6 NON-ARCHIMEDEAN ANALYTIC GEOMETRY: FIRST STEPS
Figure 2
0 |λ| 1 and 1| = 1. (The latter can be always achieved after a change of
coordinates.) Recall that the projective line
P1
has the property that any two differ-
ent points can be connected by a unique path. Let us connect the points 0, 1, λ and
in
P1.
We get one of the two graphs Γλ presented in Figure 1, which correspond
to the cases |λ| = 1 and |λ| 1. (For brevity the point p(E(0; r)) is denoted by pr.)
The complement of Γλ in
P1
is a disjoint union of open discs of the form D(a; ra)
with a k\{0, 1, λ} and ra = min{|a|,|a 1|,|a λ|}. Every such disc is glued to
its boundary which is a point of Γλ, and Γλ is a strong deformation retraction of
the whole projective line
P1.
Consider now the x-projection π :
Ean

P1
from
the analytification
Ean
of E. Since the characteristic of the residue field of k is not
two, the square root of each of the linear factors of x(x −1)(x− λ) can be extracted
at D(a; ra) and, therefore, the preimage of
π−1(D(a;
ra)) is a disjoint union of two
open discs which are glued to their boundaries at the preimage π−1(Γλ). Thus, the
latter is a strong deformation retraction of Ean. If 0 r |λ|, then the square roots
of x λ and x 1 are extracted at the open annulus D(0; r + ε)\E(0; r ε) with
0 r ε r + ε |λ|, but the square root of x is not. This means that each point
of the interval that connects 0 with p|λ| has a unique preimage in Ean. Similarly,
each point from the intervals that connect 1 with p1, λ with p|λ|, and with p1,
has a unique preimage in
Ean.
In particular, if |λ| = 1, then
π−1(Γλ)

Γλ. If now
|λ| 1, then the square roots of x 1 and of the product x(x λ) are extracted
at the open annulus D(0; r + ε)\E(0; r ε) with |λ| r ε r + ε 1. This
means that each point pr with |λ| r 1 has two preimages in
Ean,
and the graph
π−1(Γλ)
has the form presented in Figure 2.
Thus, the analytic curve
Ean
is contractible if |λ| = 1, and homotopy equivalent
to a circle if |λ| 1. It is well known that these two cases correspond to those
when the modular invariant j(E) is integral or not. But the latter case |j(E)| 1
is precisely that of a Tate elliptic curve. Wow, such a curve is homotopy equivalent
to a circle! I was always fascinated by Tate elliptic curves, but never understood
the reason for which they admit uniformization. And here I had a very elementary
explanation of this astonishing phenomenon discovered by Tate; it reminded me of
the classical construction of the Riemann surface of an algebraic function.
Of course, all this strongly lifted up my spirit and eagerness in exploration of
the new spaces. This was very timely since it distracted me from a serious health
problem I had at that time, not to mention my job in a dull institution and the
reality of a country in an advanced stage of decaying.
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