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I would like to know the structure of the fixed set of an order $p$ automorphism [Edit: induced by a matrix in $GL_2(K)$] on the Bruhat-Tits tree for a p-adic field $K$, specifically in the case where the order divides (you can assume equals) the residue characteristic.

I have convinced myself that when the order is prime to the residue characteristic there are only two cases: if the corresponding root of unity is in the residue field, then there is a fixed apartment, if the corresponding root of unity is not in the residue field, then there is an isolated fixed point.

But what happens if the order is equal to the residue characteristic? To make the situation more specific, consider the matrix $$\left(\begin{array}{cc}\zeta_p&0\\0&\zeta_p^{-1}\end{array}\right)$$ and consider its action on the Bruhat-Tits tree for $K=\mathbb{Q}_p(\zeta_p)$. What is the structure of the fixed set? Is it connected or not? The action of the above matrix on the ends of the Bruhat-Tits tree has two fixed points because it is multiplication with $\zeta_p^2$ on $\mathbb{P}^1(K)$. But if I look at the point of the tree corresponding to the lattice $\mathcal{O}_K^2\subset K^2$, then the matrix acts on the link $\mathbb{P}^1(\mathbb{F}_p)$ via the homomorphism $SL_2(\mathcal{O}_K)\to SL_2(\mathbb{F}_p)$ and the image of the matrix under this homomorphism is trivial. In particular, the whole link of the standard lattice is stabilized. The fixed set thus has to be larger than just a single geodesic.

Can somebody please clarify the structure of the fixed set in this case? I am sure that this has been discussed in the literature, can someone please point me to some relevant papers or books?

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up vote 7 down vote accepted

First a caveat: there is a difference between "an automorphism of the Bruhat-Tits tree $T$ of a $p$-adic field $K$" and "an automorphism of $T$ induced by a matrix in $GL_2(K)$". The first set is much larger: contrarily to higher rank buildings, Bruhat-Tits trees are very "soft" and have a lot of automorphisms. I believe that while you write "automorphism of $T$" you mean "automorphism of $T$ induced by a matrix" because otherwise the result you convinced yourselves of is false (while it is true for automorphisms given by matrices).

So restricting your question to automorphisms given by matrices, you can find the answer in section 3.2 of my paper with Chenevier "sous-groupes de GL(2) et arbres", in French, published at Journal of Algebra. In general, your fixed set will be either a "ball", or an "infinite band", or an "horodisc", or everything depending of whether it is semi-simple non-diagonalizeble, diagonalizable with distinct eigenvalues, unipotent, or central. In your example, it will be an infinite band.

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You are absolutely right, I wanted the automorphism to come from a matrix. I edited the question accordingly. –  Matthias Wendt Jun 11 at 16:39
    
I had a look at your paper, and from what I have understood so far it answers my question very satisfactorily. There is only one thing I am still confused about: the horodiscs are the fixed sets of unipotent elements, but the example matrix I wrote is not. It only becomes unipotent after suitable reduction, but it has two distinct eigenvalues in $\mathbb{Q}_p(\zeta_p)$. Is the horodisc case really applicable? –  Matthias Wendt Jun 11 at 16:41
    
@Matthias: you're right, your specific example is a diagonalizable matrix so the answer is an infinite band of width $v(\zeta_p - \zeta_p^{-1})$, where $v$ is the normalized valuation of your field $\Q_p(\zeta_p)$, so of width 1 if I am not mistaken. That is, the set of fixed point is the set of points at distance $\leq 1$ of the apartment defined by your matrix. I will edit my answer to make it clearer. –  Joël Jun 11 at 17:03

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