I have a subvariety $V \subset X$, and I want to compute the dimension of the connected component of $\textrm{Hilb}(X)$ containing $[V]$. I can give explicit deformations of $V$ showing that the dimension of this component is at least $m$. I can also show that $H^0(V, N_{V /X}) \leq m$. Do these facts together imply that the dimension of this component is exactly $m$, and that $\textrm{Hilb}(X)$ is smooth at $[V]$? I am a little worried about how the tangent sheaf works in the singular case.
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1$\begingroup$ So, your $X$ is singular? $\endgroup$– Francesco PolizziCommented Oct 22, 2014 at 18:02
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$\begingroup$ Let's say $X$ and $V$ are both smooth, though I'd be interested in a more general statement if there is one. (By "the singular case" I mean when $\textrm{Hilb}(X)$ is singular or non-reduced.) $\endgroup$– markCommented Oct 22, 2014 at 18:07
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In general, the Zariski tangent space of $\textrm{Hilb}(X)$ at $[V]$ is naturally isomorphic to $\textrm{Hom}_V(I_V/I_V^2, \, \mathcal{O}_V)$.
When $X$ and $V$ are both smooth and projective, this group equals $H^0(V, \, N_{V/X})$. Therefore in your case we can write $$m \leq \dim _{[V]}\textrm{Hilb}(X) \leq \dim T_{[V]} \textrm{Hilb}(X) = H^0(V, \, N_{V/X}) \leq m.$$
This means $$\dim_{[V]}\textrm{Hilb}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$ that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.
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$\begingroup$ Thanks a lot! I suspected this was OK, but worried I was missing some subtlety. $\endgroup$– markCommented Oct 22, 2014 at 18:42
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1$\begingroup$ You are welcome. Note that in this way you show that the irreducible component of $\textrm{Hilb}(X)$ passing through $X$ is (generically) smooth of dimension $m$. I do not think that you can say anything about the connected component with your assumptions: the Hilbert scheme can be very pathological, and maybe some component of bigger dimension can intersect $\textrm{Hilb}_{[V]}(X)$ away from $[V]$. $\endgroup$ Commented Oct 22, 2014 at 18:47