mixedmath

Explorations in math and programming
David Lowry-Duda



Hecke Algebras

I want to give complete details about a set of descriptions in Bump's Automorphic Forms and Representations concerning Hecke operators and Hecke algebras. I like that Bump's description is so light on annoying computation. But the problem is that I think too much is swept under the rug.

In this note, I give an alternate treatment that fills in details I found otherwise confusing.

Below, I will refer to Automorphic Forms and Representations merely as Bump.

Also available as a note

I wrote this first as a note and converted it for presentation here. The note is available here.

Double Cosets

If H is a group acting on the left on a set X, the set of orbits of X under this action is denoted by HX. If H acts on X on the right, we denote the set of orbits by X/H. When H1 acts on X on the left and H2 acts on X on the right, and if the actions are compatible in the sense that (1)(h1x)h2=h1(xh2) for all h1H1,xX, and h2H2, then we again have a nicely defined dual-action and we have orbits under this dual action. Two elements x and y in X are in the same orbit if and only if x=h1yh2 for some h1 and h2. We denote these orbits by H1X/H2.

One easy source of compatible actions are when H1 and H2 are subgroups of a group G=X. Then H1G/H2 is the set of double cosets H1gH2.

Below, we let Γ(1)=SL(2,Z).

Take αGL(2,Q)+. The double coset Γ(1)αΓ(1) is a finite union of right cosets (2)Γ(1)αΓ(1)=iΓ(1)αi,αiGL(2,Q)+. The number of right cosets is equal to [Γ(1):α1Γ(1)αΓ(1)]<.

For this proof, we follow Bump closely and add only a couple of additional details.

Multiplying on the right by α1 is a bijection of GL(2,Q)+ onto itself. This gives an obvious bijection between the sets of orbits (3)Γ(1)Γ(1)αΓ(1)Γ(1)Γ(1)αΓ(1)α1. Any element U of Γ(1)Γ(1)αΓ(1)α1 can be written as U=Γ(1)γαuα1=Γ(1)αuα1αΓ(1)α1 for some γ,uΓ(1). Observe that two orbits orbits U=Γ(1)αuα1 and V=Γ(1)αvα1 are the same when (4)Γ(1)αuα1=Γ(1)αvα1Γ(1)αu=Γ(1)αv(5)Γ(1)αuv1α1=Γ(1)(6)αuv1α1Γ(1)αΓ(1)α1. Thus we have directly seen that (7)Γ(1)Γ(1)αΓ(1)α1(Γ(1)αΓ(1)α1)αΓ(1)α1. Conjugating by α gives a bijection between this set of cosets and the set of cosets (8)(α1Γ(1)αΓ(1))Γ(1).

This shows that the cardinality is [Γ(1):α1Γ(1)αΓ(1)]. It only remains to show that this is finite. This follows from Lemma~1.4.1 of Bump, which we give below.

Let Γ be a congruence subgroup of SL(2,Z). Let αGL(2,Q)+. Then α1ΓαΓ(1) is a congruence subgroup.

(This is an easy proof).

Now let ΓSL(2,Z) be any congruence subgroup.

Take αGL(2,Q)+. The double coset ΓαΓ is a finite union of right cosets (9)ΓαΓ=Γαi,αiGL(2,Q)+. The number of right cosets is equal to [Γ:α1ΓαΓ].

Structurally, the proof is identical until we want to show that [Γ:α1ΓαΓ]<. This follows from a general result: if [G:H1]< and [G:H2]<, then [G:H1H2]<. We prove this in Lemma 4 below.

We apply this repeatedly to a lattice of subgroups formed from Γ(1) and α1Γ(1)α. Note that [Γ(1):Γ]< and [α1Γ(1)α:α1Γα]< because Γ is a congruence subgroup. And as shown above, Γ(1)α1Γ(1)α has finite index in both Γ(1) and α1Γ(1)α.

Applying Lemma 4 to Γ and Γ(1)α1Γ(1)α in Γ(1) shows that Γα1Γ(1)α has finite index in Γ(1). Applying Lemma 4 to Γ(1)α1Γ(1)α and α1Γα inside α1Γ(1)α shows that Γ(1)α1Γα has finite index in α1Γ(1)α.

Finally, applying the lemma once more to Γ(1)α1Γα and Γα1Γ(1)α (which we've now shown have finite index in either Γ(1) or α1Γ(1)α) shows that Γα1Γα has finite index in Γ(1) and α1Γ(1)α. (And thus also in Γ).

This proof is annoying. It's slightly clearer if one draws the lattice of subgroups. Alternately, here is the idea: we start with Γ(1) and α1Γ(1)α. We twice pass to Γ (and use finite index because it's a congruence subgroup) to get results on smaller groups. We apply the previous result once to understand Γ(1)α1Γ(1)α. We then repeatedly (three times — once for the first group, once for the second group, and then once together) intersect things with Γ and things with Γ(1) to pass to the smaller subgroups.

If [G:H1]< and [G:H2]<, then [G:H1H2]<.

The group G acts on the cross-product of orbits G/H1×G/H2 by multiplication in each coordinate: g(aH1,bH2)=(gaH1,gbH2). The stabilizer of the orbit of (1,1), i.e. of (H1,H2), consists of those gG such that gH1=H1 and gH2=H2. Of course, gH1=H1gH1, similarly for H2. Thus the stabilizer is H1H2.

By the orbit-stabilizer theorem, [G:H1H2]=|O(H1,H2)|, the size of the orbit of (H1,H2) under multiplication by G. As both H1 and H2 have finite index in G, the size of this orbit is at most the product of the indices. Thus [G:H1H2][G:H1][G:H2].

Hecke Algebras

We define the Hecke algebra R associated to a congruence subgroup Γ. As an abelian group, R is the free abelian group on symbols Tα=[ΓαΓ] as α runs through a complete set of representatives for ΓGL(2,Q)+/Γ.

Defining the multiplication on R is annoying. It's nonobvious how to define multiplication on R directly — a lot of the complexity of various introductions to Hecke algebras comes what path is taken to define multiplication. The approach we take here is to define an action of R on a particular set, and the necessary structure for this to be a well-defined action will give our multiplicative structure.

Defining an action

Given any group G, a (right) G-module M is an abelian group with a Z-linear G action. I emphasize that we will use right multiplication here, which (in my experience) is much less common.

For a G-module M and a subgroup ΓG, write MΓ to be the fixed points of M under the action of G, (10)MΓ:={mM:mγ=mγΓ}. Suppose that for any gG, we have [Γ:Γg1Γg]<. The argument from Proposition 1 and its corollary shows that (11)ΓgΓ=Γgi,giG. We now define the action of the element [ΓgΓ] (thought of as an element of R(G,Γ), the abelian free group on the symbols [ΓαΓ] where α ranges over elements of ΓG/Γ) on an element mMΓ as (12)m[ΓgΓ]:=imgi, where the gi are as in (11).

We prove two properties of this action. Let mMΓ and gG as above.

  1. m[ΓgΓ] depends only on the double coset ΓgΓ.
  2. m[ΓgΓ]MΓ.

To make sure this makes sense: the item [ΓgΓ] is a formal symbol, a basis element in an abelian free group on symbols. The content of this proposition is that using any symbol from the same double coset is equivalent, and the action yields an element of M stabilized by Γ.

Suppose ΓgΓ=Γgi for gi=γigi. Any such choice is an equally valid way of writing ΓgΓ as a union of right cosets. Then (13)imgi=i(mγi)gi=imgi, where we have used that mMΓ. This proves the first claim.

For the second claim, the point is that Γgi is a complete set of coset representatives for ΓgΓ. Multiplying on the right by hG doesn't change ΓgΓ, and hence {Γgih} is a permutation {Γgj} of the coset representatives. This means that (14)(m[ΓgΓ])h=(imgi)h=imgih=jmgj=m[ΓgΓ], proving the second claim.

Now let Z[ΓG] be the free abelian group on cosets [Γg]. The group G acts on the right on Z[ΓG] by right multiplication. Consider the map from R(G,Γ), the free abelian group on double cosets [ΓgΓ] with g ranging across representatives ΓG/Γ, to Z[ΓG]Γ given by (15)F:R(G,Γ)Z[ΓG]Γ[ΓgΓ][Γgi], where ΓgΓ=Γgi. The double coset [ΓgΓ] is a (right) orbit of the single coset ΓG under Γ, which are exactly those elements of Z[ΓG] invariant under right multiplication by Γ. Thus F is an isomorphism.

The free abelian group R(G,Γ) on double cosets [ΓgΓ] with g ranging across representatives ΓG/Γ is isomorphic to the submodule of Z[ΓG] fixed under right multiplication by Γ. (The isomorphism is as groups).

There is a product on R(G,Γ) making it into an associative ring, such that for any (right) G-module M we have MΓ is a right R(G,Γ) module.

Take M=Z[ΓG] initially. Write ΓgΓ=Γgi and ΓhΓ=Γhj.

We claim that (16)i,j[Γgihj]MΓ. To see this, note that F([ΓgΓ])=i[Γgi]MΓ. Now compute (17)i[Γgi][ΓhΓ]=i,j[Γgihj] by the definition of the right action of R(G,Γ). By Proposition 5, this is in MΓ and is well-defined.

The point that we've been working towards this whole time is that we can define (18)[ΓgΓ][ΓhΓ]=F1(i,j[Γgihj]), where F is the isomorphism from (15). This is clearly associative as associativity in G is associative, and the isomorphism preserves this. The multiplicative unit is [Γ1Γ].

And more generally, if M is any right G-module, then for mMΓ we have (19)m[ΓgΓ][ΓhΓ]=mgi[ΓhΓ]=mgihj=m([ΓgΓ][ΓhΓ]).

As used here, the actual multiplication is defined through F1. This is annoying.

We now give an explicit formula. To do so, we suppose we have a fixes set of representatives σ ranging over ΓG/Γ.

Write ΓgΓ=Γgi and ΓhΓ=Γhj. Then (20)[ΓgΓ][ΓhΓ]=σΓG/Γm(h,g;σ)[ΓσΓ], where m(h,g;σ) is the number of pairs (i,j) such that Γgihj=Γσ.

We define [ΓgΓ][ΓhΓ]=F1(i,j[Γgihj]), where F takes [ΓgΓ] to [Γgi] for all i in an index set I (similarly for [ΓhΓ] and [Γhj] for jJ) and where the sum is over all (i,j)I×J.

This is a matter of identifying the preimage correctly. Any element in R(G,Γ) has the form mσ[ΓσΓ] for some integers mσ. How do we determine what mσ is?

For each σ in the preimage (i.e. where mσ>0), it must be that ΓσΓ=i,jAΓgihj, for some indices i,j in some index set A. It's not necessary for A=I×J — it's only necessary that the union is stable under right multiplication by Γ, which could be in a smaller index set.

Thinking of ΓσΓ=(Γσ)Γ as a right orbit of Γσ, it suffices to count how often Γσ=Γgihj for any pair (i,j). For each pair (i,j) with Γσ=Γgihj, the preimage of i,j[Γgihj] must include at least 1 [ΓσΓ]. As σ ranges across representatives of (ΓG)/Γ (where we've added parentheses for emphasis), no other elements require any further [ΓσΓ]. Thus we find that mσ=m(h,g;σ) is the number of pairs (i,j) such that Γgihj=Γσ.

Stated differently: for any σ with Γσ=Γgihj for some (i,j) pair, the full orbit ΓσΓ will be given by a union Γgihj across some index set. Instead of counting each element of the index set, we only count the "trivial" elements of the orbits under Γ, given by Γσ1. The other elements are handled by the averaging of F, and the fact that R(G,Γ) is isomorphic to Z[ΓG]Γ means that we don't have to worry about the preimage not existing.

Write ΓgΓ=Γgi, ΓhΓ=Γhj, and ΓfΓ=Γfk. Then (21)[ΓgΓ][ΓhΓ][ΓfΓ]=σΓG/Γm(h,g,f;σ)[ΓσΓ] where m(h,g,f;σ) is the number of triples (i,j,k) such that Γgihjfk=Γσ.

As above, we define [ΓgΓ][ΓhΓ][ΓfΓ]=F1(i,j,k[Γgihjfk]). The idea is the same: it's sufficient to account for the trivial elements of orbits, Γσ1. And these give one [ΓσΓ] for each σ with Γσ=Γgihjfk, exactly as before.

The statement of Corollary 9 is stated without proof in Bump as a way to note that multiplication in a Hecke algebra is associative.

Bump defines only the Hecke algebra R(GL(2,Q)+,SL(2,Z)), while the presentation here is more general. I don't know much about Hecke algebras apart from their use in automorphic forms and representations.

Acting on Modular Forms

To end, we bring this discussion back to the context of modular forms, as in the first chapter of Bump.

Let R denote the Hecke algebra R(GL(2,Q)+,SL(2,Z)). Let Mk=Mk(SL(2,Z)) denote the space of holomorphic modular forms of level 1 and weight k. Then Mk is a (right) GL(2,Q)+ module under the slash operator (22)(fγ)(z):=(detγ)k/2(cz+d)kf(γz),γGL(2,Q)+. Verifying that (fγ)γ=f(γγ) is a straightforward direct calculation.

By Theorem 7, MkSL(2,Z) (i.e. the space of modular forms in Mk fixed by the action of SL(2,Z)) is a right R module. The definition of modularity gives immediately that MkSL(2,Z)=Mk, so this shows that Mk is a right R module. For αGL(2,Q)+, we denote the element ΓαΓ=SL(2,Z)αSL(2,Z)R by Tα. For fMk and αGL(2,Q)+, the right action is given by (23)fTα:=f[SL(2,Z)αSL(2,Z)]=ifαi, where (24)SL(2,Z)αSL(2,Z)=iαi as a disjoint union (cf. Proposition 1). Various properties claimed or shown in Bump follow from our discussion:

  1. Proposition 5 shows that fTα depends only on the equivalence class of α (and not on the choice of representatives αi), and further that fTαMk.

  2. By Proposition 8, we have that (25)(fTα)Tβ=σSL(2,Z)GL(2,Q)+/SL(2,Z)m(α,β;σ)(fTσ). This is a proof of equation~4.4 in Bump (although in Bump this falls out as a consequence of the definition).

  3. The action of R on Mk is associative! The associativity (and other properties) are baked into the formulation given in this note: multiplication in R(G,Γ) comes from the right action of R(G,Γ) on Z[ΓG]Γ, and this right action is trivially associative because group multiplication is associative. As noted above, Corollary 9 proves an unproven assertion in Bump.

I note that neither the presentation in Bump or the presentation here makes it obvious that the Hecke algebra R, or those associated to congruence subgroups, is commutative. This structural result is remarkable, but I don't discuss that further here.


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