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It is wellknown that the character of an irreducible, unitary representation of $GL(n,\mathbb{C})$ uniquely determines the isomorphism classes. I fail to construct a function for $GL(2, \mathbb{C})$, which seperates the traces for $I_q(s)$ and $I_q(-s)$ with $q\neq 0$ and $I_q(s)$ being the representation given as the induced $$ \begin{pmatrix} re^{i \theta} & x \\ 0 & r'e^{i \theta'} \end{pmatrix}\mapsto |r/r'|_v^s \cdot e^{iq\theta}$$ up to $GL_2(\mathbb{C})$. Note that $I_q(s) \cong I_{-q}(-s) \otimes e^{iq \arg \det}$ by the usual intertwiner. Note $Re\; s =0$.

I understand abstractly, why the functionals must linearly independant, because they can't be isomorphic as $C_c^\infty( G)$-modules and must have distinct traces therefor.

How can we construct a smooth, compactly supported modulo center function $\phi$ which seperates $tr I_q(s)$ from $tr I_{q}(-s)$ as functionals?

Let's focus on $q=1$. This is a little bit more simple then the general case, because it contains the usual $\rho: U(2) \mapsto U(2)$ identity-representation (not trivial;). So what I did is cosnidering $$\phi: GL_2(\mathbb{C}) \rightarrow \mathbb{C} $$ smooth with $\phi(kza_tk) = \overline{e^{i q\theta} tr \rho(kk')} \phi(a_t)$ with $k,k'\in U(2)$, $z =re^{i\theta}$ in the center, $a_t$ being diagonal with entries $e^{t/4}$ and $e^{-t/4}$, whose convolution operator projects onto the $\rho$-isotype. Assume that $\phi$ has compact support modulo the center. Then $g(t): = e^{-t/2}\int_{N(\mathbb{C})} \phi(a_tn) d n$ is an even and compactly supported function, and $$ tr I_q(s) g = \hat{g}(is),$$ hence does not distinguish between $s$ and $-s$.

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Aren't you missing the half-sum of positive roots, if your normalization has an intertwiner with $s$ and $-s$? – paul garrett Feb 20 '14 at 14:15
I am not sure how this translates in this context? Any suggestions? – Marc Palm Feb 21 '14 at 10:52
A computation in a somewhat different style is done for $GL(n)$ at which is linked-to from the page But, also, generically those intertwiners are isomorphisms (when everything's set up right), so the traces should be the same. What is your ulterior goal here? – paul garrett Feb 21 '14 at 15:18
very nice, but it sadly does not solve my problem. Our computations are not that different, my integral along $N(C)$ is precisely what you denote as the orbital integral, because $\phi$ is $U(2)$-conjugation variant. Btw, the third equality from the end you have lost the cardinality of the Weyl group? – Marc Palm Feb 21 '14 at 15:31

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