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Can someone help me find the Euler product associated to this newform? I have a product of eta functions:

$$ \eta(z)^2 \eta(11z)^2 = q \prod_{n=1}^\infty (1-q^n)^2 (1 - q^{11n})^2 = q - 2q^2 - q^3 + 2q^4 + q^5 + 2 q^6 - 2 q^7 - 2q^9 + \text{O}(q^{10})$$

is a modular form of weight $2$ over $\Gamma_0(11)$. What is a presentation of that group? What are the generators there? I think $$ \Gamma_0(11) = \langle z \mapsto z + 11, z \mapsto - \frac{1}{z} \rangle $$ so these are like continued fractions all of whose digits are multiples of 11? In any case, I did verify multiplicative in some cases: $$ a_9 \times a_5 = (-2) \times (+1) = -2 = a_{45} $$ However, in certain other cases, in particular prime powers, the coefficients were not multiplicative: $$ \quad \quad \; a_2 \times a_2 \times a_2 = (-2)\times (-2)\times (-2) = -8 \neq 0 = a_8 $$ All the number theory books prove these theorems abstractly, but I am failing to verify them for specific cases. $$ L(f,s) = 1^s -2 \times 2^s - 1 \times 3^s + 2 \times 4^s + 1 \times 5^s + 2 \times 6^s + \dots = \left( 1 - \frac{a_{11}}{11^s} \right)^{-1} \prod_{p \neq 11} \left( 1 - \frac{a_p}{p^s} \right)^{-1} $$ I would like to verify the Euler product for the L-function.

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    $\begingroup$ Multiplicativity in this game always boils down to proving it's a Hecke eigenform. In this case, this will follow from the space of weight 2 forms for $\Gamma_0(11)$ being one-dimensional. I should add that the L-function in this case will be that associated to the elliptic curve of conductor 11. $\endgroup$ – Lord Shark the Unknown Apr 15 '17 at 21:53
  • $\begingroup$ @LordShark did you search on lmfdb.org or what ? $\endgroup$ – reuns Apr 15 '17 at 22:11
  • $\begingroup$ my link is lmfdb $\endgroup$ – cactus314 Apr 15 '17 at 22:16
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    $\begingroup$ You are confusing multiplicative : $a_{nm} = a_n a_m$ when $gcd(n,m)=1$ with completely multiplicative $a_{nm} = a_n a_m$ for every $n,m$ $\endgroup$ – reuns Apr 15 '17 at 22:16
  • $\begingroup$ In the case of Hecke eigenforms (the only modular forms having multiplicative coefficients, once normalized) there is a relation more constrained than just being multiplicative : en.wikipedia.org/wiki/Hecke_operator#Explicit_formula $\endgroup$ – reuns Apr 15 '17 at 22:18

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