3
$\begingroup$

The constant term of the Eisenstein series (for an adele group $GL_2$, say) contains an intertwining operator, often written as $M(s)$. In the form given in Gelbart-Jacquet's Corvallis paper, for example, $$E_N(\varphi(s),g)=\varphi(s)(g)+M(s)\varphi(s)(g).$$ The resulting the trace formula involves the logarithmic derivative of $M(s)$. In the classical setting this is sometimes known as the determinant of the scattering matrix $\Phi(s)$, and the logarithmic derivative makes sense.

How should I think of $M'(s)M^{-1}(s)$ in general? In particular, in what way is it more than a formal expression mimicking the logarithmic derivative?

$\endgroup$

1 Answer 1

1
$\begingroup$

The functional equation gives $M(s)M(-s)=1$, so yes $M(-s)=M^{-1}(s)$. Note that $M(s)'$ is not an intertwiner, only intertwines the compact group $K$ in $GL_2(A)$. The irreducible $K$-isotypes of $GL_2(A)$-reps have all multiplicity one, meaning when you restrict to $K$ irreducible representations of $GL_2(A)$ decompose with multiplicity one, so $M(s)$ and $M'(s)$ act by a constant on irreducible $K$-isotypes by Schur's lemma. If you think of this constant as a function, it is a logarithmic derivative. I have computed some of the relevant constants at the non-archimedean places here.

$\endgroup$
2
  • $\begingroup$ Thanks! Especially for pointing out that the functional equation gives a straightforward interpretation of $M(s)^{-1}$. I am trying to compute the $M(s)$ explicitly in certain cases and am stuck with some computations. I will look through your computations and see if it can help me. $\endgroup$
    – Tian An
    Commented Apr 7, 2014 at 14:38
  • $\begingroup$ Note that my computation apply only to the highest type. For the computations at the real places and the unramified complex cases, you can have a look at my PhD thesis. There is a good reason for working with highest types/smallest weights = irreducible $K$-reps as soon as you have pinned down the local conditions. $\endgroup$
    – Marc Palm
    Commented Apr 7, 2014 at 14:45

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .