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Suppose we have a complex algebraic variety $X$ with a map $f: X \to \mathbb{C}$ with $Y=f^{-1}(0)$. Let $\overset{\sim}{\mathbb{C}}$ be the universal cover of $\mathbb{C}-\{0\}$ and consider the fiber product $\overset{\sim}{X}= X \times_{\mathbb{C}} \overset{\sim}{\mathbb{C}}$. Given a perverse sheaf $F$ on $X$, we may apply the nearby cycles functor $\psi$ to $F$ by first pulling back to $\overset{\sim}{X}$, pushing back to $X$, and then pulling back to $Y$.

It seems like the definition of nearby cycles doesn't rely on the fact that the base of $f$ is one-dimensional, so I assume there are some nice properties of $\psi$ that fail when our base has dimension $\geq 2$.

What are the properties of the functor $\psi$ that fail when we replace $\mathbb{C}$ with something higher dimensional, say $\mathbb{C}^r$, for $r>1$, and why?

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  • $\begingroup$ What are you intending to subtract and take the universal cover of? $\endgroup$
    – Will Sawin
    Commented Dec 7, 2019 at 19:46
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    $\begingroup$ Initial problems are failure of constructibility and failure of base change. That said, there is a theory of higher dimensional nearby cycles, see math.u-psud.fr/~illusie/vanishing1b.pdf. $\endgroup$
    – user125639
    Commented Dec 7, 2019 at 19:47
  • $\begingroup$ @WillSawin Well I guess I was thinking that we could just take $\mathbb{C}^r-\{0\}$, in which case the universal cover would be the space itself $\endgroup$
    – Exit path
    Commented Dec 7, 2019 at 19:48
  • $\begingroup$ @user125639 Where exactly does constructibility fail? $\endgroup$
    – Exit path
    Commented Dec 7, 2019 at 19:58
  • $\begingroup$ No universal cover means you're just taking the usual pushforward. This is a totally valid construction, and indeed preserves constructibility, but it just isn't nearby cycles. $\endgroup$
    – Will Sawin
    Commented Dec 7, 2019 at 20:17

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