If we restrict to those lines passing through the origin, we of course get $\mathbb{R}P^2$. Is there a good topological description of the space that we get when we remove the restriction that they pass through the origin? Is there a name for this space?
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1The resulting equvalence relation would postulate every point as equivalent, hence it would be a space made of one point ${p}$... – AlexR May 27 '14 at 08:22
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4@AlexR What equivalence relation? Certainly there is more than one distinct line in $R^3$. – Carl May 27 '14 at 08:27
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If all points lying on a line in $\mathbb R^3$ are defined to be equvalent, they are all equivalent (since $\mathbb R^3$ is convex). $\mathbb P^2$ is obtained by defining points to be equivalent if they lie on the same line through the origin. If you ask for a way to describe the set of all lines in $\mathbb R^3$, you should ask it differently. – AlexR May 27 '14 at 08:31
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1@AlexR: the question is completely clear. Nowhere did the OP mention an equivalence relation. – May 27 '14 at 08:32
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@AlexR Even in the case of the projective plane, defining it as an equivalence relation like that would give you only a single point, as every point is identified with the origin (although of course you can just remove the origin from each line). – Carl May 27 '14 at 08:35
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@AsalBeagDubh How do you define $\mathbb R P^2$ without implicitly using $x\sim y :\Leftrightarrow \exists \lambda \in \mathbb R\setminus{0}: x=\lambda y$ as an equivalence relation and forming $\mathbb R^3/\sim$? – AlexR May 27 '14 at 08:35
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@Carl I have written down the appropriate equivalence relation. If I understand correctly, your Space will be isomorphic to $\mathbb R^3 \oplus \mathbb RP^2$ – AlexR May 27 '14 at 08:38
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@AlexR Remember that each line passes through multiple points. So for example, the z axis would be part of a copy of $\mathbb{R}P^2$ "located" at the origin, but it would also be part of the copy of $\mathbb{R}P^2$ at every other point of the z axis. – Carl May 27 '14 at 08:40
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@Carl indeed so this is where the interesting part begins ;) I suppose you'll lose one more dimension but I'm not sure how to treat lines orthogonal to the "removed" dimension (say the z-axis) maybe $$\mathbb R^2 \oplus \mathbb RP^2 \cup \mathbb R^2 \oplus \mathbb RP^1$$ – AlexR May 27 '14 at 08:43
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One way to think about it is this. Call your space $U$.
If we think about $\mathbf R^3$ as an open submanifold of $\mathbf RP^3$, then $U$ is an open submanifold of the space of lines in $\mathbf RP^3$, which is a Grassmannian variety $G(2,4)$. (This shows that $U$ is 4-dimensional, as it should be.)
The complement $V$ of $U$ inside $G(2,4)$ consists of lines that are contained in the boundary $\mathbf RP^3 \setminus \mathbf R^3 = \mathbf RP^2$. So $V$ is an embedded copy of $G(2,3)$ inside $G(2,4)$: the standard name for a submanifold of this kind is a Schubert cycle $\sigma_{1,1}$.
So $U$ is the complement of a $\sigma_{1,1}$ inside $G(2,4)$.
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To every line in $\mathbb R^3$ there corresponds a line in $\mathbb P^3(\mathbb R)$ obtained by adding a point at infinity to it.
This way you obtain all lines in $\mathbb P^3(\mathbb R)$, except those lines completely included in the plane at infinity.
The lines in $\mathbb P^3(\mathbb R)$ form the grassmannian $\mathbb G(1,3)$ and have as moduli space under the Plücker embedding a smooth $4$-dimensional quadric $Q\subset \mathbb P^5(\mathbb R)$ (the Klein quadric).
The lines of $\mathbb P^3(\mathbb R)$ lying completely at infinity correspond to a plane $P\subset \mathbb P^5(\mathbb R)$ included in the Klein quadric $Q$: $P\subset Q$
Conclusion
The moduli space of lines in $\mathbb R^3$ is a quadric hypersurface in $Q\subset \mathbb P^5(\mathbb R)$ minus a plane included in it.
This moduli space is thus quasi-projective, but neither projective nor affine.
Georges Elencwajg
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Dear Georges, this is a nice answer. (Since the tag was "algebraic-topology", I treated my answer as an exercise in constrained writing, forbidding myself to use any words from algebraic geometry. But I think your approach was more succcessful!) – May 27 '14 at 09:14
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Dear Asal, it is nice to have two slightly different but ( reassuringly !) equivalent answers (I hadn't read yours while I typed mine). Needless to say, I upvoted you. – Georges Elencwajg May 27 '14 at 09:17
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As it seems you guys guessed, I don't know too much algebraic geometry, so it may take me a bit of work to understand these answers, but for now thanks! – Carl May 27 '14 at 09:22
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1So, it's standard to denote the Grassmanian of $k$-planes in $\mathbb{P}^n$ by $G(k,n)$, even though this can be identified with $k+1$-planes in $\mathbb{A}^{n+1}$? There must be a risk of confusing $G(k,n)$ with $G(k+1,n+1)$, as even the two answers here use both notations-unless I've missed something. – Kevin Carlson May 27 '14 at 09:40
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@KevinCarlson: good point. But note that I wrote $G(k,n)$, whereas Georges wrote $\mathbb G(k,n)$. – May 27 '14 at 09:50
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Yes-is that a standard way to distinguish, then? It's good to know, if so. – Kevin Carlson May 27 '14 at 10:18
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2@KevinCarlson: yes, I think the use of blackboard bold for the "projective" version is pretty standard. Of course, any sensible author should also make clear what their notation means when they introduce it, so in practice there shouldn't be confusion in any case. – May 27 '14 at 10:26
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Just for kicks here is another "bundle" description of the moduli space.
It is not hard to see that this is the "orthogonal complement" of the canonical bundle over $\mathbb{RP}^2$, that is it is the space
$$ \{ (l,v) \in \mathbb{RP}^2 \times \mathbb{R}^3 | v \perp w \mbox{ for any }w \in l \}$$
after all all you need to describe an arbitrary line is its slope and a point, and you can choose this point to be the one closest to the origin.
apurv
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