Proof. Parts $(1)$ and $(2)$ and the statement on smoothness of Hilbert point in part $(3)$ of the theorem follow from [Reference Bangere, Mukherjee and Raychaudhury8, Theorem $3.1(3)$ , Theorem $4.3$ , Theorem $4.4$ , Theorem $5.1$ ] and [Reference Gallego and Purnaprajna26, Proposition $1.7$ , Theorem $3.5$ , Theorem $4.1$ ].

We prove part $(3)$ . Let $Y = \mathbb {F}_e$ with $0 \leq e \leq 2$ . Consider the $K3$ double cover of $\pi :X \to Y$ branched along a very general smooth $D \in |-2K_Y|$ . Let $H_0 = \pi ^{\ast }(\mathcal {O}_{Y}(1)) = \pi ^{\ast }(aC_0+bf)$ and consider the composition

Recall the smoothing fibers $X_t \stackrel{\varphi _t}{\hookrightarrow} \mathbb {P}^M$ are obtained as the image of an embedding $\varphi _t$ which is a general deformation of $\varphi $ along a one-parameter family T and are hence embedded inside $\mathbb {P}^M$ by the complete linear series of a line bundle $H_t$ which is a deformation of $H_0$ . Since the multiplicity of a polarization remains constant on a smooth family, we have to find the multiplicity of $H_0$ on the double cover X. Now since D is very general, smooth and irreducible part $(3)$ for $Y = \mathbb {F}_e$ , $0 \leq e \leq 1$ follows from the fact that $\textrm {Pic}(X)$ is generated by $\pi ^{\ast }\mathcal {O}_Y(C_0)$ and $\pi ^{\ast }\mathcal {O}_Y(f)$ (see [Reference Artebani, Hausen and Laface1, Proposition $6.3$ ] and [Reference Artebani, Hausen and Laface1, Proposition $6.5$ (i), (iii)]).

Notation.

1. (a) Let $\widetilde {V}$ be a vector bundle on the ribbon $\widetilde {Y}$ and L is a line bundle on Y. Then L has the structure of an $\mathcal {O}_{\widetilde {Y}}$ module due to the homomorphism $\mathcal {O}_{\widetilde {Y}} \to \mathcal {O}_Y$ that defines the ribbon $\widetilde {Y}$ . We define $\widetilde {V}(L):=\widetilde {V}\otimes _{\mathcal {O}_{\widetilde {Y}}} L$ .

2. (b) By $I_{\widetilde {Y}}$ and $I_{Y}$ , we mean the ideal sheaf of $\widetilde {Y}$ and Y in $\mathbb {P}^M$ respectively.

Proof. Part $(1)$ has been proved in [Reference Bangere and Mukherjee9, Theorem $2.2$ ]. We prove part $(2)$ . We have the following exact sequences (see [Reference Gallego, González and Purnaprajna23, Lemma $4.1$ ])

(3.1) $$ \begin{align} 0 \to N_{\widetilde{Y}/\mathbb{P}^M} (K_Y) \to N_{\widetilde{Y}/\mathbb{P}^M} \to N_{\widetilde{Y}/\mathbb{P}^M} \otimes \mathcal{O}_Y \to 0 \end{align} $$

(3.2) $$ \begin{align} 0 \to \mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y) \to N_{\widetilde{Y}/\mathbb{P}^M} \otimes \mathcal{O}_Y \to \mathcal{O}_Y(-2K_Y) \to 0 \end{align} $$

(3.3) $$ \begin{align} 0 \to \mathcal{O}_Y(-K_Y) \to N_{Y/\mathbb{P}^M} \to \mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y) \to 0 \end{align} $$

Tensoring equation 3.1 by $\mathcal {O}_{\widetilde {Y}}(-\widetilde {H})$ , we have that

(3.4) $$ \begin{align} h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-\widetilde{H})) \leq h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-H+K_Y)) + h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-H)) \end{align} $$

We compute $h^0(N_{\widetilde {Y}/\mathbb {P}^M}(-H+K_Y))$ . Tensoring equation 3.2 by $\mathcal {O}_Y(-H+K_Y)$ we have

$$ \begin{align*} 0 \to \mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H+K_Y) \to N_{\widetilde{Y}/\mathbb{P}^M}(-H+K_Y) \to \mathcal{O}_Y(-H-K_Y) \to 0 \end{align*} $$

Hence

(3.5) $$ \begin{align} h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-H+K_Y)) \leq h^0(\mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H+K_Y)) + h^0(\mathcal{O}_Y(-H-K_Y)) \end{align} $$

Tensoring equation 3.3, by $\mathcal {O}_Y(-H+K_Y)$ we have

$$ \begin{align*} 0 \to \mathcal{O}_Y(-H) \to N_{Y/\mathbb{P}^M}(-H+K_Y) \to \mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H+K_Y) \to 0 \end{align*} $$

Since $H^1(-H) = 0$ , by Kodaira vanishing theorem, and $h^0(-H) = 0$ , we have that

(3.6) $$ \begin{align} h^0(\mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H+K_Y)) = h^0(N_{Y/\mathbb{P}^M}(-H+K_Y)) \end{align} $$

We now compute $h^0(N_{\widetilde {Y}/\mathbb {P}^M}(-H))$ . Tensoring equation 3.2 by $\mathcal {O}_Y(-H)$ we have

$$ \begin{align*} 0 \to \mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H) \to N_{\widetilde{Y}/\mathbb{P}^M}(-H) \to \mathcal{O}_Y(-H-2K_Y) \to 0 \end{align*} $$

Hence

(3.7) $$ \begin{align} h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-H)) \leq h^0(\mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H)) + h^0(\mathcal{O}_Y(-H-2K_Y)) \end{align} $$

Tensoring equation 3.3, by $\mathcal {O}_Y(-H)$ we have

$$ \begin{align*} 0 \to \mathcal{O}_Y(-H-K_Y) \to N_{Y/\mathbb{P}^M}(-H) \to \mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H) \to 0 \end{align*} $$

Now we claim that

Granting the claim for now, we have that

(3.8) $$ \begin{align} h^0(\mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H)) = h^0(N_{Y/\mathbb{P}^M}(-H)) - h^0(-H-K_Y) \end{align} $$

Hence we have that

$$ \begin{align*} h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-\widetilde{H})) \leq h^0(-H-K_Y) + h^0(N_{Y/\mathbb{P}^M}(-H+K_Y)) + h^0(N_{Y/\mathbb{P}^M}(-H)) - h^0(-H-K_Y) + h^0(-H-2K_Y) \end{align*} $$

(3.9) $$ \begin{align} h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-\widetilde{H})) \leq h^0(N_{Y/\mathbb{P}^M}(-H+K_Y)) + h^0(N_{Y/\mathbb{P}^M}(-H)) + h^0(-H-2K_Y) \end{align} $$

We now compute $h^0(N_{Y/\mathbb {P}^M}(-H+K_Y))$ . We have

(3.10) $$ \begin{align} 0 \to T_Y \to T_{\mathbb{P}^M}|_Y \to N_{Y/\mathbb{P}^M} \to 0 \end{align} $$

and the Euler exact sequence restricted to Y

(3.11) $$ \begin{align} 0 \to \mathcal{O}_Y \to \mathcal{O}_Y(H)^{M+1} \to T_{\mathbb{P}^M}|_Y \to 0 \end{align} $$

Tensoring equation 3.11 by $\mathcal {O}_Y(-H+K_Y)$ , we have

$$ \begin{align*} H^0(\mathcal{O}_Y(K_Y))^{M+1} \to H^0(T_{\mathbb{P}^M}|_Y(-H+K_Y)) \to H^1(\mathcal{O}_Y(-H+K_Y)) \end{align*} $$

We have that $H^0(\mathcal {O}_Y(K_Y)) = 0$ and by assumption $H^1(\mathcal {O}_Y(-H+K_Y)) = 0$ and hence $H^0(T_{\mathbb {P}^M}|_Y(-H+K_Y)) = 0$ .

Tensoring equation 3.10 by $\mathcal {O}_Y(-H+K_Y)$ , we have

$$ \begin{align*} H^0(T_{\mathbb{P}^M}|_Y(-H+K_Y)) \to H^0(N_{Y/\mathbb{P}^M}(-H+K_Y)) \to H^1(T_Y(-H+K_Y)) \end{align*} $$

Since $H^0(T_{\mathbb {P}^M}|_Y(-H+K_Y)) = 0$ ,

(3.12) $$ \begin{align} h^0(N_{Y/\mathbb{P}^M}(-H+K_Y)) \leq h^1(T_Y(-H+K_Y)) \end{align} $$

Plugging everything back into equation 3.9, we have

(3.13) $$ \begin{align} h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-\widetilde{H})) \leq h^1(T_Y(-H+K_Y)) + h^0(N_{Y/\mathbb{P}^M}(-H)) + h^0(-H-2K_Y) \end{align} $$

Proof of Claim 3.2

Consider the exact sequence given by twisting equation 3.3 by $-H$ .

(3.14) $$ \begin{align} 0 \to -K_Y-H \to N_{Y/\mathbb{P}^M}(-H) \to \mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-H) \to 0 \end{align} $$

Let $p: Y \to \mathbb {P}^1$ denote the smooth fibration. Now note that $-K_Y-H = -(b-e-2)f$ and hence $R^1p_*(-H-K_Y) = 0$ . Hence we have the following exact sequence by pushing forward the exact sequence 3.14

Proof. The Euler exact sequence is classical; see Hartshorne [Reference Hartshorne27, Theorem II.8.13] for the dual (cotangent) version. We only give a proof of part (ii) because the local coordinate formula (3.33) will be used later.

Part (ii). We compute $\rho (s)$ on $U_0$ by evaluating the derivation $\widetilde {X}=\sum _{j=0}^N s_j\,\partial /\partial x_j$ on the affine coordinate functions $y_i=x_i/x_0$ .

$$\begin{align*}\widetilde{X}(y_i) =\sum_{j=0}^N s_j\,\frac{\partial}{\partial x_j}\!\left(\frac{x_i}{x_0}\right) =\frac{s_i}{x_0}-\frac{x_i\, s_0}{x_0^2} =\frac{s_i\, x_0-x_i\, s_0}{x_0^2}. \end{align*}$$

Substituting $s_i=x_0 f_i$ and $x_i=x_0 y_i$ gives $\widetilde {X}(y_i)=f_i-f_0\, y_i$ . Since a vector field on $U_0\simeq \mathbb {A}^N$ is determined by its values on $y_1,\dots ,y_N$ , we conclude $\rho (s)=\sum _{i=1}^N (f_i-f_0\, y_i)\,\partial /\partial y_i$ .

Proof. Let $\mathbb {P}^1 = \operatorname {Proj}k[u,v], \mathbb {P}^2 = \operatorname {Proj}k[X_0,X_1,X_2]$ and $\mathbb {P}^M = \operatorname {Proj}k[X_0,X_1,X_2,X_3,\cdots , X_M]$ . Since the claim is local on $\mathbb {P}^1$ , it suffices to check it on the affine chart $U=\{v\neq 0\}\subset \mathbb {P}^1$ . Set $y=u/v$ on U.

Step 1: Local coordinates on $\mathbb {P}^M$ along $i(U)$ . On the affine chart $\{X_2\neq 0\}\subset \mathbb {P}^M$ set

$$\begin{align*}z_0=\frac{X_0}{X_2},\qquad z_1=\frac{X_1}{X_2},\qquad w_j=\frac{X_j}{X_2}\ (j=3,\dots,M). \end{align*}$$

Along $i(U)$ we have $X_0=u^2$ , $X_1=uv$ , $X_2=v^2$ , and $X_j=0$ for $j\ge 3$ , hence

$$\begin{align*}z_0=\frac{u^2}{v^2}=y^2,\qquad z_1=\frac{uv}{v^2}=y,\qquad w_j=0\ (j=3,\dots,M). \end{align*}$$

Therefore the differential $\beta =di$ satisfies

(3.34) $$ \begin{align} \beta\!\left(\frac{\partial}{\partial y}\right) = \frac{d z_0}{d y}\frac{\partial}{\partial z_0} +\frac{d z_1}{d y}\frac{\partial}{\partial z_1} +\sum_{j=3}^M \frac{d w_j}{d y}\frac{\partial}{\partial w_j} = 2y\,\frac{\partial}{\partial z_0}+\frac{\partial}{\partial z_1}, \end{align} $$

since $w_j\equiv 0$ on $i(U)$ .

Step 2: Local formula for the Euler maps. On $U=\{v\neq 0\}$ the line bundle $\mathcal {O}_{\mathbb {P}^1}(1)$ is trivialized by v, so any local section of $\mathcal {O}_{\mathbb {P}^1}(1)^{\oplus 2}$ can be written as $(vA,\;vB)$ with $A,B\in \mathcal {O}_{\mathbb {P}^1}(U)$ . By Equation (3.33), from Lemma 3.3, we have

(3.35) $$ \begin{align} \rho_{\mathbb{P}^1}(vA,\;vB)=(A-By)\,\frac{\partial}{\partial y}. \end{align} $$

Similarly, on $\{X_2\neq 0\}$ the bundle $\mathcal {O}_{\mathbb {P}^M}(1)$ is trivialized by $X_2$ . After pulling back to U, we have $X_2=v^2$ , so $\mathcal {O}_{\mathbb {P}^1}(2)$ is trivialized by $v^2$ . Hence any local section of $\mathcal {O}_{\mathbb {P}^1}(2)^{\oplus (M+1)}$ can be written as

$$\begin{align*}(v^2C_0,\;v^2C_1,\;v^2C_2,\;v^2D_3,\dots,v^2D_M) \end{align*}$$

with $C_i,D_j\in \mathcal {O}_{\mathbb {P}^1}(U)$ . The affine Euler formula on the chart $\{X_2\neq 0\}$ in Equation (3.33), Lemma 3.3, gives

(3.36) $$ \begin{align} (i^{\ast}\rho_{\mathbb{P}^M})(v^2C_0,\;v^2C_1,\;v^2C_2,\;v^2D_3,\dots,v^2D_M) = (C_0-C_2 z_0)\frac{\partial}{\partial z_0} +(C_1-C_2 z_1)\frac{\partial}{\partial z_1} +\sum_{j=3}^M (D_j-C_2 w_j)\frac{\partial}{\partial w_j}. \end{align} $$

Restricting to $i(U)$ , we may substitute $z_0=y^2$ , $z_1=y$ , and $w_j=0$ .

Step 3: The Jacobian map $\widetilde \alpha $ in local trivializations. Let $(vA,\;vB)$ be a local section of $\mathcal {O}_{\mathbb {P}^1}(1)^{\oplus 2}$ on U. Using $u=yv$ we compute

$$\begin{align*}\widetilde\alpha(vA,\;vB) = \big(2u\cdot vA,\; u\cdot vB+v\cdot vA,\;2v\cdot vB,\;0,\dots,0\big) = \big(v^2(2yA),\;v^2(A+yB),\;v^2(2B),\;0,\dots,0\big). \end{align*}$$

Thus in the notation of (3.36),

(3.37) $$ \begin{align} C_0=2yA,\qquad C_1=A+yB,\qquad C_2=2B,\qquad D_j=0\ (j=3,\dots,M). \end{align} $$

Step 4: Compare both compositions. First compute $(i^{\ast }\rho _{\mathbb {P}^M})\circ \widetilde \alpha $ using (3.36) and (3.37), and then substitute $z_0=y^2$ , $z_1=y$ , and $w_j=0$ :

$$ \begin{align*} (i^{\ast}\rho_{\mathbb{P}^M})(\widetilde\alpha(vA,vB)) &= (C_0-C_2 y^2)\frac{\partial}{\partial z_0} +(C_1-C_2 y)\frac{\partial}{\partial z_1} +\sum_{j=3}^M (D_j-C_2\cdot 0)\frac{\partial}{\partial w_j}\\ &= \big(2yA-2B y^2\big)\frac{\partial}{\partial z_0} +\big(A+yB-2By\big)\frac{\partial}{\partial z_1} +\sum_{j=3}^M 0\cdot \frac{\partial}{\partial w_j}\\ &= (A-By)\left(2y\frac{\partial}{\partial z_0}+\frac{\partial}{\partial z_1}\right). \end{align*} $$

On the other hand, using (3.35) and then (3.34),

$$\begin{align*}\beta\big(\rho_{\mathbb{P}^1}(vA,vB)\big) = \beta\!\left((A-By)\frac{\partial}{\partial y}\right) = (A-By)\,\beta\!\left(\frac{\partial}{\partial y}\right) = (A-By)\left(2y\frac{\partial}{\partial z_0}+\frac{\partial}{\partial z_1}\right). \end{align*}$$

The two expressions agree, proving $i^{\ast }\rho _{\mathbb {P}^M}\circ \widetilde \alpha =\beta \circ \rho _{\mathbb {P}^1}$ on U.

Proof. Let us first prove parts $(1)$ and $(3)$ . Let $Y = \mathbb {F}_e$ and let $p: Y \to \mathbb {P}^1$ denote the projection to $\mathbb {P}^1$ . Note that we have an exact sequence

$$ \begin{align*}0 \to T_{Y/\mathbb{P}^1} \to T_Y \to p^{\ast}T_{\mathbb{P}^1} \to 0\end{align*} $$

and that $T_{Y/\mathbb {P}^1} = 2C_0+ef$ and $p^{\ast }T_{\mathbb {P}^1} = 2f$ . So it is enough to show $H^i(2C_0+ef-H) = H^i(2f-H) = 0$ for $i = 0,1$ . Now the claim follows by combining Lemmas 3.5 and 3.6. Part $(2)$ follows similarly after observing that $H^1(-H-2f) = H^1(H-2f) = 0$ under the given hypothesis.

Proof (1)

This follows by combining Lemma 3.7, 3.8, and 3.9 and applying Theorem 3.1.

Proof (2)

Note that by Lemma 3.9, $(2)$ , $h^1(T_Y(-H+K_Y)) = 0$ if $a = 2$ and $b \geq 2e+1$ . Similarly by Lemma 3.8, $h^0(-H-2K_Y) = 0$ if $a = 2$ and $b \geq 2e+5$ . Now we show that $h^0(N_{Y/\mathbb {P}^M}(-H)) = M+1$ . Twisting the sequence

$$ \begin{align*}0 \to T_Y \to T_{\mathbb{P}^M}|_Y \to N_{Y/\mathbb{P}^M} \to 0\end{align*} $$

by $-H$ and taking cohomology, we have

(3.38) $$ \begin{align} 0 \to H^0(T_Y(-H)) \to H^0(T_{\mathbb{P}^M}|_Y(-H)) \to H^0(N_{Y/\mathbb{P}^M}(-H)) \to H^1(T_Y(-H)) \xrightarrow{s} H^1(T_{\mathbb{P}^M}|_Y(-H)) \end{align} $$

Proof. Let $k \geq 2$ . We tensor Equation (3.1) by $-k\widetilde {H}$ to obtain

$$ \begin{align*}h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-k\widetilde{H})) \leq h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-kH+K_Y)) + h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-k\widetilde{H}))\end{align*} $$

Now tensoring Equation (3.2) by $(-kH+K_Y)$ and noting that $h^0(-kH-K_Y) = 0$ we have,

$$ \begin{align*}h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-kH+K_Y)) = h^0(\mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-kH+K_Y))\end{align*} $$

Once again, tensoring Equation (3.3), by $-kH+K_Y$ and noting that $h^0(-kH) = h^1(-kH) = 0$ , we have

$$ \begin{align*}h^0(\mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-kH+K_Y)) = h^0(N_{Y/\mathbb{P}^M}(-kH+K_Y))\end{align*} $$

Now tensoring Equation (3.2) by $(-kH)$ we have,

$$ \begin{align*}h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-kH)) \leq h^0(\mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-kH))+h^0(-kH-2K_Y)\end{align*} $$

Once again, tensoring Equation (3.3), by $-kH$ and noting that $h^0(-kH-K_Y)) = 0$ , we have

$$ \begin{align*}h^0(\mathcal{H}om(I_{\widetilde{Y}}/I_{Y}^2, \mathcal{O}_Y)(-kH)) \leq h^0(N_{Y/\mathbb{P}^M}(-kH))+h^1(-kH-K_Y)\end{align*} $$

Hence we have,

$$ \begin{align*}h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-k\widetilde{H})) \leq h^0(N_{Y/\mathbb{P}^M}(-kH+K_Y)) + h^0(N_{Y/\mathbb{P}^M}(-kH)) + h^1(-kH-K_Y) + h^0(-kH-2K_Y)\end{align*} $$

Now Equation (3.10) and Equation (3.11) yield

$$ \begin{align*}h^0(N_{Y/\mathbb{P}^M}(-kH)) \leq h^1(T_Y(-kH))\end{align*} $$

and

$$ \begin{align*}h^0(N_{Y/\mathbb{P}^M}(-kH+K_Y)) \leq h^1(T_Y(-kH+K_Y))\end{align*} $$

Hence we have

$$ \begin{align*}h^0(N_{\widetilde{Y}/\mathbb{P}^M}(-k\widetilde{H})) \leq h^1(T_Y(-kH+K_Y)) + h^1(T_Y(-kH)) + h^1(-kH-K_Y) + h^0(-kH-2K_Y)\end{align*} $$

Now the rest of the proof follows from Lemma 3.7, Lemma 3.8, Lemma 3.9. For the ease of the reader we point out that in the case $a = 2, b = 2, e = 0$ , the only term that contributes in the upper bound of $h^0(N_{\widetilde {Y}/\mathbb {P}^M}(-2\widetilde {H}))$ is $h^0(-2H-2K_Y) = 1$ .

Proof. Let $B^2 = 2g-2$ . The case $g = 2$ can be worked out separately (see, e.g., [Reference Ciliberto, Lopez and Miranda17]). For $r = 2, g = 3$ , $g(C) = 9$ and $\Phi _{\omega _C}$ cannot be surjective since the domain is of dimension less than the target. For $r = 2, g \geq 4$ or $r \geq 3, g \geq 3$ , consider the Hilbert scheme $\mathcal {H}_{r,g}$ . It is enough to prove the statement for a general element of $\mathcal {H}_{r,g}$ since surjectivity of $\Phi _{\omega _C}$ is an open condition. Let $M = 1+r^2(g-1)$ . Now we have an exact sequence

$$ \begin{align*}0 \to H^0(N_{X/\mathbb{P}^M}(-2)) \to H^0(N_{X/\mathbb{P}^M}(-1)) \to H^0(N_{C/\mathbb{P}^{M-1}}(-1))\end{align*} $$

By Proposition 4.1, $H^0(N_{X/\mathbb {P}^M}(-2)) = 0$ and hence we have that $\operatorname {corank}(\Phi _{\omega _C}) = h^0(N_{C/\mathbb {P}^{M-1}}(-1))-M \geq h^0(N_{X/\mathbb {P}^M}(-1))-M \geq 1$ .

Proof. The proof of the corollary follows from Proposition 4.1, the fact that $\operatorname {dim}\mathcal {H}_{r,g} = h^0(N_{X/\mathbb {P}^M}) = 18+(M+1)^2$ and the fact that

$$ \begin{align*}\operatorname{dim}T_{C(X)}(\mathcal{V}_{r,g,1}) = \Sigma_{k \geq 0} h^0(N_{X/\mathbb{P}^M}(-k))\\[-34pt]\end{align*} $$

Proof. The proof follows from standard deformation theory arguments (see, e.g., [Reference Ciliberto, Lopez and Miranda17, Lemma $3.7$ ], or [Reference Beauville4]).

Proof of $(1)$

First note that for every r, there is a unique component $\mathcal {H}_{r,g}$ of the Hilbert scheme parameterizing $K3$ surfaces $X \subset \mathbb {P}^{h^0(rB)-1}$ embedded by the complete linear of a very ample line bundle $rB$ , where $B^2 = 2g-2$ . Now except for the case $(r = 3, g = 3)$ , part $(1)$ follows from Theorem 3.11. For the case $(r=3, g=3)$ , note that if the hyperplane section of a Fano threefold V in $\mathcal {V}_{r,g,1}$ is denoted by $rL$ and $L^3 = d$ , then computing $h^0(rL) = h^0(rB)-1$ by Riemann-Roch theorem, we have $d = 2(g-1)/r$ . But for $r = 3, g = 3$ , d cannot be an integer. This resolves the case.

Proof of $(2)$

First one notes that for each r and g, $\mathcal {H}_{r,g}$ is irreducible. Now the list of examples of Fano-Iskovskih-Mori-Mukai (see, e.g., [Reference Ciliberto, Lopez and Miranda17, Table $3.1$ ]), shows that in each of the cases mentioned by part $(2)$ , there exist $K3$ surfaces in $\mathcal {H}_{r,g}$ which are extendable to smooth Fano threefolds. Hence by Lemma 4.5, in each of the cases, $\mathcal {H}_{r,g}$ is dominated by $\mathcal {V}_{r,g,1}$ . We end the proof by showing, $\mathcal {V}_{r,g,1}$ has a unique irreducible component and that the examples of Fano-Iskovskih-Mori-Mukai form an open dense subset of $\mathcal {V}_{r,g,1}$ . To see this, we observe first as in [Reference Ciliberto, Lopez and Miranda15] and [Reference Ciliberto, Lopez and Miranda17], that a projectively Cohen-Macaulay scheme degenerates along a flat family to the cone over its hyperplane section. Hence once again using, Lemma 4.5, it is enough to show that for a general $K3$ surface $X \subset \mathbb {P}^{M}$ in $\mathcal {H}_{r,g}$ , the cone $C(X) \subset \mathbb {P}^{M+1}$ is a smooth point of $\mathcal {V}_{r,g,1}$ . For the stated values of pairs $(r,g)$ , Corollary 4.4 along with Theorem 3.15 gives an upper bound for $\operatorname {dim}T_{C(X)}(\mathcal {V}_{r,g,1})$ which is listed as follows

1. (1) $\operatorname {dim}T_{C(X)}(\mathcal {V}_{2,3,1}) \leq 139$

2. (2) $\operatorname {dim}T_{C(X)}(\mathcal {V}_{2,4,1}) \leq 234$

3. (3) $\operatorname {dim}T_{C(X)}(\mathcal {V}_{2,5,1}) \leq 363$

4. (4) $\operatorname {dim}T_{C(X)}(\mathcal {V}_{2,6,1}) \leq 525$

5. (5) $\operatorname {dim}T_{C(X)}(\mathcal {V}_{3,4,1}) \leq 889$

6. (6) $\operatorname {dim}T_{C(X)}(\mathcal {V}_{4,3,1}) \leq 1209$ .

Now for a smooth Fano threefold $V \in \mathcal {V}_{r,g,1}$ , V represents a smooth point of $\mathcal {V}_{r,g,1}$ , since embedded deformations of an anticanonically embedded Fano variety of dimension at least $3$ can be shown to be unobstructed using the vanishing of $h^1(N_{V/\mathbb {P}^{M+1}})$ by Kodaira-Nakano vanishing theorem. Therefore the dimension of $\mathcal {V}_{r,g,1}$ at a smooth point is given by $\operatorname {dim}T_V(\mathcal {V}_{r,g,1}) = h^0(N_{V/\mathbb {P}^{M+1}})$ . Now using the descriptions of the examples of Fano-Iskovskih-Mori-Mukai, one can calculate the dimension $h^0(N_{V/\mathbb {P}^{M+1}})$ (see, e.g., [Reference Ciliberto, Lopez and Miranda15, Table $2$ , pg $666$ ] and [Reference Ciliberto, Lopez and Miranda17, Table $3.1$ ]) to check that in each case, $\operatorname {dim}T_V(\mathcal {V}_{r,g,1})$ is exactly the upper bound just obtained. Hence $C(X)$ is also a smooth point of $\mathcal {V}_{r,g,1}$ .

To see the statement on the irreducibility of fibers of p, note that for a general $X \subset V$ , we have an exact sequence

$$ \begin{align*}0 \to H^0(N_V(-1)) \to H^0(N_V) \to H^0(N_X) \to 0\end{align*} $$

The surjectivity arises from the fact that from Lemma 4.5, one can show that every embedded deformation of X can be lifted to an embedded deformation of V. Since $H^1(N_{V/\mathbb {P}^{M+1}}) = 0$ , this implies $H^1(N_{V/\mathbb {P}^{M+1}}(-1)) = 0$ . This implies that V is a smooth point of the fiber and that the dimension of the fiber at V is $h^0(N_V(-1))$ . Now once again the cone $C(X)$ lies in every irreducible component of the fiber and the tangent space to the fiber at $C(X)$ is given by

$$ \begin{align*}h^0(N_X(-1))+h^0(N_X(-2))\end{align*} $$

But we already checked that

$$ \begin{align*}h^0(N_V(-1)) = h^0(N_V) - h^0(N_X) = h^0(N_X(-1))+h^0(N_X(-2))\end{align*} $$

This implies that $C(X)$ is a smooth point of the fiber and hence the fiber is irreducible.

The last statement follows from the fact that $\alpha (X) = 1$ in these cases.

Proof. By Theorem 3.11, the cases to deal with are the lower genus cases $(1)-(7)$ listed in Theorem 3.15. Now $\mathcal {V}_{n,r,g,1} = \varnothing $ for $n \geq 4$ for cases 4, 6, 7 since $\alpha (X) \leq 1$ . $\mathcal {V}_{n,r,g,1} = \varnothing $ for $n \geq 3$ for case 5 has been shown in Theorem 4.6 part $(1)$ . Similarly $\mathcal {V}_{n,r,g,1} = \varnothing $ for $n \geq 6$ for case 3 since $\alpha (X) \leq 3$ . $\mathcal {V}_{n,r,g,1} = \varnothing $ for $n \geq 4$ , for cases 1, 2, can be shown by basic adjunction theory arguments, for example, see [Reference Ciliberto, Lopez and Miranda17, Theorem $3.15$ ]. It remains to show part $(2)$ . It is enough to show the statement for $\mathcal {V}_{5,2,5,1}$ . In this case by the same arguments in Corollary 4.4, we see that dimension of the tangent space $T_{C(X)}(\mathcal {V}_{5,2,5,1})$ to the (triple) cone $C(X) \in \mathcal {V}_{5,2,5,1}$ over $X \in \mathcal {V}_{2,5,1}$ is given by

$$ \begin{align*}\operatorname{dim}T_{C(X)}(\mathcal{V}_{5,2,5,1}) = 18+2(r^2(g-1))^2+3(\alpha(X)+2+r^2(g-1)) = 405\end{align*} $$

Now the result follows by comparing with the number of parameters of Mukai’s family of example.

Proof. Except in case $(5)$ , we have by Theorem 4.6 that $\mathcal {V}_{r,g,1}$ is non-empty and that $T_{C(X)}(\mathcal {V}_{r,g,1}) = T_V(\mathcal {V}_{r,g,1})$ where V is a smooth point of $\mathcal {V}_{r,g,1}$ . Hence by Corollary 4.4, we conclude that the upper bounds for $\alpha (X)$ in Theorem 3.15, are actually equalities.

Proof. Let us prove part $(1)$ . We apply [Reference Bangere, Gallego and Mukherjee5, Theorem $2.10$ ]. First of all, there exists nowhere vanishing sections inside $H^0(N_{Y/\mathbb {P}^M} \otimes K_Y)$ corresponding to embedded ribbons $\widetilde {Y}$ by [Reference Gallego and Purnaprajna26, Proposition $1.7$ ] for $a = 1$ and by [Reference Bangere, Mukherjee and Raychaudhury8, Theorem $3.1$ ] for $a \geq 2$ . A simple computation shows that $H^1(N_{Y/\mathbb {P}^M}) = 0$ and hence Y is unobstructed in $\mathbb {P}^M$ . If $a = 1$ , then we have that $H^1(N_{Y/\mathbb {P}^M} \otimes K_Y) = 0$ by [Reference Gallego and Purnaprajna26, Proposition $1.7$ ] while if $a \geq 2$ , the same vanishing follows from [Reference Bangere, Mukherjee and Raychaudhury8, Lemma $2.7$ , $(2)$ ] and hence, the functor $F_{E, Y}$ is unobstructed. If $a = 1$ , the smoothability of $\widetilde {Y}$ inside $\mathbb {P}^M$ and the smoothness of its Hilbert point follows from [Reference Gallego, González and Purnaprajna24, Theorem $2.7$ ] and [Reference Gallego and Purnaprajna26, Theorem $4.1$ ] respectively while if $a \geq 2$ , the same follows from [Reference Bangere, Mukherjee and Raychaudhury8, Theorem $4.3$ ] and [Reference Bangere, Mukherjee and Raychaudhury8, Theorem $5.1$ ] respectively. Therefore part $(1)$ follows from [Reference Bangere, Gallego and Mukherjee5, Theorem $2.10$ ].

To show part $(2)$ we need to show that $H^1(N_{V/\mathbb {P}^M}') = 0$ for one subvariety of the form $V = Y_1 \bigcup _E Y_2$ where $Y_2 \in \mathcal {F}_{E,Y}$ . For this, first note that for any such V we have an exact sequence

$$ \begin{align*}0 \to N_{V/\mathbb{P}^M}' \to N_{V/\mathbb{P}^M} \to T_V^1 \to 0\end{align*} $$

Note that by part $(1)$ , we have a flat family $\mathcal {Y} \to T$ over a smooth irreducible curve T, such that $\mathcal {Y}_0$ is an embedded $K3$ carpet $\widetilde {Y}$ on Y and $\mathcal {Y}_t$ is a subvariety of the form $V = Y_1 \bigcup _E Y_2$ . When $a \geq 2$ , by [Reference Bangere, Mukherjee and Raychaudhury8, Theorem $5.1$ , equation $(5.9)$ ], we have that $H^1(N_{\widetilde {Y}/\mathbb {P}^M}) = 0$ , while for $a = 1$ , the same holds by [Reference Gallego and Purnaprajna26, Theorem  $4.1$ ]. Now since the family $\mathcal {Y} \to T$ is local complete intersection, we have that $N_{\mathcal {Y}/\mathbb {P}_T^M}$ is locally free and hence flat over T. Therefore by semi-continuity, $H^1(N_{\mathcal {Y}_t/\mathbb {P}^M}) = 0$ , that is, $H^1(N_{V/\mathbb {P}^M}) = 0$ . The vanishing of $H^1(N_{V/\mathbb {P}^M}')$ now follows if we show that $H^0(N_{V/\mathbb {P}^M}) \to H^0(T_V^1)$ is a surjection. Once again, we show this is a surjection by passing to the ribbon, that is, we show that $H^0(N_{\widetilde {Y}/\mathbb {P}^M}) \to H^0(T_{\widetilde {Y}}^1)$ is a surjection. First note that the sheaf $T_{\widetilde {Y}}^1 = \mathcal {O}_Y(-2K_Y)$ . Granting this, note that it sits as the last term in the exact sequence (see Equation (3.2))