Let $f:X \rightarrow T$ be a proper morphism from a projective scheme $X$ to a smooth projective curve $T$ over $\mathbb{C}$. I know that fiber dimension is upper-semicontinuous, but is the degree of fibers also upper-semicontinuous? Here I fix the embedding of $X$ into some projective space (and so all the fibers of $f$ over each closed point of $T$) and then define the degree inside this fixed projective space in an usual way.

Actually, what I have in mind is the following situation: Let $\mathbb{P}^N$ be a projective space over $\mathbb{C}$ and let $Z$ and $Z_t$ be two irreducible varieties, where $Z$ is fixed and $Z_t$ moves with $t \in T$ as a parameter on a smooth curve $T$. I want to prove upper semicontinuity of degrees of the intersection scheme of $Z$ and $Z_t$ in $\mathbb{P}^N$, when dimension of $Z$ and $Z_t$ are smaller than $N \over 2$ but they intersect. In other words, the intersection is not proper in $\mathbb{P}^N$ and the expected dimension of the intersection is negative. For example, you could consider a line L in $\mathbb{P}^N$ and a family of curves $C_t$ with t as its parameter such that $L \cap C_t$ is nonempty, when $N>2$.


1 Answer 1


There's a slight stupidity here which is that if the dimension of the fiber jumps up, the degree can go down. E.g. let $X = (T \times C) \cup (p \times \mathbb P^2) \subseteq T \times \mathbb P^2$, where $C$ is a plane conic and $p\in T$ is a point. Then over most points of $T$, the fiber is $C$ of degree $2$, but over $p$ it's $\mathbb P^2$ of degree $1$. But this is just because studying the leading coefficient of a polynomial, as the polynomial changes degree (careful! this means "dimension" here), is a silly thing to do.

Assuming that the dimension is constant not just semicontinuous, then the degree is indeed semicontinuous. Let $X' \subseteq X$ be the closure of the fiber over the generic point, i.e. the union of the components of $X$ that dominate $T$. Then since $C$ is $1$-dimensional and regular, $X' \to X$ is flat, so the degree of fibers is actually constant. Adding back any remaining components of $X$, each of which maps to a point, we increase the degree at those points.

  • $\begingroup$ Thank you for your answer. I think your counterexample works for my main question, but it does not seem to occur in my situation of 2nd paragraph. What I am thinking for my situation is the following: Consider a quadric surface $Q$ in $\mathbb{P}^3$ and cosider lines $L_t$ in $\mathbb{P}^3$ intersecting with $Q$ at 2 points when $t \neq 0$ and $L_0$ is in $Q$. My question is, is it right that $L_0 \cap Q$ is just $L_0$, not $2L_0$? (If it's $L_0$ then this can be a counterexample for my situation although it is proper intersection for general $t$, but I am not sure that this is right.) $\endgroup$
    – user55992
    Jul 14, 2014 at 9:40

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