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Jul 5, 2023 at 12:48 answer added Patrick I-Z timeline score: 5
Jul 5, 2023 at 6:10 history edited YCor CC BY-SA 4.0
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May 18, 2011 at 8:30 answer added Paolo Giordano timeline score: 7
Dec 13, 2010 at 16:36 answer added André Henriques timeline score: 5
Dec 13, 2010 at 10:57 answer added Thomas Nikolaus timeline score: 26
Dec 8, 2010 at 18:41 comment added Andrew Stacey There's lots to say on this topic, but I feel that I'd rather have a proper discussion (say, on a forum?) than use the MO engine where it's difficult to do proper replies. For example, I'd like to reverse the challenge: given that I like manifolds and would like to stay as close to manifolds as I can, why should I take sheaves just to get a topos? (Please don't answer that here! If you - or anyone else - has an answer, take it to the nForum.)
Dec 8, 2010 at 2:24 comment added Harry Gindi @David: You use the property of the hom-space $C^\infty(-,\mathbf{R})$, which turns colimits into limits. Since we know what $C^\infty(-,\mathbf{R})$ is for any cartesian space, use Yoneda's lemma to represent a sheaf $X$ as the colimit over its category of elements (which are cartesian spaces). To define $\mathbf{O}_F$, I think this is simply the product $\mathbf{O}\times F$, where we view $\mathbf{R}$ as the structure sheaf of the terminal object.
Dec 8, 2010 at 1:57 comment added David Carchedi @Harry: Alright, but, in the differential geometric setting, say $F$ is a sheaf over the site of Cartesian manifolds, how do I define $\mathcal{O}\left(F\right)$?
Dec 7, 2010 at 21:55 comment added Harry Gindi $S\to Spec\mathbf{Z}$ rather.
Dec 7, 2010 at 21:54 comment added Harry Gindi @David: At least in the algebro-geometric setting, you can pull back the affine line $Spec (\mathbf{Z}[X])\to Spec \mathbf{Z}$ by the structural morphism $S\to \mathbf{Z}$ to get the new affine line living over $S$ for any sheaf $S$ on $Aff$. This ring object is the structure sheaf for the gros topos (Zariski, etale, or fppf).
Dec 7, 2010 at 21:49 comment added David Carchedi @Konrad + Harry: You can certainly get a tangent sheaf, simply by left Kan extending the underlying manifold of the tangent bundle functor $U \circ T:Mfd \to Mfd$. Harry- How are you defining the structure sheaf of an arbitrary sheaf $F$ on the site of Cartesian manifolds? At any rate, I have to think about tangent SPACES. But for instance, you can define the tangent space of a generalized point (a la Grothendieck) of a supermanifold. But we have to appeal to derivations.
Dec 7, 2010 at 21:46 comment added Harry Gindi The cotangent complex can also be defined in some huge generality (see Illusie's books on the cotangent complex) on any ringed grothendieck topos, although I'm not sure that this will give the correct smooth cotangent complex.
Dec 7, 2010 at 21:43 comment added Harry Gindi (That is, if I remember correctly, it can be defined abstractly on the Gros-topos by pulling back the smooth affine line).
Dec 7, 2010 at 21:41 comment added Harry Gindi @Konrad: I believe that you can give a definition of the tangent sheaf by abstract nonsense. This, if I remember correctly, can be given simply by taking the sheaf of $\mathbf{R}$-derivations $\mathcal{O}_X\to \mathcal{O}_X$, where $\mathcal{O}_X$ is the structure sheaf (which can be defined abstractly).
Dec 7, 2010 at 21:24 comment added Konrad Waldorf Can you do tangent spaces with sheaves? Martin Laubinger's definition of tangent spaces goes "pointwise".
Dec 7, 2010 at 20:52 answer added arsmath timeline score: 15
Dec 7, 2010 at 20:12 answer added Andrew Stacey timeline score: 19
Dec 7, 2010 at 15:31 history asked David Carchedi CC BY-SA 2.5