Let $X$ be a smooth, projective variety over $\mathbb{C}$ for which $\chi(X) = 0$. Here by $\chi$, I mean the topological Euler characteristic of $X(\mathbb{C})$; this number can also be computed as the degree of the top Chern class of the tangent bundle of $X$, and there are various other equivalent definitions.

I want to find/construct smooth hyperplane sections $Y$ of $X$ for which $\chi(Y) \neq 0$; I do not care too much about the projective embedding. Under which additional conditions (if any) is doing such a thing possible or reasonable? What are potential obstructions, et cetera?

Motivation: I can prove an interesting statement for varieties with $\chi \neq 0$. I believe it to be true for varieties with $\chi = 0$ as well, but alas my arguments do no longer apply. I want to try to reduce it to the case $\chi \neq 0$ using suitably chosen hyperplane sections.


There is no easy way to express $\chi (Y)$ in terms of $\chi (X)$. Using $\chi (X)= c_n(T_X)$ (with $n:=\dim X$ and the identification $H^{2n}(X,\mathbb{Z})=\mathbb{Z}$), and the exact sequence $0\rightarrow T_Y\rightarrow T_{X|Y}\rightarrow N_{Y/X}\rightarrow 0\ $ you get $\ \ \ \chi (Y)=c_{n-1}\cdot h-c_{n-2}\cdot h^2+\ldots +(-1)^{n-1}h^{n}$, where the $c_i$ are the Chern classes of $X$ and $h\in H^2(X,\mathbb{Z})$ the class of a hyperplane section.

It is not easy to decide whether this is $0\,$! On the other hand if you are willing to replace your embedding by a multiple, say $h$ by $kh$, you know that $|\chi (Y)|$ will be $\gg 0$ for $k$ large enough, since $\chi (Y)$ is given by a fixed polynomial in $k$ with leading coefficient $(-1)^{n-1}h^n$.

  • $\begingroup$ I am actually doing the computation in $H^{2n}(X,\Bbb{Z}$ -- which amounts to multiply by $h$. $\endgroup$ – abx Jun 21 '16 at 14:04

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