Xuanting Cai
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 May 18 awarded Popular Question Sep 24 awarded Autobiographer Jun 25 awarded Tumbleweed May 28 comment recursion formula for odd holonomic function C[t^{\pm}] means the Laurant Polynomials of t May 8 revised recursion formula for odd holonomic function edited title Apr 26 comment recursion formula for odd holonomic function Is this question too hard? I do not have any good idea yet. Apr 18 revised recursion formula for odd holonomic function added 47 characters in body Apr 18 comment Why must a reducible flat SU(2)-connection over a homology sphere be trivial? $A^flat(M)/G(M)≅Hom(π_1(M),SU(2))/SU(2)$ is easy to prove. Just use geometric definition of connection. Apr 18 asked recursion formula for odd holonomic function Feb 6 comment What is the definition of algebro-gemetric quotient? I will check it. Thank you. Feb 6 comment What is the definition of algebro-gemetric quotient? what I am really interested is following: I have a group G, finitely presented. Consider R(G)=HOM(G,SL(2,C)), the space of all reps of G into SL(2,C). Then SL(2,C) acts on R(G) naturally, i.e. conjugation. So the usual quotient of this action is the space of orbits, which are all conjugacy classes of reps. Now the algebro-geometric quotient is not this. It is the character variety of G. Thanks. Feb 6 awarded Editor Feb 6 revised What is the definition of algebro-gemetric quotient? added 51 characters in body Feb 6 asked What is the definition of algebro-gemetric quotient? Jan 27 comment Seiberg-Witten equation on S^2\times S^1 OK. Thanks. So it is not a topological invariant? Jan 27 accepted Seiberg-Witten equation on S^2\times S^1 Jan 27 asked Seiberg-Witten equation on S^2\times S^1 Jan 27 awarded Supporter Jan 26 comment Why must a reducible flat SU(2)-connection over a homology sphere be trivial? Of Course. Thank you for your help. Jan 26 comment Why must a reducible flat SU(2)-connection over a homology sphere be trivial? I am a knot theory student. Try to learn some 4 manifold thing. So maybe this is a dummy question for expert. Thank you.