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The construction of the Hilbert scheme of a projective scheme $X$ requires us to fix an ample line bundle $L$ on $X$ in order to define the Hilbert polynomial $P$.

Suppose that $S \subset X$ is a closed subscheme of $X$. Suppose $L_1$ and $L_2$ are two different ample line bundles on $X$, and say that $S$ has Hilbert polynomial $P_1$ with respect to $L_1$ and $P_2$ with respect to $L_2$.

Then we can construct the Hilbert scheme $\operatorname{Hilb}^1(X)$ using $L_1$, and in here $S$ will give a point in the component $\operatorname{Hilb}^1_{P_1}(X)$. Alternatively we can construct $\operatorname{Hilb}^2(X)$ using $L_2$, and here $S$ is a point in $\operatorname{Hilb}^2_{P_2}(X)$.

Do we then have an isomorphism $\operatorname{Hilb}^1_{P_1}(X) \cong \operatorname{Hilb}^2_{P_2}(X)$?

I think the answer must be yes, but I would like to have a reference.

Thanks!

Schemer
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  • I think I see how the proof goes: using the universal families, we get morphisms $\operatorname{Hilb}^i_{P_i}(X) \rightarrow \operatorname{Hilb}^j_{P_j}(X)$ in both directions that are bijective on points. By an argument that escapes me for the moment, one then soups these up to show the morphisms are actually isomorphisms. But the main point of my question was to find a reference for this fact. – Schemer Oct 23 '15 at 15:27
  • I would very much expect that the answer is "no" unless $P_{1}$ and $P_{2}$ are constant. The proof you give in your comment will not work since the universal family will only help you for families, of which you know they have the same Hilbert polynomial with respect to the given ample bundle. I don't see any reason why one should be able to compare the decompositions of the Hilbert scheme with respect to the Hilbert polynomials for different ample bundles. – Rieux Oct 23 '15 at 19:01
  • @Rieux: I'm not sure I understand your comment. Let me try to rephrase mine and see if we still disagree. The universal family over $\operatorname{Hilb}^1_{P_1}(X)$ is a flat family one of whose fibres is $S$. So all members of that family have Hilbert polynomial $P_2$ with respect to $L_2$. So by the universal property, we get a morphism $\operatorname{Hilb}^1_{P_1}(X) \rightarrow \operatorname{Hilb}^2_{P_2}(X)$, and so on. Of course $P_1$ and $P_2$ need not be the same. – Schemer Oct 23 '15 at 20:15
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    The Hilbert polynomial is only guaranteed to be constant in a flat family over a CONNECTED base. So you'll have a natural map from the component of $Hilb^1$ containing $S$ to $Hilb^2$. If you knew the Hilbert scheme with polynomial $P_i$ was connected (which is true at least in the case $X=\mathbb{P}^n_{k}$ and $L_i$ ample), then you would get the natural maps you claim above. In fact, if you know you have such natural maps, it's trivial to see that they're inverse to one another. – Cass Oct 25 '15 at 19:43
  • @Cass: good point! In fact I guess I really only care about the components of either Hilbert scheme containing my point (as the sloppy wording of my question indicates). As for "trivial": ok, fair enough. Thanks! – Schemer Oct 25 '15 at 20:01

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