Can a U(1) Generator be Normalized to SU(1) through Determinant Condition?

  • Context: Undergrad 
  • Thread starter Thread starter DuckAmuck
  • Start date Start date
  • Tags Tags
    Group
Click For Summary

Discussion Overview

The discussion revolves around the relationship between U(1) and SU(1) generators in the context of their normalization and determinant conditions. Participants explore whether a U(1) generator can be normalized to fit within the framework of SU(1), examining the implications of determinant and trace conditions for these Lie algebras.

Discussion Character

  • Debate/contested

Main Points Raised

  • Some participants question if a U(1) generator can be normalized to SU(1), noting that the "S" in SU(1) implies a determinant condition of 1 or a trace of 0.
  • Others assert that elements of U(1) already satisfy the determinant condition since they lie on the unit circle (|z|=1).
  • There is a proposal that SU(1) could be equivalent to U(1), but this is met with skepticism.
  • One participant points out that SU(1) is the trivial group consisting of a single element, suggesting that U(1) and SU(1) are not equal.
  • A later reply clarifies the definitions of SU(n) and SU(1), emphasizing the determinant condition and correcting a previous misunderstanding regarding the determinant and absolute value.

Areas of Agreement / Disagreement

Participants do not reach a consensus on whether U(1) can be normalized to SU(1). There are competing views regarding their equivalence, particularly concerning the nature of SU(1) as a trivial group.

Contextual Notes

There are unresolved assumptions regarding the definitions of the groups and the implications of the determinant condition. The discussion reflects varying interpretations of the properties of U(1) and SU(1).

DuckAmuck
Messages
238
Reaction score
40
TL;DR
How are these two related?
If you have a U(1) generator, can it just be normalized to SU(1)?
 
Physics news on Phys.org
DuckAmuck said:
Summary:: How are these two related?

If you have a U(1) generator, can it just be normalized to SU(1)?
The "S" stands for determinant = 1 or trace = 0 for the Lie algebras. Elements of ##U(1)## are all ##|z|=1##, so they have already determinat =1.
 
fresh_42 said:
The "S" stands for determinant = 1 or trace = 0 for the Lie algebras. Elements of ##U(1)## are all ##|z|=1##, so they have already determinat =1.
so could one say SU(1) = U(1)? If not, why not.
 
Yes. No.
 
Last edited:
Isn't SU(1) the trivial group of one element? I don't think they are equal. U(1) is the set of complex numbers with radius 1
 
##SU(n)=\{A\in \mathbb{M}(n,\mathbb{C})\, : \,A\bar{A}^\tau =\bar{A}^\tau A = 1\, , \,\det(A)=1\}##

You are right, the determinant condition fixes the ##1##. I mistakenly thought ##\det = |\, . \,|##.
##SU(1)=\{A\in \mathbb{C}\, : \,A\bar{A}=1 \Longleftrightarrow |A|=1\, , \,\det(A)=A=1\}##
 

Similar threads

Undergrad ##SU(2)## homeomorphic with ##\mathbb S^3##
  • · Replies 61 ·
3
Replies
61
Views
7K
Undergrad ##SU(2, \mathbb C)## parametrization using Euler angles
  • · Replies 15 ·
Replies
15
Views
4K
Undergrad Dyadic cubes generate Borel sigma algebra of subset
  • · Replies 2 ·
Replies
2
Views
2K
Graduate Do Dynkin Diagrams Only Describe Lie Algebras?
  • · Replies 2 ·
Replies
2
Views
2K
Undergrad Proving SL_2(C) Homeomorphic to SU(2)xT & Simple Connectedness
  • · Replies 1 ·
Replies
1
Views
2K
Graduate Decomposing SU(4) into SU(3) x U(1)
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad ##SL(2,\mathbb R)## Lie group as manifold
  • · Replies 12 ·
Replies
12
Views
3K
Undergrad Page 183-184 of Howard George's group book
  • · Replies 1 ·
Replies
1
Views
3K
Undergrad Showing that a group acts freely and discretely on real plane
  • · Replies 5 ·
Replies
5
Views
2K
Graduate New ideas on confinement and meson masses from Brazil
  • · Replies 4 ·
Replies
4
Views
3K