I hope this question ends up having an easy answer, but I'm a little stuck on the morass of notation. I have a couple questions about this example that will probably help me greatly.

Here $L$ is a separable quadratic extension of $K$. We take the standard Hermitian form, define our split torus etc.

We find root subgroups $U_{a_{ij}}(K)$ and $U_{2a_j}(K)$ in terms of matrices involving elements $c, c^{\tau}, d, d^{\tau}$ where all of these are in $L$, not $K$. See question 3 for an example.

Question 1: Why are these "K-points"? My guess is that I should let the Galois group act on this set of matrices in some fashion and these root subgroups $U_{a_{ij}}(K)$ and $U_{2a_j}(K)$ should be pointwise fixed under this action: is this the correct idea?

Question 2): I wish to compute the $\alpha(a_i u_i(c,d))$ and the $\alpha(2a_i, u_i(0,d))$. In order to do this, one supposedly first computes the $m(u_{ij}(c))$ and the $m(u_{i}(c,d))$. These elements should all be in $N(K)$, the normalizer of the torus $S(K)$. We must identify the apartment $A$ with $X_* \otimes \mathbb{R}$ and we do so as Tits says in (4) (I do not really understand the significance of (4)) and then... TaDa, we apparently have

$\alpha(a_i u_i(c,d))a_i+1/2 \omega(d)$ and $\alpha(2a_i, u_i(0,d))=2a_i+\omega(d)$.

This second one at least seems to make sense: these should be affine functions which describe how conjugation by an element of the affine Weyl group acts on a coweight, so I generally expect the answer for $\alpha$ to be the vector part $a_i$ associated to the root and then a non-vector part associated to the valuation of $a$ for $u_i(a)$. But I do not see at all where we get $1/2 \omega(d)$ in the first of these formulas.

Question 3) To compute the function $d$, let me take the example of $SU(3)$; here $U_{a_1}(c)=$ \begin{pmatrix} 1 & -c^{\tau} & d \\ 0 & 1 & c \\ 0 & 0 & 1 \\\end{pmatrix}

And if we quotient by $U_{2a_1}$, then we should be left with matrices like \begin{pmatrix} 1 & -c^{\tau} & 0 \\ 0 & 1 & c \\ 0 & 0 & 1 \\\end{pmatrix}. In computing $d$ are we merely checking the ramified degree of the extension $L/K$ for whatever field $c$ is permitted to live in?

I've got more questions for the rest of this example but I think the answers to these three questions can help me find my footing.

I hope this question about Tits's paper "Reductive groups over local fields" in Algebraic groups and discontinuous subgroups ends up having an easy answer, but I'm a little stuck on the morass of notation. I have a couple questions about this example that will probably help me greatly.

I am considering Example 1.15, where $L$ is a separable quadratic extension of $K$. We take the standard Hermitian form, define our split torus etc.

We find root subgroups $U_{a_{ij}}(K)$ and $U_{2a_j}(K)$ in terms of matrices involving elements $c, c^{\tau}, d, d^{\tau}$ where all of these are in $L$, not $K$. See question 3 for an example.

Question 1: Why are these "$K$-points"? My guess is that I should let the Galois group act on this set of matrices in some fashion and these root subgroups $U_{a_{ij}}(K)$ and $U_{2a_j}(K)$ should be pointwise fixed under this action: is this the correct idea?

Question 2: I wish to compute the $\alpha(a_i u_i(c,d))$ and the $\alpha(2a_i, u_i(0,d))$. In order to do this, one supposedly first computes the $m(u_{ij}(c))$ and the $m(u_{i}(c,d))$. These elements should all be in $N(K)$, the normalizer of the torus $S(K)$. We must identify the apartment $A$ with $X_* \otimes \mathbb{R}$ and we do so as Tits says in (4) (I do not really understand the significance of (4)) and then … TaDa, we apparently have $$ \alpha(a_i u_i(c,d))a_i+(1/2)\omega(d) \quad\text{and}\quad \alpha(2a_i, u_i(0,d))=2a_i+\omega(d). $$

This second one at least seems to make sense: these should be affine functions which describe how conjugation by an element of the affine Weyl group acts on a coweight, so I generally expect the answer for $\alpha$ to be the vector part $a_i$ associated to the root and then a non-vector part associated to the valuation of $a$ for $u_i(a)$. But I do not see at all where we get $(1/2)\omega(d)$ in the first of these formulas.

Question 3: To compute the function $d$, let me take the example of $SU(3)$; here $$ U_{a_1}(K)= \left\{\begin{pmatrix} 1 & -c^{\tau} & d \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}\right\}. $$

And if we quotient by $U_{2a_1}$, then we should be left with matrices like $\begin{pmatrix} 1 & -c^{\tau} & 0 \\ 0 & 1 & c \\ 0 & 0 & 1 \end{pmatrix}$. In computing $d$ are we merely checking the ramified degree of the extension $L/K$ for whatever field $c$ is permitted to live in?

I've got more questions for the rest of this example but I think the answers to these three questions can help me find my footing.