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I think that the answer is that there is a flavor symmetric octet representation and a flavor antisymmetric octet representatio, while the decuplet is totally symmetric. Therefore, when you consider the spin and flavor wavefunction of a baryon for an octet baryon you have: $\chi(spin)\cdot\phi(flavor)=\frac{1}{\sqrt{2}}(\chi^{1/2}_s\cdot ... 3 Ghostly Lie algebra cohomology Let$\mathfrak{g}$be our Lie algebra and$V_\rho$a representation space with representation map$\rho : \mathfrak{g} \to \mathrm{End}(V_\rho)$.$V_\rho$is, by the action through the representation, naturally a$\mathfrak{g}$-module (people missing the ring structure in$\mathfrak{g}\$ - just embed it into the universal ...

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A quote from http://en.wikipedia.org/wiki/Symmetry_protected_topological_order : The SPT order (for both frermionic and bosonic systems) has the following defining properties: Distinct SPT states with a given symmetry cannot be smoothly deformed into each other without a phase transition, if the deformation preserves the symmetry. However, they all can ...

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Yes, this is a known problem in the condensed matter community. There are two different definitions of short range entangled (SRE) states. The key difference is whether the fermion symmetry protected topological (SPT) state belongs to the SRE state. Strictly speaking, the fermion state is not SRE, because fermion statistics is already a phenomenon of long ...

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The answer to my question is Yes. Embarrassingly, one of the simplest examples is given by the fermionic quasistring topological order I described in my paper http://arxiv.org/abs/1404.4385 . The magic is that the 5th oriented bordism group is generated by the mapping torus of complex conjugation on CP^2. Thus, if we consider the fermionic quasistring top ...

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