In the study of number theory (and in other branches of mathematics) presence of Hecke Algebra and Hecke Operator is very prominent.

One of the many ways to define the Hecke Operator $T(p)$ is in terms of double coset operator corresponding to the matrix $ \begin{bmatrix} 1 & 0 \\ 0 & p \end{bmatrix}$ .

On the other hand Hecke Algebra $\mathcal{H}(G,K)$ associated to a group $G$ of td-type ( topological group, such that every neighborhood of unity contains a compact open subgroup), where $K$ is a compact open subgroup of $G$ is defined as the space of locally constant compactly supported $K$ bi-invariant functions on $G$. Convolution product makes it an associative algebra.

I was told that the hecke algebra $\mathcal{H}(Gl(2,\mathbb{Q}_p) , Gl(2,\mathbb(Z)_p))$ corresponds to the classical algebra of hecke operators attached to $p$ via Satake Isomorphism Theorem. Using Satake Isomorphism theorem I can show $\mathcal{H}(Gl(2,\mathbb{Q}_p) ,Gl(2,\mathbb(Z)_p))$ is commutative and finitely generated over $\mathbb{C}$.

So my question is how one uses Satake Isomorphism Theorem (or otherwise) to see this? And secondly in general what is the relation between hecke operators and hecke algebra?


The fact that Hecke operators (double coset stuff coming from $SL_2(\mathbf{Z})$ acting on modular forms) and Hecke algebras (locally constant functions on $GL_2(\mathbf{Q}_p)$) are related has nothing really to do with the Satake isomorphism. The crucial observation is that instead of thinking of modular forms as functions on the upper half plane, you can think of them as functions on $GL_2(\mathbf{R})$ which transform in a certain way under a subgroup of $GL_2(\mathbf{Z})$, and then as functions on $GL_2(\mathbf{A})$ ($\mathbf{A}$ the adeles) which are left invariant under $GL_2(\mathbf{Q})$ and right invariant under some compact open subgroup of $GL_2(\widehat{\mathbf{Z}})$.

Now there's just some general algebra yoga which says that if $H$ is a subgroup of $G$ and $f$ is a function on $G/H$, and $g\in G$ such that the $HgH$ is a finite union of cosets $g_iH$, then you can define a Hecke operator $T=[HgH]$ acting on the functions on $G/H$, by $Tf(g)=\sum_i f(gg_i)$; the lemma is that this is still $H$-invariant.

Next you do the tedious but entirely elementary check that if you consider modular forms not as functions on the upper half plane but as functions on $GL_2(\mathbf{A})$, then the classical Hecke operators have interpretations as operators $T=[HgH]$ as above, with $T_p$ corresponding to the function supported at $p$ and with $g=(p,0;0,1)$. Because the action is "all going on locally" you may as well compute the double coset space locally, that is, if $H=H^pH_p$ with $H_p$ a compact open subgroup of $GL_2(\mathbf{Q}_p)$, then you can do all your coset decompositions and actions locally at $p$.

Now finally you have your link, because you can think of $T$ as being the characteristic function of the double coset space $HgH$ which is precisely the sort of Hecke operator in your Hecke algebra of locally constant functions. Furthermore the sum $f(gg_i)$ is just an explicit way of writing convolution, so everything is consistent.

I don't know a book that explains how to get from the classical to the adelic point of view in a nice low-level way, but I am sure there will be some out there by now. Oh---maybe Bump?

  • $\begingroup$ This explanation is very useful.The place where I read about how to think of hecke operators adelicly was the book Automorphic forms on Adele groups by Gelbert. I recently came up with a version of Satake Isomorphism Theorem which says $\mathcal(H)(Gl(2,\mathbb(Q)_p),Gl(2,\mathbb(Z)_p)) \cong \mathbb(C)[T,S,S^-1] $, I was told that $T$ corresponds to the hecke operator $T(p)$ when thought as an operator on automorphic form.I am not familiar with this form of Satake isomorphism theorem, so was hoping someone can explain this.But your answer clearly shows there is a much elementary connection. $\endgroup$ – Dipramit Majumdar Mar 29 '10 at 19:33
  • $\begingroup$ There are also some very nice notes of W. Gan explaining the passage from classical to adelic. They are actually slides for some talks (see the bottom of math.ucsd.edu/~wgan), but are clearer than most other sources I have seen! $\endgroup$ – David Hansen Mar 29 '10 at 19:37
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    $\begingroup$ Right: if G=GL_2(Q_p) and K is GL_2(Z_p) then H(G//K) is C[T,S,S^{-1}] and this is Satake. Unravelling the explicit isomorphism isn't too hard (see e.g. Cartier's notes in Corvallis?), and you can check that T is the char fn of K(p 0;0 1)K and S the char fn of K(p 0;0 p)K. As I say, the lemma now is to write down the explicit dictionary which starts with a modular form and produces a function on GL_2(adeles) and check that the usual T_p on modular forms induces the operator [K(p 0;0 1)K] on functions on GL_2(adeles). But this isn't Satake so I am a bit confused about what you are asking. $\endgroup$ – Kevin Buzzard Mar 29 '10 at 20:13
  • $\begingroup$ Your explanation of the relationship between Hecke Algebra and Hecke Operator is great. In fact it shows the relationship is much more basic than the Satake Isomorphism. My understanding of Statake Isomorphism was H(G//K) is the symmetric polynomials in 2 variables, rather than C[T,S,S^{-1}], as Weyl group is S_2. Probably I misunderstood,I will go back to Cartier article and unravell the definitions and isomorphisms .Thanks for this valuable comment and the answer. $\endgroup$ – Dipramit Majumdar Mar 29 '10 at 20:44
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    $\begingroup$ H(G//K)=H(T//(T intersect K))^{S_2} (T the torus) but the map is quite explicit. H(T//(T intersect K)) is C[X,Y,X^{-1},Y^{-1}] and S_2 acts by switching X and Y. One sets T_p=X+Y and S_p=XY. The exercise is to check that the characteristic function of K(p 0;0 1)K becomes identified with X+Y (possibly up to a power of p^{1/2}) via the map. The map is quite explicit: it's "restrict to B, average over N, and renormalise by 1/2 the sum of the positive roots". You can easily work it out in this case (in the sense that I could once so it can't be hard!) $\endgroup$ – Kevin Buzzard Mar 30 '10 at 6:51

Regarding your second question, the relationship between Hecke operators and algebras was discussed in Baez's This Week's Finds. For instance, take a look at David Ben-Zvi's comment on Week 254.


Sorry, the first edition of this answer was shamefully incoherent. We'll see if this attempt is any better.

Any double coset KgK (for G and K as given) has a unique representative in elementary divisor form $\binom{a0}{0d}$ where a and d are (possibly negative) powers of p and a/d is a p-adic integer (i.e., a positive power of p). The Hecke operator $T(p^n)$ is given by a sum over convolutions with KgK as g ranges over elementary divisor matrices with p-adic integer entries with determinant $p^n$. In particular, T(p) is given by convolving with the double coset corresponding to $\binom{p0}{01}$. In the notation of Buzzard's answer, the operators $T(p^n)$ generate the subalgebra of the Hecke algebra generated by $S = \binom{p0}{0p}$ and $T = \binom{p0}{01}$, and it coincides with the subalgebra generated by those double cosets whose elementary divisor representative has p-adic integer entries.

You can find a non-adelic treatment in terms of Hecke operators acting on modular forms on the upper half plane in section 1.4 of Bump's Automorphic Forms and Representations, where he introduces operators $T_\alpha$ for diagonal matrices $\alpha$ in elementary divisor form, and shows how $T(n)$ is given as a sum over double cosets with determinant n. Decomposing these double cosets into left cosets for $\Gamma(1)$ yields the usual set of representatives $\{ a,b,d| ad=n, 0 \leq b < d \}$ over which one sums when evaluating a Hecke operator.

The Satake isomorphism gives an isomorphism with the representation ring of $GL_2(\mathbb{C})$, which is commutative and finitely generated. This implies the Hecke algebra here is commutative and finitely generated, but this can be seen without invoking such machinery. In the case of $GL_2$, the isomorphism can be made very explicit. $T(p)$ corresponds to the standard 2 dimensional representation, the scalar matrices give powers of determinant, and $T(p^n)$ corresponds to the $n$th symmetric power.


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