I have a straightforward question. Let $f$ be a holomorphic cusp form of weight $k$, level $N$, and nebentypus $\chi$ that is new in the sense of Atkin-Lehner theory. Write its Fourier expansion at $\infty$ as
$$f(z) = \sum\_{n=1}^\infty \lambda\_f(n)n^{(k-1)/2}e(nz)$$
and form the $L$-function $L(f,s)$ by the Dirichlet series
$$L(f,s)=\sum\_{n=1}^\infty \frac{\lambda\_f(n)}{n^s}$$
for $\Re(s)>1$. Then $L(f,s)$ can be continued to an entire function on $\mathbf{C}$, and, by the above normalization of Fourier coefficients, obeys a functional equation relating $s$ to $1-s$. Let $\alpha,\beta$ be the Satake parameters associated to $f$; i.e.
$$L(s,f) = \prod\_p \left(1-\frac{\alpha(p)}{p^s}\right)^{-1}\left(1-\frac{\beta(p)}{p^s}\right)^{-1}$$
and then define the symmetric square $L$-function $L(\operatorname{sym}^2f,s)$ associated to $L(f,s)$ by
$$L(\operatorname{sym}^2f,s) :=
\prod\_p (1-\alpha(p)^2p^{-s})^{-1}(1-\alpha(p)\beta(p)p^{-s})^{-1}
(1-\beta(p)^2p^{-s})^{-1}.$$
My question is, when exactly is $L(\operatorname{sym}^2f,s)$ entire, and if it is not entire, what poles can it have?


Pages 136–137 of Iwaniec-Kowalski's book seem to answer this question. We know that $L(\operatorname{sym}^2f,s)$ factors like
$$L(\operatorname{sym}^2f,s) = L(f\otimes f,s)L(s,\chi)^{-1},$$
where the Rankin-Selberg convolution $L(f\otimes f,s)$ has a simple pole at $s=1$ iff $f$ is self-dual ($f=\overline f$) and is entire otherwise. $L(s,\chi)$ has a simple pole at $s=1$ iff $\chi$ is principal. Therefore, my understanding is that $L(\operatorname{sym}^2f,s)$ is entire unless $f$ is self-dual with non-principal character. This can happen if $f$ is a `CM form' arising from a Hecke grössencharacter; see <https://mathoverflow.net/a/164126/37110> for details. In the special case when $f$ has real Fourier coefficients and $\chi$ is not principal, $L(\operatorname{sym}^2f,s)$ has a simple pole at $s=1$.


Is my understanding correct? Is it complete? Thanks in advance.



Characterizing the newforms s.t. the associated symmetric square $L$-function has a pole
Yes, your understanding is correct. Here's a little bit more detail. If $f$ is a CM form, $f$ is associated to a Hecke Grössencharacter $\xi$. If $k \geq 2$, then $\xi$ is associated to an imaginary quadratic field and then $\xi$ is a homomorphism $\xi : I(\Lambda) \to \mathbb{C}^{\times}$ is a homomorphism from the group of all fractional ideals coprime to the modulus $\Lambda$, and satisfies $\xi(\alpha \mathcal{O}\_{K}) = \alpha^{k-1}$, provided $\alpha \equiv 1 \pmod{\Lambda}$.


Let $\omega\_{\xi}(n) = \xi\left(n \mathcal{O}\_{K}\right)/n^{k-1}$. This function is a Dirichlet character, and the modular form corresponding to $\xi$, which is
$$
f(z) = \sum\_{\mathfrak{a} \subseteq \mathcal{O\_{K}}} \xi(\mathfrak{a}) q^{N(\mathfrak{a})}, q = e^{2 \pi i z}
$$
has Nebentypus $\chi = \chi\_{K} \omega\_{\xi}$, where $\chi\_{K}$ is the Kronecker character associated to $K$.


Now, the symmetric power $L$-functions of $L(s,\xi)$ factor as products of degree $1$ and degree $2$ $L$-functions. In particular, $L({\rm Sym}^{2} f, s) = L(s, \xi^{2}) L(s, \omega\_{\xi})$. Self-duality of $f$ is (by the question you linked to) equivalent to the statement that $\chi\_{K} = \chi$, and hence $\omega\_{\xi}$ is the trivial character, which shows that $\zeta(s)$ is a factor of $L({\rm Sym}^{2} f, s)$ and so $L({\rm Sym}^{2} f, s)$ has a pole.


On the other hand, if $f$ is a CM form with trivial character, then $\omega\_{\xi} = \chi\_{K}$. Then, the first symmetric power $L$-function with a pole is
$$
 L({\rm Sym}^{4} f, s) = L(s, \xi^{4}) L(s, \xi^{2} \otimes \omega\_{\xi}) L(s, \omega\_{\xi}^{2}).
$$


