$\newcommand{\GL}{\operatorname{GL}} \newcommand{\diag}{\operatorname{diag}} \newcommand{\val}{\mathit{val}}$Let $F$ be a local non-Archimedean field with valuation $\val$ and $G$ be (the $F$-points of) a split connected reductive group over $F$. Let $T$ be a split maximal torus and $B$ a Borel containing $T$ with unipotent radical $U$. Let $K$ be a special parahoric subgroup of $G$ in good position relative to $(T,U)$. Let $X^*(T)$ be the lattice of $F$-characters of $T$, which contains the root system $\Phi = \Phi(G,T)$; let $X_*(T)$ be the $F$-cocharacters of $T$. The choice of $B$ defines a choice of positive roots $\Phi^+$. Let $X_*(T)^+ = \{ \lambda\in X_*(T)\mid \langle \alpha, \lambda\rangle \ge 0\text{ for all $\alpha\in \Phi^+$}\}$ be the cocharacters contained in the dominant Weyl chamber.

The Cartan decomposition gives a bijective map $$ X_*(T)^+\cong K\backslash G/K $$ given by $\lambda \mapsto K\lambda(\pi^{-1}) K$, where $\pi$ is a uniformizer in $F$. The Iwasawa decomposition gives a bijective map $$ X_*(T)\cong U\backslash G/K $$ given by $\mu\mapsto U\mu(\pi^{-1})K$.

I am interested in what happens when $K\lambda(\pi^{-1})K\cap U\mu(\pi^{-1})K \neq \varnothing$. What is well-known is that $\lambda -\mu$ is a sum of positive coroots.

My question is: What can be said about the $u\in U$ with $u\mu(\pi^{-1}) \in K \lambda(\pi^{-1})K$?

More precisely, the root groups $U_\alpha$, associated to $\alpha\in \Phi$, carry a valuation $\varphi_\alpha\colon U_\alpha\setminus\{1\} \to \mathbb R$ such that $\varphi = (\varphi_\alpha)_{\alpha\in\Phi}$ is a discrete and special valuation (in the sense of Bruhat–Tits) on the root group datum $(T,(U_\alpha)_{\alpha\in \Phi})$ on $G$ that is compatible with the valuation on $F$. Writing $u = u_{\alpha_1}\dotsm u_{\alpha_r}$ with $\alpha_i\in \Phi^+$, do we get a lower bound for $\varphi_{\alpha_i}(u_{\alpha_i})$?

Here are my thoughts on this problem, so far:

If $\Phi$ is irreducible and $\alpha_0$ is its highest root (with respect to $\Phi^+$), I suspect that $$ \varphi_{\alpha_i}(u_{\alpha_i}) \ge -\langle \alpha_0, \lambda\rangle, \quad \text{for all $i$.} $$ This is not too hard to verify for $\GL_n(F)$. [There, in fact, we even have $\val(u_{\alpha_{ij}}) \ge -m_j$, where $\mu(\pi^{-1}) = \diag(\pi^{m_1},\dotsc, \pi^{m_n})$, where we identify the root group $U_{\alpha_{ij}}$, sitting in the $(i,j)$-th spot, with $F$.]

One can get a more precise estimate in the case $u\in U_\alpha$ as follows: We may assume $\varphi_\alpha(u) < \min\{0, -\langle \alpha,\mu\rangle\}$ for otherwise we would have $u\in K$ or $\mu(\pi^{-1})^{-1}u\mu(\pi^{-1}) \in K$ in which case $\mu$ lies in the $W$-orbit of $\lambda$. There exist (unique) $u',u''\in U_{-\alpha}$ such that $u'uu''$ normalizes $T$. [Geometrically, $u'uu''$ is the reflection (in the apartment corresponding to $T$) in the hyperplane bounding the half-apartment fixed by $u$.] We have $\varphi_{-\alpha}(u') = \varphi_{-\alpha}(u'') = -\varphi_{\alpha}(u)$ and hence $\mu(\pi^{-1})^{-1}u''\mu(\pi^{-1}) \in K$. Thus, we may replace $u\mu(\pi^{-1})$ by $u'uu'' \mu(\pi^{-1})$ without changing either of the double cosets. Computing inside the apartment (with the valuation $\varphi$ taken as the origin), we see that $$ u'uu''\mu(\pi^{-1})(\varphi) = \varphi + \mu(\pi^{-1}) - (\langle \alpha, \mu\rangle + \varphi_\alpha(u))\alpha^\vee $$ lies inside the convex hull of the $W$-orbit of $\lambda(\pi^{-1})(\varphi)$. From this, we deduce $-\langle \alpha,\mu\rangle -2\varphi_\alpha(u) = \langle \alpha, \mu(\pi^{-1}) - (\langle \alpha,\mu\rangle + \varphi_\alpha(u)) \alpha^\vee\rangle \le \langle \alpha_0,\lambda\rangle$ and hence $$ \varphi_\alpha(u) \ge -\frac{\langle \alpha_0,\lambda\rangle + \langle \alpha, \mu\rangle}{2}. $$ Unfortunately, I was unable to generalize this argument to arbitrary $u\in U$, because the occuring commutators (possibly) destroy any information of the $u_{\alpha}$'s.

`$`

ending your definitions and the start of your question body. I think you mean to identify $U_{\alpha_{ij}}$ for $\operatorname{GL}_n(F)$ with $F$, not $F^\times$, but didn't change it in case I was missing your point. $\endgroup$ – LSpice Aug 10 '18 at 11:21