#  Zhiwei Yun, Lecture 2 

:::{.remark}
Today: what is rigidity?
:::

## Automorphic Data

:::{.definition title="Automorphic data"}
Given the following:

- $k = \FF_q$
- $S \subseteq \abs{X}$ a finite set, e.g. $X= \PP^1$ and $S = \ts{0, \infty}$ or $S=\ts{0,1,\infty}$.
- For each $x\in S$, a compact open $K_x \subseteq G(F_x)$
- For each $x\in S$, a character:

\begin{tikzcd}
	{K_x} && \CC\units \\
	\\
	& {L_\chi \text{ finite}}
	\arrow[from=1-1, to=3-2]
	\arrow[from=3-2, to=1-3]
	\arrow["\chi"', from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJLX3giXSxbMiwwLCJcXENDXFx1bml0cyJdLFsxLDIsIkxfXFxjaGkgXFx0ZXh0eyBmaW5pdGV9Il0sWzAsMl0sWzIsMV0sWzAsMSwiXFxjaGkiLDJdXQ==)

The pair $(K_S, \chi_S)$ is **automorphic data**.
:::

:::{.remark}
Note that $\chi_S = 1$ recovers $f\in \mca_K$ where $K = K_S \times \prod_{x\not\in S} G(\OO_x)$.
:::

:::{.definition title="Typical data"}
A map
\[
f\in C^0(\dcoset{G(F)} {G(\AA)} {\prod_{x\not\in S} G(\OO_x) } \to \CC)
\]
is **$(K_S, \chi_X)\dash$typical** iff
\[
f(gk_x) = \chi_x(k_x) f(g)\qquad \forall x\in S, k_x\in K_x, g\in G(\AA)
.\]
:::

:::{.remark}
We want to make $\dim \mca_c(K_S,\chi_S) = 1$.
In this case, the Hecke algebra $\mch_{K_y} \actson f\in \mca_c(K_S, \chi_S)$ by a character, making $f$ a Hecke eigenform.
:::

## Examples of naive ridigidy

:::{.example title="?"}
Let 

- $X=\PP^1$,
- $G = \SL_2$,
- $S=\ts{0, 1,\infty}$,
- $K_x = I_x$ Iwahori for all $x\in S$, where
\[
I_x = \ts{A\da \matt a b c d \in \Mat_{2}(\OO_X) \st c\in \mfm_x}
.\]

Then choosing characters $\chi_x: k\units\to \CC\units$ *generically* will imply $\dim \mca_c(K_S, \chi_S) = 1$.
Here generic means that $\prod \chi_i^{\pm 1}\neq 1$.
By global Langlands for $\SL_2$, any $f\in \mca_c(K_S, \chi_S)$ will yield a 2-dimensional local system on $\PP^1\smts{0,1,\infty}$ ramified at the 3 punctures.
These will be solutions to hypergeometric differential equations.

For $G=\PGL_2$ (where the example works similarly), for $\chi_0, \chi_1 = 1$ and $\chi_\infty$ quadratic, there is a cover

\begin{tikzcd}
\ts{E_t}: y^2 = x(x-1)(x-t)
  \ar[d] 
\\
\PP^1\smts{0,1,\infty}
\end{tikzcd}

Moreover $\ts{H^1(E_t)}$ will be a rank 2 local system on this base.
:::

:::{.example title="?"}
Let 

- $X = \PP^1$
- $S = \ts{0, \infty}$
- $K_0 = I_0$, $\chi_0 =1$
- $\tau$ a uniformizer at $\infty$
- $K_\infty = I_\infty^+ = \ts{\matt a b c d \st a\equiv d\equiv 1 \mod \mfm_x, b\in \OO_x, c\in \mfm_x}$, the pro $p$ part

There is a map $K_\infty \to k$ where $\matt abcd \to b+{c\over \tau}\mod\tau$.
Any character $k \mapsvia{\psi} \CC\units$ can be extended to $\chi_\infty: K_\infty \to k \mapsvia{\psi} \CC\units$, and $\dim \mca_c(K_S, \chi_S)$.
This yields a **Kloosterman local system** on $\PP^1\smts{0,\infty}$, where
\[
\Kl(a) = \sum_{x\in k\units} \psi\qty{x + a\over x}
\]
recovers the classical Kloosterman sum by taking trace of Frobenius.
:::

## Naive Rigidity

:::{.definition title="Rigidity (Naive Definition)"}
\[
(K_S, \chi_S) \text{ is rigid }\iff \dim \mca_c(K_S, \chi_S) = 1
.\]
:::

:::{.warnings}
If $\pi_1 G\neq 1$, then $\pi_0 \Bun_G\geq 2$, yielding multiple components. 
It's also not clear if this type of dimension bound will hold after a base change $k\to k'$.
:::

## Base Change

:::{.remark}
For $k'/k$ finite, write $X'\da X\tensor_k k'$ for the base change.
Let $S\to S'$ be the preimage of $S$ in $S'$, and consider $k'_x \da K_x\tensor_k k'$
How can we base change a character?
We need a norm map to fill in the following diagram:

\begin{tikzcd}
	{K_x\tensor_k k'} && {K_x} && \CC\units
	\arrow["{\chi_x}", from=1-3, to=1-5]
	\arrow["{\Norm(\wait)}", dashed, from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMyxbMCwwLCJLX3hcXHRlbnNvcl9rIGsnIl0sWzIsMCwiS194Il0sWzQsMCwiXFxDQ1xcdW5pdHMiXSxbMSwyLCJcXGNoaV94Il0sWzAsMSwiXFxOb3JtKFxcd2FpdCkiLDAseyJzdHlsZSI6eyJib2R5Ijp7Im5hbWUiOiJkYXNoZWQifX19XV0=)

:::

:::{.example title="Base-changing a character"}
\begin{tikzcd}
	{I_x} && k\units && \CC\units \\
	\\
	{I_x' = I_x(k')} && {(k')\units} && k\units && \CC\units
	\arrow["\chi", from=1-3, to=1-5]
	\arrow[from=1-1, to=1-3]
	\arrow["\chi", from=3-5, to=3-7]
	\arrow["{\Norm_{k'/k}(\wait)}", from=3-3, to=3-5]
	\arrow[from=3-1, to=3-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNyxbMCwwLCJJX3giXSxbMiwwLCJrXFx1bml0cyJdLFs0LDAsIlxcQ0NcXHVuaXRzIl0sWzAsMiwiSV94JyA9IElfeChrJykiXSxbMiwyLCIoaycpXFx1bml0cyJdLFs0LDIsImtcXHVuaXRzIl0sWzYsMiwiXFxDQ1xcdW5pdHMiXSxbMSwyLCJcXGNoaSJdLFswLDFdLFs1LDYsIlxcY2hpIl0sWzQsNSwiXFxOb3JtX3trJy9rfShcXHdhaXQpIl0sWzMsNF1d)

Here $I_x(k') = \ts{a,b,c,d\in \OO_x\hat{\tensor} k' \cong k'\fps{t}}$.
:::

:::{.example title="?"}
\begin{tikzcd}
	{I_x} && k\units && \CC\units \\
	\\
	{I_x' = I_x(k')} && {(k')\units} && k\units && \CC\units \\
	{\matt a b c d } && {b + {c\over \tau}\mod \tau}
	\arrow["\chi", from=1-3, to=1-5]
	\arrow[from=1-1, to=1-3]
	\arrow["\chi", from=3-5, to=3-7]
	\arrow["{\Trace_{k'/k}(\wait)}", from=3-3, to=3-5]
	\arrow[from=3-1, to=3-3]
	\arrow[maps to, from=4-1, to=4-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsOSxbMCwwLCJJX3giXSxbMiwwLCJrXFx1bml0cyJdLFs0LDAsIlxcQ0NcXHVuaXRzIl0sWzAsMiwiSV94JyA9IElfeChrJykiXSxbMiwyLCIoaycpXFx1bml0cyJdLFs0LDIsImtcXHVuaXRzIl0sWzYsMiwiXFxDQ1xcdW5pdHMiXSxbMCwzLCJcXG1hdHQgYSBiIGMgZCAiXSxbMiwzLCJiICsge2NcXG92ZXIgXFx0YXV9XFxtb2QgXFx0YXUiXSxbMSwyLCJcXGNoaSJdLFswLDFdLFs1LDYsIlxcY2hpIl0sWzQsNSwiXFxUcmFjZV97aycva30oXFx3YWl0KSJdLFszLDRdLFs3LDgsIiIsMCx7InN0eWxlIjp7InRhaWwiOnsibmFtZSI6Im1hcHMgdG8ifX19XV0=)

:::

:::{.remark}
We now geometrize this process to send characters to character sheaves, i.e. rank one local systems on $K_x$.
We have a way of taking $(K_S, \chi_S)$ to $(K'_S, \chi_S')$ for extensions $K\to K'$, so we can form $\mca(k'; K'_S, \chi'_S)$.
:::

:::{.definition title="Weakly rigid"}
Automorphic data $(K_S, \chi_S)$ is **weakly rigid** iff $\dim \mca_c(k'; K'_S, \chi_S')$ is uniformly bounded for all extensions $k\to k'$.
:::

## Relevant Points

:::{.remark}
Recall that there is a bijection
\[
\dcoset{G(F)}{G(\AA)}{K} \mapstofrom \Bun_G(K)(k)
,\]
so functions $f\in \mca_c(K_S, \chi_S)$ are functions on $\Bun_G(K_S^+)(k)$ where for $x\in S$, $K^+_x \normal K_x$ with $\ro{\chi_x}{K_x^+} = 1$ and $K_x/K_x^+$ are the $k\dash$points of a finite dimensional group $L_x$.

:::

:::{.example title="?"}
$I_x^+\normal I_x = K_x \to \GG_m(k)$.
:::

:::{.example title="?"}
$I_\infty^{++} = K_\infty^+ \normal I_\infty^+ = K_\infty \to k\sumpower{2}$ where $\matt abcd \mapsto \qty{b, {c\over \tau}} \mod \tau$.
:::

:::{.remark}
There is a right action
\[
C^0(\Bun_G(K_S^+)(k) \to \CC?) \actsonl \prod_{x\in S} L_x(k)
,\]
and the eigenfunctions with eigenvalues $(\chi_x)_{x\in S}$ are in $\mca_c(K_S, \chi_S)$.
As a set, $\Bun_G(K_S^+)$ has commuting left and right actions, where quotienting by the right action yields a principal homogeneous space.
The left action is by $\Aut(\mce)$ for $\mce\in \Bun_G(K_S)$, and permutes points in the fiber in $\tilde \mce \in \Bun_G(K_S^+)$.
So there is an evaluation map which is well-defined up to conjugacy 
\[
\ev_\mce: \Aut(\mce) \to \prod_{x\in S} L_x(k)
.\]
:::

:::{.definition title="Relevant points"}
A $k\dash$point $\mce\in \Bun_G(K_s)(k)$ is **$(K_S, \chi_S)\dash$relevant** iff
\[
\ro{ \ev_\mce^*\qty{ \prod_{x\in S} \chi_x } }{\Aut(\mce)^0(k) } = 1
.\]
Similarly one can define relevant $k'\dash$points for $k'/k$ a finite extension.
:::

:::{.fact}
\[
\dim \mca_c(k'; K_S, \chi_S) \leq \size \Rel(K_S), \quad \Rel(K_S) \da \ts{(K_S', \chi_S')\dash\text{relevant }k' \text{ points of } \Bun_G(K_S) }
.\]
Note that taking connected components in the definition is needed to make this stable under base change.
:::

:::{.corollary title="?"}
$(K_S, \chi_S)$ is weakly rigid $\iff \size \Rel(K_S) < \infty$.
:::


:::{.example title="?"}
Let 

- $G=\SL_2$ 
- $K_x = I_x$
- $x\in S \da \ts{0, 1,\infty}$

Note that
\[
\Bun_G(K_S)(k) = \ts{ V\in \mathsf{VectBundle}^{\rank = 2}, \iota: \Extalg^2 V\cong \OO_X, \ts{\ell_x \subseteq V_x}_{x\in S} }
,\]
where the $\ell$ are lines.
So these are bundles with extra structure at fixed places, and are parameterized by 5-tuples $\mce = (V, \iota, \ell_0, \ell_1, \ell_\infty)$.
For all $x\in S$ we have
\[
I_x &\to \GG_m = L_x \\
\matt a b c d &\mapsto a\mod \tau
,\]
and the evaluation map is $\ev: \Aut(\mce) \to \prod_{x\in S} \GG_m$.
Each $\gamma\in \Aut(\mce)$ is a map $V\to V$ where $\gamma_0 \actson V_0$ preserving $\ell_0$.
Is it the case that
\[
\ro{\prod_{x\in S} \chi_x}{\Aut(\mce)^0} \underset{?}{=} 1
.\]
For $V = \OO^2$ and $\ell_x \subseteq k^2$ in generic position, $\Aut(\mce) = \ts{\pm 1}$ so they are relevant.
Other points are irrelevant: if $V = L \oplus L'$ with $\ell_x\in L_x$ or $L_x'$, $\Aut(\mce)$ will contain a copy of $\GG_m$ that acts by scaling each $L$ which will map nontrivially to $\prod_{x\in S} \GG_m$.
Since $\prod \chi_i^{\pm 1}\neq 1$, we get $\ro{\ev_\mce^* \prod \chi_x}{\GG_m} \neq 1$.
:::