# Is there a thing so-called integral over the projective line of $\mathbb R$?

In projective geometry, a line can be viewed by its projective line, which has only one point at infinity, and if a point keeps moving to $+\infty$ it will return to $-\infty$. On the other hand many integrals are taken over the real line. I wonder if there is a connection between two concepts. Would it be an integral over a circle or something? Would this be affect if the integral is replaced with a sum over $\mathbb Z$? I understand the function under integration is a member of $\mathbb R^\mathbb R$, and the series under summation is a member of $\mathbb R^\mathbb Z$

• Yes from $[x:y ] \mapsto [y:x]$ you see the topology of $\mathbb{P}^1(\mathbb{R})$ is of a circle and you can define an integral accordingly, for example with $\int_{\mathbb{P}^1(\mathbb{R})} f(u)du = \int_{-\pi/2}^{\pi/2} f([\tan x:1])dx$ – reuns Nov 6 '17 at 8:41

The issue is choosing a metric (alternatively, a measure) on the real projective line with respect to which the integration is to be performed. It is natural to identify the real projective line with a circle in which case the natural metric is simply the least length of an arc joining the two points, i.e., the subtending angle $\theta$ in radians.
The way to define a distance between two points $A,B$ in the projective plane is to take the representing lines $a,b$ through the origin in the plane and set the distance between them be the angle between vectors $\alpha,\beta$ spanning $a,b$ respectively. Thus one would set the distance between $A$ and $B$ to be $\arccos \alpha \cdot \beta$ where "$\cdot$" is the scalar product, if $\alpha$ and $\beta$ are unit vectors.
• In the case of the metric induced from the euclidean $l^2$ norm, what exactly the circle we integrate? Can you give me an example? – Ooker Nov 6 '17 at 16:24
• To answer your question "what exactly the circle", the simplest way of putting it is to say that it is the circle parametrized by the angle $\theta$ appearing as $\arccos \alpha\cdot\beta$ as above, as $\theta$ runs from $0$ to $\pi$. – Mikhail Katz Nov 8 '17 at 15:14