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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.

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You have that locally $\mathbb{P}^2$ is like $k^2$; if you can define angles on $k^2$ then you can do it on $\mathbb{P}^2$. For example, if $k=\mathbb{C}$, then defining angles is the same as defining an angle between two lines in $\mathbb{R}^4$, you just find the point of intersection of the two lines, and see which affine piece it is in. – rfauffar Sep 9 '11 at 12:12
However, such angles are not preserved by projective transformations, so it is debatable whether they ought to count as angles "in" $\mathbb P^2$. – Henning Makholm Sep 9 '11 at 12:20
That's true, I guess it depends on what properties you're looking for. – rfauffar Sep 9 '11 at 12:31
If I wanted to know this, I might look in Norman Wildberger's writings. – Michael Hardy Sep 9 '11 at 13:43

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.)

perspective image of a building

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.

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I corrected a tiny typo (second paragraph of Yes). Hope you don't mind :) – Srivatsan Sep 9 '11 at 20:22
Thanks! Plus some more characters. – David Speyer Sep 9 '11 at 20:27
Looks like I messed up something :). But the site is behaving somewhat strangely, so I won't make further edits here. And sorry for the trouble... – Srivatsan Sep 9 '11 at 20:31

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