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Let $G=GL_n$ and $B^-$ the set of lower triangular matrices in $G$. It is said that there is an open embedding $B^- \to G/U$. What is the explicit map of $B^- \to G/U$. For example, in the case of $GL_2$. We have every element in $B^-$ is of the form $\left( \begin{matrix} a & 0 \\ c & d \end{matrix} \right)$. What is the images of elements in $B^- \to G/U$. Thank you very much.

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  • $\begingroup$ What's $U$ denote? $\endgroup$ – Randall Jun 19 '19 at 12:48
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Assuming $U$ is the set of upper-triangular unipotent matrices. Think about $B^-\to G\to G/U$.

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  • $\begingroup$ @Thank you very much. Yes, $U$ is upper triangular matrices. But according to your proof, it seems no matter what $U$ is, $B^- \to G/U$ is an open embedding which is not true. Where do you use the condition that $U$ is upper triangular? $\endgroup$ – LJR Jun 19 '19 at 13:00
  • $\begingroup$ You get embedding, and for openness you need to use this $U$ to count dimensions (or recall $LU$-factorisation of matrices). $\endgroup$ – user10354138 Jun 19 '19 at 13:04
  • $\begingroup$ thank you very much. Why LU-factorisation implies openness? For example, let $g=\left(\begin{array}{cc} a & b\\ c & d \end{array}\right)$. Then $g=b_- u = \left(\begin{array}{cc} a & 0\\ c & d - \frac{b\, c}{a} \end{array}\right) \left(\begin{array}{cc} 1 & \frac{b}{a}\\ 0 & 1 \end{array}\right)$. What is the map $B^- \to G/U$ in this case? $\endgroup$ – LJR Jun 19 '19 at 13:53
  • $\begingroup$ You think of the map $G/U\to B^-$ instead, which is the "L" part of LU factorization and we know $g\mapsto(\ell,u)$ is a diffeomorphism, so projecting (univesal property of quotient) gives the induced map $G/U\to B^-$ a diffeomorphism. The inverse map $B^-\to G/U$ maps a lower triangular matrix to its $U$-orbit in $G$. $\endgroup$ – user10354138 Jun 19 '19 at 14:29
  • $\begingroup$ thank you very much. $\endgroup$ – LJR Jun 19 '19 at 14:32

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