# What is the explicit map of the open embedding $B^- \to G/U$?

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.

• What's $U$ denote? – Randall Jun 19 '19 at 12:48

## 1 Answer

Assuming $$U$$ is the set of upper-triangular unipotent matrices. Think about $$B^-\to G\to G/U$$.

• @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? – LJR Jun 19 '19 at 13:00
• You get embedding, and for openness you need to use this $U$ to count dimensions (or recall $LU$-factorisation of matrices). – user10354138 Jun 19 '19 at 13:04
• 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? – LJR Jun 19 '19 at 13:53
• 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 $G$. – user10354138 Jun 19 '19 at 14:29
• thank you very much. – LJR Jun 19 '19 at 14:32