Fundamental and contravariant representations

  • Thread starter gentsagree
  • Start date
  • #1
96
1
The invariant of SL(2,C) is proven to be invariant under the action of the group by the following

[itex]\epsilon'_{\alpha\beta} = N_{\alpha}^{\rho}N_{\beta}^{\sigma}\epsilon_{\rho\sigma}=\epsilon_{\alpha\beta}detN=\epsilon_{\alpha\beta}[/itex]

The existence of an invariant of this form (with two indices down) tells us that the covariant (here fundamental of SL(2,C)) and contra variant reps aren't independent. This seems to me to follow directly from the fact that detN=1. Is this true? And can anyone bring forward an example where covariant and contra variant reps are truly independent?
 

Answers and Replies

  • #2
samalkhaiat
Science Advisor
Insights Author
1,721
1,002
The invariant of SL(2,C) is proven to be invariant under the action of the group by the following

[itex]\epsilon'_{\alpha\beta} = N_{\alpha}^{\rho}N_{\beta}^{\sigma}\epsilon_{\rho\sigma}=\epsilon_{\alpha\beta}detN=\epsilon_{\alpha\beta}[/itex]

The existence of an invariant of this form (with two indices down) tells us that the covariant (here fundamental of SL(2,C)) and contra variant reps aren't independent. This seems to me to follow directly from the fact that detN=1. Is this true?
Yes, covariant and contravariant spinor representations are EQUIVALENT, i.e. they belong to the same representation space of SL(2,C).
And can anyone bring forward an example where covariant and contra variant reps are truly independent?
In SL(2,C), lowering or raising the indices by the spinor metric [itex]\epsilon_{ \alpha \beta }[/itex] and [itex]\epsilon^{ \alpha \beta }[/itex] does not produce new representation. However, complex conjugation does lead to new inequivalent representation [itex]\psi^{ \dot{ \alpha } } = \epsilon^{ \dot{ \alpha } \dot{ \beta } } \psi_{ \dot{ \beta } }[/itex].
 
  • #3
96
1
Why do you reply to posts if you don't read the questions properly first?

1. I know they are EQUIVALENT. I am asking whether this is a direct consequence of the fact that detN=1, i.e. if I could generalise this, for instance, by saying that "every algebra with detN=1 possesses a 2nd rank invariant tensor", or perhaps not, I am missing something.

2. Again, here I ask whether anybody can bring forward an example where covariant and contravariant representations are truly independent of each other, not whether the conjugate representation is.
 
  • #4
samalkhaiat
Science Advisor
Insights Author
1,721
1,002
Why do you reply to posts if you don't read the questions properly first?

1. I know they are EQUIVALENT. I am asking whether this is a direct consequence of the fact that detN=1, i.e. if I could generalise this, for instance, by saying that "every algebra with detN=1 possesses a 2nd rank invariant tensor", or perhaps not, I am missing something.

2. Again, here I ask whether anybody can bring forward an example where covariant and contravariant representations are truly independent of each other, not whether the conjugate representation is.
Instead of asking MEANINGLESS questions and being RUDE to the person who is trying to help you, you should spend your time learning the ABC of group theory.
 
  • Like
Likes ChrisVer

Related Threads on Fundamental and contravariant representations

Replies
5
Views
11K
  • Last Post
Replies
4
Views
3K
  • Last Post
Replies
11
Views
4K
  • Last Post
Replies
2
Views
4K
Replies
11
Views
4K
Replies
6
Views
591
Replies
23
Views
7K
Replies
1
Views
5K
Replies
2
Views
813
Replies
13
Views
939
Top