2
$\begingroup$

In section 4.1, chapter 4 of Pierre Deligne's paper La conjecture de Weil : I (french version, translation to English) he states:

On $\mathbb{C}$ Lefshietz local results are as follows. Let $X$ be a non singular analytic space and purely of dimension $n+1$. Let $D=\{z: |z|< 1 \}$ the unit disc and $D^*=D-\{0\}$, and let $f: X\rightarrow D$ a morphism of analytic spaces such that:

  • $f$ is proper

  • $f$ is smooth outside of a point $x$ of the special fiber $X_0=f^{-1}(0)$.

  • In $x$, $f$ has a non-degenerate quadratic point.

  • Let $t\neq 0$ in $D$ and $X_t=f^{-1}(t)$ "the" general fiber.

With the above data he associate some results for the cohomology groups of the fibers.

Then, in section 4.2. He says: There is an analog of (4.1) in abstract algebraic geometry. The disk $D$ is replaced by the spectrum of a henselian discrete valuation ring $A$ with an algebraically closed residue field. Let $S$ be the spectrum, $\eta$ its generic point (spectrum of the field of fractions of $A$), $s$ the closed point (spectrum of the residue field). The role of $t$ is played by the geometric generic point $\overline{\eta}$ (spectrum of the closure of the field of fractions of $A$).

I am wondering why he replaces $t$ by the geometric generic point $\overline{\eta}$? In any sense, there is a one to one correspondence between them?

$\endgroup$

1 Answer 1

4
$\begingroup$

If you want a canonical description of the homotopy type of the general fiber, you can take the homotopy type of the space $X\times_{\mathbb{D}}\mathbb{H}$ where $\mathbb{H}:=\{z\in\mathbb{C}\mid \Im z>0\}$ is the upper half complex plane, and the map $\mathbb{H}\to\mathbb{D}$ is the exponential map $z\mapsto e^{-z}$. This is because, as a complex manifold, $\mathbb{H}$ is the universal cover of the punctured disc $\mathbb{D}^\times=\mathbb{D}\smallsetminus\{0\}$. Since $\mathbb{H}$ is contractible and $X\smallsetminus X_0\to\mathbb{D}^\times$ is a fibration, then the fiber product $X\times_\mathbb{D}\mathbb{H}$ deformation retracts on $X\times_{\mathbb{D}}\{t\}$ for every $t\neq 0$. The monodromy action is then given by the map $\mathbb{H}\to \mathbb{H}$ sending $z$ to $z-2\pi i$, which is a map over $\mathbb{D}$ (this is exactly the deck transformation corresponding to the standard generator of $\pi_1\mathbb{D}^\times$).

In the algebraic case you can think of $\eta=\operatorname{Spec}A\smallsetminus\{s\}$ as the algebraic analogue of $\mathbb{D}^\times$, and the geometric point $\bar\eta$ as the algebraic analog of the universal cover $\mathbb{H}$. Then the Galois group of $\bar \eta/\eta$ is the one inducing the monodromy action.

$\endgroup$
1
  • $\begingroup$ Thank you @Denis Nardin!... $\endgroup$
    – Roxana
    Oct 12, 2021 at 22:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.