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Change < > to \langle \rangle
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Let's see if this fixes it.
Thanks to Emmanuel Kowalski and Pieter Belmans
http://stacks.math.columbia.edu/tag/03W3#comment-867
http://stacks.math.columbia.edu/tag/03W3#comment-868
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aisejohan committed Aug 1, 2014
1 parent 3418934 commit 83e21b9
Showing 1 changed file with 4 additions and 4 deletions.
8 changes: 4 additions & 4 deletions etale-cohomology.tex
Original file line number Diff line number Diff line change
Expand Up @@ -15149,7 +15149,7 @@ \section{Preliminaries and sorites}
$$
1 =
\sum\nolimits_{
\frac{\sigma\text{-conjugacy}}{\text{classes}\left<\gamma\right>}
\frac{\sigma\text{-conjugacy}}{\text{classes}\langle\gamma\rangle}
}
\frac{1}{\# Z_\gamma} \mod{\ell}.
$$
Expand Down Expand Up @@ -17359,10 +17359,10 @@ \section{Precise form of Chebotarev}
(2g_X - 2) d_\pi^2\sqrt{q^n}
$$
Write $1_C = \sum_\pi a_\pi \chi_\pi$, then
$a_\pi = \left<1_C, \chi_\pi\right>$, and
$a_1 = \left<1_C, \chi_1\right> = \frac{\# C}{\# G}$ where
$a_\pi = \langle1_C, \chi_\pi\rangle$, and
$a_1 = \langle1_C, \chi_1\rangle = \frac{\# C}{\# G}$ where
$$
\left<f, h\right> = \frac{1}{\# G}\sum_{g \in G} f(g)\overline{h(g)}
\langle f, h\rangle = \frac{1}{\# G}\sum_{g \in G} f(g)\overline{h(g)}
$$
Thus we have the relation
$$
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