# Lecture 8 

:::{.question}
Is this representation $\pi$ smooth, admissible, and irreducible?
:::

:::{.lemma title="?"}
Let $B(K) \da \ts{\matt * * 0 * \in \GL_2(K)} \leq \GL_2(K)$, then there is a decomposition into upper triangular and integer matrices:
\[
\GL_2(K) = B(K) \cdot \GL_2(\OO_K) \da \ts{ bg \st b\in B(K), g\in \GL_2(\OO_K)}
.\]
:::

:::{.remark}
Why this is useful: $\GL_2(\OO_K)$ is compact, so locally constant functions $\phi: \GL_2(\OO_K)\to \CC$ will only take finitely many values (using that continuous images of compact sets are compact and compact subsets of discrete spaces are finite).
So $\phi(\GL_2(\OO_K))$ is a finite set, and $\phi(B(K))$ is controlled by $I(\chi_1, \chi_2)$.

:::

:::{.proof title="Specializing a proof for $\GL_n$"}
Let $\gamma \da \matt a b c d\in \GL_2(K)$, then we want to produce $\beta, \kappa$ with $\gamma = \beta\kappa$.

- Without loss of generality $\gamma\in \SL_2(K)$ by left-multiplying by $\matt{ (\det \gamma)\inv} 0 0 1\in B(K)$.
- Without loss of generality, $c, d\in \OO_K$ and at least one is a unit:
  To scale $c,d$, choose $\alpha\in K\units$ so that $ac,ad\in \OO_K$ and at least one is in $\OO_K\units$ since not both of $c,d$ are zero.
  Left-multiply by $\matt{\alpha\inv} 0 0 {\alpha}\in B(K)$ to send $c\to \alpha c, d\to\alpha d$.
- Without loss of generality, $c\in \OO_K\units$.
  If $d \in \OO_K\units$ and $c\not\in \OO_K\units$, right-multiply by $\matt 0 {-1} 1 0\in \SL_2(\OO_K)$ to swap $c, d$
- Check $\matt 1 {-a/c} 0 1 \cdot \matt a b c d = \matt{0}{-1/c}{c}{d} \in \GL_2(\OO_K)$, noting that the first matrix is in $B(K)$.

:::

:::{.remark}
More generally, for $\GL_n$ the Weyl group $S_n$ is involved.
:::

:::{.exercise title="A really good one."}
Show that $I(\chi_1, \chi_2)$ is smooth and admissible.
:::

:::{.observation}
The norm term in $I(\chi_1, \chi_2)$ is a fudge factor, and
\[
\norm{a\over d}^{1\over 2} = {\norm a ^{1\over 2}\over \norm d ^{1\over 2}}
.\]
One could redefine 
\[
\tilde\chi_1(x) &\da \chi_1(x) \norm{x}^{-{ 1\over 2} } \\
\tilde\chi_2(x) &\da \chi_2(x) \norm{x}^{1\over 2}
\]
to make the formula read
\[
\phi \matt a b c d g = \tilde \chi_1(a) \tilde \chi_2(d)\phi(g)
.\]

:::

:::{.warnings}
The fudge factor is needed here: taking the trivial characters will yield a subspace of constant functions in $I(\chi_1, \chi_2)$ when the $\chi_i$ are trivial reps.
When $\chi_1 = \chi_2$, there is an invariant 1-dimensional subrepresentation, and is thus not irreducible.
:::

:::{.remark title="on group representations"}
Recall that for $H\leq G\in \Grp$ with $[G:H] < \infty$, one can induce $H\dash$representations to $G\dash$representations.
If $\chi$ is a character of $H$, then $\Ind_H^G \chi$ is a character of $G$.
If $\chi$ is trivial, its induction will be unlikely to be irreducible since it contains the 1-dimensional trivial representation.
:::

:::{.exercise title="?"}
Show that if $\chi_1/\chi_2 = \norm{\wait}_K^{\pm 2}$, then the naive definition of $I(\chi_1, \chi_2)$ admits a 1-dimensional quotient.

:::

:::{.remark}
What's really going on with $I(\chi_1, \chi_2)$: the pair $(\chi_1, \chi_2)$ is a 1-dim representation $\psi$ of $B$, and we're writing down an explicit model for $\Ind_B^{\GL_2} \psi$.
Moreover there is a pairing involving an integral on $G$ and $B$ with respect to a Haar measure:
\[
I(\chi_1, \chi_2) \times I(\chi_1\inv, \chi_2\inv) \to \CC
.\]
Where the fudge factor comes from: at some point one changes a left Haar measure to a right one, and there is a fudge factor of $\norm{a/d}$, so one splits it between the two terms to yield a duality $I(\chi_1, \chi_2)\dual \cong I(\chi_1\inv, \chi_2\inv)$.
:::

:::{.fact}
$I(\chi_1, \chi_2)$ is irreducible if $\chi_1/\chi_2 \neq \norm{\wait}^{\pm 1}$.
:::

:::{.exercise title="?"}
If $\chi_1/\chi_2 = \norm{\wait}^{- 1}$, show that there is a 1-dimensional representation $\Psi$ of $\GL_2(K)$ given by $(\chi_1 \cdot \norm{\wait}^{1\over 2})\circ \det$ fitting into a SES
\[
0\to \psi \to I(\chi_1, \chi_2) \to S(\chi_1, \chi_2) \to 0
,\]
where $S$ is irreducible.
If instead $\chi_1/\chi_2 = \norm{\wait}^{1}$, then there is a SES
\[
0 \to S(\chi_1, \chi_2) \to I(\chi_1, \chi_2)\to (\chi_2 \cdot \norm{\wait}^{1\over 2})\circ \det \to 0
.\]
:::

:::{.remark}
We've induced characters to get reps of $\GL_2(K)$, how do we get more?
An observation due to Weil: if $K$ is any field, 
\[
\gens{ T \da \matt t 0 0 {t\inv}, U\da \matt 1 u 0 1, \matt 0 {-1} 1 0 \st \cdots} \cong \SL_2(K)
,\]
where there are some explicit relations.
So we can cook up $\SL_2(K)$ reps by specifying these generators and checking that the relations hold.
Note that $\GL_2(K)$ is an extension of $\SL_2(K)$ by an abelian group, so this can be used to construct $\GL_2$ reps.
:::

:::{.remark}
A large source of $\SL_2(K)$ and hence $\GL_2(K)$ reps: let $\SL_2(K) \actson L^2(K; \CC)$ by 
\[
(Tf)(x) &= f(tx) \\
(Uf)(x) &= f(u+x) \\
(Wf)(x) &= \hat{f}(x)
,\]
i.e. $W$ acts by the Fourier transform and is given by an explicit integral.
:::

:::{.fact}
From Jacquet-Langlands: if 

- $L/K$ is a quadratic extension, 
- $\chi: L\units\to\CC\units$ is admissible, 
- $\chi \neq \chi \circ \sigma$ for any nontrivial $\sigma\in\Gal(L/K)$ (i.e. $\chi$ is not its own Galois conjugate),

Jacquet-Langlands construct an irreducible infinite-dimensional representation of $\GL_2(K)$ using $L^2(L)$ and Fourier transforms.
We'll call this representation $\mathrm{BC}_L^K(\psi)$, for *base change*.
:::

:::{.fact}
We have a collection of infinite-dimensional smooth admissible irreducible representations of $\GL_2(K)$.

- $I(\chi_1, \chi_2) \cong I(\chi_2, \chi_1)$ which is irreducible when $\chi_1/\chi_2 \neq \norm{\wait}^{\pm 1}$
- $S(\chi, \chi\cdot \norm{\wait})$
- $\mathrm{BC}_L^K(\psi)$.

Moreover, for $\characteristic \kappa_K = p > 2$, these exhaust all such representations.
For $p=2$, there are exceptional representations.
So we know all possibilities for $\pi$.
:::

:::{.warnings}
The obvious representations of $\GL_2(K)$ like the 2-dimensional representation $K_\disc\cartpower{2}$ work, due to the \(p\dash \)adic topology -- the usual action has stabilizers which are closed but not open.
The only finite-dimensional representations are of the form $\chi \circ \det$ for $\chi: K\units\to\CC\units$ a character.
:::