9
$\begingroup$

I'm currently reading the paper "L-functions for the classical groups" by I. Piatetski-Shapiro and S. Rallis, where they introduced the doubling method over classical groups.

I'm confused at the calculation of the negligible orbits on page 5. There we wrote the inner integral of $I(\gamma)$ as $$ \mathcal{I} = \int_{M_k \backslash M_{\mathbb{A}}} \int_{N_k \backslash N_{\mathbb{A}}^\gamma} f(\gamma(n_1, n_2)(m_1, m_2)(g_1^{\prime}, g_2^{\prime})) \phi_1(n_1 m_1 g_1^{\prime}) \phi_2(n_2 m_2 g_2^{\prime}) \,\mathrm{d}n_1 \mathrm{d}n_2 \mathrm{d}m_1 \mathrm{d}m_2. $$ Then to see that this integral vanishes, the authors separate $N^{\gamma} = N_1 \times N_2$ and shift the inner integral to $\phi_1$ and $\phi_2$. They claimed $$ \mathcal{I} = \int_{M_k \backslash M_{\mathbb{A}}} f(\gamma(m_1, m_2)(g_1^{\prime}, g_2^{\prime})) \left( \int_{N_{1,k} \backslash N_{1,\mathbb{A}}}\phi_1(n_1 m_1 g_1^{\prime}) \,\mathrm{d}n_1 \right) \left( \int_{N_{2,k} \backslash N_{2,\mathbb{A}}} \phi_2(n_2 m_2 g_2^{\prime}) \mathrm{d}n_2 \right)\mathrm{d}m_1 \mathrm{d}m_2. $$

Comparing the above two lines, I'm really confused on how the element $(n_1, n_2)$ disappeared in the section $f \in \mathrm{Ind}_{P(\mathbb{A})}^{G(\mathbb{A})}(\omega \circ \delta_P)$.

My attempts: Recall $(n_1, n_2) \in N^{\gamma} \unlhd (G \times G)^\gamma$, where the latter is the stabilizer of $\gamma \in X:= P \backslash H$ under the action of $G \times G$, we can find $p \in P$ such that $\gamma (n_1, n_2) = p \gamma$.

Stuck point 1: But this particular $p \in P$ depends on $(n_1, n_2) \in H$. So why can we move the integral $\int_N$ to the right skipping $f(\cdots)$?

Stuck point 2: Even we rewrite $$ f(\gamma(n_1, n_2)(m_1, m_2)(g_1^{\prime}, g_2^{\prime})) = f(p\gamma(m_1, m_2)(g_1^{\prime}, g_2^{\prime})) = \omega(\delta(p)) f(\gamma(m_1, m_2)(g_1^{\prime}, g_2^{\prime})), $$ then how can I know that $\delta(p)=1$? In the paper, it seems that the authors were suggesting that $p$ is actually in $N$, hence the modulus character $\delta$ takes $p$ to $1$. But I cannot see why this holds.

This is where I got stucked. Sorry for maybe not providing enough background as it seems that this is quite a famous paper. Thanks all for paying attention, commenting and answering!

$\endgroup$

0

You must log in to answer this question.