Angles in $\mathbb{P}^2$ Do there exist a way to define angle between lines at $\mathbb{P}^2(k)$?, where $k$ is an algebraically closed field of characteristic zero.
 A: No and yes:
No: There is no way to define angles on $\mathbb{P}^2$ which respects all the symmetries of $\mathbb{P}^2$. Even looking at $\mathbb{RP}^2$, which is smaller than $\mathbb{CP}^2$, the symmetry group is $PGL_3$, which is sometimes called the group of projective transformations. These transformations do not preserve angles. You can see this in any perspective drawing: The windows of this building are framed with right angles, but they appear to be trapezoidal in the photo. The map which turns the position of points on the wall into positions of pixels in the image is one of these projective transforms.  (Photo courtesy of Tambako the Jaguar, Creative Commons license.)

Yes: Inside $PGL_3$ is the subgroup $PU(3)$. There is a $PU$ invariant metric on $\mathbb{CP}^2$ called the Fubini-Study metric and, once you have a metric, you can talk about angles.
Explicitly, let our lines be given by $a_1 x_1 + a_2 x_2 + a_3 x_3=0$ and $b_1 x_1+b_2 x_2 + b_3 x_3=0$. Then, I think, the angle between them should be
$$\cos^{-1} \left( \frac{a_1 \overline{b_1} + a_2 \overline{b_2} + a_3 \overline{b_3}}{\sqrt{|a_1|^2+|a_2|^2+|a_3|^2} \ \sqrt{|b_1|^2+|b_2|^2+|b_3|^2} }\right).$$ Here $\bar{z}$ is the complex conjugate.
