Why is the line at infinity one-dimensional?

Why is the line at infinity a one-dimensional manifold, i.e. why is it truly a "line" at infinity and not a plane? Is my reasoning below at all correct?

(When I say "dimension" I mean "real dimension" if talking about $\mathbb{RP}^2$ and "complex dimension" when talking about $\mathbb{CP}^2$ -- I do see why the complex line at infinity would be a two-real-dimensional manifold if the real line at infinity is a one-dimensional manifold.)

Naively, it seems like the line at infinity has two degrees of freedom, not just one. For either $\mathbb{RP}^2$ or $\mathbb{CP}^2$, it seems like one should have the following "basis" for the line at infinity: $$(1:0:0), (0:1:0) $$

Note: while writing up this question, I think I may have already thought up of an answer, so my attempt (which is long and not very rigorous) is included as a community wiki answer below, and I will add the (proof-verification) tag.

However, I more or less only want to know any elegant answers/proofs which you may have, rather than whether or not my tentative answer is correct.

Solution 1:

Here is a bit of geometric intuition for the line at infinity. One of the motivations for working with the projective plane is that any two lines on $\mathbb{P}^{2}$ intersect (unlike the affine plane). How is this related to the line at infinity?

Well, if you have two parallel lines $L_1$ and $L_2$ in the affine plane $\mathbb{A}^2$ with the same slope $m$, we know they won't intersect in $\mathbb{A}^2$! So we want to keep track of these parallel slopes. So we define $L$ to be the "line at infinity" to consist of the distinct slopes (which is inherently one-dimensional, as the slope is a value taken from the base field). Now we know that $L_1$ and $L_2$ will intersect in $\mathbb{P}^2$, in fact they will intersect at the point $m\in L$. Does this help at all?

Solution 2:

Which formulation of the projective plane?

Synthetic projective geometry

Through every pair of distinct points there is a unique line, and every pair of distinct lines intersects in a unique point. (and we also include "betweenness" axioms)

Now, pick any line L, and define:

• "Affine line" means any line other than L
• "Affine point" means any point not lying on L

Then, you can show that the affine lines and affine points satisfy the incidence and betweenness axioms of the Euclidean plane.

That is, "(projective plane) - (line) = (affine plane)".

Construction by adding points

One way to "construct" the projective plane is by adding one new point for every class of parallel lines.

Now, pick any Euclidean point $P$ and draw a circle passing through $P$. with center $P$.

If I draw a line through $P$, this line will interect the circle at a unique other than $P$ (for the tangent line, we take this other point to also be $P$; the intersection has "multiplicity" 2).

But the lines through $P$ are also in one-to-one correspondence with the points at infinity.

Thus, the points on a circle are in one-to-one correspondence with the points at infinity in a fairly natural way.

Coordinate Geometry

The points are triples $(x:y:z)$ not all zero (the colon means that $(x:y:z)$ and $(cx:cy:cz)$ refer to the same point, if $c$ is nonzero), and the lines are triples $(a:b:c)$.

A point lies on a line iff $ax + by + cz = 0$.

The points at infinity are precisely the points $z=0$, and so they are precisely the points lying on the line $(0:0:1)$.

Solution 3:

This is more obvious in some definitions than in others.

In the geometric definition of a real line at infinity as an addition to the plane, the line is a visual horizon line for the projection of 3-d space onto a fixed 2-d plane through a fixed perspective point $P$ outside that plane. The line at infinity can be identified with lines through $P$ in the plane parallel to the projection plane, which form a one-dimensional manifold equivalent to the circle. This is the classical picture known to Renaissance artists and architects.

The same projection process and what we would now call "gluing of affine charts" to make a real projective plane can then be defined algebraically in a way that works over any field. Hence the geometric picture is compatible with the projective complex-algebraic geometry definition of the line at infinity as the place where the missing points of inhomogeneous complex algebraic curves can be found.

Solution 4:

I think the key to unravelling your confusion may be in this line from the OP:

it seems like one should have the following "basis" for the line at infinity: $(1:0:0),(0:1:0)$

I think I get what you are saying: any point on the line at infinity can be written in the form $(a:b:0)$, which seems to suggest that there are two independent coordinates.

What you are forgetting is that these are homogeneous coordinates, so that the point $(1:0:0)$ is the same point as $(2:0:0)$ and $(a:0:0)$ for any $a \ne 0$. So $(1:0:0)$ does not point in a direction like a vector; any "multiple" of it is the exact same point.

More broadly, consider $(a:b:0)$. Since we can't have both $a=0$ and $b=0$, at least one of them is nonzero. Suppose that it is $b$. Then $(a:b:0)$ can be written in the form $(r:1:0)$ for some $r \in \mathbb{R}$. The set of all such points is in one-to-one correspondence with $\mathbb{R}$.

But what if $b=0$? Then we know for sure that $a \ne 0$, and in this case we can write $(a:b:0) = (1:0:0)$, a single point.

So the line at infinity consists of two parts:

• The set of points of the form $(r:1:0)$, which is in one-to-one correspondence with the elements of $\mathbb{R}$
• The single point $(1:0:0)$, which is a "point at infinity" for the line at infinity.