Algebra of paravectors

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by  Radomański Józef

Algebra of paravectors
JÓZEF RADOMAN´SKI
ul.BohaterówWarszawy9,48-300Nysa,Poland
e-mail:[email protected]
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Abstract

7 Paravectorsjust like integers havea ring structure. By introducing an integrated product we get geometric
propertieswhichmakeparavectorssimilartovectors.Theconceptsofparallelism,perpendicularityandtheangle
]
A areconceptuallysimilartovectorcounterparts,knownfromtheEuclideangeometry. Paravectorsmeettheideaof
parallelogramlaw,Pythagoreantheoremandmanyotherpropertieswell-knowntoeveryone.
R
h. Keywords:Paravector,multivector,GeometricAlgebra
t
a This article refers to professor William Baylis’ researches and shows a surprising similarity between
m propertiesofparavectorsandvectorsintheEuclideanspace. Itcanbeconfusingatfirsttogetusedtoacolumn
[ notationofparavectors,butthisformhasprovedtobethemostadequate.Themainadvantageofthisnotationis

a transparency of mathematical transformations. The author hopes that in future paravector formalism will
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replacethemultivectorsformalisminphysicsassimpler,moreintuitiveandimaginable.Theresultspresentedin
v
5 thisarticlejustifyitfully.
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1 Basicdefinitions
0
.
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0 Definition1.1. Thetermparavectormeansapairconsistingofacomplexnumber(α)andavector(βββ)belonging
6 toathree-dimensionalcomplexspace.
1
α a+id
:
v Γ=: = (1)
i βββ b+ic
X
r
a Thenumberwillbecalledascalar.Paravectorswillbedenotedwithcapitalletters,eg:A,X,Γ.Greekletters
willmeanacomplexsize,andRomanletters-arealone.AscalarcomponentofparavectorΓwillbedenotedwith
index„S”,andavectorcomponentwith„V”,i.e.:Γ =αandΓ =βββ.
S V
Definition1.2. Thereversedelementofparavector(1)istheparavector
a+id
Γ−=: (2)
 
b ic
− − 
Definition1.3. Theconjugatedelementofparavector(1)istheparavector
a id
Γ∗=: − (3)
 
b ic
 − 
1
Definition1.4. ofthesummation:
α  α  α +α 
1 2 1 2
+ =: (4)
βββ  βββ  βββ +βββ 
 1  2  1 2
Conclusion1.1. Theneutralelementunderaddition(nullelement)istheparavector
0
0=: (5)
 
0
 
Definition1.5. Theoppositeelementofparavector(1)withrespecttotheadditionis
 α
Γ=: − (6)
−  
βββ
− 
Definition1.6. ofthemultiplication:
α α   α α +βββ βββ 
1 2 1 2 1 2
=: (7)
βββ βββ  α βββ +α βββ +iβββ βββ 
 1 2  2 1 1 2 1× 2
whereβββ βββ isascalar(dot)productofvectors,andβββ βββ isavector(cross)product.
1 2 1× 2
Conclusion1.2. Theneutralelementundermultiplicationistheparavector
1
1=: (8)
 
0
 
α
Note:Thereisnodifferenceifwewritenumberαorparavector .
 
0
 
Conclusion1.3. Theoperationofmultiplicationisassociativebutnotcommutative,becausethevectorproduct
isnon-commutative.
Definition1.7. WecallanoutcomeofmultiplicationofanyparavectorΓbytheelementconjugateΓ thevigorof
∗
paravector:
vigΓ=:ΓΓ∗ (9)
Conclusion1.4. Toeachparavectorwecanassignavigorwhichisarealparavectoranditsscalarcomponentisa
positivenumber.
Proof.
ΓΓ∗=αα∗= αα∗+ββββββ∗ = (a+id)(a−id)+(b+ic)(b−ic) =
ββββββ∗∗∗ αβββ∗+α∗βββ+iβββ×βββ∗ (a+id)(b−ic)+(a−id)(b+ic)+i(b+ic)×(b−ic)
a2+b2+c2+d2
=
2(ab+dc+b c)
 × 
Definition 1.8. We call an outcome of multiplication of any paravector Γ by the reverse element Γ the
−
determinantofaparavector
detΓ=:ΓΓ−=Γ−Γ (10)
2
Conclusion1.5. Eachparavectorhasadeterminantwhichisacomplexnumber.
Proof.
α α   α2 β2  (a+id)2 (b+ic)2 a2 b2+c2 d2+2i(ad bc)
ΓΓ−= = − = − = − − −
βββ βββ αβββ αβββ iβββ βββ  i(b+ic) (b+ic)  0 
 −   − − ×  − ×   
Conclusion1.6. Thereversionandconjugationhavethefollowingproperties:
Reversionofparavector Conjugationofparavector
1 (Γ−)−=Γ (Γ∗)∗=Γ
2 (Γ+Ψ)−=Γ−+Ψ− (Γ+Ψ)∗=Γ∗+Ψ∗
3 (ΓΨ)−=Ψ−Γ− (ΓΨ)∗=Ψ∗Γ∗
4 ΓΓ C ΓΓ R R3
−∈ ∗∈ +×
5 (Γ−)∗=(Γ∗)−
Conclusion1.7. AsetofpravectorstogetherwithanoperationofsummationformsanAbeliangroup,andwith
multiplicationformsasemigroup. Therefore,wecanconcludethatasetofparavectorstogetherwithoperations
ofsummationandmultiplicationgivesaringwithmultiplicativeidentity.
Definition1.9. WecalltheparavectorΓproperifdetΓ R 0 (thedeterminantisapositiverealnumber).
∈ +\{ }
Definition1.10. WecalltheparavectorΓsingularifdetΓ=0.
Bydefinitionofthedeterminantitfollows:
Conclusion1.8. Eachproperorsingularparavector(1)mustfulfillthefollowingcondition:
ad =bc (11)
Itisadvisabletoremembertheaboveequation,becauseitwillbeusedwithmanyproofsonproperparavectors.
Conclusion1.9. LetΓ ,Γ beparavectors,thenthefollowingstatementsaretrue:
1 2
det(Γ Γ )=detΓ detΓ
• 1 2 1· 2
det(Γ )=detΓ
• −
det(Γ )=(detΓ)
• ∗ ∗
Definition1.11. Foreachnon-singularparavectorΓ,thereexistsaninverseelementundermultiplication:
Γ
Γ−1=: − (12)
detΓ
Conclusion1.10. Asetofnon-singularparavectorstogetherwithmultiplicationisanon-commutativegroup.
Conclusion1.11. Asetofproperparavectorstogetherwithmultiplicationisanon-commutativegroup.
Definition1.12. Foreachproperand/orsingularparavectorwedefinethemoduleofparavector:
Γ =:pdetΓ (13)
| |
Conclusion1.12. Themoduleofparavector(properorsingular!)satisfiesthefollowingconditions:
3
1. sΓ = s Γ wheres R
| | | || | ∈
2. Γ Γ = Γ Γ
| 1|| 2| | 1 2|
Definition1.13. WecalltheparavectorΛorthogonalifdetΛ=1,orequivalently:
Γ
Λ=: , whereΓisaproperparavector (14)
pdetΓ
Directlyfromtheabovedefinitionitfollows:
Conclusion1.13. IfΛisanorthogonalparavector,thenΛ 1=Λ .
− −
Definition1.14. WecalltheparavectorΓspecialifΓ =Γ ,orequivalently:
− ∗
a
, wherea R,andc R3
  ∈ ∈
ic
 
Definition1.15. WecalltheparavectorΓunitarifΓΓ =1
∗
Tosummarizethecurrentknowledgeaboutparavectors,wecansaythatverylittleismissingsothataset
of paravectors with summation and multiplication operations is a field: multiplication of paravectors is not
commutative,andtheroleofthenullelementundermultiplicationisplayedbysingularparavectors.Multiplying
anyparavector bysingular one, wegetasingular paravector. Notethatalthoughthere aremanynull elements
undermultiplication,onlyoneelementisneutralwithrespecttosummation.
Conclusion 1.14. The set of special paravectors together with operations summation and multiplication is
adivisionring.
Inthediagram(fig.1)aringofparavectorswithsomeofitssubstructuresispresented:
Figure1:Theringofparavectorswithsomesubstructures
-Asetofproperparavectorstogetherwithmultiplicationisanoncommutativegroup.
4
-Asetoforthogonalparavectorstogetherwithmultiplicationisasubgroupofproperparavectorsgroup.
Theblackpointmeanszero,andthebluepointisone.
Toavoidmisunderstandings,werecallthefollowingnameswhichwewilluse:
Γ -TheparavectorreverseofΓ
• −
Γ 1-TheparavectorinverseofΓ
• −
2 Integratedproductofparavectors
Synge[7]definesthescalarproductofcomplexquaternionsas€Γ1Γ−2 +Γ2Γ−1Š/2,Hestenes[6]asascalar
partoftheproductofmultivectorsΓ1Γ2.AnalyzingtheexpressionΓ1Γ−2,wecanseethatitplaysasimilarbutmore
universalroleintheparavectorsalgebraasthedotproductinvectorspace.Propertiesofitsscalarcomponentare
thesameaspropertiesofthescalarproductofvectors,anditsvectorpartasacrossproductofvectors.Thetrouble
isthattherecanbetwodifferentproductswhichhavethesameproperties(thesecondoneisΓ−1Γ2),sowedefine
twointegratedproducts:
Definition2.1. For anyparavectorsΓ and Γ , the rightintegratedproductofparavectors Γ and Γ , denoted
1 2 1 2
(Γ ,Γ ,isdefinedby
1 2〉
(Γ1,Γ2〉=:Γ1Γ−2
α  α   α α βββ βββ 
hence (Γ1,Γ2〉=Γ1Γ−2 =βββ1 βββ2 = α βββ +1α2βββ− 1iβββ2 βββ 
 1− 2 − 1 2 2 1− 1× 2
Definition2.2. ForanyparavectorsΓ andΓ theleftintegratedproductofparavectorsΓ andΓ ,denoted Γ ,Γ ),
1 2 1 2 〈 1 2
isdefinedby
〈Γ1,Γ2)=:Γ−1Γ2
 α α   α α βββ βββ 
or 〈Γ1,Γ2)=Γ−1Γ2= 1  2= 1 2− 1 2 
βββ βββ α βββ α βββ iβββ βββ
− 1 2  1 2− 2 1− 1× 2
Inbothcases,thescalarpartisthesame,then
Definition2.3. ThescalarcomponentofanintegratedproductofparavectorsΓ andΓ ,iscalledascalarproduct
1 2
ofparavectors.Thescalarproductwillbedenoted
Γ ,Γ =:(Γ ,Γ = Γ ,Γ ) .
〈 1 2〉 1 2〉S 〈 1 2 S
Definition2.4. Thevectorcomponentofanintegratedproductwecallavectorproductofparavectors.
(Γ1,Γ2}=:(Γ1,Γ2〉V =−〈Γ1,Γ2)∗V
Itisnecessarytodistinguishtheorientationofavectorproduct,thesameasatintegratedproductcase.
Therefore,wedenotetherightvectorproduct(Γ ,Γ ,andtheleftvectorproduct: Γ ,Γ )
1 2} { 1 2
Forfurtherconsiderationitdoesnotmattermuchifaproductisrightorleft,sotalkingaboutanintegrated
productwewillmeantherightproduct. Aswell,thismaybetheleftproduct,butitisimportanttouseonlyone
productconstantly.Hencetheintegratedproductcanbedenotedasfollow:
 Γ ,Γ  (Γ ,Γ 
(Γ ,Γ = 〈 1 2〉 = 1 2〉S (15)
1 2〉 (Γ ,Γ  (Γ ,Γ 
 1 2}  1 2〉V
and
det(Γ ,Γ = Γ ,Γ 2 (Γ ,Γ 2=(Γ ,Γ 2 (Γ ,Γ 2 =detΓ detΓ (16)
1 2〉 〈 1 2〉 − 1 2} 1 2〉S− 1 2〉V 1 2
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Theorem2.1. Anintegratedproductofparavectorshasthefollowingproperties:
Rightintegratedproduct Leftintegratedproduct
1 Integratedproductisaparavector
2 (Γ +Γ ,Γ =(Γ ,Γ +(Γ ,Γ Γ +Γ ,Γ )= Γ ,Γ )+ Γ ,Γ )
1 2 3〉 1 3〉 2 3〉 〈 1 2 3 〈 1 3 〈 2 3
3 (αΓ ,Γ =α(Γ ,Γ =(Γ ,αΓ αΓ ,Γ )=α Γ ,Γ )= Γ ,αΓ )
1 2〉 1 2〉 1 2〉 〈 1 2 〈 1 2 〈 1 2
4 (Γ1,Γ2〉−=(Γ2,Γ1〉 〈Γ1,Γ2)−=〈Γ2,Γ1)
5 (Γ,Γ = Γ,Γ)=detΓ C
〉 〈 ∈
6 det(Γ ,Γ =det Γ ,Γ )=detΓ detΓ
1 2〉 〈 1 2 1 2
Conclusion2.1. LetΓ ,Γ andΓ beanyparavectors,thenthefollowingtableshowsthepropertiesofthescalar
1 2 3
productofparavectorscomparedtothepropertiesofthescalarproductofvectorsinEuclideanspace:
Scalarproductofparavectors Scalarproductofvectors
1 Γ ,Γ C x ,x R
〈 1 2〉∈ 〈 1 2〉∈
2 Γ +Γ ,Γ = Γ ,Γ + Γ ,Γ same
〈 1 2 3〉 〈 1 3〉 〈 2 3〉
3 αΓ ,Γ =α Γ ,Γ = Γ ,αΓ same
〈 1 2〉 〈 1 2〉 〈 1 2〉
4 Γ ,Γ = Γ ,Γ same
〈 1 2〉 〈 2 1〉
5 Γ,Γ C x,x R
〈 〉∈ 〈 〉∈ +
6 If Γ,Γ =0,thenΓissingular If x,x =0,thenx=0
〈 〉 〈 〉
Rows5and6ofTable2.1showafundamentaldifferencebetweenthedefinitionofthescalarproductof
paravectorsandknownalgebraicdefinitions. Thesepropertiesmakemewonderwhethertogiveanothername
forscalarproduct.However,thetraditionalnameisleft,because:
1. thevalueoftheproductisascalar,
2. therolewhichthescalarproductplaysintheParavectorAlgebracorrespondscompletelytotheroleofthe
dotproductofvectorsinEuclideanGeometry,
3. forspatialparavectors(Γ =0),thescalarproductofparavectorsbecomesthescalarproductofvectors.
S
3 Geometricalpropertiesofparavectors
Bycheckingupanintegratedproductofparavectors,anyonecanseeadeepsimilaritybetweenparavectors
andvectorsinEuclideanspace.Usingtheintegratedproductweintroducegeometricconceptsintothealgebraof
paravectors,whichcontributestoitsintuitiveunderstanding.
3.1 Parallelismandperpendicularityrelations.
Definition3.1. Non-singularparavectorsΓ andΓ areparallel(Γ Γ )ifthevectorproduct(Γ ,Γ =0.
1 2 1k 2 1 2}
Theorem3.1. Twonon-singularparavectorsΓ andΓ areparallelifandonlyifthereexistsanumberλ=0that
1 2 6
Γ =λΓ .
1 2
6
Proof.
ParallelismmeansthatΓ1Γ−2 =α,whereαisacomplexnumber. Wemultiplythisequationontherightby
paravectorΓ
2
Γ1(Γ−2Γ2)=αΓ2
SinceΓ2isnon-singularthenΓ−2Γ2=β isanon-zerocomplexnumber.HencewegetΓ1=λΓ2,whereλ=α/β.
Theabovetheoremshowsthat
Conclusion3.1. Theparallelismsatisfiestheconditionsofequivalencerelation.
Definition3.2. Twonon-singularparavectorsΓ andΓ areperpendicular(Γ Γ )ifthescalarproduct Γ ,Γ =
1 2 1⊥ 2 〈 1 2〉
0.
TheperpendicularityofparavectorshasthesamepropertiesastheperpendicularityofvectorsinEuclidean
space.
Theorem3.2. Foreachnon-singularparavector:
1. ∽(Γ Γ)
⊥
2. IfΓ Γ thenΓ Γ
1⊥ 2 2⊥ 1
3. IfΓ Γ andΓ Γ thenΓ Γ
1⊥ 2 2k 3 1⊥ 3
Proof.
1. Itfollowsbydefinitionofperpendicularity.
2. Itfollowsbythefactthatthescalarproductissymmetrical: Γ ,Γ = Γ ,Γ and Γ ,Γ =0
〈 1 2〉 〈 2 1〉 〈 1 2〉
0
3. Γ Γ (Γ ,Γ =
1⊥ 2 ⇐⇒ 1 2〉  
ωωω
 
 0 
Γ2kΓ3 ⇐⇒ (Γ2,Γ3〉=λhence (Γ1,Γ3〉=Γ1Γ−3 =Γ1Γ−21Γ2Γ−3 =deλtΓ2Γ1Γ−2 = λωωω 
detΓ2
Conclusion3.2. .
1. Theparavectorisperpendiculartoitselfifandonlyifitissingular.
2. Paravectorsmutuallyreversed(inversed)nottobeparallel.
3. Orthogonalparavectorsareparallelifandonlyiftheyareequaloropposite.
Proof.
1. Itfollowsbyassumptionanddefinitionofthesingularparavector.
αα α2+β2
2. (Γ,Γ =ΓΓ= =
−〉     
βββ βββ 2αβββ
    
7
3. ParallelismmeansthatΛ1Λ−2 =λ,hencedetΛ1detΛ2 =λ2. Paravectorsareorthogonalhenceλ=±1. First
equalitygivesthatΛ =Λ orΛ = Λ .
1 2 1 − 2
Theorem3.3. Foranynon-singularparavectorsΓ andΓ itoccursthat:
1 2
1. IfΓ1⊥Γ2thenΓ∗1⊥Γ∗2
2. IfΓ1kΓ2thenΓ∗1kΓ∗2
3. IfΓ Γ thenvigΓ vigΓ
1k 2 1k 2
Proof.
1. 〈Γ1,Γ2〉=0 =⇒ 〈Γ2,Γ1〉=0,so¬Γ∗1,Γ∗2¶=〈Γ2,Γ1〉∗=0
α  α 
1 2
2. ( ) = α βββ +α βββ iβββ βββ =0
   V − 1 2 2 1− 1× 2
βββ βββ
 1− 2
Sincecomplexvectorsaregovernedbythesamelawsasrealones,soitmustbe:
α βββ α βββ =000 and iβββ βββ =0
2 1− 1 2 1× 2
 α α 
2 1
Ontheotherhand,wehave€Γ∗1,Γ∗2¶=〈Γ2,Γ1)∗=(  )∗,
βββ βββ
− 2 1
henceundertheassumption
 α α 
2 1
( ) =α βββ α βββ iβββ βββ =0
   V 2 1− 1 2− 2× 1
βββ βββ
− 2 1
3. €Γ1Γ∗1,Γ2Γ∗2¶=Γ1Γ∗1(Γ2Γ∗2)−=Γ1Γ∗1Γ∗2−Γ−2 =Γ1(Γ−2Γ1)∗Γ−2
Undertheassumption,theproductinparenthesesisanumber,sowecanmoveitinfrontoftheproduct
λ (Γ ,Γ =λ λ
∗ 1 2〉 ∗
Theorem3.4. ForanyparavectorsΓ andΓ polarizationidentityoccurs:
1 2
det(Γ +Γ )=detΓ +2 Γ ,Γ +detΓ (17)
1 2 1 〈 1 2〉 2
Proof.
det(Γ1+Γ2)=(Γ1+Γ2)(Γ1+Γ2)−=Γ1Γ−1 +Γ1Γ−2 +Γ2Γ−1 +Γ2Γ−2 =
=detΓ +(Γ ,Γ +(Γ ,Γ +detΓ =detΓ +2 Γ ,Γ +detΓ
1 1 2〉 2 1〉 2 1 〈 1 2〉 2
Conclusion3.3. LetparavectorsΓ andΓ beperpendicular,thenthedeterminantoftheseparavectorsequalsthe
1 2
sumoftheirdeterminants:
det(Γ +Γ )=detΓ +detΓ (18)
1 2 1 2
ThisconclusioncomplieswiththePythagoreantheoreminEuclideangeometry.
Theorem3.4showsthatparavectorsmeettheparallelogramlaw,whichgivesthestructureconstructedby
ussomeofthecharacteristicsofmetricspace.
8
Conclusion3.4. ForanyparavectorsΓ andΓ theparallelogramlawoccurs:
1 2
det(Γ +Γ )+det(Γ Γ )=2detΓ +2detΓ
1 2 1− 2 1 2
α  α 
1 2
Definition3.3. Twoparavectors and arespatiallyparallelifβββ βββ =0
    1× 2
βββ βββ
 1  2
Theorem3.5. Iftwonon-singularparavectorsareparallel,theyarealsospatiallyparallel.
Proof.
Γ Γ α βββ α βββ iβββ βββ =0
1k 2 ⇐⇒ 1 2− 2 1− 1× 2
Firstwemultiplytheaboveequationbyα βββ ,andthenbyα βββ .Hencewegettwoequations:
1 2 2 1
α βββ 2 α α βββ βββ =0
(cid:0) 1 2(cid:1) − 1 2 1 2
α α βββ βββ α βββ 2=0
1 2 1 2−(cid:0) 2 1(cid:1)
Thedifferenceoftheaboveequationsyields α βββ α βββ 2=0,whichistruewhenα βββ =α βββ .
(cid:0) 1 2− 2 1(cid:1) 1 2 2 1
Henceitfollowsthatβββ βββ =0
1× 2
Asaconsequence,wecansaythat
Conclusion3.5. Spatialparallelismisanequivalencerelation.
Conclusion 3.6. The spatial parallelism of paravectors is weaker than parallelism, ie.: If two paravectors are
parallel,theymustbespatiallyparallel,too,butintheoppositedirection,implicationnolongermustoccur.
Definition3.4. WecalltwosingularparavectorsΓ andΓ singularlyparallelif(Γ ,Γ =0
1 2 1 2〉
Conclusion3.7. Thefollowingconclusionsareeasytoprove:
1. Iftwoparavectorsaresingularlyparallel,thentheyaresingular.
2. Singularparallelismisanequivalencerelation.
3.2 Angles
Definition3.5. Therightanglebetweentwoproperparavectors,denotedby∠(Γ ,Γ ,wecalltheparavector:
1 2〉
cosiΦ (Γ ,Γ
Φ=∠(Γ ,Γ = =: 1 2〉, (19)
1 2〉 dexΦ |Γ1||Γ2|
wherewecallthescalarcomponentcosinis,andthevectorcomponent–dextisofangleΦ.
Definition3.6. Theleftanglebetweentwoproperparavectorswecalltheparavector:
cosiΦ Γ ,Γ )
Φ=∠ Γ ,Γ )= =: 〈 1 2 , (20)
〈 1 2 siniΦ |Γ1||Γ2|
wherewecallthescalarcomponentcosinis,andthevectorcomponent–sinisofangleΦ.
9
Note-Pleasenotethatthecomponentsarenottrigonometric(hyperbolic)functions-thesearejustnames,
givenbecauseanglecomponentshavethesamepropertiesaswell-knowntrigonometricfunctions,makingthem
easiertoimagineandtoremember.
ThesenamesarederivedfromLatin.Sinistrammeansleft,anddextrammeansright.
Inparavectorspace(whichisnotdefinedyet!) aswellasinEuclideanspace,weshouldidentifyapositive
orientation.Wedon’tdothisinthispaper,nottoimposeanyrestrictions.Itseemsthatitwillbenecessarysooner
orlater,butnotnow.
Definition3.7. WecallanangleΦ=Φ Φ thecompositionof(left)anglesΦ andΦ .
1 2 1 2
cosiΦ cosiΦ 
1 2
Φ Φ = =
1 2
siniΦ siniΦ 
 1 2
 cosiΦ cosiΦ +siniΦ siniΦ 
1 2 1 2
= =
cosiΦ siniΦ +cosiΦ siniΦ +isiniΦ siniΦ 
 1 2 2 1 1× 2
cosi(Φ Φ ) cosiΦ
1 2
= = =Φ
sini(Φ Φ ) siniΦ
 1 2   
Wecanwriteananalogouscompositionfortherightangles.
Asaconsequence,wecanseefurtheranalogywithEuclideantrigonometry:
Conclusion3.8. Theexplementofanangleis:
∠〈Γ1,Γ2)−=∠〈Γ2,Γ1) fortheleftangle
or ∠(Γ1,Γ2〉−=∠(Γ2,Γ1〉 fortherightangle, (21)
whichgives
cosiΦ =cosiΦ
• −
siniΦ = siniΦ
• − −
dexΦ = dexΦ
• − −
Theleftandrightangleshaveanoppositeorientationinspace-theyarenotexplementary.
∠ Φ,Γ)= Φ−Γ, ∠(Φ,Γ = ΦΓ−, hencethecompositionofthisanglesis:
〈 Φ Γ 〉 Φ Γ
| || | | || |
Φ ΓΦΓ
∠ Φ,Γ)∠(Φ,Γ = − − =1
〈 〉 det(ΦΓ) 6
Theexplementoftheleftangle∠ Φ,Γ)istheleftangle∠ Γ,Φ),andthesameoccursfortherightangle.
〈 〈
InTable3.1areshownthepropertiesofleftanglecomponentstosimplyandintuitivelyjustifythenames
chosenforthem.Therightanglehasanalogousformulas.
Tab.3.1 Generalrecurrenceformulasforcomponentsoftheleftangle.
10