# Automorphisms and Inner Automorphisms

Authors: Ameer  Jaber; Moh’D Yasein
Journal: Journal of Mathematics
 (2016)

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## Abstract

LetK be a field of characteristic not 2 and let A=A0+A1 be central simple superalgebra over K, and let ⁎ be superinvolution on A. Our main purpose is to classify the group of automorphisms and inner automorphisms of (A,⁎) (i.e., commuting with ⁎) by using the classical theorem of Skolem-Noether. Also we study two examples of groups of automorphisms and inner automorphisms on even central simple superalgebras with superinvolutions.


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## Body

## 1. Introduction

Anassociative super ring R=R0¯+R1¯ is nothing but a (Z/2Z)-gradedassociative ring. A (Z/2Z)-graded ideal I=I0¯+I1¯ of an associative super ring R is called asuperideal of R. An associativesuper ringR issimple if it has no nontrivial superideals. Let R be an associative super ring with 1∈R0¯; then R is said to be adivision super ring if all nonzero homogeneous elements are invertible; that is, every 0≠rα∈Rα has an inverse rα-1 necessarily in Rα.LetK be a field of characteristic not 2; anassociative(Z/2Z)-graded K-algebra A=A0¯+A1¯ is afinite dimensional central simple superalgebra over a field K, if Z^(A)=K is thecenter of A, where (1)Z^Aα=aα∈Aα∣aαbβ=-1αβbβaα∀bβ∈Aβ,and the only superideals of A are (0) and A itself. We say that A0¯∪A1¯ is the set of all homogeneous elements of A.Finite dimensional central simple associative superalgebras over a fieldK are isomorphic to End⁡V≅Mn(D), where D=D0+D1 is a finite dimensional associative division superalgebra over K; that is, all nonzero elements of Dα, α=0,1, are invertible, and V=V0+V1 is an n-dimensional D-superspace. If D1=0, the grading of Mn(D) is induced by that of V=V0+V1, A=Mp+q(D), p=dimDV0, and q=dimDV1, so p+q is a nontrivial decomposition of n. Meanwhile, if D1≠0, then the grading of Mn(D) is given by (Mn(D))α=Mn(Dα), α=0,1.For completeness, we recall the structure theorem for central simple associative division superalgebras.Theorem 1 (Division Superalgebra Theorem [1, 2]).
IfD=D0+D1 is a finite dimensional associative division superalgebra over a field K, then exactly one of the following holds, where throughout E denotes a finite dimensional associative division algebra over K:(i)
D=D0=E, and D1=0.(ii)
D=E⊗KK[u], u2=λ∈K×, D0=E⊗K1, and D1=E⊗Ku.(iii)
D=E or M2(E); u∈D such that u2=λ∈K/K2  (λ∈K/α+α2∣α∈K if Char(K)=2)D0=CD(u), D1=SD(u), where CDu=d∈D∣du=ud and SDu=d∈D∣du=uσd, for some quadratic Galois extension K[u]⊆D with Galois automorphism σ. Moreover, in the second case, u=01λ0(u=01λ1 if Char(K)=2) and K[u] does not embed in E.Following [3] we say that a division superalgebra D iseven if Z(D)∩D1=0; that is, D is even if its form is (i) or (iii), and we say that D isodd otherwise; that is, D is odd if its form is (ii). Also, if A=Mn(D) is a finite dimensional central simple superalgebra over a field K, then we say that A is an even K-superalgebra if D is even division superalgebra and A is odd K-superalgebra if D is odd division superalgebra.In [1] Racine described all types of superinvolutions on A=Mp+q(D). It appears that if ∗ is a superinvolution on A such that (A0,∗) is simple algebra, then p=q and ∗ is conjugate to the transpose involution. Otherwise, ∗ is conjugate to the orthosymplectic involution.In [4] we proved that if A is a finite dimensional central simple associative superalgebra over a field K of characteristic not 2 such that A has a superinvolution of the first kind, then A=Mp+q(D), where D is a division algebra over K.In [5] we proved that A=Mn(D), where D1≠0 has a pseudosuperinvolution of the first kind if and only if A is of order 2 in the Brauer-Wall group BW(K), where K is a field of characteristic not 2. But if K is a field of characteristic 2, and A is a central simple associative superalgebra over K, then a superinvolution (which is a pseudosuperinvolution) on A is just an involution on A respecting the grading. Moreover, if A is of order 2 in the Brauer-Wall group BW(K), then the supercenter of A equals the center of A and ⊗^K=⊗K, which means that A is of order 2 in the Brauer group Br(K). Thus, by theorem of Albert, A has an involution of the first kind, but since A is of order 2 in the Brauer-Wall group BW(K), A has superantiautomorphism of the first kind respecting the grading; therefore by [6, Chapter 8, Theorem  8.2] A has an involution of the first kind respecting the grading, which means that A has a superinvolution (which is a pseudosuperinvolution) of the first kind if and only if A is of order 2 in the Brauer-Wall group BW(K).LetK/k be a separable quadratic field extension over k with Galois group Gal(K/k), where Gal(K/k)=1,σ and θ2∈k-k2 and σ(θ)=-θ (θ2∈k/α+α2∣α∈k, σ(θ)=θ+1 if Char(k)=2). We recall a theorem ofAlbert-Reihm on the existence of K/k-involution (involution of the second kind) which states that finite dimensional central simple algebra A over K has a K/k-involution if and only if the corestriction of A splits over k. In [7] Elduque and Villa gave much better exposition and motivation for the whole theory of the existence of superinvolutions.Throughout this work we say that ifu is homogeneous element in A, then ψu:A→A is an inner automorphism on A, with conjugation by u. Also Aut(A,∗), where A is a central simple superalgebra of any type, means the set of all automorphisms on A commuting with a given superinvolution ∗ defined on A, and inAut(A,∗) means the set of all inner automorphisms on A commuting with ∗.In this paper we examine the characterization of automorphisms and inner automorphisms on(A,∗), where A is a central simple superalgebra of any type, which commute with a given superinvolution ∗ defined on A. Then we produce two examples of even superalgebras with superinvolution to investigate in detail groups of automorphisms and inner automorphisms in these examples.


## 2. Group of Automorphisms of Superalgebras

Definition 2.
LetA be any K-superalgebra; we define the map φ:A→A by (2)aαφ=-1αaα∀aα∈Aα,α=0¯,1¯.This map,φ, is a superalgebra automorphism, called the sign automorphism, since (3)aαbβφ=-1α+βaαbβ=aαφbβφfor all aα∈Aα and bβ∈Aβ. The automorphism φ has order 2, if Char(K)≠2 (unless A1¯=0), and φ=idA if Char(K)=2.Lemma 3.
LetA=A0-+A0-v be an odd superalgebra over a field K with a superinvolution ∗, and let φ be the sign automorphism. Then (4)AutA,∗=inAutA,∗∪φ·inAutA,∗.Proof.
Ifξ∈Aut(A,∗), then by [4, Theorem  3.2] Skolem-Noether theorem forces that ξ or ξφ is inner but not both of them, so, ξ∈inAut(A,∗) or ξφ∈inAut(A,∗).Now, let∗ be a superinvolution on an odd central simple superalgebra A=A0¯+A0¯v of any kind and let ξ∈inAut(A,∗). Then ξ=ψcγ for some invertible cγ∈Aγ. But ξA0¯ is inner, so ξA0¯=ψb0¯. Therefore, cγ(b0¯-1ab0¯)cγ-1=a, ∀a∈A0¯, and cγ(b0¯-1vb0¯)cγ-1=v which implies that ψcγb0¯-1=idA⇒ψcγ=ψb0¯. So, we can choose γ to be 0¯, and hence ψc0¯=ψb0¯⇔c0¯=αb0¯ for some α∈K.Now,aα∗ψb0¯=aαψb0¯∗⇔b0¯∗b0¯=λ∈K⇔b0¯∗=λb0¯-1. Let (5)G=b0¯∈A0¯:b0¯∗=αb0¯-1for  someα∈K×.Let ~ be a relation on G defined as follows: a0¯~b0¯⇔b0¯a0¯-1=α∈K×. Then one can easily show that ~ is an equivalence relation on G, and G/~={[b0¯]∣b0¯∈G}, where b0¯={αb0¯∣α∈K×}, is a group with the well-defined operation [b0¯][c0¯]=[c0¯b0¯].Theorem 4.
LetA be an odd central simple superalgebra over a field K with a superinvolution ∗; then inAut(A,∗)≅G/~.Proof.
Letδ:inAut(A,∗)→G/~ such that δ(ψb0¯)=[b0¯]; then (6)δψb0¯ψc0¯=δψc0¯b0¯=c0¯b0¯=b0¯c0¯=δψb0¯δψc0¯.So, δ is an onto homomorphism such that ker⁡(δ)=ψ1. Therefore, inAut(A,∗)≅G/~.LetA=Mn(D) be an even central simple superalgebra over K, where D is a nontrivial grading division superalgebra. Then by [1, Division Superalgebra Theorem] D=D0¯+D1¯=CD(u)+CD(u)v, where uv=-vu and Z(A0¯)=K[u] and CD(u) is the centralizer of u in D.Let∗ be a superinvolution on A, and let ξ∈Aut(A,∗). Then by the Skolem-Noether theorem there exists an invertible element cγ∈Aγ such that aξ=cγacγ-1, ∀a∈A. Since aξ∗=a∗ξ, ∀a∈A, we have cγcγ∗=λ for some λ∈K. If ξA0¯=ψcγA0¯ is inner, then ξA0¯=ψb0¯ for some invertible element b0¯∈A0¯ which implies that aξ=b0¯ab0¯-1, ∀a∈A0¯, and therefore ψb0¯-1cγ=idA0¯. Let xγ=b0¯-1cγ; if γ=1¯, then xγaxγ-1=a∀a∈A0¯. But A1¯=A0¯xγ, so xγ centralizes A1¯; thus xγ∈Z(A)=K. This is a contradiction and therefore γ=0¯. Therefore, for any ξ∈Aut(A,∗) such that ξA0¯ is inner, ξ=ψb0¯, for some invertible element b0¯∈A0¯, and hence (7)inAutA,∗=ξ∈AutA,∗:ξ=ψb0¯,b0¯∈A0¯.Corollary 5.
LetA=Mn(D) be an even central simple superalgebra over a field K, where (8)D=CDu+CDuvand uv=-vu is a nontrivial grading division superalgebra, and let ∗ be a superinvolution on A. Then(9)AutA,∗=inAutA,∗∪ψv·inAutA,∗:ifψv∗=∗ψvinAutA,∗:ifψv∗≠∗ψv.Proof.
Ifψv∗=∗ψv, then for any ξ∈Aut(A,∗) such that uξ=-u we get uξψv=-uψv=u which implies that ξψvA0¯ is inner and so ξψv=ψb0¯ for some invertible b0¯ in A0¯. Thus ξ=ψvψb0¯. But if ψv∗≠∗ψv, then it is easy to check that Aut(A,∗)=inAut(A,∗). Therefore (10)AutA,∗=inAutA,∗∪ψv·inAutA,∗:ifψv∗=∗ψvinAutA,∗:ifψv∗≠∗ψv.Theorem 6.
LetA=Mn(D), where D=D0-+D1-=CD(u)+CD(u)v, and let uv=-vu be a superalgebra with a superinvolution ∗. Then inAut(A,∗)≅G/~.Proof.
Letδ:inAut(A,∗)→G/~ such that δ(ψb0¯)=[b0¯]. Then (11)δψb0¯ψc0¯=δψc0¯b0¯=c0¯b0¯=b0¯c0¯=δψb0¯δψc0¯.So, δ is an onto homomorphism, where ker⁡δ=idA=ψ1¯. Therefore, inAut(A,∗)≅G/~.Next we introduce this example to show thatuu∗ may equal λ with λ≠αα∗ for any α∈K.Example 7.
Letk=Q(e), let K=k(2), and let u=ei, v=24j, where i2=-1, j2=-1, and ij=-ji. Then A=(K+Ku)⊕(Kv+Kuv) is a quaternion division superalgebra over field K. Let ∗ be a superinvolution on A defined by(a+b2)∗=a-b2, ∀a,b∈k; u∗=u; v∗=v; then (uv)∗=vu=-uv.Now, ψu∈Aut(A,∗), since uu∗=-e∈K. But uu∗≠αα∗, where α∈K, because if (12)-e=a+b2a+b2∗,where a,b∈k, then (13)-e=a+b2a-b2=a2-2b2=p1eq1e2-2p2eq2e2,where pi(e),qi(e)∈Q[e] and qi(e)≠0. Hence, (14)-eq1e2q2e2=p1eq2e2-2p2eq1e2.Since 2∉Q, the highest power of e in the right-hand side is even, but the highest power of e in the left-hand side is odd, a contradiction.Now we try to classify the group of automorphisms of central simple superalgebrasA=M2p(D0¯), and A=Mp+q(D0¯), where D0¯ is a division algebra over K and (p≠q).First letA=M2p(D0¯), where D0¯ is a division algebra over K, and let ∗ be any superinvolution on A. If ξ∈Aut(A,∗), then, by [1, Isomorhpism Theorem], ξ=ψMα for some invertible element Mα∈Aα and (15)MαPγ∗Mα-1=MαPγMα-1∗,∀Pγ∈Aγ;thus,(16)MαPγ∗Mα-1=-1αα+γ-1αγMα-1∗Pγ∗Mα∗=-1αMα-1∗Pγ∗Mα∗=Mα∗-1Pγ∗Mα∗,which implies that Mα∗MαPγ∗Mα-1(Mα∗)-1=Pγ∗, ∀Pγ, and therefore Mα∗Mα=λI2p, where λ∈K×. If ∗ is a superinvolution of the first kind, then(17)Mα∗Mα∗=-1α2Mα∗Mα=λI2p∗=λI2p=Mα∗Mα.Hence α2=0¯, which implies that α=0¯ and therefore (18)AutA,∗=ψM∣Mis  an  invertible  element  inA0¯,withMM∗=λI2p.Moreover, if ∗ is a superinvolution of the second kind on A, then(19)AutA,∗=ψM∣Mis  an  invertible  element  inA0¯,withMM∗=λI2p∪ψM∣Mis  an  invertible  element  inA1¯,withMM∗=γI2p,γ∗=-γ.Theorem 8.
LetA=M2p(D0-), where D0- is a division algebra over K, and let ∗ be any superinvolution on A. If ∗ is a superinvolution of the first kind, then Aut(A,∗)≅G/~, where G/~ is the group defined in Theorem 4. If ∗ is a superinvolution of the second kind, then Aut(A,∗)≅G/~, where G={b0-∈A0-∣b0-b0-∗=α∈K×}∪b1-∈A1-∣b1-b1-∗=λ∈K×,whereλ∗=-λ and the equivalence relation ~ on G is defined by aα~bα⇔bα=γaα for some γ∈K×.Proof.
If∗ is a superinvolution of the first kind, then (20)δ:AutA,∗⟶G~,ψa0¯⟼a0¯is a group isomorphism.
Similarly, if∗ is a superinvolution of the second kind, then it is easy to check that G/~ is a group under the well-defined operation aα[bβ]=[bβaα], and so (21)δ:AutA,∗⟶G~,ψaα⟼aαis a group isomorphism. Thus in all cases we get Aut(A,∗)≅G/~.Second letA=Mp+q(D0¯), where D0¯ is a division algebra over K and (p≠q). Let ∗ be any superinvolution on A of any kind. If ξ∈Aut(A,∗), then by [1, Isomorhpism Theorem] ξ=ψMα, for some invertible element Mα in Aα, since p≠q, the homogeneous invertible elements in A are the invertible elements in A0¯. Thus α=0¯, and, therefore, if ξ∈Aut(A,∗), then ξ=ψM, where M is an invertible element in A0¯. Now, if Pγ∈Aγ, then(22)MPγ∗M-1=MPγM-1∗=M-1∗Pγ∗M∗=M∗-1Pγ∗M∗.Therefore, M∗MPγ∗M-1(M∗)-1=Pγ∗, ∀Pγ∈Aγ, and hence M∗M=λIp+q, where λ∈K×. Thus it is easy to check that AutA,∗={ψM∣Mis  an  invertible  element  inA0¯,whereMM∗=λIp+q}≅G/~, where G={b0¯∈A0¯∣b0¯b0¯∗=α∈K×}.


## 3. Examples of Group of Automorphisms of Even Superalgebras

In this section we classify in detail the groups of automorphisms of superalgebrasA=M2p(D), and A=Mp+q(D), where D=H or R, where H is the Hamiltonian quaternion algebra and R is the field of real numbers.LetA=M2p(D), where D=H or R, and let ∗ be a superinvolution of the first kind on A such that (A0¯,∗A0¯) is simple; then by [1, Proposition  13] ∗ is defined by (23)abcd⟼d~-b~c~a~,where ~ is the involution on Mp(D) induced by ∗. Then H=AutA,∗={ψM∣Mis  an  invertible  element  inA0¯,whereMM∗=αI2p,andα∈R}.Ifα>0, then by replacing M by 1/αM we get MM∗=I2p.Ifα<0, then by replacing M by 1/-αM we get MM∗=-I2p.SoH={ψM∣MM∗=I2por-I2p}.Theorem 9.
In the situation described above, let(24)G=AutA0-,∗A0-.Then H/R×≅G0, where G0 is the connected component of the identity in G.Proof.
By [8, page 593], G=G0∪ψNG0, where N=0IpIp0 and G0 is the connected component of the identity. Let (25)π:H⟶G0ψM⟼ψMA0¯.Then π is onto group homomorphism because if g0∈G0, then g0=ψMA0¯ for some invertible element M∈A0¯ since G0 is the connected component of the identity. Then it is easy to check that MM∗=αI2p (α∈R) since ψM∗=∗ψM.
Ifα>0, then by replacing M by 1/αM we get MM∗=I2p.
Ifα<0, then by replacing M by 1/-αM we get MM∗=-I2p. Therefore, ψM∈H and π(ψM)=g0. Also it is easy to check that ker⁡π={ψM∣M=λIp00γIp,whereλ,γ∈R}. But (26)λIp00γIpγIp00λIp=λγIp00γλIp=I2porλIp00γIpγIp00λIp=λγIp00γλIp=-I2p.So, γ=1/λ or -1/λ. Therefore, ker⁡π={ψM∣M=λIp001/λIp, or M=λIp00-1/λIp,whereλ∈R×}≅R×, since ψM=ψ-M. Therefore G0≅H/ker⁡π≅H/R×.In the second case, letA=Mp+q(D), where D=H or R, and let ∗ be a superinvolution of the first kind on A such that (A0¯,∗A0¯) is not simple; then by [1, Proposition  14] ∗ is defined by (27)xyzw⟼x~-Nz¯tM-1My¯tN-1w~,where x~=Nx¯tN-1 (N¯t=-N∈Mp(D)) and w~=Mw¯tM-1 (M¯t=M∈Mq(D)) and – is the standard involution on D.Since∗ is a superinvolution of the first kind we have H=AutA,∗={ψM∣Mis  an  invertible  element  inA0¯,whereMM∗=αIp+q,andα∈R}. Similarly, H={ψM∣MM∗=Ip+qor-Ip+q}.We will start with the casep≠q.Theorem 10.
In the situation described above, let(28)G=AutA0-,∗A0-.Then H/ker⁡π≅π(H)≤G, where (29)π:H⟶GψN⟼ψNA0-.Proof.
SinceA0¯ is semisimple and p≠q, we get G=AutA0¯,∗A0¯={ψM∣M=A00B,whereA,Bare  invertible  matrices}. If ψM∈G, then (30)ψM∗A0¯=∗A0¯ψM⟺MM∗=λIp00γIq,λ,γ∈R;thereforeψM∗A0¯=∗A0¯ψM⟺MM∗∈V=Ip+q,-Ip+q,Ip00-Iq,-Ip00Iq.So G=ψM∣MM∗∈V. Therefore, (31)π:H⟶GψN⟼ψNA0¯is a group homomorphism and ker⁡π={ψN∣ψNA0¯=idA0¯}={idA,ψM,M=Ip00-Iq}. To see this let N=A00B such that ψN∈ker⁡π; then NN∗=Ip+q or -Ip+q (since ψN∈H), and (32)A00BX00YA-100B-1=X00Yif and only if A=λIp and B=γIq (λ,γ∈R); since (33)NN∗=λIp00γIqλIp00γIq=λ2Ip00γ2Iq∈Ip+q,-Ip+q,we get NN∗=Ip+q. Therefore, λ2=γ2=1⇒λ, γ∈1,-1, which implies that ker⁡π={ψIp+q,ψ-Ip+q,ψN,ψM}, where N=-Ip00Iq and M=Ip00-Iq and hence ker⁡π=idA,ψM since ψIp+q=ψ-Ip+q and ψN=ψM. Therefore, H/ker⁡π≅π(H)≤G.Theorem 11.
As in the theorem above, ifp≠q, then G=π(H)∪ψMA0-π(H) for a fixed invertible matrix M=A00B such that (34)MM∗∈Ip00-Iq,-Ip00Iq.Proof.
SinceMM∗≠Ip+q,-Ip+q, then ψM∉H, which implies that ψMA0¯∉π(H). Let N=X00Y such that (35)NN∗∈Ip00-Iq,-Ip00Iq;then(36)NM-1NM-1∗=NM-1M-1∗N∗=NIp00-IqN∗:MM∗=Ip00-IqN-Ip00IqN∗:MM∗=-Ip00Iq=XX~00-YY~:MM∗=Ip00-Iq-XX~00YY~:MM∗=-Ip00Iq.Therefore, (NM-1)(NM-1)∗∈{Ip+q,-Ip+q}; thus, by Theorem 10, (37)ψNM-1A0¯=ψM-1A0¯ψNA0¯∈πH≅Hker⁡π.Now, ψNA0¯=ψNM-1MA0¯=ψMA0¯ψM-1A0¯ψNA0¯=ψMA0¯ψNM-1A0¯︸∈ψMA0¯π(H). Therefore, (38)G=πH∪ψMA0¯πH.For the casep=q, G=Aut(A0¯,∗A0¯)=G0∪ψNA0¯G0, where N=0IpIp0 (it is easy to check that ψNA0¯∗A0¯=∗A0¯ψNA0¯) and G0={ψMA0¯∣MM∗∈V} is the connected component of the identity in G. Now, let π:H→G0 (ψL↦ψLA0¯), where H={ψM∣MM∗=Ip+qor-Ip+q}; then π is a group homomorphism and ker⁡π=idA,ψN′ where N′=Ip00-Ip. Therefore, H/ker⁡π≅π(H)≤G0.Theorem 12.
In the situation described above, ifp=q, then (39)G0=πH∪ψMA0-πH,for a fixed invertible matrix M=A00B, such that (40)MM∗∈Ip00-Iq,-Ip00Iq,and G0 is the connected component of the identity in G.Proof.
ConsiderψMA0¯∉π(H) since MM∗∈{Ip00-Ip,-Ip00Ip}. Let N=X00Y such that NN∗∈{Ip00-Ip,-Ip00Ip}; then(41)NM-1NM-1∗=NM-1M-1∗N∗=NIp00-IpN∗:MM∗=Ip00-IpN-Ip00IpN∗:MM∗=-Ip00Ip=XX~00-YY~:MM∗=Ip00-Ip-XX~00YY~:MM∗=-Ip00Ip.Therefore, (NM-1)(NM-1)∗∈{I2p,-I2p} and so (42)ψNM-1A0¯=ψM-1A0¯ψNA0¯∈πH≅Hker⁡π.Now, ψNA0¯=ψNM-1MA0¯=ψMA0¯ψM-1A0¯ψNA0¯=ψMA0¯ψNM-1A0¯︸∈ψMπ(H). Therefore, (43)G0=πH∪ψMA0¯πH.



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