# 10/15/2020 Lecture 17: Comparison theorems and the fundamental group

## Proving the comparison theorem

:::{.remark}
Last time: we had a span of sites $X^{\an}_\et \to X_\et, X^\an$, where $\pi^*: \Sh(X^\an) \to \Sh(X^{\an}_\et)$ is an equivalence of categories and $\an^*: H^i(X^\an; F)\iso H^i(X^\an_\et; \an^* F)$ for any constructible $F$ is an isomorphism.
As a corollary, for $F\in \Sh(X_\et)$ constructible, there is a canonical isomorphism $H^*(X_\et; F) \iso H^*(X^\an; F^\an)$ where $F^\an \da \pi_* \an^* F$.

The first part is an exercise: one can construct an inverse functor, so given $F\in \Sh(X^{\an}_\et)$, one wants $F^\an \in \Sh(X^\an)$ with isomorphisms $\pi^* F^\an \iso F$ and similarly $(\pi^* G)^\an \iso G$.
One can construct $F^\an \da \pi_* F$, and these composing the functors in either direction is naturally isomorphic to the identity.
Take $F\in \Sh(X^\an_\et)$ and push/pull, then consider its sections over an analytic étale open of $X$.
Take the corresponding local homeomorphism, cover it by small discs which are honest homeomorphisms onto their images.
The sheaf condition says that that value of $F$ is determined by its values on the discs, and these discs are in $X^\an$.
This works because any analytic étale cover can be refined to an honest open cover.
:::

:::{.proof title="of theorem, part 2"}

:::{.claim}
ETS: 

1. There is a canonical map $F\to \an_* \an^* F$ which is an isomorphism.
2. $\RR^{\geq 1} \an_* \an^* F = 0$.

:::

Why this implies the theorem: compute $H^{i+j}(X^\an_\et; \an_* F)$ using the Leray spectral sequence for the morphism of sites $\an$:
\[
H^i(X_\et; \RR^j \an_* \an^* F)\abuts
H^{i+j}(X^\an_\et; \an_* F)
.\]
The claim implies $E_2 = H^i(X_\et; F)$ for $j = 0$ and for $j \geq 1$, we get zero.
:::

:::{.proof title="of claim, part 1"}
We'll prove the special case of $F$ lcc.
Checking an isomorphism of sheaves is local, so without loss of generality we can assume $F$ is in fact constant.
Let $F = \ul L$, then we have maps
\[
\Gamma(U; \ul L)\to \Gamma(U^\an; \an^*\ul L) \cong \Gamma(U^\an; \ul L)
\]
where the second isomorphism follows from a formal computation.
We want to show that the first map is an isomorphism, so it's ETS $\pi_0(U^\an)\iso \pi_0(U)$ is a bijection of sets, since the global sections are this many copies of $L$ on either side respectively.
This is nontrivial in general.

We can first assume that $U$ fits into an elementary fibration $(U, Y, Z)$ over $S$, and by partitioning $U$ into connected components we can assume $U$ is connected.
The claim is that it's enough to prove this for a dense open subset of $U$, since $U$ is connected and smooth iff any dense open subset is connected by a point-set argument.

Given this, assume $U^\an$ and thus $Y^\an$ are not connected, since $U^\an$ is fiberwise dense in $Y^\an$.
Write $Y = Y_1 \disjoint Y_2$ with $Y_i$ disjoint unions of fibers of $h$ -- this follows from proving the theorem for curves directly, where we know that the analytification of a curve is a Riemann surface.
Now $h(Y_1), h(Y_2)$ are connected components of $S$, since $h$ is smooth thus flat thus open, and proper thus closed.
We're now done by induction on dimension.
:::

:::{.proof title="of claim, part 2"}
WLOG $F$ is constant and $U$ fits into an elementary fibration.
We want that $\RR^i \an_* \ul L = 0$ for $i \gt 0$ where $\an: U^{\an}_\et \to U_\et$, which is a significant reduction.

:::{.lemma title="?"}
If $U$ is connected and smooth over $\CC$ and $F$ is lcc (locally constant with finite fibers) on $U^\an_\et$, for $r\gt 0$, for any section $s\in H^r(U^\an_\et; F)$ there exists an étale cover $\ts{U_i\to U}$ such that $\ro{s}{U_i^\an} = 0$.
Thus analytic étale cohomology classes can be killed by honest étale covers.
:::

:::{.corollary title="?"}
The stalks $(\RR^r \an_* F)_x = 0$, since $\RR^* \an_*$ sheafifies taking cohomology of fibers, and thus $\RR^{r} \an_* F = 0$.
:::

:::

:::{.proof title="of lemma"}
WLOG $U$ fits into an elementary fibration

\begin{tikzcd}
	U && Y && Z \\
	\\
	&& S
	\arrow["j", hook, from=1-1, to=1-3]
	\arrow["i"', hook', from=1-5, to=1-3]
	\arrow["h", from=1-3, to=3-3]
	\arrow["f"', from=1-1, to=3-3]
	\arrow["g", from=1-5, to=3-3]
\end{tikzcd}
> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwwLCJVIl0sWzIsMCwiWSJdLFs0LDAsIloiXSxbMiwyLCJTIl0sWzAsMSwiaiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoidG9wIn19fV0sWzIsMSwiaSIsMix7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Imhvb2siLCJzaWRlIjoiYm90dG9tIn19fV0sWzEsMywiaCJdLFswLDMsImYiLDJdLFsyLDMsImciXV0=)

We'll now try to compute $H^r(U^\an_\et; F)$.
By the Leray spectral sequence, we have
\[
H^i(S; \RR^j f_* F)\abuts H^{r}(U^\an_\et; F), \qquad r = i+j
.\]
We can try to kill the sheaves $\RR^j f_* F$ by induction, and then kill the $H^i$ on the LHS.
Recall that $\RR^j f_* F$ is the sheafification of $V\mapsto H^j(f\inv(V); F)$, and by induction, these contributions can be killed by passing to a cover for $j \gt 0$.
For the base case, there are potentially contributions from $H^i(S; f_* F)$.
Since the fibers in an elementary fibration are topologically constant, $f_* F$ is lcc.
Since $S$ is one dimension lower, we're done by the induction hypothesis.

As a note, that $f_* F$ is lcc follows from a computation in singular cohomology: the elementary fibration is a bundle over a disc with fibers Riemann surfaces $\Sigma_{g, n}$ with the same number of points deleted, and this will be a trivial family $\DD\times \Sigma_{g, n}$ where one can compute the singular cohomology directly.

We're now done except for the base case $\dim U = 1$.
We have $F$ an lcc sheaf on $U^{\an}_\et$ a Riemann surface, and we want to kill classes in $H^i(U^\an_\et; F)$ for $i=1,2$, using vanishing for $i\geq 3$, by passing to étale covers.

For $i=2$, passing to any affine covers suffices, since any punctured Riemann surface has no cohomology.
This follows because $H^2$ is 1-dimensional and generated by $c_1(D)$ for $D$ the line bundle corresponding to the punctured point, and puncturing at this point kills the generator.
Equivalently, build a Riemann surface by gluing a polygon, and then puncturing the 2-cell in the interior of the polygon gives a retraction onto a cell complex with only 1-cells or lower.

For $i=1$, we have $s\in H^1(U^\an_\et; \ul L)$ and we want an étale cover $\ts{U_i\to U}$ with $\ro{s}{(U_i)^{\an}_\et} = 0$.
Here we need a real theorem from complex analytic geometry: $s$ corresponds to an $L\dash$torsor over $U^\an_\et$, which corresponds to a covering space with Galois group $L$.
By the **Riemann Existence theorem**, there is an equivalence of categories between finite étale covers of $U$ and finite *analytic* étale covers of $U$, where the latter corresponds to finite covering spaces.
If $U$ is projective, this is immediate from GAGA, so the content of this is showing it holds when $U$ is open.
For this, one needs to show that given a finite analytic étale cover of $U^\an$ for $U$ normal, it can be extended to a (possibly ramified) cover over the compactification.


Applying the existence theorem, $s$ corresponds to $U_s\to U$ a finite local analytic isomorphism, and this implies that $(U_s\to U) = (V\to U)^\an$ for some $V$.
We then have $\ro{s}{V} = 0$ since torsors kill their corresponding cohomology classes -- since $U_s$ is an $L\dash$torsor, and any cohomology class in $H^1$ pulled back to the torsor that defines itself becomes zero.
:::

:::{.remark}
If a complex space comes from an algebraic variety and has a finite cover, then the cover and the covering map are algebraic.
For example, for $X\da \PP^1\smts{\pt_1,\pt_2,\pt_3}$, $\pi_1 X\cong F_2$ is free on two generators.
Finite covering spaces correspond to actions $F_2 \actson A\in \Fin\Set$, and a corollary of this theorem is that every such action comes from an algebraic curve in $X$.
Thus coverings of algebraic curves can be written in terms of combinatorial, group-theoretic data.

One needs finiteness -- for $\CC\to \CC/ \Lambda$, the Weierstrass $\wp\dash$function has essential singularities at $\infty$.
A more simple situation is a map $f: \CC\to \CC\units$, and one can ask if it comes from a map $\AA^1\to \GG_m$.
It can not because it has infinite fibers, but this proof doesn't work because $f(x) = \exp(x)$.
Applying this theorem would be asking if $f$ extends to a map $\tilde f: \PP^1\to \PP^1$, but $\exp$ has an essential singularity at $\infty$.
:::

:::{.remark}
A general principle: if there is a correspondence between an analytic object (or a rigid analytic object like a formal scheme) and algebraic objects, one expects it to be true for proper schemes but it's a miracle if it's true for non-proper things! 
This is precisely due to this issue of essential singularities, and here finiteness reduces us to the proper situation.
:::


:::{.example title="?"}
A vast majority of computations of $H_\et$ are done by reduction to complex analytic spaces.
Let $X$ be a K3 surface over $\CC$, then 
\[
H^*(X_\et; C_n) = C_n t^0 + C_n^{22} t^2 + C_n t^4
,\]
which follows from computing cohomology of $X^\an$ over $\CC$.

Note that finiteness is essential, so we'll almost always take torsion coefficients.
For a non-example, for $X = \GG_m$ over $\CC$, note that $H^1(X_\et; \ul\ZZ) = 0$ while $H^1(X_\et; \ZZ) = \ZZ$.
These correspond to $\ZZ\dash$torsors over $\GG_m$, and there is an analytic cover given by $\exp$ but not an algebraic one.
:::

## Étale fundamental groups

:::{.remark}
Let $X\in \Sch$ be locally Noetherian and define $X_{\fet}$ the category $\Sch\slice{X}$ whose morphisms are given by finite étale morphisms $X'\to X$.
:::

:::{.definition title="The étale fundamental group"}
Let $\bar x$ be a geometric point of $X$, then define a **fiber functor**
\[
F_{\bar x}: X_{\fet} &\to \Fin\Set \\
Y/X &\mapsto Y_{\bar x}
.\]
The **étale fundamental group** is defined as 
\[
\pi_1^\et(X, \bar x) \da \Aut_{\Fun}(F_{\bar x})\in \Top\Grp
\]
whose topology is the coarsest topology (fewest open sets) such that the maps $\pi_1^\et(X, \bar x)\to \Aut(F_{\bar x}(Y))$ are continuous for all $Y$, where $F_{\bar x}(Y)$ has the discrete topology.
Explicitly, for any automorphism of a fiber, the preimage is open.
:::

:::{.example title="?"}
Let $X = \spec k$ and $\bar{x}: \spec \kbar \to \spec k$, then $\opcat{ X_{\fet}}$ is the category of finite étale extensions of $L/k$.
Then 
\[
\pi_1^\et(\spec k, \bar x) = \Gal(\ksep/k)
\]
for a choice of separable closure.
:::

:::{.example title="?"}
Let $X = \AA^1\slice k$ with $\characteristic k = p > 0$, $\pi_1^\et X$ is not topologically finitely-generated for any for any such field.
The reason why: there are too many covering spaces.
We know $H^1( (\AA^1\slice k)_\et; \FF_p ) = \coker \mathrm{AS} \da \coker(k[t] \mapsvia{x\mapsto x^p-x} k[t] )$, which is huge, so there are many covers of $\AA^1$ with Galois group $\FF_p$.
Later we'll see this is isomorphic to $\Hom(\pi_1^\et(\AA^1\slice k, \bar x)^\ab, \FF_p)$, and if hom from a profinite group into a finite discrete group is not finitely-generated, then the profinite group can not be topologically finitely-generated.
:::


:::{.example title="?"}
Let $E$ be an elliptic curve over $\characteristic k = p > 0$ and $k=\kbar$, then 
\[
\pi_1^\et(E) = \cocolim_n E[n] =
\begin{cases}
\ZZpadic \times \prod_{\ell\neq p} \ZZladic^2 & E \text{ ordinary} 
\\
\prod_{\ell\neq p}\ZZladic^2 & E \text{ supersingular}.
\end{cases}
.\]

Note that this isn't immediately clearly the profinite completion of a discrete group, but the rank of what you get here depends on whether you take the pro$\dash\ell$ completion or the pro$\dash p$ completion. 
:::


:::{.example title="?"}
For $X$ normal over $\CC$ and connected, 
\[
\pi_1^\et(X, \bar x) = \pro \pi_1^\Top(X, x)
.\]
where $\pro(\wait)$ is the profinite completion.
:::