# Tag Info

## Hot answers tagged vanishing-cycles

7

Because the symplectic form on $H_1(C_t, \mathbb Z)$ is a perfect pairing, it suffices to check that there is a group homomorphism $H_1(C_t,\mathbb Z) \to \mathbb Z$ that sends $\gamma$ to $1$, which follows if $\gamma$ is not divisible by any $n>1$ in $H_1(C_t,\mathbb Z)$. Because $\gamma$ is defined as the generator of the kernel of $H_1(C_t,\mathbb Z)... 5$\DeclareMathOperator{\ord}{ord}\DeclareMathOperator{\Spec}{Spec}$This negative answer to my question is a slight expansion of an email sent to me by Brian Conrad. Since I merely added some details for my own benefit this post is community wiki. As Damian Rössler pointed out in his comment the essential ingredient is the failure of the smooth base ... 4 Here's the bottom line. If$F$is a perverse sheaf or a D-module, there is an isomorphism between$\psi_fF$and the complex$[\psi \psi F \to \psi \phi F \oplus \phi \psi F]$. The reason why we can't get to it in a nice way is that it is really non canonical. To see it, consider a weight-filtrered perverse sheaf$(F,W)$or an F-filtered D-module, the ... 2 For the first item, section 3 of M. Saito's Modules de Hodge Polarizables is pretty thorough. He works in the filtered$D$-module setting, but I suppose you can ignore that aspect. For the constructible setting, perhaps you can look at Dimca's Sheaves in Topology. 2 Everything I know about gluing perverse sheaves is written in my paper Notes on Beilinson's "How to glue perverse sheaves", which deconstructs the construction sufficiently that it is possible to identify the following minimal axioms for making gluing work: The nearby cycles functor$\psi[-1]$must be t-exact for the perverse t-structure. I prove this ... 1 Every point in$X_s(k)$extends uniquely to an element of$X_{\overline{s}}(\overline k)$when we take its$\overline{k}$-points as an$\overline{k}$-scheme. This is the same as saying, if I have a variety defined over$\mathbb Q$, and I have a rational point, there is a canonical complex point associated to it. You don't need a choice of$\overline{eta}\$ at ...

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