On the invariants of root subgroups of finite classical groups
We show that invariant fields Fq(X1, . . . , Xn)G are purely transcendental over Fq if G are root subgroups of finite classical groups. The key step is to find good similar groups of our groups. Moreover, the invariant rings of the root subgroups of special linear groups are shown to be polynomial...
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nasplib_isofts_kiev_ua-123456789-1657192025-02-09T13:53:39Z On the invariants of root subgroups of finite classical groups Про інваріанти кореневих підгруп скінченних класичних груп Jizhu Nan Yufang Qin Статті We show that invariant fields Fq(X1, . . . , Xn)G are purely transcendental over Fq if G are root subgroups of finite classical groups. The key step is to find good similar groups of our groups. Moreover, the invariant rings of the root subgroups of special linear groups are shown to be polynomial rings, and their corresponding Poincare series are presented. Показано, що iнварiантнi поля Fq(X1, . . . , Xn)G є чисто трансцендентними над Fq, якщо G — кореневi пiдгрупи скiнченних класичних груп. Ключовим мiсцем доведення є знаходження гарних подiбних груп для наших груп. Крiм того, показано, що iнварiантнi кiльця кореневих пiдгруп спецiальних лiнiйних груп є полiномiальними кiльцями. Також наведено вiдповiднi ряди Пуанкаре. Supported by the National Natural Science Foundation of China. 2013 Article On the invariants of root subgroups of finite classical groups / Jizhu Nan, Yufang Qin // Український математичний журнал. — 2013. — Т. 65, № 12. — С. 1636–1645. — Бібліогр.: 15 назв. — англ. 1027-3190 https://nasplib.isofts.kiev.ua/handle/123456789/165719 512.5 en Український математичний журнал application/pdf Інститут математики НАН України |
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Статті Статті Jizhu Nan Yufang Qin On the invariants of root subgroups of finite classical groups Український математичний журнал |
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We show that invariant fields Fq(X1, . . . , Xn)G are purely transcendental over Fq if G are root subgroups of finite classical
groups. The key step is to find good similar groups of our groups. Moreover, the invariant rings of the root subgroups of
special linear groups are shown to be polynomial rings, and their corresponding Poincare series are presented. |
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Jizhu Nan Yufang Qin |
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Jizhu Nan Yufang Qin |
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Jizhu Nan |
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On the invariants of root subgroups of finite classical groups |
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On the invariants of root subgroups of finite classical groups |
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On the invariants of root subgroups of finite classical groups |
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On the invariants of root subgroups of finite classical groups |
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On the invariants of root subgroups of finite classical groups |
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on the invariants of root subgroups of finite classical groups |
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Інститут математики НАН України |
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On the invariants of root subgroups of finite classical groups / Jizhu Nan, Yufang Qin // Український математичний журнал. — 2013. — Т. 65, № 12. — С. 1636–1645. — Бібліогр.: 15 назв. — англ. |
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AT jizhunan ontheinvariantsofrootsubgroupsoffiniteclassicalgroups AT yufangqin ontheinvariantsofrootsubgroupsoffiniteclassicalgroups AT jizhunan proínvaríantikorenevihpídgrupskínčennihklasičnihgrup AT yufangqin proínvaríantikorenevihpídgrupskínčennihklasičnihgrup |
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2025-11-26T13:26:56Z |
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UDC 512.5
Jizhu Nan, Yufang Qin (Dalian Univ. Technology, China)
ON INVARIANTS OF ROOT SUBGROUPS
OF FINITE CLASSICAL GROUPS*
ПРО IНВАРIАНТИ КОРЕНЕВИХ ПIДГРУП
СКIНЧЕННИХ КЛАСИЧНИХ ГРУП
We show that invariant fields Fq(X1, . . . , Xn)
G are purely transcendental over Fq if G are root subgroups of finite classical
groups. The key step is to find good similar groups of our groups. Moreover, the invariant rings of the root subgroups of
special linear groups are shown to be polynomial rings, and their corresponding Poincaré series are presented.
Показано, що iнварiантнi поля Fq(X1, . . . , Xn)
G є чисто трансцендентними над Fq, якщо G — кореневi пiдгрупи
скiнченних класичних груп. Ключовим мiсцем доведення є знаходження гарних подiбних груп для наших груп. Крiм
того, показано, що iнварiантнi кiльця кореневих пiдгруп спецiальних лiнiйних груп є полiномiальними кiльцями.
Також наведено вiдповiднi ряди Пуанкаре.
1. Introduction. Let Fq be a finite field with charFq = p, and GL(n, Fq) be the general linear
group. For any T = (tij) ∈ GL(n, Fq), it induces an Fq-linear action σT on the rational function
field Fq(X1, . . . , Xn) defined by
σT (f(X1, . . . , Xn)) = f(σT (X1), . . . , σT (Xn))
for all f(X1, . . . , Xn) ∈ Fq(X1, . . . , Xn), where
σT (Xi) = ti1X1 + ti2X2 + . . .+ tinXn, i = 1, 2, . . . , n.
For a group G ≤ GL(n, Fq), Noether’s problem asks whether the rational invariant field
Fq(X1, . . . , Xn)G =
{
f ∈ Fq(X1, . . . , Xn) : σT (f) = f for all T ∈ G
}
is purely transcendental over Fq.
When G = GL(n, Fq), Dickson [1] gave an affirmative answer by giving the explicit transcen-
dental bases. Chu [6] considered the invariant fields of finite orthogonal groups and obtained similar
results for n = 2, 3. Cohen [7] showed the result is true when n = 4, and finally the general case
was settled by Carlisle and Kropholler [8]. But they all assumed that the characteristic of Fq is odd.
The case of characteristic two was settled by Rajaei [12] using quadratic form language and by Tang
and Wan [14] using matrix methods. Relatively recently, Chu [10] gave a unified treat on finding the
transcendental bases of the invariant fields of some finite classical groups of the form
GAρ =
{
Q ∈ GL(n, Fq) : Q′AQρ = A
}
,
where A ∈ GL(n, Fq) and ρ is an automorphism of Fq.
In the paper, we consider the root subgroups of finite classical groups by giving explicit transcen-
dental bases. The key of our method is to find good similar groups of root subgroups and consequently
obtain the explicit transcendental bases through studying the similar groups.
* Supported by the National Natural Science Foundation of China.
c© JIZHU NAN, YUFANG QIN, 2013
1636 ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 12
ON INVARIANTS OF ROOT SUBGROUPS OF FINITE CLASSICAL GROUPS 1637
Based on our results on the invariant fields of root subgroups of the special linear groups, we
show that the invariant rings of root subgroups of the special linear groups are polynomial rings
and consequently derive the Poincaré series of these invariant rings. In the modular case, examples
of groups whose invariant rings are polynomial rings are, to name a few, the general and special
linear groups GL(n, Fq) and SL(n, Fq) [1], the group of unipotent upper triangular matrices G ≤
≤ GL(n, Fq) [5], the orthogonal and unitary groups O(n,K, S) and U(n,K,H) for n ≤ 3 and
n ≤ 2, respectively [4], and the complex reflection groups G29 and G31 of Shephard and Todd [2].
Also, the root subgroup of the special linear group is such an example.
Let us recall the definitions of the root subgroups of classical groups [11]. In these definitions,
denote by K an arbitrary field.
The root subgroup of the special linear group SL(n,K) is the subgroup X̃ij = {Tij(c) : c ∈ K}
(i 6= j) or its conjugate subgroup in GL(n,K), where Tij(c) = I+cEij and Eij is the (n×n)-matrix
with the (i, j)-entry 1 and other entries 0. We denote the root subgroup P−1X̃ijP of SL(n,K) by
Xij,P with P ∈ GL(n,K).
Assume that K has an involutive automorphism φ : a 7→ ā. The unitary group U(n,K,H) is
defined to be the group {A ∈ GL(n,K) : AHĀ′ = H}, where
H =
0 I(ν) 0
−I(ν) 0 0
0 0 H0
is the congruence normal form of the nonsingular Hermitian matrix and H0 ∈ GL(n − 2ν,K) is a
definite diagonal matrix.
The long root subgroup of U(n,K,H) is the subgroup
Tu = {I +Hū′su : s ∈ TrK},
where u is a fixed n-dimensional row vector satisfying uHū′ = H and TrK = {a + ā : a ∈ K}.
And the short root subgroup of U(n,K,H) is
Tu,w =
{
I +Hw̄′au+H(au)′w : a ∈ K},
where u,w are noncollinear n-dimensional row vectors satisfying uHū′ = wHw̄′ = uHw̄′ = 0.
Let charK 6= 2. The orthogonal group O(n,K, S) is defined to be the group {A ∈ GL(n,K) :
ASA′ = S}, where
S =
0 I(ν) 0
I(ν) 0 0
0 0 ∆
is the congruence normal form of the nonsingular symmetric matrix and ∆ ∈ GL(n − 2ν,K) is a
definite symmetric matrix. The long root subgroup of O(n,K, S) is
Yu,w = {I + Sw′au− Sau′w : a ∈ K},
where u, w are noncollinear n-dimensional row vectors satisfying uSu′ = wSw′ = uSw′ = 0. The
element of Yu,w is also called an orthogonal 2-transvection. The short root subgroup of O(n,K, S)
is
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 12
1638 JIZHU NAN, YUFANG QIN
Ŷu,w = {I + Sw′au− Sau′(w +Q(w)au) : a ∈ K},
where u,w are noncollinear n-dimensional row vectors satisfying uSu′ = uSw′ = 0 and wSw′ =
= 2Q(w) 6= 0, Q is a quadratic form on the n-dimensional vector space over K. For the definitions of
the long root subgroups and short root subgroups of symplectic groups the reader is referred to [11].
2. Invariant fields of root subgroups of the special linear groups over finite fields. In this
section, we give the explicit transcendental bases of invariant fields of root subgroups of the special
linear groups over finite fields. First of all, we prove a lemma which is very useful in the paper.
Lemma 2.1. Let K[X1, . . . , Xn] be the polynomial ring over an arbitrary field K. Assume that
f1, . . . , fn ∈ K[X1, . . . , Xn] are algebraically independent over K, then for any P = (pij) ∈
∈ GL(n,K), σP (f1), . . . , σP (fn) ∈ K[X1, . . . , Xn] are algebraically independent over K.
Proof. Since for any P ∈ GL(n,K),
det
∂σP (f1(X1, . . . , Xn))
∂X1
. . .
∂σP (f1(X1, . . . , Xn))
∂Xn
...
...
∂σP (fn(X1, . . . , Xn))
∂X1
. . .
∂σP (fn(X1, . . . , Xn))
∂Xn
=
= det
∂f1
(
n∑
i=1
p1iXi, . . . ,
n∑
i=1
pniXi
)
∂X1
. . .
∂f1
(
n∑
i=1
p1iXi, . . . ,
n∑
i=1
pniXi
)
∂Xn
...
...
∂fn
(
n∑
i=1
p1iXi, . . . ,
n∑
i=1
pniXi
)
∂X1
. . .
∂fn
(
n∑
i=1
p1iXi, . . . ,
n∑
i=1
pniXi
)
∂Xn
=
= det
n∑
i=1
pi1f1i . . .
n∑
i=1
pinf1i
...
...
n∑
i=1
pi1fni . . .
n∑
i=1
pinfni
= det
f11 . . . f1n
...
...
fn1 . . . fnn
det(P ) =
= det
∂f1(X1, . . . , Xn)
∂X1
. . .
∂f1(X1, . . . , Xn)
∂Xn
...
...
∂fn(X1, . . . , Xn)
∂X1
. . .
∂fn(X1, . . . , Xn)
∂Xn
det(P ) 6= 0,
where fij =
∂fi(Y1, . . . , Yn)
∂Yj
if we let Yi =
∑n
l=1
pilXl. It follows that σP (f1), . . . , σP (fn) ∈
∈ K[X1, . . . , Xn] are algebraically independent over K.
Lemma 2.1 is proved.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 12
ON INVARIANTS OF ROOT SUBGROUPS OF FINITE CLASSICAL GROUPS 1639
Now we pay attention to discussing the root subgroup Xi0j0,P of the special linear group. For
convenience, we set i0 > j0.
Theorem 2.1. Let fi = Xi for 1 ≤ i ≤ n, i 6= i0, fi0 = Xi0X
q−1
j0
−Xq
i0
and gi = σP−1(fi) for
1 ≤ i ≤ n. Then we have Fq(X1, . . . , Xn)Xi0j0,P = Fq(g1, g2, . . . , gn).
Proof. It is clear that
det
(
∂fi
∂Xj
)
1≤i,j≤n
6= 0,
whence g1, g2, . . . , gn are algebraically independent over Fq by Lemma 2.1.
Suppose that Fq(X1, . . . , Xn) is Galois over Fq(f1, . . . , fn) with Galois group G1. We claim
that G1 = X̃i0j0 . The inclusion X̃i0j0 ⊂ G1 is trivial. Conversely, for any Q = (qij) ∈ G1, the
assumption σQ leaves f1 invariant implies that q11 = 1 and q12 = . . . = q1n = 0. Similarly, we
know that qii = 1 and qij = 0, for 1 ≤ i 6= j ≤ n and i 6= i0. Furthermore, σQ leaves fi0 invariant,
i.e.,
(∑n
k=1
qi0kXk
)(∑n
k=1
qj0kXk
)q−1
−
∑n
k=1
qi0kX
q
k = Xi0X
q−1
j0
− Xq
i0
, which shows that
qi0i0 = 1, qi0j0 is arbitrary and qi0j = 0 for j 6= i0, j0. Hence G1 ⊂ X̃i0j0 , and this proves the claim.
Let Fq(X1, . . . , Xn) be Galois over Fq(g1, . . . , gn) with Galois group G2. To prove the theorem,
it suffices to prove that Xi0j0,P = G2. For any Q = P−1TP ∈ Xi0j0,P with T ∈ X̃i0j0 , the fact σT
leaves f1, . . . , fn invariant implies that σQ leaves g1, . . . , gn invariant. Consequently, Xi0j0,P ⊂ G2.
For the inverse inclusion, since for every Q ∈ G2, σQ leaves g1, . . . , gn invariant, i.e., σQ(gi) = gi
and σPQP−1(fi) = fi(1 ≤ i ≤ n), we conclude that PQP−1 ∈ X̃i0j0 by the previous claim. Hence,
Q ∈ Xi0j0,P and Xi0j0,P ⊂ G2, as required.
Theorem 2.1 is proved.
Remark 2.1. (1) Let us present two examples to understand Theorem 2.1. When P = I,
Xi0j0,I = X̃i0j0 . Then Fq(X1, . . . , Xn)Xi0j0,I = Fq(X1, . . . , Xi0X
q−1
j0
−Xq
i0
, . . . , Xn); Assume that
P = (pij) ∈ GL(3, Fq) and P−1 = (rij). By Theorem 2.1, we know that Fq(X1, X2, X3)
X21,P =
= Fq(r11X1 + r12X2 + r13X3, (r21X1 + r22X2 + r23X3)(r11X1 + r12X2 + r13X3)
q−1 − (r21X1 +
+ r22X2 + r23X3)
q, r31X1 + r32X2 + r33X3).
(2) For any two root subgroups Xi0j0,P1 and Xi0j0,P2 with 1 ≤ i0 6= j0 ≤ n, if
Fq(X1, . . . , Xn)Xi0j0,P1 = Fq(h1, h2, . . . , hn), then Fq(X1, . . . , Xn)Xi0j0,P2 = Fq(σT−1(h1), . . .
. . . , σT−1(hn)) with P2 = P1T.
(3) More generally, if a group G ≤ GL(n,K) satisfies G = QG̃Q−1 with Q a fixed inverse
(n×n)-matrix, then the transcendental basis of the invariant field of G̃ reduces a transcendental basis
of the invariant field of our group G.
(4) If we let f̃i = Xi for 1 ≤ i ≤ n, i 6= i0, f̃i0 =
∏
c∈Fq
(cXj0 + Xi0) and g̃i = σP−1 f̃i for
1 ≤ i ≤ n, then following arguments similar to Theorem 2.1 we can prove that g̃1, . . . , g̃n form a
second transcendental basis of Fq(X1, . . . , Xn)Xi0j0,P over Fq.
3. Invariant fields of root subgroups of finite unitary, orthogonal and symplectic groups. In
this section, we begin with discussing the long root subgroups of finite unitary groups. Let us define
polynomials
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 12
1640 JIZHU NAN, YUFANG QIN
Pnk =
(
X1, . . . , Xn
)
H
Xq2k+1
1
...
Xq2k+1
n
=
∑
1≤i,j≤n
hijXiX
q2k+1
j , k = 0, 1, . . . .
Before we prove the main result on the long root subgroup of U(n,K,H), we need the following
lemmas.
Lemma 3.1 [10]. For Q ∈ GL(n, Fq), the following statements are equivalent:
(1) Q ∈ U(n, Fq, H);
(2) σQ fixes Pnk for all k;
(3) σQ fixes Pnk for some k ≥ 1.
Lemma 3.2 [3]. If X and Y are (m × n)-matrices with rank m, then there exists a unitary
matrix U ∈ U(n,K,H) such that X = UY if and only if XHX̄ ′ = Y HȲ ′.
Remark 3.1. The analogues of Lemmas 3.1, 3.2 for orthogonal and symplectic groups are also
true (see [3, 10]).
Lemma 3.3 [3]. Any unitary transvection in U(n,K,H) can be represented as I + Hū′su,
where uHū′ = 0 and s̄ = s. And any unitary transvection is unitary similar to the normal form
I(ν)
K I(ν)
I(n−2ν)
, (3.1)
where
K =
(
s
0(ν−1)
)
.
By Lemma 3.3, there exists a unitary matrix R such that
R−1(I +Hū′su)R =
I(ν)
K I(ν)
I(n−2ν)
, (3.2)
where
K =
(
s
0(ν−1)
)
and the matrix R is independent on s. Let T̃u be the group consisting of all the matrices of the
form (3.2). Then we have that Tu = RT̃uR
−1 and T̃u is the similar group of Tu for which we are
searching.
Lemma 3.4. Let f2 =
∑
1≤i,j≤n
hijXiX
q3
j , fν+1 = Xν+1X
q−1
1 − Xq
ν+1 and fi = Xi for all
1 ≤ i ≤ n, i 6= 2, ν + 1. Then we have Fq(X1, . . . , Xn)T̃u = Fq(f1, f2, . . . , fn).
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 12
ON INVARIANTS OF ROOT SUBGROUPS OF FINITE CLASSICAL GROUPS 1641
Proof. It is trivial that f1, f2, . . . , fn are algebraically independent over Fq by the fact
det
(
∂fi
∂Xj
)
1≤i,j≤n
6= 0.
Suppose that Fq(X1, . . . , Xn) is Galois over Fq(f1, . . . , fn) with Galois group G. We only need
to prove that G = T̃u. The inclusion T̃u ⊂ G is obvious. Conversely, for any Q ∈ G, one can easily
conclude that pii = 1 and pij = 0 (1 ≤ i, j ≤ n, i 6= 2, ν + 1, i 6= j) from the fact that σQ leaves fi
(1 ≤ i ≤ n, i 6= 2, ν + 1).
Moreover, σQ leaves fν+1 invariant, i.e., (pν+1,1X1 + . . .+ pν+1,nXn)Xq−1
1 − (pν+1,1X1 + . . .
. . .+pν+1,nXn)q = Xν+1X
q−1
1 −Xq
ν+1, then we know that pν+1,j = 0 for 2 ≤ j ≤ n, j 6= ν+1 and
pν+1,ν+1 = 1. According to Lemma 3.1, the fact σQ leaves f2 invariant shows that Q ∈ U(n, Fq, H),
which implies that p22 = 1, p2j = 0 (j 6= 2) and pν+1,1 = p̄ν+1,1. Therefore, we have that Q ∈ T̃u
and G = T̃u.
Lemma 3.4 is proved.
From Lemmas 3.1, 3.3 and 3.4, we deduce the following theorem.
Theorem 3.1. Let fi, 1 ≤ i ≤ n, be as in Lemma 3.4 and gi = Rfi for 1 ≤ i ≤ n. Then we
have Fq(X1, . . . , Xn)Tu = Fq(g1, g2, . . . , gn).
Proof. By Lemmas 2.1 and 3.4, we know that g1, . . . , gn are algebraically independent over Fq.
Suppose that Fq(X1, . . . , Xn) is Galois over Fq(g1, . . . , gn) with Galois group G. According to
Lemmas 3.3 and 3.4, we have that G = Tu by using the same arguments as in Theorem 2.1, and the
proof of this theorem is complete.
Now we come to the short root subgroup Tu,w of U(n,K,H). By the normal form of Tu,w, we
obtain the similar group of Tu,w.
Lemma 3.5. The element I + Hw̄′au + H(au)′w of the short root subgroup Tu,w is unitary
similar to the normal form
I(ν)
K I(ν)
I(n−2ν)
, (3.3)
where
K =
0 a
ā 0
0(ν−2)
.
Proof. We represent the matrix I +Hw̄′au+H(au)′w as the following form:
I +H
(
(au)′, w̄′
)(0 1
1 0
)(
au
w
)
. (3.4)
Observe that (
au
w
)
H
(
(au)′, w̄′
)
= 0
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 12
1642 JIZHU NAN, YUFANG QIN
from the conditon that uHū′ = wHw̄′ = uHw̄′ = 0. Clearly,(
a 0 0 . . . 0
0 1 0 . . . 0
)
H
(
a 0 0 . . . 0
0 1 0 . . . 0
)′
= 0.
So by Lemma 3.2, there exists a unitary matrix R0 such that(
au
w
)
=
(
a 0 0 . . . 0
0 1 0 . . . 0
)
R0. (3.5)
Substituting (3.5) into (3.4), we get
R0(I +Hw̄′au+H(au)′w)R−10 =
I(ν)
K I(ν)
I(n−2ν)
,
where
K =
0 a
ā 0
0(ν−2)
.
Lemma 3.5 is proved.
Remark 3.2. In Lemma 3.5, it is proved that there exists a unitary matrix R0 such that R0(I +
+Hw̄′au+H(au)′w)R−10 has the form (3.3). We remark that the matrix R0 here is independent on
the choice of a.
Theorem 3.2. Let fν+1 = Xν+1X
q−1
2 − Xq
ν+1, fν+2 =
∑
hijXiX
q3
j and fi = Xi for all
1 ≤ i ≤ n, i 6= ν+1, ν+2. Let gi = R−10 fi with 1 ≤ i ≤ n. Then we have that Fq(X1, . . . , Xn)Tu,w =
= Fq(g1, g2, . . . , gn).
Proof. Similar to the proof of Theorem 3.1.
In the following, let us pay attention to the discussion of the long root subgroups and the short
root subgroups of the finite orthogonal groups.
Lemma 3.6 [3]. Any orthogonal 2-transvection is orthogonal similar to the following normal
form:
I(ν)
K I(ν)
I(n−2ν)
, (3.6)
where
K =
0 1
−1 0
0(ν−2)
.
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ON INVARIANTS OF ROOT SUBGROUPS OF FINITE CLASSICAL GROUPS 1643
Lemma 3.7. The element I + Sw′au− Sau′(w + Q(w)au) of the short root subgroup Ŷu,w is
orthogonal similar to the normal form
K2
I(ν−2)
K1 K3
I(n−ν−2)
, (3.7)
where
K1 =
(
0 −aQ(w)
−aQ(w) −a2Q(w)
)
, K2 =
(
1 a
0 1
)
, and K3 =
(
1 0
−a 1
)
.
Proof. We prove the normal form of I + Sw′au− Sau′(w +Q(w)au) through considering the
normal form of its transpose matrix. Note that I + au′wS − (aw′ + a2Q(w)u′)uS can be written as
I +
(
w′, u′
)(0 −a
a −a2Q(w)
)(
w
u
)
S.
Let
T =
(
Q(w) 0 0(1,ν−2) 1 0(1,ν−1) 0(n−2ν)
0 1 0 0 0 0
)
.
Then we have
TST ′ =
(
2Q(w) 0
0 0
)
.
Moreover, (
w
u
)
S
(
w
u
)′
=
(
2Q(w) 0
0 0
)
from the assumption that uSu′ = uSw′ = 0 and wSw′ = 2Q(w). So by Remark 3.1 there exists an
orthogonal matrix R2 such that (
w
u
)
= TR2.
Consequently, we know that
R
′−1
2
[
I +
(
w′, u′
)(0 −a
a −a2Q(w)
)(
w
u
)
S
]
R′2 =
K ′2 K ′1
I(ν−2)
K ′3
I(n−ν−2)
, (3.8)
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1644 JIZHU NAN, YUFANG QIN
where
K ′1 =
(
0 −aQ(w)
−aQ(w) −a2Q(w)
)
, K ′2 =
(
1 0
a 1
)
, and K ′3 =
(
1 −a
0 1
)
.
Taking the transpose of both sides of equality (3.8), then we have that I+Sw′au−Sau′(w+Q(w)au)
is orthogonal similar to
K2
I(ν−2)
K1 K3
I(n−ν−2)
,
as desired.
Lemma 3.7 is proved.
By Lemma 3.6, there exists an orthogonal matrix R1 such that
R−11 (I + Sw′au− Sau′w)R1 =
I(ν)
K I(ν)
I(n−2ν)
,
where
K =
0 a
−a 0
0(ν−2)
.
Using similar arguments as in the proof of Theorem 3.1, we get the following theorem. The details
are omitted.
Theorem 3.3. (1) Let fν+1 = Xν+1X
q−1
2 − Xq
ν+1, fν+2 =
∑
sijXiX
q
j and fi = Xi
for all 1 ≤ i ≤ n, i 6= ν + 1, ν + 2. Assume that gi = R1fi with 1 ≤ i ≤ n. Then we have
Fq(X1, . . . , Xn)Yu,w = Fq(g1, g2, . . . , gn).
(2) Let f1 = X1X
q−1
2 − Xq
1 , fν+1 = Xν+1X
q−1
2 − Xq
ν+1, fν+2 =
∑
sijXiX
q
j and fi = Xi
for all 1 ≤ i ≤ n − 1, i 6= 1, ν + 1, ν + 2. Assume that gi = R−12 fi with 1 ≤ i ≤ n. Then we have
Fq(X1, . . . , Xn)Ŷu,w = Fq(g1, g2, . . . , gn).
Remark 3.3. Note that the techniques we use can be applied to the case of the root subgroups
of finite sympletic groups.
4. Invariant rings of root subgroups of the special linear groups over finite fields and
Poincaré series. In the section, we show that the invariant rings of the root subgroups Xi0j0,P
of the special linear groups over finite fields are polynomial rings, and we give the Poincaré series of
the invariant rings Fq[X1, . . . , Xn]Xi0j0,P .
Here is an algorithm to check if the invariant ring is a polynomial ring.
Lemma 4.1 [9]. Let I be the invariant ring of a finite group G ≤ GL(n,K) over an arbitrary
field K and f1, . . . , fn ∈ I be homogeneous invariants of degrees d1, . . . , dn. Then the following
statements are equivalent:
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ON INVARIANTS OF ROOT SUBGROUPS OF FINITE CLASSICAL GROUPS 1645
(1) I = K[f1, . . . , fn];
(2) f1, f2, . . . , fn are algebraically independent over K and
∏n
i=1
di = |G|.
Theorem 4.1. Let g1, . . . , gn and g̃1, . . . , g̃n be as in Theorem 2.1 and in Remark 2.1. Then
Fq[X1, . . . , Xn]Xi0j0,P = Fq[g1, g2, . . . , gn] = Fq[g̃1, g̃2, . . . , g̃n].
Proof. This assertion follows form Lemma 4.1 and the fact
|Xi0j0,P | =
n∏
i=1
deg(gi) =
n∏
i=1
deg(g̃i) = q.
Theorem 4.1 is proved.
Let M = K[x1, . . . , xn] be a polynomial ring over an arbitrary field with deg(xi) = ki. Then the
Poincaré series of M is equal to
∏n
i=1
1
1− tki
(see [13], Ch. 16.1, Ch. 7.1). From this assertion and
Theorem 4.1, we have the following theorem.
Theorem 4.2. The Poincaré series of the invariant ring Fq[X1, . . . , Xn]Xi0j0,P of the root sub-
group Xi0j0,P is equal to
1
(1− t)n−1(1− tq)
.
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Received 08.03.12
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