Recall that the three stranded ribbon diagrams are elements of either $B_3$ or $B_6$. We consider the special elements of $B_6$ that group the six strands into pairs, to form ribbons. Today's diagram shows how the three positive generators of a $B_4$ segment of $B_6$ can create a twisted ribbon that crosses over a straight ribbon segment.

Thus a neutral neutrino braid might be represented by the length $12$ $B_6$ word

$( \sigma_1 \sigma_2 \sigma_3 )^2 ( \sigma_{3}^{-1} \sigma_{4}^{-1} \sigma_{5}^{-1} )^2$

In this example, the ribbon twist is undone by canceling both the central $\sigma_{3} \sigma_{3}^{-1}$ standard $B_6$ term and the end crossings, $\sigma_1$ and $\sigma_{5}^{-1}$, leaving a word

$\sigma_2 \sigma_3 \sigma_1 \sigma_2 \cdot \sigma_{4}^{-1} \sigma_{5}^{-1} \sigma_{3}^{-1} \sigma_{4}^{-1}$

where a block of form $\sigma_2 \sigma_3 \sigma_1 \sigma_2$ represents one flat ribbon crossing another in $B_4$. That is, all flat ribbon diagrams are generated by these cyclic blocks of $B_n$ generators.

7 years ago

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