topological types of conics in real projective space of dimension 2

Question

I know that all the conics have the same topological type in ${\mathbb{P}^2}_{\mathbb{R}}$. But I can not see why. Someone help me please.

for example, how do hyperbola and circle have the same topological type?

Answer 1

We can arrive to the projective plane if we add the line at infinity.

Then we can remove any line and we will arrive back to the Euclidean plane.

Now, a parabola has one point in infinity, (so the line at infinity is tangent to it), and a hyperbola has two points in infinity.
If we choose a line that avoids our conic, and 'project it to the infinity', then we the curve becomes ellipse.

Answer 2

Think of the projective plane as what you get from the Euclidean plane by adding one point "at infinity" for each line through the origin. Topologically this means you have taken a closed disc and identified pairs of opposite points on the boundary. Given a hyperbola, like the graph of $y = 1/x$, the identifications at infinity glue the "extensions to infinity" of the two connected components in the Euclidean plane into a topological circle ("at infinity" for the graph of $y = 1/x$, there is just one point with $x = 0$ and one point with $y = 0$).

This may become clearer if you look at a diagram of the fundamental polygon of the projective plane (which shows you how to obtain the projective plane from a square by identifying points on its boundary) and draw some ellipses, hyperbolas and parabolas on it. When you do this, remember that a parabola in the Euclidean plane has a single point at infinity in the projective plane while a parabola has two. E.g., the point at infinity on the $y$-axis for the parabola $y = x^2$ and the points at infinity at the ends of the two coordinate axes in the case of the hyperbola $y = \frac{1}{x}$.

Answer 3

All concise in the projective space are topologically identical to an ellipse(in the projective plane). One of the ways to do it is to go case by case. Also note that a projective plane is just like a normal $\Bbb R^2$ plane however with a point at infinity in every direction

I will present pictures from Arnolds book Real Algebraic Geometry

We have $3$ types of conics: ellipses, parabolas, and hyperbolas.

Ellipse is topologically equivalent to an ellipse so thats done.

Now parabola

Here $C'=C$ means that both $C$ and $C'$ are the same point at infinity. This might seem odd at first since the parabola branches out in two opposite directoin but if you play with it(draw line through origins an parabola) you will convince yourself that $C=C'$

similar story here, however the difference here is that there are two points at infinity, each long their respective asymptotes.

So all conics in the projective space are topologically equivalent to an ellipse. Note that the pictures presented are not what the conics actually look like in the projective space, but is a useful illustration. You can see how, for example, in the case of a parabola the point at infinity is what makes the branhces connect