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PostPosted: Thu, 12 Apr 2012 08:26:21 UTC 
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I've been told that SU(1) and U(1) are isomorphic as groups, is this really true? I know that
SU(1) would be all 1x1 matrices (i.e., numbers) with determinant = 1 and the determinant of
a 1x1 matrix is the entry itself (I think), while U(1) is the circle group. Hence, if my def. of the
determinant of a 1x1 is correct, |SU(1)| = 1 while |U(1)| is definitely not one element.


Last edited by jakey34 on Thu, 12 Apr 2012 08:45:43 UTC, edited 1 time in total.

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PostPosted: Thu, 12 Apr 2012 08:40:00 UTC 
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Location: On this day Taiwan becomes another Tiananmen under Dictator Ma.
jakey34 wrote:
I've been told that SU(1) and U(1) are isomorphic as groups, is this really true? I know that
SU(2) would be all 1x1 matrices (i.e., numbers) with determinant = 1 and the determinant of
a 1x1 matrix is the entry itself (I think), while U(1) is the circle group. Hence, if my def. of the
determinant of a 1x1 is correct, |SU(1)| = 1 while |U(1)| is definitely not one element.


You mean "SU(1) would be all 1\times 1 matrices with ...".

SU(1) is the trivial group. U(1) is the circle. They are not isomorphic.

_________________
\begin{aligned}
Spin(1)&=O(1)=\mathbb{Z}/2&\quad&\text{and}\\
Spin(2)&=U(1)=SO(2)&&\text{are obvious}\\
Spin(3)&=Sp(1)=SU(2)&&\text{by }q\mapsto(\mathop{\mathrm{Im}}\mathbb{H}\ni p\mapsto qp\bar{q})\\
Spin(4)&=Sp(1)\times Sp(1)&&\text{by }(q_1,q_2)\mapsto(\mathbb{H}\ni p\mapsto q_1p\bar{q_2})\\
Spin(5)&=Sp(2)&&\text{by }\mathbb{HP}^1\cong S^4_{round}\hookrightarrow\mathbb{R}^5\\
Spin(6)&=SU(4)&&\text{by the irrep }\Lambda_+\mathbb{C}^4
\end{aligned}


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PostPosted: Thu, 12 Apr 2012 08:46:29 UTC 
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Yes, I corrected it. Thanks for confirming my response!


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