Interpolation of Gini Index? Is there a conventional curve used to interpolate the Lorenz curves that model the distribution for calculating the Gini coefficient? Is it polynomial? But why? Which degree?
A Lorenz curve is a P-P plot, so the shape of the curve depends on the underlying probability distribution. Depending on what probability distribution you're using for your model (uniform, normal, exponential, etc.), you'll end up with a different functional form for the Lorenz curve.
Given an underlying cumulative distribution function $ F(x) $ with inverse (if it exists) $ x(F) $, the Lorenz curve is given by
$$ L\big(F(x)\big) = \frac{\int_0^F{x(G)}dG}{\int_0^1{x(G)}dG} $$
In practice, when you're looking at Lorenz curves and Gini coefficients, it's common to use a Pareto distribution as the underlying probability distribution.
Here's a derivation for the functional form of the Lorenz curve for the Pareto distribution as an example of the process. Starting from the CDF of the Pareto distribution:
$$ F(x) = 1 - \bigg(\frac{x_m}{x}\bigg)^\alpha , $$
we can invert it to
$$ x(F) = x_m (1-F)^{-\frac{1}{\alpha}} . $$
Then, inserting the inverse of the CDF into the definition of the Lorenz curve above, we have
$$ L(F) = \frac{\int_0^F{x_m (1-G)^{-\frac{1}{\alpha}}}dG}{\int_0^1{x_m (1-G)^{-\frac{1}{\alpha}}}dG} $$
$$ = 
\frac{\bigg[-\frac{x_m}{1-\frac{1}{\alpha}}\big(1-G\big)^{1-\frac{1}{\alpha}}\bigg]^F_0}{\bigg[-\frac{x_m}{1-\frac{1}{\alpha}}\big(1-G\big)^{1-\frac{1}{\alpha}}\bigg]^1_0} $$
$$ = 
\frac{\frac{x_m}{1-\frac{1}{\alpha}}\bigg[1-\big(1-F\big)^{1-\frac{1}{\alpha}}\bigg]}{\frac{x_m}{1-\frac{1}{\alpha}}} . $$
So, for the Pareto distribution with parameters $x_m$ and $\alpha$, the corresponding Lorenz curve has the functional form
$$ L(F) = 1 - (1 - F)^{1-\frac{1}{\alpha}} . $$
Now, if you want to find the functional form of the Lorenz curve given data on some measure, like wealth or income, you would first have to fit the distribution of your data to the Pareto distribution first and estimate its two parameters, $x_m$ and $\alpha$ (see here).
Once you have your estimate of $\alpha$ ($\hat{\alpha}$), you can simply plug that into the Lorenz curve for the Pareto distribution we found above:
$$ L(F) = 1 - (1 - F)^{1-1/\hat{\alpha}} . $$