# The reductive $p$-adic group $^{2}\!A_3''$ via Galois decent

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.

• (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.) Aug 26, 2019 at 21:09
• 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. Aug 27, 2019 at 11:53
• @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. Aug 27, 2019 at 14:29
• @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. Aug 27, 2019 at 14:29

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.