# Characterizing the newforms s.t. the associated symmetric square $L$-function has a pole

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.

• Note that the definition of $L(\operatorname{sym}^2f,s)$ is not quite what you display, because at a ramified prime $p\mid N$ the Euler factor might be different. See the discussion in the middle of p.133 and the bottom of p.137 in Iwaniec-Kowalski's book. – GH from MO Aug 16 '14 at 10:48
• Note also that even when $N$ is square-free, the Euler factors at $p\mid N$ are different, contrary to what Iwaniec-Kowalski say on p.133, namely it would contradict Theorem 6 in Ogg's 1969 paper "On a convolution of L-series". – GH from MO Aug 16 '14 at 11:47
• by the way, @GHfromMO, I mention what I believe to be another error in IK regarding the symmetric and adjoint square $L$-functions in my comment to Jeremy on his answer. You are probably already aware of this mistake, but I thought I'd notify you in case you were not. – Owen Barrett Aug 18 '14 at 19:33

## 1 Answer

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 $$L({\rm Sym}^{4} f, s) = L(s, \xi^{4}) L(s, \xi^{2} \otimes \omega_{\xi}) L(s, \omega_{\xi}^{2}).$$

• moreover, it must be that IK make an oversight on p137 when they claim that if $f$ has real coefficients, $L(\operatorname{sym^2}f,s)$ and $L(\operatorname{ad^2}f,s)$ coincide, since if the definitions (5.97) and (5.98) are valid, in the CM form case, the symmetric square would have a pole, whereas the adjoint square would not. Even in the case of principal character, though, it seems the two would differ by a local zeta function, since $L(\chi,s)$ is missing the local factors at primes dividing the level. Sadly, Iwaniec only treats the level 1 case in his Topics' GSM book. – Owen Barrett Aug 17 '14 at 20:48
• This is the issue of what, exactly, the definition of the Rankin-Selberg convolution of two $L$-functions should be. (The answer is that it is safest to compute it using the local Langlands correspondence. This is mentioned in the errata here. In fact, I don't think that your Euler product for the symmetric square is correct in all cases.) – Jeremy Rouse Aug 19 '14 at 0:41
• @Owen: I agree. The symmetric square lift is the adjoint square lift twisted by the nebentypus of $f$. Assuming the form $f$ has real coefficients (i.e. it is self-dual), there are two possibilities. If $f$ has trivial nebentypus, then the two lifts are the same (and they are cuspidal). If $f$ has quadratic nebentypus, then $f$ is a CM form, and the two lifts are different (as they have different nebentypus) and neither of them is cuspidal. – GH from MO Aug 19 '14 at 11:56
• I don't think there's anything wrong with (5.97) and (5.98) (as long as one interprets $\chi$ as a primitive Dirichlet character), and I concur that the statement that the lifts are the same in the self-dual case isn't right. I think that IK might have temporarily forgot that $f$ can be self-dual if $\chi$ is non-trivial (although they explicitly point this out on page 131). – Jeremy Rouse Aug 19 '14 at 19:27
• @Owen: Indeed, (5.97) is not right if $L(s,\chi)$ means the traditional Dirichlet $L$-function. Instead, as Jeremy Rouse suggested, one needs to take the underlying automorphic $L$-function (i.e. replace $\chi$ by the primitive character that induces $\chi$). It is best to consult the original source, see Definition 3.1.3 in Gelbart-Jacquet: A relation between automorphic representations of GL(2) and GL(3). – GH from MO Aug 19 '14 at 22:08