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Oct 3, 2018 at 17:14 history edited Bombyx mori CC BY-SA 4.0
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Oct 2, 2018 at 7:23 vote accept Bombyx mori
Oct 3, 2018 at 17:13
Oct 1, 2018 at 14:02 answer added Aftermath 12345 timeline score: 2
Apr 7, 2018 at 4:13 comment added Bombyx mori @ThomasKojar: My main confusion from reading his paper (arxiv.org/pdf/math/0605337.pdf) is that I do not see how the weights for the edges and vertices are being assigned, and how the DGFF relates to the metric from GFF. Do you mean that he just consider an element in GFF, then act it on a function in the subspace of $H^{1}$ functions that are affine on the triangles, and claim it is a way of defining DGFF? But then some hypothesis has to be put onto the triangles.
Apr 7, 2018 at 2:56 comment added Bombyx mori @ThomasKojar: I spent sometime reading through his papers. However it is not entirely clear to me how the projecting procedure works even for 2D, where he used TG domains and did impose boundary conditions. Do you mind to turn your comment into an answer?
Apr 6, 2018 at 20:58 comment added Thomas Kojar Projecting is made precise in their contour lines paper(arxiv.org/pdf/1008.2447.pdf). Basically we are projecting GFF on the subspace of H1 functions that are affine on triangles. This is how one shows convergence of DGFF to GFF.
Apr 6, 2018 at 3:13 comment added Bombyx mori @ThomasKojar: I think this is the same idea as used in Sheffied's paper. My reservation is that I do not know how "by projecting..." works. If I put a triangulation on $M$, and I define GFF using Green functions via distributions, what does "projecting..." tell me? It is very unclear to me.
Apr 5, 2018 at 21:21 comment added Thomas Kojar In "LQG AND Bosonic 2d string "( arxiv.org/pdf/1607.08467.pdf) , Vargas etal mention the construction of continuous GFF on compact manifolds. Then by projecting you obtain the DGFF. Now beyond compact, you lose the nice eigenfunction decomposition.
Apr 4, 2018 at 11:49 history asked Bombyx mori CC BY-SA 3.0