I am reading a paper by Diaconis and Graham and in page 219, in equation 2.5, Fourier transform of a probability measure $Q$ on $\mathbb{Z}^d_2$ is defined as:
For $y\in \mathbb{Z}^d_2$, the Fourier transform of $Q$ at $y$ is defined by:
$$ \hat{Q}(y) = \sum_{x\in \mathbb{Z}_2^d} (-1)^{y^t x}Q(x) $$
where $x^ty$ is a dot product of the vectors $x,y$ and here the dot product is taken mod $2$.
May I know why the Fourier transform is defined this way? Let's say $Q$ is a probability measure on $\mathbb{Z}_k^d$, $k>2$ is an integer, then would the Fourier transform be defined the same way where we sum over $x\in \mathbb{Z}^d_k$ and the dot product is taken mod $k$?
I know that the Fourier transform of a probability measure generally is $\hat{Q}(y)=\int e^{iyx}dQ(x)$ but because the state space is $\mathbb{Z}^d_2$, somehow $e^{iyx}$ translates to $(-1)^{y^tx}$ which is don't understand?
In general, the Fourier transform of $Q$ at an element $\boldsymbol y\in(\Bbb Z/n\Bbb Z)^d$ is $$\widehat Q(\boldsymbol y)=\sum_{\boldsymbol x\in(\Bbb Z/n\Bbb Z)^d}Q(\boldsymbol x)\overline{\chi_{\boldsymbol y}(\boldsymbol x)}=\sum_{\boldsymbol x\in(\Bbb Z/n\Bbb Z)^d}Q(\boldsymbol x)e^{-2\pi i\boldsymbol x\cdot \boldsymbol y/n}$$ as defined in the Fourier transform on finite groups. The case $n=2$ gives us $\chi_{\boldsymbol y}(\boldsymbol x)=(-1)^{\boldsymbol x\cdot \boldsymbol y}$.
Here we simply have the root of unity transformation (character) $\overline\chi_{\boldsymbol y}$ which trivially satisfies $\varrho:(\Bbb Z/n\Bbb Z)^d\to\operatorname{GL}(1,\Bbb C)$, and is a group homomorphism (see the section on finite abelian groups).