Download Evaluating the BCH formula

Evaluating the BCH formula
Serena Cicalò (joint work with Willem de Graaf )
We know that
1 2
1
x + x3 + . . . .
2!
3!
If x and y are two noncommuting variables we have that ex ey = ez , where z is a formal
series in x and y with rational coefficients.
If u is such that
1
1
ez = 1 + z + z 2 + z 3 + . . . = 1 + u
2!
3!
we have that
u2 u3 u4
z = ln(1 + u) = u −
+
−
+ ....
2
3
4
Let sm (r) be the sum of all elements of degree r obtained by multiplication of m factors
1
of the form ki !h
xki y hi that appear in ez and that we call blocks.
i!
Now, let w be a word in x and y, that is w = w1 . . . wn where wi = xki y hi with ki , hi > 0
all but k1 , hn ≥ 0. The main our purpose is to determine a formula to calculate the coefficient
zw of the word w in z.
For this, firstly, we determine the coefficient of wi = xki y hi in sm (ki + hi ). We denote this
coefficient with fm (ki , hi ) or fm (wi ).
ex = 1 + x +
Lemma 1. For all m ≤ h + k, fm (k, h) is the coefficient of xk y h in sm (k + h) and we have
m−2
m
1 X
j
(−1) ϕm−j (k, h)
fm−1 (k, h) =
k!h!
j
(1)
j=0
where
X
ϕm (k, h) =
pk q h .
p+q=m
Remark 1. We can easily see that fm (k, h) = 0 for all m > k + h because m counts the
number of blocks in which we can divide xk y h in sm (k + h). Then it not makes sense to divide
xk y h in a number of blocks greater than its degree, that is k + h.
We can now give the formula to calculate the coefficient zw of a word w.
ki hi
Theorem
Pn 1. Let w = w1 . . . wn where wi = x y with ki , hi > 0 all but k1 , hn ≥ 0 and let
N = i=1 (ki + hi ). Then the coefficient of w in z is
N
X
(−1)m−1
Fm (w)
zw =
m
m=n
(2)
where
Fm (w) =
X
fm1 (w1 ) · · · fmn (wn ).
(3)
m1 +...+mn =m
Now we present
for zw using the Stirling numbers of the second kind
an equivalentformula
n
n
denoted by
. We say that
stands for the number of ways to partition a set of n
m
m
things into m nonempty subsets.
1
For example, there seven ways to split a four-element set into two parts:
{1, 2, 3} ∪ {4}, {1, 2, 4} ∪ {3}, {1, 3, 4} ∪ {2}, {2, 3, 4} ∪ {1},
{1, 2} ∪ {3, 4}, {1, 3} ∪ {2, 4}, {1, 4} ∪ {2, 3}.
4
Thus
= 7.
2
Let again w = w1 . . . wn , where wi = xki y hi with ki , hi > 0 all but k1 , hn ≥ 0. We put
n
X
Y
1
k
hi
si !ti ! i
.
(4)
Gm (w) =
s
ti
P Pn
i
i=1 (si +ti )=m
where P =
i=1
Qn
i=1 ki !hi !.
Remark 2. We can remark that Gm (w) depends only by the exponents that appear really
in w.
Example 1. Let w = x3 y 2 x4 . We want to calculate G5 (w).
We have v = (3, 2, 4) and P = 3!2!4! = 288. Then
1
3
2
4
3
2
4
3
2
4
3!
1!
1!
+ 2!
2!
1!
+ 2!
1!
2!
G5 (w) =
3
1
1
2
2
1
2
1
2
288
1
4
2
3
4
2
3
3!
1!
+ 1!
2!
2!
1!
+
3
1
1
2
2
1
288
1
=
[6 + 12 + 84 + 28 + 36]
288
1
=
· 166.
288
Let now d defined as

if k1 , hn 6= 0;
 n,
n − 1, if k1 = 0, hn 6= 0 (or vice versa);
d=

n − 2, if k1 = hn = 0.
We have showed that:
P
Theorem 2. Let N = ni=1 (ki + hi ), we have that
N
X
1
Gm (w)
zw =
(−1)m−d−1 m .
d+1
d+1
m=d+n
Example 2. Let w = y 4 xy 2 . We have n = 2, d = 1 and N = 7 then
zw =
7
1X
Gm (w)
(−1)m−2 m .
2
2
m=3
Also we have P = 4!1!2! = 48 then
m
3
4
5
6
7
Hence
48 · Gm (w)
1
16
64
96
48
"
#
1 1
1
16
64
96
48
1
zw = ·
− 3 + 4 − 5 + 6 − 7 =
.
2 48
2016
2
2
2
2
2
2
(5)