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Linear
Algebra
and
its
Applications
www.elsevier.com/locate/laa
Matrix
representations
of
the
real
numbers
Yu Chen
DepartmentofMathematics,UniversityofTorino,ViaCarloAlberto10, 10123Torino,Italy
a r t i c l e i n f o a bs t r a c t
Article history: Received28June2017 Accepted14September2017 Availableonline18September2017 SubmittedbyP.Semrl MSC: 15A04 12D99 Keywords: Matrixrepresentation Realnumberfield Complexnumberfield Discontinuoushomomorphisms
Themainpurposeofthispaperistodetermineallmatrix rep-resentationsoftherealnumbers.Itisshownthateverysuch representationiscompletelyreducible,whileallnon-trivial ir-reduciblerepresentationsmustbeof2-dimensionalandcanbe expressedina uniqueform.Itisfoundthat those represen-tationsareessentiallydeterminedbythewaysofembedding therealnumbersintothecomplexnumbers.Thisresultsina one-to-one correspondencebetweentheequivalentclasses of irreduciblerepresentationsandtheequivalentclassesof homo-morphismsfromtherealnumberfieldtothecomplexnumber field.Thematrixrepresentationsofthecomplexnumbersare alsodetermined.
©2017ElsevierInc.Allrightsreserved.
1. Introductionandmaintheorem
The main purpose of this paper is to answer the question: for an arbitrary posi-tive integer n, in how many different ways the real numbers can be represented by n× n real matrices such that the two basic operations, addition and multiplication, are preserved. In other words, we aim to determine all matrix representations of R.
E-mailaddress:yu.chen@unito.it.
http://dx.doi.org/10.1016/j.laa.2017.09.012
Here a matrixrepresentation meansa ring homomorphism between R andthe ring of n× n matrices Mn(R) for an arbitrary n ∈ N. For example, let ρ : R → M2(R) be
a matrix representation of R. The restriction of ρ to R+, which is the multiplicative
subgroup of R consisting of all positive real numbers, induces a group representa-tion of R+. Note that R+ is a locally compact abelian group,of which all continuous
representations are well known (cf. [5]). Hence if ρ is continuous, from the restricted grouprepresentation ofR+ it iseasy to deducethatupto conjugacy ρ mustbe of the form ρ(a) = τ (a) 0 0 τ(a) , ∀a ∈ R
whereτ andτ arethezeromaportheidentitymapofR.Ontheotherhandif discon-tinuityisgranted,there existsahuge familyofrepresentations ofform ρσ :R→ M2(R)
definedby ρσ(a) = σ(a)r −σ(a)i σ(a)i σ(a)r , ∀a ∈ R (1)
where σ isadiscontinuoushomomorphismfrom R toC while σ(a)r and σ(a)i arereal
andimaginarypartsofσ(a) respectively.Thecardinalityofthisfamilyofrepresentations is2cwherec isthecardinalityofC (cf.Corollary 3.1).Wewillseethatuptoequivalence
theaboverepresentationsactuallyprovideacompletelistforalltwo-dimensionalmatrix representationsofR,whichisaconsequenceofourmaintheorem.
In order to state ourmain results we recall some elementary notions. Let ρ : R → Mn(R) be a matrix representation of R for an arbitrary n ∈ N. By identifying Mn(R)
with theR-linearendomorphisms EndR(Rn),the image ρ(R) of R under ρ canbe
con-sideredas asubset oflineartransformationsover thevectorspaceRn.Then ρ iscalled irreducibleifthereisnonon-trivialpropersubspaceofRn stabilizedbyρ(R).Amatrix
representationofR iscalledcompletelyreducibleifitisadirectsumofsomeirreducible representations.Twon-dimensionalrepresentationsρ andρ aresaidequivalent ifthere existsanon-singularmatrixX∈ Mn(R) suchthat
ρ(a) = Xρ(a)X−1, ∀a ∈ R.
WedenotebyιX theinner automorphismofMn(R) viatheconjugationbyX.
Denote by Hom(R,C) the set of ring homomorphisms from R to C. The cardinality of this set is infinite (cf. [3] and [7]). Two homomorphisms σ,τ ∈ Hom(R,C) are said equivalent if σ = ατ , where α is either the identity map or the complex conjugation ofC.ThisisobviouslyanequivalentrelationofHom(R,C).Forconveniencewecallboth the zeromap andtheidentitymap ofR thetrivialrepresentationsofR.
Theorem 1.Letρ:R→ Mn(R) beamatrix representationof R wheren∈ N.
1. The representation ρ iscompletelyreducible.
2. If ρ is irreducible and non-trivial, then n = 2 and there exist a homomorphism σ∈ Hom(R,C) and anon-singularmatrixX ∈ M2(R) such that
ρ = ιX· ρσ (2)
where ρσ isdefinedby theidentity (1).
3. The above σ is uniqueuptoequivalence. Moreover, thereexists aone-to-one corre-spondence betweentheequivalent classes ofirreduciblerepresentations ofR and the equivalent classesof Hom(R,C).
The proof of this theorem is developed inthe following sections. Mean while, from thistheoremwedeterminealsothematrixrepresentationsforthecomplexnumbers(cf.
Theorem 2).
2. Preliminaries
Throughout thispaper for arepresentation we alwaysmean areal matrix represen-tation unless otherwise explained. Let ρ : R → Mn(R) be a representation. Then by
identifying Mn(R) withEndR(Rn),Rn is aR-ρ(R) bimodule. Equivalently, ifwe denote
by Rρ the subringof Mn(R) generated by ρ(R) and RI where I is the n× n identity
matrix,then Rn isaR
ρ-moduleinnaturalway.Thefollowingpropertyis obvious.
Proposition 2.1.A representation ρ :R → Mn(R) is irreducibleif and only if Rn is an
irreducibleRρ-module.
RecallthattheradicalofamoduleM isbydefinitiontheintersectionofallmaximal submodules of M . Given a representation ρ: R→ Mn(R),we denote byRadρ(Rn) the
radical oftheRρ-moduleRn.
Proposition 2.2. Let ρ: R → Mn(R) be a representation. The following statements are
equivalent.
1. ρ iscompletelyreducible. 2. Rn is asemisimpleRρ-module.
3. Radρ(Rn)={0}.
Proof. Theequivalenceofthe firsttwostatementsisobvious.Theequivalenceofthelast twostatementscomesfromthefactthat,sinceRnisanartinianmodule,itissemisimple
We need some elementary properties from linear algebra. Recall that an element x∈ Mn(R) is nilpotent if xm = 0 for some m ∈ N and that x isunipotent if x− I is
nilpotent, where I is the identity matrix. Anelement x ∈ Mn(R) is semisimple ifit is
conjugatedtoadiagonalmatrixinMn(C).
Lemma2.1(Jordan–Chevalleydecomposition).Letx beanon-zero element ofMn(R).
1. Thereexistauniquesemisimpleelementxs∈ Mn(R) and auniquenilpotentelement
xn∈ Mn(R) suchthat x= xs+ xn andxsxn= xnxs.Moreover, there exist
polyno-mialswithout constantterm f (t),g(t)∈ R[t] suchthat xs= f (x) and xn= g(x).
2. If x is non-singular, there exists aunique unipotent element xu ∈ Mn(R) such that
x= xsxu andxsxu = xuxs.More over, everysubspace of Rn stabilizedby x isalso
stabilizedby xu.
Theelementsxs,xnandxuarecalledsemisimple,nilpotentandunipotentpartsofx
respectively.Wereferto[2,p. 80]for moredetailsoftheJordan–Chevalley decomposi-tions.
Lemma 2.2. Let {x1,x2,...,xm} be a commutative subset of Mn(R) and let xi,s be the
semisimplepartof xi for1≤ i≤ m.Thenan element
m
i=1aixi∈ Mn(R),where ai∈ R
for1≤ i≤ m,isnilpotentifand onlyif
m
i=1
aixi,s= 0
Proof. ItfollowsfromthepolynomialpropertyofJordandecompositionthat{x1,s,x2,s,
...,xm,s} is a commutative set since so is {x1,x2,...,xm}, therefore {a1x1,s,a2x2,s,...,
amxm,s} is also acommutative set. Note that the sum and the multiplication of a
fi-nite number of commutative semisimple elements are still semisimple. Then aixi,s is
semisimple forall 1≤ i ≤ m and therefore mi=1aixi,s isalso semisimple. Let xi,n be
thenilpotent partof xi forall1≤ i≤ m.Then {x1,n,x2,n,...,xm,n} is acommutative
setsinceso is{x1,x2,...,xm}.Since aixi,n isnilpotent forall1≤ i≤ m,
m
i=1aixi,n is
alsonilpotent.Wehave
m i=1 aixi= m i=1 ai(xi,s+ xi,n) = m i=1 aixi,s+ m i=1 aixi,n,
meanwhileobviously mi=1aixi,s and mi=1aixi,n commutewith each other. Then the
uniquenessof Jordandecompositionimpliesthat
( m i=1 aixi)s= m i=1 aixi,s.
Consequently the lemma follows from thefact thatanelement of Mn(R) is nilpotent if
and onlyifthesemisimplepartoftheelementvanishes. 2
Lemma2.3. Let G be acommutative subgroupofGLn(R) andGu bethesetofthe
unipo-tent partsofallx∈ G.If V isasubspaceofRn stabilizedbyG,thenV isalso stabilized by Gu.Moreover, there exists anon-zero vectorv∈ V which is fixedby every element
of Gu.
Proof. It is a consequence of Lemma 2.1 thateach subspace V ⊆ Rn stabilized by G must be also stabilized by Gu. More over, since G is commutative, it is easy to check
that Gu is aunipotent subgroup of GLn(R).Then the existence of anon-trivialvector
of V fixed byGuisaconsequenceofLie–KolchinTheorem (cf.[2,p. 87]). 2
Lemma 2.4.Given arepresentation ρ:R→ Mn(R) and anon-zeroelement v∈ Rn,let
Rρv ={xv ∈ Rn|∀x ∈ Rρ}
which isaRρ-module.Thentheradical Rad(Rρv) of Rρv is trivial.
Proof. Note thattheset of non-zeroelements of ρ(R),denotedby ρ(R)∗,is a commu-tative multiplicative subgroup of GLn(R), which obviously stabilizes thesubspace Rρv
ofRn.Letρ(R)
ubethegroupconsistingoftheunipotentpartsofallelementsofρ(R)∗.
Then byLemma 2.3, Rρv isalso stabilizedbyρ(R)u. Againitfollows fromLemma 2.3
thatthereexists anon-zeroelement v0∈ Rρv suchthat
xuv0= v0, ∀xu∈ ρ(R)u. (3)
LetRρv0 bethesubmoduleof Rρv generatedbyv0.Weclaimthat
Rρv0= Rρv. (4)
In fact,consider Rρv asaR-spaceand letα be anon-singularR-lineartransformation
of Rρv such thatα(v)= v0.Since v0 belongs to Rρv, we havev0 = yv for an element
y ∈ Rρ.Letλα: Rρv→ Rρv beamapdefinedby
λα(xv) = yxv, ∀x ∈ Rρ.
ClearlyλαisanendomorphismofRρ-modulesinceRρisacommutativering.Moreover,
since
yxv = xyv = xv0, ∀x ∈ Rρ
wehaveλα(Rρv)= Rρv0.Sinceα−1isalsoaR-lineartransformationofRρv,inasimilar
waywedefineanendomorphismλα−1 ∈ EndRρ(Rρv).Itiseasytocheckthatλα−1 isthe
Rρv = λα(Rρv) = Rρv0.
Now we show that the radical Rad(Rρv0) of the module Rρv0 is trivial. Let J be the
Jacobsonradical ofthering Rρ. SinceRρv is afinitelygenerated module, wehave (cf.
[1,p. 172])
Rad(Rρv0) = J· Rρv0= RρJ v0= J v0. (5)
Foranarbitraryx∈ J,wecanwrite
x =
m
i=1
aixi, for ai∈ R, xi∈ ρ(R), 1 ≤ i ≤ m.
Let xi,s and xi,u be the semisimple and the unipotent parts of xi respectively. Then
xi,u∈ ρ(R)uforall1≤ i≤ m.WehavebyLemma 2.1andtheidentity (3)that
xv0= m i=1 aixi,sxi,uv0= m i=1 aixi,sv0
Note thatallelements of J arenilpotent sinceRρ is finitely generated.It followsfrom
Lemma 2.2andthenilpotencyofx∈ J that
m i=1 aixi,s= 0. Hence xv0= 0, ∀x ∈ J.
Thisimpliesbytheidentities(4)and (5)that
Rad(Rρv) = Rad(Rρv0) ={0}. 2
Lemma 2.5.Let R bea R-subalgebra of Mn(R). If R is an integral domain, then itis a
divisionring.
Proof. Weshowthateverynon-zeroelements∈ R hasaninverseinR.DenotebyR[s] thesubalgebraofR generatedbyRI ands. Notethats isalgebraic overR because
dimRR[s] ≤ dimRMn(R) < ∞.
Letf (x)∈ R[x] be theminimal polynomialofs over R and denoteby (f (x)) theideal ofR[x] generatedbyf (x).Then wehaveanisomorphism
R[s] ∼=R[x]/(f(x)).
We claim thatf (x) is irreducible, which implies by above isomorphism thatR[s] is a field and consequently s−1 ∈ R[s] ⊆ R. In fact if otherwise there exist non-constant polynomials g(x),h(x)∈ R[x] suchthatf (x)= g(x)h(x).Then
g(s)h(s) = 0.
However,sincebothg(x) andh(x) havedegreestrictlylessthanthatoff (x),
g(s) = 0, h(s) = 0.
ThisiscontrarytotheassumptionthatR isanintegraldomain.ThusweobtainthatR is adivisionring. 2
Lemma 2.6.Let ρ : R → Mn(R) be an irreducible representation. Then Rρ is a field.
In particular, if ρ is non-trivial, there exists an isomorphism β : C → Rρ such that
β(a)= aI foralla∈ R.
Proof. Since Rn isanirreducibleR
ρ-module,itcomesfrom Proposition 2.1andSchur’s
lemma thatEndRρ(R
n) is adivisionring.It isobvious thatbothRI and ρ(R) are
con-tainedinEndRρ(R
n).Hencewehave
Rρ ⊆ EndRρ(R
n).
ThismeansthatRρisanintegraldomain.ThenitfollowsfromLemma 2.6thatRρ isa
fieldsinceitiscommutative.Inparticular, ifρ isnon-trivial,thenρ(R) isnotcontained in RI. Hence Rρ is a non-trivial fieldextension of finite degree over RI. This implies
thatRρhastobeisomorphictoC whichistheuniquenon-trivialfieldextensionoffinite
degreeoverR.Theexistenceofanisomorphismasβ isobvious. 2 Denote byCMn(R)Rρ thecentralizerofRρ inMn(R).
Lemma 2.7.Letρ:R→ Mn(R) beanon-trivialand irreduciblerepresentation. Then
CMn(R)Rρ= Rρ. (6)
Proof. ByidentifyingMn(R) withEndR(Rn) wehave
Rρ⊆ CMn(R)Rρ= EndRρ(R
n).
On the otherhand, foran arbitrarys∈ CMn(R)Rρ, thesubringRρ(s) of CMn(R)Rρ
gen-erated by Rρ and s is a commutative R-subalgebra. Note that CMn(R)Rρ is a division
ring sinceso is EndRρ(R
n) by Schur’s lemma. It follows from Lemma 2.5 thatR ρ(s) is
afield.Moreover,sinces mustbe algebraicoverRI,Rρ(s) isanalgebraicextension of
thefield Rρ. Thisimplies thatRρ(s)= Rρ sinceRρ isisomorphicto C byLemma 2.6.
Thens belongstoRρ.Hencewehave
CMn(R)Rρ ⊆ Rρ
fromwhichfollowstheidentity (6). 2
Lemma2.8.If arepresentation ρ:R→ Mn(R) isnon-trivialandirreducible,thenn= 2.
Proof. Itfollowsfromthedoublecentertheoremofsimplealgebras(cf.[4,p. 232])that
dimRRρ· dimRCMn(R)Rρ= dimRMn(R) = n
2.
NotethatbyLemma 2.6andLemma 2.7wehave
dimRCMn(R)Rρ= dimRRρ= 2.
Thisimpliesthatn= 2. 2
3. ProofofTheorem 1andmiscellaneousresults
Proof ofTheorem 1.1. Givenarepresentation ρ:R→ Mn(R),we havetheR-space Rn
becomingaRρ-modulewhere Rρ isthesubalgebraofMn(R) definedatthebeginningof
Section2.Let{v1,v2,. . . ,vn} beaR-basisofRn.Thenwehave
Rn = n i=1 Rvi= n i=1 Rρvi.
NotethatsinceRρviisafinitelygeneratedmodulewithtrivialradicalbyLemma 2.4,it
hasto besemisimple(cf.[1,p. 129])forall1≤ i≤ n.Thereforeasasumofsemisimple submodules,Rn is also semisimple.This is equivalent tothe complete reducibilityof ρ
byProposition 2.2. 2
Proof ofTheorem 1.2. Supposethatρ:R→ Mn(R) isirreducibleand non-trivial.Then
itfollows from Lemma 2.8 thatn= 2. Moreover, it comesfrom Lemma 2.6 thatthere existsanisomorphismβ :C→ Rρ suchthatβ(a)= aI foralla∈ R.Letμ:C→ M2(R)
bethecanonicalringhomomorphismdefinedby
μ(z) = zr −zi zi zr , ∀z ∈ C (7)
wherezrandzi arerealandimaginarypartsofz respectively.Thecompositionμβ−1 is
M2(R) byLemma 2.6. ThenbyNoether–SkolemTheorem (cf. [4,p. 230])there exists a
non-singular matrixY ∈ M2(R) suchthatμβ−1 isjusttherestrictionof theinner
auto-morphismιY ofM2(R) whichistheconjugationviaY .Denotebyσ thecomposition β−1ρ.
It iseasytocheck thatthefollowingdiagram iscommutative.
R ρ −−−−→ M2(R) σ ⏐ ⏐ ⏐⏐ιY C μ −−−−→ M2(R)
Then wehave,foralla∈ R,
ρ(a) = ι−1Y μσ(a) = Y−1 σ(a)r −σ(a)i σ(a)i σ(a)r Y.
BysettingX = Y−1,weobtaintheidentity(2)ofTheorem 1. 2
Inorder toproveTheorem 1.3weneedsomepropertiesofendomorphismsas wellas representationsofcomplexnumbers.Notethatthepropertiesandthediscussionusedfor proving Theorem 1.1and1.2areextendableto therepresentationsofcomplexnumbers bysimplysubstitutingC forR intheargument.Thesubstitutionleadstoadescriptionof thematrixrepresentationsofcomplexnumbers,whichissimilartothatofrealnumbers.
Theorem 2.Every matrix representation ρ:C → Mn(R) is completelyreducible.
More-over, if ρ isirreducible,then n= 2 and thereexistsaσ∈ End(C) suchthat ρ = ιXμσ
where ιX is the inner automorphism of M2(R) via the conjugation by a non-singular
matrix X andμ isthehomomorphismdefined by theidentity(7). Proof. Thisisanalogousto theproofsofTheorem 1.1and1.2. 2
WeconsiderHom(R,C).Itisatrivialfactthateachhomomorphismσ∈ Hom(R,C) can be extendtoanendomorphismofC.
Lemma 3.1. Let C be afield which is isomorphic to C. Then each homomorphism σ∈ Hom(R,C) has exactlytwoextensions σ,¯ σ¯∈ Hom(C,C) and,more over,σ = ¯¯ σ where is thecomplexconjugation.
Proof. Thiscomesfromthefactthat,fortheimaginaryi,theextensionsofσ haveonly two possibleimages{i,−i} sincetheminimalpolynomialofi isx2+ 1∈ Q[x]. 2
Twoendomorphismsβ,τ ∈ End(C) aresaid equivalent ifβ = ατ α, where α and α arethe identity map orthe complexconjugation.This isclearly anequivalent relation forEnd(C).Thefollowingproperty isobvious.
Proposition 3.1.The extension of each element of Hom(R,C) to an element of End(C) inducesaone-to-onecorrespondencebetweenthefamilyofequivalentclassesofHom(R,C) andthat ofEnd(C).
Proposition 3.2. Each irreducible representation ρ : R→ M2(R) has exactly two
exten-sions ρ :¯ C→ M2(R). In precise,if ρ= ιXρσ forsome X ∈ GL2(R) and σ∈ Hom(R,C),
then ρ is¯ either ιXμ¯σ or ιXμ¯σ , where ¯σ ∈ End(C) is an extension of σ and is the
complex conjugation.
Proof. Theexistenceof extensions forρ isobvious. Note thateveryextension ρ of¯ ρ is irreduciblesinceso isρ.Denote byRρ¯thesubringof M2(R) generatedbyRI andρ(¯C).
ThenR2 isanirreducibleR ¯
ρ-module.Since,bySchur’slemma, EndRρ¯(R
2) isadivision
ringwhich contains theR-subalgebra Rρ¯ ofM2(R),it follows from Lemma 2.5 thatRρ¯
is a field extension of degree 2 over RI. Note that by the same lemma Rρ is also an
extension of degree 2 over RI and thatRρ¯ ⊇ Rρ. We obtain thatRρ¯= Rρ. Then by
Lemma 3.1ρ has onlytwo possible extensions. Inconsequence,the second assertionof thelemmaisobvious. 2
Proof ofTheorem 1.3. Givenanon-trivialandirreduciblerepresentationρ:R→ M2(R),
supposethatthereexist X,Y ∈ GL2(R) andσ,τ∈ Hom(R,C) suchthat
ρ = ιXρσ = ιYρτ.
SetT = Y−1X.Then theaboveidentityimpliesthat
ιTρσ = ρτ. (8)
NotethateachextensionofanelementofHom(R,C) eitherfixtheimaginaryi orsend i to −i.For conveniencewe denote byσ¯∈ End(C) the extension ofσ such thatσ(i)¯ = i and similarly by τ the¯ extension of τ which fixes the imaginary i. Since ρτ has two
extensionsμ¯τ and μ¯τ byProposition 3.2,theidentity(8)impliesthateither
ιTσ = ¯¯ τ (9)
or
ιTσ = ¯¯ τ . (10)
Supposethatthe first identity (9)holds. Byapplying to theimaginary i from both sidesoftheidentitywehaveanequation
T 0 −1 1 0 T−1 = 0 −1 1 0 . This impliesthat
T =
a −b b a
for somea,b∈ R.Inparticular, sinceallmatricesofρσ(R) havethesameformas that
of T ,itiseasytocheckthat
ιTρσ(a) = ρσ(a), ∀a ∈ R
whichmeansthatρσ= ρτ bytheidentity(8).Weobtainhenceσ = τ inthis case.
Ifthesecondidentity(10)holds,byapplyingtoi frombothsideswehaveanequation
T 0 −1 1 0 T−1 = 0 1 −1 0 . This impliesthat
T = a b b −a = 0 1 1 0 · b −a a b
forsomea,b∈ R.Ifwedenote byS thematrix
0 1 1 0
, then
ιTρσ(a) = ιSρσ(a) = ρσ(a), ∀a ∈ R.
Weobtainagainbytheidentity(8)that σ = τ ,whichmeansthatbothσ andτ belong to thesameequivalentclass. ThuswecompletetheproofofTheorem 1. 2
We observe from Theorem 1.3 that the cardinality of the equivalent classes of the irreducible representations ofR is thesameas thatof Hom(R,C),whichis equalto the cardinality ofEnd(C) byProposition 3.1. Notethattheset offunctionsfrom C toitself hascardinalitycc since|C|= c andthatthecardinalityof automorphismsAut(C) is2c.
Wehave
2c=|Aut(C)| ≤ |End(C)| ≤ cc= 2ℵ0c= 2c
which meansthatthecardinality ofEnd(C) is2c.Hencetheirreduciblerepresentations
of R canbequantifiedas follows.
Corollary 3.1. Letc represent2ℵ0.The cardinality of thefamily of equivalent classesof
Remark.Amongvariousapplications ofthematrixrepresentationsofR,wementionits connectionwithso-called“abstract”representationsof Liegroupswhere thecontinuity isnotan assumptionimposedon thegroupactionon realspaces(see forinstance [6]). An exampleisthateachdiscontinuousirreduciblerepresentationρ:R→ M2(R) inducesa
discontinuousirreduciblegrouprepresentationρ : SL¯ 2(R)→ GL4(R) whereρ is¯ obtained
byapplyingρ totheentriesofeachmatrixinSL2(R),whichshedslightondetermining
alldiscontinuousactionsofSL2(R) onrealspaces sincebynowallrepresentations ofR
areknown.
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