2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thank you very much. But it seems that $wu=w'u'$ and $w = w'$ imply that $u=u'$ (not $u'=wuw^{-1}$). $\endgroup$ – Jianrong Li Jun 9 '15 at 10:24
  • $\begingroup$ You're right. I believe this is the correct argument now. $\endgroup$ – WSL Jun 9 '15 at 11:31
  • $\begingroup$ Are you currently working through this book? I ask because I am doing the same would be interested in discussing it. $\endgroup$ – WSL Jun 10 '15 at 10:25
  • $\begingroup$ 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 $u \in w^{-1} U w$. $\endgroup$ – Jianrong Li Jun 10 '15 at 10:31
  • $\begingroup$ 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. $\endgroup$ – Jianrong Li Jun 10 '15 at 10:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.