# Tangent space to spaces of maps

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.

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.
• Thank you very much. Just a question. In genera,l for morphisms $f:X\rightarrow Y$ and $g:B\rightarrow Y$, is it true that if $h^1(X,f^{*}T_Y\otimes\mathcal{I}_B)$ then $Hom(X,Y,f_{|B} = g)$ is smooth at $f$ and has dimension $h^0(X,f^{*}T_Y\otimes\mathcal{I}_B)$?
• I suppose you mean "if $h^1=0$", and that your $X$ is a curve; then yes, I think what you write is true.