Uncountably many non-isomorphic nilpotent real n-Lie algebras
There are an uncountable number of non-isomorphic nilpotent real Lie algebras for every dimension greater than or equal to 7. We extend an old technique, which applies to Lie algebras of dimension greater than or equal to 10, to find corresponding results for n-Lie algebras. In particular, for n...
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irk-123456789-1573702019-06-21T01:25:25Z Uncountably many non-isomorphic nilpotent real n-Lie algebras Stitzinger, E. Williams, M.P. There are an uncountable number of non-isomorphic nilpotent real Lie algebras for every dimension greater than or equal to 7. We extend an old technique, which applies to Lie algebras of dimension greater than or equal to 10, to find corresponding results for n-Lie algebras. In particular, for n ≥ 6, there are an uncountable number of non-isomorphic nilpotent real n-Lie algebras of dimension n + 4. 2006 Article Uncountably many non-isomorphic nilpotent real n-Lie algebras / E. Stitzinger, M.P. Williams // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 1. — С. 81–88. — Бібліогр.: 5 назв. — англ. 1726-3255 2000 Mathematics Subject Classification: 17A42. http://dspace.nbuv.gov.ua/handle/123456789/157370 en Algebra and Discrete Mathematics Інститут прикладної математики і механіки НАН України |
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There are an uncountable number of non-isomorphic nilpotent real Lie algebras for every dimension greater
than or equal to 7. We extend an old technique, which applies
to Lie algebras of dimension greater than or equal to 10, to find
corresponding results for n-Lie algebras. In particular, for n ≥ 6,
there are an uncountable number of non-isomorphic nilpotent real
n-Lie algebras of dimension n + 4. |
| format |
Article |
| author |
Stitzinger, E. Williams, M.P. |
| spellingShingle |
Stitzinger, E. Williams, M.P. Uncountably many non-isomorphic nilpotent real n-Lie algebras Algebra and Discrete Mathematics |
| author_facet |
Stitzinger, E. Williams, M.P. |
| author_sort |
Stitzinger, E. |
| title |
Uncountably many non-isomorphic nilpotent real n-Lie algebras |
| title_short |
Uncountably many non-isomorphic nilpotent real n-Lie algebras |
| title_full |
Uncountably many non-isomorphic nilpotent real n-Lie algebras |
| title_fullStr |
Uncountably many non-isomorphic nilpotent real n-Lie algebras |
| title_full_unstemmed |
Uncountably many non-isomorphic nilpotent real n-Lie algebras |
| title_sort |
uncountably many non-isomorphic nilpotent real n-lie algebras |
| publisher |
Інститут прикладної математики і механіки НАН України |
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2006 |
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http://dspace.nbuv.gov.ua/handle/123456789/157370 |
| citation_txt |
Uncountably many non-isomorphic nilpotent real n-Lie algebras / E. Stitzinger, M.P. Williams // Algebra and Discrete Mathematics. — 2006. — Vol. 5, № 1. — С. 81–88. — Бібліогр.: 5 назв. — англ. |
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Algebra and Discrete Mathematics |
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AT stitzingere uncountablymanynonisomorphicnilpotentrealnliealgebras AT williamsmp uncountablymanynonisomorphicnilpotentrealnliealgebras |
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2025-07-14T09:48:40Z |
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Algebra and Discrete Mathematics RESEARCH ARTICLE
Number 1. (2006). pp. 81 – 88
c© Journal “Algebra and Discrete Mathematics”
Uncountably many non-isomorphic nilpotent real
n-Lie algebras
Ernest Stitzinger and Michael P. Williams
Communicated by V. M. Futorny
Abstract. There are an uncountable number of non-iso-
morphic nilpotent real Lie algebras for every dimension greater
than or equal to 7. We extend an old technique, which applies
to Lie algebras of dimension greater than or equal to 10, to find
corresponding results for n-Lie algebras. In particular, for n ≥ 6,
there are an uncountable number of non-isomorphic nilpotent real
n-Lie algebras of dimension n + 4.
Classifying nilpotent real Lie algebras has been an often studied sub-
ject since Engel. In 1962, Chao [1] proved that there are uncount-
ably many such Lie algebras of dimension 10 and greater that are non-
isomorphic. We shall prove an n-Lie algebra analogue of this theorem.
Before we proceed we recall the identities of n-Lie algebras as intro-
duced by Fillipov [2]. An n-Lie algebra, is an algebra equiped with an
n-linear, skew-symmetric bracket with the identity
[[x1, x2, . . . , xn]y2 . . . , yn] =
n
∑
i=1
[x1, . . . , xi−1, [xi, y2, . . . , yn], xi+1, . . . , xn]
which we call the n-Jacobi identity. For further resuts, see [2], [3] and [4].
Theorem 1. There are uncountably many non-isomorphic n-Lie algebras
of dimensions d and nilpotent of length 2 when
1) n = 2 and d = 10.
2) n = 3 and d = 10.
2000 Mathematics Subject Classification: 17A42.
Key words and phrases: n-Lie algebras, nilpotent, algebraically independent,
transcendence degree.
82 Uncountably many non-isomorphic...
3) n = 4 and d = 9.
4) n = 5 and d = 10.
5) n ≥ 6 and d = n + 4.
Definition 2. Let F be a subfield of R. An n-Lie algebra A over R is
said to be an F-algebra if its structure constants with respect to some basis
of A lie in F.
Let F be a subfield of R and Ck
i1,i2,...,in
be real numbers in F such
that σ(Ck
i1,i2,...,in
) = Ck
iσ(1),iσ(2),...,iσ(n)
= sgnσ(Ck
i1,i2,...,in
) for all σ ∈ Sn,
the symmetric group. Let A be an n-Lie algebra over F with a basis
(x1, x2, . . . , xℓ, y1, y2, . . . , ym) where ℓ ≥ n and multiplication given by
[xi1 , xi2 , . . . , xin ] =
∑n
k=1
Ck
i1,i2,...,in
yk and all other products 0. Note
that this fits the anti-symmetric condition as,
σ([xi1 , xi2 , . . . , xin ]) = [xiσ(1)
, xiσ(2)
, . . . , xiσ(n)
]
= σ(
n
∑
k=1
Ck
i1,i2,...,in
yk)
=
n
∑
k=1
σ(Ck
i1,i2,...,in
)yk
=
n
∑
k=1
sgnσ(Ck
i1,i2,...,in
)yk
= sgnσ
( n
∑
k=1
Ck
i1,i2,...,in
yk
)
= sgnσ
(
[xi1 , xi2 , . . . , xin ]
)
.
Lemma 3. If the numbers Ck
i1,i2,...,in
, for 1 ≤ i1 < i2 < . . . < in ≤ ℓ, and
1 ≤ k ≤ m are algebraically independent over F and if
(
ℓ
n
)
m > m2 + ℓ2,
then A is not an F-algebra.
Proof of Lemma 3
We want to show that A2 =< y1, . . . , ym >= Z(A). First we show that
A2 =< y1, . . . , ym >. Since
(
ℓ
n
)
> m we can pick m distinct sets of n
integers between 1 and ℓ. Each such set determines a set of vectors from
x1, . . . , xℓ and we label these sets Sj , j = 1, . . . , m. Let zj be the product
of the elements in Sj where the indices are arranged in increasing order
in the product. As a result zj =
∑m
k=1
Ck
j yk for j = 1, . . . , m where Ck
j =
Ck
j1,j2,...,jn
if xj1 , xj2 , . . . , xjn
∈ Sj . The polynomial det(xij) for i, j =
1, . . . , m has integer coefficients and thus lies in F[x11, . . . , xkj , . . . , xmm].
If det(Ck
j ) = 0, then the Ck
j ’s are not algebraically independent, which is
E. Stitzinger, M. P. Williams 83
a contradiction. Therefore (Ck
j ) is a non-singular matrix which generates
< y1, . . . , ym > and hence A2 =< y1, . . . , ym >.
Now we show that Z(A) =< y1, . . . , ym >. Since the only non-zero
products in A are products of the xi’s it is clear that < y1, . . . , ym >⊂
Z(A).
Let z =
∑ℓ
j=1
ajxj +
∑m
j=1
bkyk ∈ Z(A) and let Rπ =
[_, xi2 , xi3 , . . . , xin ] then,
0 = zRπ
=
( ℓ
∑
j=1
ajxj
)
Rπ +
( m
∑
k=1
bkyk
)
Rπ
=
ℓ
∑
j=1
aj(xjRπ) + 0
=
ℓ
∑
j=1
m
∑
k=1
ajC
k
jπyk
where Ck
jπ = Ck
ji2i3...in
.
By virtue of the linear independence of the yk’s we obtain
∑ℓ
j=1
aiC
k
jπ = 0. For each 1 ≤ t ≤ ℓ choose πt = t2, . . . , tn such
that tr 6= ts for r 6= s and t 6= t2, . . . , tn. Then
∑ℓ
j=1
aiC
k
jπt
= 0.
We observe that Ck
tπt
6= 0 and Ck
tπt
is in the algebraically indepen-
dent set. Repeating this process for each t, 1 ≤ t ≤ ℓ gives us a
system of ℓ equations and ℓ unknowns. The coefficient matrix C has
non-zero elements on the diagonal and hence are algebraically indepen-
dent. Considering det(xij) as in the last paragraph, gives us a poly-
nomial in ℓ2 variables with coefficients +1. If C is singular then the
elements of C satisfy det(xij). The non-zero elements of C satisfy a
polynomial obtained from det(xij) by deleting terms if necessary, from
any elements of C that are 0. The resulting polynomial is non-zero be-
cause of the non-zero diagonal of C. This non-zero polynomial is satisfied
by a set of algebraically independent elements. This is a contradiction
and hence C is non-singular. As a result a1 = a2 = . . . = aℓ = 0
and z =
∑ℓ
j=1
ajxj +
∑m
j=1
bkyk =
∑m
j=1
bkyk ∈< y1, . . . , ym >. Thus
Z(A) =< y1, . . . , ym >= A2.
Now we prove lemma 3. Suppose, to the contrary, that A satisfies
the conditions of the lemma. Namely, A is an F-algebra with basis
(z1, . . . , zℓ, zℓ+1, . . . , zℓ+m) and structure constants Dk
i1,i2,...,in
’s for 1 ≤
i1, i2, . . . , in ≤ ℓ and 1 ≤ k ≤ ℓ + m. We can assume without loss of
generality that (z1, . . . , zℓ) form a basis for C a compliment of A2. We
84 Uncountably many non-isomorphic...
can write zℓ+i = vi+ti for all i where vi ∈ C and ti ∈ A2 for i = 1, . . . , m.
We observe
[zi1 , zi2 , . . . , zin ] =
=
ℓ
∑
r=1
Dr
i1,i2,...,in
zr +
ℓ+m
∑
s=ℓ+1
Ds
i1,i2,...,in
vs−ℓ +
ℓ+m
∑
s=ℓ+1
Ds
i1,i2,...,in
ts−ℓ
Since [zi1 , zi2 , . . . , zin ] ∈ A2 we see the first two summands must be 0 and
we obtain
[zi1 , zi2 , . . . , zin ] =
ℓ+m
∑
s=ℓ+1
Ds
i1,i2,...,in
ts−ℓ
=
m
∑
u=1
Du+ℓ
i1,i2,...,in
tu.
As a result (z1, . . . , zℓ, tℓ+1, . . . , tℓ+m) is a new basis for A whose structure
coefficients are a subset of the structure coefficients for the old basis. We
observe that (x1, . . . , xℓ) is a basis for C ′ a compliment of A2. Now let si
be such that si−zi ∈ A2 and si ∈ C ′ for 1 ≤ i ≤ ℓ. We obtain yet another
basis (s1, . . . , sℓ, t1, . . . , tm) which has the same structure coefficients as
(z1, . . . , zℓ, tℓ+1, . . . , tℓ+m).
Indeed,
[si1 , si2 , . . . , sin ] = [zi1 + A2, zi2 + A2, . . . , sin + A2]
= [zi1 + Z(A), zi2 + Z(A), . . . , zin + Z(A)]
= [zi1 , zi2 , . . . , zin ].
Since (s1, . . . , sℓ) and (x1, . . . , xℓ) both form a basis for C ′ there exists
a non-singular matrix, B = (bip) such that si =
∑ℓ
p=1
bipxp for all 1 ≤
i ≤ ℓ. Likewise there exists a non-singular G = (gur) such that tu =
∑m
r=1
guryr for all 1 ≤ u ≤ m. Substituting into
[zi1 , . . . , zin ] = [si1 , . . . , sin ] =
m
∑
u=1
Dℓ+u
i1,...,in
tu
E. Stitzinger, M. P. Williams 85
we observe for all 1 ≤ i1, i2, . . . , in ≤ ℓ
[si1 , si2 , . . . , sin ] =
[ ℓ
∑
p1=1
bi1p1xp1 ,
ℓ
∑
p2=1
bi2p2xp2 , . . . ,
ℓ
∑
pn=1
binpn
xpn
]
=
ℓ
∑
p1=1
ℓ
∑
p2=1
, . . . ,
ℓ
∑
pn=1
(
bi1p1bi2p2 , . . . , binpn
[xp1 , xp2 , . . . , xpn
]
)
=
ℓ
∑
p1=1
ℓ
∑
p2=1
, . . . ,
ℓ
∑
pn=1
(
bi1p1bi2p2 , . . . , binpn
m
∑
r=1
Cr
p1,p2,...,pn
yr
)
=
m
∑
u=1
Dℓ+u
i1,i2,...,in
tu
=
m
∑
u=1
m
∑
r=1
Dℓ+u
i1,i2,...,in
guryr.
This implies that for fixed i1, i2, . . . , in and r we obtain
ℓ
∑
p1=1
ℓ
∑
p2=1
. . . ,
ℓ
∑
pn=1
bi1p1bi2p2 , . . . , binpn
Cr
p1,p2,...,pn
=
m
∑
u=1
Dℓ+u
i1,i2,...,in
gur.
We claim that this in turn implies that,
Cr
p1,p2,...,pn
=
ℓ
∑
p1=1
ℓ
∑
p2=1
. . . ,
ℓ
∑
pn=1
m
∑
u=1
Dℓ+u
i1,i2,...,in
gurbp1i1bp2i2 , . . . , bpnin
where B−1 = [bip].
We show the tth step. Suppose for
1 ≤ p1, p2 . . . pt−1 ≤ ℓ and 1 ≤ it, it+1 . . . in ≤ ℓ and r fixed that
ℓ
∑
pt=1
ℓ
∑
pt+1=1
, . . . ,
ℓ
∑
pn=1
bitpt
bit+1pt+1 , . . . , binpn
Cr
p1,p2,...,pn
=
ℓ
∑
i1=1
ℓ
∑
i2=1
. . . ,
ℓ
∑
it−1=1
bp1i1bp2i2 . . . bpt−1it−1
m
∑
u=1
Dr
i1i2...in
gur.
Let
Apt
=
ℓ
∑
pt+1=1
ℓ
∑
pt+2=1
. . . ,
ℓ
∑
pn=1
bit+1pt+1bit+2pt+2 , . . . , binpn
Cr
p1,p2,...,pn
86 Uncountably many non-isomorphic...
for pt = 1, . . . , ℓ and
Eit =
ℓ
∑
i1=1
ℓ
∑
i2=1
. . . ,
ℓ
∑
it−1=1
bp1i1bp2i2 . . . bpt−1it−1D
r
i1i2...in
gur
for it = 1, 2, . . . , ℓ.
This implies that
bit1A1 + bit2A2 + . . . + bitℓAℓ = Eit
or
b11 b12 . . . b1ℓ
b21 b22 . . . b2ℓ
...
...
. . .
...
bℓ1 bℓ2 . . . bℓℓ
A1
A2
...
Aℓ
=
E1
E2
...
Eℓ.
So
Apt
= bpt1E1 + bpt2E2 + . . . + bptnEn
and
Apt
=
ℓ
∑
i1=1
ℓ
∑
i2=1
. . . ,
ℓ
∑
it=1
bp1i1bp2i2 . . . bptitD
r
i1i2...in
.
Finally,
ℓ
∑
pt+1=1
ℓ
∑
pt+2=1
. . . ,
ℓ
∑
pn=1
bit+1pt+1bit+2pt+2 , . . . , binpn
Cr
p1,p2,...,pn
=
ℓ
∑
i1=1
ℓ
∑
i2=1
. . . ,
ℓ
∑
it=1
bp1i1bp2i2 . . . bptitD
r
i1i2...in
.
This proves the claim.
The claim implies that Cr
p1,p2,...,pn
∈ E = F(bip, gur). But the degree
of transcendence of E over F is at most ℓ2 + m2 which is less than,
(
ℓ
n
)
m,
the number of Cr
p1,p2,...,pn
’s. This a contradiction and hence A is not an
F-algebra, proving the lemma.
Proof of Theorem 1
It is known that there exists a set S of uncountably many real numbers
E. Stitzinger, M. P. Williams 87
that are algebraically independent over Q. We can divide S into un-
countably many disjoint subsets {Ck
i1,i2,...,in
}α of size
(
ℓ
n
)
m where α dis-
tinguishes subsets. Define the n-Lie algebra Aα with basis (x1, x2, . . . , xℓ,
y1, y2, . . . , ym) and multiplication given by
[xi1 , xi2 , . . . , xin ] =
n
∑
k=1
Ck
i1,i2,...,in
yk
and all other products 0 where Ck
i1,i2,...,in
∈ {Ck
i1,i2,...,in
}α for all 1 ≤
i1, i2, . . . , in ≤ ℓ and 1 ≤ k ≤ m. For α 6= β we claim that Aα and
Aβ are non-isomorphic. Indeed, since the (Ck
i1,i2,...,in
)α’s are algebraically
independent over Q[{Ck
i1,i2,...,in
}β ], if we apply lemma 3 to Aα, we see that
it is not a Q[(Ck
i1,i2,...,in
)β ]-algebra. Hence Aα and Aβ are non-isomorphic
as claimed.
To prove the theorem it remains to find for each given n and d in 1-4,
an m and k where d = n + m + k such that f(k, m, n) =
(
n+k
n
)
m − (n +
k)2 − m2 > 0. We do this case by case.
1) When k = m = 4, we obtain f(4, 4, n) = 1/6n4 + 5/3n3 + 29/6n2 +
1/3n − 28. The only positive root is approximately n = 1.807126451.
Hence f(4, 4, n) > 0 if n ≥ 2. Setting n = 2 gives d = n + m + k = 10.
This coincides with Chao’s result.
2) When k +m = 3+4 = 7, we obtain f(3, 4, n) = 2/3n3 +3n2 +4/3n−
21. The only positive root is approximately n = 2.046172397. Hence
f(3, 4, n) > 0 if n ≥ 3. Hence f(3, 4, n) > 0 if n ≥ 3. Setting n = 3 gives
d = n + m + k = 10.
3) When k = 3, m = 2, we obtain f(3, 2, n) = 1/3n3 + n2 − 7/3n − 11.
The only positive root is n = 3. Hence f(3, 2, n) > 0 if n ≥ 4. Setting
n = 4 gives d = 9 and setting n = 5 gives d = 10.
4) When k = 3, m = 1, we obtain f(3, 1, n) = 1/6n3−25/6n−9. The only
positive root is approximately n = 5.850622760. Hence f(3, 1, n) > 0 if
n ≥ 6. Thus d = n + m + k = n + 4.
Note that if n ≥ 6, then d − n cannot be less than 4. That is to say,
we have found the minimal k + m such that f(k, m, n) > 0. If we set
k + m < 4, we get no solutions. Indeed, if k = 0 or m = 0, we obtain
f(k, m, n) = m − m2 − (n)2 ≤ 0 and f(k, m, n) = −(n + k)2 ≤ 0. For
m + k = 1 + 1 = 2 we obtain f(k, m, n) = −n − n2 − 1 which has no
real roots. For m + k = 2 + 1 = 3 and m + k = 1 + 2 = 3 we obtain
f(k, m, n) = −1/2n2 − 5/2n − 4 and f(k, m, n) = −3 − n2 neither of
which have real roots and are always negative.
88 Uncountably many non-isomorphic...
References
[1] Chao, Chong-Yun. Uncountably many nonisomorphic nilpotent Lie algebras. Proc.
Amer. Math. Soc. 13 1962 903–906.
[2] Filippov, V. T. n-Lie algebras. Sibirsk. Mat. Zh. 26 (1985), no. 6, 126–140, 191.
(in Russian)
[3] Kasymov, Sh. M. On a theory of n-Lie algebras. Algebra i Logika 26 (1987), no.
3, 277–297, 398. (in Russian)
[4] Kasymov, Sh. M. Nil-elements and nil-subsets in n-Lie algebras. Sibirsk. Mat. Zh.
32 (1991), no. 6, 77–80, 204 (in Russian); translation in Siberian Math. J. 32
(1991), no. 6, 962–964 (1992)
[5] Williams, Michael P. Nilpotent n-Lie algebras, in preparation.
Contact information
E. Stitzinger,
M. P. Williams
North Carolina State University, Box 8205,
Raleigh, NC 27695
E-Mail: stitz@math.ncsu.edu,
skew1823@yahoo.com
|