Lifting of quadrics containing a curve

Question

Let $C \subset \mathbb{P}^r$ be a projective curve (over $k=\mathbb{C}$), smooth, irreducible and nondegenerate of degree $d$, ie the embedding line bundle $\mathcal{O}_C(1)=(\mathcal{O}_{\mathbb{P}^r}(1) )\vert _C$ has degree $d$.

Assume that $d > 2r − 2$. Let $\mathbb{P}^{r-1} \cong H \subset \mathbb{P}^{r}$ be a general hyperplane, such that the hyperplane section $H \cap C$ consists of $d$ points in general position.

Let $Q \in \Gamma(H,\mathcal{O}_H(2)) $ be any quadric in $H$ containing a hyperplane section $H \cap C$, or in other words, it's image/ "pullback" under canon restriction map $p^*_{H\cap C}$ in following diagram is zero:

$$ \require{AMScd} \begin{CD} \Gamma(\mathbb{P}^r, \mathcal{O}_{\mathbb{P}^r}(2)) @>{p^*_{C,2}} >> \Gamma(C, \mathcal{O}_{C}(2)) \\ @Vp^*_HVV @VVV \\ \Gamma(H, \mathcal{O}_{H}(2)) @>{p^*_{H\cap C}}>> \Gamma(H \cap C) =\mathbb{C}^d \end{CD} $$

Question: Under which conditions it is known that the quadric $Q$ admits a lift to a quadric $\overline{Q} \in \Gamma(\mathbb{P}^r, \mathcal{O}_{\mathbb{P}^r}(2))$ containing the curve $C$, ie that which is mapped to zero under $p^*_{C,2}$ in the diagram?
More generally, what about if we pose the same question but with cubics, quartics, etc, instead...

Motivation: This can be regarded as generalization of the question posed here, where is was asked - refering the a claim on page 75 in these lecture notes - why projective normality of $C$ gives a sufficient condition for the stated claim in the question.

(Recall, "projectively normal", means that the canonical pullback maps

$$ p^*_{C,s}:\Gamma(\mathbb{P}^r, \mathcal{O}_{\mathbb{P}^r}(s))\to\Gamma(C,\mathcal{O}_C(s)) $$

are surjective.
That looks cumbersome from viewpoint of naive dimension count. In general, the dimension of the kernel of $p^*_{C,s}$ which is $\dim_k \Gamma(\mathbb{P}^r, \mathcal{O}_{\mathbb{P}^r}(s)) -\dim_k\text{Image}(p^*_{C,s})$ is bounded from below by $\dim_k \Gamma(\mathbb{P}^r, \mathcal{O}_{\mathbb{P}^r}(k)) -\dim_k\Gamma(C,\mathcal{O}_C(s)) $, and as the restriction of $p^*_{C,s}$ to kernel of $p^*_{H}$ is injective by nondegeneracy of $C$, it looks more naturally from dimensional reasons considering the diagram above that we could find a lift of $Q$ that maps zero, if the kernel of $p^*_{C,s}$ is big, or $\dim_k\text{Image}(p^*_{C,s})$ is small, but for $p^*_{Cs}$ surjective, the dimension of image gets maximal.

That looks rather counter intuitive if we only focus linear algebra in the quesr tion.
So I'm woundering if *projective normality" really gives a sufficient criterion for the existence of such lift of $Q$ containing the curve $C$, and if not, what is the "right" reasonable criterion? Does it has some "geometric" intuition?

Of course, more generally one feels to be tempted to pose same question for arbitrary $k$ instead of $s=2$, so eg for cubics, quartics, etc, presumably we inpose approp restrictions on degree $d$ of the curve.

Answer 1

1-normality is sufficient, i.e. it suffices that $H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(1))\to H^0(C, \mathcal O_C(1))$ is surjective.

Indeed, choose class $q \in H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(2))$ lifting the equation defining the quadric $q$, and a class $a \in H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(1))$ vanishing on $H$. Then $a \mid_C$ vanishes only on $H \cap C$, and vanishes to multiplicity $1$ at each point of $H \cap C$ (since the intersection is transverse, which is the only assumption we need on $H$).

Since $q|_C$ vanishes on $H \cap C$, it follows that $a \mid_C \in H^0(C, \mathcal O_C(1))$ divides $q\mid_C \in H^0(C, \mathcal O_C(2))$ producing a quotient in $\overline{r}\in H^0(C, \mathcal O_C(1))$.

Now lifting $\overline{r}$ to a class $r\in H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(1))$ using 1-normality, we see that $q-ra$ vanishes on $C$, and thus the vanishing locus of $q-ra$ gives the desired lift $\overline{Q}$.