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It is said that $GL_n/U$ is birationally isomorphic to $B^-$. Here $U$ acts by right multiplication on $GL_n$. I think that $GL_n/U$ consisting of cosets. Two matrices in the same coset if any two minors of these two matrices which have the same row indices and column indices and contains the first column are the same. But I don't know how to prove that $GL_n/U$ is birationally isomorphic to $B^-$. Thank you very much.

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For $GL_n$ you have the Gauss decomposition $G' = L \cdot D \cdot U$, where $L$ are the lower triangular matrices with $1$ on the diagonal, $D$ the diagonal matrices and $U$ the upper triangular matrices with $1$ on the diagonal. $G'$ is the open subset of $GL_n$ consisting of all matrices with nonzero leading principal minors $M_1$, $\ldots $ , $M_n \ne 0$. You can explicitely calculate the decomposition $$g = l \cdot d \cdot u$$ using quotients of minors of $g$. The components $l$ and $u$ will be expressed in terms of minors invariant under right $U$ multiplication. For instance, the diagonal elements are

$$d_p = M_p/M_{p-1}$$

and the off diagonal elements are again quotients of minors

\begin{equation} n_{qp}= \frac {M_{(1,\ldots, p-1, q) ( 1,\ldots,p-1, p)}(g) } { M_p(g)} \end{equation}

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They key to this observation is the Bruhat decomposition $G=\bigcup_{w\in W} BwB$, where $W$ is the Weyl Group of $G=\operatorname{GL}_n$ with respect to a maximal torus $T$ of $B$. For $w_0\in W$ the longest Weyl element, $Bw_0B$ is an affine open subset of $G$ which is isomorphic to $B^-\times U$. A good reference for this is the book Linear algebraic groups by Jim Humphreys.

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