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Francesco Polizzi
Francesco Polizzi
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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 $$\dim \textrm{Hilb}_{[V]}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$$$\dim_{[V]}\textrm{Hilb}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$ that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.

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 $$\dim \textrm{Hilb}_{[V]}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$ that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.

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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Francesco Polizzi
Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283

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 $$\dim \textrm{Hilb}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$$$\dim \textrm{Hilb}_{[V]}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$ that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.

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 $$\dim \textrm{Hilb}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$ that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.

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 $$\dim \textrm{Hilb}_{[V]}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$ that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.

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Francesco Polizzi
Francesco Polizzi
  • 66.3k
  • 5
  • 180
  • 283

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 $$\dim \textrm{Hilb}(X) = \dim T_{[V]} \textrm{Hilb}(X)=m,$$ that is, $\textrm{Hilb}(X)$ is smooth at $[V]$, of dimension $m$.