FINDING N-TH ROOTS IN NILPOTENT GROUPS AND ...

580 S. Sze, D. Kahrobaei, R. Dambreville, M. Dupas
Let
N = Jn,0
c =
⎛
⎞
1 0 c 0 · · · 0
⎜ .
⎜0
.. . .. . ..
⎟
. ⎟
⎜
.
⎜.
.. . .. . ..
⎟
0⎟
.
⎟
⎜ .
⎝.
.. . .. 1⎟
⎠
c
0 · · · · · · 0 0
Then T = c(I + N) and N is nilpotent with N n = 0.
Now recall the binomial expansion for √ 1 + x:
√ 1 + x = 1 + 1
2
and squaring both sides yields
x + 1/2(−1/2)
2!
x 2 + 1/2(−1/2)(−3/2)
x
3!
3 + · · ·
1 + x = 1 + 1 1/2(−1/2)
x + x
2 2!
2 + 1/2(−1/2)(−3/2)
x
3!
3 2 + · · · .
Substituting I for 1 and N for x,
I + N = I + 1 1/2(−1/2)
N + N
2 2!
2 + 1/2(−1/2)(−3/2)
N
3!
3 2 + · · · .
Since N n = 0, this series has at most n + 1 terms. If
A = √
c
I + 1
2
N + 1/2(−1/2)
2!
N 2 + 1/2(−1/2)(−3/2)
N
3!
3
+ · · · ,
then A2 = c(I + N) = T and we have found a square root of T.
Example 4.2. Let N = J3,0
16 =
⎛ ⎞ 1 0 16 0
⎝ 1 0 0 ⎠. 16 Here, c = 16, N
0 0 0
2 =
⎛ ⎞ 1 0 0 256
⎝0
0 0 ⎠ and N
0 0 0
3 ⎛ ⎞
16 1 0
= 0. We define T = 16(I + N) = ⎝ 0 16 1 ⎠.
0 0 16
Then using the above computations, we have √ T = A = 4(I + 1 1
2N − 8N2 ) =
⎛ ⎞ 1 1 4 8 −512 ⎝ 1 0 4 ⎠.
8
0 0 4