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 is a group acting on the left on a set , the set of orbits of under this action is
denoted by .
If acts on on the right, we denote the set of orbits by .
When acts on on the left and acts on on the right, and if the actions are
compatible in the sense that
for all , and , then we again have a nicely defined dual-action
and we have orbits under this dual action.
Two elements and in are in the same orbit if and only if for some
and .
We denote these orbits by .
One easy source of compatible actions are when and are subgroups of a group .
Then is the set of double cosets .
Below, we let .
Take .
The double coset is a finite union of right cosets
The number of right cosets is equal to .
For this proof, we follow Bump closely and add only a couple of additional details.
Multiplying on the right by is a bijection of onto itself.
This gives an obvious bijection between the sets of orbits
Any element of can be written as
for some .
Observe that two orbits orbits and are the same when
Thus we have directly seen that
Conjugating by gives a bijection between this set of cosets and the set of cosets
This shows that the cardinality is .
It only remains to show that this is finite.
This follows from Lemma~1.4.1 of Bump, which we give below.
Let be a congruence subgroup of .
Let .
Then is a congruence subgroup.
(This is an easy proof).
Now let be any congruence subgroup.
Take .
The double coset is a finite union of right cosets
The number of right cosets is equal to .
Structurally, the proof is identical until we want to show that .
This follows from a general result: if and , then .
We prove this in Lemma 4 below.
We apply this repeatedly to a lattice of subgroups formed from and .
Note that and because is a congruence subgroup.
And as shown above, has finite index in both
and .
Applying Lemma 4 to and in shows that has finite index in
.
Applying Lemma 4 to and
inside shows that has finite index in .
Finally, applying the lemma once more to
and (which we've now shown have finite index in either
or ) shows that
has finite index in and .
(And thus also in ).
This proof is annoying.
It's slightly clearer if one draws the lattice of subgroups.
Alternately, here is the idea: we start with and .
We twice pass to (and use finite index because it's a congruence subgroup) to get results
on smaller groups.
We apply the previous result once to understand .
We then repeatedly (three times — once for the first group, once for the second group, and then
once together) intersect things with and things with to pass
to the smaller subgroups.
The group acts on the cross-product of orbits by multiplication in
each coordinate: .
The stabilizer of the orbit of , i.e. of , consists of those such
that and .
Of course, , similarly for .
Thus the stabilizer is .
By the orbit-stabilizer theorem, , the
size of the orbit of under multiplication by .
As both and have finite index in , the size of this orbit is at most the product of
the indices.
Thus .
Hecke Algebras
We define the Hecke algebra associated to a congruence subgroup .
As an abelian group, is the free abelian group on symbols as runs through a complete set of representatives for
.
Defining the multiplication on is annoying.
It's nonobvious how to define multiplication on 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 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 , a (right) -module is an abelian group with a -linear action.
I emphasize that we will use right multiplication here, which (in my experience) is much less
common.
For a -module and a subgroup , write to be the fixed points of
under the action of ,
Suppose that for any , we have .
The argument from Proposition 1 and its corollary shows that
We now define the action of the element (thought of as an element of
, the abelian free group on the symbols where
ranges over elements of ) on an element as
where the are as in .
We prove two properties of this action.
Let and as above.
- depends only on the double coset .
- .
To make sure this makes sense: the item 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 stabilized by .
Suppose for .
Any such choice is an equally valid way of writing as a union of right cosets.
Then
where we have used that .
This proves the first claim.
For the second claim, the point is that is a complete set of coset
representatives for .
Multiplying on the right by doesn't change , and hence
is a permutation of the coset representatives.
This means that
proving the second claim.
Now let be the free abelian group on cosets .
The group acts on the right on by right multiplication.
Consider the map from , the free abelian group on double cosets with ranging across representatives , to given by
where .
The double coset is a (right) orbit of the single coset
under , which are exactly those elements of invariant under right
multiplication by .
Thus is an isomorphism.
The free abelian group on double cosets with
ranging across representatives is isomorphic to the submodule of
fixed under right multiplication by . (The isomorphism is as
groups).
There is a product on making it into an associative ring, such that for
any (right) -module we have is a right module.
Take initially.
Write and .
We claim that
To see this, note that .
Now compute
by the definition of the right action of .
By Proposition 5, this is in and is well-defined.
The point that we've been working towards this whole time is that we can define
where is the isomorphism from .
This is clearly associative as associativity in is associative, and the isomorphism preserves
this.
The multiplicative unit is .
And more generally, if is any right -module, then for we have
As used here, the actual multiplication is defined through .
This is annoying.
We now give an explicit formula.
To do so, we suppose we have a fixes set of representatives ranging over
.
Write and .
Then
where is the number of pairs such that .
We define ,
where takes to for all in an index set (similarly for and
for ) and where the sum is over all .
This is a matter of identifying the preimage correctly.
Any element in has the form for
some integers .
How do we determine what is?
For each in the preimage (i.e. where ), it must be that , for some indices in some index set .
It's not necessary for — it's only necessary that the union is stable under
right multiplication by , which could be in a smaller index set.
Thinking of as a right orbit of ,
it suffices to count how often for any pair .
For each pair with , the preimage of must include at least .
As ranges across representatives of (where we've added
parentheses for emphasis), no other elements require any further .
Thus we find that is the number of pairs such that .
Stated differently: for any with for some pair,
the full orbit will be given by a union across
some index set.
Instead of counting each element of the index set, we only count the "trivial" elements of the
orbits under , given by .
The other elements are handled by the averaging of , and the fact that
is isomorphic to means that we don't have to worry about the
preimage not existing.
Write
,
, and
.
Then
where is the number of triples such that .
As above, we define .
The idea is the same: it's sufficient to account for the trivial elements of orbits, .
And these give one for each with , 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.
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 denote the Hecke algebra .
Let denote the space of holomorphic modular forms of level and weight
.
Then is a (right) module under the slash operator
Verifying that is a straightforward direct
calculation.
By Theorem 7, (i.e. the space of modular forms in fixed by the
action of ) is a right module.
The definition of modularity gives immediately that , so this shows that
is a right module.
For , we denote the element by .
For and , the right action is given by
where
as a disjoint union (cf. Proposition 1).
Various properties claimed or shown in Bump follow from our discussion:
-
Proposition 5 shows that depends only on the
equivalence class of (and not on the choice of representatives ), and further
that .
-
By Proposition 8, we have that
This is a proof of equation~4.4 in Bump (although in Bump this falls out as
a consequence of the definition).
-
The action of on is associative!
The associativity (and other properties) are baked into the formulation given in this note:
multiplication in comes from the right action of
on , 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 , 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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