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Let $X$ be a smooth (complex) variety, and $V\subset X$ a reduced, normal subvariety. Fix $k\geq 0$. Then there exists an $n$ such that: for a generic point $v\in V$, we can intersect $V$ with the order $k$ nilpotent neighborhood $\rm{Spec} \mathcal{O}_X/m_{X,v}^k$ of $v$ in $X$ to get a point in the punctual Hilbert scheme $X^{[n]}$ of $X$. We thus get a map from an open subset of $V$ to $X^{[n]}$.

Question: Is there a way to extend this map to all of $V$? If not, is there a larger/different space of "local germs on $X$" to which $V$ maps?

I am willing to make the target space as singular as neccessary, and $k$ as large as neccessary but I want $V$ to map there, not just a blowup of $V$.

Why I think V has to be normal An example I have in mind is a nodal curve sitting in the plane. Then it seems like I really need to go to the normalization to get a reasonable map. Intuitively, it seems like it might be enough to take $V$ such that its "everywhere locally analytically irreducible', or in other words the completion of $\mathcal{O}_{V,v}$ is a domain for all $v\in V$.

Thanks!

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    $\begingroup$ For $X$ a smooth variety, the moduli space of points of $X$ plus a $d$-dimensional subspace of their tangent space is proper - a Grassmanian bundle on $X$. So any such moduli space of local germs in the case $k=1$ must either (1) not include this space as a subvariety, (2) not map the smooth points of $V$ into this subvariety, (3) map all the points of $V$ into this subvariety, or (4) not be separated. One can rule out (3) by an explicit example, and then I have a hard time seeing how any of the others could be viable. $\endgroup$
    – Will Sawin
    Commented Jun 3, 2018 at 4:53
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    $\begingroup$ Yeah, I think one has to take a large enough $k$ dependent on $V$. Namely, I think $k$ has to at least be big enough such that at every point of $v$, the defining ideal for $V$ does not lie entirely in the $k$'th infinitesimal neighborhood of $v$. $\endgroup$
    – jacob
    Commented Jun 3, 2018 at 5:10
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    $\begingroup$ If you let $V$ be the vanishing locus of a nondegenerate homogeneous quadratic form $f$ in $m$ variables, it has an isolated singularity at the origin. If for some $k$ we can extend the map to that point then it must define an $O_m$-invariant ideal of codimension ${m + k -2 \choose m-1}$ inside the quotient $\mathbb C[x_1,\dots,x_m]/f$. As a representation of $O_m$ this splits into the spherical harmonic representations which are irreducible and nonisomorphic. $\endgroup$
    – Will Sawin
    Commented Jun 3, 2018 at 5:55
  • $\begingroup$ So the only such ideals are the ideals generated by homogeneous polynomials of degree $d$, which have codimension ${m+d-1 \choose m} - {m + d-3 \choose m}$. So at minimum you have to solve some weird Diophantine equation to find an example. $\endgroup$
    – Will Sawin
    Commented Jun 3, 2018 at 5:57
  • $\begingroup$ Ok, so youre suggesting that the map doesn't extend to $X^{[n]}$. But is there some slightly different space it DOES extend to? $\endgroup$
    – jacob
    Commented Jun 3, 2018 at 22:12

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