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By Daniel Bump (auth.)

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Extra resources for Automorphic Forms on GL (3,ℝ)

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0 as t .... > 0 The sum: ~. EXERCISE: of equations left, M w . ( v l , V 2 ). NI that the one m a y to the constants is b o u n d e d the differential of i n t e g r a t i o n - t o of the Vl,V2, under EXERCISE: right useful. the of i n t e g r a t i o n exis~ Differentiating Let to the appear satisfies lines of the t i n u o u s f a s h i o n on n~ n2 "W(YI'Y2) Yl ±'Y2 partial W at the p o l e s combination THEOREM eventually that Moving residues linear taken and: factor We are fSl+O~ fSl+4 (Sl+~ fs2-~ fs2-~ (s2-d ~--~-/~ t-=z-/~ t--~-/~ t--z-/~ t-T-/~ \ - ~ - / ....................

K. Thus, A AI ~-basis on G. 31) by an element of the left ideal generated X2-Z 2. 13). 29) that: 2 + 2+ 2 2 2 2 ~2 - 3(-HIH2 HIH2 2HI-HIH2-2HI+ZoX2XI+ZoXIX2-XIH2+X2HI-ZoHI + ~ 2+~2. ~. 36) We find that: A2 = _y2y 2 3 2 ~3 YlY2 3x23yl + 3 33 2 33 3y23y2 + YlY2 ~yl3y ~ 33 2+ 2, 2 + (-x2 Y2)YlY2 - 2y~Y2x 2 33 2 3x33Y 2 3Xl~X33Y 2 2 2 + 2YlY 2 33 3x13x23x 3 32 32 + Y~ 3y~ 2. 13). 15). all occurrences procedure volving . 13). It is easy to see that this ~Y2 will yield a sixth degree differential equation in F in- only ; the exact expression is hard to find because ~Yl length of the computation.

We claim that if that is, a function which is right invariant by is also right invariant by K. Indeed, the right K implies that spect to the Lie algebra of K a c t i n g as an a l g e b r a of d i f f e r e n t i a l Since D this implies that at least Df component of the identity of intersect sequently Df the center is right We see, therefore, operators on the a c t i o n of ~. @ is invariant with re- commutes with the Lie algebra of invariant under the Lie algebra of K f invari- with respect to operators.

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