Lefschetz Hyperplane Theorem says: Let $X\subset\mathbb{C}P^n$ be a smooth projective variety of complex dimension $m$ and let $Y = X \cap H$ be a generic hyperplane section. Then the natural morphism $\pi_i(Y)\to \pi_i(X)$ is an isomophism for $i\le m-2$ and is onto for $i=m-1$.

Now $X\subset\mathbb{C}P^n$ be a simply connected smooth projective variety of complex dimension $m$ and $Y$ is the intersection of $X$ with hyperplanes s.t. $\dim Y=m-d$.

Then why does Lefschetz Hyperplane Theorem show that $Y$ is connected and simply connected if $\dim(Y) = \dim(X)-d = m-d\ge 2$?

I think it used the fundamental group of $X$ is trivial, however, how is that related to the dimension of $Y$ in Lefschetz Hyperplane Theorem?

  • $\begingroup$ Well, this certainly require an assumption on $X$. If $d=0$, then $Y=X$, so we need $X$ to be simply connected. $\endgroup$ – Roland Dec 6 '18 at 8:35
  • $\begingroup$ @Roland Thanks, I have fixed $\endgroup$ – 6666 Dec 6 '18 at 8:41
  • $\begingroup$ @Roland now how to apply Lefschetz Hyperplane Theorem to get the result? $\endgroup$ – 6666 Dec 6 '18 at 8:47
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    $\begingroup$ Well then this is easy : let $X=Y_0\supset Y_1\supset...\supset Y_d=Y$ where $Y_i$ is $X$ intersected by $i$-hyperplane, so that $\dim Y_i=m-i$. Then $\pi_1(Y_d)\to \pi_1(Y_{d-1})\to...\to \pi_1(Y_0)$ are isomorphisms : indeed for $1\leq i\leq d$, we have $1\leq m-(i-1)-2$ by assumption. So $\pi_1(Y_i)\to\pi_1(Y_{i-1})$ is an isomorphism. $\endgroup$ – Roland Dec 6 '18 at 8:52
  • $\begingroup$ @Roland can you explain why simply connected rules out the case of $d=0$? $\endgroup$ – 6666 Dec 6 '18 at 18:43

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