Functional equation of twisted triple product L-function Let $\mathbb{E}=E_1\times E_2\times E_3$ denote the product of three elliptic curves over $\mathbb{Q}$ of prime level $p$ and consider  the $p$-adic Galois representation $$V_p(\mathbb{E})=H^1_{et}(E_{1/\bar{\mathbb{Q}}}, \mathbb{Q}_p)\otimes H^1_{et}(E_{2/\bar{\mathbb{Q}}}, \mathbb{Q}_p)\otimes H^1_{et}(E_{3/\bar{\mathbb{Q}}}, \mathbb{Q}_p).$$ Denote by $L(\mathbb{E}, s)=L(V_p(\mathbb{E}),s)$ the associated triple product $L$-function. It has a functional equation centered at $s=2$ with global sign equal to $a_p(E_1)a_p(E_2)a_p(E_2)\in \{ \pm 1 \}$ (cf. Gross-Kudla '92). Here, $a_p(E_i)$ denotes the $p$-th Fourier coefficient of the weight 2 normalized newform of level $\Gamma_0(p)$ associated to $E_i$ by modularity. 
Let $\chi$ be a Dirichlet character modulo $p$ and denote by $L(\mathbb{E}\otimes \chi, s)$ the $L$-function attached to the Galois representation $V_p(\mathbb{E})\otimes \chi$. My question is: what is the functional equation of $L(\mathbb{E}\otimes \chi, s)$ and what is its global sign?   
Thank you in advance for any help.
 A: There is a functional equation for $L(\mathbb{E} \otimes \chi, s)$, but it relates $L(\mathbb{E} \otimes \chi, s)$ to $L(\mathbb{E} \otimes \bar\chi, 4-s)$. If $\chi$ is not trivial or quadratic, then $L(\mathbb{E} \otimes \chi, s)$ and $L(\mathbb{E} \otimes \bar\chi, s)$ are different functions, so you cannot use this to deduce anything in particular about vanishing at the central value. The Langlands $\varepsilon$-factor is still defined, and it is a complex number of absolute value 1, but it isn't $\pm 1$, so you can't reasonably call it a "sign"; at a guess it is probably something like $\tau(\chi)^4 / p^{2n}$, where $p^n$ is the conductor of $\chi$ and $\tau(\chi)$ is the Gauss sum.
(You can try to cheat by considering the function $M(s) = L(E \otimes \chi, s) L(E \otimes \bar\chi, s)$, which does satisfy a functional equation relating $M(s)$ and $M(4-s)$, but unfortunately the order of vanishing of $M(s)$ at $s = 2$ is automatically even anyway, since $L(E \otimes \bar\chi, s) = \overline{L(E \otimes \chi, \overline{s})}$, so both factors have the same order of vanishing. So although $M(s)$ does have a functional equation, you can't get any nontrivial vanishing information out of it.)
The moral here is that the whole story of L-values vanishing because of "sign" phenomena is specific to self-dual settings.
