Let $X$ be a projective smooth variety, $D$ a Cartier divisor on $X$. If for any Cartier divisor $B$ on $X$, $\mathcal{O}_X(B)$ and $\mathcal{O}_X(B+D)$ have the same Euler characteristic $$\chi(X,\mathcal{O}_X(B))=\chi(X,\mathcal{O}_X(B+D)),$$ then is $D$ numerically trivial?

## 1 Answer

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Take $B$ ample and $m\gg1$ so that $\chi(X, mB)=h^0(X, mB)$ and $\chi(X, mB+D)=h^0(X, mB+D)$.

Identifying the leading term in the polynomial expansion in $m$, you get $(mB+D)^n=(mB)^n$.

This implies that $(B^{n-1}\cdot D)=0$ for any ample $B$. A polarization argument shows that $D$ is therefore numerically trivial.