# Submodule of $R$-Invariants for a Hecke Pair $(R,S)$

I'll just define a Hecke pair here for completeness:

Definition: Let $$S$$ be a monoid and $$R$$ a group contained in $$S$$. We call the pair $$(R,S)$$ a Hecke pair if every $$R$$-double coset of $$S$$ is a disjoint union of finitely many $$R$$-left cosets of $$S$$.

As an example, we have $$(\operatorname{SL}_2(\Bbb Z), \operatorname{GL}_2(\Bbb Q)^+)$$ is a Hecke pair (where $$\operatorname{GL}_2(\Bbb Q)^+$$ are those rational matrices with positive determinant).

Denote by $$\mathcal{L}(R,S)$$ the $$\Bbb Z$$-module of finite formal $$\Bbb Z$$-linear cominations of $$R$$-left cosets of $$S$$ and by $$\mathcal{H}(R,S)$$ the $$\Bbb Z$$-module of finite formal $$\Bbb Z$$-linear combinations of $$R$$-double cosets of $$S$$.

By definition there exist $$s_1, \dots, s_\ell \in S$$ so that $$RsR = \bigsqcup_{j=1}^\ell Rs_j$$ so we set $$\iota(RsR) := \sum_{j=1}^\ell Rs_j \in \mathcal{L}(R,S)$$ and extend this linearly to get a $$\Bbb Z$$-linear map

$$\iota : \mathcal{H}(R,S) \to \mathcal{L}(R,S).$$

Note that $$S$$ acts by right multiplication on the set of $$R$$-left cosets and by linear extension on $$\mathcal{L}(R,S)$$. Let's consider the submodule of $$R$$-invariants $$\mathcal{L}(R,S)^R$$ under this action. Then:

The map $$\tilde{\iota} : \mathcal{H}(R,S) \to \mathcal{L}(R,S)^R$$ is in fact an isomorphism of $$\Bbb Z$$-modules.

My confusion comes from the explanation for such an isomorphism: an element $$\sum_{s \in R\setminus S} a_sRs \in \mathcal{L}(R,S)$$ is $$R$$-right invariant if and only if for all $$s, t$$ with $$RsR = RtR$$ the coefficients $$a_s$$ and $$a_t$$ are equal. Why is this true?

Note that $$\mathcal{L}(R,S)$$ is a free $$\mathbb{Z}$$-module with basis $$R\backslash S$$. In identities regarding elements of $$\mathcal{L}(R,S)$$, we may thus compare coefficients.
Suppose first that $$x = \sum_{Rs \in R\backslash S} a_{Rs} Rs \in \mathcal{L}(R,S)$$ is $$R$$-invariant and $$RsR = RtR$$. Then there exists $$r \in R$$ such that $$Rsr = Rt$$. The coeeficient of $$x$$ at $$Rt$$ is $$a_{Rt}$$ and the coefficient of $$xr$$ at $$t$$ is $$a_{Rs}$$. By $$R$$-invariance, they have to coincide.
Coversely, suppose that $$x = \sum_{Rs \in R\backslash S} a_{Rs} Rs$$ is such that $$a_{Rs} = a_{Rt}$$ whenever $$RsR = RtR$$. Then you find $$xr = \sum_{Rs \in R\backslash S} a_{Rs} Rsr = \sum_{Rs \in R\backslash S} a_{Rsr^{-1}} Rs = \sum_{Rs \in R\backslash S} a_{Rs} Rs = x$$ for all $$r \in R$$.
• Thanks for the answer. I assume in the final line you meant to leave the factor of $r$ out of the last sum :) Commented Jan 23, 2020 at 11:04