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Let $f$ be a primitive holomorphic cusp form of weight $k$, level $N$ and nebentypus $\chi$, with its $L$-function $L(s,f)=\displaystyle\sum_{n\geq1}\lambda_f(n)n^{-s}$ for $\mathrm{Re}(s)>1$. Let $\alpha, \beta$ be the Satake parameters such that $$L(s, f)=\prod_{p}\left(1-\frac{\alpha(p)}{p^{s}}\right)^{-1}\left(1-\frac{\beta(p)}{p^{s}}\right)^{-1}.$$ One can then define the symmetric square $L$-function to be $$L(s,\mathrm{sym}^2f)=L(s,f\otimes f)L(s,\chi)^{-1}.$$ Such a $L$-function has Euler product $L(s,\mathrm{sym}^2f)=\displaystyle\prod_p L_p(s,\mathrm{sym}^2f)$, and we know that for $p$ not dividing $N$, $$L_p(s,\mathrm{sym}^2f)=\left(1-\frac{\alpha(p)^2}{p^{s}}\right)^{-1}\left(1-\frac{\chi(p)}{p^s}\right)^{-1}\left(1-\frac{\beta(p)^2}{p^s}\right)^{-1}$$ $$=\left(1-\lambda_{f}\left(p^{2}\right) p^{-s}+\lambda_{f}\left(p^{2}\right)\chi(p) p^{-2 s}-\chi(p)^3p^{-3 s}\right)^{-1}.$$

If $N$ is square free, we know that $$L_p(s,\mathrm{sym}^2f)=\left(1-\frac{1}{p^{s+1}}\right)^{-1}$$ and $$L(s,\mathrm{sym}^2f)=L(2s,\chi^2)\sum_{n\geq1}\lambda_f(n^2)n^{-s}.$$

My questions are: Are the two equalities above still true when $N$ is not square free? If it is not:

  1. Can I still write $L(s,\mathrm{sym}^2f)$ in a simple manner using the coefficients $\lambda(n^2)$?

  2. Does $\displaystyle\prod_p L_p(s,\mathrm{sym}^2f)$ still define a $L$-function with $L_p(s,\mathrm{sym}^2f)=\left(1-\frac{1}{p^{s+1}}\right)^{-1}$ for $p|N$?

  3. If I apply the approximate functional equation (Thm 5.3 in Iwaniec-Kowalski's book), do I still get something that looks like $$L\left(s,\mathrm{sym}^2f\right)\ll c(N)\mathop{\sum\sum}_{n,m}\lambda_f(n^2)\chi^2(m)n^{-s}m^{-2s}V_s\left(\frac{nm^2}{XN}\right)+ \text{ something of similar structure },$$ with some $c(N)$ depending only on the primes dividing $N$?

Added: It appears that the answer may be long for full generality. What if we restrict our attention to trivial nebentypus $\chi$?

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    $\begingroup$ Are you sure you want to be working with $L(s,\operatorname{sym}^2 f)$ and not $L(s,\operatorname{ad} f)$? These agree when $\chi$ is principal but not in general, and the latter is what is usually more relevant for analytic applications. $\endgroup$ Feb 22, 2022 at 3:37
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    $\begingroup$ In any case, the answer to (1) is no in general. See e.g. my answer here (mathoverflow.net/questions/410459/…). The key point is at the end: Gelbart and Jacquet tell you what the local factors $L_p(s,\operatorname{sym}^2 f)$ or $L_p(s,\operatorname{ad} f)$ should be, and these depend delicately on the local component of the underlying automorphic representation. This cannot be determined alone by the conductor and nebentypus. $\endgroup$ Feb 22, 2022 at 3:39
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    $\begingroup$ (2) is also not true in general, since these need not be the correct local $L$-functions, so you will get something that isn't quite correct at certain primes. So the functional equation won't quite be right. Think of this as a generalisation of how imprimitive Dirichlet $L$-functions don't have "correct" functional equations. $\endgroup$ Feb 22, 2022 at 3:41
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    $\begingroup$ (3) is wrong even if $N$ is squarefree: the correct version of the approximate functional equation for this is $$L(s,\operatorname{sym}^2 f) = \sum_{n,m = 1}^{\infty} \frac{\lambda_f(n^2) \chi^2(m)}{n^s m^{2s}} V_s\left(\frac{nm^2}{XN}\right) + \cdots.$$ You really cannot just neglect the presence of $L(2s,\chi^2)$ in the definition of $L(s,\operatorname{sym}^2 f)$ as a Dirichlet series when using approximate functional equations! $\endgroup$ Feb 22, 2022 at 3:44
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    $\begingroup$ Nope, you still run into the problem that if $N$ is not squarefree, it can be the case that $\lambda_f(n^2) = 0$ whenever $n \mid N^{\infty}$ but that $L_p(s,\operatorname{ad} f)$ is not uniquely determined by $N$. $\endgroup$ Feb 22, 2022 at 14:29

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