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I am running into some confusion when trying to explicitly describe the group $^{2}\!A_3''$ (using the naming convention that Tits gives in his Corvallis notes). If anyone can give me any advice, I would greatly appreciate it.

Set-up: Let $k$ is a non-Archimedian local field with: \begin{align*} \mathfrak o&=\text{ring of integers in $k$}\\ \mathfrak p&=\text{$\mathfrak o \pi$ the maximal ideal in $\mathfrak o$}\\ \mathfrak f&=\text{$\mathfrak o/\mathfrak p$ the residue field of $k$}\\ \end{align*} Let $K/k$ be a maximal unramified extension with: \begin{align*} \mathfrak O&=\text{ring of integers in $K$}\\ \mathfrak P&=\text{$\mathfrak O\pi$ the maximal ideal in $\mathfrak O$}\\ \mathfrak F&=\text{$\mathfrak O/\mathfrak P$ the residue field of $K$}\\ \end{align*} I will denote by $F$ the Frobenius automorphism which cyclically generates ${\rm Gal}(K/k)\cong{\rm Gal}(\mathfrak F/\mathfrak f)$.

The group: I will be constructing the group $^{2}\!A_3''$ via Galois decent from the group $G={\rm SL}_4(K)$. Let $I$ be the standard Iwahori subgroup of $G$ $$ I=\begin{bmatrix}\mathfrak O&\mathfrak O&\mathfrak O&\mathfrak O\\\mathfrak P&\mathfrak O&\mathfrak O&\mathfrak O\\\mathfrak P&\mathfrak P&\mathfrak O&\mathfrak O\\\mathfrak P&\mathfrak P&\mathfrak P&\mathfrak O\end{bmatrix}\cap G $$ We want to give an action of ${\rm Gal}(K/k)$ on $G$, giving rise to a $k$-structure. Moreover, I will choose my ${\rm Gal}(K/k)$-action so that it leaves stable the Iwahori subgroup $I$. For this, I will let the Frobenius automorphism $F$ act on $G$ via $$ F:X\mapsto Q^{-1}(^{t}\!X^F)^{-1}Q\qquad\text{with }Q=\begin{bmatrix}1\\&&&1/\pi\\&&1/\pi\\&1/\pi\end{bmatrix}, $$ where $^{t}[x_{ij}]^F=[F(x_{ji})]$. Sure enough, the Iwahori subgroup $I$ is preserved by this action, since $F$ permutes the corresponding simple affine root groups of $G$: $$ F\begin{bmatrix}1&x\\&1\\&&1\\&&&1\end{bmatrix}=\begin{bmatrix}1\\&1\\&&1\\-\pi F(x)&&&1\end{bmatrix} $$ and $$ F\begin{bmatrix}1\\&1&x\\&&1\\&&&1\end{bmatrix}=\begin{bmatrix}1\\&1\\&&1&-F(x)\\&&&1\end{bmatrix} $$ If I understand things correctly, the fixed point group $G^F$ should be the group $^{2}\!A_3''$ given in Tits' Corvallis table.

My Confusion: My confusion comes from the definition of the/an Iwahori subgroup of $G^F$. On one hand, it would make sense to me that I would define the fixed point group $I^F$ to be the Iwahori subgroup of $G^F$. But on the other hand, it would also make sense that the Iwahori subgroup of $G^F$ is the stabilizer of the appropriate alcove in the apartment of $G^F$. Unfortunately, these two objects do not appear to coincide. In particular, if we consider the matrix $$ g=\begin{bmatrix} a&&b\\&1\\-\pi F(b)&&F(a)\\&&&1 \end{bmatrix} $$ with $a,b\in K$ such that $a\,F(a)+\pi b\, F(b)=1$ and $F^2(a)=a$ and $F^2(b)=b$. Therefore $g\in G^F$. We have that this $g$ stabilizes the fundamental alcove of $G^F$, but $g$ belongs to $I^F$ only if $a,b\in\mathfrak O$.

If anything is unclear or you would like me to provide any more information, let me know. I'd be happy to.

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    $\begingroup$ (My sympathies: there are not many good sources for Galois-twisted, not-necessarily-quasi-split, etc., cases, despite their being understood "in principle". So you/we are left to double-check whether some "prescribed" object really is what it is claimed to be... or maybe there's a typo somewhere!?!?! Amid endless notational and normalization conventions... No way to avoid being sure first-hand.) $\endgroup$ Aug 26, 2019 at 21:09
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    $\begingroup$ If $a F(a) + \pi b F(b) = 1$, then it is not too difficult to show that $a$ and $b$ have to be in $\mathfrak{O}$; otherwise you get a contradiction by considering valuations. $\endgroup$ Aug 27, 2019 at 11:53
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    $\begingroup$ @paulgarrett Thank you for the sympathies. This has been the most consistent source of struggles of mine as a grad student. Its comforting to know that I'm not the only one. $\endgroup$ Aug 27, 2019 at 14:29
  • $\begingroup$ @DavidLoeffler Thank you! This is exactly what I needed. I should have suspected that this was the case. Thank you for the quick, and simple answer. $\endgroup$ Aug 27, 2019 at 14:29

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(Expanded version of my earlier comment, reposted as an answer):

If $F^2(a) = a$, $F^2(b) = b$, and $a F(a) + \pi b F(b) = 1$, then it is not too difficult to show that $a$ and $b$ have to be in $\mathfrak{O}$; otherwise you get a contradiction by considering valuations, because the valuation of $a F(a)$ has to be even and the valuation of $\pi b F(b)$ has to be odd.

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