Hilbert scheme of projected Veronese surfaces

Question

Let $\mathbb{P}^2_\mathbb{K}$ be the projective space, with $\mathbb{K}=\bar{\mathbb{K}}$. Let $n\geq 7$ be odd and $f_1, \dotsc, f_n$ be $n$ general forms of degree $\frac{n-1}{2}$ in $\mathbb{K}[x_0,x_1,x_2]$ and let $\mathbb{V}$ be the Veronese surface projected in $\mathbb{P}^{n-1}$ obtained as the image of the regular map $[f_1:f_2:\dotso:f_n]:\mathbb{P}^2 \rightarrow \mathbb{P}^{n-1}$.

Define $\mathscr{H}$ to be the union of the irreducible components in the Hilbert scheme containing the general choices of $\mathbb{V}$ (i.e. the general choices of $f_i$).

Can I show that the general point in $\mathscr{H}$ is indeed a projected Veronese surface? In other words, that the rational map $$ H^0\left(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}\left(\frac{n-1}{2}\right)\right)^n \dashrightarrow \mathscr{H} $$ is dominant?

Answer 1

Yes, one can show that map is domiant, and even generically smooth. For a map $[f] = [f_1,\dots,f_n]$ that is a closed immersion (automatic for general $f$ once $n>5$), your rational transformation is (regular and) smooth at $[f]$ if both (a) $h^1(\mathbb{P}^2,f^*T_{\mathbb{P}^{n-1}}/T_{\mathbb{P}^2})$ equals $0$, and (b) the following natural map is surjective, $$H^0(\mathbb{P}^2,f^*T_{\mathbb{P}^{n-1}}) \to H^0(\mathbb{P}^2,f^*T_{\mathbb{P}^{n-1}}/T_{\mathbb{P}^2}).$$ Using the long exact sequence in cohomology associated to the short exact sequence, $$ 0 \to T_{\mathbb{P}^2} \to f^*T_{\mathbb{P}^{n-1}} \to f^*T_{\mathbb{P}^{n-1}}/T_{\mathbb{P}^2} \to 0,$$ it suffices to prove that $h^1(\mathbb{P}^2,T_{\mathbb{P}^2}) = h^2(\mathbb{P}^2,T_{\mathbb{P}^2}) = 0$ and $h^1(\mathbb{P}^2,f^*T_{\mathbb{P}^{n-1}})=0$. Using the Euler sequence and computation of cohomology of invertible sheaves on $\mathbb{P}^2$, this holds for every $[f]$ that is a closed immersion.