It is a pity you don't read French. For those who do (or after you've learnt to read it), I would recommend Annexe B in the undergraduate text Éléments d’analyse et d’algèbre (et de théorie des nombres) by Pierre Colmez, and also his popular article Un autre monde est possible.

Addendum. The basic idea for computing the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is that this group can be identified with $$ (\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})) \times \mathrm{SL}_2(\mathbf{Z}_2) \times \mathrm{SL}_2(\mathbf{Z}_3) \times \mathrm{SL}_2(\mathbf{Z}_5) \times\cdots $$ so the volume in question is the product of the volumes of the various factors. The computation of a difficult integral shows that the volume of $\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})$ is $\zeta(2)$, and an easy computation shows that the volume of $\mathrm{SL}_2(\mathbf{Z}_p)$ is $1-{1\over p^2}$, for every prime $p$. Finally, the eulerian product $\zeta(2)=\prod_p(1-{1\over p^2})^{-1}$ allows you to conclude that the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is $1$.

It is a pity you don't read French. For those who do (or after you've learnt to read it), I would recommend Annexe B in the undergraduate text Éléments d’analyse et d’algèbre (et de théorie des nombres) by Pierre Colmez, and also his popular article Un autre monde est possible.

Addendum. The basic idea for computing the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is that this space can be identified with $$ (\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})) \times \mathrm{SL}_2(\mathbf{Z}_2) \times \mathrm{SL}_2(\mathbf{Z}_3) \times \mathrm{SL}_2(\mathbf{Z}_5) \times\cdots $$ so the volume in question is the product of the volumes of the various factors. The computation of a difficult integral shows that the volume of $\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})$ is $\zeta(2)$, and an easy computation shows that the volume of $\mathrm{SL}_2(\mathbf{Z}_p)$ is $1-{1\over p^2}$, for every prime $p$. Finally, the eulerian product $\zeta(2)=\prod_p(1-{1\over p^2})^{-1}$ allows you to conclude that the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is $1$.

It is a pity you don't read French. For those who do (or after you've learnt to read it), I would recommend Annexe B in the undergraduate text Éléments d’analyse et d’algèbre (et de théorie des nombres) by Pierre Colmez, and also his popular article Un autre monde est possible.

It is a pity you don't read French. For those who do (or after you've learnt to read it), I would recommend Annexe B in the undergraduate text Éléments d’analyse et d’algèbre (et de théorie des nombres) by Pierre Colmez, and also his popular article Un autre monde est possible.

Addendum. The basic idea for computing the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is that this group can be identified with $$ (\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})) \times \mathrm{SL}_2(\mathbf{Z}_2) \times \mathrm{SL}_2(\mathbf{Z}_3) \times \mathrm{SL}_2(\mathbf{Z}_5) \times\cdots $$ so the volume in question is the product of the volumes of the various factors. The computation of a difficult integral shows that the volume of $\mathrm{SL}_2(\mathbf{R})/\mathrm{SL}_2(\mathbf{Z})$ is $\zeta(2)$, and an easy computation shows that the volume of $\mathrm{SL}_2(\mathbf{Z}_p)$ is $1-{1\over p^2}$, for every prime $p$. Finally, the eulerian product $\zeta(2)=\prod_p(1-{1\over p^2})^{-1}$ allows you to conclude that the volume of $\mathrm{SL}_2(\mathbf{A})/\mathrm{SL}_2(\mathbf{Q})$ is $1$.