# Secant Lines contained in Hypersurfaces

If X is a hypersurface of degree d in $\mathbb{P}^n$ and S is the singular locus of X then when is it true that the secant line of two points in S is contained in X? I think this has something to do with intersection theory but I really have no idea about how to think about this question. For cubic hypersurfaces I've seen that the secant line connecting two singular point of the hypersurface lies in the hypersurface but is this true for higher degree hypersurfaces?

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## 2 Answers

The statement that you mention about cubic hypersurface is a consequence of Besout theorem. The intersection of L with S counted with multiplicity is 3. And every (isolated) multiple point contributes at least 2. 2+2>3, this is why the line is in the cubic. But for any other degree this will not be true, you can construct a hypersurface of degree 4 with two double points and generically the secant line in not indide. A very degenerate such an example of degree 4 will be a union of two quadric. The higher is the degree the easier is to construct examples

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The condition you need is the following. Take two points in the singular locus $x,y\in S$. Assume that the multiplicities $m_x,m_y$ of $x,y$ as points of $X$ satisfy $m_x+m_y > deg(X)$ then the line $L_{x,y}:=\left\langle x,y\right\rangle$ is contained in $X$. Indeed the intersection $L\cdot X\geq m_x+m_y > deg(X)$ implies, by Bezout's theorem, $L_{x,y}\subset X$.

Take $X$ an hypersurface of degree $d$ with a points $x\in X$ of multiplicty $d$. For any other point $y\neq x$ in $X$ we have $$L_{x,y}\cdot X\geq d+1 >d.$$ Then any line through $x$ and any other point of $X$ is contained in $X$, and $X$ is a cone of vertex $x$ over an hypersurface $Y\subset\mathbb{P}^{n-1}\subset\mathbb{P}^n$ of degree $d$.

In your case $d = 3,m_x = m_y = 2$, and $m_x+m_y = 4 >d$. Indeed an irreducibile cubic surface $X\subset\mathbb{P}^3$ can have at most $4$ nodes. In this case any of the $6$ lines spanned by two of the nodes is contained in $X$.

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+1 for a clear answer. – P Vanchinathan Apr 13 '14 at 4:25