Non-archimedean Real Estate

Making our way up some buildings

The Bruhat-Tits tree

3 Comments Posted by on June 9, 2011

In this post I’ll specialize Marc’s discussion to the building of {mathbf{PGL}_2(mathbf{Q}_p)}. It turns out that the building {mathcal{T}} is an infinite tree in this case.

1. Definition

Recall that the vertices of {mathcal{T}} are lattices in {mathbf{Q}_p^2} taken up to rescaling. If {L subseteq mathbf{Q}_p^2} is a lattice, then we’ll write {[L]} for the homothety class of {mathbf{Q}_p^times}-multiples of {L}. Two vertices bound a {1}-simplex of {mathcal{T}} if and only if there are representative lattices {L}, {L'} for the corresponding homothety classes such that

displaystyle L supsetneq L' supsetneq pL.

Note that this is actually a symmetric relation: for {[L'] = [pL']} by definition of the square-brackets notation, and one deduces from the lined formula above that also

displaystyle L' supsetneq pL supsetneq pL'.

Since {L} is a lattice in {mathbf{Q}_p^2}, it is isomorphic with {mathbf{Z}_p^2}, and thus

displaystyle L/pL cong mathbf{Z}_p^2/pmathbf{Z}_p^2 cong (mathbf{Z}_p/pmathbf{Z}_p)^2 cong (mathbf{Z}/pmathbf{Z})^2.

The lattice {L'} above corresponds to a line in {(mathbf{Z}/pmathbf{Z})^2}, and there are {p+1} such lines. This computation also shows that there are no higher dimensional simplices in {mathcal{T}}, for such a simplex would correpsond to a sequence of lattices

displaystyle L supsetneq L' supsetneq L'' supsetneq pL.

But additive subsets sandwich between {L} and {pL} as above correspond one to one with vector subspaces of {(mathbf{Z}/pmathbf{Z})^2}, and this plane is too small to admit such inclusions. To summarize, we have shown that {mathcal{T}} is a graph such that each vertex is adjacent to exactly {p+1} other vertices.

It does not take much more work to show that the graph {mathcal{T}} is a tree: given two vertices {u} and {v}, one can use the elementary divisors theorem to find representative lattices {L} and {L'} such that there is a basis {(e_1,e_2)} for {L} with the property that {(p^ne_1, e_2)} is a basis for {L'}, for some integer {n geq 1}. If we let {L_i} denote the {mathbf{Z}_p}-span of {p^ie_1} and {e_2}, then the inclusions

displaystyle L_0 supsetneq L_1 supsetneq cdots supsetneq L_n

describe the unique nonbacktracking path from {u} to {v}. Hence every pair of vertices in {mathcal{T}} are joined by a unique nonbacktracking path, so that {mathcal{T}} is connected and acyclic, and hence a tree.

2. Distances and contiguity

Note that the integer {n} above describes the distance between {u} and {v} as defined in Marc’s previous post; that is, the distance is simply the number of edges between the two vertices. The distance between two edges is the maximum distance between any two of the endpoints of the edges. So for example, if two distinct edges share a vertex, then their distance from one another is {2}. The distance from an edge to itself is {1}, as Marc remarked last time, which seems a little pathological. the distance from a vertex to an edge is the maximum distance from the vertex to the endpoints of the edge.

Recall that two simplices are said to be contiguous if and only if they are at distance at most {1} from one another. Hence vertices are contiguous if and only if they are joined by an edge. A vertex is contiguous with an edge if and only if it is an endpoint of the edge. Finally, the previous paragraph shows that an edge is contiguous with another if and only if they’re equal. This is one instance of things being simpler in the case of the Bruhat-Tits tree: contiguity is not very exciting.

3. Apartments

In this low-dimensional case, apartments are also very simple: given a basis {(e_1,e_2)} for {mathbf{Q}_p^2}, the corresponding appartment is described by the vertices which correspond with the lattices {(p^ne_1,e_2)} for {n} an integer, either positive or negative. Hence apartments are nothing but paths in the tree {mathcal{T}} in both directions. They can be given a natural euclidean topology which makes them homeomorphic with the real line.

Affine buildings, PGL(2), PGL(n+1)

3 responses to “The Bruhat-Tits tree”

  1. Marc June 9, 2011 at 9:45 am

    To make it more consistent with the previous posts, I’d suggest:
    1) Change Lambda to L.
    2) Change Qp to an arbitrary extension K (and p to pi, etc).

    Also, I would add a word or two about types. The fact that there is a single type in this case and more in higher dimensions will cause us some challenges in the future, actually! For example, the action of G will stop being transitive (it preserves the type, and it is transtive on cells of the same type).

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  2. cfranc June 9, 2011 at 5:14 pm

    I agree that I should have used L instead of \Lambda, but I’m not going to change it. šŸ™‚ As for Qp, I liked the idea of keeping the example as simple as possible.

    Why don’t you add a short post about types and how things will be different in the case of higher dimensional buildings?

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  3. Marc June 14, 2011 at 1:19 pm

    I changed the Lambda’s to L’s (took 1minute, counted it!). I still think that K would be better than Qp because in this way we’d do a particular case instead of an example. But I don’t want to impose my criterion on this, so let’s leave as it is šŸ™‚

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