over Q(sqrt 5) with the correct ramification
properties and a few Frobenius traces using Steve Donnelly's search function, and did not find
any. This suggests that the HNF over Q(sqrt 5) that we are looking for may have a nontrivial
character, even though there is no guarantee that the search was complete.
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I'll have to think further about @davidfarmer 's suggestion. It may be compatible with the Galois-representation-theoretic viewpoint, since the polynomials which appear in the Euler product expansion are the same as the characteristic polynomials of Frobenius I was talking about, after you group together primes of the number field above each rational prime. |
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@davidfarmer It sounds like a reasonable guess. But how do you define the bad factors of the convolution of two L-functions? |
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Short answer: assuming standard conjectures, the bad factors
exist and are unique, but there (usually) is not a simple recipe
only using the data of the L-function.
Longer answer:
Given two L-functions, a simple recipe tells you the good factors
of the convolution, and another simple recipe tells you the
Gamma-factors.
However, it is not even a theorem that the convolution is an L-function!
That is, you don't know that it has an analytic continuation,
and you also don't know that it satisfies a functional equation.
(If the original L-functions are known to be modular,
then you do have those properties.)
But, if you have those properties, then by the strong multiplicity
one theorem for L-functions, the conductor and bad factors are uniquely
determined.
If you know enough about the underlying automorphic object, then
you can use it to find the bad factors of the convolution. Only in
a few cases can you do that calculation directly from the L-function.
…On Wed, 1 Mar 2017, AurelPage wrote:
@davidfarmer It sounds like a reasonable guess. But how do you define the bad factors of the
convolution of two L-functions?
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@davidfarmer OK, thanks for the explanation! This is what I expected, but I wanted to be sure of what you meant. |
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@davidfarmer, like @JohnCremona, I'd be curious to know what your definition of zeta_{K/F}(s) is. I guess it can't be zeta_{K/F}(s) = zeta_K(s)/zeta_F(s), since then the degree of zeta_{K/F}(s) is equal to [K:QQ] - [F:QQ] <> [K:F], for example if F = QQ and K/F is quadratic, you'd be getting the twist by the quadratic character of K/F. On the other hand, if F = QQ, then you could take zeta_{K/QQ}(s) = zeta_K(s) the Dedekind zeta function, and then I think my definition agrees with yours (which I only gave in the quadratic case, but extends in a natural way). But for this to work more generally, you need to make a degree [K:F] L-function. Just so this doesn't get lost: it's completely clear what "base change" means when we're talking about a geometric object or a Galois representation--and in these contexts, there is nothing conjectural. Putting our "motivic hat"s on, the new Euler factors of the base change are given by counting points on the base change, and this can be expressed in terms of symmetric functions relating the trace to appropriate powers of the Frobenius. For Galois representations, you basically end up with the same thing. But converting this procedure into something about modular forms or L-functions is where it gets tricky, and I think any good answer must be consistent with the motivic one. |
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Also, what is the content of this issue now? Is it that we need a knowl explaining base change, and we're discussing what that knowl will say? We could put in a placeholder from my paper with Steve (a_{sigma(nn)} = a_{nn} for all sigma in Gal(F/F_0) for some proper subfield F_0), and discuss this at length in June when we are together in Warwick to hash out the mathematics in detail? |
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When the base field is Q, then by zeta_{K/Q} I mean zeta_K,
the Dedekind zeta function, and I am pretty sure everything is
as I described. In the quadratic case the "base change"
L-function is the original L-function times itself twisted by
a character.
I don't know what happens when the base field is not Q, but
I see your point about degrees. I assumed the more algebraically
minded among us would already know about zeta_{K/F}.
…On Fri, 3 Mar 2017, jvoight wrote:
@davidfarmer, like @JohnCremona, I'd be curious to know what your definition of zeta_{K/F}(s) is.
I guess it can't be zeta_{K/F}(s) = zeta_K(s)/zeta_F(s), since then the degree of zeta_{K/F}(s) is
equal to [K:QQ] - [F:QQ] <> [K:F], for example if F = QQ and K/F is quadratic, you'd be getting
the twist by the quadratic character of K/F. On the other hand, if F = QQ, then you could take
zeta_{K/QQ}(s) = zeta_K(s) the Dedekind zeta function, and then I think my definition agrees with
yours (which I only gave in the quadratic case, but extends in a natural way, and is modelled
after thinking of the Euler factors as being given by counting points on the base change of an
arithmetic object). But for this to work more generally, you need to make a degree [K:F]
L-function.
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Isn't there a way of cheating in this tensor product construction? If we see both the L-function of the object over K and zeta_L as Euler products over the primes of K rather than those of QQ, then the tensor product using the usual definition but for Euler products over the primes of K should give the right answer. In particular, as an L-function over K, zeta_L has the right degree [L:K]. |
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@AurelPage: Yes, if you can interpret Rankin-Selberg convolution "over F", then that would do it, and you should do this formal operation with the Dedekind zeta function zeta_K. I looked but couldn't find a construction of this sort before, but maybe it's probably a special case of something more general? (I don't know how to do this with a classical Rankin-Selberg, except over QQ.) |
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@jvoight @davidfarmer Actually, I am starting to doubt that it is possible in general. Here is the reason. Let's restrict to Artin L-functions for simplicity (by replacing Artin representations with Langlands parameters and adapting the arguments this should say something in general, and this case is enough to find a counter-example). Here is my problem: if the construction suggested by David exists, then the L-function of the base-change from K to L of an object X (defined over K) is uniquely determined by the L-function of X. That does not sound right to me. If one looks at Artin representations rho : G_K -> GL(V) then by Artin formalism the L-function of rho caracterises exactly Ind_{G_Q/G_K}(rho), but it may not characterise rho itself as a representation of G_K. David's construction would imply that for every Artin representation rho of G_K, Ind_{G_Q/G_L}(Res_{G_K/G_L}rho) is uniquely determined by Ind_{G_Q/G_K}(rho). At this point this is only a question about representations of finite groups! Let K<H<G be finite groups. If we can find two representations rho and chi of H such that Ind_{G/H}rho and Ind_{G/H}chi are isomorphic but Ind_{G/K}Res_{H/K}rho and Ind_{G/K}Res_{H/K}chi are not isomorphic, then (assuming the Galois inverse problem for G) we have a counter-example. I am not sure whether such an example exists (need to think more about it), so I don't know what to think about David's question now. |
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I hope we'll have a chance to talk about this tomorrow, since we left this hanging. I can imagine a world where we lose information in the base change of an object from F to a finite extension K, so I don't see a contradiction; and I don't see any problems with definitions of base change, only what theorems could be proven about it. In any case, what are we trying to decide in this issue anymore? As far as I can tell, we could close the issue if we wrote a knowl. |
AndrewVSutherland
added
the
critical
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@jvoight Agreed. Knowls for CM and base change are both still missing, once they are added I think we can close this. |
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@jvoight @JohnCremona Can we get these knowls written tomorrow or Friday so that we can close this? |
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OK, I added I think this duplicates We need the more general definitions for HMFs, so maybe we keep them both as long as they refer to one another? |
http://beta.lmfdb.org/ModularForm/GL2/TotallyReal/4.4.13725.1/holomorphic/4.4.13725.1-1.1-a
on this page, it is claimed that the orbit is a base change, but it should not be (there is no definition of base change given, maybe I am misinterpreting it).
The isogeny class of elliptic curves corresponding (http://beta.lmfdb.org/EllipticCurve/4.4.13725.1/1.1/a/)
contains curves which are not Q-curves nor base change, so the form should not be base change. Another argument it that If the form was base change then there should exists certain forms over Q(sqrt 5) which do not exist.
I think the problem comes from the fact that the base field (http://beta.lmfdb.org/NumberField/4.4.13725.1) is not Galois over Q.
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