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I am working on a problem related to representations of the Weil group of a local field $\mathcal{W}_F$. In many articles one introduces the set $\hat{\mathcal{W}}_F$ of all equivalence classes of irreducible representations of $\mathcal W_F$.

It seems to me that strictly speaking this does not exist (due to set-theoretical reasons). Instead one thinks of $\hat{\mathcal{W}}_F$ as a set that contains for every irreducible representation of $\mathcal{W}_F$ an isomorphic copy of it. So elements of $\hat{\mathcal{W}}_F$ are true representations (and not some collection of representations). In order to create $\hat{\mathcal{W}}_F$ we need to be able to make a universal choice to pick a 'canonical model' from every class of isomorphic irreducible representations.

My questions are the following:

  • Which axioms people assume in order to define a true set like $\hat{\mathcal{W}}_F$?
  • Does it following from the axioms used for example in Bourbaki? (to answer this one needs to be familiar with their foundations as described in 'Theory of Sets'.

Other considerations/remarks related to my question are also welcome. Thanks in advance.

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closed as off-topic by Fernando Muro, Will Sawin, Chris Godsil, Benjamin Steinberg, Andrés E. Caicedo Aug 15 '15 at 19:24

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    $\begingroup$ You don't need to make one universal choice from every class, you can make a nonempty set of choices from each class. Scott's Trick is a systematic way of doing this in ZF and in Bourbaki's system: en.wikipedia.org/wiki/Scott%27s_trick $\endgroup$ – François G. Dorais Aug 15 '15 at 12:47
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    $\begingroup$ @FrançoisG.Dorais: Scott's trick is overkill here (and does not give a strong enough result anyways, because you want a set of equivalence classes), because you can just bound the cardinality of all the irreducible representations. $\endgroup$ – Eric Wofsey Aug 15 '15 at 12:51
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    $\begingroup$ Why does the question restrict to the Weil group? By the way, I think that such a question has already appeared on MO. $\endgroup$ – Martin Brandenburg Aug 15 '15 at 13:24
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    $\begingroup$ This is more or less answered in the second paragraph of Borcherds's reply to the "Wiles's theorem uses inaccessible cardinals" question mathoverflow.net/a/35762/121 Your reasoning also eliminates the possibility of classifying groups of order 4 up to isomorphism. $\endgroup$ – S. Carnahan Aug 15 '15 at 16:19
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It is easy to show that the cardinality of the underlying set of any irreducible representation is bounded in terms of the cardinality of $\mathcal{W}_F$, so can you just take your favorite set $S$ of sufficiently large cardinality and consider only irreducible representations whose underlying set is a subset of $S$. More generally, whenever you can get a cardinality bound on some kind of object, you can use this to get a set of such objects that represents every isomorphism class.

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