Proof that the Bruhat-Tits tree is regular #
Let R be a discrete valuation ring. Assume that the residue field k = R ⧸ 𝓂 R of R
is finite.
In this file we show that the Bruhat-Tits tree associated to R is regular, i.e. that every
vertex has the same finite number of neighbours. Furthermore we show that this number is #k+1.
Main result #
BruhatTits.btgraph_regular: The Bruhat-Tits Tree isq + 1-regular, whereqis the cardinality ofR ⧸ 𝓂 R.
A basis putting a neighbour of L into standard form.
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The uniformizer used in the standard representative of a neighbour.
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The standard lattice representative of a neighbour of L.
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The neighbours of ⟦L⟧ are in one to one correspondence to standard neighbours of L.
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Restrict scalars from the residue field quotient to an R-submodule.
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Relate quotient submodules to submodules of the lattice containing the maximal ideal multiple.
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Relate residue-field subspaces of the quotient to intermediate R-submodules of the lattice.
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Lines in the residue quotient correspond to proper nonzero intermediate submodules.
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R-submodules of K^2 that are strictly between ϖ L and L are in one-to-one
correspondence to standard neighbours of L.
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The standard neighbours of ⟦L⟧ are in one-to-one correspondence to
one-dimensional subspaces of L ⧸ ϖ L, i.e. lines.
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Neighbors of a vertex ⟦L⟧ correspond to lines in L ⧸ ϖ L.
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Neighbors of a vertex ⟦L⟧ correspond the projectivization of L ⧸ ϖ L.
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If the residue field of R is finite, every vertex has finitely many neighbors.
The degree of the Bruhat-Tits tree is q + 1 where q is the cardinality of R ⧸ 𝓂 R.
If R ⧸ 𝓂 R is finite, the Bruhat-Tits Tree is q + 1-regular, where q is the
cardinality of R ⧸ 𝓂 R.