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Let $\begin{pmatrix} P_1 & 0 \\ P_3 & P_4 \end{pmatrix}$ be a matrix in $SL(m+n,\mathbb R)$ with $P_1$ an $m\times m$ matrix, $P_3$ an $n\times m$ matrix and $P_4$ an $n\times n$ matrix (the block for $P_2$ is zero). In the quotient homegeneous space $SL(m+n,\mathbb R)/SL(m+n,\mathbb Z)$

I wonder if $\begin{pmatrix} P_1 & 0 \\ P_3 & P_4 \end{pmatrix} \begin{pmatrix} I_m & A \\ 0 & I_n \end{pmatrix}SL(m+n, \mathbb Z) = \begin{pmatrix} I_m & B \\ 0 & I_n \end{pmatrix} SL(m+n, \mathbb Z)$, where $A,B$ are real $m\times n$ matrices, would imply that $$\begin{pmatrix} I_m & A \\ 0 & I_n \end{pmatrix}SL(m+n, \mathbb Z) = \begin{pmatrix} I_m & B \\ 0 & I_n \end{pmatrix} SL(m+n, \mathbb Z)$$ as two left cosets? (or in other words $A-B$ is a matrix with integer entries)

I am inclined to believe this is correct but can't prove it formally.

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1 Answer 1

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No. $\begin{pmatrix} 2 & 0 \\ 3 & 1/2 \end{pmatrix} \begin{pmatrix} 1 & 1/2 \\ 0 & 1 \end{pmatrix}SL(2, \mathbb Z) = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} SL(2, \mathbb Z)$. But

$$\begin{pmatrix} 1 & 1/2 \\ 0 & 1 \end{pmatrix}SL(2, \mathbb Z) \ne \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} SL(2, \mathbb Z).$$

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