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Dec 26, 2016 at 14:11 comment added Will Sawin @DenisNardin I will think about how to justify it precisely...
Dec 26, 2016 at 11:51 comment added Denis Nardin For this to work (that is to provide a functor from the $\mathbb{A}^1$-homotopy category) you need to deal with iterated homotopies too (that is for any sequence of composable maps of varieties $X_0→X_1→\cdots→X_n$ you need to provide higher homotopies $|X_i|×(\Delta^1)^{j-i}→|X_j|$ (where to the vertices of the cube correspond to the composition of partial maps $|X_{i_0}|→\cdots→|X_{i_k}|$). I think your strategy, using more complicated Hilbert schemes, will work, but it is not completely trivial.
Dec 26, 2016 at 4:16 comment added David Treumann Will that is a very appealing approach.
Dec 25, 2016 at 23:47 history answered Will Sawin CC BY-SA 3.0