TITLE: Dimension of Hilbert scheme?
QUESTION [5 upvotes]: 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.

REPLY [6 votes]: 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 $m$.