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I am following Wiles' paper On class groups of imaginary quadratic fields, and it's my first time learning about Shimura curves. There is part of the setup that I don't understand, concerning the definition of the Hecke correspondences on Shimura curves over $\mathbf{Q}$.

Let $B$ be an indefinite quaternion algebra over $\mathbf{Q}$, let $G/\mathbf{Q}$ be the corresponding algebraic group, and let $K \subset G(\mathbf{A}_f)$ be an open compact subgroup (the level). Then Wiles defines the Shimura curve corresponding to this data, $$M_K = G(\mathbf{Q}) \backslash G(\mathbf{A}_f) \times \mathbf{H}^{\pm} / K,$$ where $\mathbf{H}^{\pm} = \mathbf{C} - \mathbf{R}$. In fact, Wiles further restricts to the case where $K$ is given by the product of local components $K_p$ where $K_p$ is the unit group of a maximal order in $B \otimes \mathbf{Q}_p$ for finite primes $p$ where $B$ is ramified, and otherwise (when $B$ is split at $p$) is so that $K_p \subset B \otimes \mathbf{Q}_p = M_2(\mathbf{Q}_p)$ has determinants surjective onto $\mathbf{Z}_p^\times$. So far so good.

I am a little confused about the definition of the Hecke correspondences on $M_K$ which are given next. It is as follows:

Let $\mathcal{O}_B$ be a maximal order in $B$, and $\widehat{\mathcal{O}}_B = \mathcal{O}_B \otimes \widehat{\mathbf{Z}}$. Let $m$ be a positive integer such that for every $p | m$, $K_p$ is a maximal order. We allow the case where $B$ is ramified at $p$. Let $G_m$ be the set of elements $g \in \widehat{O}_B$ which have component $1$ at primes not dividing $m$ and such that $\mathrm{det}(g)$ generates the ideal $m\mathbf{Z}_p$ at each prime $p | m$. In particular, $G_1$ is a subgroup of $K$, and $G_m$ is a union of cosets of $G_1$ in $\widehat{\mathcal{O}}_B$. We define an correspondence $T_m$ on $M_K$ by the formula $$T_m(y) = \sum_{\gamma \in G_m \setminus G_1}[(g \gamma, x)]$$ where $y \in M_K$ is represented in $G(\mathbf{A}_f) \times \mathbf{H}^{\pm}$ by $(g, x)$.

I am confused for the following reasons. First, isn't $G_1$ just $1$, since its elements have component $1$ at all primes? Also, when $m = 1$, regardless of whether I am right or wrong about what $G_1$ is, the formal sum here is empty. Shouldn't I expect $T_1$ to be the identity instead?

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  • $\begingroup$ They meant unit component instead of component $1$ ? And $G_m \setminus G_1$ is the quotient of a set by the group action of $G_1$, not a setminus $\endgroup$
    – reuns
    Dec 17, 2020 at 7:51
  • $\begingroup$ @reuns: thanks ! So the \ is a typo for a forward slash, then? $\endgroup$
    – babu_babu
    Dec 17, 2020 at 16:46

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