# About the proof of Weil-Langlands theorem

The statement of the theorem is as follows: Let $\rho$ be an irreducible two-dimensional representation of $G_\mathbb{Q}=Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ with Artin conductor $N$. Suppose that $\epsilon = \det(\rho)$ is odd and $\rho$ satisfies condition $(A)$. Let $L(s,\rho) = \sum a_n n^{-s}$. Then the function $f(z) = a_n q^n$ is a normalized newform on $\Gamma_0(N)$ of type $(k, \epsilon)$.

The condition (A) is: There exists a positive integer $M$ such that, for all one-dimensional representation $\chi$ of with Artin conductor prime to $M$, $\Lambda(s, \rho \otimes \chi)$ is entire, in particular, is holomorphic at $s = 0, 1$.

For proving it, I need to consider the Artin root number $W(\rho \otimes \chi)$. In the article ''Local Constants'' by Tate in the Durham symposium (MR0457408), he has proved that $$W(\rho \otimes \sigma) = (-1)^a \det{}_\rho(\mathfrak{f}(\sigma)) \det{}_\sigma (\mathfrak{f}(\rho)) W(\rho)^{\dim(\sigma)} W(\sigma)^{\dim(\rho)}$$ where $\mathfrak{f}$ is the Artin conductor, and $a$ is the number of archimedean places such that both $\det_\rho$ and $\det_\sigma$ are non-trivial (at the decomposition group of that archimedean place).

In the setting of Weil-Langlands, the term $W(\rho \otimes \chi)$ then reads

$$(-1)^a \epsilon(m) \chi(N) W(\rho) W(\chi)^2$$ $$=(-1)^a W(\rho) \epsilon(m) \chi(-N) \frac{G(\chi)}{G(\overline{\chi})}$$ where $m$ is the conductor of $\chi$ and $G(\chi)$ is the Gauss sum.

It now suffices to show that the function $f$ is a newform, for this, I wish to apply theorem 8 of Wen-Ch'ing Winnie Li (Newforms and functional equations Math. Ann. 212 (1975) 285-315 doi:10.1007/BF01344466).

However, the term $(-1)^a$ quite bothers me, which I need it to be $1$ to apply the theorem of Li. The assumption that $\rho$ is odd then implies $\det(\rho)$ is non-trivial at the only archimedean place of $\mathbb{Q}$. For the character $\chi$, it may or may not be non-trivial at the archimedean place, so $a = 0$ or $1$ respectively. But then I can not apply the theorem in the latter case.

So how could I handle this obstruction? What have I missed?

I managed to confuse myself trying to answer your question, but I think if it does have an answer it's likely to be as follows. According to [1], the Artin root number of a Dirichlet character $\chi$ with conductor $f_\chi$ is equal to $$W(\chi) = \frac{\tau(\chi)}{\sqrt{f_\chi}i^a},$$ where $a=0,1$ is as in your question, and $\tau(\chi)$ is the usual Gauss sum $$\tau(\chi) = \sum_{i=1}^f \chi(a)\exp\left(\frac{2\pi i a}{f}\right).$$ Using the relation $\tau(\chi)\tau(\overline{\chi})=\chi(-1)f_\chi$ One has $$W(\chi)^2 = \frac{\tau(\chi)}{{f_\chi}(-1)^a}\frac{\chi(-1) f_\chi}{\tau(\overline{\chi})} = (-1)^a \chi(-1)\frac{\tau(\chi)}{\tau(\overline{\chi})},$$ whereas you're substituting $\chi(-1)\frac{\tau(\chi)}{\tau(\overline{\chi})}$ for $W(\chi)^2$ in the calculation of your question. The extra $(-1)^a$ factor would then cancel out with itself.
[1]: Ehud de Shalit, Artin $L$-functions In: Bernstein J., Gelbart S. (eds) An Introduction to the Langlands Program. Birkhäuser, Boston, MA doi:10.1007/978-0-8176-8226-2_4, (free version, dvi file)
• Thanks for the help. It seems that I have ignored this fact. Somehow I have been assuming that it is simply $W(\chi) = \tau(\chi)/ \sqrt{f}$. – Ting-Han Huang May 16 at 6:28