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

[MathematicalPhysicist]

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?

 

[fresh_42]

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.