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Let $V$ be a complex projective variety and $L$ a nef line bundle on $V$ (i.e., $L$ is non-negative on every curve in $V$). Denote, as usual, $\deg_LX = c_1(L)^{\dim{X}}.[X]$ for $X$ a subvariety of $V$, considered as a prime cycle of dimension $\dim{X}$.

Question. For subvarieties $X$ and $Y$ of $V$ with $\deg_LX = \deg_LY = 0$, does it follow that all positive-dimensional components $Z$ of $X \cap Y$ necessarily have $\deg_LZ = 0$?

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    $\begingroup$ Sorry, I misread the question! $\endgroup$
    – dorebell
    Commented Sep 17, 2017 at 3:29

2 Answers 2

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Consider $V=\mathbb{P}^1\times \mathbb{P}^2$ with projections $p_1\colon V \rightarrow \mathbb{P}^1 \text{ and } p_2\colon V \rightarrow\mathbb{P}^2.$ Let $L = p_1^*(\mathcal{O}_{\mathbb{P}^1}(1))$, let $X = p_2^{-1}(\ell_1)$ and $Y=p_2^{-1}(\ell_2)$ for two distinct lines $\ell_1$ and $\ell_2\subset \mathbb{P}^2$. Then $Z = X\cap Y \cong \mathbb{P}^1$ is a section of $p_1$.

Finally we have $\deg_L(X) = \deg_L(Y) = 0$, but $\deg_L(X\cap Y) = 1$.

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  • $\begingroup$ This is a constant family of projective planes, parametrized by the base $B = \mathbb{P}^1$ under $p_1$. In such a case, $\deg_L = 0$ precisely on the constant subvarieties. The section $X \cap Y$ is constant, and has zero intersection with $L$. The line bundle $L$ is relatively ample over $B$, but these intersection numbers take place on the total space $V$, and not in a fibre. In fact $\deg_L(X \cap Y) = 0$ on that example. $\endgroup$
    – Vesselin Dimitrov
    Commented Sep 17, 2017 at 3:53
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    $\begingroup$ The line bundle L is not relatively ample over $\mathbb{P}^1$ but rather over $\mathbb{P}^2$. $\endgroup$
    – Chiles
    Commented Sep 17, 2017 at 4:03
  • $\begingroup$ That makes a difference! I had misread the indexing in the definition of $L$, my apology - and thank you for the simple and illuminating example! This question was in too much generality. $\endgroup$
    – Vesselin Dimitrov
    Commented Sep 17, 2017 at 4:10
  • $\begingroup$ Oops, I missed this - sorry for posting the same answer! They must have gone up around the same time. $\endgroup$
    – dorebell
    Commented Sep 18, 2017 at 7:42
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I think this is a counterexample. Let $V = \mathbf P^2 \times \mathbf P^1$, $L= \mathscr{O}_{P^2} \boxtimes \mathscr{O}_{P^1}(1)$. Let $\ell_1, \ell_2$ be two distinct lines in $\mathbf{P}^2$ and let $X_i=\ell_i \times \mathbf P^2$. Then $Z=X_1 \cap X_2 = \{\mathrm{pt}\}\times \mathbf P^1$, $L|_Z$ has degree $1$ but $L|_{X_i}$ has degree $0$

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