Groups Fields Vector paces Homework #3 (2014-2015 Answers Q1: Tensor products: concrete examples Let V W be two-dimensional vector spaces with bases { 1 2} v v { } w w o { vi wj} 1 2 Ä is a basis for V Ä W ay xi Î V has the basis expansion x = a1v1+ a2v2 yi Î W has the basis expansion y = b1w1+ b2w2 A Exp x y vi Ä wj x Ä y = ( a1v1+ a2v2 Ä ( b1w1+ b2w2 = ( a1v1+ a2v2 Ä ( b1w1 + ( a1v1+ a2v2 Ä( b2w2 = av Ä b w + a v Ä b w + av Ä b w + a v Ä b w = ab v Ä w + ab v Ä w + ab v Ä w + ab v Äw Ä in the basis { } ( 1 1 ( 1 1 ( 2 2 ( 1 1 ( 1 1 ( 2 2 ( 2 2 ( 2 2 ( ( ( ( 1 1 1 1 2 1 2 1 1 2 1 2 2 2 2 2 B Now say V = W we are using the same basis for x y so that x = a1v1+ a2v2 y = bv + b v Exp x y 1 1 2 2 Ä in the basis { vi Ä Taking wi = vi in part A x Ä y= ab v Ä v + ab v Ä v + ab v Ä v + ab v Ä v ( ( ( ( 1 1 1 1 2 1 2 1 1 2 1 2 2 2 2 2 C Exp x y y x Ä + Ä in the basis { vi Ä ( xä y + ( yäx 1 1( v1ä v1 + ab 2 1( v2ä v1 + ab 1 2( v1ä v2 + ab 2 2( v2äv2 + ( ba 1 1( v1ä v1 + ba 2 1( v2ä v1 + ba 1 2( v1ä v2 + ba 2 2( v2äv2 = 2ab ( v Ä v + ( ab + ba( v Ä v + ( ab + ba ( v Ä v + 2ab ( v Äv 1 1 1 1 2 1 2 1 2 1 1 2 1 2 1 2 2 2 2 2 D Exp x y y x Ä - Ä in the basis { vi Ä ( xäy -( yäx 1 1( v1ä v1 + ab 2 1( v2ä v1 + ab 1 2( v1ä v2 + ab 2 2( v2äv2 -( ba 1 1( v1ä v1 + ba 2 1( v2ä v1 + ba 1 2( v1ä v2 + ba 2 2( v2äv2 2 1-ba 2 1( v2ä v1 + ( ab 1 2-ba 1 2( v1äv2 -ba (( v Äv -( v Äv 2 1 2 1 2 1 1 2
Q2: Free vector spaces direct sums tensor products Let V be the free vector space on a set namely the set of all functions v on with addition defined pointwise ( v1+ v2( s = v1( s + v2( s scalar multiplication defined by ( av( s = a ( v( s imilarly let W be the free vector space on a set T namely the set of all functions w on T with addition multiplication defined in an analogous fashion A how that the direct-sum vector space V Å W is the same as (ie canonically isomorphic to the free vector space on È T the union of the sets T That is construct an isomorphism between the two spaces without resorting to choosing a basis This is mostly an exercise in keeping straight what operates on what Elements in V Å W by definition consist of pairs of elements in V W added element-by-element while elements in the free vector space on È T consist of all functions on È T We construct a homomorphism F from V Å W to the free vector space on È T a homomorphism Q from the free vector space on È T back to V Å W show that they are inverses ay z= ( v w is in V Å W To find its corresponding element in the free vector space on È T we need to re-interpret it as a function F ( z in T F ( z ( u for every u in È T The natural choice is È that is assign a value to ( ìvu ( if uî ( F ( z ( u =í We ll skip some of the details of showing that F preserves vector-space î wu ( if uît structure but for example here are the details for showing F ( z1+ z2 =F ( z1 +F ( z2 : zi = ( vi wi then z1+ z2 = ( v1+ v2 w1+ w2 ì v1( u v2( u if u ( ( z1 z2 ( u + Î ìvi ( u if uî F + =í while ( F ( z i ( u =í so î w1( u + w2( u if uît î wi ( u if u Î T ( F ( z1+ z2 ( u = ( F ( z1 ( u + ( F ( z2 ( u for all u hence F ( z1+ z2 =F ( z1 +F ( z2 Conversely say x is an element of the free vector space on È T We need to find a Q ( x that maps x into an ordered pair of elements ( vw with v in the free vector space on w in the free vector space on T o we take Q ( x = ( vw where vs ( = xs ( similarly wt ( = xt ( noting that for x( u is defined for all uîè T so it is defined both on T
To show that Q is a homomorphism we need to show Q ( x1+ x2 =Q ( x1 +Q ( x2 Q ( lx1 = lq ( x1 We show the first in detail Q ( x1+ x2 is an ordered pair ( vw where v is defined by vs ( =Q ( x1+ x2( s =Q ( x1( s+q ( x2( s wt ( =Q ( x1+ x2( t =Q ( x1( t +Q ( x2( t On the other h each Q ( x i is an ordered pair ( vi w i with vi( s =Q ( xi( s wi( t =Q ( xi( t o vs ( = v1( s+ v2( s for all sî wt ( = w1( t + w2( t for all tî T o ( vw = ( v1 w1 + ( v2 w2 Q ( x + x = ( v w = ( v w + ( v w =Q ( x +Q ( x 1 2 1 1 2 2 1 2 Finally we need to show that the above two constructions are inverses of each other namely that QF ( z = z for z= ( v w in V W FQ ( x = x for x in free vector space on È T ( Å that ( QF According to the definition of Q this is the element ( vw of V Consider first ( ( z which vs ( = ( F (( z s for sî wt ( =F ( ( z( t for tî T But ( ( F ( z ( t = w( t according to the definition of F o ( ( z z (unilluminating Å W for F ( z ( s = v( s QF = The other way around is equally B As free vector spaces recall that V has the delta-function basis consisting of the vectors d defined by d ( s 1 = for s = s 0 otherwise W has the analogous delta-function basis consisting of the vectors d defined by d ( t 1 = for t = t 0 otherwise Display the delta-function basis for V Å W These are the delta-functions on È T namely d ( u u defined by d ( u 1 u = for u = u 0 otherwise Equivalently they are extensions of d d to È T giving them a value of 0 beyond the set on which they were originally defined C (optional how that the tensor-product vector space V Ä W is the same as (ie canonically isomorphic to the free vector space on T ie the set of all ordered pairs ( st of elements sî tî T That is construct an isomorphism between the two spaces without resorting to choosing a basis As in A we construct a homomorphism F from V Ä W to the free vector space on T a homomorphism Q from the free vector space on T back to V Ä W show that they are inverses First we construct the mapping F for the elementary tensor products vä w ie we construct F( vä w show that it obeys the tensor-product rules o we have to specify the value of vä w on a typical element ( st in T We define ( F( vä w ( s t = v( s w( t where the multiplication on the right is in the base field We have to show consistency with the tensor-product rules namely that F( lvä w =F( vä lw F (( v1+ v2 Ä w =F( v1ä w +F( v2ä w We do this by evaluating both
sides on elements ( st Î T For the first ( ( ( ( ( ( F( lvä w( st = lvs ( wt ( = l vs ( wt ( = vs ( lwt ( =F( vä lw( st For the second F (( v1+ v2 Ä w( s t = (( v1+ v2( s ( w( t = (( v1( s + v2( s ( w( t = ( v1( s( wt ( + ( v2( s( wt ( =F( v1ä w( st +F( v2äw( st where we have used the definition of F in the first equality then the definition of addition in the free vector space on then the distributive law in the base field in the final step the definition of F again Note that showing consistency with tensor-product rules also shows that F is a homomorphism Next we construct a homomorphism Q from the free vector space on T back to V Ä W Motivated by the way F is defined we work on elements in the free vector space on T that are the images of elementary tensor products These are the separable elements of the free vector space on T namely the functions x( ( st for which x( ( st = vswt ( ( For these functions we define Q ( x = vä w But we need to check that this is a self-consistent definition: since æ x( st ( vs (( wt ( ( vs ( 1 ö = = a ç wt ( çèa we need to be sure that applying Q to the second factorization ø æ1 ö yields the same result as the first The second yields ( av Ä ç w çèa this is guaranteed equal to ø æ1 ö 1 a vä w by the rules for the tensor product in V Ä W : ( av Ä ç w = (( av Ä w = ( vä w = väw a çè ø a a Now we need to extend this definition to the entire free vector space on T not just the separable elements We note that the separable elements contain a basis the delta-function basis functions dv w( st = dv( sdw( t Each element of the entire free vector space on T can be written uniquely as x( ( st = å ( xs ( t d ( st ie x( s t is the coefficient of the basis element d in the s t Q ( x = å x( s t d d s Ä t expansion of x o Q extends to the entire free vector space by ( Finally note that we didn t need to introduce the basis to define the homomorphism Q but we did need it to show that it extended to the whole free vector space on T o we need to check that this extended homomorphism coincides with our original definition Q ( x = vä w for x( ( st = vswt ( ( x ( st = vswt ( ( then its coefficients in the basis representation also factor: This holds because if ( x( s t = v( s w( t o Q ( x = x( s t d Ä d = v( s w( t d Äd å å ( ( s t s t s t s t æ ö æ ö = ( vs ( d ( wt ( d vs ( d wt ( d å Ä = Ä = väw çå è ø çå è ø s t s t s t s t s t
where we ve used the rules for the tensor product for the third equality we separated the sum on the fourth equality we used the delta-basis representation for v w for the final equality F is the function on T for To show that QF ( ( z = z we take z= vä w in V Ä W Then ( z which F ( z (( s t = v( s w( t (by the definition of F ince this is separable ( ( the definition of Q as required imilarly to see that FQ ( ( x = x take an x for which ( the definition of Q ( ( x the definition of F FQ is the function on T QF z = vä w (from x ( st = vswt ( ( Then Q ( x = vä w (from FQ ( x ( st = vswt ( ( (by for which ( (