How can the generators of ##U(1)## be traceless?

  • #1
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In Peskin and Schroeder on page 681 they write:
...if the gauge group of the theory contains ##U(1)## factors, the theory cannot be consistently coupled to gravity unless each of the ##U(1)## generators is traceless.
Now, as far as I can tell the generators of ##U(1)## are just exponents, i.e ##e^{2\pi ix}##, so how can they have a zero trace if it's just one number which never vanishes?

Perhaps I am missing something here, can anyone clear this matter to me?
Ah, wait perhaps this contradiction means that you cannot couple gravity to this theory?
 

Answers and Replies

  • #2
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Only ##0 \in \mathfrak{u}(1)= i \cdot \mathbb{R}## is traceless. But there could be another representation in which it is traceless. E.g. we could take ##\mathfrak{u}(1)\cong \left\{\begin{bmatrix}i \varphi & 0 \\0 & -i \varphi\end{bmatrix}\right\}##. So either we conclude that these generators vanish, or we choose another representation.
 
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