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Let $X$ be a very general hypersurface of degree $d$ in $\mathbb{P}^n.$ Does $X$ contain a smooth surface $S$ with $c_1(T_S)^2>2c_2(T_S)$?

For $d<<\sqrt{n}$ the answer is yes, as $X$ will contain a plane. I'm mainly interested in understanding what happens when $d>>\sqrt{n}$. If the answer is no then this seems likely to be very hard to show. But maybe somebody knows a simple construction of such a $S$?

Some more motivation: Bogomolov-Miyaoka-Yau tells us that (in characteristic $0$) we have $c_1^2\leq 3c_2$ for a surface of nonnegative Kodaira dimension. (I believe smooth ruled surfaces that are not $\mathbb{P}^2$ automatically satisfy $c_1^2\leq 2c_2$ but not BMY, as $c_2$ is negative.) Several natural constructions (e.g. complete intersections, products of curves) give surfaces of general type with $\frac{c_1^2}{c_2}$ close to or equal to $2$, but to the best of my (very limited) knowledge all constructions of general type surfaces with $c_1^2>2c_2$ are rather complicated.

On the other hand, it is a folklore conjecture (see e.g. Conjecture 7.5 in http://arxiv.org/abs/1407.7478v1) that very general hypersurfaces with $d>>\sqrt{n}$ do not contain any rational surfaces. More generally, I'm curious what surfaces one should expect in our hypersurface $X$, and in particular what pairs of numerical invariants $(c_1^2,c_2)$ one should expect.

Finally, this is also motivated by techniques used by Bogomolov to bound rational curves in general type surfaces with $c_1^2>c_2$.

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    $\begingroup$ It is worth noting that the condition $c_1(T_S)^2 > 2c_2(T_S)$ is equivalent to $S$ having positive signature. $\endgroup$
    – Michael Albanese
    Commented Sep 4, 2015 at 16:38
  • $\begingroup$ What happens when $n=4$?. Then, all surfaces in such an $X$ are hypersurface intersections and calculations should be easy. $\endgroup$
    – Mohan
    Commented Sep 4, 2015 at 16:45
  • $\begingroup$ @Mohan: I'm pretty sure $n=4$ is fine; IIRC, all smooth complete intersection surfaces of nonnegative Kodaira dimension satisfy $c_1^2\leq 2c_2$. $\endgroup$
    – dhy
    Commented Sep 4, 2015 at 17:01
  • $\begingroup$ Of course you know this, but you do not need $d^2$ "much less" than $n$ to find surfaces with $\text{ch}_2$ positive. Already $(d+2)(d+1)\leq 6(n-2)$ is enough to guarantee existence of a linear $2$-plane. Riedl proved flatness results for the evaluation map from pointed 2-planes to the hypersurface in an appropriate degree range. $\endgroup$
    – Jason Starr
    Commented Sep 4, 2015 at 17:07
  • $\begingroup$ @Mohan: "What happens when $n=4$?" dhy already answered this, but for complete intersections, $\text{ch}_r$ equals $(n+1-\sum_i d_i^r)/r!$. So the only complete intersections surfaces with $\text{ch}_2$ positive are linear $2$-planes. $\endgroup$
    – Jason Starr
    Commented Sep 4, 2015 at 17:10

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