Deformation theoretic argument on dimension counting of naive Hurwitz scheme

Question

I'm reading the Atanas Atanasov's course notes of Joe Harris' course Geometry of Algebraic Curves and have a question about a suggested modification of an dimension countinging argument applying methods from deformation theory.

On page 22 one consideres a version of Hurwitz scheme

$$ V_{d,g}:= \{(X, f: X \to \mathbb{P}^2) \ \vert \ X \text{ curve of genus } g, f \text{ has degree } d \text{ and is birational } \\ \text{ onto a plane curve with } \delta \text{ nodes } \} $$

together with two canonical canonical projection maps $V_{d,g} \to M_g $ (to the 'naive' moduli set) and $V_{d,g} \to \mathbb{P}^{\delta} \backslash \Delta$.

Rather elementary considerations in the script show that $\dim V_{d,g}=3+g−1$ if $d(d+3)/2 \ge 3 \delta$ but the Remark 4.2 says:

There is a serious problem with this argument if $3 \delta> d(d + 3)/2 $ but this can be fixed using deformation theory.

Namely, the counting method in the script used as intermediate equalities $\dim V_{d,g}=d(d+3)/2−3\Delta+2\Delta=3+g−1$. Of course for $3 \delta> d(d + 3)/2 $ these considerations make no any sense, but the final equality between left and right is known to be still true.

How to fix this gap using deformation theoretic arguments as suggested in the remark 4.2 in detail?

Answer 1

You can get a lower bound on the dimension of $V_{d,g}$ using deformation theory as follows. The deformation obstruction theory of a map $f : X \to Y$ between smooth varieties (where $f$ and $X$ are allowed to deform by $Y$ is held fixed) is governed by the complex $$ \mathbb{L}_f = \left[f^*\Omega_Y \to \Omega_X\right] $$ concentrated in degrees $[-1,0]$. The infinitessimal automorphisms, first order deformations and obstructions respectively lie in the groups $\mathrm{Ext}^i(\mathbb{L}_f, \mathcal{O}_X)$ for $i = 0,1,2$. In particular, if $\mathrm{Ext}^0(\mathbb{L}_f, \mathcal{O}_X) = 0$, then we have the lower bound $$ \dim \mathrm{Def}(f : X \to Y) \geq \mathrm{ext}^1(\mathbb{L}_f, \mathcal{O}_X) - \mathrm{ext}^2(\mathbb{L}_f, \mathcal{O}_X). $$

In the example $f : X \to \mathbb{P}^2$ where $X$ is a smooth curve mapping birationally onto a nodal curve, the map $f$ is an immersion so $f^*\Omega_Y \to \Omega_X$ is surjective. Let $K = \operatorname{ker} f$ so $$ \mathbb{L}_f = K[1] $$ is $K$ in degree $-1$. Then $\mathrm{Ext}^i(\mathbb{L}_f, \mathcal{O}_X) = \mathrm{Ext}^{i - 1}(K, \mathcal{O}_X)$ which is $0$ for $i = 0$ so we have the dimension inequality above. Moreover, by the exact sequence $$ 0 \to K \to f^*\Omega_{\mathrm{P}^2} \to \Omega_X \to 0 $$ we see that $\mathcal{E}xt^i(K,\mathcal{O}_X) = 0$ for $i > 0$. Denoting by $K^\vee = \mathcal{H}om(K, \mathcal{O}_X)$, we have that $$ H^i(X, K^\vee) = \mathrm{Ext}^i(K, \mathcal{O}_X) = \mathrm{Ext}^{i+1}(\mathbb{L}_f, \mathcal{O}_X) $$ by the local-to-global spectral sequence.

Putting this together, we have the lower bound $$ \dim_{[f : X \to \mathbb{P}^2]} V_{d,g} \geq \chi(X, K^\vee). $$ Using the exact sequence $$ 0 \to T_X \to f^*T_{\mathbb{P}^2} \to K^\vee \to 0 $$ we get $$ \chi(K^\vee) = \chi(f^*T_{\mathbb{P}^2}) - \chi(T_X) $$ and pulling back the Euler sequence $$ 0 \to O_X \to O_X(1)^{\oplus 3} \to f^*T_{\mathbb{P}^2} \to 0 $$ we have $$ \chi(f^*T_{\mathbb{P}^2}) = 3\chi(\mathcal{O}_X(1)) - \chi(O_X). $$ Putting this together and using Riemann-Roch, we conclude $$ \dim_{[f]}V_{d,g} \geq \chi(K^\vee) = 3(d-1+g) - (1-g) - (3g-3) = 3d + g - 1 $$

To make this an equality, we would need to know that the obstructions $H^1(X, K^\vee)$ vanish, which follows if $H^1(X, f^*T_{\mathbb{P}^2}) = 0$ and more generally is equivalent to the surjectivity of $$ H^1(X, T_X) \to H^1(X, f^*T_{\mathbb{P}^2}). $$ I don't see any reason why this has to hold in general but I'm not sure.