a and generators

One can express $\vec{\tau}(\mathcal{G}_{\vec{\alpha}})$, the vector $\vec{\tau}$ of a PCE generator $\mathcal{G}_{\vec{\alpha}}$, in terms of $a(\alpha_{j_{k}})$, the rows or columns of matrix

$\displaystyle a=\left(\begin{array}[]{cccc}1&1&1&1\\ 1&1&-1&-1\\ 1&-1&1&-1\\ 1&-1&-1&1\\ \end{array}\right).$

(5)

The aforementioned expression is

$\displaystyle\vec{\tau}(\mathcal{G}_{\vec{\alpha}})=\frac{a(\alpha_{j_{1}})\otimes\ldots\otimes a(\alpha_{j_{N}})+\mathbb{1}_{4^{N}}}{2},$

(6)

where $\mathbb{1}_{4^{N}}$ is a $4^{N}$-length vector with all its components equal to 1.

Note that the rows or columns $a(\alpha_{j_{k}})$ of matrix $a$ are eigenvectors of the reflection matrix

$\displaystyle R=\left(\begin{array}[]{cccc}0&0&0&1\\ 0&0&1&0\\ 0&1&0&0\\ 1&0&0&0\\ \end{array}\right)$

(11)

with eigenvalues $r_{0,1,2,3}=1,-1,-1,1$. In other words, we can identify the ‘symmetric’ (‘anti-symmetric’) eigenvectors as those with eigenvalues 1 (-1).

JA Note: probar que (2) si es la tau de un generador We shall start proving that $\vec{\tau}(\mathcal{G}_{\vec{\alpha}})$ has $4^{N-1}$ components equal to 1 and the rest 0.

The vector $a(\alpha_{j_{1}})\otimes\ldots\otimes a(\alpha_{j_{N}})$ has $n_{-}$ and $n_{+}$ number of components with minus and plus sign, respectively.

$\displaystyle n_{\pm}$

$\displaystyle=\sum_{k}\mathbb{n}_{\pm}\big{[}a(\alpha_{j_{k}})\big{]},$

(12)

with $\mathbb{n}_{\pm}\big{[}a(\alpha_{j_{k}})\big{]}$ the number of components with sign $\pm$ in vector $a(\alpha_{j_{k}})$. JA Note: asumiendo que ya se probó que el número de +1 es $4^{N-1}$ o $4^{N}$ y de -1 $4^{N-1}$ o 0:

Now we shall prove the symmetry.

$\displaystyle\Sigma^{(k)}\big{[}\vec{\tau}(\mathcal{G}_{\vec{\alpha}})\big{]}=\frac{a(\alpha_{j_{1}})\otimes\ldots\otimes R\big{[}a(\alpha_{j_{k}})\big{]}\ldots\otimes a(\alpha_{j_{N}})+\mathbb{1}_{4^{N}}}{2}$

(13)