Why use class multiplication to describe topological entangling and merging?

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I'm studying some references about topological defects in ordered media like *Soft matter physics: An introduction* by Kleman and the Review modern physics paper *The topological theory of defects in ordered media* by Mermin. In both of them, the authors emphasize the class multiplication instead of elements multiplication for describing disclination merging and entangling, and then admit the arbitrariness of the class multiplication. However, I have some problem to understand this. E.g., for a bi-axial liquid crystal, disclinations are classified by $\pi_1(SO(3)/D_2) = Q_8$. This is a quaternion group who has five conjugacy classes $\{1\}, \{-1\},\{i,-i\},\{j,-j\},\{k,-k\}$. I understand the fact that elements $i$ and $-i$ describing , e.g., the disclination and anti-disclination in $yz$ plane, so it is reasonable to group them together. However, when I do defects merging or entangling using the class multiplication, as suggest in the references, I would have problem to predict the result: $\{i,-i\}$ mutiplities $\{i,-i\}$ can either give me $\{1\}$ or $\{-1\}$ who are different defects. Why doesn't one use the elements multiplication directly which doesn't lead to the arbitrary? Put differently, why is it necessary to use class multiplication? Can any one give me any hints?
edited Jan 31, 2015

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 Disclaimer: I'm not at all learnt on the subject, in fact this is the first time I read something on topological defects, my answer will be tentative.  After briefly reading through the relevant sections of The topological theory of defects in ordered media, mostly section VI.B, here is what the discussion looks like to me: 1. The classification is most naturally done in terms of conjugacy classes, since it's physically more natural to consider free homotopy rather than base-point-preserving homotopy. 2. Surely we can do multiplication on (fundamental group)elements instead of conjagcy classes, it's not forbidden, and it indeed has the merit of being unique. (But again, the result of the multiplication is still most naturally classified by conjugacy classes) 3. But since the classification is done in terms of conjugacy classes, it's reasonable to think about what multiplication of classes means, despite the ambiguities OP mentioned. Even further, the ambiguities have very clear physical meanings(still in section VI.B), which certainly makes a good reason to study class multiplication. In short, it is not necessary but desirable to consider class multiplication.
answered Jan 31, 2015 by (2,640 points)
edited Jan 31, 2015

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