this paper would go through with very minor modifications, had we chosen a different definition of ${\mathcal{K}}({\mathfrak{c}})$.

### Measures

On $K_\infty$ we use the normalized Lebesgue measure $|D_K|^{-1/2}dx_1\cdots dx_d$. On $K_{\mathfrak{p}}$ we normalize the Haar measure so that ${\mathfrak{o}}_{\mathfrak{p}}$ has measure $1$. On $\mathbb{A}$ we use the Haar measure $dx$ which is the product of these measures, this induces the Haar probability measure on $K\backslash\mathbb{A}$. On $K_\infty^\times$ we use the Haar measure $(dy_1/|y_1|)\cdots(dy_d/|y_d|)$. On $K_{\mathfrak{p}}^\times$ we normalize the Haar measure so that ${\mathfrak{o}}_{\mathfrak{p}}^\times$ has measure $1$. On $\mathbb{A}^\times$ we use the Haar measure $d^{\times}y$ which is the product of these measures, this induces some Haar measure on $K^\times\backslash\mathbb{A}^\times$. On ${\mathcal{K}}$ and its factors we use the Haar probability measures. On $Z(K_\infty)\backslash\operatorname{GL}_2(K_\infty)$ we use the Haar measure which satisfies $$\int_{Z(K_\infty)\backslash\operatorname{GL}_2(K_\infty)}f(g)\,dg=
\int_{K_\infty^\times}\int_{K_\infty}\int_{\operatorname{SO}_2(K_\infty)}
f\left(\begin{pmatrix}y &x\\0
&1\end{pmatrix}k\right)\,dk\,dx\,\frac{d^\times y}{|y|}.$$ On $\operatorname{GL}_2(K_{\mathfrak{p}})$ we normalize the Haar measure so that ${\mathcal{K}}({\mathfrak{o}}_{\mathfrak{p}})$ has measure $1$. On $Z(K_\infty)\backslash\operatorname{GL}_2(\mathbb{A})$ we use the product of these measures, this induces the Haar measure on $Z(\mathbb{A})\backslash\operatorname{GL}_2(\mathbb{A})$ satisfying (cf.[@GJ (3.10)]) $$\int_{Z(\mathbb{A})\backslash\operatorname{GL}_2(\mathbb{A})}f(g)\,dg=
\int_{\mathbb{A}^\times}\int_\mathbb{A}\int_{\mathcal{K}}f\left(\begin{pmatrix}y &x\\0
&1\end{pmatrix}k\right)\,dk\,dx\,\frac{d^\times y}{|y|}.$$

Spectral decomposition and Eisenstein series {#spectralsection}
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Let $\omega : K^{\times} \backslash \mathbb{A}^{\times} \to S^1$ be a Hecke character, regarded also as a character of $Z(K)\backslash Z(\mathbb{A})$. Without loss 