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YCor
YCor
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Commensurator of $SL_2$\mathrm{SL}_2(\mathbb{Z})$ on $GL_2^+$\mathrm{GL}_2^+(\mathbb{Q})$

I$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$I am trying to find the commensurator of $SL_{2}(\mathbb{Z})$$\SL_{2}(\mathbb{Z})$ on $GL_{2}^+(\mathbb{R})$$\GL_{2}^+(\mathbb{R})$. So far I have been able to prove that $GL_{2}^+(\mathbb{Q})$$\GL_{2}^+(\mathbb{Q})$ is included in the commensurator by looking at the congruence subgroups, and my intuition tells me that this inclusion could in fact be an equality, but I am not able to prove it. Is there a way to prove the other inclusion, or to find (in case it is not $GL_2^+(\mathbb{Q})$$\GL_2^+(\mathbb{Q})$) the commensurator in another way?

By $GL_2^+$$\GL_2^+$ I mean invertible and positive determinant matrices.

Commensurator of $SL_2(\mathbb{Z})$ on $GL_2^+(\mathbb{Q})$

I am trying to find the commensurator of $SL_{2}(\mathbb{Z})$ on $GL_{2}^+(\mathbb{R})$. So far I have been able to prove that $GL_{2}^+(\mathbb{Q})$ is included in the commensurator by looking at the congruence subgroups, and my intuition tells me that this inclusion could in fact be an equality, but I am not able to prove it. Is there a way to prove the other inclusion, or to find (in case it is not $GL_2^+(\mathbb{Q})$) the commensurator in another way?

By $GL_2^+$ I mean invertible and positive determinant matrices.

Commensurator of $\mathrm{SL}_2(\mathbb{Z})$ on $\mathrm{GL}_2^+(\mathbb{Q})$

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\GL{GL}$I am trying to find the commensurator of $\SL_{2}(\mathbb{Z})$ on $\GL_{2}^+(\mathbb{R})$. So far I have been able to prove that $\GL_{2}^+(\mathbb{Q})$ is included in the commensurator by looking at the congruence subgroups, and my intuition tells me that this inclusion could in fact be an equality, but I am not able to prove it. Is there a way to prove the other inclusion, or to find (in case it is not $\GL_2^+(\mathbb{Q})$) the commensurator in another way?

By $\GL_2^+$ I mean invertible and positive determinant matrices.

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Siegmeyer of Catarina
Siegmeyer of Catarina
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Commensurator of $SL_2(\mathbb{Z})$ on $GL_2^+(\mathbb{Q})$

I am trying to find the commensurator of $SL_{2}(\mathbb{Z})$ on $GL_{2}^+(\mathbb{R})$. So far I have been able to prove that $GL_{2}^+(\mathbb{Q})$ is included in the commensurator by looking at the congruence subgroups, and my intuition tells me that this inclusion could in fact be an equality, but I am not able to prove it. Is there a way to prove the other inclusion, or to find (in case it is not $GL_2^+(\mathbb{Q})$) the commensurator in another way?

By $GL_2^+$ I mean invertible and positive determinant matrices.