On uniformizable 𝑇 -modules,
an example
Hans-Georg Rück (with Oliver Bangert)
Institut für Mathematik
Universität Kassel
Drinfeld-modules:
The ring 𝔽𝑞 [𝜃], the field 𝐾 = 𝔽𝑞 (𝜃),
ˆ .
the valuation 𝑣∞ (𝜃) = −1, 𝐾∞ = 𝔽𝑞 (( 1𝜃 )) and 𝐶∞ = 𝐾
∞
In 𝐶∞ consider 𝔽𝑞 [𝜃]-lattice Λ of rank 𝑟 and corresponding torus 𝐶∞ /Λ,
parameterize it by
𝜓 : 𝐶∞ /Λ → 𝐶∞
∞
𝑧
∏
7→ 𝑒(𝑧) = 𝑧
∑
𝑧
𝑖
𝑒𝑖 𝑧 𝑞 .
(1 − ) =
𝜆
𝑖=0
0∕=𝜆∈Λ
𝑒(𝑧) converges for each 𝑧, 𝑒(𝑧) is surjective,
ker(𝑒(𝑧)) = Λ, hence 𝜓 is an isomorphism.
2
The 𝔽𝑞 [𝑇 ]-module structure of 𝐶∞ :
there is an additive polynomial
𝑃 (𝑋) = 𝜃 𝑋 + 𝑔1 𝑋 𝑞 + . . . + 𝑔𝑟 𝑋 𝑞
𝑟
(𝑔𝑟 ∕= 0) with 𝑒(𝜃 𝑧) = 𝑃 (𝑒(𝑧)).
Consider
𝑇 = 𝜃 + 𝑔1 𝜏 + . . . + 𝑔𝑟 𝜏 𝑟 ,
where 𝜏 : 𝑋 7→ 𝑋 𝑞 , then 𝜓 induces 𝔽𝑞 [𝑇 ]-module structure on 𝐶∞ .
Take 𝑀 = 𝐾 ∞ {𝜏 } as an 𝐾 ∞ [𝑇 ]-module, it has rank 𝑟 with basis (1, 𝜏, . . . , 𝜏 𝑟−1 )
(𝑔𝑟 𝜏 𝑟 = (𝑇 − 𝜃) ⋅ 1 − 𝑔1 𝜏 − . . . − 𝑔𝑟−1 𝜏 𝑟−1 )
3
Generalize this (Anderson):
𝑑 and a polynomial
Consider 𝐶∞
𝑇 = (𝜃𝐼𝑑 + 𝑁 ) + 𝐺1 𝜏 + . . . + 𝐺𝑠 𝜏 𝑠
with 𝑁, 𝐺1 , . . . , 𝐺𝑠 ∈ M𝑑×𝑑 (𝐾 ∞ ), 𝑁 is nilpotent,
𝑑 is an 𝔽 [𝑇 ]-module, called 𝑇 -module,
then 𝐶∞
𝑞
and 𝑀 = 𝐾 ∞ {𝜏 }𝑑 is an 𝐾 ∞ [𝑇 ]-module, called corresponding 𝑇 -motive.
𝑀 should be finitely generated, free of rank 𝑟.
There is an exponential
⎛
⎞
𝑧1
𝐸(𝑧) = 𝐸(⎝ ... ⎠) =
𝑧𝑑
∞
∑
⎛
𝑞𝑖
𝑧1
⎞
𝐸𝑖 ⎝ ... ⎠
𝑖
𝑖=0
𝑧𝑑𝑞
converging everywhere with 𝐸((𝜃𝐼𝑑 + 𝑁 ) 𝑧) = 𝑇 𝐸(𝑧).
4
Properties of 𝐸: Is 𝐸(𝑧) surjective? What is ker(𝐸(𝑧))?
The following statements are equivalent:
∙ kernel(𝐸(𝑧)) is an 𝔽𝑞 [𝜃]-module of rank 𝑟,
∙ 𝐸(𝑧) is surjective.
Then the 𝑇 -module is called uniformizable.
5
How can we handle this? There is a third condition (due to Anderson):
∑
𝑛
𝐾 ∞ < 𝑇 >= { ∞
𝑛=0 𝑎𝑛 𝑇 ∈ 𝐾 ∞ ((𝑇 )) ∣ lim𝑛→∞ 𝑣∞ (𝑎𝑛 ) = ∞, 𝑎𝑛 ∈ finite ext. of 𝐾∞ },
let again 𝑀 = 𝐾 ∞ {𝜏 }𝑑 ,
consider
𝑀 < 𝑇 >:= 𝑀 ⊗𝐾 ∞ [𝑇 ] 𝐾 ∞ < 𝑇 > .
𝜏 operates on 𝑀 < 𝑇 > by
𝜏 (𝑚 ⊗
∑
𝑛
𝑎𝑛 𝑇 ) = 𝜏 (𝑚) ⊗
∑
𝑎𝑞𝑛 𝑇 𝑛 .
Consider
𝜑 : 𝑀 < 𝑇 >𝜏 ⊗𝔽𝑞 [𝑇 ] 𝐾 ∞ < 𝑇 >→ 𝑀 < 𝑇 > ,
then uniformizable is equivalent to ”𝜑 is an isomorphism” (rigid analytically trivial).
6
Our Examples (motivated by an example by Coleman-Anderson):
(
𝑑 = 2, 𝑇 =
𝜃 0
0 𝜃
)
(
+
𝑎 𝑏
𝑐 𝑑
)
(
𝜏+
1 0
0 1
)
𝜏 2 with 𝑐 ∕= 0.
We study it in the sense of Böckle-Hartl.
A basis of 𝑀 = 𝐾 ∞ {𝜏 }2 as 𝐾 ∞ [𝑇 ]-module is (1, 0), (0, 1), (𝜏, 0), (0, 𝜏 ),
an element in 𝑀 < 𝑇 > is given as
∑
∑
∑
∑
( 𝑛 𝑤𝑛 𝑇 𝑛 )(1, 0) + ( 𝑛 𝑥𝑛 𝑇 𝑛 )(0, 1) + ( 𝑛 𝑦𝑛 𝑇 𝑛 )(𝜏, 0) + ( 𝑛 𝑧𝑛 𝑇 𝑛 )(0, 𝜏 )
7
ℕ0
Evaluating invariants, some calculations, yields sequences (𝑦𝑛 )𝑛∈ℕ0 ∈ 𝐾 ∞
given by recurrence relations
4
3
2
𝑃 (𝑦𝑛 , 𝑦𝑛−1 , 𝑦𝑛−2 ) = 𝐴4 𝑦𝑛𝑞 + 𝐴3 𝑦𝑛𝑞 + 𝐴2 𝑦𝑛𝑞 + 𝐴1 𝑦𝑛𝑞 + 𝐴0 𝑦𝑛 + 𝐵(𝑦𝑛−1 , 𝑦𝑛−2 ) = 0,
where
𝐴0
𝐴1
𝐴2
𝐴3
𝐴4
𝐵(𝑦𝑛−1 , 𝑦𝑛−2 )
=
=
=
=
=
=
2
𝜃𝑞+1 𝑐𝑞 −1
2
2
𝑑𝑞 𝜃𝑞 𝑐𝑞 −𝑞 + 𝜃𝑞 𝑎𝑐𝑞 −1
2
2
2
2
𝜃𝑞 + 𝑑𝑞 𝑎𝑞 𝑐𝑞 −𝑞 − 𝑏𝑞 𝑐𝑞 + 𝜃𝑞 𝑐𝑞 −1
2
2
𝑎𝑞 + 𝑑𝑞 𝑐𝑞 −𝑞
1
𝑞2
𝑞4
𝑞3
−𝑞 3
−𝑞 2
−𝑞 2
𝑞 2 −1
(−𝜃
− 𝜃 )𝑦𝑛−1 + (−𝜃 𝐴3 )𝑦𝑛−1 + (−𝑐
− 1)𝑦𝑛−1
−𝑞 2 −𝑞 3 𝑞 4
+𝜃
𝑦𝑛−2
8
Result: The 𝑇 -module is uniformizable if and only if
{(𝑦𝑛 )𝑛∈ℕ0 ∣ 𝑃 (𝑦𝑛 , 𝑦𝑛−1 , 𝑦𝑛−2 ) = 0 and lim𝑛→∞ 𝑣∞ (𝑦𝑛 ) = ∞}
is an 𝔽𝑞 [𝑇 ]-module of rank 4.
In other words:
it is uniformizable if and only if we find for each solution of 𝑃 (𝑦0 , 0, 0) = 0
a sequence (𝑦𝑛 )𝑛∈ℕ0 with and 𝑣∞ (𝑦𝑛 ) = ∞.
9
We have the initial equation:
4
3
2
𝑃 (𝑦0 , 0, 0) = 𝐴4 𝑦0𝑞 + 𝐴3 𝑦0𝑞 + 𝐴2 𝑦0𝑞 + 𝐴1 𝑦0𝑞 + 𝐴0 𝑦0 = 0 ,
and the recurrence equation:
4
3
2
𝑃 (𝑦𝑛 , 𝑦𝑛−1 , 𝑦𝑛−2 ) = 𝐴4 𝑦𝑛𝑞 + 𝐴3 𝑦𝑛𝑞 + 𝐴2 𝑦𝑛𝑞 + 𝐴1 𝑦𝑛𝑞 + 𝐴0 𝑦𝑛 + 𝐵(𝑦𝑛−1 , 𝑦𝑛−2 ) = 0 .
We consider their Newton polygon:
let
𝑚=
min (
𝑖=1,2,3,4
−𝑣∞ (𝐴𝑖 )
𝑣∞ (𝐴0 ) − 𝑣∞ (𝐴𝑖 )
)
and
𝑀
=
max
(
).
𝑖=0,1,2,3
1 − 𝑞𝑖
𝑞4 − 𝑞𝑖
10
11
Conditions that large values 𝑣∞ (𝑦𝑛−2 ) and 𝑣∞ (𝑦𝑛−1 ) imply also large 𝑣∞ (𝑦𝑛 ):
Lemma: Suppose
𝑣∞ (𝑐𝑞
2
−1
+ 1) − 𝑞 2 𝑚 > 𝑣∞ (𝐴0 ) − 𝑚 .
If 𝑣∞ (𝑣𝑛−2 ) ≥ −𝑚 + 𝜖 and 𝑣∞ (𝑣𝑛−1 ) ≥ −𝑚 + 𝑞 2 𝜖,
then there is 𝑦𝑛 with 𝑃 (𝑦𝑛 , 𝑦𝑛−1 , 𝑦𝑛−2 ) = 0 and 𝑣∞ (𝑣𝑛 ) ≥ −𝑚 + 𝑞 4 𝜖.
12
Now try to find for each 𝑦0 values 𝑦1 , 𝑦2 with 𝑣∞ (𝑦1 ), 𝑣∞ (𝑦2 ) > −𝑚:
Lemma: Suppose
(𝑣∞ (𝐴0 ) − 𝑚) − (−𝑞 4 𝑀 ) < 𝑞 2 and 𝑣∞ (𝑐𝑞
2
−1
+ 1) − 𝑞 2 𝑀 > 𝑣∞ (𝐴0 ) − 𝑚 ,
then for each 𝑦0 with 𝑃 (𝑦0 , 0, 0) = 0
we find 𝑦1 with 𝑃 (𝑦1 .𝑦0 , 0) = 0 and 𝑣∞ (𝑦1 ) > −𝑚
and 𝑦2 with 𝑃 (𝑦2 .𝑦1 , 𝑦0 ) = 0 and 𝑣∞ (𝑦2 ) > −𝑚.
13
(𝑣∞ (𝐴0 ) − 𝑚) − (−𝑞 4 𝑀 ) < 𝑞 2
14
Result: The 𝑇 -module with
(
) (
)
(
)
𝜃 0
𝑎 𝑏
1 0
𝑇 =
+
𝜏+
𝜏2
0 𝜃
𝑐 𝑑
0 1
(𝑐 ∕= 0)
is uniformizable if
(𝑣∞ (𝐴0 ) − 𝑚) − (−𝑞 4 𝑀 ) < 𝑞 2 and 𝑣∞ (𝑐𝑞
2
−1
+ 1) − 𝑞 2 𝑀 > 𝑣∞ (𝐴0 ) − 𝑚 .
15
Can these conditions be fulfilled?
Example:
If
𝑣∞ (𝑎), 𝑣∞ (𝑑) ≥ −
𝑞
,
𝑞+1
𝑣∞ (𝑏) > −𝑞,
0 ≤ 𝑣∞ (𝑐) <
1
,
𝑞+1
then it is uniformizable.
16
Special case:
If the Newton polygon of 𝑃 (𝑦0 , 0, 0) is a line, i.e. 𝑚 = 𝑀 , then the 𝑇 -module is
uniformizable if and only if the least slope of 𝑃 (𝑦1 , 𝑦0 , 0) is smaller.
17
When is it not uniformizable? – if we find 𝑦0 with 𝑃 (𝑦0 , 0, 0) = 0 which cannot be
extended to a converging sequence.
Result: If
𝑣∞ (𝑐𝑞
2
−1
+ 1) ≤ −𝑞 2 (𝑞 2 − 1)𝑀 ,
then the 𝑇 -module is not uniformizable.
18
Example:
If
𝑞2 + 𝑞
𝑣∞ (𝑎), 𝑣∞ (𝑑) ≥ − 2
,
𝑞 +1
𝑞
−𝑞 2 − 𝑣∞ (𝑎) − 𝑣∞ (𝑑)
−
≥ 𝑣∞ (𝑐) >
,
𝑞−1
𝑞−1
−1
𝑏 = 𝜃𝑐−𝑞 ,
then it is not uniformizable.
19