# Special linear sections of a hypersurface

We discuss on the field of complex numbers. Let $X$ be a smooth projective hypersurface of dimension $n \geq 4$ in $\mathbb{P}^{n+1}(\mathbb{C})$. Assume that for a general point $x \in X$, there exists a 4-dimensional linear subspace $L_x$ in $\mathbb{P}^{n+1}$ passing through $x$ such that the set-theoretic intersection of $L_x$ and $X$ is an irreducible cubic hypersurface in $L_x$. Then does $X$ have to be a cubic hypersurface?

We know that if $L_x$ is a general 4-dimensional linear subspace passing through $x$, then $L_x \cap X$ is a hypersurface with the same degree as $X$. The trouble is that if $L_x$ is special, then maybe the scheme-theoretic intersection of $L_x$ and $X$ is reduced (for example, the restriction on $L_x$ of the equation $f$ of $X$ in $\mathbb{P}^{n+1}$ maybe a square of a cubic equation and in this case $\deg(X)=6$). Hence, without the generality assumption of $x \in X$, the answer of my question is negative.

Now we have assumed the point $x$ is general in $X$ and that $X$ is smooth. Probably these assumptions can imply that $\deg(X)=3$. But I do not how to prove it.

The answer is "no" if the degree $d$ is a multiple of $3$, say $d=3e$. I sketched the argument in the comments, but now that comment has disappeared. For a point $x$ in $X$, there is a Grassmannian $G= \mathbb{G}(\mathbb{P}^3,\mathbb{P}^{n-1})$ parameterizing linear spaces $L_x$ of dimension $4$ that contain $x$. If $x$ is a general point, then there is a dense open subset $U$ of $G$ parameterizing $L_x$ that are not contained in $X$. There is a morphism, $$\zeta: U \to [\mathbb{P}k[x_0,x_1,x_2,x_3,x_4]_d / \textbf{PGL}_5],$$ that sends each $L_x$ to the moduli point of the degree d hypersurface $L_x\cap X$ in $L_x \cong \mathbb{P}^4$. There is a subvariety $S$ of the target parameterizing hypersurfaces $[Y]$ that are of the form $e[Z]$ for a cubic hypersurface $Z$ (using additive notation). If the morphism $\zeta$ is surjective, then the image will intersect $S$, even if $e$ does not equal $1$.

It is a straightforward parameter count to determine when $\zeta$ should be dominant. That parameter count can be made into a rigorous proof. An example of this is in my preprint, "Fano Varieties and Linear Sections of Hypersurfaces".

Edit. I was asked to provide more details. Let $F(x_0,x_1,x_2,x_3,x_4)$ be any nonzero homogeneous polynomial of degree $d=3e$, e.g., $(x_0^3+x_1^3+x_2^3+x_4^3)^e$. Consider $F$ as an element in both $k[x_0,\dots,x_3]_d$ and in $k[x_0,\dots,x_n]_d$, the finite dimensional $k$-vector space of homogeneous polynomials $G$ of degree $d$ on $\mathbb{P}^4$, resp. on $\mathbb{P}^n$. Let $V_F \subset k[x_0,\dots,x_n]_d$ denote the subspace of those $G$ such that $G(x_0,x_1,x_2,x_3,0,\dots,0)$ is a scalar multiple of $F$. Note that $V_F$ contains a subspace, call it $W$, of polynomials $G$ such that $G(x_0,x_1,x_2,x_3,0,\dots,0)$ is the zero polynomial. Moreover $W$ does not equal all of $V_F$, since $F$ is in $V_F\setminus W$. Thus $W$ is a codimension $1$ linear subspace of $V_F$. Thus also the projective linear subspace $\mathbb{P}W$ of $\mathbb{P}V_F$ has codimension $1$.

Now let $A$ denote the linear algebraic group of all projective linear automorphisms $\alpha$ of $\mathbb{P}^n$ that map the point $p=[1,0,\dots,0]$ to itself. Let $U\subset k[x_0,\dots,x_n]_d$ denote the subspace of all $G$ such that $G(1,0,\dots,0)$ equals $0$.
Since $\mathbb{P}W$ is a Cartier divisor in $\mathbb{P}V_F$, also $A\times \mathbb{P}W$ is a Cartier division in $A\times\mathbb{P}V_F$. There is an induced morphism, $$f : A\times \mathbb{P} V_F \to \mathbb{P}U, \ \ f(\alpha,[G]) = G\circ \alpha^{-1}.$$ Denote by $f_W$ the restriction of $f$ to $A\times \mathbb{P}W$.

Hypothesis. The positive integers $(n,d)$ satisfy the inequality $4(n-4) \geq (d+3)(d+2)(d+1)/6$.

Under this hypothesis, the claim is that $f$ is surjective. To prove this, it suffices to prove that the restriction $f_W$ is surjective. This new claim is equivalent to the claim that every point $q$ on every degree $d$ hypersurface $\text{Zero}(G)$ in $\mathbb{P}^n$ is contained in a linear $\mathbb{P}^4$ that is contained in $\text{Zero}(G)$. Assuming this claim, then we are done: let $S \subset \mathbb{P}U$ be the open subset parameterizing smooth hypersurfaces. Then $f^{-1}(S)$ is an open subset of $A \times \mathbb{P} V_F$. It is nonempty by the claim. This nonempty open subset cannot be contained in the Cartier divisor $A\times \mathbb{P}W$. Thus, this nonempty open subset intersect $A\times \mathbb{P}V_F \setminus A \times \mathbb{P}W$. That means that for every sufficiently general smooth hypersurface and for every general point on that hypersurface, there is a is a linear $\mathbb{P}^4$ containing that point whose intersection with the hypersurface is projectively linearly equivalent to $\text{Zero}(F)$.

So now it suffices to prove that $f_W$ is surjective. This can be a bit delicate if we want to get precisely the correct inequality above. However, if you allow me to make $n$ much larger, then it is trivial. We can make the normal bundle of a given $\mathbb{P}^4$ in a given hypersurface "sufficiently general", via appropriate choices of the partial derivatives $\partial G/\partial x_{4+i}|_{\mathbb{P}^4}$, so that even the derivative of $f_W$ is explicitly surjective.

• Thanks for your answer! But there are two questions about your answer. Firstly, I doubt that hypersurfaces in P^{4} that have multiplicity>=2 are semistable elements or not (if not, then the surjectivity of \zeta has no help to my question). Secondly, we can only get that \zeta is dominate instead of that it is surjective. Hence, your answer can not convince me. Mar 12, 2015 at 4:17
• That quotient is a stack, not a geometric invariant theory quotient. Also, the point of my answer, and of the article I linked, is that it is easier to prove the conclusion than to understand the morphism $\zeta$. This is just an incidence correspondence computation: form the incidence correspondence of pairs of a hypersurface in $\mathbb{P}^n$ together with a linear $\mathbb{P}^4$ section. If you do not (yet) know how to make incidence correspondence arguments, then you have more fundamental questions to answer than the one above. Mar 12, 2015 at 6:09
• From the answer above, it can be seen that when $n$ is large enough comparing to $d$, then the counterexample appears. Then my new question is that: let $X$ be as in the first question, then whether can we show that $d(d+1)\leq 2n-2$? Or equivalently, is the variety parameterizing the lines on $X$ passing through a fixed general point covered by lines ? If this is true, then we can say that the degrees of this kind of special hypersurfaces $X$ are small enough. May 23, 2015 at 7:10