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$\DeclareMathOperator\SO{SO}$This might be an old question. But since I have not found an explicit answer to this question, I put the question here.

The background is that we need to use a similar technique when we construct the Euler class for $S^1$-bundle. It is written in Bott, Tu, Differential Forms in Algebraic Topology, page 73.

To construct the $S^1$-bundle, we use the transition function $g_{\alpha\beta}: U_{\alpha} \cap U_{\beta} \rightarrow \SO(2)$. By identifying $\SO(2)$ with the unit circle in the complex plane via $e^{i\theta}$, we consider the change of angle when we change the chart as $\theta_{\alpha} - \theta_{\beta} = \pi^*(1/i)\log(g_{\alpha\beta})$. So the Euler class $e(E) = -(1/2{\pi}i)\sum_{\gamma} d(\rho_{\gamma}d{\log}(g_{\gamma s}))$ with $\rho_{\gamma}$ the partition of unity.

For the tautological $C$-bundle over $CP^1$, what is the explicit differential form of its Euler class? Is there any reliable result and detailed computing process that we can check?

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    $\begingroup$ It might be a good idea to explain all the notation here. I do find it in Bott and Tu, but one shouldn't have to have the book to hand to answer the question. In particular, it seems quite confusing to me (although consistent with Bott and Tu) to use $\pi$ both as an angle measure of $3.14\ldots$, and as a bundle map $E \to M$. $\endgroup$
    – LSpice
    Commented May 14 at 14:52
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    $\begingroup$ The Euler class of the canonical line bundle is a generator on $H^2(CP^1,\mathbb Z)\cong\mathbb Z$ corresponding to the orientation given by the complex structure, so you just need to find a form that represents the generator (after changing coefficients to $\mathbb Z$. By the way, the Euler class of the canonical line bundle is the same as that of the associated principal circle bundle, so maybe your form already works. $\endgroup$
    – Igor Belegradek
    Commented May 14 at 16:19

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