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Contents lists available atScienceDirect

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

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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.

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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 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 ofthe-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

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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.

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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.

Clearlyλαisanendomorphismof-modulesinceisacommutativering.Moreover,

since

yxv = xyv = xv0, ∀x ∈ Rρ

wehaveλα(Rρv)= Rρv0.Sinceα−1isalsoaR-lineartransformationofRρv,inasimilar

waywedefineanendomorphismλα−1 ∈ EndRρ(Rρv).Itiseasytocheckthatλα−1 isthe

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Rρv = λα(Rρv) = Rρv0.

Now we show that the radical Rad(Rρv0) of the module Rρv0 is trivial. Let J be the

Jacobsonradical ofthering . 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 since 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

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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 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 thatEnd(R

n) is adivisionring.It isobvious thatbothRI and ρ(R) are

con-tainedinEnd(R

n).Hencewehave

⊆ EndRρ(R

n).

Thismeansthatisanintegraldomain.ThenitfollowsfromLemma 2.6that isa

fieldsinceitiscommutative.Inparticular, ifρ isnon-trivial,thenρ(R) isnotcontained in RI. Hence is a non-trivial fieldextension of finite degree over RI. This implies

thathastobeisomorphictoC whichistheuniquenon-trivialfieldextensionoffinite

degreeoverR.Theexistenceofanisomorphismasβ isobvious. 2 Denote byCMn(R) thecentralizerof 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)= End(R

n).

On the otherhand, foran arbitrarys∈ CMn(R), thesubringRρ(s) of CMn(R)

gen-erated by and s is a commutative R-subalgebra. Note that CMn(R) is a division

ring sinceso is End(R

n) by Schur’s lemma. It follows from Lemma 2.5 thatR ρ(s) is

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afield.Moreover,sinces mustbe algebraicoverRI,Rρ(s) isanalgebraicextension of

thefield Rρ. Thisimplies thatRρ(s)= Rρ since isisomorphicto C byLemma 2.6.

Thens belongsto.Hencewehave

CMn(R) ⊆ 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)= dimRMn(R) = n

2.

NotethatbyLemma 2.6andLemma 2.7wehave

dimRCMn(R)= dimRRρ= 2.

Thisimpliesthatn= 2. 2

3. ProofofTheorem 1andmiscellaneousresults

Proof ofTheorem 1.1. Givenarepresentation ρ:R→ Mn(R),we havetheR-space Rn

becominga-modulewhere 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

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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

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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 by¯thesubringof M2(R) generatedbyRI andρ(¯C).

ThenR2 isanirreducibleR ¯

ρ-module.Since,bySchur’slemma, End¯(R

2) isadivision

ringwhich contains theR-subalgebra ¯ ofM2(R),it follows from Lemma 2.5 that¯

is a field extension of degree 2 over RI. Note that by the same lemma is also an

extension of degree 2 over RI and that¯ ⊇ Rρ. We obtain that¯= 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

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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= 20c= 2c

which meansthatthecardinality ofEnd(C) is2c.Hencetheirreduciblerepresentations

of R canbequantifiedas follows.

Corollary 3.1. Letc represent20.The cardinality of thefamily of equivalent classesof

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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.

References

[1]F.W. Anderson, K.R. Fuller, Rings and Categories of Modules, Grad. Texts in Math., vol. 13, Springer-Verlag,NewYork,1974.

[2]A.Borel,LinearAlgebraicGroups,2nded.,Grad.TextsinMath., vol. 126,Springer-Verlag,New York,1991.

[3]H.Kestelman,Automorphismsofthefieldofcomplexnumbers,Proc.Lond.Math.Soc.53(1951) 1–12.

[4]R.S.Pierce,AssociativeAlgebra,Grad.TextsinMath.,vol. 88,Springer-Verlag,NewYork,1982. [5]A.Robert, Introductionto theRepresentation TheoryofCompact andLocallyCompact Groups,

LondonMath.Soc.LectureNoteSer.,vol. 80,CambridgeUniv.Press,1983.

[6]J.Tits,Homomorphismes“abstrait”degroupesdeLie,in:Symp.Math.Inst.Naz.AltaMat.,vol. 13, Bologna,1974,pp. 479–499.

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