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Additional info for A Cauchy Harish-Chandra integral, for a real reductive dual pair

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By taking the transpose we get CD = 0. This verifies (a). Let C, D ∈ SMm (R) \ {0}. 8], and (a), (D, C) ∈ WF(µ) ˆ if and only if (−C, D) ∈ WF(µ), which happens if and only if C ∈ −supp µ, D ∈ SMm (R) and CD = 0. Also, the fiber of WF(µ) ˆ over zero coincides with −supp µ. This verifies (b). Notice that for D ∈ SMm (R) and w ∈ Rm , Dwwt = 0 if and only if Dw = 0. Indeed, both sides of the equivalence are invariant under the action of the orthogonal group Om (R). Hence we can assume that D is diagonal.

Let X = X 1 ⊕ X 2 ⊕ ... and Y = Y1 ⊕ Y2 ⊕ ... be the decomposition of X , Y into A -isotypic components. , Y = Y1 ⊕ Y2 ⊕ ... 1), is non-degenerate. Let Ws = Hom(Vs , V ), Wc = Hom(Vc , V ), W j = Hom(V j , V ). Then we have the following direct sum orthogonal decompositions W = Wc ⊕ Ws , Ws = W1 ⊕ W2 ⊕ ... 3) W j = Hom(X j , V ) ⊕ Hom(Y j , V ) ( j ≥ 1). 4) a = sp(Wc ) ⊕ EndR (Hom(X 1 , V )) ⊕ EndR (Hom(X 2 , V )) ⊕ ... A = Sp(Wc ) × G L R (Hom(X 1 , V )) × G L R (Hom(X 2 , V )) × ... Let A be the centralizer of A in Sp.

X (1 + z 0 ) − x(z 0 + z)) dx d x 2 354 T. Przebinda = const |det(1 − z0 )|−1 G \X max δ(x z 0 + z) X . δ(x (1 + z 0 ) − x(1 − z0 )−1 (z 0 + z)) dx d x = const |det(1 − z0 )|−1 . G \X max δ(x (1 + z 0 + z 0 (1 − z 0 )−1 (z 0 + z))) d x = const |det(1 − z0 )|−1 |det(1 − z)|−1 . G \X max δ(x (1 + z 0 + z 0 (1 − z 0 )−1 (z 0 + z))(1 − z)−1 ) d x = const |det(1 − z0 )|−1 |det(1 − z)|−1 . G \X max δ(x (1 + (1 − z0 )−1 (z 0 + z)(1 − z)−1 )) d x = const |det(g0 − 1)det(g − 1)| = const |det(g0 − 1)det(g − 1)| G \X max δ x 1 .

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