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This question is inspired by some others on MathOverflow. Hecke operators are standardly defined by double cosets acting on automorphic forms, in an explicit way.

However, what bother me is that Hecke operators are also mentioned in the automorphic representations language. Those remain very mysterious to me: what are they exactly? Is there a way to define them more explicitly than using the coefficients appearing in the Dirichlet series of the L-function associated to the representation? (What is, even if it appears explicit, unpractical).

My main aim is to grasp the Hecke eigenvalues of a representation, for instance using trace formulas. In the "function" language, since Hecke operators are defined as the action of double classes, we can take the Hecke operators themselves as test-functions and obtain some "Eichler-Selberg"-type trace formulas.

But on the "representation" language, is there something doing the similar work?

More precisely my question could be: is there a function $f_p$ in the Hecke algebra of a local/global group $G(F)$ such that $\mathrm{tr} \ \pi(f_p) = \lambda_\pi(p)$ where $\pi$ is an admissible/automorphic representation of $G(F)$ and $\lambda_p(\pi)$ the Hecke eigenvalue of $\pi$?

Any clue or reference is welcome!

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  • $\begingroup$ Thanks for the question, I also would like to understand deeper what the Hecke operators are on representations, and what Hecke eigenvalues of representations are! $\endgroup$ – Wolker Oct 17 '17 at 19:53
  • $\begingroup$ In the book "Automorphic forms" by Anton Deitmar the case of GL(2) is worked out and the connection between the classical and rep-theoretic Hecke Ops is made explicit. $\endgroup$ – user1688 Oct 18 '17 at 1:43
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If $K$ is a local field and $\mathcal O$ is its ring of integers, we say an irreducible representation of $G(K)$ is unramified if it contains a vector invariant under $G(\mathcal O)$. It is known that such a representation naturally has a unique vector invariant under $G(\mathcal O)$. The Hecke eigenvalue of a double coset in $G(\mathcal O)\backslash G(K)/G(\mathcal O)$ on this representation is simply the eigenvalue of the operator defined by averaging over that double coset on this vector. (Because the vector is the only $G(\mathcal O)$-invariant vector in the representation, it is necessarily sent to a multiple of itself by this operator).

In fact, this operator vanishes on the orthogonal complement of this $G(\mathcal O)$-invariant vector, so its trace is equal to this eigenvalue. On any ramified representation, the trace of this operator is zero. I think this answers your last question.

Of course when we say "the" Hecke operator we mean a particular double coset. I think this is only ambiguous in $GL_n$, where we take the coset generated by a diagonal matrix where one entry is a uniformizer and the rest are units - as I'm guessing you know, based on your familiarity with the double coset theory in the functions world.

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    $\begingroup$ Ok for the unramified case. When you say "on any unramified representation, the trace of this operator is zero", should it read "ramified" instead? Also, that would mean that ramified representations always have zero Hecke eigenvalues? $\endgroup$ – Gory Oct 17 '17 at 20:22
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    $\begingroup$ @Gory In the ramified case, the space V of K-invariant vectors may have dimension greater than 1, and typically the local Hecke algebra surjects to End(V). I see no reason why the trace should be 0. $\endgroup$ – François Brunault Oct 17 '17 at 21:34
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    $\begingroup$ Aren't we talking about an irreducible representation, where the K-invariant vectors are at most 1-dimensional? Then isn't it the case that V^K = 0 if V is ramified? $\endgroup$ – user84144 Oct 17 '17 at 21:41
  • $\begingroup$ In the case K is the maximal compact open subgroup, yes you are right. I was thinking of the case K is an arbitrary compact open. In this case the Hecke algebra doesn't need to be commutative. $\endgroup$ – François Brunault Oct 18 '17 at 1:33
  • $\begingroup$ @Gory Yes, that's precisely what I mean - I meant ramified representations, which have zero Heccke eigenvalues (for the Hecke algebra of the maximal compact). $\endgroup$ – Will Sawin Oct 18 '17 at 11:38
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To keep things simple, let $G$ be a finite group and $K$ a subgroup of it. The simplest definition of the Hecke algebra associated to this pair $(G, K)$ is that it is the algebra of $G$-endomorphisms $\text{End}_G(\mathbb{C}[G/K])$ of the permutation representation of $G$ on $G/K$.

The significance of this algebra is that $\mathbb{C}[G/K]$ represents the functor sending a $G$-representation $V$ to its $K$-fixed points $V^K$, and hence the Hecke algebra is the endomorphism ring of this functor: it naturally acts on $V^K$ for each $V$ and is the largest thing to do so.

Say that the pair $(G, K)$ is a Gelfand pair if any of the following equivalent conditions hold:

  • whenever $V$ is irreducible, $\dim V^K \le 1$;
  • the representation $\mathbb{C}[G/K]$ is multiplicity-free;
  • the Hecke algebra is commutative.

Then we can diagonalize the action of the Hecke algebra on $V^K$ and hence associate to any $K$-fixed vector its Hecke eigenvalues, which together give a character of the Hecke algebra. If $V$ is irreducible and $\dim V^K = 1$ (which, by Frobenius reciprocity, is equivalent to $V$ being a component of $\mathbb{C}[G/K]$) then we only have one $K$-fixed vector to choose from up to scale and hence only one collection of Hecke eigenvalues.

One thing that is really nice about the Gelfand pair condition is that it can sometimes be verified "geometrically," by thinking explicitly about multiplication in the Hecke algebra in terms of double cosets and "relative positions" as described here and here. For example, $(S_n, S_{n-1})$ is a Gelfand pair because double cosets in this case correspond to relative positions of two elements of $\{ 1, 2, ... \dots n \}$ under the action of $S_n$, and there are two such relative positions: "equal" and "not equal." "Equal" is the identity so the Hecke algebra is generated by "not equal," and in particular must be commutative. This simple argument already implies that branching for the symmetric groups is multiplicity-free, allowing us to construct Gelfand-Tsetlin bases etc.

More generally, relative positions have a natural involution which switches the order of the two cosets involved; this gives an involution on the Hecke algebra, and the Hecke algebra is commutative iff every element is fixed by this involution. To check this condition we just need to check whether each relative position is the same when the two cosets involved are switched.

A more explicitly geometric example involving Lie groups is $(SO(3), SO(2))$: here the Hecke algebra involves relative positions of two points on the sphere $S^2$, which are labeled by the length of the shortest path between them. All of these relative positions are invariant under switching the two points involved, and so the Hecke algebra, whatever exactly that means, is also commutative in this case.

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