Does $SL(2,\mathbb{Z})$ have a finitedimensional faithful unitary representation? No such representation exists for $SL(2,\mathbb{R})$, but I don't see a reason why one shouldn't exist for $SL(2,\mathbb{Z})$.

1$\begingroup$ The answer is certainly yes; there should even be a faithful representation into $SU(2)$. On the other hand there's no faithful finitedimensional unitary representation of $SL(3,\mathbf{Z})$. $\endgroup$ – YCor Aug 24 '18 at 19:31

$\begingroup$ Thanks. Do you mind saying something about how such representations are constructed or providing a reference? $\endgroup$ – DanielHarlow Aug 24 '18 at 19:38

4$\begingroup$ $SL(2,Z)$ is the amalgam $<u,v,zu^3=v^2=z,z^2=1>$. So I'd take $v$ the diagonal matrix $(i,i)$ (so $z=1$) and I'd take for $v$ a random conjugate (in $SU(2)$) of the diagonal matrix $(t,t^{1})$, where $t=\exp(2i\pi/6)$. This is not a complete proof but an expectation, that for a generic conjugate they indeed generate the amalgam. $\endgroup$ – YCor Aug 24 '18 at 19:42

4$\begingroup$ @YCor's suggestion works  one can prove a faithful action of $SL(2,Z)$ on the BruhatTits tree associated to $SU(2)$ for the representation variety, and then restrict to a generic value to get a faithful representation. But I'll point out that lowdimensional representations are classified in this paper: mathscinet.ams.org/mathscinetgetitem?mr=1815266 $\endgroup$ – Ian Agol Aug 24 '18 at 20:21
Here a nonexplicit proof of the existence of a faithful representation of $\mathrm{SL}_2(\mathbf{Z})$ in $\mathrm{SU}(2)$, using basic algebraic geometry and topology, and relying on the amalgam decomposition of $\mathrm{SL}_2(\mathbf{Z})$.
[The basic idea is that if all representations in $\mathrm{SU}(2)$ were nonfaithful, by Zariskidensity this would also be the case for representations into $\mathrm{SL}_2$. We need to use the particular form of the presentation of $\mathrm{SL}_2(\mathbf{Z})$, since the argument will not carry over representations of $\mathrm{SL}_3(\mathbf{Z})$ in $\mathrm{SU}(3)$.]
Let $P_t$ be the set of $2\times 2$ matrices with determinant 1 and trace $t$. Both $P_0$ and $P_1$ are irreducible as algebraic varieties (being $\mathrm{SL}_2$ conjugacy classes). Then for $K$ a field of characteristic zero, $P_0(K)$ is the set of elements of order 4 in $\mathrm{SL}_2(K)$, and $P_1(K)$ is the set of elements of order 6 in $\mathrm{SL}_2(K)$.
For every $(g,h)\in P_0\times P_1$, $g^2=h^3$ equals $I_2$. Hence the set of representations of $$\mathrm{SL}_2(\mathbf{Z})=\langle u,v\mid u^4=v^6=[u^2,v]=[u,v^3]=1\rangle$$ (restricting to those for which the image of $u$ has order 4 and the order of $v$ has order 6) into $\mathrm{SL}_2(K)$ can be naturally identified to $(P_0\times P_1)(K)$. Note that $P_0\times P_1$ is irreducible.
Write $P_t^\sharp=P_t(\mathbf{C})\cap\mathrm{SU}(2)$. Then using that $\mathrm{SU}(2)$ is Zariskidense in $\mathrm{SL}_2(\mathbf{C})$ and describing $P_t(\mathbf{C})$ as a conjugacy class, one deduces that $P_t^\sharp$ is Zariskidense in $P_t(\mathbf{C})$. So $P_0^\sharp\times P_1^\sharp$ (which is a 4dimensional real manifold) is Zariskidense in $(P_0\times P_1)(\mathbf{C})$.
For every given nontrivial element $w$ in $\mathrm{SL}_2(\mathbf{Z})$ the set of representations vanishing on $w$ is a proper subvariety of $P_0\times P_1$, hence has dimension $\le 3$. By Zariski density of $P_0^\sharp\times P_1^\sharp$, we deduce that its intersection with $P_0^\sharp\times P_1^\sharp$ is a proper Zariski closed subset (because $\mathrm{SL}_2(\mathbf{Z})$ admits one faithful representation into $\mathrm{SL}_2(\mathbf{C})$, the standard one); in particular it has empty interior (in the ordinary topology). By the Baire theorem, the union over all $w$ is still a proper subset. Hence we deduce the existence of an element of $P_0^\sharp\times P_1^\sharp$ defining a faithful representation.

6$\begingroup$ Just to state the obvious: this answer describes the extra mileage you need to go for having a faithfu two dimensional unirep. Just to get a faithful f.d unirep it is a lot easier to induce one from a finite index free group... $\endgroup$ – Uri Bader Aug 25 '18 at 8:54

$\begingroup$ @UriBader thanks for noticing. I guess the smallest index of a free subgroup is 12, so this yields a 24dimensional representation that has finite kernel, hence kernel of order $\le 2$ (I guess 2). Using a 1dimensional representation (through $C_{12}$ that's faithful on $I_2$, we eventually get a faithful 25dimensional unitary representation. $\endgroup$ – YCor Aug 25 '18 at 9:25