Can $T$ act trivial in a repn of SL$_2(\mathbb{Z}_N)$? I am confronted with the following problem:
If $\rho : \text{SL}_2(\mathbb{Z}) \to \text{GL}_{\mathbb{C}}(V)$ is a finite dimensional representation such that $\text{ker}(\rho)$ contains the principal congruence group
$$\Gamma(N) = \{ M \in \text{SL}_2(\mathbb{Z}) : M \equiv \text{id} \mod N\}$$
(i.e. $\rho$ can be viewed as a representation of the finite group $\text{SL}_2(\mathbb{Z}_N)$ where I write $\mathbb{Z}_N = \mathbb{Z}/N\mathbb{Z}$ ) then we can endow $V$ with a scalar product making this representation unitary (or we can assume this right away). Hence, if we put $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ then $\rho(T)$ can be diagonalized. As $T^N \in \Gamma(N)$, the eigenvalues are of the form $e^{2 \pi i s/N}, s=0,1,...,N-1$. 
Problem: What eigenspaces do really occur?
Let us take $N$ above to be minimal then I suspect that there should be at least one nontrivial eigenspace correpsonding to $e^{2 \pi i s/N}$ for some $\text{gcd}(s,N) = 1$.
* Question 1: Is this assertion true? *
I think that I know how to prove it but it seems like cheating to me: Let us take $N=p$ to be a prime for brevity. If the eigenspaces for all $e^{2 \pi i s/N}$ with $s\neq 0$ disapper then in fact, $T$ acts trivial. I want to conclude that $p$ was not the level (but in fact, $1$ is the level). We can view the representation as a continuous representation
$$ \text{SL}_2(\mathbf{Z}_p) \times V \to V$$
(here $\mathbf{Z}_p$ are the $p$-adic integers). In $\text{SL}_2(\mathbf{Z}_p)$, the closure of the normal subgroup generated by $T^M$ is the $p$-adic version of $\Gamma(M)$, i.e. since $T$ acts trivial, also the $p$-adic version of $\Gamma(1)$ acts trivial (which is all of $\text{SL}_2(\mathbf{Z}_p)$).
* Question 2: Does this proof make sense or am I overlooking something? *
I am confused, because I am not sure whether in general, the normal subgroup generated by $T^M$ is $\Gamma(M)$, so all that makes the proof work out is that we introduced some new structural "things" (like topology)...
Thanks in advance.
FW
 A: I answer the question in the title: $T$ can act trivial iff the rep is trivial.
Here is the proof:
We have
$$ SL_2(Z/N) = \oplus_{p^k || N} SL(Z/p^k)$$
by the Chinese remainder theorem and because SL(2) is an algebraic group. So you have to work with $G=SL_2(Z/p^k)$ only.
Have understood this: Assume wlog that the rep $\sigma$ is irreducible and $T$ acts trivial, then $1 \subset Res_{N} \sigma$ for $N$ strict upper triangular matrices so $\sigma \subset Ind_{N} 1$ by Frobenius. So $\sigma \subset Ind_{B} \mu$ for $B$ upper triangular and some $\mu: B \rightarrow \mathbb{C}^\times$ trivial on $N$. It is explictly known how to decompose $Ind_N \mu$ into irreducible representaion $\sigma'$ for $SL_2(Z_p)$ (see Casselman :Restriction of representation of GL_2(F) to GL_2(o)). For those, you can compute that $Hom_N(1,Res_{N} \sigma') =  Hom_G(Ind_{N}^G 1, \sigma') = \mathbb{C} \neq \mathbb{C}^{\dim(\sigma')}$ unless $\sigma'$ and hence $\sigma$ is trivial. 
Determining the eigenspaces, you need to compute
$$ Res_N \sigma.$$
This is easy with Mackey induction restriction formula with expensive bookkeeping for those induced from $B$. For the supercuspidals, I think things become much worse. I also suggest working with GL(2) and study Bushnell-Kutzko if you can. I don't see an easier way. Usually $N$ squarefree is easier, because you don't need primes powers, and can work with a finite field. 
A: Let us assume $N = p^e$ for some prime $p$ (not necessarily odd). By this paper Lemma 3.1, 
$$\overline{\langle \langle T^M \rangle \rangle} = \Gamma(M, \mathbf{Z_p}) = \{ \alpha = \begin{pmatrix} a & b \\ c & d \end{pmatrix} : a \equiv d \equiv 1 \mod p^{\nu_p(M)} ~\text{and}~ c \equiv b \equiv 0 \mod p^{\nu_p(M)} \}$$
where $\nu_p(M)$ is the $p$-order of $p$ in $M$, for any set $X$ inside a group $G$ I put $\langle \langle X \rangle \rangle$ to be the normal subgroup generated by $X$ (i.e. this is just $\langle g^{-1} x g : x \in X, g \in G \rangle$) and $\overline{\cdot}$ means the topological closure inside $\text{SL}_2(\mathbf{Z}_p)$.
We have a ring homomorphism
  $$r_{p^e} : \mathbf{Z}_p \to \mathbf{Z}_{p^e}$$
given by
  $$r_{p^e}(\alpha_0 + \alpha_1p + ...) = \alpha_0 + \alpha_1 p + ... + \alpha_{e-1} p^{e-1} + p^e \mathbf{Z}$$
and its matrix valued version
  $$R_{p^e}\begin{pmatrix} a & b \\ c & d \end{pmatrix} := \begin{pmatrix} r_{p^e}a & r_{p^e}b \\ r_{p^e}c & r_{p^e}d \end{pmatrix}$$
The map
  $$\Phi : \text{SL}_2(\mathbf{Z}_p) \to \text{GL}(V), ~~~~ \Phi(M) := \rho(R_{p^e}(M))$$
is continuous (in fact, with respect to every topology on GL because around every point $\alpha \in \text{SL}_2(\mathbf{Z}_p)$ we find an open nbh $U$ such that $R_{p^e}(U) = \{ R_{p^e}(\alpha)\}$ [just vary in every coordinate maximally by $\delta := p^{-{e+1}}$]).
If there was no "unit" eigenspace, then $\rho(T^{p^{e-1}})$ acts trivial. Hence, $T^{p^{e-1}} \in \ker(\Phi)$ so that also $\overline{\langle \langle T^M \rangle \rangle} = \Gamma(M, \mathbf{Z_p}) \subset \ker(\Phi)$ and this precisely translated back to $\Gamma(p^{e-1}) \subset \ker(\rho)$ in contradiction to the assumption that the level was $N=p^e$.
For the general case $N = p_1^{e_1} \cdot ... \cdot p_r^{e_r}$, decompose $V$ into irreducibles $W_1 \oplus ... \oplus W_n$. Renew terminology to $V=W_1$. Using the chinese remainder theorem we can view $\rho$ as a representation of $\text{SL}_2(p_1^{e_1}) \times ... \times \text{SL}_2(p_r^{e_r})$. By Serres Linear representations of Finite Groups, Thm. 10, p.27,
$$ \rho = \rho_1 \otimes ... \otimes \rho_r$$
Then one shows that $\rho_i$ really has level $p_i^{e_i}$ so that by the case above, there are $a_i \in \mathbb{Z}_{p_i^{e_i}}, v_i \in V_i\setminus \{0\}$ such that $\rho_i(T) v_i = e^{2 \pi i a_i/{p_i^{e_i}}}$. Now $v_1 \otimes ... \otimes v_r$ is a nontrivial vector in the eigenspace of $\rho(T)$ with eigenvalue $e^{2 \pi i a/N}$ where
$$a = N_1 a_1 + ... + N_r a_r$$
where $N_i = \prod_{j \neq i} p_j^{e_j}$ and $a$ is in $\mathbb{Z}_N^\times$.
