When speaking of a sheaf $E$ on a complex surface $X$, if an author writes


should I understand this to really mean

$$\int_X ch_2(E) < \int_X c_1(E)\wedge c_1(E)?$$

Because I think of chern characters and chern classes as living in $H^\bullet(X,Q)$ and $H^\bullet(X,Z)$ respectively, I don't know how else to interpret the inequality.

I apologize ahead of time for this naive question, as I'm new to algebraic geometry.

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    $\begingroup$ @GeorgesElencwajg: it seems a little strong (to me) to assert that the inequality doesn't make sense. Identifying top cohomology with the coefficient ring is hardly the naughtiest abuse of notation one sees in algebraic geometry on a daily basis; am I missing something? $\endgroup$ – user64687 Sep 25 '13 at 10:42
  • $\begingroup$ Dear @Asal, the problem is the use of Chern character on the left hand side and of a Chern class on the right. If $ch_2(E)$ means $\frac {c_1^2(E)}{2}-c_2(E)$, the inequality makes sense. Is that the interpretation? $\endgroup$ – Georges Elencwajg Sep 25 '13 at 11:07
  • $\begingroup$ Anyway, I have suppressed that first comment. $\endgroup$ – Georges Elencwajg Sep 25 '13 at 11:19
  • $\begingroup$ Dear @Georges, thanks for elucidating. Yes, I was taking $ch_2$ to mean precisely that. $\endgroup$ – user64687 Sep 25 '13 at 12:33
  • $\begingroup$ Hi! So is it correct that the inequality should be interpreted as $(\int_X c_1 \wedge c_1 / 2 - \int_X c_2) < \int_X c_1 \wedge c_1$? I just want to make sure I know the convention. I suppose that, equivalently, since both $ch_2$ and $c_1^2$ live in $H^\bullet(X,Q)$ and we have $H^{top}(X,Q) \cong Q$, one could inherit the ordering from $Q$ to make sense of the inequality. $\endgroup$ – AlgGeomQ Sep 25 '13 at 13:03

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