# How to visualize the pole/polar of a line/point in projective space PG(2, R) not in contact with the curvature?

So, I understand that given a curvature $$\mathbb{C}$$. The polar of a point on the conic is tangent to the conic at that point and the pole of a line touching the conic only once and tangent to the conic is the point where it touches.

I'm failing to understand what the polar of internal and external points are and the poles of secant lines and external lines are. It is stated that an external point has a secant polar and an internal point has an external line as a polar.

Could someone explain why this is? How does this work intuitively?

• Is your “curvature” a conic? – amd Oct 17 '18 at 20:36
• Yes, it is a conic. – Neo Oct 18 '18 at 5:29

Take a point external to the conic. Draw the tangents from this point to the conic. There will be two of these. They touch the conic in two points of contact. Their connection is the polar line of the initial point. Conversely for a secant you can intersect with the conic, draw tangents and so obtain the external pole.

For internal pole and passing polar it is harder to picture. Algebraically it's the same, except many of the coordinates are complex-valued. But perhpas it's better to think of this in terms of incidences preserved by polarity. Start with your external line. Pick two different points on it. Construct their polar lines. The point where these polars intersect is the pole of the initial line. And conversely, starting with an inner point, draw two arbitrary lines to it, construct their poles and the line joining these poles will be the polar of the initial point.

In this figure, $$A$$ is the pole of $$a$$, $$B$$ of $$b$$ and $$C$$ of $$c$$. Both $$(B,b)$$ and $$(C,c)$$ have the point outside and the polar line a secant and thus can be paired via the tangents (dark green) and points of contact (light green). The pair $$(A,a)$$ then follows from that due to incidence conservation.