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  1. Consider a smooth projective surface $S\subset\Bbb P^N_{\Bbb C}$ which is a complete intersection of hypersurfaces of degrees $(d_1,..,d_{k\ge2})$ with $d_i\ge2$ for all i. Is it true that for such surfaces $c^2_1\le 2c_2$? (i.e. much better than BMY) At least asymptotically (i.e. for high enough $d_i$'s)?

Let $td_2$ be the top-dimensional Todd class, i.e. $td_2=\frac{c^2_1+c^2}{12}$. The inequality as above can be written as $c_2\ge 2^2td_2$.

  1. More generally, let $X\subset\Bbb P^N_{\Bbb C}$ be a smooth complete intersection of dimension $n$. Let $c_n$ and $td_n$ be its top-dimensional Chern and Todd classes. What are the known inequalities on $c_n$ and $td_n$? (I would like to have smth like $c_n\ge 2^n td_n$)
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I'd expect that there'd be a formula for $c_1$ and $c_2$ in terms of the degrees $d_i$ and then one could just check the resulting inequality directly. – Noam D. Elkies Jul 8 '11 at 17:35
You can find these formulas in Bart-Peters-Van de Ven, Chapter V – Francesco Polizzi Jul 8 '11 at 21:10
Yes, certainly, for a complete intersection in $\Bbb P^N$ the Chern and Todd classes can be computed explicitly. But the formulas are quite messy. I need the bound: $(-1)^n(c_n(X)-1)\ge 2^n(-1)^n\Big( td(T_X) ch(\mathcal{O}_X(-1))\Big)_{top.dim.}$ After computation of both sides I get some messy school-type inequality. No idea how to prove it in general. But in many numerical cases (of low dimension and codimension) I verified it. I wonder, is there some (indirect?) way to prove the bound without proving this numerical inequality? – Dmitry Kerner Jul 9 '11 at 12:03
An additional question. Suppose $S\subset\Bbb P^N$ is a smooth surface, "close to being a complete intersection". For example it is ACM or better arithmetically Gorenstein or even smth better. Do such surfaces satisfy the bound as above? An example not satisfying the bound would be of interest too. – Dmitry Kerner Jul 9 '11 at 12:05
up vote 5 down vote accepted

By the formulae in [Barth-Peters-Van de Ven, Chapter V] one has, for a surface which is complete intersection of type $(d_1, \ldots, d_{n-2})$ in $\mathbb{P}^n$: $$c_1^2(X)= \big(\sum d_i-(n+1)\big)^2 \prod d_i,$$ $$c_2(X)=\bigg[\binom{n+1}{2}-(n+1)\sum d_i+\sum d_i^2 +\sum_{i < j} d_id_j \bigg]\prod d_i.$$

Then one obtains that the inequality $$c_1^2(X) \leq 2 c_2(X)$$ is equivalent to $$n+1 \leq \sum d_i^2,$$ and this is of course almost always true, since the right-hand term is $\geq 4(n-2)$. So the answer to $1.$ seems to be yes.

ADDED. Actually, this is also written in the book by Barth-Peters- Van de Ven. In Chapter V, at the beginning of the Section "The Geography of Chern Numbers", they say:

" The simplest examples, like complete intersections and double coverings of $\mathbb{P}^2$, pratically always yield a point of $D_1$ [where $D_1$ is the region in the $(c_1, c_2)$-plane given by $c_1^2 \leq 2c_2$]. Indeed, for a long time only few examples were known of surfaces with Chern pairs $(c_1^2, c_2)$ in $D_2$ [i.e., such that $2c_2 < c_1^2 \leq 3c_2$]."

For your question in the last comment, instead, the answer is clearly no if $S$ is ACM. In fact, every smooth surface $S$ with $H^1(S, \mathcal{O}_S)=0$ is ACM for some embedding in the projective space. Now take for instance a fake projective plane. It satisfies $p_g(S)=q(S)=0$, so it is ACM, but $$c_1^2(S)=3c_2(S).$$

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The square should be inside the parentheses, not outside. This is precisely the inequality in the work of Tsen and Lang. – Jason Starr Jul 9 '11 at 17:16
Uhm, I've made the computation again and I find the same result: just use $\sum d_i^2 +\sum_{i \neq j} d_id_j = (\sum d_i)^2$. But maybe I'm making some silly mistake... – Francesco Polizzi Jul 9 '11 at 17:50
The expression $(c_1^2-2c_2)/2$ is the second graded piece of the Chern character. The Chern character is both additive for short exact sequences and multiplicative for tensor products. Combined with the Euler sequence, it is easy to compute that the Chern character of a complete intersection $X$ in $P^n$ of type $(d_1,\dots,d_c)$ equals $(n+1)e^H - 1 - \sum_i e^{d_iH}$, where $H = c_1(O(1))$. In particular, $(c_1^2-2c_2)/2$ equals $1/2(n+1-\sum_i d_i^2) H^2$. So $c_1^2-2c_2$ is negative if and only if $n+1 < \sum_i d_i^2$. This is all done explicitly in my papers with de Jong. – Jason Starr Jul 11 '11 at 12:50
An easier way to see that the square has to be inside is to consider the case when the integers $d_i$ equal $1$. Then the "complete intersection" is really just a projective space of lower dimension. So you can compare your formula to the Chern classes of projective space (which are easy to compute from the Euler sequence). -- Best. – Jason Starr Jul 11 '11 at 12:53
Ok, thank you very much. I will check again the computation. Best. – Francesco Polizzi Jul 11 '11 at 12:55

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