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Nov 26, 2016 at 22:34 comment added LSpice @dulalnarugopal, the relevant point is that the representation is supercuspidal if it is irreducible (since then the extension of a matrix coefficient of the inducing subgroup is a compactly supported matrix coefficient); but Mackey theory allows you to test for irreducibility by looking at intertwining.
May 6, 2016 at 18:06 comment added PL. Oh, something that it occurs to me I should have said: when you look at $GL_n$ for $n$ prime, you can forget about simple strata and $k_0$. Instead, you just consider fundamental strata of the form $[\mathfrak{A},n,n-1,\beta]$, which are much easier to get used to (in particular, the definition of "fundamental" is easy). I think that this is done for $n=2$ in Bushnell--Henniart, but I can't quite remember how much they prove. Maybe one good, although not quite precise, way of understanding $k_0$ is that it measure how far a stratum is from being fundamental.
May 6, 2016 at 16:46 vote accept MathStudent
May 5, 2016 at 18:50 comment added PL. The one I know is Carayol, Représentations cuspidales du groupe linéaire, Proposition 1.5. He cites Howe's paper on tamely ramified supercuspidals, which I've never read, so I'm not sure where this originates.
May 5, 2016 at 18:25 comment added MathStudent @P.L Can you give me a reference for the fact that if the interwining of $\lambda$ is $J$ then the induced representation is supercuspidal?
May 5, 2016 at 18:02 history edited PL. CC BY-SA 3.0
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May 5, 2016 at 17:52 history answered PL. CC BY-SA 3.0