Why does the refractive index not mirror the Lorentzian peak shape of the absorption index? In the characterization of materials, there are many methods used: One of them is infrared spectroscopy. In a lab we saw the indices of refraction and absorption of a certain (semiconductor) solid material for some infrared frequencies. 
I understand that we expect some absorptions because of the phonons' interaction and functional groups and I can see that clearly in the absorption index of the Lorentzian curve.
My question is: Both indices had the same morphology (as expected) but while the absorption index had that Lorentzian curve the refraction index had that kind of loop (for the same wavenumber region)
Why does the refractive index have that "loop" instead of a Lorentzian dip?

 A: Absorption and refraction indices are the real and imaginary parts of the propagation constant for a medium. This means, given certain mild assumptions on the material's physics, they are the real and imaginary parts of a meromorphic (holomorphic with poles) function of the complex frequency that is holomorphic in the right half plane. They are therefore intimately related to one another through the Hilbert transform or the Kramers Kronig relationships.
The blips in the refractive index plots are actually the Hilbert transforms of the Lorentzian peaks in the absorption index plots. Indeed, I recommend working out and plotting the real and imaginary parts of the function:
$$f:\hat{\mathbb{C}}\to\hat{\mathbb{C}};\,f(z) = \frac{1}{z-z_0}-\frac{1}{z-z_0^*}$$
when $z_0,\,z_0^*$ are the the position of a pole in the left half plane when the complex frequency $z=i\,\omega$. You'll find the real part is the Lorentzian and the imaginary part has the same shape as the blips in the refractive index curves.
