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Generally, I think of a line in $\mathbb R^2$ as characterized by any two points. Yet, this question proves that any line not through the origin is characterized by a single point. And a line through the origin can likewise be characterized through a single point. This implies that any line can be characterized by a single point and a single bit (indicating whether the line is through the origin or not).

This seems mystifying: To define a line, can we choose any one point (plus a bit) or any two points?

More precisely: Is there a continuous transformation $(\mathbb R^2 \times \mathbb R^2) \longleftrightarrow (\mathbb R^2 \times \{0,1\})$ that is injective? How does this fit with the concept of dimension and degrees of freedom?

I may indeed be struggling to ask the right question here. Indeed, turning this into a well formed question may almost provide the answer. So help turning this baffling (at least to me) situation into a rigorous question is a very good way to start.

SRobertJames
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  • I don't think this is the right question to ask. In the situation you describe for lines not through the origin, there are many many elements of $\Bbb R^2\times\Bbb R^2$ that give rise to the same line; so even that map by itself isn't injective. – Greg Martin Dec 28 '22 at 03:37
  • @GregMartin Excellent point. I've edited the OP to note that while I've identified something baffling, I've struggled to formulate it as a rigorous question. – SRobertJames Dec 28 '22 at 03:42
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    The right domain would be the set of all lines in $\Bbb R^2$, which is the punctured projective plane or a Möbius band. You've already described a continuous surjective (indeed bijective) map from $\Bbb R^2\times{0,1}$ to this set, so the question is really whether there's a continuous injective map in the other direction. – Greg Martin Dec 28 '22 at 03:47
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    You have to specify which point is the origin. That's a second point. You've still got an interesting, non-standard way to parameterize the space of all lines, but you're not getting away with just one point. – JonathanZ Dec 28 '22 at 06:15
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    The closest point on a line to the origin gives you the slope of a perpendicular line, hence the lines slope. Is is a mirage that it is only one piece of information – Paul Dec 28 '22 at 09:35
  • Moreover, the distance from any other point will do (you find another perpendicular,) not necessarily the origin. So, it isn’t necessary to distinguish between line through the origin or not. – BakerStreet Dec 28 '22 at 13:08
  • IMO, you’re bit vague. It depends on how to define word “characterized” and dimension. There are lots of representation of line in $\Bbb{R}^2$, like $y=mx+c$ where $m$ is slope of line and $c$ is $y$-intercept. One can also write in vector form. We need “two info” to precisely specify a line, rotation & translation about $x$ axis. So it’s dof $=2$. In linear algebra (undergrad), we define dimension on vector space object. Since vector space contain zero vector, in this case $(0,0)$, so line passing through origin are consider vector space. – user264745 Dec 28 '22 at 16:57

2 Answers2

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There is more than one way to characterize a line in two dimensions. One of those ways is to use projective geometry. Consider Euclidean three dimensional space with an origin. Fix a plane which does not contain the origin. Any line in that plane uniquely determines a new plane that contains that line and also the origin. Now suppose that new plane is not perpendicular to the fixed plane. Then the unique line from the origin which is perpendicular to the new plane intersects the fixed plane in a unique point. This is known as the "polar" of the original line in the fixed plane. Thus every line in the fixed plane (with exceptions) is associated with a unique point in the fixed plane.

The question you linked to is closely related. In that case, the closest point of the line to the origin of the plane is reciprocally associated with the polar point in the previous paragraph. They are equivalent ways to associate a unique point to every line (with exceptions) in a plane. They also share the exceptional lines which are without any associated points. The main difference is that in this case only points in the plane are needed, while in the "polar" case, the plane must be embedded in a three dimensional Euclidean space.

Somos
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As Greg Martin has already pointed out in the comments, two points of a plane is actually too much information for determining a line in the plane, since many pairs of points determine the same line.

Two points on a plane is 2+2=4 dimensions. How much redundancy is there? Well, any two pairs of points on a fixed line determine that same line. This gives 1+1=2 redundant dimensions, leaving 4-2=2 dimensions, which is the number of dimensions for a single point. This agrees with the observation you made.

This also agrees with other standard representations of a line. For example, a line can be expressed in terms of two real numbers $m$ and $b$ as $y = mx + b$, so this is another way to see that the set of lines on a plane is 2 dimensional. In this latter representation, the vertical lines need to be treated differently, in the same way that in your example, lines through the origin are a special case that needs to be treated differently. These exceptions go away if we go into the projective plane rather than the affine plane.

Ted
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