Assume $X$ is a hyper surface in $\mathbb{P}^n$, can one always find a closed immersion $i:\mathbb{P}^1 \rightarrow X$?
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2$\begingroup$ Not for the general $X$ of degree at least $n+1$. $\endgroup$– Felipe VolochCommented Oct 9, 2014 at 20:11
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$\begingroup$ So, what is the exact statement, the way curve sits inside X is important for me, It should be a closed immersion. @Felipo Voloch $\endgroup$– Sina BaghalCommented Oct 9, 2014 at 20:16
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2$\begingroup$ Think about the first case, that is $n=3$. By Noether-Lefschetz theorem, the very general surface $X$ of degree $\geq 5$ in $\mathbb{P}^3$ has the Picard group which is generated by the hyperplane section. In particular, $X$ contains no lines (more generally, no smooth rational curves). $\endgroup$– Francesco PolizziCommented Oct 9, 2014 at 21:00
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1 Answer
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More details since my comment was off by 1. Assume $X$ is given by $F=0$ where $F$ is a homogeneous polynomial of degree $d$. Now let $f_0,\ldots,f_n$ be homogeneous polynomials in two variables of some degree $m$. To have the map defined by $(f_0:\ldots :f_n)$ define a map from the line to $X$ we need $F(f_0,\ldots,f_n)=0$. If we regard the coefficients of the $f_i$ as variable we have $(n+1)(m+1)$ variables and the condition to lie on $X$ translates into $dm+1$ equations in these variables. If $d>n+1$ we expect no solution (for any $m$) unless $F$ is special in some way.
Edit: There is a mistake in my calculation which is pointed out by dhy in the comments, particularly the third one. Since that comment has a formatting problem, I repeat his calculation here. To have $dm+1 > (n+1)(m+1)$ one has to have $d > n+1 +n/m$. Since one can take $m=1$, it's only true that a general hypersurface of degree $d$ has no rational curves if $d>2n+1$. If $d>n+1$ one can only assert that on a general hypersurface of degree $d$, the degree of the rational curves is bounded.
answered Oct 9, 2014 at 21:21
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3$\begingroup$ This is only the case for $m$ large. In fact for $d < 2n-1$ there will always be lines on your hyper surface. It is nontrivial that for $d \geq 2n-1$ there are no rational curves on hypersurfaces. This was first shown by Clemens, see for example Claire Voisin's paper "On a conjecture of Clemens on rational curves on hypersurfaces." (There are later contributions trying to bound the degree of rational curves that can appear for hypersurfaces with $d<2n-1.$) The reason why the naive argument fails is that it is very hard to control the behavior of the natural incidence correspondence between $\endgroup$– dhyCommented Oct 10, 2014 at 0:03
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$\begingroup$ (continued) rational curves in $P^n$ and hypersurfaces. For example you don't even know the dimension of this incidence correspondence. The way Clemens's result is proven is by looking at your hypersurfaces as hyperplane sections of higher-dimensional hypersurfaces, arguing that the rational curves from these sections must cover this higher dimensional hypersurface, and then using that rational curves cannot cover a Calabi-Yau variety. The proof (in "Curves on generic hypersurfaces") is relatively short but not completely trivial. $\endgroup$– dhyCommented Oct 10, 2014 at 0:10
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$\begingroup$ (continued) There is a calculational error in this answer that gives you the wrong expectation (this heuristic should work, at least for large n): For $dm+1>(n+1)(m+1)$ to hold, you need $d>\frac{nm+n+m}{m}=\frac{n+1}+\frac{n}{m}$, so for $m=1$ you get $d>2n+1.$ I'm pretty sure there's some error here in the constant term, but I'm not sure what; in any case, you will get something with a $2n$ term. Also I realized that I forgot to say that everything I said was for a generic hypersurface; none of this works at all if your hypersurface is not generic. $\endgroup$– dhyCommented Oct 10, 2014 at 0:13
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3$\begingroup$ @dhy I think you should post an answer. $\endgroup$ Commented Oct 10, 2014 at 1:28
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1$\begingroup$ You guys should listen to dhy. Felipe's answer is wrong. Every degree $d$ hypersurface in $\mathbb{P}^n$ contains lines so long as the degree of the hypersurface is at most $2n-3$. $\endgroup$ Commented Oct 11, 2014 at 14:50