Fourier expansion of boolean functions

[Dragonfall]

Any boolean function on n variables can be thought of as a function

$$f : \mathbb{Z}_2^n \rightarrow \mathbb{Z}_2$$

which can be written as

$$f(x) = \sum_{s \in \mathbb{Z}_2^n} \hat{f}(s) \prod_{i : x_i = 1} (-1)^{x_i}$$

where

$$\hat{f}(s) = \mathbb{E}_t \left[ f(t) \prod_{i : s_i = 1} (-1)^{t_i} \right]$$

This is the Fourier expansion of a boolean function. But this uses the group $\mathbb{Z}_2^n$. Why not use $\mathbb{Z}_{2^n}$? Characters of the latter would be nice looking roots of unity on the complex circle, instead of points on an $n$-cube.

EDIT: You know what, nevermind. I don't even understand my own question anymore.

 

[Mark44]

This is the Fourier expansion of a boolean function. But this uses the group $\mathbb{Z}_2^n$. Why not use $\mathbb{Z}_{2^n}$?

Because the function maps vectors of length n whose components are all either 0 or 1, to a set containing 0 and 1 ($\mathbb {Z_2}$). Your alternate version would map a single number in the set $\{0, 1, \dots, 2^n - 1\}$ to $\mathbb {Z_2}$.