I am reading the paper Double Bruhat cells and total positivity.
Let $G$ be an algebraic group and $\omega_i$'s the fundamental weights.
I am trying to prove the identity (2.14) on page 12: \begin{align} \Delta^{\omega_i}(ax) = \Delta^{\omega_i}(xa)=a^{\omega_i}\Delta^{\omega_i}(x) \end{align} using the definition: $\Delta^{\omega_i}(x) = [x]_0^{\omega_i}$, where $x=[x]_-[x]_0[x]_+$ is the Gauss Decomposition of $x \in G_0=N_-HN_+ \subset G$.
My solution: Suppose that $x = [x]_-[x]_0[x]_+$ is the Gauss decomposition of $x$. Then \begin{align} \Delta^{\omega_i}(ax) = \Delta^{\omega_i}(a [x]_-[x]_0[x]_+) = \Delta^{\omega_i}(a[x]_-a^{-1} \cdot a[x]_0 \cdot [x]_+) = \Delta^{\omega_i}(a[x]_0), \end{align} since $a[x]_-a^{-1} \in N_-$ and $[x]_+ \in N_+$. Do we have $\Delta^{\omega_i}(t t') = \Delta^{\omega_i}(t) \Delta^{\omega_i}(t')$ for $t, t' \in H$? If so, then $\Delta^{\omega_i}(a[x]_0) = \Delta^{\omega_i}(a) \Delta^{\omega_i}([x]_0) = \Delta^{\omega_i}(a)\Delta^{\omega_i}(x)$. The identity is proved.
Is my prove correct? Thank you very much.