# The Automorphism Group of the Lie Ring of Real Skew-Symmetric Matrices

Authors: Jinli Xu; Baodong Zheng; Li Yang
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

Denote by𝔄n the set of all n×n skew-symmetric matrices over the field of real numbers, which forms a Lie ring under the usual matrix addition and the Lie multiplication as
[A,B]=AB-BA,  A,B∈𝔄n. In this paper, we characterize the automorphism group of the Lie ring 𝔄n.


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## Body

## 1. Introduction and Main Result

A Lie ring is defined as a nonassociative ring with multiplication that is anticommutative and satisfies the Jacobi identity. More specifically, we can define a Lie ringL to be an abelian group with an operation [,] that has the following properties:(i)
biadditive:(1)[x+y,z]=[x,z]+[y,z],[z,x+y]=[z,x]+[z,y],
for all x,y,z∈L;(ii)
the Jacobi identity:(2)[[x,y],z]+[[y,z],x]+[[z,x],y]=0,
for all x,y,z∈L;(iii)
for allx in L,
(3)[x,x]=0.It is well known that a Lie algebra can be viewed as a Lie ring. So, the theory of Lie ring can be used in the theory of the Lie algebra. Recall that an automorphism of a Lie ring(L,[,]) is a bijective map ϕ form L onto itself such that ϕ(x+y)=ϕ(x)+ϕ(y) and ϕ([x,y])=[ϕ(x),ϕ(y)] for all x,y∈L. There are a lot of papers that studied the automorphism groups of some fixed Lie rings (or, more for the Lie algebras), see [1–7].Note that any associative ringL can be made into a Lie ring (L,+,[,]) by defining a bracket operator [x,y]=xy-yx. Let ℝ be the real number field and denote by ℝ× the rest ℝ∖{0}. Let Mn be the algebra of all n×n matrices over ℝ. We denote by 𝔄n the subset of Mn consisting of all n×n skew-symmetric matrices, that is,
(4)𝔄n={[aij]∈Mn:[aij]t=-[aij]}.It is well known that the set𝔄n forms a Lie ring under the usual matrix addition and the Lie multiplication as [A,B]=AB-BA,  A,B∈𝔄n. In the same way, we know that Mn or Tn (the set of all n×n upper triangular matrices) as well as forms a Lie ring.Hua [8] gives the form of any automorphism of the Lie ring Mn over a skew field by using the fundamental theorem of geometry of matrices; Dolinar [1] studies the automorphism of the Lie ring of triangular matrices Tn over any field. Jacobson [9] considers the Lie algebra 𝔄n over any algebraically closed field; he gives the form of the automorphism of the Lie algebra 𝔄n for the case n≥5 if n is odd or n≥10 if n is even. Now, let us see a general result on isomorphism of some Lie rings as follows.Proposition 1 (see [10, 11]).
LetA′ and A be prime rings with involutions of the first kind and of characteristic not 2. Let K′ and K denote, respectively, the skew elements of A′ and A. Assume that the dimension of the central closure of A′ over CA′ is different from 1, 4, 9, 16, 25, and 64. Then, any Lie isomorphism θ of K′ onto K can be extended uniquely to an associative isomorphism of 〈K′〉 onto 〈K〉, the associative subrings generated by K′ and K, respectively.Note that the Lie ring𝔄n is a particular class of the previous setting of skew elements of Mn. So, the previous proposition in fact partially solved the problem to characterize the automorphism group of 𝔄n. However, the problem is still open when n takes any positive integer.The purpose of this paper is to characterizeAut𝔄n, the automorphism group of the Lie ring 𝔄n, for n≥2. Our main result is the following.Theorem 2.
Suppose thatn≥2 is an integer, then ϕ∈Aut𝔄n if and only if there is a real orthogonal matrix Q such that
(5)ϕ(X)=QXQt,∀X∈𝔄n.
Further, one hasAut𝔄n≅On(R), where On(R) is the real orthogonal group.


## 2. Preliminary Results

Now, let us start this section by denoting some notations. Denote by[n/2] the maximal integer number no more than n/2. Let Eij(n) be the n×n matrix which has 1 in the (i,j) entry and is 0 elsewhere. Set D=E12(2)+E21(2), and denote by In=Σi=1nEii(n) the n×n identity matrix. Note that the notation I0 means that the matrix vanished. Let J=E11(2)-E22(2), K=E12(2)-E21(2).Suppose that𝔖⊂𝔄n. We call 𝔖 to be commutative if [x,𝔖]=0,  for all x∈𝔖, and call 𝔖 to be maximal commutative if 𝔖 is not only commutative but [y,𝔖]≠0, for all y∉𝔖. Clearly, the maximal commutative subset is a subring of 𝔄n. Suppose that A∈𝔄n. Set
(6)ℭ(A):={X∈𝔄n:[A,X]=0}.We denote byA⊕B and A⊗B the direct sum and the Kronecker product of A and B, respectively.Definition 3.
A matrixA∈𝔄n is called regular if it satisfies the following conditions: (i) when n is an even number, there is an orthogonal matrix Q, and the real numbers ε1,…,εn/2∈ℝ with different absolute values, such that
(7)A=Qdiag(ε1K,…,εn/2K)Qt.
(ii) Whenn is an odd number, there is an orthogonal matrix Q, and the nonzero real numbers ε1,…,ε[n/2]∈ℝ with different absolute values, such that
(8)A=Qdiag(ε1K,…,ε[n/2]K⊕0)Qt.Now, a subring𝔥 of 𝔄n is called a regular subring if 𝔥 is maximal commutative and there is a regular matrix in 𝔥.Forx∈𝔄n, denote ℭ(x)={A∈𝔄n:[A,x]=0}.Lemma 4 (see [12, 2.5.14]).
Suppose thatA∈𝔄n. Then, there are an orthogonal matrix Q and real numbers a1,…,a[n/2] such that
(9)A=Qdiag(a1K⊕⋯⊕a[n/2]K⊕0)Qt.Lemma 5 (see [13, 14]).
Let𝔽 be any field, and let Kn(𝔽) denote the space of all n×n alternate matrices over 𝔽. Then, ϕ is an additive surjective mapping of Kn(𝔽) (n≥2) to itself that preserves rank 2 matrices if and only if ϕ is of the following forms:(i)
n≥4, ϕ((aij))=αPt(f(aij))P, for all (aij)∈Kn(𝔽), where α∈𝔽∖{0}, P is an n×n invertible matrix, and f is a field automorphism of 𝔽;(ii)
whenn=4, ϕ is of the form
(10)ϕ((aij))=αPt(f(aij))*P,∀(aij)∈K4(𝔽),where α, P, and f have the same meaning as before, and (aij)↦(aij)* is either the identity map or the map:
(11)[0a12a13a14-a120a23a24-a13-a230a34-a14-a24-a340]⟼[0a12a13a23-a120a14a24-a13-a140a34-a23-a24-a340].In the next text, we always assume thatϕ∈Aut𝔄n is arbitrary.Lemma 6.
Suppose that𝔥 is a regular subring of 𝔄n. Then, there is an orthogonal matrix Q and maps ηi:𝔄n→ℝ, i=1,…,[n/2] such that
(12)QXQt=η1(X)K⊕⋯⊕η[n/2](X)K⊕0,∀X∈𝔥.Proof.
For everyX∈𝔥, note that QXQt is commutative with every regular matrix in 𝔥. So, one can obtain the conclusion by Lemma 4.Corollary 7.
Suppose that𝔥 is a regular subring of 𝔄n, and H∈𝔥 is a regular matrix. Then,
(13)X∈𝔥⟺[X,H]=0.Lemma 8.
Suppose that𝔥1,  𝔥2 are both regular subrings of 𝔄n, and that there is a regular matrix H∈𝔥1∩𝔥2. Then,  𝔥1=𝔥2.Proof.
Note that a regular subring is maximal; the conclusion follows by Corollary7.Lemma 9.
Both mapsϕ and ϕ-1 preserve the regular subring. Expressly, for H∈𝔄n, one has that H is a regular matrix if and only if ϕ(H) is so.Proof.
Take any regular subring𝔥 of 𝔄n and a regular matrix H∈𝔥. By Lemma 6, we can assume that Qϕ(H)Qt=ε1K⊕⋯⊕ε[n/2]K⊕0, where Q is an orthogonal matrix.
Suppose thatA∈𝔄n satisfying Qϕ(A)Qt=1K⊕2K⊕⋯⊕[n/2]K⊕0. Then, ϕ(A) is a regular matrix in 𝔄n. Since [ϕ(H),ϕ(A)]=0, [H,A]=0. This means that A∈𝔥. Let 𝔥1 be a regular subring containing ϕ(A). By Lemma 8, we only need to prove that ϕ(𝔥)=𝔥1. Take any ϕ(X)∈𝔥1. Then, we see by Lemma 6 that there are x1,…,x[n/2]∈ℝ such that
(14)Qϕ(X)Qt=x1K⊕⋯⊕x[n/2]K⊕0.
So,[ϕ(X),ϕ(H)]=0, and [X,H]=0. Hence, X∈𝔥. This shows that 𝔥1⊂ϕ(𝔥). Note that 𝔥1 is maximal, so we obtain that ϕ(𝔥)=𝔥1.
Now, we prove thatϕ preserves the regular matrix. Otherwise, suppose that ϕ(H) is not a regular matrix, then we will get a contradiction. By the definition, we see that one of the following cases holds.
Case
1.
n is odd and there is εi=0.
Case
2.There is some εi∈{±εj}.
If Case 1 happens, we assume without loss the generality thatε1=0. We take X∈𝔄n such that
(15)Qϕ(X)Qt=E1n(n)-En1(n).
If Case 2 happens, we assume without loss the generality thatε2∈{±ε1}. When ε1=ε2, we take X∈𝔄n such that
(16)Qϕ(X)Qt=K⊗I2⊕0.
Whenε1=-ε2, we take X∈𝔄n such that
(17)Qϕ(X)Qt=K⊗J⊕0.
On one hand, it is clear that[ϕ(X),ϕ(A)]≠0, so we have ϕ(X)∉𝔥1. On the other hand, [ϕ(X),ϕ(H)]=0; hence, [X,H]=0. Thus, X∈𝔥, and so ϕ(X)∈ϕ(𝔥)=𝔥1; this is impossible. Note that ϕ is an automorphism; we see that ϕ-1 also preserves the regular matrix. The proof is completed.Lemma 10.
Suppose thatn≥5 and A,B∈𝔄n. If rankA=2 and B∉ℝA, then there is C∈𝔄n such that
(18)[A,C]=0,[B,C]≠0.Proof.
We can assume without loss the generality by Lemma4 that A=aK⊕0,  a≠0. Hence, we have ℭ(A)=0⊕𝔄n-2. If any matrix C cannot satisfy the conclusion, then one has [B,0⊕𝔄n-2]=0. Note that n≥5, so we have n-2≥3. This implies that B∈ℝK⊕0, which contradicts with B∉ℝA.Lemma 11.
LetA∈𝔄n. Then, ϕ(ℭ(A))=ℭ(ϕ(A)).Proof.
As[A,ℭ(A)]=0, we deduce that [ϕ(A),ϕ(ℭ(A))]=0. Farther, we have ϕ(ℭ(A))⊂ℭ(ϕ(A)). The desired result follows from the following:
(19)ℭ(ϕ(A))=(ϕϕ-1)ℭ(ϕ(A))=ϕ(ϕ-1(ℭ(ϕ(A))))⊂ϕ(ℭ(ϕ-1(ϕ(A))))=ϕ(ℭ(A)).Lemma 12.
Suppose thatA∈𝔄4 is not a regular matrix. Then, ϕ(ℝA)=ℝϕ(A).Proof.
It follows from Lemma6 that there is an orthogonal matrix Q such that
(20)ϕ(rA)=Q(ε1(r)K⊕ε2(r)K)Qt,∀r∈ℝ.
SinceA is not a regular matrix and so is rA, we see that ε2(r)∈{±ε1(r)}, for all r∈ℝ. If ε1(1)=ε2(1), then we will see that ε1=ε2. Otherwise, there is r0∈ℝ× such that ε1(r0)=-ε2(r0), and so Q(K⊗I2)Qt∈ℭ(ϕ(A)). But we know that Q(K⊗I2)Qt∉ℭ(ϕ(r0A)), this, together with Lemma 11, gives that
(21)ℭ(ϕ(A))=ϕ(ℭ(A))=ϕ(ℭ(r0A))=ℭ(ϕ(r0A)).
This is impossible. Similarly, we can show that ifε1(1)=-ε2(1), then ε1=-ε2, and then we get the conclusion.Lemma 13.
Suppose thatn≥5 and A∈𝔄n such that rankA=2. Then, ϕ(ℝA)=ℝϕ(A).Proof.
If there isB∉ℝA such that ϕ(B)∈ℝϕ(A), then by Lemma 10 we can choose C∈𝔄n such that [A,C]=0, [B,C]≠0. Thus, [ϕ(A),ϕ(C)]=0; [ϕ(B),ϕ(C)]≠0. But we see that ϕ(B)∈ℝϕ(A); this is impossible. Furthermore,
(22)ℝϕ(A)⊂ϕ(ℝA).
For any nonzero real numberr, we replace A by rA in the previous equation. It follows that
(23)ℝϕ(ℝA)⊂ϕ(ℝA)⊂ℝϕ(ℝA),
that is, ϕ(ℝA)=ℝϕ(ℝA). Note that ϕ is additive. So, ϕ(ℝA) is a subspace.
Suppose that(24)ϕ(rA)=Qdiag(ε1(r)K⊕⋯⊕ε[n/2](r)K⊕0)Qt,
where Q is an orthogonal matrix, εi:ℝ→ℝ,  i=1,…,[n/2]. We first prove the following.
Assertion. If there is an index i0 such that εi0(r)≠0, for all r∈ℝ×, then ϕ(ℝA)⊂ℝϕ(A).
In fact, for any givenr∈ℝ×, suppose that εi0(r)=ρεi0(1), then ρ≠0. Now, we assume that the assertion is not true; then there is some index s such that εs(r)≠ρεs(1). Then, we see by ϕ(A)∈ϕ(ℝA),  ϕ(rA)∈ϕ(ℝA), and the fact ϕ(ℝA) is a space that ρϕ(A)-ϕ(rA)∈ϕ(ℝA). This tells us that there is some c∈ℝ such that ρϕ(A)-ϕ(rA)=ϕ(cA). Thus, εs(c)=ρεs(1)-εs(r)≠0, and so we have c≠0. But we know that εi0(c)=ρεi0(1)-εi0(r)=0, which contradicts with the conditions of the assertion. This gives that ϕ(rA)=ρϕ(A). The assertion is proved.
AsA is not a regular matrix, one has that ϕ(rA) is not a regular matrix too. Next, the proof of the lemma is divided into the following cases with respect to n.
Case 1. When n is odd, note that A≠0, so we can assume without loss the generality that ε1(1)≠0. If for some r0∈ℝ× such that ε1(r0)=0, then it follows by Lemma 11 that Q(E1n(n)-En1(n))Qt∈ℭ(ϕ(r0A))=ℭ(ϕ(A)), which is a contradiction. Now, the lemma follows by using the previous assertion for the index i0=1.
Case 2. When n is even, assume without loss of the generality that ε1(1)∈{±ε2(1)}. If ε1(1)=ε2(1), then ε1=ε2. In fact, if there is some r0∈ℝ× such that ε1(r0)≠ε2(r0), then one has by Lemma 11 that Q(K⊗I2⊕0)Qt∈ℭ(ϕ(A))=ℭ(ϕ(r0A)) is a contradiction. If ε1(1)=-ε2(1), then ε1=-ε2. In fact, if there is some r0∈ℝ× such that ε1(r0)≠-ε2(r0), then we see by Lemma 11 that Q(K⊗J⊕0)Qt∈ℭ(ϕ(A))=ℭ(ϕ(r0A)); this is impossible.
Whenε1(1)≠0, if there is r0∈ℝ× such that ε1(r0)=0, then ε2(r0)=0. Thus, 𝔄4⊕0⊂ℭ(ϕ(r0A))=ℭ(ϕ(A)), which is a contradiction. Now, we get the lemma by using the previous assertion for the index i0=1.
Whenε1(1)=0, if there is r0∈ℝ× such that ε1(r0)≠0, then since ε2(1)=ε1(1)=0, 𝔄4⊕0⊂ℭ(ϕ(A))=ℭ(ϕ(r0A)). This is absurd. As A≠0, it is clear that n≥6. Hence, we can assume without loss of the generality that ε3(1)≠0. If for some r0∈ℝ× such that ε3(r0)=0, then we have by Lemma 11 that
(25)Q((E15(5)-E51(5))⊕0)Qt∈ℭ(ϕ(r0A))=ℭ(ϕ(A)),
which is a contradiction. The lemma can be shown by using the previous assertion for the index i0=3.Corollary 14.
Suppose thatn≥5 and W≤𝔄n is a subspace with bases which are formed by rank 2 matrices. Then, we have
(26)dimW=dimϕ(W).Proof.
Suppose that rank 2 matricese1,…,es form bases of W. Then,
(27)ℝe1⊕⋯⊕ℝes=W.
It follows immediately that(28)ϕ(ℝe1)+⋯+ϕ(ℝes)=ϕ(W).
If there isi such that ϕ(ℝei)∩Σj≠iϕ(ℝej)≠0, then we can choose λ1,…,λs∈ℝ, not all zero, such that ϕ(λiei)=Σj≠iϕ(λjej), which is absurd. We see by Lemma 13 that ϕ(ℝei)=ℝϕ(ei), for all i=1,…,s. Thus,
(29)ℝϕ(e1)⊕⋯⊕ℝϕ(es)=ϕ(W).
The proof is completed.Lemma 15.
Suppose thatn≥5 and A∈𝔄n is of  rank 2. Then,
(30)dimℭ(ϕ(A))=12(n-2)(n-3)+1.Proof.
Note thatdimℭ(A)=dim𝔄n-2+1=(1/2)(n-2)(n-3)+1 and ℭ(A) has bases which are formed by rank 2 matrices. This, together with Corollary 14, proves the conclusion.Lemma 16.
Suppose thata1,…,as are positive real numbers, which are different from one another. Let x=a1In1⊗K⊕-a1In-1⊗K⊕⋯⊕asIns⊗K⊕-asIn-s⊗K⊕0. Then, we have
(31)dimℭ(x)=(Σi(ni+n-i))2+2-1×(n-2Σi(ni+n-i))(n-2Σi(ni+n-i)-1).
In particular, if we letd=Σi(ni+n-i), then
(32)dimℭ(x)≤d2+2-1(n-2d)(n-2d-1),
and the equation holds if and only if s=1.Proof.
It follows by a direct computation.Lemma 17.
Suppose thatϕ∈Aut𝔄n preserves the rank 2 matrix subset of 𝔄n. Then, there is a real orthogonal matrix Q such that
(33)ϕ(X)=QXQt,∀X∈𝔄n.Proof.
The proof under the casen=2 is obvious. It is not difficult to see that, if n=3, then a surjective map preserving rank 2 matrices still is of the form (i) of Lemma 5. Next, we assume that n≥3 and assume that ϕ has the form (i) of Lemma 5. For distinct i, j, k, it is clear that
(34)[Eij-Eji,Eik-Eki]=Ekj-Ejk,[Eij-Eji,Ejk-Ekj]=Eik-Eki.
Consider the image ofϕ; then, it follows by the form (i) of Lemma 5 that
(35)α((Eij-Eji)PtP(Eik-Eki)-(Eik-Eki)PtP(Eij-Eji))=Ekj-Ejk,α((Eij-Eji)PtP(Ejk-Ekj)-(Ejk-Ekj)PtP(Eij-Eji))=Eik-Eki.
Hence, by a direct computation and the arbitrariness ofi, j, k, it follows that PtP=α-1In. Clearly, α>0. Note that ℝ is the field of real numbers, so we have f=1. Let Q=αP; then, the conclusion is obtained.
Whenn=4 and ϕ is of the form (ii) of Lemma 5, then we let i=1, j=2,  and k=3. Thus, we have by taking the images under ϕ for the previous two equations that 1=0, which is a contradiction. So, the form (ii) of Lemma 5 does not occur.


## 3. The Proof of the Main Result

The proof of the main theorem is divided into the following three propositions.Proposition 18.
Suppose thatn=2 or 3 and ϕ∈Aut𝔄n. Then, there is an orthogonal matrix Q such that
(36)ϕ(X)=QXQt,∀X∈𝔄n.Proof.
Sinceϕ is bijective, ϕ preserves the rank 2 matrices of 𝔄2 or 𝔄3. If n=2, the conclusion is clear. If n=3, then we also can get the conclusion by Lemma 17.Proposition 19.
Suppose thatϕ∈Aut𝔄4. Then, there is an orthogonal matrix Q such that
(37)ϕ(X)=QXQt,∀X∈𝔄4.Proof.
It is clear that[K,D]=2J, [J,K]=2D, and [J,D]=2K. Note that K⊕0 is regular, so we can assume that ϕ(K⊕0)=Q(aK⊕bK)Qt, where a≠±b and Q is an orthogonal matrix. Without loss of generality, one can assume that
(38)ϕ(K⊕0)=aK⊕bK.
Since the regular subring containing the nonregular matrixϕ(I2⊗K) is determined by ϕ(K⊕0), there are c∈ℝ× and ε∈{±1} such that
(39)ϕ(I2⊗K)=c(K⊕εK).
Therefore,(40)ϕ(0⊕K)=ϕ(I2⊗K)-ϕ(K⊕0)=(c-a)K⊕(εc-b)K.
Suppose that(41)ϕ(K⊗I2)=[X1Y1-Y1tZ1],
where X1 is a 2×2 matrix. It follows by [I2⊗K,K⊗I2]=0 that
(42)[c(K⊕εK),ϕ(K⊗I2)]=0.
So, we haveKX=XK, KZ=ZK, KY1=εY1K, and KYt=εYtK. Note that
(43)[K⊕0,[K⊕0,K⊗I2]]=-K⊗I2.
Thus,(44)[[aK00bK],[[aK00bK],[X1Y1-Y1tZ1]]]=-[X1Y1-Y1tZ1].
This, together withKY1=εY1K, gives that (a-εb)2=1, X1=0, Z1=0. We deduce that
(45)ϕ(K⊗I2)=[0Y1-Y1t0].
Similarly, we see by[I2⊗K,D⊗K]=0 and [K⊕0,[K⊕0,D⊗K]]=-D⊗K that
(46)ϕ(D⊗K)=[0Y2-Y2t0],
where Y2K=εKY2. We also have by [K⊗I2,D⊗K]=2J⊗K that
(47)ϕ(J⊗K)=2-1[[0Y1-Y1t0],[0Y2-Y2t0]]=2-1[Y2Y1t-Y1Y2t00Y2tY1-Y1tY2].
Note thatYiK=εKYi, so we can assume that
(48)Y1=[y1y2-εy2εy1],Y2=[y3y4-εy4εy3].
Further,(49)Y2Y1t-Y1Y2t=2ε(y1y4-y2y3)K,Y2tY1-Y1tY2=-2(y1y4-y2y3)K,ϕ(J⊗K)=(y1y4-y2y3)(εK⊕-K).
On the other hand, we know byJ⊗K=K⊕0-0⊕K that
(50)ϕ(J⊗K)=(2a-c)K⊕(2b-εc)K.
By a direct computation with (49) and (50), we have 2a-c=c-2εb=ε(y1y4-y2y3) and then a+εb=c. Noting that (a-εb)2=1, we can assume without loss of the generality that a-εb=1 (for the case a-εb=-1, the proof is similar). We deduce that
(51)ϕ(J⊗K)=J⊗K.
Ifε=-1, then ϕ(I2⊗K)=c(K⊕-K)∈ℝϕ(J⊗K), which contradicts Lemma 12. This tells us that ε=1.
Again by[K⊕0,K⊗I2]=D⊗K, we get Y2=KY1.
Suppose that(52)ϕ(K⊗J)=[X3Y3-Y3tZ3].
It follows by[J⊗K,K⊗J]=0 that
(53)[J⊗K,[X3Y3-Y3tZ3]]=0.
Therefore, we haveKY3+Y3K=0. Note that [I2⊗K,[I2⊗K,K⊗J]]=-4K⊗J. This tells us that c2=1. We can assume without loss of the generality that c=1. Then, we get that a=1 and b=0. Thus,
(54)ϕ(K⊕0)=K⊕0.
For any but fixedr∈ℝ×, we assert that ϕ(rK⊕0)∈ℝK⊕0.
In fact, firstly, by Lemma12, we can assume that ϕ(rI2⊗K)=sI2⊗K, ϕ(rK⊕0)=uK⊕vK. So, we have that ϕ(0⊕rK)=(s-u)K⊕(s-v)K.
Secondly, noting that[I2⊗K,rK⊗I2]=0 and [rK⊕0,[K⊕0,K⊗I2]]=-rK⊗I2, we deduce that
(55)ϕ(rK⊗I2)=(u-v)ϕ(K⊗I2).
Furthermore, we see byrJ⊗K=(1/2)[rK⊗I2,[K⊕0,K⊗I2]] and rJ⊗K=rK⊕0-0⊕rK that
(56)ϕ(rJ⊗K)=(u-v)J⊗K,ϕ(rJ⊗K)=(2u-s)K⊕(2v-s)K.
Thus,(2u-s)=u-v=(s-2v). It follows that u+v=s.
Finally, due to[rK⊕0,[K⊕0,K⊗J]]=-rK⊗J and [rI2⊗K,[I2⊗K,K⊗J]]=-4rK⊗J, one can obtain that
(57)ϕ(rK⊗J)=sϕ(K⊗J)=(u-v)ϕ(K⊗J).
This tells us thatu-v=s, and so we have that u=s,  v=0. In other words, ϕ(rK⊕0)=uK⊕0, which proves the assertion. Now, we prove that ϕ preserves the set of rank 2 matrices on 𝔄4. By applying Lemma 17, we finish the proof.Proposition 20.
Suppose thatn≥5 and ϕ∈Aut𝔄n. Then, there is an orthogonal matrix Q such that
(58)ϕ(X)=QXQt,∀X∈𝔄n.Proof.
Take any rank 2 matrixA∈𝔄n. By Lemma 4, we can assume that
(59)ϕ(A)=Q(a1In1⊗K⊕-a1In-1⊗K⊕⋯⊕asIns⊗K⊕-asIn-s⊗K⊕0)Qt.
LetΣi(ni+n-i)=d. Now, we assert that d=1 and so that the rank of ϕ(A) is 2; that is, we will assert that ϕ is a preserver of rank 2 on 𝔄n; then, we can finish the proof by Lemma 17.
It follows by Lemmas15 and 16 that
(60)12(n-2)(n-3)+1≤d2+2-1(n-2d)(n-2d-1).
Moreover, we see that(d-1)(3d-2n+4)≥0. Hence, we have either d≤1 or d≥3-1(2n-4). The former means that d=1, as desired. If the latter holds, then it is clear that
(61)n≥2d≥2·3-1(2n-4).
In this case, we deduce thatn≤8 and n≠7. Hence, the remainder of the proof is the cases (i) n=5, d=2, (ii) n=6, d=3, and (iii) n=8, d=4.
Suppose thatB=Q(0⊕K⊕0)Qt. We consider the rank of ϕ(B).
Whenrankϕ(B)=2, it is clear that there is an orthogonal matrix P such that ϕ(A)=P(εIp⊗K⊕-εIq⊗K⊕0)Pt and ϕ(B)=P(ηK⊕0)Pt. Without loss of the generality, we can assume that p≠0. Note that η≠0. If η≠-2ε, then one has ε+η≠-ε. Let C=Q(K⊗I2⊕0)Qt. As [A+B,C]=0, we can find a matrix X∈𝔄n-4 such that
(62)ϕ(C)=P(0⊕X)Pt.
Ifη=-2ε, then ε-η≠-ε. Let C=Q(K⊗J⊕0)Qt. Since [A-B,C]=0, there is a matrix X∈𝔄n-4 such that
(63)ϕ(C)=P(0⊕X)Pt.
Thanks to[B,[B,C]]=-C, we deduce ϕ(C)=0, which is a contradiction.
Whenrankϕ(B)≠2, then for the previous three cases of n and d, one always has rankϕ(B)=rankϕ(A). Note that ϕ(A) and ϕ(B) are in a common regular subring, and s=1. It follows by Lemma 6 that there is an orthogonal matrix P such that ϕ(A)=P(ε1K⊕⋯⊕εdK⊕0)Pt and ϕ(B)=P(η1K⊕⋯⊕ηdK⊕0)Pt, where ηi∈{±η1}, εi∈{±ε1}. Due to dimℭ(A+B)=dimℭ(A-B), we see by Lemma 11 that
(64)dimℭ(ϕ(A)+ϕ(B))=dimℭ(ϕ(A)-ϕ(B))=dimℭ(A-B)=dim𝔄n-4+4.
Case 1.
n
=
5. We first prove that rank(ϕ(A)±ϕ(B))≠2.
Ifrank(ϕ(A)+ϕ(B))=2, then we may as well assume that ϕ(A)=εP(K⊕K⊕0)Pt and ϕ(B)=εP(K⊕-K⊕0)Pt. Let E=K⊗I2⊕0, F=K⊗J⊕0. It is easy to see that [E,F]=0. Now, we want to show that [ϕ(E),ϕ(F)]≠0, which is a contradiction. Note the following:
(65)[A,[A,E]]=-E,[A,[A,F]]=-F,[B,[B,E]]=-E,[B,[B,F]]=-F.
So, we know that bothϕ(E) and ϕ(F) satisfy an equation about the matrix X=[xij]∈𝔄5 as follows:
(66)[ϕ(A),[ϕ(A),X]]=-X,[ϕ(B),[ϕ(B),X]]=-X.
That is,(67)[002ε2(x24-x13)-2ε2(x14+x23)-ε2x1500-2ε2(x14+x23)2ε2(x13-x24)-ε2x252ε2(x13-x24)2ε2(x14+x23)00-ε2x352ε2(x14+x23)2ε2(x24-x13)00-ε2x45ε2x15ε2x25ε2x35ε2x450]=-X,[00-2ε2(x24+x13)2ε2(x23-x14)-ε2x15002ε2(x14-x23)-2ε2(x13+x24)-ε2x252ε2(x24+x13)2ε2(x23-x14)00-ε2x352ε2(x14-x23)2ε2(x13+x24)00-ε2x45ε2x15ε2x25ε2x35ε2x450]=-X.
Hence, we get that(68)X=[0000x150000x250000x350000x45-x15-x25-x35-x450],
and ε2=1. Note that [A+B,E]=0,  [A-B,F]=0. After taking the image, we can assume by ε≠0 that
(69)ϕ(E)=[00000000000000a0000b00-a-b0],ϕ(F)=[0000c0000d0000000000-c-d000].
Again by[E,F]=0, we see that
(70)[ϕ(E),ϕ(F)]=[00acbc000adbd0-ac-ad000-bc-bd00000000]=0.
We deduce thatac=0, bc=0. It follows by ϕ(E)≠0 that c=0. Due to ad=0, bd=0, one has d=0. This tells us that ϕ(F)=0, which is a contradiction. Similarly, we know that rank(ϕ(A)-ϕ(B))≠2.
Sincen=5, it is clear that rank(ϕ(A)±ϕ(B))=4. When
(71)ϕ(A)=εP(K⊕K⊕0)Pt,ϕ(B)=ηP(K⊕-K⊕0)Pt,
we have that ε≠η. Note that A±B is not a regular matrix; hence, ϕ(A)±ϕ(B) is not too. Further, one has ε+η∈{±(ε-η)}. This implies that ε=0 or η=0, which is impossible. Similarly, we deduce that
(72)ϕ(A)=εP(K⊕-K⊕0)Pt,ϕ(B)=ηP(K⊕K⊕0)Pt,
which is also a contradiction.
Case 2.
n
=
6,8. We first prove that rank(ϕ(A)±ϕ(B))≠2. Otherwise, if  rank(ϕ(A)+ϕ(B))=2, then we have
(73)dimℭ(ϕ(A)+ϕ(B))=dim𝔄n-2+1≠dim𝔄n-4+4,
which is a contradiction. In a similar way, we get rank(ϕ(A)-ϕ(B))≠2.
Ifn=6, we assert that rank(ϕ(A)±ϕ(B))≠4. In fact, if rank(ϕ(A)+ϕ(B))=4, then by ηi∈{±η1}, εi∈{±ε1}, we deduce that ηi∈{±ε1}. Without loss of the generality, we can assume that εi=ηi, i=1,2, and εj=-ηj, j=3. Hence, we see that rank(ϕ(A)-ϕ(B))=2, which is impossible. Similarly, we deduce that rank(ϕ(A)-ϕ(B))≠4.
Next, we prove whenn=6 that rank(ϕ(A)±ϕ(B))≠6. Otherwise, by (64) we can assume without loss of the generality that ε2+η2∈{±(ε1+η1)} and ε3+η3∉{±(ε1+η1)}. Note that ηi∈{±η1},  εi∈{±ε1}, so we have ε2-η2∈{±(ε1-η1)} and ε3-η3∉{±(ε1-η1)}. Thus,
(74)ℭ(ϕ(A+B))=ℭ(ϕ(A-B)).
But it is clear thatℭ(A+B)≠ℭ(A-B), which contradicts with ℭ(ϕ(X))=ϕ(ℭ(X)), forallX∈𝔄n.
Similarly, we have whenn=8 that rank(ϕ(A)±ϕ(B))≠6,8.
Finally, we prove whenn=8 that rank(ϕ(A)±ϕ(B))≠4. Let Z=Q(E33(4)⊗K)Qt. If rankϕ(Z)=2, then we can find a contradiction similar to the case of rankϕ(B)=2. Otherwise, if rankϕ(Z)=8, then there is an orthogonal matrix P such that
(75)ϕ(A)=P(ε1K⊕⋯⊕εdK)Pt,ϕ(B)=P(η1K⊕⋯⊕ηdK)Pt,ϕ(Z)=P(λ1K⊕⋯⊕λdK)Pt,
where ηi∈{±η1}, εi∈{±ε1}, and λi∈{±λ1}. It is easy to see that the three cases rank(ϕ(A)±ϕ(B))=4, rank(ϕ(A)±ϕ(Z))=4, and rank(ϕ(Z)±ϕ(B))=4 cannot simultaneously hold. This means that rank(ϕ(A)±ϕ(B))=4 is impossible.
To sum up the previous arguments, we get thatrankϕ(A)=2. The proof is completed.



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*Source: 638230-2013-08-18.xml*