Stabilizer group of generalized determinant

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by  Ryo Yamamoto

STABILIZER GROUP OF GENERALIZED DETERMINANT
RYO YAMAMOTO
7
1 Abstract. In this paper, we introduce the notion of generalized determinant and de-
0 termine the stabilizer group in GL(Matn(K)) of the generalized determinant.
2

n
a
J
1. Introduction.
5
2
For an n×n matrix A := (a ) , we will define the even determinant det (A) :=
ij 1≤i,j≤n An
A] σ∈An ni=1aiσ(i) and the odd determinant detA¯n(A) := σ∈A¯n ni=1aiσ(i) where A¯n is
Pthe setQSn\An. Let K be a field of characteristic 0. LetPα,β ∈QK. We introduce the
R
generalized determinant det(α,β)(A) of an n×n matrix A by
.
h
at det(α,β)(A) := αdetAn(A)+βdetA¯n(A).
m Let X := (x ) where x are the standard basis of Mat∗. Write det := det(X),
ij 1≤i,j≤n ij n n
[ perm := perm(X) and det(α,β) := det(α,β)(X). The stabilizer group Stab(f) of f ∈
n n
1 Symn(Mat∗) is defined as Stab(f) := {T ∈ GL(Mat (K)) | T ·f = f }. If α = −β 6= 0,
v n n
0 thenStab(det(α,−α)) = Stab(detn), which hadbeendetermined byFrobenius[1]asfollows.
3
4 Theorem 1 (Frobenius). It holds that Stab(det ) = {X 7→ PXQ or PtXQ |
n
7 detP detQ = 1}. Here P,Q ∈ GL (K).
0 n
.
1 On the other hand, if α = β 6= 0, then Stab(det(α,α)) = Stab(perm ), which was
n
0
determined by Marcus and May [2] as follows.
7
1
Theorem 2 (Marcus and May). Let n ≥ 3. It holds that Stab(perm ) = {X 7→ LPXQR
: n
v
or LPtXQR | detLdetR = 1} where P and Q are permutation matrices, L and R are
i
X
diagonal matrices.
r
a Our purpose is to relate the two results by P1(K)-family {det(α,β) | [α : β] ∈ P1(K)}.
n
Our main result is the following.
Theorem 3. Let n ≥ 5 . If α 6= ±β, then Stab(det(α,β)) = Stab(det )∩Stab(perm ).
n n n
2. Notations.
We define a submatrix Xk1···kr of X by Xk1···kr := (x ) . Let P(α,β)(X) :=
l1···lr l1···lr kilj 1≤i.j≤r r
(det(α,β)(Xk1···kr)) be a n × n matrix. Specially, we write P (X) :=
l1···lr 11≤≤kl11<<······<<lkrr≤≤nn (cid:0)r(cid:1) (cid:0)r(cid:1) r
P(1,0)(X) and P (X) := P(0,1)(X).
r r r
Lemma 4. Let n ≥ 4. If det(α,β)(T(X)) = det(α,β)(X), then det(α,β)(T(X)k1k2)
and det(β,α)(T(X)k1k2) are expressible as a linear combination of det(α,β)(Xk1′k2′)l1al2nd
det(β,α)(Xk1′k2′)(1 ≤l1kl2′,k′,l′,l′ ≤ n) respecticely. l1′l2′
l′l′ 1 2 1 2
1 2
1
2 RYOYAMAMOTO
Proof. Let Y := T(X). We can write each entry of X = T−1(Y) as x = n gpqy .
st p,q=1 st pq
We first prove the case det(α,β)(T(X)k1k2). Since n ≥ 4, there exists σ ∈ PA such that
l1l2 n
σ(k ) = l , σ(k ) = l . The permutation (l l )σ ∈ A¯ satisfies (l l )σ(k ) = l and
1 1 2 2 1 2 n 1 2 1 2
(l l )σ(k ) = l . We compute
1 2 2 1
y y
det(α,β)(Yk1k2) = det(α,β) k1l1 k1l2
l1l2 (cid:18)yk2l1 yk2l2(cid:19)
∂n−2
= det(α,β)(Y)
\ \
∂y ···∂y ···∂y ···∂y
1σ(1) k1σ(k1) k2σ(k2) nσ(n)
∂n−2
= det(α,β)(X).
\ \
∂y ···∂y ···∂y ···∂y
1σ(1) k1σ(k1) k2σ(k2) nσ(n)
To compute it, we use
n
∂ ∂x ∂
det(α,β)(Xk1···kr) = st det(α,β)(Xk1···kr)
∂y l1···lr ∂y ∂x l1···lr
pq sX,t=1 pq st
and
n
∂ ∂x ∂
det(β,α)(Xk1···kr) = st det(β,α)(Xk1···kr).
∂y l1···lr ∂y ∂x l1···lr
pq sX,t=1 pq st
We have ∂xst = gpq ∈ K and ∂ det(α,β)(Xk1···kr) is equal to det(α,β)(Xk1···sˆ···kr),
∂ypq st ∂xst l1···lr l1···tˆ···lr
det(β,α)(Xk1···sˆ···kr) or 0. Hence, differentiating n−2 times, the lemma follows.
l1···tˆ···lr
We now turn to the case det(β,α)(T(X)k1k2). Since n ≥ 4, there exists σ ∈ A such
l1l2 n
that σ(k ) = l , σ(k ) = l . The permutation (l l )σ ∈ A¯ satisfies (l l )σ(k ) =
1 2 2 1 1 2 n 1 2 1
l ,(l l )σ(k ) = l . Hence the same proof works for det(β,α)(T(X)k1k2). (cid:3)
1 1 2 2 2 l1l2
Lemma 5. Let A ∈ Mat (K). If P (A) = P (A) = 0, then A is 0, a row matrix or a
n 2 2
column matrix.
Proof. Consider A 6= 0. Without loss of generality we can assume a 6= 0. Then P (A) =
11 2
0 implies a = 0 (2 ≤ ∀i, j ≤ n). If there is j ≥ 2 such that a 6= 0, then a = 0(2 ≤
ij 1j i1
∀i ≤ n) from P (A) = 0. Thus A is a row matrix. If there is i ≥ 2 such that a 6= 0,
2 i1
then a = 0(2 ≤ ∀j ≤ n) from P (A) = 0. Thus A is a column matrix. (cid:3)
1j 2
Lemma 6. Define F := T(E ). If α 6= ±β, then the number of non-zero entries in F
ij ij ij
is one.
Proof. By Lemma 4, both det(α,β)(F k1k2) and det(β,α)(F k1k2) are linear combination of
det(α,β)(E k1′k2′) and det(β,α)(E k1′k2′)ij(l11l2≤ k′, k′, l′, l′ ≤ijl1nl2) respecticely. We thus get
ijl′l′ ijl′l′ 1 2 1 2
1 2 12
P(α,β)(F ) = αP (F )+βP (F ) = 0 and P(β,α)(F ) = βP (F )+αP (F ) = 0. Since
2 ij 2 ij 2 ij 2 ij 2 ij 2 ij
α2 −β2 6= 0, P (F ) = P (F ) = 0. Applying Lemma 5, we see that F is a row matrix
2 ij 2 ij ij
or a column matrix.
Suppose that the number of non-zero entries in F is two or more. Let us assume
ij
that F is a row matrix with non-zero entries in the i′th row. Since F + F =
ij ij it
T(E +E ), F +F = T(E +E ), wehaveP (F +F ) = P (F +F ) = P (F +F ) =
ij it ij tj ij tj 2 ij it 2 ij it 2 ij tj
P (F + F ) = 0(1 < ∀t ≤ n). By Lemma 5, F + F and F + F are row matrices
2 ij tj ij it ij tj
or column matrices, so that F and F are row matrices lying in the i′th row. However
it tj
STABILIZER GROUP OF GENERALIZED DETERMINANT 3
dimspan{Ei1,Ei2,...,Ein,E1j,...,Enj} > dimspan{Ei′1,Ei′2,...,Ei′n}, which contra-
dicts the fact that T is non-singular. The same proof works in the case that F is a
ij
column matrix. (cid:3)
By Lemma 6, we have T(Eij) = cijEi′j′. Since T is non-singular, cij 6= 0 and (i,j) 6=
(s,t) implies (i′,j′) 6= (s′,t′). Hereafter, we always assume α 6= ±β and define maps µ,λ
by T(E ) = c E .
ij ij µ(i,j)λ(i,j)
Lemma 7. There exist permutation matrices P := (δ ) , Q := (δ )
iσ(j) 1≤i,j≤n iτ(j) 1≤i,j≤n
where sgn(σ)sgn(τ) = 1, and a matrix C := (c ) ∈ M (K) with ∀c 6= 0 such
ij 1≤i,j≤n n,n ij
that T(X) = C ∗PXQ or T(X) = C ∗PtXQ (the operation ∗ is the Hadamard product).
Proof. We may assume that µ(1,1) = 1 and λ(1,1) = 1 by swapping rows or columns
even number of times, that is, the number of row and column transpositions are both even
or both odd. Since rank(E +E ) = 2, we have P (F +F ) 6= 0 or P (F +F ) 6= 0.
11 22 2 11 22 2 11 22
It follows that µ(2,2) ≥ 2 and λ(2,2) ≥ 2. Therefore, swapping rows or columns even
number of times properly, we may assume that µ(2,2) = 2 and λ(2,2) = 2. By continuing
thesameargument, wecanassume µ(i,i) = i(1 ≤ ∀i ≤ n) andλ(i,i) = i(1 ≤ ∀i ≤ n−2).
Therearetwopossibilities: (i)λ(n−1,n−1) = n−1andλ(n,n) = n, (ii)λ(n−1,n−1) = n
and λ(n,n) = n − 1. However, the case (ii) never happens because the coefficients
of x ...x x in det(α,β)(T(X)) and in det(α,β)(X) are different. Therefore, we
11 n−1n−1 nn
conclude that
∗
x
11
 ... 
T(X) = C ∗P Q
∗ xn−1,n−1 
 
 xnn
 
where sgn(σ)sgn(τ) = 1. To continue the argument, we may assume P = Q = I that is
n
µ(i,i) = i and λ(i,i) = i(1 ≤ ∀i ≤ n) without loss of generality.
By P (E + E ) = P (E + E ) = 0, we get µ(1,2) = 1 or λ(1,2) = 1. By
2 11 12 2 11 12
P (E + E ) = P (E + E ) = 0, we also get µ(1,2) = 2 or λ(1,2) = 2. Combining
2 22 12 2 22 12
these, we obtain two possibilities: (I) µ(1,2) = 1 and λ(1,2) = 2, (II) µ(1,2) = 2 and
λ(1,2) = 1.
Suppose first that (I) holds. Let 3 ≤ γ ≤ n. By P (E +E ) = P (E +E ) = 0,
2 11 1γ 2 11 1γ
we get µ(1,γ) = 1 or λ(1,γ) = 1. By P (E + E ) = P (E + E ) = 0, we also get
2 12 1γ 2 12 1γ
µ(1,γ) = 1 or λ(1,γ) = 2. Combining these gives µ(1,γ) = 1. By P (E + E ) =
2 γγ 1γ
P (E +E ) = 0, we obtain λ(1,γ) = γ.
2 γγ 1γ
Let δ 6= 1. By P (E + E ) = P (E + E ) = 0, we have µ(δ,1) = 1 or λ(δ,1) =
2 11 δ1 2 11 δ1
1.However µ(1,γ) = 1(1 ≤ γ ≤ n) gives µ(δ,1) 6= 1 as T is non-singular. Hence λ(δ,1) =
1. By P (E +E ) = P (E +E ) = 0, we obtain µ(δ,1) = δ.
2 δδ δ1 2 δδ δ1
Let 1 < γ 6= δ ≤ n. By P (E + E ) = P (E + E ) = 0, we get µ(δ,γ) = δ or
2 δ1 δγ 2 δ1 δγ
λ(δ,γ) = 1 but the latter is impossible. By P (E +E ) = P (E +E ) = 0, we also
2 1γ δγ 2 1γ δγ
get λ(δ,γ) = γ.
By the above argument, we obtain µ(i,j) = i and λ(i,j) = j(1 ≤ ∀i, j ≤ n) that is
T(X) = C ∗PXQ (sgn(σ)sgn(τ) = 1).
The same proof works for the case (II) and we also obtain
T(X) = C ∗PtXQ (sgn(σ)sgn(τ) = 1),
4 RYOYAMAMOTO
if (II) holds. (cid:3)
Lemma 8. If n ≥ 5, then rank(C) = 1.
Proof. Comparing the coefficients of det(α,β)(X) and det(α,β)(C ∗PXQ), we obtain
(1) c ...c = 1 (∀w ∈ A ) (β = 0)
1w(1) nw(n) n
(2) c ...c = 1 (∀w ∈ A¯ ) (α = 0)
1w(1) nw(n) n


(3) c ...c = 1 (∀w ∈ S ) (α 6= 0,β 6= 0).
1w(1) nw(n) n

Let us solve these simultaneous equations.
For preperation, we first consider the case n = 4. Let β = 0. By c c (c c ) =
11 22 33 44
1, c c (c c ) = 1 and c c (c c ) = 1, we can write
12 23 31 44 13 21 32 44
c c c a a a
11 12 13 11 12 13
 c21 c22 c23  = a21 u v .
c c c c c c 1 1 1
 31 44 32 44 33 44 a12v a13a21 a11u
Set a := c . Then we have
31 31
1 a a v a a v
c = , c = 12 31 , c = 12 31 .
44 a a v 32 a a 33 a u
12 31 13 21 11
Set a := c . By c c c c = 1, we have
14 14 14 23 32 41
a a v a a
a ·v · 12 31 ·c = 1 ∴ c = 13 21 .
14 a a 41 41 a a a v2
13 21 12 14 31
By c c c c = 1 and c c c c = 1, we have
12 24 33 41 13 22 34 41
a a v a a a a uv
a ·c · 12 31 · 13 21 = 1 ∴ c = 11 14
12 24 a u a a a v2 24 a a a
11 12 14 31 12 13 21
a a a a a v2
a ·u·c · 13 21 = 1 ∴ c = 12 14 31 .
13 34 a a a v2 34 a2 a u
12 14 31 13 21
By c c c c = 1 and c c c c = 1, we have
13 24 31 42 14 21 33 42
a a uv a a
a · 11 14 ·a ·c = 1 ∴ c = 12 21
13 31 42 42
a a a a a a uv
12 13 21 11 14 31
a a v a u
a ·a · 12 31 ·c = 1 ∴ c = 11 .
14 21 a u 42 42 a a a a v
11 12 14 21 31
Combining these yields
a a
u = ± 12 21.
a
11
By c c c c = 1 and c c c c = 1, we have
11 24 32 43 14 22 31 43
a a uv a a v a2 a2
a · 11 14 · 12 31 ·c = 1 ∴ c = 13 21
11 a a a a a 43 43 a2 a a uv2
12 13 21 13 21 11 14 31
1
a ·u·a ·c = 1 ∴ c = .
14 31 43 43 a a u
14 31
Combining these yields
a a
v = ± 13 21.
a
11
STABILIZER GROUP OF GENERALIZED DETERMINANT 5
a2
Now, set a := c = 11 . Summarizing the above, we obtain
41 41 a12a13a14a21a31
1 1 1 1
(4) C = 1 εu εv εuεv∗ ai1a1j
1 εv εuεv εu  (cid:18) a11 (cid:19)1≤i,j≤4
1 ε ε ε ε 
u v u v
 
where ε ,ε ∈ {+1,−1} and a ···a a ···a = a4 . Conversely, the matrix C is a
u v 11 14 11 41 11
solution of the simultaneous equation (1).
Let α = 0. Interchanging the 3rd and the 4th row of (4), we obtain
1 1 1 1
(5) C = 1 εu εv εuεv∗ ai1a1j
1 εuεv εu εv  (cid:18) a11 (cid:19)1≤i,j≤4
1 ε ε ε ε 
v u v u
 
where ε ,ε ∈ {+1,−1} and a ···a a ···a = a4 as a solution of the simultaneous
u v 11 14 11 41 11
equation (2).
Let α 6= 0 and β 6= 0. Combining (4) and (5), we obtain
a a
(6) C = i1 1j
(cid:18) a (cid:19)
11 1≤i,j≤4
where a ···a a ···a = a4 as a solution of the simultaneous equation (3).
11 14 11 41 11
Let us consider the simultaneous equations for n ≥ 5. We prove that the solution of
the each simultaneous equations (1),(2),(3) is expressible as
a a
(7) C = i1 1j
(cid:18) a (cid:19)
11 1≤i,j≤n
where a ···a a ···a = an respectively by induction of the matrix size n.
11 1n 11 n1 11
In the case n = 5 and β = 0, using the fact that any solution of (1) may be written as
(4), we can write
a a a a
 11 12 13 14 
c c c c a a a a a a
11 12 13 14  a ε 21 12 ε 21 13 ε ε 21 14
 c c c c   21 u a v a u v a 
21 22 23 24 =  11 11 11 
c c c c  a a a a a a 
 31 32 33 34   a ε 31 12 ε ε 31 13 ε 31 14 
c41c55 c42c55 c43c55 c44c55  31 v a u v a u a 
   11 11 11 
 za za za 
 z ε ε 12 ε 13 ε 14 
 u v a u a v a 
 11 11 11 
where ε ,ε ∈ {+1,−1} and a ···a a ···a z = a4 . Set a := c . Then c =
u v 11 14 11 31 11 41 41 55
z , c = ε ε a41a12, c = ε a41a13, c = ε a41a14. Set a := c . If ε = −1 or ε =
−a411, t4h2ere euxisvt aw11 ∈ A43 sucuh ta1h1at w4(41) =v5a,1w1 (5) = 1,15c 1=5 ai1a1wu(i) (2 ≤ i ≤v 4)
and w′ ∈ An such thatnw′(1) = 5, w′(5) = 1, ciw′(i) = −aiwi1(aai)11w1′(i) (2a≤11 i ≤ 4). Then
c15c2w(2)c3w(3)c4w(4)c51 6= c15c2w′(2)c3w′(3)c4w′(4)c51 and one of the two cannot be equal to 1,
a contradiction. Thus ε = ε = +1. The similar consideration applies to the case n = 5
u v
and α = 0.
6 RYOYAMAMOTO
Therefore, assuming (7) to hold for n−1, we can write
a ··· ··· a
11 1,n−1
 
. a a a a
c11 ··· c1,n−1  .. 21 12 ... 21 1,n−1 
 a a 
 ... ... ...  =  .. ..11 .. ..11 
 . . . . 
 
 
 c ··· c 
cn−n1−,12c,1nn ··· cn−n1−,n2−,n1−c1nn an−2,1 an−a2,1a12 ... an−2a,1a1,n−1
 11 11 
 za za 
 z 12 ... 1,n−1 
 a a 
 11 11 
where a ···a a ···a z = an−1. Set a := c . Then we have
11 1,n−1 11 n−2,1 11 n−1,1 n−1,1
z a a
c = , c = n−1,1 1j (1 ≤ j ≤ n−1).
nn n−1,j
a a
n−1,1 11
Set a := c . By c (c ···c )c = 1 for w ∈ A such that w(1) = n, w(n) =
1n 1n 1n 2w(2) n−1w(n−1) n1 n
1,
a ···a a ···a
a · 12 1,n−1 21 n−1,1 ·c = 1
1n an−2 n1
11
an−2 a z
∴ c = 11 = 11 .
n1 a ···a a ···a a a
12 1n 21 n−1,1 1n n−1,1
Set a := c . Then
n1 n1
a a a
z = 1n n−1,1 n1.
a
11
It follows that c = ai1a1j for 1 ≤ i, j ≤ n − 1 and (i,j) = (1,n), (n,1), (n,n) where
ij a11
a ···a a ···a = an . By c (c ···c ···c )c = 1, we have
11 1n 11 n1 11 in 1w(1) in n−1w(n−1) n1
a ···a a ···a
c · 12 1,n−1 c21 n−1,1 ·a = 1
in an−3a n1
11 i1
an−3a a a
∴ c = 11 i1 = i1 1n.
in
a ···a a ···a a
12 1,n−1 21 n1 11
By (c ···c )c = 1, we have
1w(1) n−1w(n−1) nj
a ···a a ···a
12 1n 11 n−1,1 ·c = 1
an−1a nj
11 1j
an−1a a a
∴ c = 11 1j = n1 1j.
nj
a ···a a ···a a
12 1n 11 n−1,1 11
From the above, we obtain (7).
Each of 2×2 minor determinants of C is 0, so that rank(C) = 1 follows. (cid:3)
Proof of Theorem 3. Let T−1 ∈ Stab(det(α,β)) and α 6= ±β. By Lemma 6 and 7, We can
n
write T(X) = C ∗ PXQ or T(X) = C ∗ PtXQ where sgn(σ)sgn(τ) = 1. By Lemma 8,
we can write c = l r (1 ≤ i, j ≤ n) where l ···l r ···r = 1. Set L := diag(l ,...,l )
ij i j 1 n 1 n 1 n
and R := diag(r ,...,r ). Then we obtain T(X) = LPXQR or T(X) = LPtXQR where
1 n
sgn(σ)sgn(τ) = 1, which proves the theorem. (cid:3)
STABILIZER GROUP OF GENERALIZED DETERMINANT 7
References
[1] G. Frobenius, U¨ber die Darstellung der endlichen Gruppen durchlineare Substitutionen, Sitzungs-
ber Deutsch. Akad. Wiss. Berlin (1897), 994–1015.
[2] Marvin Marcus and F. C. May, The permanent function, Canad. J. Math., 14 (1962), 177-189.
Department of Pure and Applied Mathematics, Graduate School of Information Sci-
ence and Technology, Osaka University, Suita, Osaka 565-0871, Japan
E-mail address: [email protected]