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?


1 Answer 1


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 $r \in R$.

  • $\begingroup$ Thanks for the answer. I assume in the final line you meant to leave the factor of $r$ out of the last sum :) $\endgroup$ Commented Jan 23, 2020 at 11:04
  • 1
    $\begingroup$ Yes, you are right. Thank you, I corrected it. $\endgroup$ Commented Jan 23, 2020 at 11:16

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