18
$\begingroup$

Let $K$ be a non-arch local field (I'm only interested in the char 0 case), let $\mathbb{G}$ be a connected reductive group over $K$ and let $G=\mathbb{G}(K)$. If $V$ is a smooth irreducible complex representation of $G$ and if $H$ is the Hecke algebra of locally constant complex-valued functions on $G$ with compact support (fix a Haar measure on $G$ to make $H$ an algebra under convolution), then $V$ is naturally an $H$-module, and every $h\in H$ acts on $V$ via a finite rank operator and hence has a trace.

It is my understanding that in this connected reductive situation, a theorem of Harish-Chandra says that this trace function $t:H\to\mathbb{C}$ can actually be expressed as

$$t(h)=\int_G tr(g)h(g) dg$$

for $tr:G\to\mathbb{C}$ an $L^1$ function, called the trace of $V$.

If $V$ is finite-dimensional then $tr$ is the usual trace. However I realised earlier this week that I do not know one single explicit example of this function if $V$ is infinite-dimensional. I just spent 20 minutes trying to fathom out what I guessed was probably the simplest non-trivial example: if $G=GL(2,K)$ and $V$ is, say, an unramified principal series representation. I failed :-( I could compute the trace of $h$ for various explicit $h$ (typically supported in $GL(2,R)$, $R$ the integers of $K$) but this didn't seem to get me any closer to an actual formula: in particular, although I could figure out $t$ on various functions I couldn't figure out $tr$ on any elements of $G$. On the other hand I imagine that this sort of stuff is completely standard, if you know where to look.

If $\mu_1$ and $\mu_2$ are unramified characters of $K^\times$ and $V$ is the associated principal series representation of $GL(2,K)$, then what is $tr(g)$ for $g$, say, a diagonal matrix? Or $g$ a unipotent matrix?

[EDIT: Alexander Braverman points out that I have over-stated Harish-Chandra's result: $t$ is only locally $L^1$. Furthermore one has to be a little careful---more careful than I was at least---because $t$ is only defined via some integrals so one could change it on a set of measure zero---hence in some sense asking to evaluate $t$ at an explicit point makes no sense. However he, in his answer, shows how to make sense of my question anyway, as well as answering it.]

[EDIT: Loren Spice points out that my paranthetical char 0 comment is actually an assumption in Harish-Chandra's result, and that apparently local integrability is still open in char $p$. I didn't make a very good job of stating H-C's theorem at all!]

$\endgroup$
2
  • 2
    $\begingroup$ Thanks to everyone for the extensive list of references, all of which seem to have appeared within an hour of two of my asking the question. Much obliged to all of you. $\endgroup$ Commented May 9, 2011 at 20:31
  • $\begingroup$ Concerning positive characteristic, there has been progress by Cluckers, Gordon and Halupczok ("Local integrability results in harmonic analysis on reductive groups in large positive characteristic", Annales ENS 47, 2014), using ideas from motivic integration. $\endgroup$ Commented Jul 22, 2017 at 16:36

6 Answers 6

13
$\begingroup$

First, let us formulate the theorem of Harish-Chandra in a little more precise manner: it is a priori obvious that the character of $V$ is well-defined as a distribution. Now the theorem says that this distribution is given by a locally $L^1$-function which is well defined and is locally constant on an open dense subset of $G$ (but there is no good way to define on the whole of $G$). For example for principal series the function will be well defined on the open subset of regular semi-simple elements. What you can prove is that if such an element $g$ is not split, then the value of the character of an unramified principal series representation at $g$ is just equal to $0$. If $g$ is split, then up to conjugacy it lies in the standard split torus and the character is equal to the Weyl group average of the original character $\chi:T\to {\mathbb C}^*$ from which the principal series representation was induced (up to the standard "$\rho$-shift").

$\endgroup$
5
  • $\begingroup$ Thanks for both the making-precise of my assertion about Harish-Chandra's result, and also the explicit answer. Where does one read a proof of the result you're stating here? $\endgroup$ Commented May 9, 2011 at 20:28
  • 3
    $\begingroup$ The result is proven in the van Dijk notes (“Harmonic analysis on reductive, $p$-adic groups”, SLN162, springerlink.com/content/978-3-540-05189-3). The theorem about characters of (irreducibly) induced principal-series representations is due to van Dijk (“Computation of certain induced characters …”, springerlink.com/index/H155476K15868139.pdf). Notice the Harish-Chandra result, at least, is peculiar to characteristic $0$; as far as I know, local integrability in positive characteristic is still unknown in general. $\endgroup$
    – LSpice
    Commented May 9, 2011 at 21:02
  • $\begingroup$ Thanks Loren. When I was writing I realised I was a bit unsure about the char p situation, hence the vague paranthetical comment! But I hadn't realised that the question was still open in that setting. $\endgroup$ Commented May 9, 2011 at 21:11
  • 2
    $\begingroup$ Well, for principal series the result is valid in any characteristic - it is a simple exercise. The "vanishing" part is basically the statement that if $g$ is regular semi-simple and not split then it has no fixed points on $G/B$ where $B$ is a Borel subgroup. For split $G$ the statement follows from the fact that the fixed points of $G$ on $G/B$ are in one-to-one correspondence with elements of $W$. I can try to write a more detailed proof later today -- I have to run now. $\endgroup$ Commented May 9, 2011 at 21:19
  • 1
    $\begingroup$ Dear @Alexander Braverman : Have you finished running and been ready to "write a more detailed proof " ? $\endgroup$
    – user4245
    Commented Aug 28, 2012 at 12:08
7
$\begingroup$

Although you didn't mention them, it seems appropriate to bring up the so called ‘reducible principal series’—more precisely, the irreducible components of a full induced representation off a Borel. These were computed for the case of $\operatorname{SL}_2$ by Paul Sally's student Stephen Franklin. As far as I know, the thesis was never published; but the formula is announced in Sally–Shalika 3. There must be other explicit calculations out there in this setting, but I don't know them.

$\endgroup$
1
  • 1
    $\begingroup$ anādimadhyānta (mathoverflow.net/questions/64419/…) points out that there is also work of Assem on principal series for bigger special linear groups. $\endgroup$
    – LSpice
    Commented May 10, 2011 at 2:12
6
$\begingroup$

The trace of a principal series representation can be easily computed from the main result of "Computation of Certain Induced Characters of p-Adic Groups" by van Dijk, Math. Ann. 199 229-240 (1972), doi: 10.1007/BF01429876, eudml. It gives you a formula to compute the character of a parabolically induced character based on that of the inducing character. It is of course supported on conjugates of Levi, and for $\gamma$ belonging to the Levi M, the character of the induced representation at gamma equals the weighted sum of characters of the inducing representation at W(G, M)-conjugates of $\gamma$, the weight being an appropriate discriminant factor. The formula also occurs somewhere in Kazhdan's "Cuspidal geometry of p-adic groups".

In particular, in the p-adic case if you induce $(\mu_1, \mu_2)$ to $GL_2(K)$, the character is supported on the conjugates of diagonals, and the value of the character at $(\lambda_1, \lambda_2)$ is equal to something like (I might get some factor wrong) $\displaystyle\frac{\mu_1(\lambda_1) \mu_2(\lambda_2) + \mu_1(\lambda_2) \mu_2(\lambda_1)}{|\lambda_1 - \lambda_2|}$. Here the $|\lambda_1 - \lambda_2|^2$ factor is the reason why it does not extend nicely to $GL_2(K)$, and is also why it is not bounded. However, if you multiply it by $|\lambda_1 - \lambda_2|$, which is what is denoted $|D(g)|^{1/2}$ where $g = (\lambda_1, \lambda_2)$ then you do get a nice function on $GL_2(K)$ (it extends nicely to the non-regular set).

Note that the computation of characters of subquotients of principal series is a totally different ballgame. These guys are not supported on conjugates of the Levi. I doubt anyone has ventured seriously into that problem.

$\endgroup$
5
  • $\begingroup$ I think that factors like $D_{G/M}(\gamma)$ are probably too universal to have a reference attached to them. Indeed, Harish-Chandra's proof that $\lvert D(g)\rvert^{1/2}\Theta_\pi(g)$ is bounded can probably be adapted to explain, at least philosophically, why any reasonable character formula must involve ‘relative discriminants’ $D_{G/H}(g)$, where $H = C_G(g)$. By the way, for subquotient characters, there is the Franklin result mentioned above (mathoverflow.net/questions/64419/…). $\endgroup$
    – LSpice
    Commented May 9, 2011 at 23:49
  • $\begingroup$ Yes, there is the Franklin result and also Magdi Assem's computation of characters of reducible principal series for SL_l, l prime. I should have said "except for some specific results" or so - as in, no such thing has, for instance, been developed in parallel with R-group computations. $\endgroup$ Commented May 10, 2011 at 2:04
  • $\begingroup$ Good point on both counts! It's absurd that I forgot Assem's lovely work, since it was a crucial ingredient in my thesis. $\endgroup$
    – LSpice
    Commented May 10, 2011 at 2:11
  • $\begingroup$ Just noticed that you had given the van Dijk reference above (and Braverman had given the formula modulo $|\lambda_1 - \lambda_2|$). Should have read the comments fully before venturing out with what came to my mind :-( $\endgroup$ Commented May 10, 2011 at 2:43
  • $\begingroup$ I don't think it's a bad thing; your expanded explanation, including the significance of the square root of the discriminant, is very nice. $\endgroup$
    – LSpice
    Commented May 10, 2011 at 3:11
5
$\begingroup$

computing the trace of a smooth irreducible representation is a very difficult problem which is far from being totally solved.

For a nice overview, you may read :

Sally, Paul J., Jr.; Spice, Loren Character theory or reductive $p$-adic groups. Ottawa lectures on admissible representations of reductive $p$-adic groups, 103–111, Fields Inst. Monogr., 26, Amer. Math. Soc., Providence, RI, 2009.

For GL(2) you already have computations in Jacquet-Langlands LN 414 (at least for principal series, at certain elements). I think you may find similar things in Gelfand-Graev-Piateski-Shapiro as well.

Kutzko and Pantoja have determined the Harish-Chandra characters of all smooth irreducible representations of GL(2)

P. C. Kutzko, Amer. J. Math. 109 (1987), no. 2, 201--221, doi: 10.2307/2374571.

Kutzko, Phil; Pantoja, José Character formulas for supercuspidal representations of the groups ${\rm GL}_2,\ {\rm SL}_2$. Comm. Algebra 26 (1998), no. 6, 1679–1697, doi: 10.1080/00927879808826232.

You have more general results by Sally, Corwin on GL(n) but in the tame case (see Math Sci Net). See also the computations by Bushnell and Henniart in their explicit version of the Jacquet-Langlands correspondence in the tame case.

Of course I must forget to cite a lot of contributors.

For supercuspidal representations, you may compute its character as soon as you get it as a compactly induced representation. Indeed you may apply an adapted form of Mackey's formula. On that subject, I recommand the appendix of Bushnell-Henniart, Publication IHES. All explicitely known supercuspidal representations are obtained as compactly induced representations.

You have more modern tools as well. Schneider and Stuhler have associated to any smooth irreducible representation $\pi$ of a $p$-adic reductive group $G$ a $G$-equivariant coefficient system $C(\pi )$ on the affine building $X$ of $G$ (see their IHES publication). They proved that if $\gamma$ is a regular elliptic element of $G$, then the the value of the Harish-Chandra character at $G$ is given by the trace of $\pi (\gamma )$ in the Euler-Poincaré module of the restriction of $C(\pi )$ to $X^{\gamma}$, the fixed point set of $\gamma$ in $X$.

Unfortunately this coefficient system as well as the set of fixed points in $X$ are difficult to work out for explicit representations. In simple cases like that of level zero represenatations of $GL(m,D)$ and $\gamma$ minimal in the sense of Bushnell-Kutzko (those elements have a single fixed point in the building) you may obtain a simple formula. For example, as an exercice you may compute this way the character of the Steinberg representation at regular elliptic elements (it is constant, equals to $\pm 1$.

$\endgroup$
7
  • $\begingroup$ Thanks so much for all these references. In fact I got hold of a copy of the Ottawa lectures recently because I wanted to read another of the articles: I should have kept hold of it :-/ I hadn't really realised the question was so subtle either. $\endgroup$ Commented May 9, 2011 at 20:29
  • 2
    $\begingroup$ “All explicitly known supercuspidal representations are obtained as compactly induced representations.” In fact, for some groups, ‘all’ without qualification: arxiv.org/abs/math/0607262. $\endgroup$
    – LSpice
    Commented May 9, 2011 at 21:09
  • 1
    $\begingroup$ “For supercuspidal representations, you may compute its character as soon as you get it as a compactly induced representation. Indeed you may apply an adapted form of Mackey's formula.” I also feel that, while true, this may sweep aside some of the difficulties. Even given very concrete and pleasant inducing data, evaluating the Harish-Chandra integral is no straightforward exercise! $\endgroup$
    – LSpice
    Commented May 9, 2011 at 21:11
  • $\begingroup$ Hearing these comments makes me feel far better about the fact that I couldn't fathom much out at all this afternoon when I was trying to develop the theory from first principles with no references! $\endgroup$ Commented May 9, 2011 at 21:17
  • 2
    $\begingroup$ P Broussous : Minor nitpick : Kutzko-Pantoja haven't described the characters of all irreducible admissible (admissibility is necessary for characters to be defined) representations of GL_2. They have considered only supercuspidals, eg. not the reducible principal series (I think Franklin has only considered the tame case). And among representations induced from the normalizer of $GL_2(o)$ they have only considered depth 0 ones. They have only computed character values in a small neighborhood around identity. AFAIR Kutzko can write all characters of GL_2, he just hasn't written/published them. $\endgroup$ Commented May 10, 2011 at 2:20
4
$\begingroup$

I'm not sure about principal series representations, but I know that there is a Frobenius formula for supercuspidal representations of $p$-adic groups that helps in calculating these trace characters $tr$. More precisely :

$\mathbf{Theorem}$: Let $G$ be a connected reductive $p$-adic group, and suppose (for simplicity) that $Z(G)$ is compact. Let $K$ be an open compact subgroup of $G$ and let $(\sigma,W)$ be an irreducible representation of $K$ such that $$\pi := Ind_K^G \sigma$$ is a supercuspidal representation of $G$ (the induction is compact induction). If $g \in G$ is regular and if $tr_{\pi}$ denotes the distribution character you described above for $\pi$,

$$tr(g) = \displaystyle\sum_{x \in K \setminus G / K} \ \ \displaystyle\sum_{y \in K \setminus KxK} \dot{tr}_{\sigma}(ygy^{-1})$$

where $\dot{tr}_{\sigma}(h)$ is defined to be

$tr_{\sigma}(h)$ if $h \in K$, and zero if $h \notin K$. Here, $tr_{\sigma}$ is the distribution character of the representation $\sigma$.

This is Theorem 1.9 in Paul Sally's paper "Some remarks on discrete series characters for reductive p-adic groups". Note: If $Z(G)$ is not compact, you get a similar formula but with a central character out front.

In principle, one starts with understanding $K \setminus G / K$. Then, one knows that $KxK$ can be written as a finite union of right cosets. To see this, consider $K \cap x^{-1} Kx$. Then $K / (K \cap x^{-1} Kx)$ is finite since $K$ is compact open. Then, $$KxK = \displaystyle\bigcup_{z \in K / (K \cap x^{-1} K x)} Kxz$$ since $K \cap x^{-1} Kx$ is "all the stuff in $K$ that you can move to the left of $x$ in $KxK$". Therefore, the inner sum becomes $$\displaystyle\sum_{z \in K / (K \cap x^{-1} Kx)} \dot{tr}_{\sigma}(xzgz^{-1} x^{-1}),$$ a finite sum. Then, one goes about calculating all these terms and simplifying and organizing the result.

This is difficult in general. I'm sure Loren Spice or Jeffrey Adler will answer/comment at some point, since they have computed these things in large generality, but here is a toy example :

One place to start is with depth zero supercuspidal representations. Let $G$ be a split group and suppose that $Z(G)$ is compact. If $K = G(\mathfrak{o})$, then we have the Cartan decomposition $G = KA^- K$, and the bijection $A^- / A(\mathfrak{o}) \leftrightarrow K \setminus G / K$, given by $a \mapsto KaK$, where $A$ is the maximal split torus of $G$, and where $A^-$ is the set of all $a \in A : |\alpha(a)| \leq 1 \ \forall \alpha \in \Delta$. Here, $\Delta$ is a set of simple roots. So we now have simplified the outer sum. Then, one goes about calculating $K / (K \cap x^{-1} Kx)$, and then the elements $xzgz^{-1} x^{-1}$, and tallies everything together.

You can try $SL(2,F)$ and fix some concrete $x \in A^-$. Then literally calculate $K \cap x^{-1} K x$, and then $K / (K \cap x^{-1} K x)$, and see what happens, and you will have calculated the inner sum. Then you can calculate $tr_{\pi}$ in terms of $tr_{\sigma}$. Moreover, since we are in depth zero, $tr_{\sigma}$ will be given in terms of information over the finite field, which is known in principle.

I should mention Loren Spice and Jeffrey Adler's paper http://lanl.arxiv.org/abs/0707.3313 where they compute these distribution characters in large generality. It is quite complicated in general, just by looking at their paper.

$\endgroup$
3
  • $\begingroup$ So just like when analysing Bernstein components, the principal series ones in some sense turn out to be the hardest rather than the easiest, it seems! Thanks a lot for these references. $\endgroup$ Commented May 9, 2011 at 20:30
  • 2
    $\begingroup$ I should mention that the Frobenius formula in the form you mention has enjoyed only limited success; I know only of its use to compute supercuspidal characters of $\operatorname{SL}_2$ by Paul, and by Julia to analyse, though not to ‘compute’ in the sense of this answer, depth-$0$ supercuspidal characters of classical groups (xxx.lanl.gov/abs/math.RT/0403529). It seems to be more tractable to work directly with Harish-Chandra's integral (of which Paul's Frobenius formula is a ‘discretisation’), as Jeff and I do. $\endgroup$
    – LSpice
    Commented May 9, 2011 at 21:06
  • $\begingroup$ (Sorry, I should have said Paul and Joe Shalika. Perhaps Phil Kutzko's calculation also proceeds this way; I don't remember.) $\endgroup$
    – LSpice
    Commented May 9, 2011 at 21:12
2
$\begingroup$

I am not sure, if you are still interested in this, but here is the general computation:

Let $\phi \in C_c^\infty (GL_n(F))$, and let $\pi$ be a super-cuspidal representation of a Levi subgroup $M$ of a parabolic $P$ with unipotent radical $N$, and let $\pi_0 = Ind_{P}^{GL_n(F)} \pi$ be the normalized induced representation (assume unitary, irreducible for safety). Let $K$ be compact open subgroup with $GL_n(F) = P K$.

  1. Define $ \phi^K (x) = \int\limits_K \phi(k^{-1}xk) d k,$
  2. Define $ A\phi^K(m) = \Delta_P(m)^{1/2} \int\limits_{N} \phi^K(mn)d n$ for $m \in M$
  3. Then we have that $A\phi^K \in C_c^\infty(M)$ and the formula $$ tr \pi_0(\phi) = tr \pi( A \phi^K).$$

The same formula is also useful, if the representation is not irreducible (unitarizability is not really an issue, and admissibility follows from the Iwasawa decomposition), but one has to normalize and decompose according to the $K$-isotypes.

So in your situation, you get a Fourier transform of $A \phi^K$. This in addition with Moshe Adrian answer computes all the irreducible, unitary principal series representation, at least in prinicple.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .