R.K. Kerimbaev , K.A. Dosmagulova , Zh.Kh. Zhunussova
∗Al-Farabi Kazakh National University, Almaty, Kazakhstan
∗
e-mail: [email protected]
ALGORITHMIC COMPLEXITY OF LINEAR NONASSOCIATIVE ALGEBRA
One of the central problems of algebraic complexity theory is the complexity of multiplication in algebras. For this, first, the concept of algebra is defined and the class of algebras under study is fixed. Then the concept of the algorithm and its complexity are clarified. in the most general sense, an algebra is a set with operations. An operation is defined, as a rule, as a function of one or more elements of a set, the set of values of which is the original set or some of its subset.
Usually, a set of elementary operations is fixed, for example, a Boolean operation on two bits, addition or multiplication of two numbers, after which a computation model is fixed, for example, a sequential algorithm, at each step of which one elementary operation is performed on some inputs and the results of intermediate calculations, the result of which can be used to enter an elementary operation at subsequent steps of the algorithm. The most significant ones are the column-by-column multiplication algorithm, which has quadratic complexity (along the input length) and the row-by-column matrix multiplication algorithm, which has O(mnp) complexity for multiplying m×n by n×p matrices. Estimation of the complexity of algebras from other more complex classes is relevant. in this paper, we derive an estimate for the complexity of a nondegenerate symmetric bilinear form over an algebraic closed field for a simple Jordan algebra, as well as an estimate for a Cayley-Dixon body and for a simple Lie algebra over a characteristic field.
Key words: algebra complexity, optimal algorithm, simple algebra, Cayley-Dixon body, Lie algebra, characteristic field.
Р.К. Керимбаев, Қ.А. Досмағұлова, Ж.Х. Жунусова
∗Әл-Фараби атындағы Қазақ Ұлттық университетi, Алматы қ., Қазақстан
∗e-mail: [email protected]
Кейбiр сызықты ассоциативтiк емес алгебраның алгоритмдiк күрделiлiгi
Алгебралық күрделiлiк теориясының негiзгi мәселелерiнiң бiрi - алгебралардағы көбейтудiң күрделiлiгi. Ол үшiн, бiрiншiден, алгебра ұғымы анықталып, зерттелетiн алгебралар класы бекiтiледi. Содан кейiн алгоритм түсiнiгi және оның күрделiлiгi нақтыланады.
Жалпы мағынада алгебра - амалдардан тұратын жиынтық. Операция, ереже бойынша, жиынның бiр немесе бiрнеше элементтерiнiң функциясы ретiнде анықталады, оның мәндерi жиынтығы бастапқы жиын немесе оның кейбiр бөлiгi болып табылады. Әдетте, қарапайым операциялардың жиынтығы, мысалы, екi битке логикалық операция, екi санды қосу немесе көбейту, содан кейiн есептеу моделi бекiтiлген, оған мысал - кезектi алгоритм, оның әр сатысында бiр элементар амал орындалады. Кейбiр кiрiстер мен аралық есептеулердiң нәтижелерi, олардың нәтижесi алгоритмнiң келесi кезеңдерiнде элементар әрекеттi енгiзу үшiн қолданыла алады. Ең маңыздыларына баған бойынша көбейту алгоритмi, жәнеm×n өлшемдi матрицаныn×pөлшемдi матрицаға көбейтетiн квадраттықO(mnp)күрделiлiгi бар (кiрiс ұзындығы бойынша) көбейту алгоритмi болып табылады. Басқа күрделi кластардан алгебралардың күрделiгiн бағалау өте маңызды. Бұл жұмыста қарапайым йордан алгебрасы үшiн алгебралық жабық өрiстен азайтылмайтын симметриялы билинарлы форманың күрделiлiгiнiң бағасы, сонымен қатар Кейли-Диксон денесi мен қарапайым өрiс үшiн Ли алгебрасы үшiн баға алынған.
Түйiн сөздер: алгебраның күрделiлiгi, оңтайлы алгоритм, қарапайым алгебра, Кейли- Диксон денесi, Ли алгебрасы, характеристикалық өрiс.
c
2020 Al-Farabi Kazakh National University
Р.К. Керимбаев, Қ.А. Досмағұлова, Ж.Х. Жунусова
∗Казахский национальный университет имени аль-Фараби, г. Алматы, Казахстан
∗e-mail: [email protected]
Алгоритмическая сложность некоторых линейных неассоциативных алгебр Одной из центральных задач алгебраической теории сложности является сложность умно- жения в алгебрах. Для этого сначала определяется понятие алгебры и фиксируется класс изучаемых алгебр. Затем уточняется понятие алгоритма и его сложности. В наиболее общем смысле алгеброй называется множество с операциями. Операция определяется, как правило, как функция одного или нескольких элементов множества, множеством значений которой является исходное множество или некоторое его подмножество. Обычно фиксируется неко- торое множество элементарных операций, например, булева операция над двумя битами, сложение или умножение двух чисел, после чего фиксируется модель вычислений, например, последовательный алгоритм, на каждом из шагов которого выполняется одна элементарная операция над некоторыми входами и результатами промежуточных вычислений, результат которой может быть использован для входа элементарной операции на последующих шагах алгоритма. К наиболее значимым следует отнести алгоритм умножения чисел "в столбик имеющий квадратичную сложность (по длине входа) и алгоритм умножения матриц
"строка на столбец имеющий сложность O(mnp) для умножения матриц размера m×n на n×p. Оценка сложности алгебр из других более сложных классов актуальна. В данной статье мы выводим оценку сложности невырожденной симметрической билинейной формы над алгебраическим замкнутым полем для простой йордановой алгебры, а также оценку для тела Кэли-Диксона и простой алгебры Ли над характеристическим полем.
Ключевые слова: сложность алгебры, оптимальный алгоритм, простая алгебра, тело Кэли- Диксона, алгебра Ли, характеристическое поле.
1 Introduction
The complexity L(A) of a finite-dimensional algebra A is a multiplication number (non- scalar:), divisions of the optimal algorithm, computing the production of two elements of algebra.
In the work [1] for associated algebra the results are obtained:
1)L(A) ≥ L(A/radA) + 2 ∗ dim(radA), where radA is a radical of algebra.
2)L(A) ≥ 2 ∗ dimA − 1,
where A is a simple algebra. More general: for simple algebra A and arbitrary algebra B it is proved, that
L(A ⊕ B) ≥ 2 ∗ dimA − 1 + L(B)
As a result, for arbitrary finite-dimensional associative algebra A the final estimation is obtained:
3)L(A) ≥ 2 ∗ dimA − t,
where t is a number of maximal two-sided ideal of algebra A. Naturally, the issue about complexity of algebra from other classes is arisen.
We note, that the 1-st result, as a sequence of structural theorems [2], are true for jordan alternative algebra.
The 2-nd result for jordan algebra, in general, is not true. We show in §2 of this work, that for simple jordan algebra B (f ) = K ∗ 1 + V nondegenerate symmetric bilinear form f over algebraic closed field K the complexity
L(B(f)) = 2 ∗ dimB(f) − 2.
In §3 we show for keli-Dixon body C, that
15 ≤ Z(C) ≤ 30
In §4 we show for simple Li algebra sl(2, K) over filed K the characteristics 6= 2, that L(sl(2, K)) = 5.
2 The main definitions
We present some definitions from [1] below. Let K be infinite filed, x
1, . . . , x
nare variables over K.
Definition 1 The sequence of rational functions g
1, ..., g
r∈ K(x
1, ..., x
n) are called by computing sequence, if for any number ρ ≤ r there exist
u
ρ, v
ρ∈ K + Kx
1+ ... + Kx
n+ Kg
1+ ...Kg
ρ−1, such, that
g
ρ= u
ρ∗ v
ρor
g
ρ= u
ρ/v
ρ, v
ρ6= 0.
Definition 2 Let f
1, ..., f
q∈ K (x
1, ..., x
n). The complexity L(f
1, ..., f
q) of set f
1, ..., f
qis a least r with property: there exists a computing sequence g
1, ..., g
rsuch, that for all i ≤ q
f
i∈ K + Kx
1+ ... + Kx
n+ Kg
1+ ...Kg
rLet E, W be finite-dimensional vector K - spaces with basis, accordingly e
1, ..., e
rand ˆ e
1, ...ˆ e
q. Definition 3 Mapping f : E → W is called by quadratic, if there exist quadratic forms f
1, ..., f
qof K[x
1, ..., x
n] such that for all ξ
1, ..., ξ
n∈ K
f(
n
X
i=1
ξ
ie
i) =
q
X
j=1
f
j(ξ
1, ..., ξ
n)ˆ e
jL(f) = L(f
1, ..., f
q) is called by complexity f, where f
1, ..., f
qare considered as elements K(x
1, ..., x
n).
Let A is a finite-dimensional algebra with unit, e
1, ..., e
nis a basis of vector space.
e
i∗ e
j=
n
X
m=1
τ
ijme
m, where τ
ijm∈ K, i, j, = 1, ..., n. Then we get
(
n
X
i=1
ξ
ie
i) ∗ (
n
X
j=1
η
je
j) =
n
X
m=1
(
n
X
i,j=1
τ
ijmξ
iη
j)e
mThe elements x, y ∈ A are considered as vector-columns with coordinates x
1, ..., x
nand y
1, ..., y
naccordingly. Then for P
ni,j=1
τ
ijmx
iy
jwe get the following:
n
X
i,j=1
τ
ijmx
iy
j=
x
1x
2.. . x
n
t
τ
11mτ
12mτ
1nmτ
21mτ
22mτ
2nmτ
n1mτ
n2mτ
nnm
y
1y
2.. . y
n
= x
tT
my,
where x
tis a vector-line, T
m∈ M
n(K ), m = 1, . . . , n. For any x, y ∈ A we get xy =
n
X
m=1
(x
tT
my)e
mLet T = {T
1, . . . , T
n}. We consider M ⊆ M
n(K) - subset, for which power |M | = r, r >
0, linM is a linear membrane M .
Definition 4 M is called by the algorithm of the length r of algebra A, if the conditions are held:
1) M consists of linearly independent matrixes, 2) for any m ∈ M, rang = 1,
3) T ≤ linM.
Algorithm M is called by optimal, if its length does not exceed of the length of any algorithm of algebra A.
Multiplication in algebra A is a bilinear mapping f : A ⊕ A → A. For any x, y ∈ A we get f(x + y) = xy =
n
X
m=1
x
tT
mye
m=
n
X
m=1
f
m(x
1, ..., x
n, y
1, ..., y
n)e
m=
n
X
m=1
f
m(x, y)e
m, where x
tT
my = f
m(x
1, ..., x
n, y
1, ..., y
n) = f
m(x, y) ∈ k[x
1, ..., x
n, y
1, ..., y
n]- are quadratic forms, m = 1, ..., n. In fact, f : A L
A → A is quadratic mapping.
Definition 5 Complexity of algebra A is called complexity of quadratic mapping f : A L
A → A, defined by the rule:
f(x + y) = xy.
We denote the complexity of algebra A by L(A).
Proposal 1 L(A) is equal to the optimal algorithm of algebra A.
Proof. If M is optimal algorithm of the length r of algebra A, then f
m(x
1, . . . , x
n, y
1, . . . , y
n) =
r
X
j=1
d
mjx
tM
jy,
where M
j∈ M, j = 1, . . . , r. Any matrix C of rang 1 is represented in the form
C =
a
1.. . a
n
(b
1, . . . , b
n).
Therefore for each matrix M
jthere exist such vector-lines u
j, v
jthat M
j= u
tjv
j; moreover x
tM
jy = x
tu
tjv
jy = u
j(x)v
j(y), j = 1, . . . , r,
where u
j(x) = (x
t, u
j), v
j(y) = (y
t, v
j) are linear forms.
As a result we get
f
m(x, y) =
r
X
j=1
d
mju
j(x)v
j(y) hence, that L(A) = L(f) ≤ r.
Let L(f ) = L(f
1, . . . , f
n) = k. In the work [1] is proved, that the set of quadratic forms is optimally calculated without division. There exists
u
ρ∈ K + Kx
1+ · · · + Kx
n, v
ρ∈ K + Ky
1+ · · · + Ky
n, ρ = 1, . . . , k, such that
f
m(x, y ) =
k
X
ρ=1
d
mρu
ρ(x)v
ρ(y).
If u
ρ, v
ρ, ρ = 1, . . . , k, are linear dependent, then the number k can be decreased, consequently u
ρ, v
ρare linear independent.
Let u
ρ(x) = (u
ρ, x
t), v
ρ(y) = (v
ρ, y
t), for some vectors u
ρ, v
ρ. Then the matrix M
ρ= u
tρv
ρhas a rang 1, and we get
u
ρ(x)v
ρ(y) = x
tM
ρy.
Then f
m(x, y) = P
kj=1
d
mjx
tM
jy = x
t( P
kj=1
d
mjM
j)y, on the other hand f
m(x, y) = x
tT
my, hence T
m∈ lin(M
1, . . . , M
k). We have constructed the algorithm, the length of which is equal to L(A) Now it is clear, that L(A) ≥ r and finally, we get, that L(A) = r.
Example 1 Let C are field of complex numbers. For any x, y ∈ C we get xy = (x
1y
1− x
2y
2) + (x
1y
2− x
2y
2)i,
i.e. xy = x
t1 0 0 −1
y + x
t1 0 0 −1
yi, For C T = n
T
1=
1 0 0 −1
, T
2=
0 1 1 0
o
. We take M =
n 1 0 0 0
,
0 0 0 1
,
1 1 1 1
o .
Then L(C) ≤ 3. We suppose, that M = {M
1, M
2} is an algorithm of the length 2 for C. Then there exist d
1, d
2, d
3, d
4∈ R such that d
1d
4− d
2d
36= 0 and
d
1M
1+ d
2M
2=
1 0 0 −1
, d
3M
1+ d
4M
2=
0 1 1 0
.
Let d
16= 0, then M
1=
d 11d4−d2d3
−d
3d
1d
1d
3and d
21+ d
23= 0, i.e. d
1= 0, contradiction, L(C) = 3.
3 Complexity of B(f)-jordan algebra of nondegenerated symmetric bilinear form B(f ) = K1 + V is jordan algebra [2] of symmetric nondegenerated bilinear form f : V × V → K, where K is a field, V is the vector space over K dimensionality n. Multiplication in B (f ) is defined:
(x
0∗ 1 + x)(y
0∗ 1 + y) = (x
0y
0+ f (x, y))(x
0y
0+ y
0x).
Theorem 1 a) If K is algebraic closed field, then L(B(f)) = 2n.
b) If K = < is a field of the real numbers, then L(B(f )) =
2n + 1, if f is negative def ined;
2n, in the other case.
Proof. By choosing the canonical basis l
1, . . . , l
nin V with respect to f for any a, b ∈ B (f ) we obtain, that
ab = (x
01 + x)(y
01 + y) = (x
0y
0+ x
1y
1+ · · · + x
ky
k−
−x
k+1y
k+1− · · · − x
ny
n)1 + (x
0y
1+ x
0y
1)l
1+ · · · + (x
0y
n+ x
ny
0)l
n,
where K = 0, 1, . . . , n for K = < and k = n, if the field K is algebraic closed. If K = 0, then we take
f
0= x
0y
0, f
i= r 1
n x
0+ x
ir 1
n y
0+ y
i, g
i=
r 1
n x
0+ x
ir 1
n y
0− y
i, i = 1, . . . , n.
Then
ab =
2f
0−
n
X
i=1
f
i+ g
i2
1 +
n
X
i=1
f
i− g
i2 l
i. If 0 < k < n, then we take
f
i= r
2
k x
0+ x
ir 2
k y
0+ y
i, g
i=
r 2
k x
0− x
ir 2
k y
0− y
i, i = 1, . . . , k,
f
i= r 2
n − k x
0+ x
ir 2
n − k y
0+ y
i, i = k + 1, . . . , n, g
i= r
2
n − k x
0− x
ir 2
n − k y
0− y
i, i = k + 1, . . . , n.
Then
ab = X
ki=1
f
i+ g
i2 −
n
X
i=k+1
f
i+ g
i2
1 +
k
X
i=1
r k 2
f
i− g
i2 l
i+
k
X
i=k+1
√ n − k f
i− g
i2 l
i. If k = n, then
f
i= r 1
n x
0+ x
ir 1
n y
0+ y
i, g
i=
r 1
n x
0− x
ir 1
n y
0− y
i, i = 1, . . . , n and we get
ab =
k
X
i=1
f
i+ g
i2 1 +
n
X
i=1
√ n f
i− g
i2 l
i. Thus, we obtain, that
L(B(f)) ≤ 2n, (1)
Under 0 < K < n or algebraic closed K,
L(B(f)) ≤ 2n + 1, K = 0. (2)
Now the elements a, b ∈ B (f) are considered as vector-columns.
Then we get
ab = a
tE
k+10 0 −E
n−kb · 1 +
n+1
X
i=2
a
t(E
1i+ E
2i)b · l
i−1=
= a
tC
0b +
n+1
X
i=2
a
tC
i−1b · l
i−1,
where C
0, C
1, . . . , C
n∈ M
n+1(K). Let M ⊆ M
n+1(K) is the algorithm of the length r for B(f ). Then we get
C
i=
r
X
j=0
d
ijx
j,
where i = 0, 1, . . . , n, x
j∈ M, d
ij∈ K. We consider a matrix of order n × r,
d
11· · · d
1n· · · d
1rd
21· · · d
2n· · · d
2r· · · · · · · · · · · · · · · d
n1· · · d
nn· · · d
nr
.
Since C
1, . . . , C
nare linear independent, that there exists a minor M
1of order n × n, that
|M
1| 6= 0. Without limitation of generality, we can assume, that
M
1=
d
11· · · d
1n· · · · · · · · · d
n1· · · d
nn
.
Then from the system of equations
d
11x
1+ · · · + d
1nx
n= C
1− P
rj=n+1
d
1jx
j= B
1d
n1x
2+ · · · + d
nnx
n= C
n− P
rj=n+1
d
njx
j= B
nWe find x
i;
x
i= 1
|M
i|
n
X
j=1
(−1)
i+jM
1ijB
j, i = 1, . . . , n.
where M
1ijis a minor of the element d
ijof the matrix M
1. By substituting x
i, i = 1, . . . , n to C
0= P
rj=1
d
0jx
iwe obtain C
0= 1
|M
1| (γ
1B
1− γ
2B
2+ · · · + (−1)
n+1γ
nB
n) +
r
X
j=n+1
d
0jx
j, where
γ
1=
d
01· · · d
0nd
21· · · d
2n· · · · · · · · · d
n1· · · d
nn, . . . , γ
n=
d
01· · · d
0nd
21· · · d
2n· · · · · · · · · d
n−11· · · d
n−1n.
Hence
∆ = C
0+
n
X
i=1
(−1)
iγ
i|M
1| C
i=
r
X
j=n+1
d
j|M
1| x
i, (3)
where
d
j=
d
01· · · d
0nd
0jd
21· · · d
2nd
1j· · · · · · · · · · · · d
n−11· · · d
n−1nd
nj, j = n + 1, ..., r. (4)
We assume δ
1=
(−1)|Miγi1|
, i = 1, . . . , n. Then in (3) for ∆ we get
∆ =
1 δ
1· · · δ
kδ
k+1· · · δ
nδ
11 · · · 0 0 · · · 0
· · · · · · · · · · · · · · · · · · · · · δ
k0 · · · 1 0 · · · 0 δ
k+10 · · · 0 −1 · · · 0
· · · · · · · · · · · · · · · · · · · · · δ
n0 · · · 0 0 · · · −1
. (5)
Let T
12(−δ
1), . . . , T
1k+1(−δ
k), T
1k+2(δ
k+1), . . . , T
1n+1(δ
n) ∈ M
n+1(K) be transfection. We assume
U = T
1n+1(δ
n) . . . T
12(−δ
1). (6)
Then
U ∆U
t=
1 − δ
12− · · · − δ
k2+ · · · + δ
n20 0
0 E
k0
0 0 −E
n−k
hence we get, that rang∆ ≥ n. It is shown, that in (3) r ≥ 2n. Together with (1) we obtain L(B(f)) = 2n,
if 0 < k < n or K is algebraic closed. If k = 0 and K = R is a field of the real numbers, then rang∆ = n + 1.
Consequently r ≥ 2n + 1. Together with (2) it leads, that L(B(f)) = 2n + 1.
The theorem is proved.
4 Complexity of the Keli-Dixon body
Let C be a Keli-Dixon body, 1, l
1, . . . , l
7is a basis C of the multiplication table: Let x, y ∈ C.
Table 1 – The multiplication table
0 l
1l
2l
3l
4l
5l
6l
7l
1-1 l
3−l
2l
5−l
4−l
7l
6l
2−l
3-1 l
1l
6l
7−l
4−l
5l
3l
2−l
1-1 l
7−l
6l
5−l
4l
4−l
5−l
6−l
7-1 l
1l
2l
3l
5l
4−l
7l
6−l
1-1 −l
3l
2l
6l
7l
4−l
5−l
2l
3-1 −l
1l
7−l
6l
5l
4−l
3−l
2l
1-1
Using the table, we compute the production x, y by its coordinates.
xy = (x
1y
1− x
2y
2− x
3y
3− x
4y
4− x
5y
5− x
6y
6− x
7y
7− x
8y
8) · 1 +(x
1y
2+ x
2y
1+ x
3y
4− x
4y
3− x
5y
6− x
6y
5− x
7y
8+ x
8y
7) · e
1+(x
1y
3− x
2y
4+ x
3y
1+ x
4y
2+ x
5y
7+ x
6y
8− x
7y
5− x
8y
6) · e
2+(x
1y
4+ x
2y
3− x
3y
2+ x
4y
1+ x
5y
8− x
6y
7+ x
7y
6− x
8y
6) · e
3+(x
1y
5− x
2y
6− x
3y
7− x
4y
8+ x
5y
1+ x
6y
2+ x
7y
3+ x
8y
4) · e
4+(x
1y
6+ x
2y
5− x
3y
8+ x
4y
7− x
5y
2+ x
6y
1− x
7y
4+ x
8y
3) · e
5+(x
1y
7+ x
2y
8+ x
3y
5− x
4y
6− x
5y
3+ x
6y
4+ x
7y
1− x
8y
2) · e
6+(x
1y
8− x
2y
7+ x
3y
6+ x
4y
5− x
5y
4− x
6y
3+ x
7y
2+ x
8y
1) · e
7= C
1· 1 + C
2· e
1+ C
3· e
2+ C
4· e
3+ C
5· e
4+ C
6· e
5+ C
7· e
6+ C
8· e
7. Now we assume
f
1= (x
1+ x
2+ x
3+ x
4+ x
5+ x
6+ x
7+ x
8)(y
1+ y
2+ y
3+ y
4+ y
5+ y
6+ y
7+ y
8), f
2= (x
1+ x
2+ x
3+ x
4− x
5− x
6− x
7− x
8)(y
1+ y
2+ y
3+ y
4− y
5− y
6− y
7− y
8), f
3= (x
1− x
2+ x
3− x
4+ x
5− x
6+ x
7− x
8)(y
1− y
2+ y
3− y
4+ y
5− y
6+ y
7− y
8), f
4= (x
1− x
2+ x
3− x
4− x
5+ x
6− x
7+ x
8)(y
1− y
2+ y
3− y
4− y
5+ y
6− y
7+ y
8), f
5= (x
1+ x
2− x
3− x
4+ x
5+ x
6− x
7− x
8)(y
1+ y
2− y
3− y
4+ y
5+ y
6− y
7− y
8), f
6= (x
1+ x
2− x
3− x
4− x
5− x
6+ x
7+ x
8)(y
1+ y
2− y
3− y
4− y
5− y
6+ y
7+ y
8), f
7= (x
1− x
2− x
3+ x
4+ x
5− x
6− x
7+ x
8)(y
1− y
2− y
3+ y
4+ y
5− y
6− y
7+ y
8), f
8= (x
1− x
2− x
3+ x
4− x
5+ x
6+ x
7− x
8)(y
1− y
2− y
3+ y
4− y
5+ y
6+ y
7− y
8),
f
9= x
1y
1, f
10= x
4y
3, f
11= x
6y
5, f
12= x
7y
8, f
13= x
2y
4, f
14= x
7y
5, f
15= x
8y
6, f
16= x
3y
2, f
17= x
6y
7, f
18= x
8y
5, f
19= x
2y
6, f
20= x
3y
7, f
21= x
4y
8,
f
22= x
3y
8, f
23= x
5y
2, f
24= x
7y
4,
f
25= x
4y
6, f
26= x
5y
3, f
27= x
8y
2, f
28= x
2y
7, f
29= x
6y
3, f
30= x
5y
4Then
C
1= 1
8 (16f
9− f
1− f
2− f
3− f
4− f
5− f
6− f
7− f
8), C
2= 1
8 (f
1+ f
2+ f
5+ f
6− f
3− f
4− f
7− f
8) − 2(f
10+ f
11+ f
12), C
3= 1
8 (f
1+ f
2+ f
3+ f
4− f
5− f
6− f
7− f
8) − 2(f
13+ f
14+ f
15), C
4= 1
8 (f
1+ f
2+ f
7+ f
8− f
3− f
4− f
5− f
6) − 2(f
16+ f
17+ f
18), C
5= 1
8 (f
1− f
4+ f
5− f
8− f
2+ f
3− f
6+ f
7) − 2(f
19+ f
20+ f
21), C
6= 1
8 (f
1+ f
4+ f
5+ f
8− f
2− f
3− f
6− f
7) − 2(f
22+ f
23+ f
24), C
7= 1
8 (f
1+ f
4+ f
5+ f
8− f
2− f
4− f
5− f
7) − 2(f
25+ f
26+ f
27), C
8= 1
8 (f
1+ f
4+ f
6+ f
7− f
2− f
3− f
5− f
8) − 2(f
28+ f
29+ f
30),
Hence we get, that L(C) ≤ 30. Let L(C) = r and the elements x, y ∈ C are considered as vector-columns, then C
i= x
tC
iy, i = 1, . . . , 8 where C
i∈ M
8(K) and there exists an algorithm M of the length r for which
C
i=
r
X
j=1
d
ijx
i, i = 1, . . . , 8,
where x
j∈ M ⊆ M
8(K), i = 1, . . . , r, d
ij∈ K. For Keli-Dixon body C the matrix ∆ has the form
∆ =
1 δ
1δ
2δ
3δ
4δ
5δ
6δ
7δ
1−1 δ
3−δ
2δ
5−δ
4−δ
7δ
6δ
2−δ
3−1 δ
1δ
6δ
7−δ
4−δ
5δ
3δ
2−δ
1−1 δ
7−δ
6δ
5−δ
4δ
4−δ
5−δ
6−δ
7−1 δ
4δ
2δ
3δ
5δ
4−δ
7δ
6−δ
1−1 −δ
3δ
2δ
6δ
7δ
4−δ
5−δ
2δ
3−1 −δ
1δ
7−δ
6δ
5δ
4−δ
3−δ
2δ
1−1
=
A
11A
12A
21A
22.
Then detA
11· det(A
22− A
21A
−111A
12) = d
41(1 + d
1d
2) 6= 0, where d
1= (1 + d
21+ d
22+ d
23+ d
24), d
2= d
24+ d
25+ d
26+ d
27. On the other hand ∆ = P
rj=8 dj
|M1|
x
j, consequently r − 7 ≥ 8, i.e.
L(C) ≥ 15. By this way it is proved the following theorem.
Theorem 2 C is a Keli-Dixon body. Then 15 ≤ L(C) ≤ 30.
5 Complexity of Li algebra
In this section we calculate the complexity of the least simple Li algebra sl(2, K) over field K of characteristics 6= 2. We mention, that sl(2, K) consists of 2 × 2 matrixes over K with zero trace: as its basis the elements can be taken l
1= l
11− l
22, l
2= l
12, l
3= l
21where l
ijare ordinary matrix units. For any x = P
3i=1
x
il
i= P
3i=1
y
il
iwe have
[x, y] = 2(x
1y
2− x
2y
1)l
2− 2(x
1y
3− x
3y
1)l
3+ (x
2y
3− x
2y
3)l
1=
= C
1l
1+ 2C
2l
2+ 2C
3l
3, where
C
1(x, y) = x
2y
3− x
3y
2, C
2(x, y) = x
1y
2− x
2y
1, C
3(x, y) = x
3y
1− x
1y
2. We assume
f
1= (x
1− x
2− x
3)(y
1− y
2+ y
3), f
2= (x
1+ x
2+ x
3)(y
1+ y
2− y
3), f
3= (x
1+ x
2− x
3)(y
1+ y
2+ y
3), f
4= (x
1− x
2+ x
3)(y
1− y
2− y
3),
f
5= x
1y
2. Then
C
1= − 1
4 (f
1+ f
2− f
3− f
4), C
3= 1
2 (f
2− f
1) + C
1C
2= 1
2 (f
2− f
1) − C
3− 2f
5.
Therefore
sl(2, K ) ≤ 5.
Let x = (x
1, x
2, x
3), y = (y
1, y
2, y
3); then we get
C
i= xC
iy
t, i = 1, 2, 3, where
C
1= l
23− l
32, C
2= l
12− l
21, C
3= l
31− l
13.
Let L(sl(2, K)) = r then there exists an algorithm of the length r for sl(2, K) i.e. there exist the matrixes X
1, . . . , X
rof M
3(K) rang 1 such that
C
i=
r
X
i=1
α
ijX
j, i = 1, 2, 3. (7)
The procedure is repeated for B(f), and we get
∆ = C
2+ δ
1C
3+ δ
2C
1=
r
X
j=2
α
j|M
1| X
j, (8)
where
|M
1| =
α
11α
12α
31α
32, α
j=
α
21α
22α
2jα
11α
12α
1jα
31α
32α
3j.j = 3, . . . , r.
Since det∆ = δ
1δ
2− δ
1δ
2= 0, then rang∆ = 2, i.e. r ≥ 4. Let r = 4, C
i= P
4j=1
