In fact in order to get the modular invariance you need to throw in an non-holomorphic term like $\frac{C}{y}$ where $y=\Im(z)$.

You are right there is a missing correcting factor that I need.

Of course if one projects $V_3$ on $V_2$ or if one tries to embed $V_2$ in $V_3$ then these maps won't be holomorphic since $V_3$ (which is of real dimension $5$) does not even support a complex structure.

Dear anonymous, in the case of $SL_n$ the symmetric space is $V_n=\{A\in M_{n x n}(\mathbf{R}): A=A^t, A>0, det(A)=1\}$. So this is a space of real dimension $n(n+1)/2-1$. So are you saying that there won't be a way of projecting $V_3$ to $V_2$ that would allow us to relate $f(q)$ to an automorphic form on $V_3$?

I see, thanks for setting me on the right path!

Dear Johan, this is quite an interesting result! It seems to me that this result suggests that given $c>0$, if we look for zeros with $1<Re(s)<1+c$ up to some large imaginary part $T$ then "most of them" will have a real part close to $1$. Are analytic number theorists close to prove anything like that?

Also @Joro, note that $\zeta(s;6,2)$ is not primitive in the sense that $gcd(6,2)=2$ and therefore $\zeta(s;6,2)=2^{-s}\zeta(s;3,1)$ where $\zeta(s;3,1)$ is now primitive.