Tangent space to spaces of maps

Question

Let $B = \{x_1,\dots,x_{d-2},y_1,\dots,y_k\}$ be a subscheme of $d-2+k$ distinct points of $\mathbb{P}^1$, and $g:B\rightarrow \mathbb{P}^2$ be a morphism mapping $x_1,\dots,x_{d-2}$ to a fixed point $p_0$ and $y_1,\dots,y_k$ to general points $q_1,\dots,q_k\in\mathbb{P}^2$.

Assume that there exists a morphism $f:\mathbb{P}\rightarrow\mathbb{P}^2$ of degree $d$ restricting to $g$ on $B$, so that the space $Hom_d(\mathbb{P}^1,\mathbb{P}^2,f_{|B} = g)$, of morphisms restricting to $g$ on $B$, is non empty.

Do we have that $h^1(\mathbb{P}^1,f^{*}T_{\mathbb{P}^2}\otimes\mathcal{I}_B) = 0$ for $d\gg 0$? Would this imply that $Hom_d(\mathbb{P}^1,\mathbb{P}^2,f_{|B} = g)$ has dimension $h^0(\mathbb{P}^1,f^{*}T_{\mathbb{P}^2}\otimes\mathcal{I}_B) = d-2k+6$ and that it is smooth at $f$?

Thank you very much.

Answer 1

I think this is not true, at least if $k\geq 6$. The Euler exact sequence pulled back to $\mathbb{P}^1$ is $$0\rightarrow \mathscr{O}_{\mathbb{P}^1}\rightarrow \mathscr{O}_{\mathbb{P}^1}(d)^3\rightarrow f^*T_{\mathbb{P}^1}\rightarrow 0\,.$$Thus $H^1(f^*T_{\mathbb{P}^1}\otimes \mathscr{I}_B)=H^1(f^*T_{\mathbb{P}^1}(-d-k+2))$ is zero iff $\alpha :H^1(\mathscr{O}_{\mathbb{P}^1}(-d-k+2)\rightarrow H^1(\mathscr{O}_{\mathbb{P}^1}(-k+2)^3$ is surjective, or equivalently, by Serre duality, $\alpha^{\vee} :H^0(\mathscr{O}_{\mathbb{P}^1}(k-4))^3\rightarrow H^0(\mathscr{O}_{\mathbb{P}^1}(k-4+d) )$ is injective. Now $\alpha ^{\vee}$ is given by $\alpha ^{\vee}(A,B,C)=AF+BG+CH$, where $f(x)=(F(x),G(x),H(x))$.

Choose the coordinates on $\mathbb{P}^2$ so that $p_0=(0,0,1)$; then $F$ and $G$ vanish at each $x_i$, so $F(x)=Q(x)\prod (x-x_i)$ and $G(x)=R(x)\prod (x-x_i)$, with $\deg Q=\deg R=2$. But this implies $RF-QG=0$, so as soon as $k\geq 6$ $\alpha ^{\vee}$ is not injective, whatever $d$ is.