Let $f$ be an automorphic form for $\Gamma_0(N)\subset SL(3,\mathbb{Z})$.

$\Gamma_0(N)=(a,b,c;d,e,f;g,h,i)\in SL(3,\mathbb{Z})|g=h=0(mod N)$

Is there any Atkin-Lehner operator for $\Gamma_0(N)$ which will give a functional equation for L-function of $f$?

  • 4
    $\begingroup$ Do you happen to know the normalizer of this group in $SL_3(\mathbb{R})$? $\endgroup$
    – S. Carnahan
    Mar 26 '12 at 2:58
  • $\begingroup$ I think you mean normalizer of the group $\Gamma_0(N)$, don't you? $\endgroup$
    – 7-adic
    Mar 26 '12 at 5:39
  • $\begingroup$ More precisely this operator should be called fricke involution. Can anyone give a reference? $\endgroup$ Apr 14 '12 at 16:53
  • $\begingroup$ I wrote an answer to the related post mathoverflow.net/questions/307442/fricke-involution-on-gl3, since it was the newer question. There is indeed a Fricke involution on GL(3). I define it in the answer and for more details I wrote a note that's also linked in the answer. $\endgroup$
    – Radu T
    Apr 15 at 16:26

You will have to decompose

$$ Endo_{SL(3, \mathbb{Z})} ( Ind_{\Gamma_0(N)}^{SL(3, \mathbb{Z})} 1 ) .$$

This will give you the analog of the Atkin-Lehner theory. However, I have some doubts that this exists in the literature. This question of mine gives you the resaon why it has not been done for $d,g,h = 0 \bmod N$:

Parabolic induction GL(n,Zp)

I am happy, if somebody proves me wrong though.

Edit: I forgot to mention, that the case for $N$ square free is in general possible, since you can rely on the representation theory of reductive groups over residue fields.

  • 1
    $\begingroup$ In general, the Atkin-Lehner are expected to be messier than in the GL(2) setting, since their definition will depend upon the residue characteristic;( $\endgroup$
    – Marc Palm
    Mar 27 '12 at 10:50
  • 1
    $\begingroup$ why is atkin-lehner operator related to the decomposition of your induced representation in GL(2)? $\endgroup$
    – 7-adic
    Mar 29 '12 at 18:06
  • $\begingroup$ Decompose $Ind_{\Gamma_0(p)}^{\SL_2(\mathbb{Z})} 1 = 1 \oplus V$ into irreducibles. You can identify $Hom_{SL_2(\mathbb{Z})}( nd_{\Gamma_0(p)}^{\SL_2(\mathbb{Z})} 1 )$ with an algebra of function on $SL_2( \mathbb{Z} ) // \Gamma_0(p)$. This is a commutative algebra with two elements. If you write these explicitely, you will see the connection. $\endgroup$
    – Marc Palm
    Apr 2 '12 at 11:38
  • $\begingroup$ ... with two generators. $\endgroup$
    – Marc Palm
    Apr 2 '12 at 11:39

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.