Consider a proper algebraic map between complex varieties $f : X \to D$ ($D$ is the unit disk), which is a submersion over $D^*$. I would like to know if they are any condition on $f$ such that the specialisation map $H^*(X_0) \to H^*(X)$ is injective ?
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asked Apr 13, 2022 at 18:37
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2$\begingroup$ Do you really mean $H^*(X)$ and not the cohomology of the nearby fiber $X_t$ ($t\neq 0$)? In that case, assuming that $f$ is proper, $X_0$ is a deformation retract of $X$, and hence the cohomology groups are the same. $\endgroup$– Piotr AchingerCommented Apr 13, 2022 at 19:24
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1$\begingroup$ Since you mention "vanishing cycles" in the title, let me assume so (and that the morphism is projective). In this case, the map will in general not be injective in the top degree, as $X_0$ may be reducible. But one can still say something, e.g. the local invariant cycle theorem tells us that if $X$ is semistable, then the image of $H^*(X_0)\to H^*(X)$ consists precisely of the classes fixed by the monodromy operator. I suggest consulting Illusie's wonderful survey "Autour du theoreme de monodromie locale" in Asterisque 223 "Periodes p-adiques", section 2.4. $\endgroup$– Piotr AchingerCommented Apr 13, 2022 at 19:30
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$\begingroup$ @PiotrAchinger : yes I meant $X_t$, thank you! So are you saying it's always injective other than possibly in top degree? $\endgroup$– Nicolas HemelsoetCommented Apr 13, 2022 at 19:41
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2$\begingroup$ No, I'm not implying that. Rather, in the projective semistable case the maps in question are the edge homomorphisms of the nearby cycles spectral sequence $$ E_2^{pq} = H^p(X_0, R^q\Psi \mathbf{Q}) \quad \Rightarrow \quad H^{p+q}(X_t, \mathbf{Q}). $$ The weight-monodromy theorem implies that it degenerates on $E_3$, and that the abutment filtration is given by kernels of $(T-1)^{p+1}$ where $T$ is the monodromy (Thm 2.4.4 in Illusie). This gives you quite a lot of information since $R^q\Psi\mathbf{Q}$ are easy to compute (Section 1). E.g. for $H^1$ of curves the map is injective (Ex 2.4.6). $\endgroup$– Piotr AchingerCommented Apr 13, 2022 at 20:09
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$\begingroup$ @PiotrAchinger : I see, thanks a lot! Unfortunately in my case the family is not semistable (the generic fiber is actually singular), and I was just hoping for a general criterion. I began to read Illusie survey, it looks very good! $\endgroup$– Nicolas HemelsoetCommented Apr 13, 2022 at 20:13
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