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Forums
Physics
Beyond the Standard Models
Cohomology and fermions in supersymmetry
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[QUOTE="kakaho345, post: 6554039, member: 695242"] I see, thanks for replying. Your answer is super relevant and helpful. He didn't explicitly stated Hodge theorem, but he did talked about the Harmonic forms previously. For future people interested, it is in his lecture 3 when he talks about representatives of cohomologies. He mentioned the codifferential ##\delta## and the Laplacian ##\Delta = \delta d + d \delta## and that we can choose a solution requiring ##d\omega = 0## and it implies ##\Delta \omega = 0##, hence ##d\omega = 0## implying an element in the cohomology implies also a member of the harmonic form ##\Delta \omega = 0##. And the Hamiltonian is exactly the Laplacian. Hence proven the claim. Then of course, I need to review why the Hamiltonian is the Laplacian. The lecture did not show in details, but the general idea is that one can identify some supercharge ##Q# and ##Q^\bar## from the Lagrangian, and those supercharge are identified with ##\delta## and ##d##. And the anticommutator between the differential gets you the Hamiltonian. P.S. anyone know why my latex code does not work properly? [/QUOTE]
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Physics
Beyond the Standard Models
Cohomology and fermions in supersymmetry
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