# Does the following first order approximation of the Kapranov-Vasserot infinitesimal loops still do any job?

Let $X$ be a scheme over, say, a field $k$. Let us denote $\mathrm{Spec}(k[\varepsilon])$ by $T$ and its (unique) $k$-point by $0\in T$. Call the first order infinitesimal cone $C_{T,0}(X)$ over $X$ the quotient $X\times T/\sim$ obtained by collapsing $X\times0$ to a point. Define $$\mathscr L^1(X):=\textrm{the scheme of sections of the projection }C_{T,0}(X)\to T.$$ Unfortunately I failed to calculate this even in the simplest cases. My question is

Has anybody encountered anything like this $\mathscr L^1(X)$? Is it nontrivial at all? Can its functor of points be represented in some convenient way? Does it have any remnants of the factorization semigroup structure (see the motivational explanations below)?

Now the motivation. In the (in my opinion) illuminating paper Vertex algebras and the formal loop space Kapranov and Vasserot introduced a very attractive algebro-geometric model $\mathscr L(X)$ of the space of infinitesimal loops of a scheme $X$. Modulo some subtleties, $\mathscr L(X)$ is defined by $$\hom(\mathrm{Spec}(R),\mathscr L(X)):=\hom(\mathrm{Spec}(R((t))^\sqrt{}),X)$$ where $R((t))^\sqrt{}$ is the ring of the series $n_\nu t^{-\nu}+\dots+n_1t^{-1}+a_0+a_1t+a_2t^2+\dots$ with coefficients in $R$ such that $n_1$,..., $n_\nu$ are nilpotent.

Kapranov and Vasserot use this model to explain some of the phenomena around chiral algebras in the sense of Beilinson-Drinfeld. In particular, they show that $\mathscr L(X)$ has the structure of factorization semigroup which gives rise to a factorization algebra structure (which is more or less equivalent to the chiral algebra structure).

Intuitively, $\mathrm{Spec}(R((t))^\sqrt{})$ plays the rôle of something like $\mathrm{Spec}(R)\times$ "infinitesimally punctured formal neighborhood of the origin". The present question arose from an attempt to find a model $\ell$ of the first-order approximation of the latter, the "first order neighborhood of the origin with the origin removed" in such a way that $\mathscr L(X)$ becomes approximated by $X^\ell$.

For convenience, let us work in the big Zariski topos $\mathscr Z_k$ over a field $k$. The first order neighborhood of the origin is $T:=\mathrm{Spec}(k[\varepsilon])$, with $0\in T$ represented by the (unique) point $\mathrm{Spec}(k)\to\mathrm{Spec}(k[\varepsilon])$ corresponding to the augmentation $k[\varepsilon]\to k$. I propose to consider the (closed) complement $\ell$ of the corresponding open subtopos $\mathscr Z_k\hookrightarrow(\mathscr Z_k)/T$ as a model of the "first order infinitesimal loop around origin".

To actually compute $X^\ell$ we may as well work with any objects $T$, $X$ in any topos $\mathscr Z$. Given a point $o:1\to T$ in $\mathscr Z$, one may consider $\ell:=$ closed complement of the open subtopos $\mathscr Z=\mathscr Z/1\hookrightarrow\mathscr Z/T$. Then, in toposes over $\mathscr Z$, one may identify $(\mathscr Z/X)^\ell$ as follows. For $f:\mathscr E\to\mathscr Z$, geometric morphisms from $\mathscr E$ to $(\mathscr Z/X)^\ell$ over $\mathscr Z$ turn out to be in one-to-one correspondence with sections of the projection $f^*(C_{T,o}(X))=C_{f^*(T),f^*(o)}(f^*(X))\to f^*(T)$, in the notation from the beginning of this question.

• Are you sure your construction yields a scheme? I could only make sense of it as a sheaf of sets. – S. Carnahan Mar 9 '14 at 14:27
• @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? – მამუკა ჯიბლაძე Mar 9 '14 at 19:16
• 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. – მამუკა ჯიბლაძე Mar 9 '14 at 19:42
• 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. – S. Carnahan Mar 10 '14 at 5:29
• @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). – მამუკა ჯიბლაძე Mar 10 '14 at 6:04