Tangent space at the identity - S. N. Bose National Centre ...

1 C Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 2011 Chapter 21 Tangent space at the identityA point on the Lie group is a group element. So a vector field onthe Lie group selects a vector at eachg G .Since left and righttranslations are diffeomorphisms, we can consider the pushforwardsdue to them. Aleft-invariant vector fieldXis invariant under left ttrans-lations, ,X=lg (X) g G .( )In other words, the vector (field) atg is pushed forward bylgto thesame vector (field) atlg(g ):lg (Xg ) =Xgg g, g G .( ) Similarly, aright-invariant vector fieldXis defined byX=rg (X) g G , rg (Xg ) =Xg g g, g G .( )A left or right invarian vector field has the important propertythat it is completely determined by its value at the identity elementeof the Lie group, sincelg (Xe) =Xg g G ,( )and similarly for right-invariant vector the set of all left-invariant vector fields onGasL(G).Sincethe push-forward is linear, we getlg (aX+Y) =alg X+lg Y ,( )81c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 201182 Chapter 21.

2 Tangent space at the identityso that if bothXandYare left-invariant,lg (aX+Y) =aX+Y ,( )so the set of left-invariant vector fields form a real vector also know that push-forwards leave the Lie algebra invariant, , forlg ,[lg X , lg Y] =lg [X , Y].( )Thus ifX, Y L(G),lg [X , Y] = [lg X , lg Y] = [X , Y],( )so [X , Y] L(G).Thus the set of all left-invariant vector fields onGforms a Lie algebra. ThisL(G) is called theLie algebra ofG .2 The dimension of this Lie algebra is the same as that ofGbecauseof theTheorem:L(G) as a real vector space is isomorphic to the tan-gent spaceTeGtoGat the identity ofG .Proof:We will show that left translation leads to an TeG ,define the vector fieldLXonGbyLX g LXg:=lg X g G( )Then for allg, g G ,l g (LXg) =lg (lg X) =lg g X=LXg g.( )Note that for two diffeomorphisms 1, 2,we can write( 1 ( 2 v))(f) = ( 2 v)(f 1)=v(f 1 2)= (( 1 2) v)(f) 1 ( 2 v) = ( 1 2) v( )Since left translation is a diffeomorphism,lg (lg X) = (lg lg) X= (lg g )X( )c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 201183So it follows thatLXis a left-invariant vector field, and we have amapTeG L(G).

3 Since the pushforward is a linear map, so is themapX need to prove that this map is 1-1 and ,we haveLXg=LYg g G ,( )solg 1 LXg=lg 1 LYg X=Y( TeG).( )So the mapX LXis givenLX,defineXe TeGbyXe=lg 1 LXgfor anyg G .( )We can also writeXe=LXe.( )Thenlg Xe=lg lg 1 LXg=LXg.( )So the mapX7 LXis we can define a Lie bracket onTeGby[u , v] = [Lu, Lv] e.( )The Lie algebra of vectors inTeGbased on this bracket is thus theLie algebra of the groupG .It follows thatdimL(G) = dimTeG= dimG .( )Note that since commutators are defined for vector fields and notvectors, the Lie bracket onTeGhas to be defined using the com-mutator of left-invariant vector fields onGand the isomorphismTeG L(G). If for ann-dimensional Lie groupG ,{t1, , tn}is a set of basisvectors onTeG'L(G),the Lie bracket of any pair of these vectorsmust be a linear combination of them, so[ti, tj] = kCkijtk( )c Amitabha Lahiri: Lecture Notes on Differential Geometry for Physicists 201184 Chapter 21.

4 Tangent space at the identityfor some set of real numbers are known as thestructure constantsof the Lie group or (G) is a Lie algebra, with the Lie bracket as the product,the Lie bracket is antisymmetric,[ti, tj] = [tj, ti] kCkijtk= kCkjitk Ckij=Ckji,( )and the structure constants satisfy the Jacobi identity [ti,[tj, tk]] + [tj,[tk, ti]] + [tk,[ti, tj]] = 0 ClijCmkl+CljkCmil+ClkiCmjl= 0.( )A similar construction can be done using a set of right-invariantvector fields defined byRXg:=rg XforX TeG( )and its inverse Xe=rg 1 RXg.