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Leaky Nun
Leaky Nun
12:57 PM
the self-duality of $\Bbb A_\Bbb Q$ means that the dual of $\Bbb Q$ is $\Bbb A/\Bbb Q$
however one can also calculate from $\Bbb Q = \varinjlim \frac1n\Bbb Z$ that the dual of $\Bbb Q$ is $\varprojlim \Bbb R/n\Bbb Z$
so $\Bbb A/\Bbb Q$ and $\varprojlim \Bbb R/n\Bbb Z$ must be isomorphic
let's find out what that isomorphism is
in fact let's start with $\Bbb Z$
so $\Bbb A/\Bbb Z$ is isomorphic to $\Bbb R/\Bbb Z$?
the standard character on $\Bbb A$ sends $(a_2, a_3, a_5, a_7, \cdots, a_\infty)$ to $\prod_{p \le \infty} \exp(2 \pi i a_p)$ by heavy abuse of notation
or if the codomain is $\Bbb R/\Bbb Z$ then just $\sum_{p \le \infty} a_p$
and that is supposed to be the isomorphism guys!
is it even injective
maybe that isn't the isomorphism
maybe I take the orthogonal complement of $\Bbb Z$
so $\{ a \in \Bbb A \mid \sum a_p \in \Bbb Z\}$?
then I now claim that $a \mapsto a_\infty$ is an isomorphism $\{a \in \Bbb A \mid \sum a_p = 0 \in \Bbb R/\Bbb Z\} \to \Bbb R/\Bbb Z$
 
Leaky Nun
Leaky Nun
1:36 PM
no this is completely wrong
$0 \to \Bbb Z \to \Bbb A \to \Bbb A/\Bbb Z \to 0$
$0 \to \widehat{\Bbb A/\Bbb Z} \to \Bbb A \to \widehat{\Bbb Z} \to 0$
so we first need an isomorphism $\widehat{\Bbb A/\Bbb Z} \to \Bbb Z$
well $\widehat{\Bbb A/\Bbb Z} = \{\chi \in \Bbb A \mid \chi|_\Bbb Z = 0\}$
what is the map $\Bbb A \to \widehat{\Bbb Z}$
well that's the restriction
sends $a$ to $1 \mapsto \sum a_p$
so the map $\Bbb A/\Bbb Z \to \widehat{\Bbb Z}$ sends $a$ to $1 \mapsto \sum a_p$
and so the isomorphism $\Bbb A/\Bbb Z \to \Bbb R/\Bbb Z$ sends $a = (a_2, a_3, a_5, a_7, \cdots, a_\infty)$ to $\sum_{p \le \infty} a_p$
 
Leaky Nun
Leaky Nun
1:59 PM
is this even injective
doesn't this send $(1, 0, 0, 0, \cdots, 0)$ to $0$ for example
is that completely wrong
I thought if $G$ is self-dual then $\widehat{H} = G/H$
but maybe it should be $\widehat{H} = G/H^\bot$ instead
yeah that sounds right
 
Leaky Nun
Leaky Nun
2:26 PM
yeah so it should be $\Bbb A/\Bbb Z^\bot \to \Bbb R/\Bbb Z$
 
Leaky Nun
Leaky Nun
3:05 PM
so it suffices to show that $\bigcap n\Bbb Z^\bot = \Bbb Q \subseteq \Bbb A$
 
Leaky Nun
Leaky Nun
3:31 PM
we have a norm map $\Bbb A^\times \to \Bbb R^\times_{>0}$
this splits $\Bbb A^\times$ into $\Bbb A^\times_f \times \Bbb R_{>0}^\times$
no it doesn't
how does the isomorphism $\Bbb A^\times = \Bbb Q^\times \times \widehat{\Bbb Z}^\times \times \Bbb R^\times_{>0}$ work
I'm not sure where $(1, 1, \cdots, -1)$ gets sent to
I guess the extra sign goes to $\widehat{\Bbb Z}^\times$ or $\Bbb Q^\times$
yeah that's right
it goes to $\Bbb Q^\times$
so a more canonical decomposition is $\Bbb Q^\times_{>0} \times \widehat{\Bbb Z}^\times \times \Bbb R^\times$
how does this even work
don't I get all the unit fractions instead
oh no
I get all the positive fractions ok this works
 
 
2 hours later…
Leaky Nun
Leaky Nun
5:49 PM
let's compute an example of a canonical divisor of a function field
take $K = \Bbb F_q(t)$, $X = \Bbb A^1_{\Bbb F_q}$
 
Leaky Nun
Leaky Nun
6:05 PM
Then $\omega := \mathrm dx$ is a non-zero meromorphic 1-form
 
 
2 hours later…
Leaky Nun
Leaky Nun
8:08 PM
let $x_t := (1, 1, 1, \cdots, t) \in \Bbb A^\times$ for $t \in \Bbb R_{>0}$
then multiplication by $x_t$ is a homeomorphism $\Bbb A^\times_1 \to \Bbb A^\times_t$ that scales the Haar measure by $t$
or does it
VSauce music plays
Let $f$ be the indicator function of $\prod \Bbb Z_p^\times \times (1,2)$
the measure of this is $\log 2$
maybe use $e$ instead of $2$
ok if $S \subset \Bbb A^\times_t$ then what is $\mu(x_t^{-1} S)$?
no this doesn't make sense
oh this is convenient
the intersection of that set with $\Bbb A^\times_1$ is just $\prod \Bbb Z_p^\times \times \{1\}$ (let's replace $(1,2)$ with $[1,e)$)
from the compatibility of Haar measures we obtain $\mu(S) = \int_{t=1}^{t=e} \mu(S \cap \Bbb A^\times_1) \ \mathrm dt/t$
since $S \cap \Bbb A^\times_t = x_t (S \cap \Bbb A^\times_1)$
so $\prod \Bbb Z_p^\times \subset \Bbb A^\times_1 = \Bbb Q^\times \times \prod \Bbb Z_p^\times$ gets measure $1$
this is interesting
so it's the counting measure on $\Bbb Q^\times$ with the normalized Haar measure on $\prod \Bbb Z_p^\times$
 
Leaky Nun
Leaky Nun
8:32 PM
wait this means $\mu(\Bbb A^\times_1/\Bbb Q^\times) = 1$
 
 
2 hours later…
Leaky Nun
Leaky Nun
10:26 PM
let's look at the curve $y^2 = x^3 + 1$ near $\infty$
so $y^2 z = x^3 + z^3$
$\infty = (0:1:0)$
so look at the patch $y=1$
so we have $z^3 + x^3 - z =0$
so $x^3 = z-z^3 = (1-z)(1+z)(z)$
$z$ vanishes at $(0,0)$ in this patch
 

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