Why Must the Parameter p be an Integer in U(1) Representations?

  • Thread starter Thread starter Pietjuh
  • Start date Start date
  • Tags Tags
    Representations
Click For Summary
The discussion centers on the requirement for the parameter p to be an integer in the irreducible representations of the Lie group U(1). Participants clarify that the irreducible representations are derived from the Lie algebra u(1), which is one-dimensional and spanned by the basis vector i. Due to the nature of U(1) being not simply connected, the exponential map from u(1) to U(1) leads to the conclusion that different values of theta that differ by integer multiples of 2π correspond to the same element in U(1). This periodicity necessitates that p be an integer to maintain consistency in the representation. The conversation emphasizes the importance of understanding the topological properties of U(1) in relation to its representations.
Pietjuh
Messages
75
Reaction score
0
Hello, perhaps this is the most dumb question ever, but I don't see why it holds.
I'm looking at the irreducible representations of the Lie group U(1). To find them I considered the irreps of the lie algebra u(1). These irreps are obviously 1 dimensional and are given by f(a i ) = p a i for some real number p. If I now exponentiate this result I find that the irreducible representations of U(1) are given by g( exp(i theta) ) = exp(i p theta). But I read that p must be an integer. I cannot see however why this must be true! :(

Thanks in advance
 
Physics news on Phys.org
What is "a i"?

Of course, have you worked out the lie algebra of U(1) correctly? Are you allowing for the fact that you really only want unitary reps, and that even if as claimed the lie algebra is just the complex numbers that this is not a semi-simple lie algebra over C?
 
Last edited:
For every integer n, exp( 2 n pi i theta) is the same element of U(1) as exp(i theta). Hence, g( exp(i theta)) = g( exp( 2 n pi i theta)) for every integer n.
 
matt grime said:
What is "a i"?

Of course, have you worked out the lie algebra of U(1) correctly? Are you allowing for the fact that you really only want unitary reps, and that even if as claimed the lie algebra is just the complex numbers that this is not a semi-simple lie algebra over C?

Let me clarfiy my derivation of the Lie algebra of U(1). Let exp(i theta) be an arbitrary element of U(1). Then the Lie algebra is the tangent space at the identity element, so u(1) is spanned by the basis vector i. This means that u(1) = { ai | a in R }.
 
Since U(1) is not simply connected the correspondence between representations of U(1) and u(1) fails to hold generally.
 
matt grime said:
Since U(1) is not simply connected the correspondence between representations of U(1) and u(1) fails to hold generally.

I'm not sure what you mean. Because U(1) is not simply connected, any Lie Group that is a covering of U(1) shares a Lie algebra with U(1).

However, U(1) is connected and compact, and therefore the exponential map from u(1) to U(1) is onto.

Isn't this enough foe what Pietjuh wants?
 
I am not entirely shure what it is pietjuh actually wants, to be honest.
 
Sorry, Pietjuh, I didn't see my typo until now

George Jones said:
For every integer n, exp( 2 n pi i theta) is the same element of U(1) as exp(i theta). Hence, g( exp(i theta)) = g( exp( 2 n pi i theta)) for every integer n.

Cleary, this should be g( exp(i theta)) = g( exp( 2 n pi i + i theta)).

Now do you see why p must be an integer?
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
496
  • · Replies 2 ·
Replies
2
Views
2K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K