Linear sections of Segre varieties and rational normal scrolls

Question

In a projective space $\mathbb{P}^{k+2}$ consider two complementary subspaces $\mathbb{P}^1,\mathbb{P}^k$, and let $C\subset\mathbb{P}^k$ be a degree $k$ rational normal curve. Fixed an isomorphism $\phi:\mathbb{P}^1\rightarrow C$ we consider the rational normal scroll

$$S_{(1,k)} = \bigcup_{p\in \mathbb{P}^1}\left\langle p, \phi(p)\right\rangle\subset \mathbb{P}^{k+2}$$

where $\left\langle p, \phi(p)\right\rangle$ is the line through $p$ and $\phi(p)$.

Now, let $\Sigma_{(1,k)}$ be the Segre embedding of $\mathbb{P}^1\times \mathbb{P}^k$ in $\mathbb{P}^{2k+1}$.

Can $S_{(1,k)}$ be recovered intersecting $\Sigma_{(1,k)}$ with a linear subspace of dimension $k+2$ of $\mathbb{P}^{2k+1}$? This is known to be true if $k = 2$. Indeed a general hyperplane section of $\Sigma_{(1,2)}\subset\mathbb{P}^5$ is a scroll surface of the form $S_{(1,2)}$.

Answer 1

The scroll $S_{(1,k)}$ can be locally parametrized by the map $$ \begin{array}{cccc} \phi: & \mathbb{A}^1\times\mathbb{P}^1 & \longrightarrow & \mathbb{P}^{k+2}\\ & (u,[\alpha_0:\alpha_1]) & \mapsto & [\alpha_0 u:\alpha_0:\alpha_1 u^k:\alpha_1 u^{k-1}:\dots:\alpha_1 u:\alpha_1]. \end{array} $$ Now, consider the Segre embedding $$ \begin{array}{cccc} \sigma: & \mathbb{P}^1\times\mathbb{P}^k & \longrightarrow & \mathbb{P}^{2k+1}\\ & ([u:v],[\alpha_0:\dots:\alpha_k]) & \mapsto & [\alpha_0u:\dots:\alpha_ku:\alpha_0v:\dots :\alpha_kv]. \end{array} $$ and let $\Sigma_{(1,k)}$ be its image. Note that $\Sigma_{(1,k)}$ is locally parametrized by $$ \begin{array}{cccc} \widetilde{\sigma}: & \mathbb{A}^1\times\mathbb{P}^k & \longrightarrow & \mathbb{P}^{2k+1}\\ & ([u:1],[\alpha_0:\dots:\alpha_k]) & \mapsto & [\alpha_0u:\dots:\alpha_ku:\alpha_0:\dots :\alpha_k]. \end{array} $$ and that $\deg(\Sigma_{(1,k)}) = \deg(S_{(1,k)}) = k+1$. Now, take $\alpha_{i} = \alpha_1u^{i-1}$ for $i=2,\dots,k$. Then $$\widetilde{\sigma}(u,[\alpha_0:\alpha_1:\alpha_1u:\dots :\alpha_1u^{k-1}]) = [\alpha_0u:\alpha_1u:\alpha_1u^2:\dots:\alpha_1u^{k-1}:\alpha_1u^k:\alpha_0:\alpha_1:\alpha_1u:\dots\alpha_1u^{k-1}]$$ and the coordinates functions of this last map are exactly the ones appearing in the above expression of $\phi$.

Therefore, if $[Z_0:\dots:Z_{2k+1}]$ are the homogeneous coordinates on $\mathbb{P}^{2k+1}$ and $$H^{k+2} = \{Z_j-Z_{k+j+2}=0,\: j = 1,\dots,k-1\}\cong\mathbb{P}^{k+2}$$ then we have $$S_{(1,k)} = \Sigma_{(1,k)}\cap H^{k+2}\subset\mathbb{P}^{2k+1}$$

Answer 2

Consider a linear section of codimension $k-1$ of $\mathbb{P}^1 \times \mathbb{P}^k$. It is equal to the projectivization over $\mathbb{P}^1$ of the vector bundle $$ \mathcal{O}(-1)^{\oplus k+1} \to \mathcal{O}^{\oplus k-1}. $$ In general the kernel is $\mathcal{O}(-(k+1)/2)^{\oplus 2}$ or $\mathcal{O}(-k/2) \oplus \mathcal{O}((k+2)/2)$ (depending on parity of $k$), but for special sections it may degenerate to any $\mathcal{O}(a) \oplus \mathcal{O}(b)$ with $a + b = -k-1$ and $a,b \le -1$. In particular, $\mathcal{O}(-k) \oplus \mathcal{O}(-1)$ is a possible degeneration. In this case, the linear section is $$ \mathbb{P}_{\mathbb{P}^1}(\mathcal{O}(-1) \oplus \mathcal{O}(-k)) \cong S_{(1,k)}. $$ To achieve this you should pick a general linear section that contains a line $\mathbb{P}^1 \times P \subset \mathbb{P}^1 \times \mathbb{P}^k$ for some point $P \in \mathbb{P}^k$.