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I use Magma to calculate the L-value, yields

E:=EllipticCurve([1, -1, 1, -1, 0]); E; Evaluate(LSeries(E),1),RealPeriod(E),Evaluate(LSeries(E),1)/RealPeriod(E);

Elliptic Curve defined by y^2 + x*y + y = x^3 - x^2 - x over Rational Field 0.386769938387780043302394751243 3.09415950710224034641915800995 0.125000000000000000000000000000

$#torsionsubgroup = 4, c_{17}(E)=1.$

But the strong BSD predicts that

$L(E,1)/\Omega_{\infty}$= $(#Sha(E)/#tor(E)^2)*c_{17}(E)$

We will get $L(E,1)/\Omega_{\infty}=1/16$, not $1/8$. Why does that happen? Thanks a lot.

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By the way, how could I compute Sha(E) and the Tamagawa factor c_p(E) by Magma? –  Carl Aug 16 '13 at 11:29
    
I get 1/8 in sage too using the equivalent of your commands. –  joro Aug 16 '13 at 11:30
    
What is going on with the BSD? I am deeply puzzled. –  Carl Aug 16 '13 at 11:35
    
For lseries you need to import eulerprod.py. The code is: sage: e0=EllipticCurve(QQ,[1, -1, 1, -1, 0]);pl=e0.period_lattice();l0=LSeries(e0);l0(1) / pl.real_period() 0.125000000000000 –  joro Aug 16 '13 at 11:38
10  
For elliptic curves with positive discriminant, $E({\mathbb R})$ has two connected components, and the BSD period is twice the real period. If you divide by that, you get $1/16$. In Magma you can use ConjecturalRegulator(E) to get the conjectural order of sha times the regulator - this is 1 in this case. (It takes the correct BSD period, and divides by that and the product of Tamagawa numbers and multiplies by torsion^2) –  Tim Dokchitser Aug 16 '13 at 12:14

2 Answers 2

up vote 26 down vote accepted

To expand my comment, there are at least 3 subtle ways to get BSD wrong:

1) The BSD period over ${\mathbb Q}$ is the real period $\Omega_\infty$ when $E({\mathbb R})$ is connected ($\Delta(E)<0$) and $2\Omega_\infty$ when it has two connected components ($\Delta(E)>0$). The same thing happens over number fields, at every real place. So in your example you have to divide the $L$-value by $2\Omega_\infty$ to get $1/16$.

Another way (alternative) of phrasing this: is that the Tamagawa number at an Archimedean place is $2$ if real and split ($\Delta>0$), and $1$ otherwise. It is less easier to forget to include it then, and you can always use $\Omega_\infty$. Magma does not have Tamagawa numbers at infinite places directly.

The other two concern BSD over number fields:

2) The height pairing in BSD depends on the ground field. For instance, $E=37A1$ has $E({\mathbb Q})={\mathbb Z}\cdot P$ and $E({\mathbb Q(i)})={\mathbb Z}\cdot P$, where $P$ is the same point $(0,0)$. The regulator is $\approx 0.051$ over ${\mathbb Q}$ but $\approx 2\cdot 0.051=0.102$ over ${\mathbb Q(i)}$; the factor $2$ is $[{\mathbb Q(i)}:{\mathbb Q}]$.

3) Over ${\mathbb Q}$ there is this luxury of having a global everywhere minimal model, so there is a canonical differential to integrate. Over number fields you cannot do this, so one usually takes any invariant differential and introduces a correction term that measures its failure to be minimal at all primes. The point is that if you start with a curve over ${\mathbb Q}$ with additive reduction at $p$ and go up to a number field $K$ where $p$ ramifies, e.g. $50A3$ over $K={\mathbb Q}(\sqrt 5)$, the minimal model might stop being minimal, and this correction factor comes in for BSD over $K$.

The functions ConjecturalRegulator and ConjecturalSha in Magma take care of these normalizations - it's actually quite nice to experiment with them.

Hope this helps.

P.S. You would not believe how many times each of these mistakes was made!

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Tim's answer is great, but I want to mention one other place where people have lost a power of 2. The canonical height is often defined relative to the divisor (O), as I do in my books. So it is given by $$ \hat h(P) = \frac12 \lim_{n\to\infty} 4^{-n} h\bigl(x([2^n]P)\bigr). $$ Here the $\frac12$ is inserted because $x$ has a double pole at $\infty$. However, in computing the height regulator for BSD, one should not use the $\frac12$, i.e., one should compute heights relative to the divisor $2(O)$.

Why, one might ask, use a weird divisor like $2(O)$. The answer is that when BSD is properly formulated for abelian varieties, it uses the height pairing $A(K)\times \hat A(K)$ that pairs points of $A$ with points on its dual, and the pairing is done relative to the Poincare divisor on $A\times \hat A$. If one traces through the definitions and identifies an elliptic curve with its dual, the Ponicare divisor on $E\times E$ is $(O)\times E + E\times (O)$, which eventually shows that one should use the height on $E$ relative to $2(O)$.

Note that this means that if you compute BSD using the wrong height on a curve of rank $r$, then your answer will be off by a factor of $2^r$.

(I learned about this potential error from Dick Gross many years ago.)

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