• <tr id="yyy80"></tr>
  • <sup id="yyy80"></sup>
  • <tfoot id="yyy80"><noscript id="yyy80"></noscript></tfoot>
  • 99热精品在线国产_美女午夜性视频免费_国产精品国产高清国产av_av欧美777_自拍偷自拍亚洲精品老妇_亚洲熟女精品中文字幕_www日本黄色视频网_国产精品野战在线观看 ?

    The Group of Automorphisms of Total Orthogonal Graphs of Odd Characteristic

    2019-03-30 08:20:48GENGTianzhen耿天真MAXiaobin馬曉玢
    應(yīng)用數(shù)學(xué) 2019年2期
    關(guān)鍵詞:天真

    GENG Tianzhen(耿天真),MA Xiaobin(馬曉玢)

    (School of Mathematics and Big Data,Anhui University of Science and Technology,Huainan 232000,China)

    Abstract: Let Fnq be the n-dimensional row vector space over a finite field Fq of odd characteristic,and let Sn be a non-singular n×n symmetric matrix over Fq.The total orthogonal graph O(Sn,q) with respect to Sn is defined to be the graph with vertices of all 1-dimensional subspace of Fnq ,two vertices [α] and [β] are adjacent if and only if αSnβT≠ 0.The three types of isotropic orthogonal graphs studied in [3] are induced subgraphs of O(Sn;q) with vertices of all isotropic lines.In this paper,the automorphism group of O(Sn,q) is determined.Not like the case of isotopic orthogonal graph studied in[3] (where the graph is vertex transitive),it turns out that the vertex set of O(Sn,q) has three orbits if n is odd,and it has only two orbits if n is even.

    Key words: Orthogonal graph;Graph automorphism;Transitivity

    1.Introduction

    Recently,some graphs based on geometry of classical groups have attracted more attention.We now list some papers concerning with such graphs.TANG and WAN[9]studied the symplectic graph over an arbitrary finite field and determined the full automorphism group.LI and WANG[6]further showed that the subconstituents of the symplectic graph are strictly Deza graphs except the trivial case whenυ= 2.Meemark and Prinyasart[8]introduced the symplectic graphSp(2υ,Zpn) and showed that it is strongly regular whenυ= 1 and it is arc transitive whenpis an odd prime.LI et al.[5]continued the research onSp(2υ,Zpn),showing thatSp(2υ,Zpn) is arc transitive for any primep,and it is a strictly Deza graph ifυ ≥2 andn ≥2.GU and WAN[2]studied the subconstituents of the symplectic graphs modulopn.WAN and ZHOU[11?12]studied the(isotropic)unitary graphs and the(isotropic)orthogonal graphs of Characteristic 2.We here particularly mention [3] due to GU,which seems much related to our topic.Forn=2υor 2υ+1 or 2υ+2,GU[3]defined the(isotropic)orthogonal graphO(n,q) with respect to a symmetric matrixoverFq,where ?=?or ?=1 or ?=diag{1,?z},in such a way: the vertex set of the graph is all isotropic lines ofFnq,two isotropic lines[α]and[β]are adjacent if and only ifαHβT≠0.GU[3]showed thatO(n,q) is strongly regular forυ ≥2,and determined the group of automorphisms ofO(n,q).By the automorphism group of the graph it is easy to see thatO(n,q) is vertex transitive and edge transitive.For more results about the graphs based on geometry of classical groups the reader may refer to [4,7,13].

    One might find that the vertex set of the graphs introduced in [2-9,11-13] are taken to be isotropic lines,or sometimes totally isotropic subspaces of the space in question.In this paper,we shall use all 1-dimensional subspace (isotropic or not) of a orthogonal space to define a new graph,which includes the isotropic orthogonal graph as an induced subgraph.The vertex set of the graph chosen in such a way seems much natural.Indeed,the vertex set of the earliest graph studied by TANG and WAN,i.e.,the symplectic graph,is taken to be all 1-dimensional subspace (although,it can be also viewed consisting of all isotropic lines).

    LetFqbe a finite field withqelements,whereqis a power of an odd primepandn ≥2 be an integer.LetFnq={(a1,a2,...,an) :ai ∈Fq}be then-dimensional row vector space overFq.Denote byei(1≤i ≤n)the row vector inFnqwhoseith coordinate is 1 and all others are 0.For a nonzero vectorα=(a1,...,an)∈Fnq,let [α] denote the 1-dimensional subspace ofFnqspanned byα.Sometimes,[α] = [(a1,...,an)] is directly written as [a1,...,an] for simplicity.Since,[α] = [kα] for anyk ∈F?q=Fq {0},the representation of a vertex in the form [α] is not unique.However if we write the vertex in the form [a1,a2,...,an] such that the first (from left) nonzero coordinate is 1,then the representation is unique.A vertex[a1,a2,...,an] written in such way is said with standard form.

    For ann×nmatrixM,we denote byMTand|M|the transpose and the determinant ofM,respectively.LetGbe the subgroup ofF?qof all square elements.The index ofGinF?qis 2.ThusF?qis exactly a disjoint union ofGandzG,wherezis an arbitrarily given non-square element inFq.Let (α,β)=αSnβTbe a non-degenerate symmetric bilinear form defined onFnqwith respect to a non-singularn×nsymmetric matrixSnoverFq.A vectorαis called isotropic if (α,α) = 0,otherwise,it is called non-isotropic.If a subspace ofFnqconsists of isotopic vectors,it is said to be total isotropic.

    We now define the so-called total orthogonal graphO(Sn,q),with respect to a nonsingularn×nsymmetric matrixSnoverFqin such a way: the vertex set is taken to be all 1-dimensional subspaces (or lines) ofFnq,two lines [α] and [β] are adjacent if and only if (α,β)≠0.Obviously,O(Sn,q) is not a simple graph since for a non-isotropic line [α],[α]~[α].By Theorem 1.23 in[10]we know that ann×nnonsingular symmetric matrix overFqis congruent to diag(In?1,θ),whereIn?1refers to the identity matrix of ordern ?1,andθ= 1 orθ=z.If two non-singular symmetric matricesSnandHnare congruent,that is,Sn=PHnPTfor an invertible matrixP,thenO(Sn,q) is isomorphic toO(Hn,q) under the mapping [α][αP],?[α]∈V(O(Sn,q)).In view of this point,we will directly assume thatSn=InorSn=diag(In?1,z),withza fixed non-square element ofFq.

    A similar discussion as Theorem 2.1 in [9] shows thatO(Sn,q) is also a strongly regular graph with parameters

    In this paper,the automorphism group ofO(Sn,q) is determined,and the partition ofV(O(Sn,q)) as the union of different orbits is done.

    Remark 1.1(i)The three types of isotropic orthogonal graphs studied in[3]are induced subgraphs ofO(Sn,q) with vertices of all isotropic lines.

    (ii) For certain unknown reasons,the case when ?is a non-square element ofFqis left by [3] without consideration.However,the orthogonal graphs studied in this paper cover all cases,especially the case whenSnis congruent towhich is left without consideration in [3].

    2.Automorphisms of O(Sn,q)

    In this section,we determine the automorphism group,denoted by Aut(O(Sn,q)),ofO(Sn,q).Some standard automorphisms are constructed firstly.Based on them every automorphism ofO(Sn,q) is described explicitly.Alln×ninvertible matrices overFqforms a group with respect to matrix multiplication,which is called the general linear group of degreenoverFqand is denoted by GLn(Fq).

    LetP ∈GLn(Fq).We defineσP:V(O(Sn,q))→V(O(Sn,q)) by [α][αP].Then using a similar proof as that of Proposition 3.1 in [3] we obtain:

    Lemma 2.11) LetP ∈GLn(Fq).ThenσPis an automorphism ofO(Sn,q) if and only if PSPT=kSfor somek ∈F?q.

    2) ForP1,P2∈GLn(Fq),σP1=σP2if and only ifP1=kP2for somek ∈F?q.

    IfP ∈GLn(Fq) satisfiesPSnPT=kSnfor somek ∈F?qwe callPa generalized orthogonal matrix.All generalized orthogonal matrices of ordernwith respect toSnforms a subgroup of GLn(Fq),called the generalized orthogonal group of degreenoverFqand denoted by GOn(Fq).The center of GOn(Fq) consists of allkEnwithk ∈F?q.The factor group of GOn(Fq) with respect to its center is called the projective generalized orthogonal group of degreenoverFqand is denoted by PGOn(Fq).According to Lemma 2.1,PGOn(Fq) can be viewed as a subgroup of Aut(O(Sn,q)).

    Letn= 2,and supposeπis a bijective mapping onFqsatisfyingπ(?x?1θ?1) =?θ?1π(x)?1for anyx ∈F?qandπ(0) = 0.DefineσπfromV(O(S2,q)) to itself by sending each vertex [a1,a2],with the standard form,to [a1,π(a2)].

    Lemma 2.2σπis an automorphism ofO(S2,q).

    ProofEach vertex [α] is written in the standard form.Obviously,σπis a bijective mapping onV(O(S2,q)).Suppose that [α] = [a1,a2][b1,b2] = [β].Equivalently,a1b1+θa2b2= 0.Ifa1= 0,thena2=b1= 1 andb2= 0.Clearly,σπ([α])σπ([β]).Ifb1= 0,the result is similar.Now suppose that botha1andb1are 1.Thusb2=?θ?1a?12.By the hypothesis forπ,we haveπ(b2) =π(?θ?1a?12) =?θ?1π(a2)?1,which follows that 1+π(a2)π(b2)θ= 0,showing thatσπ([α])σπ([β]).Consequently,[α][β]?σπ([α])σπ([β]).It follows thatd(σπ([α]))≤d([α])for any vertex[α],whered([α])denotes the degree of [α].By

    we haved(σπ([α]))=d([α]) for any vertex [α].Then we conclude that [α]~[β]?σπ([α])~σπ([β]).Thus,σπis an automorphism.

    Remark 2.1The bijective mappingπonFq,for whichπ(?x?1θ?1)=?θ?1π(x)?1andπ(0)=0,needs not to be an automorphism ofFq.For example,letFqbe the field Z/3Z,and letθ=.Defineπon Z/3Z by fixing,permutingand.Thenπ(?x?1) =?π(x)?1forx=1 and 2.But clearly,πis not an automorphism of Z/3Z.

    Forn ≥3,letπbe an automorphism of the fieldFqand letSn= diag(In?1,θ) withθ= 1 orz.Ifθ=z,thenπ(z)/∈F2qandz?1π(z)∈F2q.Supposer ∈F?qsuch thatπ(z) =zr2.Lety=rifθ=z,and lety= 1 ifθ= 1.Then,π(θ) =θy2.Defineσπ,y:V(O(Sn,q))→V(O(Sn,q)) by

    where [a1,a2,...,an] is written in the standard form.

    Lemma 2.3σπ,yis an automorphism ofO(Sn,q).

    ProofObviously,σπ,yis a bijective mapping onV(O(Sn,q)).The following derivation shows thatσπ,ypreserves adjacency relation of vertices in both directions.

    Hence,σπ,yis an automorphism ofO(Sn,q).

    Letω= (w1,w2,...,wn)∈ Fnq,wherew2i= 1 for eachiandw1= 1.We defineσωfromV(O(Sn,q)) to itself by sending any [a1,a2,...,an],with standard form,to[w1a1,w2a2,...,wnan].

    Lemma 2.4σωis an automorphism ofO(Sn,q).

    ProofObviously,σωis a bijective mapping onV(O(Sn,q)).The procedure

    shows thatσωpreserves adjacency relation of vertices in both directions.Thus the result follows.

    With the known automorphisms in hands,we now describe every automorphism ofO(Sn,q).Letn ≥3 and let a ={α1,α2,...,αn}be a non-isotropic orthogonal basis ofFnq,and suppose that (αi,αi) =a≠ 0 fori= 1,2,...,n ?1 and (αn,αn) =b≠ 0.Letσbe an automorphism ofO(Sn,q).Assumeσ([αi]) = [βi],and assume (βi,βi) =ci.Thenci≠0 fori= 1,2,...,nand (βi,βj) = 0 fori≠j.Namely,{β1,β2,...,βn}also forms a non-isotropic orthogonal basis ofFnq.For a nonzero vectorsuppose thatIt is easy to see thatbi≠0 if and only ifai≠0.Indeed,ifai≠0,then [α]~[αi].Applyingσ,we havewhich leads tobici≠0.Thusbi≠0.Similarly,bi≠0 impliesai≠0.Sinceσis a bijection fromV(O(Sn,q)) to itself,we define permutationsπi,i=2,3,...,nonFqsuch that

    It seems more accurate if one likes to use the symbolπi,σ,ainstead ofπi,as the definition ofπidepends onσand a.

    Next,we useπito characterizeσand give some properties forπiandci.

    Lemma 2.5(i){β1,β2,...,βn}forms a non-isotropic orthogonal basis ofFnq;

    Proof(i) was proved just before this lemma.Assume thatσ([α1+n i=2aiαi]) =Denotebyα,andTo complete the proof of (ii) we need to show thatbi=πi(ai) for eachi ≥2.The case whenai= 0 is trivial.Assume thatai≠0.For the case when 2≤i ≤n ?1,bywe havewhich follows thatThus

    On the other hand,by applyingσtowe havewhich leads to

    Comparing (2.1) and (2.2),we havebi=πi(ai).Next,we consider the case wheni=n.Applyingσto [α][α1?ab?1a?1n αn],we obtain [β][β1+πn(?ab?1a?1n)βn],which follows thatc1+bnπn(?ab?1a?1n)cn=0.Thus

    On the other hand,by applyingσto[α1+anαn][α1?ab?1a?1n αn]we have[β1+πn(an)βn][β1+πn(?ab?1a?1n)βn],which follows that

    Comparing (2.3) and (2.4),we also havebn=πn(an).The discussions above say that for anya2,a3,...,an ∈Fq,

    This completes the proof of (ii).

    Let 2≤i ≤n ?1 and denoteαi+ai+1αi+1+...+anαnalso byα.Assume that

    To complete the proof of (iii) we need to show thatb′j=c?11ciπi(1)πj(aj) fori+1≤j ≤n.Denoteβi+b′i+1βi+1+...+b′nβnalso byβ.For the case wheni+1≤j ≤n ?1,applyingσto [α][α1+αi ?a?1j αj] we have

    which leads to

    Following from (2.2) and (2.6),we obtainb′j=c?11ciπi(1)πj(aj).For the case whenj=n,applyingσto [α][α1+αi ?ab?1a?1n αj] we have

    which leads to

    From (2.4) and (2.7) we obtainb′n=c?11ciπi(1)πn(an).Thus,

    withqi=c?11ciπi(1),which completes the proof of (iii).

    Now we are in a position to complete the proof of (iv).Considering the action ofσon[α2+αn][α2?ab?1αn],we have

    which follows that

    Applyingσto [α1+α2+αn][α2?ab?1αn] we have

    leading to

    Comparing (2.9) and (2.10),we have that

    Applyingσto [α1+α2][α1?α2] we have

    which implies that

    From (2.11) and (2.12) we have

    Ifn=3,then the proof of(iv)is completed.Ifn ≥4,applyingσto[α1?α2+αj][α2+αj]for 3≤j ≤n ?1,we have

    which shows that

    Recalling thatπ2(?1)=?π2(1),we obtain

    On the other hand,the action ofσon [α1+α2+αj][α2?αj] leads to

    showing that

    (2.15) and (2.16) imply that

    The first assertion of (iv) is confirmed by (2.13) together with (2.17),and the second one is confirmed by (2.11) together with (2.15).

    For the proof of (v),considering the action ofσon [α1?α2+αn][α2+ab?1αn],we have

    which implies that

    Comparing (2.10) with (2.19),we have

    The proof of (v) is completed.

    Lemma 2.6Letσ,a,b,πibe defined as in Lemma 2.3.Ifσ([αi])=[αi] for all values ofi,then the following assertions hold.

    ProofBy (iv) of Lemma 2.3,assertion (i) of this lemma is obvious.

    Forx ∈Fqand 2≤j ≤n ?1,applyingσto [α1+xα2+xαj][α2?αj],we have[α1+π2(x)α2+πj(x)αj][α2+π2(1)πj(?1)αj],which leads toπ2(x) =π2(1)πj(1)πj(x).Sinceπj(1)2=1,we have

    For the case whenj=n,applyingσto [α1+xα2+xαn][α2?ab?1αn],we have [α1+π2(x)α2+πn(x)αn][α2+π2(1)πn(?ab?1)αn],which leads to

    By takingx=1,(2.22) becomes

    Substituting this equation into (2.22),we have that

    (2.21) and (2.24) together confirm (ii).

    Now we prove (iii).Ifxy=0,the result is obvious.Suppose thatxy≠0.By [α1+α2+ab?1xαn][α2?x?1αn] we have [α1+π2(1)α2+πn(ab?1x)αn][α2+π2(1)πn(?x?1)αn],which follows that

    Applyingσto [α1+xyα2+yαn][α2?ab?1xαn] we have that

    which implies that

    Substituting (2.25) into (2.26),we obtain

    Multiplying two sides of (2.27) byπ2(1) and using the result of (ii),we have

    Replacingywithxy,andxwithx?1,then we have

    Multiplying two sides of (2.29) byπj(1)2and using the result of (ii),we have

    By (2.30),we easily obtain

    The proof of (iii) is completed.

    For the proof of (iv),the case wheny=0 is trivial.Suppose thaty≠0.Applyingσto[α1?(x+y)α2+xαn][α1+y?1α2+ab?1y?1αn] we have

    which leads toa ?π2(x+y)π2(y?1)a+πn(x)πn(ab?1y?1)b=0.Or equivalently,

    Multiplying two sides of (2.32) byπn(1),we have

    By(iii)of this lemma and(v)of Lemma 2.5,πn(y)πn(ab?1y?1)a?1b=πn(ab?1)πn(1)a?1b=1.Thus (2.33) becomes

    By multiplying two sides of (2.34) byπ2(1),it follows that

    Finally,multiplying two sides of (2.35) byπj(1),we have

    The proof of (iv) is completed.

    It follows from (v) of Lemma 2.5 thatab?1=πn(1)πn(ab?1).(iii) of this lemma shows thatπn(ab?1)=πn(a?1b)?1πn(1)2.Consequently,πn(a?1b)=a?1b·πn(1)3.

    Now,we are in position to announce the main result of this paper.

    Theorem 2.1(i) Ifn= 2,thenσis an automorphism ofO(S2,q) if and only if there exists someK ∈GOn(Fq) and a bijective mappingπonFq,fixing 0 and satisfyingπ(?x?1θ?1)=?θ?1π(x)?1,such thatσ=σK ·σπ.

    (ii) Ifn ≥3,thenσis an automorphism ofO(Sn,q) if and only ifσ=σK ·σω ·σπ,y,whereπis an automorphism of the fieldFqsuch thatπ(θ)=y2θ,ω=(w1,w2,...,wn)∈Fnqwithw2i=1,K ∈GOn(Fq).

    ProofLetn= 2.Lemma 2.1 and 2.2 together confirm the sufficiency of (i).For necessity,letσbe an automorphism ofO(S2,q),and assumeσ([ei])=[fi] fori=1,2.Let

    ThenKS2KT= diag(c1,c2),whereci= (fi,fi) =fiS2fTi.Ifθ=z,we havez|K|2=c1c2.Thus only one ofciis a square element,sayc1.By replacingfiwith suitable multiple we may assume thatc1=1 andc2=z.Thus,KS2KT=S2,showing thatK,K?1∈GOn(Fq).ThusσKandσK?1are automorphisms ofO(S2,q).FromKK?1=I2,it follows thatfiK?1=ei.Thus,(σK?1·σ)([ei]) =σK?1([fi]) = [fiK?1] = [ei] fori= 1,2.Ifc2is a square element,similarly,by replacingfiwith suitable multiple we may assume thatc1=zandc2=z2.ByKS2KT=zS2,we also haveK,K?1∈GO2(Fq).Similarly,we know thatσK?1·σfixes [e1]and [e2].Ifθ= 1,by similar discussions as above,we can choose someK ∈GOn(Fq) such that (σK?1·σ)([ei])=[ei] fori=1,2.The procedure is omitted.

    Now,letσ1=σK?1·σ.For [α] = [1,x]∈V(O(S2,q)),the first coordinate ofσ1([α])cannot be zero sinceσ1fixes[e1].Thus,we can define a mappingπonFqsuch thatσ1([1,x])=[1,π(x)].Obviously,πis bijective andπ(0) = 0.Forx≠ 0,by applyingσ1to [1,x][1,?x?1θ?1],we have [1,π(x)][1,π(?x?1θ?1)],which leads to 1+θπ(x)π(?x?1θ?1) = 0,showing thatπ(?x?1θ?1) =?θ?1π(x)?1.Thusπcan be used to define the automorphismσπofO(S2,q).Clearly,σ1([α])=σπ([α]) for each [α]∈V(O(S2,q)),with the standard form.Finally,we haveσ1=σπandσ=σK ·σπ.

    For the proof of (ii),sufficiency is obvious.Now we consider necessity.Letσbe an automorphism ofO(Sn,q).Obviously,(ei,ei) =eiSeTi= 1 fori= 1,2,...,n ?1,(en,en) =enSeTn=θand (ei,ej) = 0 fori≠j.Thus,a ={e1,e2,...,en}forms a non-isotropic orthogonal basis ofFnq.Assume thatσ([ei])=[fi] and (fi,fi)=ci.Let

    ThenKSKT= diag(c1,c2,...,cn).As what we did just before Lemma 2.3,we define the permutationsπi(i=2,3,...,n),with respect toσand a,onFqsuch that

    By Lemma 2.3,we havecj=c1πj(1)?2forj=2,3,...,n ?1,andcn=c1πn(1)?1πn(θ?1)?1.

    Setπ1(1)=1 andπ1(0)=0.Letfi′=πi(1)fifori=1,2,...,n,and let

    Then we also haveσ([ei])=[f′i],however

    For simplity,f′iis also written asfiandK1is also written asK.

    Now we consider the case whenθ= 1.ThenS=InandKKT=c1In.ThusK,K?1∈GOn(Fq),andσK,σK?1∈Aut(O(Sn,q)).It follows fromKK?1=InthatfiK?1=eifori=1,2,...,n.Thus,(σK?1·σ)([ei])=σK?1([fi])=[fiK?1]=[ei]for eachi.DenoteσK?1·σbyσ1.Thenσ1fixes each [ei].Define permutationsπi,with respect toσ1and a,onFqsuch that

    Let 0≠α ∈Fnq.We write [α] in the standard form [α]=[a1,a2,...,an].Then by using (ii)and (iii) of Lemma 2.3 and recalling thatπi(1)2=1 for 1≤i ≤n,we have

    DefineπonFqbyπ(x) =π2(1)?1π2(x) forx ∈Fq.Sinceπi(1)?1πi(x) =π2(1)?1π2(x) fori= 3,4,...,n,we knowπ=πi(1)?1πifor eachi(≥2).By (iii) and (iv) of Lemma 2.6,it is easy to see thatπ(xy)=π(x)π(y) andπ(x+y)=π(x)+π(y) for allx,y ∈Fq,showing thatπis an automorphism ofFq.Now,(2.37) becomes

    Letω=(π1(1),π2(1),...,πn(1)).Sinceπi(1)2=1 for eachi(thanks to(iv)and(v)of Lemma 2.3),we can useωto define the automorphismσωofO(Sn,q).Now,

    This equation shows thatσω·σ1=σπ,1.Finally,we haveσ=σK ·σω·σπ,1.

    For the left caseθ=z(a fixed non-square element inFq),KSKT=diag(c1,c1,...,c1θ1)withθ1=πn(1)πn(z?1)?1.We claim thatθ1cannot be a square element.Otherwise,sayθ1=r2.By replacingfnwithr?1fn,we haveKSKT=diag(c1,c1,...,c1).Thus(fi,fi)=c1for alli.Sinceσ?1([fi])=[ei] for eachi,applying (v) of Lemma 2.3 toσ?1(viewingfiasαiand viewingeiasβi),we have that 1=(e1,e1)=(en,en)πn(1)2=zπn(1)2,absurd.So we may assume thatθ1=zs2.Replacingfnbys?1fn,we have thatKSKT=diag(c1,c1,...,c1z)=c1S.ThusK?1is a generalized orthogonal matrix.UsingK?1we define the automorphismσK?1onO(Sn,q) and denoteσK?1· σbyσ1.Similar as the case whenθ= 1,we haveσ1([ei])=σK?1([fi])=[fiK?1]=[ei] for eachi.

    Define permutationsπi,with respect toσ1and a,onFqsuch that

    Then by (ii) and (iii) of Lemma 2.3,

    for [α]∈V(O(Sn,q)) with standard form,whereπ1fixes 1 and 0.DefineπonFqbyπ(x)=π2(1)?1π2(x)forx ∈Fq.Also,πis an automorphism ofFq,andπ=πi(1)?1πifor eachi(≥2).Now,(2.39) becomes

    Letω=(π1(1),π2(1),...,πn?1(1),1).By(i)of Lemma 2.4,πi(1)2=1 for each 1≤i ≤n?1.So we can useωto construct the automorphismσωofO(Sn,q).Then

    Lety=πn(1).(v) of Lemma 2.4 shows thatπ(z) =πn(1)?1πn(z) =y2z.Thus we can useπandyto define the automorphismσπ,yofO(Sn,q).It is easy to see thatσω ·σ1=σπ,y.Finally,σ=σK ·σω·σπ,y,as desired.

    3.Orbits Partition of V(O(Sn,q)) Under the Automorphisms

    The study of orbits partition ofV(O(Sn,q)) needs an elementary lemma on GOn(Fq).

    Lemma 3.1LetSn= diag(In?1,θ) withθ= 1 orz.Ifnis odd,then for eachK ∈GOn(Fq) withKSnKT=kSn,kmust be a square element.Ifnis even,then there existsK ∈GOn(Fq) such thatKSnKT=zSn.

    ProofLetnbe odd,and supposeK ∈GOn(Fq) such thatKSnKT=kSn.Thenθ|K|2=knθ,showing thatkn=|K|2,andkis a square element.Now,assume thatnis even.A well known result about finite fields is that the equationX2+Y2=zhas solutions inFq.Suppose thata,b ∈Fqsuch thata2+b2=z.Ifθ=1,for 1≤2k+1<2k+2≤n,set

    Ifθ=z,we set

    for 1≤2k+1<2k+2≤n ?2,and setfn?1=en,fn=zen?1.Let

    Then,we haveQSnQT=zSn.LetV0be the subset ofV(O(Sn,q)) of all isotropic lines and let

    where (F?q)2refers the set of all non-zero square element ofFq.

    Theorem 3.1Ifnis odd,thenV(O(Sn,q)) has three orbits,they areV0,V1andV2.Ifnis even,thenV(O(Sn,q)) has only two orbits,they areV0andV1∪V2.

    ProofFor [α],[β]∈V(O(Sn,q)),if they belong to the sameVi,then by replacingαwith its suitable multiple we may assume that (α,α) = (β,β).Using Lemma 6.8 in [10],we can find an orthogonal matrixQsuch thatαQ=β.ThusσQ([α]) = [β],showing that all elements lying in the sameVibelong to one orbit.If [α]∈V0,clearly,σ([α])∈V0for allσ ∈Aut(O(Sn,q)).It implies thatV0forms an orbit.

    Ifnis odd,we want to prove thatV2is stable under the action of Aut(O(Sn,q)).Let[α]=[a1,a2,...,an]∈V2(with the standard form),and letσbe an automorphism ofO(Sn,q).We consider the action ofσon[α].By Theorem 2.1,σcan be decomposed asσ=σK·σω·σπ,y,whereK ∈GOn(Fq) withKSnKT=kSn,ω= (w1,w2,...,wn)∈Fnqwithw1= 1 andw2i= 1,andπis an automorphism ofFqsatisfyingπ(θ) =θy2fory ∈Fq.Assume thatσ([α])=[β].Then,

    and

    Since (α,α)∈(F?q)2,it follows thatπ(α,α)∈(F?q)2.By Lemma 3.1,k ∈(F?q)2,thusσ([α])=[β]∈V2.It implies thatV2forms an orbit,which immediately follows thatV1forms an orbit.

    Ifnis even,by Lemma 3.1,we can chooseQ ∈GOn(Fq) such thatQSnQT=zSn.Then for [α]∈V2,we have [αQ]∈V1sinceαQSnQTαT=z(α,α)∈z(F?q)2.Consequently,σQ([α])=[αQ]∈V1.It follows thatV1∪V2forms an orbit.

    猜你喜歡
    天真
    彩墨渾成,天真自然
    天真真好
    天真組詩
    滇池(2022年5期)2022-04-30 21:44:36
    天真熱
    天真童年
    小讀者(2020年4期)2020-06-16 03:34:10
    雪天真快樂
    天真給你最美的夢
    雪天真快樂
    古淡天真之美——倪瓚《淡室詩》
    丹青少年(2017年1期)2018-01-31 02:28:21
    年啊年
    久久韩国三级中文字幕| 全区人妻精品视频| 最近2019中文字幕mv第一页| 综合色av麻豆| 22中文网久久字幕| 免费观看精品视频网站| 成人无遮挡网站| 国产亚洲91精品色在线| 亚洲熟妇中文字幕五十中出| 成人午夜高清在线视频| 国产精品久久久久久久久免| 国产伦精品一区二区三区视频9| 国产成人freesex在线| 晚上一个人看的免费电影| 深爱激情五月婷婷| 精品久久久久久久久亚洲| 插逼视频在线观看| 亚洲三级黄色毛片| 简卡轻食公司| 国产精品爽爽va在线观看网站| 欧美bdsm另类| 日本免费在线观看一区| 亚洲在线自拍视频| 一区二区三区四区激情视频| 精品久久久久久久久亚洲| 人人妻人人澡人人爽人人夜夜 | 精品人妻一区二区三区麻豆| 亚洲av免费高清在线观看| 成年女人永久免费观看视频| 只有这里有精品99| 三级男女做爰猛烈吃奶摸视频| 成人综合一区亚洲| 干丝袜人妻中文字幕| av国产久精品久网站免费入址| 99久久精品热视频| 亚洲av日韩在线播放| 日日摸夜夜添夜夜添av毛片| 99热这里只有是精品50| 国产精品久久久久久精品电影小说 | 免费观看性生交大片5| 国产高清视频在线观看网站| 3wmmmm亚洲av在线观看| 午夜福利网站1000一区二区三区| 久久久久免费精品人妻一区二区| 国产精品久久电影中文字幕| 2021少妇久久久久久久久久久| 色哟哟·www| 五月伊人婷婷丁香| 国产熟女欧美一区二区| 国产激情偷乱视频一区二区| 久久精品夜色国产| 18禁在线无遮挡免费观看视频| 国产亚洲最大av| 色噜噜av男人的天堂激情| 看十八女毛片水多多多| 成人美女网站在线观看视频| 超碰97精品在线观看| 直男gayav资源| 亚洲国产成人一精品久久久| 综合色丁香网| 国产成人91sexporn| 91aial.com中文字幕在线观看| 熟女电影av网| 成人午夜精彩视频在线观看| 婷婷色av中文字幕| 日本欧美国产在线视频| 插阴视频在线观看视频| 欧美性猛交╳xxx乱大交人| 日韩大片免费观看网站 | 精品久久久久久电影网 | av国产久精品久网站免费入址| a级毛片免费高清观看在线播放| 国产伦精品一区二区三区视频9| 日韩一本色道免费dvd| 国语自产精品视频在线第100页| 国产亚洲av片在线观看秒播厂 | 美女黄网站色视频| 久久久午夜欧美精品| 中文亚洲av片在线观看爽| 不卡视频在线观看欧美| 久久久久久久亚洲中文字幕| 日韩一区二区三区影片| 日日摸夜夜添夜夜爱| 欧美精品国产亚洲| 国产午夜精品久久久久久一区二区三区| 九九在线视频观看精品| 国产午夜精品论理片| 亚洲人成网站高清观看| 99热这里只有精品一区| 成人av在线播放网站| 91精品伊人久久大香线蕉| 天美传媒精品一区二区| 国产国拍精品亚洲av在线观看| 久久国产乱子免费精品| 欧美一区二区精品小视频在线| av福利片在线观看| 在线播放无遮挡| 看黄色毛片网站| 最近最新中文字幕大全电影3| 老司机福利观看| 亚洲精品456在线播放app| 超碰97精品在线观看| 乱系列少妇在线播放| 午夜精品在线福利| 日本爱情动作片www.在线观看| 国产午夜精品一二区理论片| av专区在线播放| 日韩欧美国产在线观看| 国产视频内射| 国产69精品久久久久777片| 麻豆成人午夜福利视频| 两个人视频免费观看高清| 久久久精品94久久精品| www.色视频.com| 亚洲欧美精品专区久久| 久久99精品国语久久久| 纵有疾风起免费观看全集完整版 | 天堂√8在线中文| 国产av在哪里看| 波多野结衣巨乳人妻| 男女那种视频在线观看| 天堂网av新在线| 婷婷色av中文字幕| 麻豆精品久久久久久蜜桃| 精品人妻熟女av久视频| 国产精品久久视频播放| 久久6这里有精品| 国产亚洲精品久久久com| 青春草国产在线视频| 欧美极品一区二区三区四区| 精品久久久久久成人av| 丝袜喷水一区| a级一级毛片免费在线观看| 秋霞伦理黄片| 99久久九九国产精品国产免费| 午夜福利高清视频| av免费观看日本| 久久精品久久久久久噜噜老黄 | 蜜桃亚洲精品一区二区三区| 亚洲天堂国产精品一区在线| 亚洲国产日韩欧美精品在线观看| 国产精品国产高清国产av| 三级经典国产精品| 成人毛片a级毛片在线播放| 日本与韩国留学比较| 亚洲av成人精品一二三区| 国产亚洲91精品色在线| 噜噜噜噜噜久久久久久91| 天天躁日日操中文字幕| 亚洲四区av| 99热6这里只有精品| 直男gayav资源| 18+在线观看网站| 成人亚洲欧美一区二区av| 22中文网久久字幕| 最新中文字幕久久久久| 少妇的逼好多水| 听说在线观看完整版免费高清| 丰满少妇做爰视频| 秋霞在线观看毛片| 如何舔出高潮| 成人综合一区亚洲| www.av在线官网国产| av.在线天堂| 国产三级在线视频| 亚洲18禁久久av| 一个人免费在线观看电影| 美女cb高潮喷水在线观看| 午夜激情福利司机影院| 久久草成人影院| 可以在线观看毛片的网站| 亚洲色图av天堂| 看黄色毛片网站| 中文资源天堂在线| 精品国产一区二区三区久久久樱花 | 大又大粗又爽又黄少妇毛片口| 国产伦一二天堂av在线观看| 夫妻性生交免费视频一级片| 久久久久久九九精品二区国产| 草草在线视频免费看| 亚洲精品国产成人久久av| 亚洲国产精品成人综合色| 国产真实乱freesex| 精品人妻一区二区三区麻豆| 18+在线观看网站| av福利片在线观看| 国产av在哪里看| 亚州av有码| 一级毛片电影观看 | 亚洲国产精品久久男人天堂| 欧美97在线视频| 国产精品.久久久| 又粗又爽又猛毛片免费看| 午夜a级毛片| 日本猛色少妇xxxxx猛交久久| 亚洲五月天丁香| 欧美3d第一页| 免费一级毛片在线播放高清视频| 免费av观看视频| 99视频精品全部免费 在线| 日韩av在线免费看完整版不卡| 五月伊人婷婷丁香| 国产精品三级大全| 国产免费视频播放在线视频 | 国产精品久久久久久久电影| 2021天堂中文幕一二区在线观| 日韩欧美三级三区| 国产精品久久视频播放| 久久久国产成人精品二区| 亚洲三级黄色毛片| av福利片在线观看| 少妇人妻一区二区三区视频| 97超视频在线观看视频| 国产成人a∨麻豆精品| 成人av在线播放网站| 在线播放国产精品三级| av在线亚洲专区| 精品人妻偷拍中文字幕| 九九久久精品国产亚洲av麻豆| 国国产精品蜜臀av免费| 午夜福利在线观看吧| 成人二区视频| 精品久久久噜噜| 看十八女毛片水多多多| 亚洲精品成人久久久久久| 久久久久久大精品| 简卡轻食公司| 国产又黄又爽又无遮挡在线| 亚洲无线观看免费| 91久久精品电影网| 国产伦精品一区二区三区视频9| 我要搜黄色片| 亚洲欧美一区二区三区国产| 天堂影院成人在线观看| 九草在线视频观看| 婷婷色麻豆天堂久久 | 美女高潮的动态| 亚洲人成网站在线观看播放| 中文亚洲av片在线观看爽| 成人国产麻豆网| 日本免费a在线| 国产精品一区www在线观看| 国产综合懂色| 一个人看的www免费观看视频| 免费观看a级毛片全部| 97在线视频观看| 国产精华一区二区三区| 亚洲图色成人| 九草在线视频观看| 久久热精品热| 亚洲国产欧美在线一区| 69人妻影院| 啦啦啦啦在线视频资源| 亚洲国产日韩欧美精品在线观看| 一区二区三区高清视频在线| 国产精品人妻久久久影院| 1024手机看黄色片| 少妇熟女欧美另类| 国产欧美日韩精品一区二区| 免费看光身美女| 亚洲精品乱码久久久久久按摩| 久久久成人免费电影| 天堂√8在线中文| 免费观看的影片在线观看| 欧美另类亚洲清纯唯美| 久久精品国产亚洲av涩爱| 蜜臀久久99精品久久宅男| 一边亲一边摸免费视频| 国产精品久久久久久精品电影小说 | 国产免费福利视频在线观看| 久久精品久久久久久噜噜老黄 | 久久久欧美国产精品| 国产av在哪里看| 欧美性猛交╳xxx乱大交人| 午夜爱爱视频在线播放| 午夜视频国产福利| 欧美一区二区精品小视频在线| 99在线人妻在线中文字幕| 美女cb高潮喷水在线观看| 久久人妻av系列| 欧美激情久久久久久爽电影| 欧美bdsm另类| 免费搜索国产男女视频| 午夜精品国产一区二区电影 | 国产精品无大码| 一夜夜www| 中文乱码字字幕精品一区二区三区 | 日日干狠狠操夜夜爽| 老司机福利观看| 看十八女毛片水多多多| 99久久精品热视频| 国产精品久久视频播放| 免费黄网站久久成人精品| 国产精品99久久久久久久久| 亚洲av中文av极速乱| 中国国产av一级| 狂野欧美白嫩少妇大欣赏| 波多野结衣巨乳人妻| 久久久午夜欧美精品| 一级二级三级毛片免费看| 欧美丝袜亚洲另类| 国产伦精品一区二区三区四那| 免费观看在线日韩| 国产精品av视频在线免费观看| 在线观看美女被高潮喷水网站| a级毛色黄片| 亚洲,欧美,日韩| av国产免费在线观看| 美女xxoo啪啪120秒动态图| 亚洲最大成人手机在线| 黄色欧美视频在线观看| 全区人妻精品视频| 3wmmmm亚洲av在线观看| av卡一久久| videos熟女内射| 国产白丝娇喘喷水9色精品| 草草在线视频免费看| 亚洲婷婷狠狠爱综合网| 国产高清三级在线| 波多野结衣高清无吗| 麻豆精品久久久久久蜜桃| 99热精品在线国产| 在线观看av片永久免费下载| 国产精品一及| 日韩亚洲欧美综合| 久久久a久久爽久久v久久| 亚洲成色77777| 美女国产视频在线观看| 中文乱码字字幕精品一区二区三区 | 成年女人永久免费观看视频| a级毛色黄片| 尾随美女入室| 午夜视频国产福利| 男女啪啪激烈高潮av片| 91久久精品国产一区二区三区| 狂野欧美白嫩少妇大欣赏| 日韩在线高清观看一区二区三区| 午夜福利在线在线| 美女xxoo啪啪120秒动态图| 久久精品久久久久久久性| 人人妻人人看人人澡| 中文欧美无线码| 免费搜索国产男女视频| 99久久成人亚洲精品观看| 国产精品久久久久久久电影| 国产高清视频在线观看网站| 精品人妻熟女av久视频| 欧美变态另类bdsm刘玥| 汤姆久久久久久久影院中文字幕 | 三级国产精品欧美在线观看| 亚洲av免费在线观看| 亚洲欧美一区二区三区国产| 国产伦理片在线播放av一区| av播播在线观看一区| 一二三四中文在线观看免费高清| 亚洲国产精品国产精品| 永久免费av网站大全| 久久久精品94久久精品| 国产黄色视频一区二区在线观看 | 久久精品综合一区二区三区| 干丝袜人妻中文字幕| 人妻制服诱惑在线中文字幕| 美女国产视频在线观看| 老司机影院成人| 亚洲人成网站在线观看播放| 黄片无遮挡物在线观看| 69av精品久久久久久| 精品久久久久久久久久久久久| 亚洲成人av在线免费| 精品久久久久久久久av| 国产69精品久久久久777片| 又爽又黄a免费视频| 少妇高潮的动态图| 七月丁香在线播放| 1024手机看黄色片| 国产午夜精品论理片| 久久精品国产自在天天线| 男插女下体视频免费在线播放| 男女那种视频在线观看| 我的老师免费观看完整版| 一级二级三级毛片免费看| 国产91av在线免费观看| 国产探花极品一区二区| 网址你懂的国产日韩在线| 中文资源天堂在线| 久久久精品大字幕| 韩国av在线不卡| 国产黄片视频在线免费观看| 久久久久久久久大av| 村上凉子中文字幕在线| 中文天堂在线官网| 亚洲国产欧美人成| 亚洲国产成人一精品久久久| 波野结衣二区三区在线| 一级毛片电影观看 | 男女那种视频在线观看| 男的添女的下面高潮视频| 一区二区三区四区激情视频| 国产亚洲av片在线观看秒播厂 | 国产91av在线免费观看| 久久精品91蜜桃| 国产亚洲午夜精品一区二区久久 | 亚洲精品自拍成人| 91在线精品国自产拍蜜月| 国产又黄又爽又无遮挡在线| 午夜日本视频在线| 亚洲欧美日韩卡通动漫| 青春草视频在线免费观看| 黄片无遮挡物在线观看| 亚洲精品国产av成人精品| 中国美白少妇内射xxxbb| 美女黄网站色视频| 国产精品一二三区在线看| 最近中文字幕2019免费版| 日韩强制内射视频| 日本三级黄在线观看| 国产爱豆传媒在线观看| 丝袜美腿在线中文| a级一级毛片免费在线观看| 在线观看一区二区三区| 一本一本综合久久| 日本熟妇午夜| 国产亚洲午夜精品一区二区久久 | 少妇被粗大猛烈的视频| 好男人在线观看高清免费视频| 内射极品少妇av片p| 国产一区亚洲一区在线观看| 国产亚洲一区二区精品| 99视频精品全部免费 在线| 激情 狠狠 欧美| 亚洲av成人精品一区久久| 亚洲乱码一区二区免费版| 精品久久久久久久末码| 国产在线男女| 男女下面进入的视频免费午夜| 国产精品一区www在线观看| 精品酒店卫生间| 亚洲欧洲日产国产| 中文字幕制服av| 一区二区三区高清视频在线| 色播亚洲综合网| kizo精华| 少妇猛男粗大的猛烈进出视频 | 日本色播在线视频| 最近手机中文字幕大全| 欧美人与善性xxx| 国产精品一区二区三区四区久久| 国产亚洲一区二区精品| 免费观看在线日韩| 岛国在线免费视频观看| 国产色婷婷99| 日韩精品有码人妻一区| 一级毛片我不卡| 欧美人与善性xxx| 久久人人爽人人爽人人片va| 精品99又大又爽又粗少妇毛片| 午夜福利高清视频| 日韩欧美精品免费久久| 特级一级黄色大片| 国产成人免费观看mmmm| 日韩 亚洲 欧美在线| 精品不卡国产一区二区三区| 我要搜黄色片| 少妇人妻一区二区三区视频| 一区二区三区乱码不卡18| 欧美97在线视频| 精品久久久久久久久亚洲| 久久久久网色| 一级黄色大片毛片| 国产黄片视频在线免费观看| 成人综合一区亚洲| 亚洲伊人久久精品综合 | 精品99又大又爽又粗少妇毛片| 国内精品一区二区在线观看| 色哟哟·www| 青春草国产在线视频| 欧美精品一区二区大全| 国模一区二区三区四区视频| 亚洲无线观看免费| 床上黄色一级片| 天堂影院成人在线观看| 亚洲欧美精品专区久久| 久99久视频精品免费| 精品熟女少妇av免费看| 免费搜索国产男女视频| 99热精品在线国产| 七月丁香在线播放| 国产精品蜜桃在线观看| ponron亚洲| 亚洲va在线va天堂va国产| 伊人久久精品亚洲午夜| 青春草亚洲视频在线观看| 级片在线观看| 亚洲av中文av极速乱| 在现免费观看毛片| 亚洲性久久影院| 欧美精品国产亚洲| 尤物成人国产欧美一区二区三区| 婷婷色综合大香蕉| 水蜜桃什么品种好| 亚洲第一区二区三区不卡| 99久久无色码亚洲精品果冻| 尾随美女入室| 一级毛片电影观看 | 中文资源天堂在线| 国产高清视频在线观看网站| 国产在线一区二区三区精 | 成人午夜高清在线视频| 国产伦一二天堂av在线观看| АⅤ资源中文在线天堂| 亚洲天堂国产精品一区在线| 国产精品人妻久久久久久| 麻豆久久精品国产亚洲av| 亚洲综合精品二区| 搞女人的毛片| 国产视频首页在线观看| 秋霞伦理黄片| 天堂√8在线中文| 国产精品麻豆人妻色哟哟久久 | 日日啪夜夜撸| 日韩 亚洲 欧美在线| 午夜福利网站1000一区二区三区| 最近手机中文字幕大全| 在线a可以看的网站| 国国产精品蜜臀av免费| 男女那种视频在线观看| 亚洲伊人久久精品综合 | 精品久久久久久久人妻蜜臀av| 中文精品一卡2卡3卡4更新| av线在线观看网站| 久久精品国产自在天天线| 国产v大片淫在线免费观看| 亚洲成人精品中文字幕电影| 99热这里只有是精品在线观看| 国产免费福利视频在线观看| 一本久久精品| 人体艺术视频欧美日本| 一区二区三区免费毛片| www.av在线官网国产| 我要搜黄色片| 久久亚洲国产成人精品v| 色综合站精品国产| 中文字幕av在线有码专区| 国产精品日韩av在线免费观看| 成人一区二区视频在线观看| 精品午夜福利在线看| 男女那种视频在线观看| av又黄又爽大尺度在线免费看 | 三级经典国产精品| 岛国在线免费视频观看| 黄色配什么色好看| 女人十人毛片免费观看3o分钟| 中国美白少妇内射xxxbb| or卡值多少钱| 久久久久免费精品人妻一区二区| 91久久精品国产一区二区三区| 久久精品国产亚洲av天美| 中文字幕av成人在线电影| 热99re8久久精品国产| 日韩一本色道免费dvd| 国产黄色视频一区二区在线观看 | 欧美色视频一区免费| 在线天堂最新版资源| 午夜精品在线福利| 国内精品美女久久久久久| 午夜精品在线福利| 中文天堂在线官网| 热99re8久久精品国产| 婷婷色麻豆天堂久久 | 日韩人妻高清精品专区| 美女cb高潮喷水在线观看| 国产高清不卡午夜福利| 久久韩国三级中文字幕| 欧美最新免费一区二区三区| 天天躁日日操中文字幕| 国产老妇女一区| 美女cb高潮喷水在线观看| 免费观看在线日韩| 亚洲av免费高清在线观看| 全区人妻精品视频| 欧美色视频一区免费| 欧美精品一区二区大全| 久久精品夜色国产| 人人妻人人澡欧美一区二区| 久久草成人影院| 久久婷婷人人爽人人干人人爱| 高清毛片免费看| 99久久精品热视频| 夜夜爽夜夜爽视频| 日韩强制内射视频| 日本熟妇午夜| 国产精品久久视频播放| 国产老妇女一区| 男插女下体视频免费在线播放| av在线蜜桃| 最近2019中文字幕mv第一页| 国产极品天堂在线| 男人狂女人下面高潮的视频| 日韩精品有码人妻一区| 尤物成人国产欧美一区二区三区| 亚洲自偷自拍三级| 99久久精品一区二区三区| 欧美性猛交╳xxx乱大交人| 日本熟妇午夜| 午夜福利在线观看吧| 免费观看在线日韩| 简卡轻食公司| 久久99精品国语久久久| 免费看av在线观看网站| 亚洲av成人精品一二三区| 久久久久久久亚洲中文字幕| 国产精品日韩av在线免费观看| 日韩在线高清观看一区二区三区| 日本免费a在线| 国产高清不卡午夜福利|