Equivariant generalization of $\mathcal{O}(1)$ on $\mathbb{CP}^1$ The connection of the line bundle $\mathcal{O}(1)$ on $\mathbb{CP}^1$ is given by 
\begin{equation}
A=\frac{i}{2}\frac{\overline{z} \, dz-z\,d\overline{z}}{1+|z|^2}
\end{equation}
This follows since the first Chern class is represented by the curvature 
\begin{equation}
\frac{i}{2\pi}idA=\frac{i}{2\pi}\frac{dz\wedge d\overline{z}}{(1+|z|^2)^2}
\end{equation}
which is the positive unit volume form of $\mathbb{CP}^1$.
My question is, what is the connection on an equivariant version of $\mathcal{O}(1)$. For example, because of the $U(1)$ isometry of $\mathbb{CP}^1$ which rotates its circle fibers, one can consider the $U(1)$-equivariant version of $\mathcal{O}(1)$ on $\mathbb{CP}^1$. What is its connection? Does it involve equivariant differential forms, etc.?
 A: First of all, "equivariant version of $\mathcal{O}(1)$" sounds a bit confusing to me. What we actually do, we endow $\mathcal{O}(1)$ with equivariant structure. $U(1)$ acts as follows. Point of total space is a pair $(x, \xi)$, where $x \in \mathbb{P}^1$ i.e. $x$ is line in $\mathbb{C}^2$. $\xi$ is a linear function on this line $l$. $U(1)$ acts on hole $\mathbb{C}^2$, therefore, it acts on such pairs.
Remark 1 $PSL_2 (\mathbb{C})$ acts on $\mathbb{P}^1$. But one can not endow $\mathcal{O}(1)$ with $PSL_2 (\mathbb{C})$ equivariant structure (it is not entirely obvious). What is obvious, that $PSL_2$ does not act on $\mathbb{C}^2$, then one can not do it as I did for $U(1)$ action.
Remark 2 Equivariant structure is not unique. You can twist the action of $U(1)$ by a character. More precisely, $U(1)$ acts tautologically on the fibers of hermitian line bundle (denote this action by $\rho (u)$). If you have an equivariant structure on line bundle, you may consider composition of original action and $\rho (u)$. This composition is a new action (i.e. gives new equivariant structure).
An equivariant connection is just a connection (in usual sense), satisfying some extra property. Namely, $\nabla$ commutes with the group action (you know, group acts on sections of equivariant bundle). It is easy to check, that your connection satisfies this condition (nevertheless, Ted Shifrin is right, you should not right $i$ in that formula.). 
The correct question would be what is the "equivariant curvature". I will follow notation of classical paper by Atiyah Bott (paragraph 8). I may consider coordinates $x_0$, $x_1$ on $\mathbb{C}^2$, $z = \frac{x_1}{x_0}$. 
$x_0$ is a section of $\mathcal{O}(1)$, which trivializes the bundle, restricted to the chart $z \neq \infty$. Consider $U(1)$ action $e^{i \phi} (x_0, x_1) = (x_0 , e^{i \phi} x_1)$. Then group acts trivially on section $s=x_0$ (i.e. $L(s)=0$). Denote by $X$ vector field, which generates $U(1)$ action on $\mathbb{P}^1$
$$X = i z \partial_z- i \bar{z} \partial_{\bar{z}}$$
$$\theta = \frac{\bar{z}dz - z d \bar{z}}{2(1+|z|^2)}$$
$$\iota_X \theta = i \frac{|z|^2}{1+|z|^2}$$ 
Now we use formula $(8.8)$ from the paper and get
$$c_1 = \frac{i}{2\pi} \Big( -\frac{dz\wedge d\overline{z}}{(1+|z|^2)^2}+ i \frac{|z|^2}{1+|z|^2} u \Big)$$
You can check that this form is equivariantly closed. I.e. it is killed by equivariant differential $d + u \ \iota_X$
