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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 emphasis the class multiplication instead of elements multiplication for describing disclination merge and entangle, 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, what is the necessary of using class multiplication? Can any one give me any hits?

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    $\begingroup$ "ordered media" means symmetry breaking. This question is related to symmetry breaking order which is not related to topological order and topological phase. $\endgroup$ – Xiao-Gang Wen Feb 10 '15 at 21:11
  • $\begingroup$ @Xiao-GangWen I am still quite puzzled by the definition of the topological ordered phase. The topological order does not require symmetry breaking. However, does just this only mean symmetry breaking is not necessary for a topological ordered phase, but a topological phase can also break the symmetry, though there may be massless excitations as a result? Moreover, if the picture of trapping a vison in a 2-torus can be interpreted as topological order, might a classical system, as the example in this post, also be topological ordered, since it can trapped flux of disclinations under PBC? $\endgroup$ – R.Wigner Aug 5 '15 at 8:25
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The topological defect is the whole (homotopy) class you mentioned. Every element in the class is homotopicaly equivalent to each other. Concretely, it means that if you deform a circle continuously, the defect inside it won't change. So there's a redundancy when one considers the elements because you have different elements for the same geometrical (or physical) object. You can always perform elements multiplication if you want, obtaining an element of the multiplied elements' class ($[a]*[b]=[a*b]$), but mathematically it doesn't have a structure. An elegant way to resolve this is considering the equivalence classes of the homotopies (named fundamental group).

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  • $\begingroup$ Thanks for your answer. However, I think it is not true that "multiplying two elements in the same class, the result will also belongs to the class". Take an example from the question itself, if I multiply a defect ordered two with himself, the product will be the identity which is clearly in the different class. $\endgroup$ – R.Wigner Aug 5 '15 at 8:14
  • $\begingroup$ @hongchaniyi, you're right, my mistake, I'll edit it. Anyway, still there's the redundancy when one considers the elements because you have different elements for the same geometrical (or physical) object. You can always perform elements multiplication if you want, obtaining an element of the multiplied elements' class ([a]*[b]=[a*b], that is the correct answer for my mistake), but mathematically it doesn't have a structure. An elegant way to do this is considering the equivalence class of the homotopies. $\endgroup$ – Mr. K Aug 5 '15 at 13:39

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