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

[DuckAmuck]

TL;DR

How are these two related?

If you have a U(1) generator, can it just be normalized to SU(1)?

 

[fresh_42]

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.

 

[DuckAmuck]

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.

 

[fresh_42]

Yes. No.

 

[Office_Shredder]

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

 

[fresh_42]

$SU(n)=\{A\in \mathbb{M}(n,\mathbb{C})\, : \,A\bar{A}^\tau =\bar{A}^\tau A = 1\, , \,\det(A)=1\}$

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