In Ralf Schmidt's appendix to "Jacquet-Langlands-Shimizu correspondence for theta lifts to $\mathrm{GSp}(2)$ and its inner forms" by Narita and Okazaki , he computes the representations of $\mathrm{GSp}(1,1)$ by examining the tables in his book and discarding the representations whose Weil-Deligne representations fall in the Siegel parabolic. He obtains the answer of IIa, IVa, IVc, Va, Vb, Vc, and VIc.

When I try to replicate the calculation I run into the following problem: representations of type VIa and VIb look almost identical to those of Va. But Va is included and VIa and VIb are not. Furthermore, as written almost all the nilpotent matrices seem to fall neatly into the Siegel parabolic, and the representations are diagonal.

My guess is that we have to compute the image of the Galois representation, and that this cannot be easily read off from the Weil-Deligne representation. Unfortunately I've not succeeded in seeing how to do this either: the discussion in chapter $2$ of Local Newforms for $\mathrm{GSp}(4)$ is not helping.


1 Answer 1


I've never really got my head around ``relevant $L$-parameters'', but the condition on the $L$-parameter (i.e. the associated WD-rep) which ensures transfer to a given inner form should look something like the condition for being in the discrete series. Roughly speaking, representations whose $L$-parameters have fairly large image should show up on inner forms. I think here you want to discard the Weil--Deligne representations which land in the Levi of the Siegel parabolic. This should explain the third sentence in your second paragraph.

For the VIa/b versus Va issue, we want to show that the Weil--Deligne representation for type VIa/b lands inside the Levi of a Siegel parabolic so we can discard it. It doesn't look like it does as written, but we are allowed to conjugate the image by a symplectic matrix, so it might be contained in a conjugate of the Levi of the standard Siegel parabolic. To show that the image is contained in a Siegel parabolic it suffices to show that it leaves stable a two-dimensional isotropic subspace. You can try to cook up one as follows: take an isotropic vector $v$ such that $N(v)$ is also isotropic, and moreover $\langle v, N(v) \rangle = 0$, then take the span of $\{v, N(v)\}$ as your isotropic subspace. Here we can take $v = \sqrt{-1}e_3 + e_4$. The reason that this works in the case in VIa/b but not in case Va is that in VIb/c the Frobenius acts on $e_1$ and $e_2$ (resp. $e_3$ and $e_4$) by the same way so it preserves the isotropic subspace, but in Va the character $\xi$ is necessarily non-trivial, so Frobenius does not preserve the isotropic subspace.


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