# Is there a "Langlands philosophy" reason for the fact that the L-function of the Jacobi theta function is (almost) the Riemann zeta function?

The Jacobi theta function $\theta(z) = 1 + 2 \sum_{n = 1}^\infty q^{n^2}$, with $q = e^{\pi i z}$ is a (twisted) modular form with weight $1/2$. It has an associated $L$-function $L(\theta, s) = \sum_{m =1}^\infty \frac{1}{(m^2)^s} = \sum_{m=1}^\infty \frac{1}{m^{2s}} = \zeta(2s)$, which stands roughly in the same relation to $\theta(z)$ as the $L$-function of a modular form $f$ stands to $f$. (They are both essentially Mellin transforms).

(Of course, there are a few issues here - $\theta$ is only a twisted modular form, the Fourier expansion for $\theta$ is in terms of $e^{\pi i z}$ instead of $e^{2\pi i z}$, there is the factor of $2$ appearing in the argument, etc.)

In the case of a weight $2k$ modular form $f$ (at least in the case that $f$ is a Hecke eigenform), there is an associated Galois representation with the same $L$-function by the Eichler-Shimura construction, and the Langlands conjectures for $GL_2(\mathbb{Q})$ predict that every Galois representation of a certain type arises this way.

Is there a similar explanation for the case of $\theta$? Of course, the $\zeta$ function is the $L$-function for the trivial one-dimensional Galois representation, but modular forms should be related to $2$-dimensional Galois representations...

(I've only thought much about the Langlands philosophy quite recently, so I'm probably missing some very basic point!)

EDIT: Thanks to GH from MO for pointing out that the trivial one-dimensional representation corresponds via (a trivial case of) class field theory to the $GL(1)$-automorphic representation defined by the trivial Hecke character.

• If you think adelically, then you can talk about unitary automorphic forms on $\mathrm{GL}_1$. Over $\mathbb{Q}$, these are in bijection with the characters $n\mapsto\chi(n)n^{it}$, where $\chi$ is a primitive Dirichlet character and $t\in\mathbb{R}$. The associated $L$-function is then $L(s+it,\chi)$. So the simplest automorphic form directly yields $\zeta(s)$. (Note that most automorphic $L$-functions do not come from Galois representations, but the $L(s,\chi)$'s do, by class field theory.) Of course, this does not answer your question, which is a good one. Mar 26, 2017 at 15:41
• Which automorphic $L$-functions do not come from Galois representations? What is the explanation for them? Mar 27, 2017 at 7:45
• This is a difficult question, so let me just say that no Maass form of level one is expected to come from a Galois representation. In fact, all the parameters (e.g. the Hecke eigenvalues) of such a Maass form are expected to be transcendental. It is probably very special for an automorphic form to come from a Galois representation, these automorphic forms are expected to be geometric (cohomological) in nature. See e.g. the slides (especially pages 10-11 and 17) math.columbia.edu/~harris/resarticles/Yalecolloquium.pdf Mar 27, 2017 at 11:33

Let $f=\sum a_n q^n$ be a modular form of weight $k+1/2$, nebentypus $\chi$. Assuming $k \geq 1$, the Shimura correspondence attaches to $f$ a modular form of integral weight $2k$, $g = \sum b_n q^n$ such that $$L(g,s) = L(\chi',s-k+1) \sum \frac{a_{n^2}}{n^s},$$
where $\chi'(n)=\chi(n) \left(\frac{-1}{n} \right)^{k}$. The Shimura correspondence is well understood in terms of the Langlands program, thanks to famous work of Waldspurger, namely functoriality from the metaplectic group. [I took this statement of the Shimura correspondence from Ono's book, the web of modularity]
Now for some reason (convergence, I suppose) the Shimura correspondence does not work as stated in weight $1/2$ (that is $k=0$) but let suppose it does and apply it boldly to $\Theta$ (neglecting the $q^{1/24}$-missing factor), which has trivial nebentypus. The character $\chi'$ would be trivial in this case. The function $g$ would be a weight $0$ modular form, that is a constant. The Galois representation attached to the constant modular form $g=1$ is the sum of the trivial character and the cyclotomic character inverted, and its $L$-function is thus $\zeta(s) \zeta(s+1)$. So the displayed formula becomes $$\zeta(s)\zeta(s+1) = \zeta(s+1) \sum a_{n^2}/n^{s}$$ where $f=\theta(z)=\sum_{n \in \mathbb Z} q^{n^2} = \sum_{n \geq 0} a_n q^n$. Hence we get that $\sum \frac{a_{n^2}}{n^s}$ is the $\zeta(s)$,or that the Mellin transform of $\Theta$ is $\zeta(2s)$. This in some sense "explains" why it is so in terms of the Langlands program.
• Great, this feels like the "right" reason. I've never heard of the Shimura correspondence or thought about automorphic forms for the metaplectic group - can you say a little more about how this interpretation goes? Also, why does the theta function have trivial nebentypus? I thought the functional equation involved some power of $i$? Mar 26, 2017 at 22:59