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Apr 25, 2010 at 15:56 answer added Dylan Thurston timeline score: 6
Apr 23, 2010 at 13:38 answer added user1504 timeline score: 5
Apr 23, 2010 at 7:08 answer added Kelly Davis timeline score: 7
Apr 23, 2010 at 5:34 comment added Theo Johnson-Freyd One can "explain" why such graphs show up in path-integral QFT in the following way: they exactly parametrize the $\hbar \to 0$ asymptotics of finite-dimensional integrals of the form $\int\exp\bigl((i\hbar)^{-1}f(x)\bigr)dx$, where $f(x)\in C^\infty(\mathbb R^n)[[\hbar]]$. (More generally, the asymptotics only depend on the Taylor coefficients of $f$ near critical points of $[f]\in C^\infty(\mathbb R^n)=C^\infty(\mathbb R^n)[[\hbar]]/\hbar$.) But this isn't the kind of explanation you are looking for, since it does not explain the coincidence with geometry. And I don't know any geometry.
Apr 23, 2010 at 2:51 history asked Kevin H. Lin CC BY-SA 2.5