# Any hyperplanes is covered by non-Lefschetz pencils?

Let $$X\subset \mathbb P^n$$ be a smooth hypersurface over base field $$\mathbb C$$. A pencil of hyperplanes is just a projective line $$(X_t)$$ in $$\mathbb P^{n*}$$. It is called a Lefschetz pencil if it satisfies the following: (I followed the definition in SGA)

(i) The axis (intersection of $$X_t$$'s) intersects with $$X$$ transversally.

(ii) Almost all (enough for one) $$X_t$$ intersects with $$X$$ transversally.

(iii) The non-transversal intersection in (ii) contains one node.

And on SGA it also proves that almost all pencils are Lefschetz. Now I want to know how many are those non-Lefschetz pencils, more precisely:

Does there exist some $$X$$, on which all the non-Lefschetz pencils cover the $$\mathbb P^{n*}$$? i.e. for any hyperplane section $$H$$, we can find non-Lefschetz pencil $$(X_t)$$ pass through it.

I guess it does not exist, but have no idea how to prove it. Any hints or references would be helpful.

Remark

I should remove the condition (i) otherwise it is trivial.

Assume there is a single hyperplane section $$X_0$$ of $$X$$ which has worse singularity than just a node. Then any pencil containing $$X_0$$ is non-Lefschetz, and these pencils cover all the $$\mathbb{P}^{n*}$$.
Of course, if there is no such $$X_0$$ as above, this construction does not work. However, I think the only case when such $$X_0$$ does not exist is that of $$X$$ being a smooth quadric. In that case the projective dual variety to $$X$$ is yet another smooth quadric, and lines in $$\mathbb{P}^{n*}$$ tangent to it provide non-Lefschetz pencils that cover all the $$\mathbb{P}^{n*}$$.
• A precise reference should include expose and section numbers. In fact "axis" ("l'axe") in English is usually called "base locus". It surely doesn't follow from (ii). For instance, take $X \subset \mathbb{P}^2$ to be a smooth conic, and the pencil of lines through a point ON $X$. – Sasha Nov 17 '18 at 14:44