Considering the Arthur trace formula, why are the sort of convolution operators, whose "normalized traces" are given in geometric terms and spectral terms, actually able to distinguish all automorphic representations? Can they seperate the cuspidal representation? What about the rest? Both the local and global picture are interesting for me.

Isn't this question not the starting point, which justifies to study functioriality via trace formulas?

  • $\begingroup$ Dear pm, I thought about your question a bit, and again after reading Arno Kret's answer and your comments to it, and I'm not sure what you are asking. It seems that you are not asking about the comprasion of geometric sides of the trace formulas for different groups (which is what motivates the stablization of the trace formula, and which is the basic tool in proving (some limited number of!) functorialities), but rather asking how, on the spectral side, one can choose test functions which pin down particular constituents of the L_2-automorphic forms. Is this correct? Regards, Matthew $\endgroup$ – Emerton Apr 26 '11 at 12:02
  • $\begingroup$ P.S. Note that if a given irrep. $\pi$ appears with multiplicity in $L_2$, then one won't be able to distinguish the multiple copies with test functions. Is this part of what you are concerned about in your question? Or is it that you are concerned about how to pick out a given $\pi$ by local means (which is related to the fact that $\pi$ is a restricted tensor product of local $\pi_v$s)? Or is is that $L_2$ has a continuous as well as a discrete spectrum? Perhaps you could edit the question to indicate in more detail which of these things you are asking about. $\endgroup$ – Emerton Apr 26 '11 at 12:06
  • $\begingroup$ I meant of course $spec(H) \subset spec(G)$. $\endgroup$ – Marc Palm Apr 26 '11 at 12:18

I am not sure if I understood your question well. I try to give an answer to the following,

"Why is it expected that the trace formula implies cases of Langlands functoriality?"

hoping that it is the right question.

Very roughly I think the idea is as follows.

Let $G, H$ be two connected reductive groups, and suppose given an L-morphism $LG$ to $LH$, where $LG$ (resp. $LH$) is the Langlands dual of $G$ (resp. $H$), with certain properties. Then one expects (conjecture) to be able to relate automorphic representations of G to automorphic representations of $H$.

With the trace formula one attempts to prove (cases of) this conjecture.

To simplify, assume that the groups are anisotropic.

Let $\pi$ be a cuspdial automorphic representation of G. Assume now, that we are in the following very fortunate situation. Suppose that we have a function g on G(adeles) and h on H(adeles) which have the following properties.

(0) For convenience only: assume that the functions are elementary tensors $g = \bigotimes_v g_v$, and $h = \bigotimes_v h_v$ ($v$ ranges over the places of $Q$).

(1) The trace of the function g acting on an automorphic representation of G is zero unless it is isomorphic to $\pi$.

(2) The trace of the function h acting on an automorphic representation of H is zero unless it is isomorphic to a certain fixed representation $\pi'$.

Unfortunately I do not know how to formulate the third, and most important condition in a precise manner in the above generality. But I will try to give an idea anyway.

(3) The pair of functions (g, h) is associated. This means the following. From the map of L-groups one should be able to transfer conjugacy classes in $H(Q_v)$ to conjugacy classes in $G(Q_v)$. One then asks that the orbital integral of $h_v$ at a conjugacy class $c$ is equal to the orbital integral of $g_v$ at the $G(Q_V)$-conjugacy class obtained from $c$ by transfer. Moreover, the orbital integral at any $G(Q_v)$-conjugacy class which is not a transfer is demanded to be $0$.

Then, using (3) the geometric side of the trace formula of $G$ can be compared with the geometric side of the trace formula for $H$. So applying the trace formula for $G$ and $H$ at the same time, we see that the spectral side of the trace formula for $G$ is equal to the spectral side of the trace formula for $H$. But, by (1) and (2), on these spectral sides only the representations $\pi$, $\pi'$ remain.

The point is then that $\pi \rightarrow \pi'$ is the ''candidate'' for the automorphic representation predicted by Langlands functoriality.

There are many problems with the above. We already saw that it is not clear how to define (3) in general. Moreover, (1) and (2) are often not possible, only for finite sets called packets. Next, there is a problem with non-stable conjugacy; in an algebraic group there are several notions of conjugacy, for example rational conjugacy (two elements are conjugate by an element of G(Q)) and conjugacy over the algebraic closure (two elements are conjugate by an element of G(Q)). There is also a third notion, "stable conjugacy", and one has to work with this one if one wants to have a chance to give a meaning to (3). This also implies that you have to work with a different trace formula instead, "stable trace formula" ... complicated!

An example of (proved) Langlands functoriality is Jacquet-Langlands. One then knows how to give a precise definition of (3), because in this case G and H are inner forms. In my opinion this is explained very well in section 2 of


Another case where (3) is defined is when $H$ is an endoscopic group. This is also explained in the Paris book project by Harris,


  • $\begingroup$ Thanks for this elaborated answer. I have some questions. Concerning (1) and (2), you probably want to say that the trace of the convolution operator associated with $h$ resp. $g$ vanishes, if and only if the projection to $\pi$ respective $\pi'$ is zero? Finite set of places? The case where $G$ anisotropic corresponds to the classical case of cocompact lattices, isn't it? My main motivation behind the question was actually why is the algebra of convolution operators big enough to determine the spectral decomposition globally and locally! $\endgroup$ – Marc Palm Apr 26 '11 at 11:53
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    $\begingroup$ Dear pm, Local admissible smooth representations have a character --- more precisely, a Harish-Chandra character, which is a distribution on test functions. The value of this distribution on a test function is precisely obtained by computing the trace of the convolution operator given by the test function on the representation. These characters satisfy the usual formalism, and in particular serve to distinguish non-isomorphic irreducible reps. Regards, Matthew $\endgroup$ – Emerton Apr 26 '11 at 12:12
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    $\begingroup$ With (1) and (2) I mean to say the following. There are lots of automorphic representations, and on each one of them you can let your function act via the convolution operator. I ask the trace of this operator to be 0 for ALL automorphic representations except one: $\pi$ (up to isomorphism). $\endgroup$ – mnr Apr 26 '11 at 12:17
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    $\begingroup$ The reason to assume that $G$ is anisotropic is because it implies that the corresponding double quotient space $G(Q)\G(A)/K_\infty K$ is compact. This simplifies the trace formula, otherwise you have to ask more properties of the functions $(g, h)$ and use the "simple trace formula". $\endgroup$ – mnr Apr 26 '11 at 12:20
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    $\begingroup$ Regarding your second comment, I am not sure what you mean by "distinguish at the local places". But if you want a proof that functions f on G which satisfy (1) exist, here is the argument. Let $A$ be the space of automorphic forms on $G$. So $\pi \subset A$ is given. Pick a compact open subgroup $K \subset G(A_f)$ ($A_f = $ finite adeles) such that $\pi_f^K \neq 0$. Then an important point is that $A^K$ is finite dimensional and thus it has only finitely many irreducible sub-representations of the hecke algebra $H(G(A_f)//K)$. $\endgroup$ – mnr Apr 26 '11 at 12:33

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