A few months ago, I was curious about some properties of Maass cusp forms, of nonabelian arithmetic origin. As a result, I went through a somewhat predictable process of finding a totally real $A\_4$ extension of $Q$, lifting the resulting projective Galois representation to an honest Galois representation, and writing a short program to compute as many coefficients of the Artin L-function (thus coefficients of the Maass form) as needed.


Well, as often happens, I didn't find anything particularly surprising in the end. 


But now I "have a Maass form". Its a pretty Maass form -- the simplest one of eigenvalue 1/4, of "nonabelian" origin (not arising from a dihedral Galois representation). Its conductor is 163 -- a very attractive prime number (though its appearance here seems coincidental). Some class number 1 coincidences make the computation of its coefficients extremely quick and simple. 


So, does anyone want the Maass form (i.e. code to output coefficients quickly)? It's fun to play with, and doesn't take up too much space. I guarantee its modularity. If not, any suggestions where to put it (a little journal that publishes such cute examples)? 



Does anyone want a pretty Maass form?
[these are comments, not an answer, but there were too many for the comments box]


Hey---I wrote that code too! I did it to teach myself "practical Maass forms". I wrote in pari, not sage. I didn't do the example you did. Here's what I did, for what it's worth. First I tried a dihedral example. I used the Hilbert class field of $\mathbf{Q}(\sqrt{145})$; the class group is cyclic of order 4, giving a $D\_8$ extension of $\mathbf{Q}$ with a faithful 2-dimensional representation. It's an easy exercise in factoring polynomials mod $p$ to compute traces of Frobenius, and I got the Hecke eigenvalues with little trouble.


But here's the big question: how do you know you got them right? Here's how I did it. I computed the first 200,000 Hecke eigenvalues, created the formal power series defining a function on the upper half plane as per usual, and then I evaluated it to many decimal places at lots of random points $z$ and $\gamma.z$ with $\gamma$ in the level. In all the cases I tried, the answers were the same to within experimental error. I concluded that probably I'd got everything working.


I also did an $S\_3$ extension (the class group of $\mathbf{Q}(\sqrt{79})$) and a non-algebraic example coming from a Grossencharacter---this one had conductor 8 but eigenvalue not $1/4$. 


I also tried to do an $A\_5$ example! As you probably know from the $A\_4$ example, an issue that needs resolving here is that the image of Galois in $GL\_2(\mathbf{C}$) isn't $A\_4$, it's the central extension, so you need to know how primes split in that last extension, which is computationally more expensive. Bjorn Poonen showed me a wonderful trick though, so I could do it. I computed $a\_n$ for hundreds and thousands of $n$, built the function on the upper half plane, and checked to see if it was invariant by the level group. It wasn't :-( I concluded that either the Langlands program was wrong or my code was wrong, and I had a good idea which. [EDIT May 2012: for what it's worth I did actually get the code working in the end (in April 2011 in fact) -- my code *was* wrong (stupid error: all the "hard" code was fine but I had miscalculated $a\_{1951}$!), and Langlands' programme still looks fine. AFAIK one still cannot prove that the candidate Maass form whose power series expansion I can compute a very long way is actually a Maass form.]


Here is the pari script for the 145 example, by the way:



```
N=200000;
f=x^4 - x^3 - 3*x^2 + x + 1;
ap(p)=if(p==5,-1,if(p==29,-1,if(issquare(Mod(p,5))&&issquare(Mod(p,29)),2*(matsize(factormod(f,p))[1]-3),0)));
chi(n)=kronecker(n,145);
v=vector(N,i,0);
v[1]=1;
for(i=2,N,fac=factor(i);k=matsize(fac)[1];\
 if(k>1,v[i]=prod(j=1,k,v[fac[j,1]^fac[j,2]]),\
 if(fac[1,2]==1,v[i]=ap(i),\
 p=fac[1,1];e=fac[1,2];v[i]=v[p]*v[p^(e-1)]-chi(p)*v[p^(e-2)]))\
);
F(z)=local(x,y,M);x=real(z);y=imag(z);M=ceil(11/y);if(M>N,error("y too small."));sqrt(y)*sum(n=1,M,if(v[n]==0,0,v[n]*besselk(1e-30*I,2*Pi*n*y)*cos(2*Pi*n*x)))

```

That's it! It's pretty self-explanatory. ap(p) returns the coefficient $a\_p$ of the form. chi is the character of the form. If you run this the computer will pause for a few seconds while it computes the first 200,000 coefficients of the Maass form. After that it will give you a function $F$ on the upper half plane, which is defined by a Fourier expansion, and the miracle will be that it will be $\Gamma\_1(145)$-invariant. For example, after running the code above, you can try this:



```
gp > z=-0.007+0.08*I
%5 = -0.007000000000000000000000000000 + 0.08000000000000000000000000000*I
gp > F(z)
%6 = 0.2101332751524672135753981488 + 0.E-30*I
gp > F(z/(145*z+1))
%7 = 0.2101332751524672135753981489 + 0.E-30*I
gp > %6-%7
%8 = -5.67979851 E-29 + 0.E-30*I

```

What this says is that for $z$ a random element of the upper half plane such that $z$ and $\gamma.z$ both have imaginary part which is not too small (if the im part is too small you need more Fourier coeffts), where here $\gamma=(1,0;145,1)$, $F$ evaluates, to within experimental error, to the same value at $z$ and $\gamma.z$.


Note that Marty's example is of $A\_4$ type, so more interesting than this example, but a theorem of Langlands tells us that Marty's example really will be a cusp form.


