gp in Sage

[Surendran Karippadath]

Hi,

     I evaluated the j-invariant in Pari/gp In SageMathCell

? \p 50

? ellj(sqrt(163.0)*I)
%1 = 68925893036109279891085639286944512.000000000163739

Furthermore the Cube-root of the j-invariant I obtained

? (ellj(sqrt(163.0)*I))^(1/3)
%2 = 410009702400.00077461269365317226812447191214259043

Is it possible to check in Sage with High Precision if the values are Integers.

Thanking you in advance

[Dima Pasechnik]

On Sun, Aug 30, 2020 at 5:50 AM Surendran Karippadath
<kksin...@gmail.com> wrote:
> I evaluated the j-invariant in Pari/gp In SageMathCell
> ? \p 50
> ? ellj(sqrt(163.0)*I)
> %1 = 68925893036109279891085639286944512.000000000163739

Sage has this function too (it calls Pari, so that's not an
independent confirmation that this number is (not) an integer:
sage: elliptic_j(sqrt(163)*I,prec=500)
6.89258930361092798910856392869445120000000001637386442092346075751855217523117650690239250072955532985645916831850173541132959651401661828116253839333e34

The output in Pari is a bit easier to read:

? \p 500
? ellj(sqrt(163)*I)
%4 = 68925893036109279891085639286944512.00000000016373864420923460757518552[...]

Is it one of these "almost integers" (unless it's a bug, and this
number must be an integer, I don't know - number theorists, please
step forward!), such as
? \p 500
? exp(sqrt(163)*Pi)
%3 = 262537412640768743.99999999999925007259719818568887935385633733699086270[...]

Or a bug in Pari/GP ?

>
> Furthermore the Cube-root of the j-invariant I obtained
> ? (ellj(sqrt(163.0)*I))^(1/3)
> %2 = 410009702400.00077461269365317226812447191214259043

the closest integer to (ellj(sqrt(163.0)*I))^(1/3) is 410009702400,
and so you can check
that its cube is quite far from ellj(sqrt(163)*I)

>
> Is it possible to check in Sage with High Precision if the values are Integers.
>
> Thanking you in advance
>

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[Dima Pasechnik]

On Sun, Aug 30, 2020 at 9:24 AM Dima Pasechnik <dim...@gmail.com> wrote:
>
> On Sun, Aug 30, 2020 at 5:50 AM Surendran Karippadath
> <kksin...@gmail.com> wrote:
> > I evaluated the j-invariant in Pari/gp In SageMathCell
> > ? \p 50
> > ? ellj(sqrt(163.0)*I)
> > %1 = 68925893036109279891085639286944512.000000000163739
>
> Sage has this function too (it calls Pari, so that's not an
> independent confirmation that this number is (not) an integer:
> sage: elliptic_j(sqrt(163)*I,prec=500)
> 6.89258930361092798910856392869445120000000001637386442092346075751855217523117650690239250072955532985645916831850173541132959651401661828116253839333e34
>
> The output in Pari is a bit easier to read:
>
> ? \p 500
> ? ellj(sqrt(163)*I)
> %4 = 68925893036109279891085639286944512.00000000016373864420923460757518552[...]
>
> Is it one of these "almost integers" (unless it's a bug, and this
> number must be an integer, I don't know - number theorists, please
> step forward!), such as
> ? \p 500
> ? exp(sqrt(163)*Pi)
> %3 = 262537412640768743.99999999999925007259719818568887935385633733699086270[...]
>
> Or a bug in Pari/GP ?

If I read the discussion after Cor. 42 in
http://people.maths.ox.ac.uk/greenbj/papers/ramanujanconstant.pdf
right, this is not an integer.
Indeed, there are 9 imaginary quadratic extensions of Q for which one
gets integer j-invariant, one of them
Q[sqrt(-163)], but as 163 mod 4 = 3, one has to compute its j-invariant as
ellj((1+sqrt(163)*I)/2)
getting -262537412640768000

[Surendran Karippadath]

Yes, I knew the point regarding

>>

ndeed, there are 9 imaginary quadratic extensions of Q for which one
gets integer j-invariant, one of them
Q[sqrt(-163)], but as 163 mod 4 = 3, one has to compute its j-invariant as
ellj((1+sqrt(163)*I)/2)
getting -262537412640768000

<<

However on the boundary of the fundamental domain, my calculation shows only j-invariants ( positive) which are perfect cubes as

Q(sqrt(-1)).....12^3

Q(sqrt(-2))......20^3

Q(sqrt(-4))......66^3

Q(sqrt(-7))......255^3

and the above almost integer. Are there any others on the Imaginary axis?

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[John Cremona]

On Wednesday, September 2, 2020 at 5:00:07 PM UTC+1 kks wrote:

Yes, I knew the point regarding

>>

ndeed, there are 9 imaginary quadratic extensions of Q for which one
gets integer j-invariant, one of them
Q[sqrt(-163)], but as 163 mod 4 = 3, one has to compute its j-invariant as
ellj((1+sqrt(163)*I)/2)
getting -262537412640768000

<<

However on the boundary of the fundamental domain, my calculation shows only j-invariants ( positive) which are perfect cubes as

Q(sqrt(-1)).....12^3

Q(sqrt(-2))......20^3

Q(sqrt(-4))......66^3

Q(sqrt(-7))......255^3

and the above almost integer. Are there any others on the Imaginary axis?

No. cm_j_invariants_and_orders(QQ) gives all 13 imaginry quadratic orders of class number 1, from which you can recover the 13 associated imaginary discriminants D.  Most of these are congruent to 1 mod 4 so the j-value is j((1+sqrt(d))/2) , only those which are 0 mod 4 are on the imaginary axis with values j(sqrt(D)/2) as in your list.

THere is a big theory of complex multiplcation behind these facts, but I don'y think that "gp in Sage" is an accurate sub                       ject line for discussion about that.

John Cremona

[Surendran Karippadath]

Thank you all for the temporary solution to my problem arising from an ambitious effort to understand Table 12.12 in David Cox's book Primes of the form x^2+ny^2. As Prof Cremona has stated the existence of only four perfect cubes on the imaginary axis is  to be discussed under an appropriate topic head.

Sincerely

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