# Positive geometry and log singularities

In order to define a positive geometry it is a requirement that has to be a logarithmic singularities on the boundaries, for example for an interval (endpoints $$a$$ and $$b$$) the canonical form is $$\frac{dx(b-a)}{(b-x)(x-a)}$$ My question is: Why do we need that logarithm singularities requirement in order to get that differential form?

I'm not sure I understood exactly what you want to know.

Anyhow, a positive geometry is an oriented geometry that has a canonical form. A canonical form is a differential form with dlog divergences, which is like d$$x/x$$ for a boundary $$x=0$$, on all the boundaries of the geometry. Moreover, the residue of the canonical form on a boundary must be equal to the canonical form of the boundary. Finally, the canonical form of a point is equal to 1 for a positively oriented point and to -1 for a negatively oriented point.

Your example is the canonical form of a segment $$a with an orientation form equal to d$$x$$. Following the definition, to verify that the expression is indeed correct we need to verify that the residue in $$x=a$$ is equal to 1, since $$x=a$$ is a positively oriented point, and the residue in $$x=b$$ is equal to -1, since $$x=b$$ is a negatively oriented point.

Your expression for the canonical form can also be rewritten as $$\Omega(x)= \text{dlog}(x-a)-\text{dlog}(x-b)=\frac{\text{d} x}{x-a}-\frac{\text{d} x}{x-b}$$

The residue $$Res_{x=a}(\Omega(x))$$ for example can be computed as $$Res_{x=a}(\Omega(x))=\lim_{x\to a}(x-a)(\frac{1}{x-a}-\frac{1}{x-b})=1$$

The idea of the canonical form can be generalized to non-logarithmic singularities for polytopes, as proposed in https://arxiv.org/abs/2005.03612, but the topic remains still in development for more general geometries.