Timeline for Does the following first order approximation of the Kapranov-Vasserot infinitesimal loops still do any job?
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| Mar 11, 2014 at 5:35 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Mar 10, 2014 at 6:12 | comment | added | მამუკა ჯიბლაძე | I mean inside the infinitesimal cone $C_{T,0}(X)$ since the latter "has no base"... | |
| Mar 10, 2014 at 6:04 | comment | added | მამუკა ჯიბლაძე | @S.Carnahan Oh I see the problem now, thank you very much! This seems to be crucial. My $C_{T,0}(X)$ really ought to be the quotient object in the topos of big Zariski sheaves (or maybe in some other gros topos, but necessarily the quotient sheaf there). You know examples when it is not representable by a scheme? As for the case when $C_{T,0}(X)$ is the first neighborhood of the affine cone vertex: I was hoping that taking sections will detect something more about $X$. In any case the whole $X$ sits inside $\mathscr L^1(X)$ as "constant loops" (although it cannot be placed inside the cone). | |
| Mar 10, 2014 at 5:29 | comment | added | S. Carnahan♦ | Yes, the Hom scheme $S^T$ is a scheme when $S$ is, since it is the relative spectrum of the symmetric algebra on $\Omega^1$. However, I do not see why your definition of $C_{T,0}(X)$ yields a scheme in general. Moreover, when $X$ is projective, $C_{T,0}(X)$ is the first-order neighborhood of the vertex in the affine cone, and you seem to lose almost all information about $X$ this way. For example, if $X$ is the Fermat curve $x^3 + y^3 + z^3 = 0$, the defining equation cuts out nothing new in the first-order neighborhood. | |
| Mar 9, 2014 at 19:42 | comment | added | მამუკა ჯიბლაძე | Probably I should mention that Kapranov-Vasserot's $\mathscr L(X)$ is not a scheme, it is an ind-scheme. And I would not mind if $\mathscr L^1(X)$ would be only an ind-scheme too, but I think it is in fact a scheme. | |
| Mar 9, 2014 at 19:16 | comment | added | მამუკა ჯიბლაძე | @S.Carnahan Maybe I fail to take something into account but does not $S^T$ exist as a scheme for any (not necessarily smooth) $S$? This granted, $\mathscr L^1(X)$ is the inverse image of the point of $T^T$ corresponding to the identity by the map $\textrm{projection}^T:C_{T,0}(X)^T\to T^T$. Certainly 0 is almost always a singular point of $C_{T,0}(X)$, even for smooth $X$, but I think this is not an obstacle, is it? | |
| Mar 9, 2014 at 14:27 | comment | added | S. Carnahan♦ | Are you sure your construction yields a scheme? I could only make sense of it as a sheaf of sets. | |
| Mar 9, 2014 at 8:47 | history | edited | მამუკა ჯიბლაძე |
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| Mar 9, 2014 at 8:09 | history | edited | მამუკა ჯიბლაძე | CC BY-SA 3.0 |
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| Mar 9, 2014 at 8:02 | history | asked | მამუკა ჯიბლაძე | CC BY-SA 3.0 |