Lemma 7

— a) Let $U = (u_{ij}) \in \mathbf{GL}(n, K)$. In order that $U \in \Omega$, it is necessary and sufficient that $\det(u_{ij})_{k \leq i, j \leq n} \neq 0$ for $k = 2, 3, \ldots, n$. b) $\Omega$ is an open subset of $\mathbf{GL}(n, K)$. c) The mapping $\varphi$ is a homeomorphism of $T_1(n, K) \times {}^t(T(n, K)^*)$ onto $\Omega$.

In order that $U \in \Omega$, it is necessary and sufficient that there exist a $Z = (z_{ij}) \in T_1(n, K)$ such that $ZU \in {}^t(T(n, K))$ (then necessarily

$ZU \in {}^t(T(n, K)^*)$ since $U$ and $Z$ are invertible). By what we saw earlier, if $Z$ exists then it is unique. Thus, in order that $U \in \Omega$, it is necessary and sufficient that the linear system

(where $(z_{ij}) \in T_1(n, K)$) admit a unique solution. Now, this system may be written

For fixed $i$, one has a system of $n-i$ equations in the unknowns $z_{i,i+1}, z_{i,i+2}, \ldots, z_{i,n}$; for these systems to admit unique solutions, it is necessary and sufficient that

for $i = 1, 2, \ldots, n-1$. This proves a). From this it follows that $\Omega$ is open in $\mathbf{GL}(n, K)$. On the other hand, on solving the system (13) by means of Cramer’s formulas, the $z_{ij}$ are obtained as rational functions of the $u_{ij}$ with nonzero denominators in $\Omega$, therefore $Z$ depends continuously on $U$ in $\Omega$, which proves c).

Now let $G'_1 \subset {}^t(T(n, K)^*)$ be the lower special triangular group. The mapping $(M, N) \mapsto M \cdot N$ is a continuous bijection $\psi$ of $T_1(n, K) \times G'_1$ onto a subset $\Omega'$ of $\mathbf{SL}(n, K)$.