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

Proposition 1 (see [10], [11].)Let A^′ 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 $$ 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.

Theorem 2. Suppose that n ≥ 2 is an integer, then ϕ ∈ Aut𝔄_n if and only if there is a real orthogonal matrix Q such that

Further, one has Aut𝔄_n≅O_n(R), where O_n(R) is the real orthogonal group.

Definition 3. A matrix A ∈ 𝔄_n 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, …, ε_n/2 ∈ ℝ with different absolute values, such that

(ii) When n 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

Lemma 4 (see [12], 2.5.14.)Suppose that A ∈ 𝔄_n. Then, there are an orthogonal matrix Q and real numbers a₁, …, a_[n/2] such that

Lemma 5 (see [13], [14].)Let 𝔽 be any field, and let K_n(𝔽) denote the space of all n × n alternate matrices over 𝔽. Then, ϕ is an additive surjective mapping of K_n(𝔽) (n ≥ 2) to itself that preserves rank 2 matrices if and only if ϕ is of the following forms:

• (i)

  n ≥ 4, ϕ((a_ij)) = αP^t(f(a_ij))P, for all (a_ij) ∈ K_n(𝔽), where α ∈ 𝔽∖{0}, P is an n × n invertible matrix, and f is a field automorphism of 𝔽;

• (ii)

  when n = 4, ϕ is of the form

  $$ ()

where α, P, and f have the same meaning as before, and (a_ij)↦(a_ij) ^* is either the identity map or the map:

Lemma 6. Suppose that 𝔥 is a regular subring of 𝔄_n. Then, there is an orthogonal matrix Q and maps η_i : 𝔄_n → ℝ, i = 1, …, [n/2] such that

Proof. For every X ∈ 𝔥, note that QXQ^t is commutative with every regular matrix in 𝔥. So, one can obtain the conclusion by Lemma 4.

Corollary 7. Suppose that 𝔥 is a regular subring of 𝔄_n, and H ∈ 𝔥 is a regular matrix. Then,

Lemma 8. Suppose that 𝔥₁, 𝔥₂ are both regular subrings of 𝔄_n, 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 Corollary 7.

Lemma 9. Both maps ϕ and ϕ⁻¹ preserve the regular subring. Expressly, for H ∈ 𝔄_n, one has that H is a regular matrix if and only if ϕ(H) is so.

Proof. Take any regular subring 𝔥 of 𝔄_n and a regular matrix H ∈ 𝔥. By Lemma 6, we can assume that Qϕ(H)Q^t = ε₁K ⊕ ⋯⊕ε_[n/2]K ⊕ 0, where Q is an orthogonal matrix.

Suppose that A ∈ 𝔄_n satisfying Qϕ(A)Q^t = 1K ⊕ 2K ⊕ ⋯⊕[n/2]K ⊕ 0. Then, ϕ(A) is a regular matrix in 𝔄_n. 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

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 that ϕ 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 ε_i = 0.

Case 2. There is some ε_i ∈ {±ε_j}.

If Case 1 happens, we assume without loss the generality that ε₁ = 0. We take X ∈ 𝔄_n such that

If Case 2 happens, we assume without loss the generality that ε₂ ∈ {±ε₁}. When ε₁ = ε₂, we take X ∈ 𝔄_n such that

When ε₁ = −ε₂, we take X ∈ 𝔄_n such that

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 that n ≥ 5 and A, B ∈ 𝔄_n. If rankA = 2 and B ∉ ℝA, then there is C ∈ 𝔄_n such that

Proof. We can assume without loss the generality by Lemma 4 that A = aK ⊕ 0, a ≠ 0. Hence, we have ℭ(A) = 0 ⊕ 𝔄_n−2. If any matrix C cannot satisfy the conclusion, then one has [B, 0 ⊕ 𝔄_n−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. Let A ∈ 𝔄_n. 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:

Lemma 12. Suppose that A ∈ 𝔄₄ is not a regular matrix. Then, ϕ(ℝA) = ℝϕ(A).

Proof. It follows from Lemma 6 that there is an orthogonal matrix Q such that

Since A 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)Q^t ∈ ℭ(ϕ(A)). But we know that Q(K ⊗ I₂)Q^t ∉ ℭ(ϕ(r₀A)), this, together with Lemma 11, gives that

This is impossible. Similarly, we can show that if ε₁(1) = −ε₂(1), then ε₁ = −ε₂, and then we get the conclusion.

Lemma 13. Suppose that n ≥ 5 and A ∈ 𝔄_n such that rankA = 2. Then, ϕ(ℝA) = ℝϕ(A).

Proof. If there is B ∉ ℝA such that ϕ(B) ∈ ℝϕ(A), then by Lemma 10 we can choose C ∈ 𝔄_n 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,

For any nonzero real number r, we replace A by rA in the previous equation. It follows that

that is, ϕ(ℝA) = ℝϕ(ℝA). Note that ϕ is additive. So, ϕ(ℝA) is a subspace.

Suppose that

where Q is an orthogonal matrix, ε_i : ℝ → ℝ, i = 1, …, [n/2]. We first prove the following.

Assertion. If there is an index i₀ such that $$, for all r ∈ ℝ^×, then ϕ(ℝA) ⊂ ℝϕ(A).

In fact, for any given r ∈ ℝ^×, suppose that $$, then ρ ≠ 0. Now, we assume that the assertion is not true; then there is some index s such that ε_s(r) ≠ ρε_s(1). Then, we see by ϕ(A) ∈ ϕ(ℝA), ϕ(rA) ∈ ϕ(ℝA), and the fact ϕ(ℝA) is a space that ρϕ(A) − ϕ(rA) ∈ ϕ(ℝA). This tells us that there is some c ∈ ℝ such that ρϕ(A) − ϕ(rA) = ϕ(cA). Thus, ε_s(c) = ρε_s(1) − ε_s(r) ≠ 0, and so we have c ≠ 0. But we know that $$, which contradicts with the conditions of the assertion. This gives that ϕ(rA) = ρϕ(A). The assertion is proved.

As A 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 $$, 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)Q^t ∈ 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)Q^t ∈ ℭ(ϕ(A)) = ℭ(ϕ(r₀A)); this is impossible.

When ε₁(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.

When ε₁(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

which is a contradiction. The lemma can be shown by using the previous assertion for the index i₀ = 3.

Corollary 14. Suppose that n ≥ 5 and W ≤ 𝔄_n is a subspace with bases which are formed by rank 2 matrices. Then, we have

Proof. Suppose that rank 2 matrices e₁, …, e_s form bases of W. Then,

It follows immediately that

If there is i such that ϕ(ℝe_i)∩Σ_j≠iϕ(ℝe_j) ≠ 0, then we can choose λ₁, …, λ_s ∈ ℝ, not all zero, such that ϕ(λ_ie_i) = Σ_j≠i  ϕ(λ_je_j), which is absurd. We see by Lemma 13 that ϕ(ℝe_i) = ℝϕ(e_i), for all i = 1, …, s. Thus,

The proof is completed.

Lemma 15. Suppose that n ≥ 5 and A ∈ 𝔄_n is of rank 2. Then,

Proof. Note that dimℭ(A) = dim𝔄_n−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 that a₁, …, a_s are positive real numbers, which are different from one another. Let $$. Then, we have

In particular, if we let d = Σ_i(n_i + n_−i), then

and the equation holds if and only if s = 1.

Proof. It follows by a direct computation.

Lemma 17. Suppose that ϕ ∈ Aut𝔄_n preserves the rank 2 matrix subset of 𝔄_n. Then, there is a real orthogonal matrix Q such that

Proof. The proof under the case n = 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

Consider the image of ϕ; then, it follows by the form (i) of Lemma 5 that

Hence, by a direct computation and the arbitrariness of i, j, k, it follows that P^tP = α⁻¹I_n. Clearly, α > 0. Note that ℝ is the field of real numbers, so we have f = 1. Let $$; then, the conclusion is obtained.

When n = 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.

Proposition 18. Suppose that n = 2 or 3 and ϕ ∈ Aut𝔄_n. Then, there is an orthogonal matrix Q such that

Proof. Since ϕ 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 that ϕ ∈ Aut𝔄₄. Then, there is an orthogonal matrix Q such that

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) Q^t, where a ≠ ±b and Q is an orthogonal matrix. Without loss of generality, one can assume that

Since the regular subring containing the nonregular matrix ϕ(I₂ ⊗ K) is determined by ϕ(K ⊕ 0), there are c ∈ ℝ^× and ε ∈ {±1} such that

Therefore,

Suppose that

where X₁ is a 2 × 2 matrix. It follows by [I₂ ⊗ K, K ⊗ I₂] = 0 that

So, we have KX = XK, KZ = ZK, KY₁ = εY₁K, and KY^t = εY^tK. Note that

Thus,

This, together with KY₁ = εY₁K, gives that (a−εb)² = 1, X₁ = 0, Z₁ = 0. We deduce that

Similarly, we see by [I₂ ⊗ K, D ⊗ K] = 0 and [K ⊕ 0, [K ⊕ 0, D ⊗ K]] = −D ⊗ K that

where Y₂K = εKY₂. We also have by [K ⊗ I₂, D ⊗ K] = 2J ⊗ K that

Note that Y_iK = εKY_i, so we can assume that

Further,

On the other hand, we know by J ⊗ K = K ⊕ 0 − 0 ⊕ K that

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

If ε = −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

It follows by [J ⊗ K, K ⊗ J] = 0 that

Therefore, we have KY₃ + 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,

For any but fixed r ∈ ℝ^×, we assert that ϕ(rK ⊕ 0) ∈ ℝK ⊕ 0.

In fact, firstly, by Lemma 12, 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

Furthermore, we see by rJ ⊗ K = (1/2)[rK ⊗ I₂, [K ⊕ 0, K ⊗ I₂]] and rJ ⊗ K = rK ⊕ 0 − 0 ⊕ rK that

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

This tells us that u − 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 that n ≥ 5 and ϕ ∈ Aut𝔄_n. Then, there is an orthogonal matrix Q such that

Proof. Take any rank 2 matrix A ∈ 𝔄_n. By Lemma 4, we can assume that

Let Σ_i(n_i + 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 𝔄_n; then, we can finish the proof by Lemma 17.

It follows by Lemmas 15 and 16 that

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

In this case, we deduce that n ≤ 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 that B = Q(0 ⊕ K ⊕ 0)Q^t. We consider the rank of ϕ(B).

When rankϕ(B) = 2, it is clear that there is an orthogonal matrix P such that ϕ(A) = P(εI_p ⊗ K ⊕ −εI_q ⊗ K ⊕ 0)P^t and ϕ(B) = P(ηK ⊕ 0)P^t. 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)Q^t. As [A + B, C] = 0, we can find a matrix X ∈ 𝔄_n−4 such that

If η = −2ε, then ε − η ≠ −ε. Let C = Q(K ⊗ J ⊕ 0)Q^t. Since [A − B, C] = 0, there is a matrix X ∈ 𝔄_n−4 such that

Thanks to [B, [B, C]] = −C, we deduce ϕ(C) = 0, which is a contradiction.

When rankϕ(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 ⊕ ⋯⊕ε_dK ⊕ 0)P^t and ϕ(B) = P(η₁K ⊕ ⋯⊕η_dK ⊕ 0)P^t, where η_i ∈ {±η₁}, ε_i ∈ {±ε₁}. Due to dimℭ(A + B) = dimℭ(A − B), we see by Lemma 11 that

Case 1. n = 5. We first prove that rank(ϕ(A) ± ϕ(B)) ≠ 2.

If rank(ϕ(A) + ϕ(B)) = 2, then we may as well assume that ϕ(A) = εP(K ⊕ K ⊕ 0)P^t and ϕ(B) = εP(K ⊕ −K ⊕ 0)P^t. 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:

So, we know that both ϕ(E) and ϕ(F) satisfy an equation about the matrix X = [x_ij] ∈ 𝔄₅ as follows:

That is,

Hence, we get that

and ε² = 1. Note that [A + B, E] = 0, [A − B, F] = 0. After taking the image, we can assume by ε ≠ 0 that

Again by [E, F] = 0, we see that

We deduce that ac = 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.

Since n = 5, it is clear that rank(ϕ(A) ± ϕ(B)) = 4. When

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 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

which is a contradiction. In a similar way, we get rank(ϕ(A) − ϕ(B)) ≠ 2.

If n = 6, we assert that rank(ϕ(A) ± ϕ(B)) ≠ 4. In fact, if rank(ϕ(A) + ϕ(B)) = 4, then by η_i ∈ {±η₁}, ε_i ∈ {±ε₁}, we deduce that η_i ∈ {±ε₁}. Without loss of the generality, we can assume that ε_i = η_i, i = 1,2, and ε_j = −η_j, 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 when n = 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 η_i ∈ {±η₁}, ε_i ∈ {±ε₁}, so we have ε₂ − η₂ ∈ {±(ε₁ − η₁)} and ε₃ − η₃ ∉ {±(ε₁ − η₁)}. Thus,

But it is clear that ℭ(A + B) ≠ ℭ(A − B), which contradicts with ℭ(ϕ(X)) = ϕ(ℭ(X)), for  all  X ∈ 𝔄_n.

Similarly, we have when n = 8 that rank(ϕ(A) ± ϕ(B)) ≠ 6,8.

Finally, we prove when n = 8 that rank(ϕ(A) ± ϕ(B)) ≠ 4. Let $$. If rank ϕ(Z) = 2, then we can find a contradiction similar to the case of rank ϕ(B) = 2. Otherwise, if rank ϕ(Z) = 8, then there is an orthogonal matrix P such that

where η_i ∈ {±η₁}, ε_i ∈ {±ε₁}, and λ_i ∈ {±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 that rank ϕ(A) = 2. The proof is completed.