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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.

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    $\begingroup$ As to "why they are $K$-points", it's the same reason that the unitary group $\{g \in \operatorname{GL}_n(\mathbb C) \mathrel: g g^* = 1\}$ is the $\mathbb R$-points of an $\mathbb R$-group, not the $\mathbb C$-points. A very easy check: the Lie-algebra version of this gives a $K$-vector space, not an $L$-vector space. (Why does this seem to behave differently from more familiar situations? Because, unlike those more familiar situations, the Galois group doesn't simply act coördinatewise.) $\endgroup$
    – LSpice
    Commented Aug 4, 2020 at 1:50
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    $\begingroup$ Oh, $d$ like $d_\alpha$, not $d$ like $u_i(c, d)$. Yes, it is more or less as you say (although I'd argue we really care about the residual degree—but the two pieces of information are complementary). Note, by the way, that, though it's fine for casual discussion, your discussion of the quotient group is more like a discussion of a transversal of the collection of cosets: the set of matrices you describe isn't a group! $\endgroup$
    – LSpice
    Commented Aug 4, 2020 at 1:58
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    $\begingroup$ So I don't know how you can draw this 1/2 out of (4). In particular, (4) seems to be identical to the formula for example 1.14, at the bottom of page 39; we let the normalizer act on the apartment by the natural action on the vectors plus a term corresponding to the valuation. I am probably being slow here but I still don't see where the 1/2$\omega(d)$ comes from. I would simply expect $a_i+\omega(d)$. $\endgroup$
    – Marc Besson
    Commented Aug 4, 2020 at 2:01
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    $\begingroup$ You are right; I pointed to the wrong thing. Note $\nu(m(u_i(c, d))v = v$ means $-v_i = v_{-i} = v_i + \omega(d)$, i.e., $2v_i + \omega(d) = 0$. This can be expressed as the vanishing of $a_i + \frac1 2\omega(d)$ or, when $c = 0$ and so we are in $U_{2a_i}$, of $2a_i + \omega(d)$. $\endgroup$
    – LSpice
    Commented Aug 4, 2020 at 2:09
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    $\begingroup$ By the way, as you go on, it may help to know that the $\delta$ that occurs later in this section is $0$ unless $p = 2$. (I thought that that was remarked somewhere, but I can't find it now.) Of course, Tits's Corvallis article is much easier reading than BT; but probably more pleasant than either is Yu's lovely article "Bruhat–Tits theory and buildings" in the Ottawa lectures on admissible representations of reductive, $p$-adic groups. It used to be on J.-K.'s web page, but anyway you can find it on Semantic Scholar. $\endgroup$
    – LSpice
    Commented Aug 4, 2020 at 2:13

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