# A computation about Whittaker functions and Eisenstein series

I have some questions about the computation of Eisenstein series and Whittaker functions in the book. The question is on page 29, Theorem 4.3.

My questions are in the following.

(1) I think that $B(F) \backslash G(F) = \cup_{w \in W} B \backslash BwB = \cup_{w \in W} B \backslash BwU = \cup_{w \in W} \{B wu : u \in U\}$. But why representatives for $B(F) \backslash G(F)$ map be taken to be the set of $w^{-1}v$ where $w \in W$ and for each $w$, $v$ runs through a set of representatives for $$(U(F)\cap wU(F)w^{-1})\backslash U(F)?$$ Why here $v$ runs through a set of representatives for $(U(F)\cap wU(F)w^{-1})\backslash U(F)$ but not $U(F)$?

(2) Why we have $$\sum_{w \in W} \int_{(U(F) \cap w U(F) w^{-1})\backslash U(A)} f_{\zeta}(w^{-1} u g) \psi(u)^{-1} du = \\ \sum_{w \in W} \int_{(U(A) \cap wU(A)w^{-1})\backslash U(A)} \int_{(U(F) \cap wU(F)w^{-1})\backslash (U(A) \cap wU(A)w^{-1})} \psi(u')du' f_{\zeta}(w^{-1}ug) \psi(u)^{-1} du?$$

Thank you very much.

I am not entirely sure what you are asking for question 2, but let me take a crack at (1). At verious points I will be sloppy about distiguishing between a coset and its chosen representative, but it shouldn't cause any confusion.

(1) We have $B(F)\backslash G(F) = \cup_{w\in W}B\backslash BwU$. To get the coset reps, we need to see when we can have $$bwu=b'w'u'\quad \text{or, more simply} \quad wu=bw'.$$

Since we know that the Bruhat decomposition is a disjoint union, for this to occur we must have $w'=w$, reducing the identity to $wuw^{-1}=b$. Note that this forces $b\in U(F)$, since $b\in B(F)$ and $wuw^{-1}\in G(F)$ is unipotent.

Thus, the elements of $U(F)$ such that $B(F)wu = B(F)w$ is precisely $(U(F)\cap wU(F)w^{-1})$. Hence, to get coset representatives for the Bruhat cell corresponding to $w\in W$, we need to select representatives of the quotient $$(U(F)\cap wU(F)w^{-1})\backslash U(F).$$

For (2), it seems that all they are doing is the standard change of variables $u\mapsto u(u')^{-1}$ with $u$ a representative in the quotient $$(U(A)\cap wU(A)w^{-1})\backslash U(A),$$ and $u'$ a representative of the quotient $$(U(F)\cap wU(F)w^{-1})\backslash (U(A)\cap wU(A)w^{-1}).$$

The reason $u'$ does not appear in the argument for $f$ is that $u'\in (U(A)\cap wU(A)w^{-1})$ allows you to push it past $w^{-1}$ and use the invariance properties of $f$ as an element of the parabolocally induced representation to get rid of it.

• thank you very much. But it seems that $wu=w'u'$ and $w = w'$ imply that $u=u'$ (not $u'=wuw^{-1}$). – Jianrong Li Jun 9 '15 at 10:24
• You're right. I believe this is the correct argument now. – WSL Jun 9 '15 at 11:31
• Are you currently working through this book? I ask because I am doing the same would be interested in discussing it. – WSL Jun 10 '15 at 10:25
• yes, I am reading the book. It is great to discuss the book with you. It seems that $Bwu=Bw$ implies that $u \in w^{-1}Uw$ (not $wUw^{-1}$): suppose that $Bwu=Bw$. Then $b w u = b' w$ for some $b, b' \in B$. By your proof, we have $b^{-1}b' \in U$. We also have $u = w^{-1} b^{-1}b' w$. Therefore $u \in w^{-1} U w$. – Jianrong Li Jun 10 '15 at 10:31
• it seems that in the book they use the decomposition $G = \cup_{w \in W} Bw^{-1}B=\cup_{w \in W} Bw^{-1}U$. Then we have the representatives of $B \backslash G = \cup_{w \in W} B \backslash Bw^{-1}U$ are of the form $w^{-1}u$. We also have $B w^{-1} u = B w^{-1}$ implies that $u \in w U w^{-1}$. So we have verified the statement in the book. – Jianrong Li Jun 10 '15 at 10:37