Rigid Metric Spaces Robert Coleman

Homometries

Suppose K is a complete subfield of C_{p} and R is the ring of integers in K. Let (X, d_{X}) and (Y, d_{Y}) be rigid metric spaces over K and f: X → Y a finite rigid morphism of metric spaces, i.e., a finite morphism of rigid spaces such that

d_{Y} (f(x), f(y)) ≤ d_{X}(x, y)

for all x, y ∈ X(C_{p}). Suppose X has finite diameter (it follows that the same is true for Y). For y ∈ Y set

D_{f}(y) = \{ \begin{matrix}∏_{f(x)=f(x′)=y} ; x≠x′ d_{X}(x, x′) & if f is unramified at y \\ 0 & otherwise. \end{matrix}

Let δ(X/Y) = sup_{y} |D_{f}(y)|. We say f is a homometry if it is finite and

δ(Y/X) d_{Y} (f(x), z) ≥ ∏_{f(y)=z} d_{X}(x, y)^{e(y/z)}.

for all x ∈ X and all z ∈ Y. Let n be the degree of f. Then f is a homometry of degree one if and only if it is an isometry.

For all α_{>0} ∈ R let U_{α} be the subspace of Y × Y {(x, y): d_{Y} (x, y) ≤ α}. We will suppose if f ∈ A(U_{α}) and g ∈ A(Y), |g|_{Y} = δ < ∞, then if |f(w, y)| ≤ β for all (w, y) ∈ U_{α} with |g(w)| > δ − ϵ then |f(w, y)| ≤ β for all (w, y) ∈ U_{α}. Examples of such include affinoids with good connected reduction in weight space and wide open disks in weight space.

Theorem. Suppose f: X → Y is a homometry such that if H = Λ(X) is the ring of Lipshitz functions on X,

d_{X}(x, y) = sup_{h∈H} |h(x) − h(y)|.

Then if h ∈ H, the coefficients of the characteristic polynomial P_{h}(T) of h over Y are Lipshitz functions on Y. (Note that it follows that X and Y have diameter at most one.) Proof. We can extend scalars and assume the residue field of K is algebraically closed. First, if h ∈ H, let D^{2}(h) denote the rigid function on X which is the square of the “discriminent” of h,

D^{2}(h)(y) = \{ \begin{matrix}∏_{f(x)=f(x′)=y} ; x≠x′(h(x) − h(x′))^{2} & if f is unramified at y \\ 0 & otherwise. \end{matrix}

Clearly, |D(h)(z)| := \sqrt{|D^{2}(h)(z)|} ≤ δ =: δ(X/Y) for all z ∈ Y. We claim, for each ϵ > 0, there exists an h ∈ H and a w ∈ Y such that

|D(h)(w)| ≥ δ − ϵ.

We can suppose δ − ϵ > 0. Suppose D_{f}(w) > δ − ϵ/2. We know for each ρ > 0 and each pair {x, x′} such that f(x) = f(x′) = w, there exists an h_{x,x′} ∈ H such that

d_{X}(x, x′) − ρ < |h_{x,x′}(x) − h_{x,x′}(x′)|.

Now we can choose λ_{x,x′} ∈ R so that if

h = ∑_{x,x′} λ_{x,x′}h_{x,x′}

|D_{h}(w)| = ∏ |h_{x,x′}(x) − h_{x,x′}(x′)|

> ∏(d_{X}(x, x′) − ρ) > D_{f}(w) − ϵ/2 > δ − ϵ,

for ρ sufficiently small.

Now suppose D_{f}(w) > δ − ϵ > 0 and h(x) = w. For g ∈ H. let P_{g}(T) = ∑_{i=0}^{n} e_{i}(g)T^{i}. We have

| ∏_{f(y)=v}(h(x) − h(y))| ≤ ∏_{f(y)=v} d_{X}(x, y) ≤ δ · d_{Y} (w, v).

The first term equals |P_{h}(v)(h(x))| which equals

|P_{h}(w)(h(x)) − P_{h}(v)(h(x))| = | ∑_{0≤i<n} E_{n−i}(w, v)h(x)^{i}|

where E_{i}(w, v) := E_{i}(h)(w, v) = e_{i}(h)(w) − e_{i}(h)(v). Now observe, if x_{1}, …, x_{n} are the distinct points above w

( \begin{matrix}E_{0}(w, v) & E_{1}(w, v)h(x_{1}) & … & E_{n−1}(w, v)h(x_{1})^{n−1} \\ E_{0}(w, v) & E_{1}(w, v)h(x_{2}) & … & E_{n−1}(w, v)h(x_{2})^{n−1} \\ … & … & … & … \\ E_{0}(w, v) & E_{1}(w, v)h(x_{n}) & … & E_{n−1}(w, v)h(x_{n})^{n−1} \end{matrix} )

equals

( \begin{matrix}1 & h(x_{1}) & … & h(x_{1})^{n−1} \\ 1 & h(x_{2}) & … & h(x_{2})^{n−1} \\ … & … & … & … \\ 1 & h(x_{n}) & … & h(x_{n})^{n−1} \end{matrix} ) ( \begin{matrix}E_{0}(w, v) & 0 & … & 0 \\ 0 & E_{1}(w, v) & … & 0 \\ … & … & … & … \\ 0 & 0 & … & E_{n−1}(w, v) \end{matrix} ) .

As the determinant of the first matrix has absolute value > δ − ϵ by hypothesis, we see that

|E_{i}(h)(w, v)| < (1 + ρ(ϵ)) d_{Y} (w, v)

where ρ(ϵ) = ϵ/(δ − ϵ) for all v and all w such that |D_{h}(w)| > δ − ϵ. In particular,

|E_{i}(h)(w, v)| < (1 + ρ(ϵ))α  (∗)

for all (w, v) ∈ U_{α} for all α_{>0} ∈ R where |D_{h}(w)| > δ − ϵ. The hypotheses imply it must hold on all of U_{α}. Since, it is true for all ϵ > 0, it follows that e_{i}(h) is Lipshitz.

Now, if g ∈ H, |D_{h+λg}(w)| = δ for almost all roots of unity of order prime to p in R and so e_{i}(h + λg) is Lipshitz for these λ. As

e_{i}(h + λg) = ∑_{j=0}^{i−1} λ^{j}ℓ_{j}(h, g) + λ^{i}e_{i}(g)

for some functions ℓ_{j}(h, g), it follows that e_{i}(g) is Lipshitz.

Example. Let X = A[r, r] and Y = B[0, r] with the usual metrics and suppose |a| = r ≤ 1. Suppose

f(T) = T + ba + \frac{ca^{2}}{T}

where 1 = |c| ≥ |b|. Then f: X → Y is a finite morphism of rigid metric spaces and δ(X/Y) = sup_{|u|=r} |u − ca^{2}/u| = r, and if u, v ∈ X,

d_{Y} (f(u), f(v)) = |u − v + ca^{2}(\frac{1}{u} − \frac{1}{v})|

= |\frac{1}{u}|||u − v||u − \frac{ca^{2}}{v} |

= r^{−1} d_{X}(u, v) d_{X}(u, a^{2}/v).

As the function T is a Lipshitz function on X, f satisfies the conditions of the theorem and so the trace and norm of a Lipshitz function on X are Lipshitz functions on Y.