12
$\begingroup$

I recently asked this question about Euler subgroups and generalizing the automorphic theory of $\mathrm{GL}_n$ to a more general setting. My question here is more specific.

As mentioned there, the important factor from the $\mathrm{GL}_n$ setting is the existence of a subgroup, the mirabolic subgroup, $\mathrm{P}_n\subset \mathrm{GL}_n$ of matrices with bottom row $(0\; 0 \ldots 0\; 1)$.

The existence of this subgroup seems to be one of the most important facts in the theory of automorphic representations of $\mathrm{GL}_n$:

Globally, it allows for useful "Fourier-like" expansions leading towards "multiplicity one"-type results, and enable period integrals to be rewritten as Euler products. This latter property if crucial for the development of the theory of standard L-functions for $\mathrm{GL}_n\times \mathrm{GL}_m$.

Locally, one can restrict Kirillov models to this subgroup to prove statements leading to "strong multiplicity one," as well as the local functional equations for local L-factors.

My question is, as the title suggests,

What is so miraculous about the mirabolic subgroup?

Structually, one has $ \mathrm{P_n} \cong \mathbb{G}_a^{n-1} \rtimes \mathrm{GL}_{n-1}, $ which pops up in certain induction arguments. But it seems like other forces need to converge for its utility. In particular, I am curious about why there don't seem to be effective generalizations (see the Euler subgroup question).

$\endgroup$
2
  • 1
    $\begingroup$ Are you aware of the recent work of Roberts and Schmidt on multiplicity one for the group $GSp_4$? This uses a subgroup called the "paramodular group", which plays a role that is in some ways analogous to the mirabolic subgroup for $GL_n$. $\endgroup$ Feb 8, 2015 at 21:50
  • 1
    $\begingroup$ Section 4.4 of Cogdell's Fields notes tell you a couple suprising things about $P_n$, and sketch how these are used in the proof of Strong Multiplicity One, which of course fails for most other groups. $\endgroup$
    – Kimball
    Feb 9, 2015 at 8:52

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.