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I am often annoyed by the presence of the contragredient $\tilde{\pi}$ of the representation we consider, when we write the functional equation of its L-function: $$L(s, \pi) = \varepsilon(s, \pi) L(1-s, \tilde{\pi})$$

I am interested in the $GL(2)$, and maybe $GL(n)$ case: when do we have $\pi = \tilde{\pi}$? Are there explicit examples where this is not the case and where we can compute $\tilde{\pi}$?

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    $\begingroup$ If $\pi \cong \widetilde{\pi}$, then $\pi$ is said to be self-dual or self-contragredient. For $\mathrm{GL}_2$, this is particularly easy to classify: we have that $\widetilde{\pi} \cong \pi \otimes \omega_{\pi}^{-1}$, where $\omega_{\pi}$ is the central character of $\pi$, so self-contragredient automorphic representations are precisely those with trivial central character. $\endgroup$
    – Peter Humphries
    Commented Nov 30, 2017 at 13:57
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    $\begingroup$ For $n \geq 3$, it is no longer necessarily the case that $\widetilde{\pi} \cong \pi \otimes \omega_{\pi}^{-1}$. For $n = 3$, self-contragredient representations $\Pi$ are nonetheless "easy" to classify: they are, up to a quadratic twist, of the form $\operatorname{ad} \pi$ for some automorphic representation $\pi$ of $\mathrm{GL}_2$. This is a result of Ramakrishnan. For $n \geq 4$, however, there's no nice classification of self-contragredient representations that I know of. $\endgroup$
    – Peter Humphries
    Commented Nov 30, 2017 at 14:00
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    $\begingroup$ For general $n$, self-dual automorphic representations of $GL(n)$ have been classified by Arthur. They are the ones arising as functorial transfers from orthogonal or symplectic groups (according to whether it is $L(Sym^2 \pi)$ or $L(\bigwedge^2 \pi)$ that has a pole). This is one of the major results of Arthur's book on automorphic representations of the classical groups. $\endgroup$
    – David Loeffler
    Commented Nov 30, 2017 at 22:17
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    $\begingroup$ Having $\tilde\pi$ on the RHS of the functional equations is just the analogue of having $1-s$ there. So, please, don't be annoyed by it. It serves harmony! In fact twisting $\pi$ by $|\det|^{it}$ has the same effect as shifting $s$ by $it$. On the right hand side this corresponds to twisting $\tilde\pi$ by $|\det|^{-it}$, or shifting $1-s$ by $-it$. It is as nice as it can get. $\endgroup$
    – GH from MO
    Commented Dec 2, 2017 at 9:32
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    $\begingroup$ Another comment: The functional equation is for the completed version (with the gamma factors inclued). The complete $L$-function should be denoted by $\Lambda(s,\pi)$, while the notation $L(s,\pi)$ should be reserved for the product of local factors over the finite (non-archimedean) places. Just to pay respect to our ancestors and not to confuse analytic number theorists (who discovered these objects). $\endgroup$
    – GH from MO
    Commented Dec 2, 2017 at 9:35

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[Expanding my earlier comment into an answer]:

Self-dual cuspidal automorphic representations of $GL(n)$ have been classified by Jim Arthur, as one of the main results of his monograph on automorphic forms for classical groups.

The general shape of the classification is this. If $\pi$ is self-dual, then the Rankin--Selberg $L$-function $L(\pi \times \pi, s)$ has a simple pole at $s = 1$. On the other hand, this $L$-function factors as $L(Sym^2 \pi, s) \times L(\wedge^2 \pi, s)$, and neither can vanish at $s = 1$. So if $\pi$ is self-dual, exactly one of these functions has a pole. We say a selfdual $\pi$ is of orthogonal type if $Sym^2 \pi$ has a pole, and of symplectic type if $\wedge^2 \pi$ has a pole.

If $\pi$ has a Galois representation attached, then you can easily convince yourself that orthogonal / symplectic type is precisely saying that the Galois representation lands in an orthogonal, resp. symplectic, group; and this leads one to expect that $\pi$ should be a functorial transfer from the Langlands duals of these groups. This, if I understand correctly, is exactly what Arthur proves. More precisely:

  • if $\pi$ is symplectic, then $n = 2m$ has to be even, and $\pi$ should be a functorial transfer from the dual group of $Sp(2m)$, which is $SO(2m+1)$.
  • if $\pi$ is orthogonal and $n = 2m+1$ is odd, then, up to a quadratic twist coming from the quotient $O(2m+1)/SO(2m+1)$, $\pi$ must be a functorial transfer from the dual group of $SO(2m+1)$, which is $Sp(2m)$.
  • if $\pi$ is orthogonal and $n = 2m$ is even, then either the Langlands parameter of $\pi$ factors through $SO(2m)$, in which case $\pi$ comes from the Langlands dual of $SO(2m)$ which is $SO(2m)$ again; or there is a non-trivial quadratic character intervening, in which case $\pi$ comes from a twisted form of $SO(2m)$.

Let me make this explicit in some small cases.

  • If $n = 2$ then we have $\tilde\pi = \pi \otimes \omega_{\pi}^{-1}$, as @PeterHumphries has pointed out. Hence there are two ways $\pi$ can be selfdual: we may have $\omega_{\pi} = 1$, in which case $\pi$ is of symplectic type and is a functorial lift from $PGL_2 \cong SO(3)$; or we may have $\omega_\pi \ne 1$ but still $\pi \cong \pi \otimes \omega_\pi^{-1}$, so $\omega_{\pi}$ has to be quadratic and $\pi$ is a dihedral representation induced from the corresponding quadratic field -- which is precisely saying that $\pi$ is of orthogonal type and comes from a twisted form of $SO(2)$.

  • If $n = 3$, then this recovers Ramakrishnan's theorem that $\pi$ has to come from $SL(2)$ via the adjoint square lift, up to twisting by a quadratic character.

  • If $n = 4$, then there are two possibilities again: either $\pi$ is symplectic, so it comes from $SO(5) \cong PGSp(4)$; or $\pi$ is orthogonal, in which case it comes from a (possibly) twisted form of $SO(4)$. The split form of SO(4) is a quotient of $SL(2) \times SL(2)$, so we get $\pi$'s that are transfers from $GL_2 \times GL_2$ via the Rankin--Selberg lifting, and the non-split forms correspond to Asai transfers from $GL_2$ over a quadratic field extension.

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