Sunday, May 30, 2010

M Theory Lesson 334

For the usual choice of $3 \times 3$ Fourier matrix, the transformed matrix $M = R_{23}(a) R_{12}(b) R_{31}(c)$ takes the form where we note that the first component is fully fixed by $F_3$. This neatly illustrates the $(1,2)$ block form of transformed magic mixing matrices, which is basically a combination of the diagonal of $1$-circulant eigenvalues and a $2$-circulant codiagonal. For each $R_2$ factor, there is an $F_3$ transform that fixes it. It follows that for a suitable twisted Fourier transform, the whole triple product will remain fixed.

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