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Let $X$ be a smooth projective variety over $\mathbf{C}$ and denote by $\mathcal{L}_m(X)$ the $m$-th jet space, a smooth $\mathbf{C}$-scheme representing the functor on $\mathbf{C}$-algebras $$A\mapsto \text{Hom}_{\mathbf{C}}(\text{Spec}(A[x]/(x^{m+1})), X).$$ We call $\pi_m : \mathcal{L}_m(X)\to \mathcal{L}_{m-1}(X)$ the projection, an affine surjective map. We denote by $\mathcal{L}(X)$ the inverse limit scheme. We also have that the topological space of complex points of $\mathcal{L}(X)$ endowed with the analytic topology agrees with the inverse limit topological space of the spaces $\mathcal{L}_m(X)(\mathbf{C})$.

We can also associate to $X$ the path space $X^I := \text{Map}_{\text{Top}}(I, X(\mathbf{C}))$ with $I$ the interval $[0,1]$ and $X^I$ is endowed with the compact-open topology.

Question 1 Are $\mathcal{L}(X)(\mathbf{C})$ and $X^I$ related, and if so, how?

Question 2 $X^I$ is an H-space (via concatenation of paths). Is $\mathcal{L}(X)(\mathbf{C})$ an H-space?

Question 3 Is $\mathcal{L}_m(X)$ proper?

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  • $\begingroup$ I imagine that $\mathbb C[\![x]\!]$ is more similar to a formal disk than an interval $I=[0,1]$. $\endgroup$
    – Z. M
    Commented Feb 25, 2022 at 11:58
  • $\begingroup$ @Z.M Of course. In the literature, however, $\mathcal{L}(X)$ is often thought of as an algebraic analog of $X^I$. My question asks to make the analogy precise if possible. For example, by constructing a map between the two spaces, perhaps analyzing its fibers, etc. $\endgroup$
    – user178246
    Commented Feb 25, 2022 at 12:12
  • $\begingroup$ The question of whether $\mathcal{L}(X)(\mathbf{C})$ is an H-space stands on its own: perhaps there's a way to link germs of arcs (a way I'm not aware of) so as to endow $\mathcal{L}(X)(\mathbf{C})$ with a (possibly non associative/commutative) operation with a two-sided identity? (in a way that is similar to the construction of the H-product on $X^I$. $\endgroup$
    – user178246
    Commented Feb 25, 2022 at 12:16
  • $\begingroup$ $X^I$ does not seem to be an $H$-space (you cannot concatenate arbitrary two paths without the condition that the end point of one is the start point of the other), and the evaluation map $X^I\to X$ is a homotopy equivalence. Another difference that I see is that the formal disk is very local, but a path $I\to X(\mathbb C)$ could be very "long" (e.g. a cycle on a Riemann surface). $\endgroup$
    – Z. M
    Commented Feb 25, 2022 at 13:00
  • $\begingroup$ There is a description which somehow relates these two in Lurie's notes, but I am not aware of any precise statement to relate. If I am not mistaken, the $\mathcal L(X)$ is the completion $(X\times X)^\wedge$ along the diagonal $\Delta_X\colon X\to X\times X$ (see the argument in these notes), so there is a sort of "concatenation" of $\mathcal L(X)$ since it is an equivalence relation. For the third, the jet space does not seem to be proper in general: consider the simplest case of the tangent bundle. $\endgroup$
    – Z. M
    Commented Feb 25, 2022 at 13:19

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