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Trying to understand the proof of Corollary 2.3 in the following paper,

http://arxiv.org/pdf/1312.3884.pdf

I would like to be able to justify that the root number of the quadratic twist $E^{(-l_0)}$ is $-w_E$, where $w_E$ is the root number of $E$. To spare the readers some time, the set up is the following

$E$ is an elliptic curve over $\mathbb Q$ with conductor $C$ and $K=\mathbb{Q}(\sqrt{-l_0})$, where $l_0>3$ is a prime congruent to $3$ modulo $4$. It is given that $K$ satisfies Heegner hypothesis for $E$, namely every prime factor of $C$ splits in $K$. Does this imply that $C$ is a square or something? If yes, then I can prove my assertion about the root number.

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  • $\begingroup$ As far as I can tell, the argument uses Gross--Zagier: over $K$ one deduces from the fact that Heegner point has infinite order that the $L$-function has a simple zero (this is Gross--Zagier), thus the the sign in the fun'l equation over $K$ is $-1$. This is the product of the signs for each of $E$ and its twist. Earlier in the argument the sign for $E$ was shown to be $1$, and so the sign for the twist must be $-1$. As to why the root number over $K$ is always $-1$, you can look in GZ for a proof. Why should the Heegner hypothesis imply that $C$ is a square? E.g. if $C$ is prime, ... $\endgroup$
    – tracing
    Commented Feb 26, 2015 at 0:41
  • $\begingroup$ ... you can just choose $l_0$ such that $C$ splits in $K$. $\endgroup$
    – tracing
    Commented Feb 26, 2015 at 0:42
  • $\begingroup$ Thank you very much for your response, but as far as I understand, GZ just tells us that the $L'(E/K,1) \neq 0$. To be able to say that $L(E/K,s)$ vanishes at $s=1$, we need to look at something else. Is there something I miss? Thank you again $\endgroup$
    – George Turcas
    Commented Feb 26, 2015 at 7:45
  • $\begingroup$ In the context of GZ, the root number over $K$ is always $-1$. This is how one knowns that the $L$-function vanishes at $s = 1$, and why it makes sense to try to find a formula for its derivative (which is then given by the GZ formula). $\endgroup$
    – tracing
    Commented Feb 27, 2015 at 0:59

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It is a general fact that if $K$ is a number field, $K'/K$ is a quadratic extension, $E \rightarrow \mathrm{Spec}\, K$ is an elliptic curve, and $E' \rightarrow \mathrm{Spec}\, K'$ is its quadratic twist by $K'/K$, then the global root numbers are related by $$w(E_{K'}) = w(E)w(E'),$$ even though the corresponding formula for local root numbers fails (there is an extra term $(-1, K'/K)$ that disappears globally due to the product formula). This may be derived from standard properties of root numbers or, in your situation, from looking at $L$-functions (I can provide you with a reference, if you want).

Once you have the above formula, it remains to note that $w(E_{K'}) = -1$ because there is a single infinite place and all the "bad" finite places come in pairs (due to the Heegner hypothesis), so the corresponding local root numbers kill each other in the product.

See also the related question Root number of a quadratic twist of an elliptic curve, whose answers also settle your present question.

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  • $\begingroup$ Thank you very much! I had the wrong impression that the character $\chi_{-l_0}(p)=-1$ if $p$ splits, and that's why I was desperately trying to see if $C$ happens to be a square. But $\chi_{-l_0}(p)$ in fact $1$ in this case. $\endgroup$
    – George Turcas
    Commented Feb 26, 2015 at 7:52

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