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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 $f$?

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    $\begingroup$ Do you happen to know the normalizer of this group in $SL_3(\mathbb{R})$? $\endgroup$
    – S. Carnahan ♦
    Commented Mar 26, 2012 at 2:58
  • $\begingroup$ I think you mean normalizer of the group $\Gamma_0(N)$, don't you? $\endgroup$
    – 7-adic
    Commented Mar 26, 2012 at 5:39
  • $\begingroup$ More precisely this operator should be called fricke involution. Can anyone give a reference? $\endgroup$
    – Jeep Wrangler
    Commented Apr 14, 2012 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
    Commented Apr 15, 2021 at 16:26

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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.

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    $\begingroup$ 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
    Commented Mar 27, 2012 at 10:50
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    $\begingroup$ why is atkin-lehner operator related to the decomposition of your induced representation in GL(2)? $\endgroup$
    – 7-adic
    Commented Mar 29, 2012 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
    Commented Apr 2, 2012 at 11:38
  • $\begingroup$ ... with two generators. $\endgroup$
    – Marc Palm
    Commented Apr 2, 2012 at 11:39

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