2020
DOI: 10.1007/s11139-020-00284-9
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Klingen $${\mathfrak {p}}^2$$ vectors for $$\mathrm{GSp}(4)$$

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Cited by 3 publications
(5 citation statements)
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“…Proof (i)For H=normalKfalse(frakturpfalse)$H=\mathrm{K}(\mathfrak {p})$, see [27, Section A.8]. For H=Γ0(frakturp2)$H=\Gamma _0^{\prime }(\mathfrak {p}^2)$, see [47, Table 1]. For every other H , there exists a conjugate subgroup H$\tilde{H}$ such that normalΓ(p)trueHK$\Gamma (\mathfrak {p})\subset \tilde{H}\subset K$.…”
Section: Local Dimensionsmentioning
confidence: 99%
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“…Proof (i)For H=normalKfalse(frakturpfalse)$H=\mathrm{K}(\mathfrak {p})$, see [27, Section A.8]. For H=Γ0(frakturp2)$H=\Gamma _0^{\prime }(\mathfrak {p}^2)$, see [47, Table 1]. For every other H , there exists a conjugate subgroup H$\tilde{H}$ such that normalΓ(p)trueHK$\Gamma (\mathfrak {p})\subset \tilde{H}\subset K$.…”
Section: Local Dimensionsmentioning
confidence: 99%
“…We thus get the desired dimensions from the rK(π)$r_K(\pi )$ listed in Table 4. (ii)If HΓ0(frakturp2)$H\ne \Gamma _0^{\prime }(\mathfrak {p}^2)$, then a conjugate of H contains normalΓfalse(frakturpfalse)$\Gamma (\mathfrak {p})$, so that rK(π)=VnormalΓfalse(frakturpfalse)VH0$r_K(\pi )=V^{\Gamma (\mathfrak {p})}\supset V^H\ne 0$. If H=Γ0(frakturp2)$H=\Gamma _0^{\prime }(\mathfrak {p}^2)$, then [47, Lemma 4] shows that VnormalΓfalse(frakturpfalse)0$V^{\Gamma (\mathfrak {p})}\ne 0$.$\Box$…”
Section: Local Dimensionsmentioning
confidence: 99%
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