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Consider a $\textbf{flat}$ surjective map $f: X \rightarrow \mathbb{A}^1$. The general fibers $F_{\epsilon}$ are canonically isomorphic, and the special fiber $F_0$ above $0 \in \mathbb{A}^1$ is not isomorphic to the general fibers.

Given a closed subscheme $B \subset X$, we define its special fiber limit $\widetilde{B}$ to be the intersection of the special fiber $F_0$ with the closure of $B \times \mathbb{A}^1 \backslash \{ 0 \}$ in $X$, $F_0 \cap \overline{B \times \mathbb{A}^1 \backslash \{ 0 \}}$.

Let $B_1$ and $B_2$ be two closed subschemes of $X$. When does the limit of their intersection equal to the intersection of their limits, i.e. $\widetilde{B_1} \cap \widetilde{B_2} = \widetilde{B_1 \cap B_2}$? Is it enough for the dimension of $\widetilde{B_1} \cap \widetilde{B_2}$ to be equal to the dimension of $B_1 \cap B_2$? Any relevant comments and references are welcome. Thanks!

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Obviously $\widetilde{B_1} \cap \widetilde{B_2} \subseteq \widetilde{B_1\cap B_2}$. I'll discuss a sufficient condition for the reverse.

If $B_1\cap B_2$ is equidimensional, then so is $\widetilde{B_1\cap B_2}$. It seems like you know that its dimension is that of $\widetilde{B_1}\cap \widetilde{B_2}$. By any chance is $\widetilde{B_1}\cap \widetilde{B_2}$ reduced, and the family projective? Then if $\deg(B_1\cap B_2) = \deg(\widetilde{B_1}\cap \widetilde{B_2})$ (w.r.t. the projective embedding), your $\widetilde{B_1\cap B_2}$ has to be all of $\widetilde{B_1}\cap \widetilde{B_2}$.

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  • $\begingroup$ Thanks! This sounds very nice. Why is it true? Is the proof/reasoning written somewhere? In my case, it is hard for me to figure out the degree of the projective embedding, but maybe I can find some other equivalent conditions... $\endgroup$ – Qiao Mar 7 '16 at 4:44
  • $\begingroup$ By flatness, $\deg(B_1\cap B_2) = \deg(\widetilde{B_1\cap B_2)}$. Then use Lemma 1.7.5 of my paper with Ezra Miller on Schubert polynomials. arxiv.org/pdf/math/0110058.pdf $\endgroup$ – Allen Knutson Mar 7 '16 at 12:34

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