# Can space be augmented with a plane at infinity so that parallel planes intersect at a line at infinity?

The real plane can be augmented with a line at infinity such that two parallel lines intersect at a point at infinity, and the set of all such points forms a line.

In space (3 dimensional solid geometry), can a similar augmentation be made? Can we augment space, adding a plane at infinity, such that parallel planes intersect in a line at infinity?

I ask because augmenting space accordingly would seem to cause a contradiction: In general, if line $$\ell$$ is parallel to plane $$P$$, and $$\ell$$ lies within plane Q, then $$\ell$$ is also parallel to any line $$m$$ formed by the intersection of planes $$P$$ and $$Q$$. How would that apply to the line at infinity? That is, if $$P$$ and $$Q$$ are parallel, and $$\ell_1, \ell_2$$ are distinct intersecting lines which are both parallel to $$P$$, how can they both be parallel to the line at infinity?

To clarify the geometric objection, we have in Euclidean geometry:

Let planes $$P, Q$$ intersect in line $$m$$. Let line $$\ell$$ be parallel to $$P$$ and lie within $$Q$$. Then $$\ell \parallel m$$.

Now, let's assume $$P, Q$$ are parallel and hence intersect at line $$m$$ at infinity. Then by the theorem above, any line $$\ell$$ parallel to $$P$$ must be parallel to $$m$$.

But consider two intersecting lines $$\ell_1, \ell_2$$, both of which are parallel to $$P$$ and lie within $$Q$$. By the above, they must both be parallel to $$m$$. Since parallelism is transitive, then $$\ell_1 \parallel \ell_2$$. But this is impossible, since they are assumed to intersect, giving us a contradiction.

Without the line at infinity, this contradiction cannot arise, because two intersecting lines, both lying in plane $$Q$$, cannot both be parallel to plane $$P$$ if $$P$$ intersects $$Q$$.

• Of course; this will be explained in any textbook covering higher dimensional projective spaces. Commented Feb 22 at 21:18
• Not sure I understand the geometric objection in the final paragraph, but note that two lines in space can fail to meet without being parallel, i.e., by being skew. Commented Feb 22 at 21:26
• Might want to look into Grassmanians Commented Feb 22 at 21:32
• You seem to be making an assumption that the concept of parallelism, which is a binary relation on the set of lines in Euclidean space, can be extended in some manner to obtain a binary "parallelism" relation on the set of lines in your extended space. One conclusion to perhaps draw from your argument is that this assumption leads to a contradiction and is thererefore false. That is to say, there does not exist a concept of parallelism for lines in the extended space. Commented Feb 22 at 23:26
• By the way, this extended space does exist, it is called 3-dimensional projective space, and it does not have any concept of parallelism of lines. Commented Feb 22 at 23:29

Yes, you can extend Euclidean 3-dimensional space to projective 3-space by adding a plane at infinity. A projective plane at infinity, to be precise.

The conflict you observed is due to an incorrect assumption of transitivity. Let me explain in more detail, and do so in the projective plane because that's enough to illustrate this point.

What does it mean for two distinct lines in the plane to be parallel? In projective terminology, it means that the unique point of intersection lies on the line at infinity. Now what happens if you intersect the line at infinity with a finite line? You get a point if intersection that lies on the line at infinity. So by the definition I just gave, you have to consider them parallel. The line at infinity is parallel to every finite line, in this sense.

That of course breaks transitivity. If you say finite line $$a$$ is parallel to the line at infinity, and the line at infinity is parallel to finite line $$b$$, that doesn't mean $$a$$ and $$b$$ share the same point of intersection with the line at infinity, so they don't have to be parallel to one another.

In a way this problem is very similar to some basic arithmetic. For positive integers $$a,b,c$$, if $$ac=bc$$ then you would conclude that $$a=b$$. If you slightly increase your domain of consideration to include zero, then suddenly that's no longer the case: you have the case $$c=0$$ to consider, which plays a special role for multiplication. It forces you to do case distinctions: either $$a=b$$ or $$c=0$$ (or maybe both).

The same happens when you go from Euclidean to projective geometry. If you have $$a\Vert c$$ and $$b\Vert c$$ then in Euclidean geometry you conclude that $$a\Vert b$$. But as you extend the scope to the projective plane, suddenly you have a case distinction, with $$c$$ being the line at infinity as the second alternative. And again that's because the newly added element plays a special role for the operation you consider.

Your 3d setup is just a more complicated setup where this lack of transitivity defies your initial intuition.

• This is a great explanation, especially the analogy to $ac = bc$. Should "You get a punt if intersection" be "You get a point of intersection"? Also, it seems you disagree with the comment above that "3-dimensional projective space... does not have any concept of parallelism of lines" - it has parallel lines, only that parallelism isn't transitive. Is that correct? Commented Feb 23 at 21:22
• @SRobertJames: Thanks, corrected typo. Projective geometry in its pure sense doesn't have a concept of parallelism. Nor of infinity. All the points are the same, all the lines are the same, ask the planes the same. If you want to talk about Euclidean concepts like parallelism, then you do that by designating one plane as special, and giving it the role of the plane at infinity. And it's in such a space with a designated element at infinity that I formulated my answer. See also math.stackexchange.com/a/1815758/35416 where I'm discussing some of these relationships.
– MvG
Commented Feb 23 at 21:47