# lower and upper block matrices in $SL(m+n,\mathbb R)/SL(m+n,\mathbb Z)$

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.

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