Page 415 - Advanced Linear Algebra
P. 415
Tensor Products 399
manner that holds regardless of the characteristic of the base field. Namely,
rather than restricting the domain of in order to get an isomorphism, we can
factor out by the kernel of .
Consider a tensor
p s
#~ : ²#³ ~ 4 ! !
4 4 q !. ²#³ t
4
Since sends elements of different groups .²#³ ~ ¸! Á Ã Á ! ¹ to different
4
c
monomials in - ´ Á Ã Á µ or - ´ Á Ã Á µ , it follows that # ker ² ³ if and
only if ²: ²#³³ ~ for all 4 , that is, if and only if
4
! !b Ä ! b ! ~
In the symmetric case, is constant on .²#³ and so # ker ² ³ if and only if
4
bÄb ! ~
!
In the antisymmetric case, Á !~ ²c ³ where ! Á ²! ³ ~ ! and so
# ker ² ³ if and only if
²c ³ bÄb²c ³ Á Á ! ~
!
In both cases, we solve for ! and substitute into 4 :²#³ . In the symmetric case,
~c ! ! c Ä c !
and so
: ² # 4 ³ ~ ! b ! Ä b ! ~ ! ² ! ! c ! ³ b Ä b ² ! ! c ! ³
In the antisymmetric case,
~ c²c ³ Á ! Á cÄc²c ³ !
!
and so
:²#³ ~ 4 ! ! b Ä b ! !
~ ! ²²c ³ Á ! c ! ³ b Ä b ! ²²c ³ Á ! c ! ³
Since ! 8 , it follows that : ²#³ and therefore , is in the span of tensors of
#
4
the form in the symmetric case and ²c ³ ²!³ c ! ²!³ c ! in the
and ! . 8
antisymmetric case, where :
Hence, in the symmetric case,
ker² ³ 0 º ²!³ c !! Á 8 : »
and since ²!³ c !³~ , it follows that ker ² ³~0 . In the antisymmetric
²
case,
