As far as I have known, for irreducible admissible representation $\pi$ of $p$-adic group $G$, the matrix coefficient is defined as follows:

For $v\in \pi$ and $w \in \pi ^\vee$, the contragredient representation of $\pi$, there is a canonical pairing $B:\pi \times \pi ^\vee \to \mathbb{C}$ such that $B(v,w):=w(v)$.

Then the matrix coefficient of $\pi$ is a function of $G$ defined like this;

$\{B(gv,w):G \to \mathbb{C}\}_{v\in \pi, w\in \pi ^\vee}$.

But some papers seems to define matrix coefficient not using this canonical pairing but some inner product of $\pi$.

For example, in the global Gan-Gross-Prasad conjecture tells about tempered representations, that is tempered at all places.

In Ichino-Ikeda's paper 'On the periods of automorphic form and Gross-Prasad conjecture',or Neal Harris's paper 'The refined Gross-Prasad conjecture on unitary group' they define the notion of matrix coefficient of $\pi_v$, the local represenation of global automorphic representation $\pi$, as the $<g\cdot v,w>_{v,w\in \pi}$ where $<,>$ is some inner product of $\pi_v$. And then they define the notion 'temperedness' when all the matrix coefficient are $L^{2+\epsilon}(G)$. But obviously, this definition of temperedness seems depend on the choice of inner product $<,>$ and it seems different with the usual tempered notion, because the usual tempered notion concerns the canonical inner product of $\pi_v$ and $\pi^{\vee}_v$.

Is they are the same?

I am also wondering that if a given representation $\pi$ is unitary, can we use hermitian form as the natural local inner product of $V$ and define matrix coefficient using it?

Please deliver me from this confusing.

Thank you for reading my question.

  • $\begingroup$ I suggest that this comment be left as an answer (the point is a basic one, but perhaps it is not immediately obvious to people who are unused to unitary representations) $\endgroup$
    – Yemon Choi
    Sep 8, 2015 at 20:09
  • $\begingroup$ @Venkataramana, Thank you! But I also wondering that irreducible smooth representations are all unitary? I heard that tempered representation is unitary. But I don't know why it is. For square integrable representation, it is possible to make a invariant hermition form. But is it also tru for tempered representations? $\endgroup$
    – Andrew
    Sep 9, 2015 at 17:18
  • 2
    $\begingroup$ actually, when you define a tempered rep, you assume it is unitary. There are (many) non-tempered unitaru representations" these are the reps of the complementary series (and their limits). $\endgroup$ Sep 10, 2015 at 1:15
  • $\begingroup$ @Venkataramana, Then when we deal with irreducible smooth representation or admissible representation, is it tacitly assumed that it is unitary? $\endgroup$
    – Andrew
    Sep 13, 2015 at 14:18
  • 1
    $\begingroup$ @Venkataramana, I see. Since tempered is unitary, Ichino and Neal Harris defined the matrix coefficient using the canonical inner product from the unitarity of given representtion. But for the general admissible representation, to define matrix coefficient, we should associate it with dual representation. $\endgroup$
    – Andrew
    Sep 15, 2015 at 3:42

1 Answer 1


Given a representation π you can define the complex conjugate representation by taking the same vector space on which the new mult by a complex number is the old multiplication by the conjugate of the complex number. An invariant Hermitian form on π gives an identification of the contragredient of π with the complex conjugate of π. Therefore, the two matrix coefficients that you have defined agree.

(This was initially given as a comment; since Yemen Choi suggested that it be given as an answer, I do so!


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