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

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SUMMARY

The discussion clarifies the relationship between the U(1) and SU(1) groups in the context of Lie algebras. U(1) consists of complex numbers with a modulus of 1, inherently satisfying the determinant condition of SU(1), which requires a determinant of 1. However, SU(1) is defined as the trivial group containing only the identity element, thus U(1) and SU(1) are not equivalent. The determinant condition is crucial in distinguishing these groups, as SU(1) is constrained to a single element.

PREREQUISITES
  • Understanding of Lie groups and algebras
  • Familiarity with the properties of U(1) and SU(1)
  • Knowledge of complex numbers and their properties
  • Basic understanding of determinants in linear algebra
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  • Study the properties of Lie groups, focusing on U(n) and SU(n)
  • Explore the implications of determinant conditions in group theory
  • Learn about the structure and representation of SU(1) in mathematical physics
  • Investigate the applications of U(1) in quantum mechanics and gauge theories
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This discussion is beneficial for mathematicians, physicists, and students studying group theory, particularly those interested in the applications of U(1) and SU(1) in theoretical physics and mathematics.

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TL;DR
How are these two related?
If you have a U(1) generator, can it just be normalized to SU(1)?
 
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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\}##
 

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