I present three realizations about the SIC problem which excited me several years ago but which did not - unsurprisingly - lead anywhere. 1. In odd dimensions d, the metaplectic representation of SL(2,Z_d) decomposes into two irreducible components, acting on the odd and even parity subspaces respectively. It follows that if a fiducial vector | Psi> possesses some Clifford-symmetry, the same is already true for both its even and its odd parity components |Psi_e>, |Psi_o>. What is more, these components have potentially a larger symmetry group than their sum. Indeed, this effect can be verified when looking at the known numerical solutions in d=5 and d=7. A finding of remarkably little consequence! 2. In composite dimensions d=p_1^r_1 ... p_k^r_k, all elements of the Clifford group factor with respect to some tensor decomposition C^d=C^(p_1^r_1) x ... x C^(p_k^r_k) of the underlying Hilbert space. This structure may potentially be used to simplify the constraints on fiducial vectors. My optimism is vindicated by the following, ground-breaking result: In even dimensions 2d not divisible by four, the Hilbert space is of the form C^2 x C^d. So it makes sense to ask for the Schmidt-coefficients of a fiducial vector with respect to that tensor product structure. They can be computed to be 1/2(1 +/- sqrt{3/(d+1)}), removing one (!) parameter from the problem and establishing that, asymptotically, fiducial vectors are maximally entangled. 3. Becoming slightly more esoteric, I could move on to talk about discrete Wigner functions and show in what sense finding elements of a set of MUBs corresponds to imposing that a certain matrix be positive, while a similar argument for fiducial vectors requires a related matrix to be unitary. Now, positivity has \'local\' consequences: it implies constraints on small sub-matrices. Unitarity, on the other hand, seems to be more \'global\' in that all algebraic consequences of unitarity involve \'many\' matrix elements at the same time. This point of view suggests that SICs are harder to find than MUBs (in case anybody wondered). If we solve the problem by Wednesday, I\'ll talk about quantum expanders.