Let $X \subseteq \mathbb{P}^n$ be a irreducible complex projective variety. It is called nondegenerate if it is not contained in a hyperplane in $\mathbb{P}^n$.

Assuming $X$ is nondegenerate and irreducible, I am wondering whether a hyplernane section $H \cap X$ of $X$ is also nondegenerate in $H$, assuming the intersection of $X$ and $H$ is "transverse" meaning it has non components with multiplicity greater than one.

I think I have found a proof of this but this seems like a very strong statement and I couldn't find it anywhere in the literature so I'm a bit skeptical.

My proof(?):

Consider the diagram $$ \require{AMScd} \newcommand{\O}{\mathcal{O}} \begin{CD} X \cap H @>j>> X\\ @Vi'VV & @ViVV \\ H @>k>> \mathbb{P}^n \end{CD} $$ where all arrows are immersions. We get a corresponding diagram of global sections

\begin{CD} H^0(X \cap H, j^* i^* \O(1)) @<j^*<< H^0(X, i^*\O(1))\\ @Ai'^*AA & @Ai^*AA \\ H^0(H, k^*\O(1)) @<k^*<< H^0(\mathbb{P}^n, \O(1)) \end{CD}

Since $k^*\O(1) = \O_H(1)$ is the hyperplane bundle of $H \cong \mathbb{P}^{n-1}$ and $j^* j^* \O(1) \cong i'^* \O_H(1)$ saying that $X \cap H$ is nondegenerate is equivalent to $i'^*$ being injective.

We know that $k^*$ is surjective and has a one-dimensional kernel and since we're assuming $X \hookrightarrow \mathbb{P}^n$ is nondegenerate, $i^*$ is injective. Finally, consider the short exact sequence of sheaves $$ 0 \to \mathscr{I}_{X \cap H / X} \otimes i^*\O(1) \to i^*O(1) \to j_* j^* i^* \O(1) \to 0 \label{eq:ses}\tag{1} $$ We have $X \cap H \subsetneq X$ by nondegeneracy, so by irreducibility, $X \cap H$ is a hypersurface in X and $$ \mathscr{I}_{X \cap H / X} \cong \O_X(-[X \cap H]) \cong i^*\O(-1) $$ Therefore the left term in the short exact sequence (\ref{eq:ses}) is $\O_X$. (Here I think we either need to work with the scheme-theoretic intersection $X \cap H$ or assume that the intersection of $X$ and $H$ has no components with multiplicity greater than one.) Taking global section of (\ref{eq:ses}) gives us $$ 0 \to H^0(X, \O_X) \to H^0(X, i^*\O(1)) \xrightarrow{j^*} H^0(X \cap H, j^* i^* \O(1)) $$ Since $H^0(X, \O_X) \cong \mathbb{C}$ by irreducibility we see that $j^*$ has a one-dimensional kernel. This gives us $1 \geq \dim \ker (j^* i^*) = \dim \ker (i'^* k^*) = 1 + \dim\ker i'^*$ so $i'^*$ must be injective

Let $X \subseteq \mathbb{P}^n$ be a irreducible complex projective variety. It is called nondegenerate if it is not contained in a hyperplane in $\mathbb{P}^n$.

Assuming $X$ is nondegenerate and irreducible, I am wondering whether a hyplernane section $H \cap X$ of $X$ is also nondegenerate in $H$, assuming the intersection of $X$ and $H$ is "transverse" meaning it has non components with multiplicity greater than one.

I think I have found a proof of this but this seems like a very strong statement and I couldn't find it anywhere in the literature so I'm a bit skeptical.

My proof(?):

Consider the diagram $$ \require{AMScd} \newcommand{\O}{\mathcal{O}} \begin{CD} X \cap H @>j>> X\\ @Vi'VV & @ViVV \\ H @>k>> \mathbb{P}^n \end{CD} $$ where all arrows are immersions. We get a corresponding diagram of global sections

\begin{CD} H^0(X \cap H, j^* i^* \O(1)) @<j^*<< H^0(X, i^*\O(1))\\ @Ai'^*AA & @Ai^*AA \\ H^0(H, k^*\O(1)) @<k^*<< H^0(\mathbb{P}^n, \O(1)) \end{CD}

Since $k^*\O(1) = \O_H(1)$ is the hyperplane bundle of $H \cong \mathbb{P}^{n-1}$ and $j^* i^* \O(1) \cong i'^* \O_H(1)$ saying that $X \cap H$ is nondegenerate is equivalent to $i'^*$ being injective.

We know that $k^*$ is surjective and has a one-dimensional kernel and since we're assuming $X \hookrightarrow \mathbb{P}^n$ is nondegenerate, $i^*$ is injective. Finally, consider the short exact sequence of sheaves $$ 0 \to \mathscr{I}_{X \cap H / X} \otimes i^*\O(1) \to i^*O(1) \to j_* j^* i^* \O(1) \to 0 \label{eq:ses}\tag{1} $$ We have $X \cap H \subsetneq X$ by nondegeneracy, so by irreducibility, $X \cap H$ is a hypersurface in X and $$ \mathscr{I}_{X \cap H / X} \cong \O_X(-[X \cap H]) \cong i^*\O(-1) $$ Therefore the left term in the short exact sequence (\ref{eq:ses}) is $\O_X$. (Here I think we either need to work with the scheme-theoretic intersection $X \cap H$ or assume that the intersection of $X$ and $H$ has no components with multiplicity greater than one.) Taking global section of (\ref{eq:ses}) gives us $$ 0 \to H^0(X, \O_X) \to H^0(X, i^*\O(1)) \xrightarrow{j^*} H^0(X \cap H, j^* i^* \O(1)) $$ Since $H^0(X, \O_X) \cong \mathbb{C}$ by irreducibility we see that $j^*$ has a one-dimensional kernel. This gives us $1 \geq \dim \ker (j^* i^*) = \dim \ker (i'^* k^*) = 1 + \dim\ker i'^*$ so $i'^*$ must be injective

Let $X \subseteq \mathbb{P}^n$ be a irreducible complex projective variety. It is called nondegenerate if it is not contained in a hyperplane in $\mathbb{P}^n$.

Assuming $X$ is nondegenerate and irreducible, I am wondering whether a hyplernane section $H \cap X$ of $X$ is also nondegenerate in $H$, assuming the intersection of $X$ and $H$ is "transverse" meaning it has non components with multiplicity greater than one.

I think I have found a proof of this but this seems like a very strong statement and I couldn't find it anywhere in the literature so I'm a bit skeptical.

My proof(?):

Consider the diagram $$ \require{AMScd} \newcommand{\O}{\mathcal{O}} \begin{CD} X \cap H @>j>> X\\ @Vi'VV & @ViVV \\ H @>k>> \mathbb{P}^n \end{CD} $$ where all arrows are immersions. We get a corresponding diagram of global sections

\begin{CD} H^0(X \cap H, j^* i^* \O(1)) @<j^*<< H^0(X, i^*\O(1))\\ @Ai'^*AA & @Ai^*AA \\ H^0(H, k^*\O(1)) @<k^*<< H^0(\mathbb{P}^n, \O(1)) \end{CD}

Since $k^*\O(1) = \O_H(1)$ is the hyperplane bundle of $H \cong \mathbb{P}^{n-1}$ and $j^* j^* \O(1) \cong i'^* \O_H(1)$ saying that $X \cap H$ is nondegenerate is equivalent to $i'^*$ being injective.

We know that $k^*$ is surjective and has a one-dimensional kernel and since we're assuming $X \hookrightarrow \mathbb{P}^n$ is nondegenerate, $i^*$ is injective. Finally consider the short exact sequence of sheaves $$ 0 \to \mathscr{I}_{X \cap H / X} \otimes i^*O(1) \to i^*O(1) \to j_* j^* i^* O(1) \to 0 \label{eq:ses}\tag{1} $$ We have $X \cap H \subsetneq X$ by nondegeneracy, so by irreducibility, $X \cap H$ is a hypersurface in X and $$ \mathscr{I}_{X \cap H / X} \cong \O_X(-[X \cap H]) \cong i^*\O(-1) $$ Therefore the left term in the short exact sequence (\ref{eq:ses}) is $\O_X$. (Here I think we either need to work with the scheme-theoretic intersection $X \cap H$ or assume that the intersection of $X$ and $H$ has no components with multiplicity greater than one.) Taking global section of (\ref{eq:ses}) gives us $$ 0 \to H^0(X, \O_X) \to H^0(X, i^*O(1)) \xrightarrow{j^*} H^0(X \cap H, j^* i^* O(1)) $$ Since $H^0(X, \O_X) \cong \mathbb{C}$ by irreducibility we see that $j^*$ has a one-dimensional kernel. This gives us $1 \geq \dim \ker (j^* i^*) = \dim \ker (i'^* k^*) = 1 + \dim\ker i'^*$ so $i'^*$ must be injective

Let $X \subseteq \mathbb{P}^n$ be a irreducible complex projective variety. It is called nondegenerate if it is not contained in a hyperplane in $\mathbb{P}^n$.

Assuming $X$ is nondegenerate and irreducible, I am wondering whether a hyplernane section $H \cap X$ of $X$ is also nondegenerate in $H$, assuming the intersection of $X$ and $H$ is "transverse" meaning it has non components with multiplicity greater than one.

I think I have found a proof of this but this seems like a very strong statement and I couldn't find it anywhere in the literature so I'm a bit skeptical.

My proof(?):

Consider the diagram $$ \require{AMScd} \newcommand{\O}{\mathcal{O}} \begin{CD} X \cap H @>j>> X\\ @Vi'VV & @ViVV \\ H @>k>> \mathbb{P}^n \end{CD} $$ where all arrows are immersions. We get a corresponding diagram of global sections

\begin{CD} H^0(X \cap H, j^* i^* \O(1)) @<j^*<< H^0(X, i^*\O(1))\\ @Ai'^*AA & @Ai^*AA \\ H^0(H, k^*\O(1)) @<k^*<< H^0(\mathbb{P}^n, \O(1)) \end{CD}

Since $k^*\O(1) = \O_H(1)$ is the hyperplane bundle of $H \cong \mathbb{P}^{n-1}$ and $j^* j^* \O(1) \cong i'^* \O_H(1)$ saying that $X \cap H$ is nondegenerate is equivalent to $i'^*$ being injective.

We know that $k^*$ is surjective and has a one-dimensional kernel and since we're assuming $X \hookrightarrow \mathbb{P}^n$ is nondegenerate, $i^*$ is injective. Finally, consider the short exact sequence of sheaves $$ 0 \to \mathscr{I}_{X \cap H / X} \otimes i^*\O(1) \to i^*O(1) \to j_* j^* i^* \O(1) \to 0 \label{eq:ses}\tag{1} $$ We have $X \cap H \subsetneq X$ by nondegeneracy, so by irreducibility, $X \cap H$ is a hypersurface in X and $$ \mathscr{I}_{X \cap H / X} \cong \O_X(-[X \cap H]) \cong i^*\O(-1) $$ Therefore the left term in the short exact sequence (\ref{eq:ses}) is $\O_X$. (Here I think we either need to work with the scheme-theoretic intersection $X \cap H$ or assume that the intersection of $X$ and $H$ has no components with multiplicity greater than one.) Taking global section of (\ref{eq:ses}) gives us $$ 0 \to H^0(X, \O_X) \to H^0(X, i^*\O(1)) \xrightarrow{j^*} H^0(X \cap H, j^* i^* \O(1)) $$ Since $H^0(X, \O_X) \cong \mathbb{C}$ by irreducibility we see that $j^*$ has a one-dimensional kernel. This gives us $1 \geq \dim \ker (j^* i^*) = \dim \ker (i'^* k^*) = 1 + \dim\ker i'^*$ so $i'^*$ must be injective