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.

  • 2
    $\begingroup$ Do you understand why the projective plane itself is only $2$-dimensional? $\endgroup$ – Eric Wofsey Sep 4 '16 at 3:05
  • 1
    $\begingroup$ How would you describe a line in the projective plane? $\endgroup$ – Jacob Bond Sep 4 '16 at 3:13
  • 1
    $\begingroup$ @JacobBond I just realized that a line in the projective plane is homeomorphic to $\mathbb{CP}^1$ or $\mathbb{RP}^1$, not to $\mathbb{C}^1$/$\mathbb{R}^1$ respectively -- this has begun to alleviate my confusion substantially, since the former is a quotient of a two-dimensional object. $\endgroup$ – Chill2Macht Sep 4 '16 at 3:19
  • 3
    $\begingroup$ The line at infinity in the projective plane is a projective line which is specified by two points. For $(1:0:0)$ and $(0:1:0)$ the line at infinity in this direction is exactly the line through those two points $\endgroup$ – Jay Sep 4 '16 at 4:49
  • 1
    $\begingroup$ I guess I'm confused as to what the question is. If you have the description of $\mathbb P^2$ then I have confidence that you can verify a non-canonical decomposition into $\mathbb A^2$ and $\mathbb P^1$. And most introductions will tell you some story about making parallel lines meet. Here is the question I anticipate: why the centrality of $\mathbb P^n$ and its subvarieties? Why not some other compactification of affine space, or why a focus on a particular ambient setting for geometry at all? $\endgroup$ – Hoot Sep 4 '16 at 21:03

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?

  • $\begingroup$ Yes I think it does help, because it gives a nice intuitive explanation for why the line at infinity should be homeomorphic to $\mathbb{P}^1$, which is as you say inherently one-dimensional -- the point at infinity corresponds to the lines with infinite slope, and all of the other points correspond to lines with non-infinite slope. So this both answers my question plus gives me new intuition for thinking about $\mathbb{P}^1$, in terms I would have understood even in middle school -- this is a really nice answer actually, the more I think about it -- I appreciate it! $\endgroup$ – Chill2Macht Sep 4 '16 at 3:24
  • $\begingroup$ It's one thing to brute force a solution via formulas, but this answer gives nice geometric insights for how to (1) visualize the space in question (2) understand why these objects are/should be natural (3) understand how the construction of $\mathbb{P}^2$ does and should serve as the "intersection completion" of the regular affine plane. Personally I prefer being able to explain formal methods of proof with visual intuition, and this answer is perfect for that. $\endgroup$ – Chill2Macht Sep 4 '16 at 3:27
  • 1
    $\begingroup$ You are welcome! :) Yes, that is a good way to think about $\mathbb{P}^1$. Namely, $\mathbb{P}^1$ is just the collection of slopes of distinct lines in $\mathbb{A}^2$. This is how you define $\mathbb{P}^1$ algebraically as well (namely the quotient of $\mathbb{C}^2 - 0$ by the scalar action, which just says geometrically that you should identify the points that lie on the same line as a single point; and this single point is basically the slope of the line!). $\endgroup$ – Prism Sep 4 '16 at 3:30
  • 1
    $\begingroup$ @William: As a student myself, sometimes it feels good to brute force and be absolutely sure that the things work as they should! But yes, the best part is when we can understand the concept geometrically, because then the algebraic proofs also come naturally. Personally I am curious about your answer below and will read it! $\endgroup$ – Prism Sep 4 '16 at 3:33

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 no 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 $(0:0:1)$.


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.


$\newcommand{\span}{\operatorname{span}}$TL;DR version: Specifically, (if we let $\mathbb{K}$ equal $\mathbb{R}$ or $\mathbb{C}$ as appropriate), then $\mathbb{K}^3$ equals $$\left(\span\{(1,0,0),(0,1,0),(0,0,1) \} - \span\{(1,0,0),(0,1,0) \} -\span\{(1,0,0) \}\right) \cup (\span\{(1,0,0),(0,1,0) \} -\span\{(1,0,0)\})\\ \cup (\span\{(1,0,0)\})$$ So when we take the equivalence relation $\sim$ to get the projective space $\mathbb{KP}^2$, we get that $$\mathbb{KP}^2 = \left(\span\{(1,0,0),(0,1,0),(0,0,1) \} - \span\{(1,0,0),(0,1,0) \} -\span\{(1,0,0) \}\right)/\sim \\ \cup (\span\{(1,0,0),(0,1,0) \} -\span\{(1,0,0)\})/\sim \\ \cup (\span\{(1,0,0)\})/\sim$$ and ostensibly (hopefully) we have that $$\left[\left(\span\{(1,0,0),(0,1,0),(0,0,1) \} - \span\{(1,0,0),(0,1,0) \} -\span\{(1,0,0) \}\right)/\sim\right] \approx \mathbb{K}^2 $$ $$\left[(\span\{(1,0,0),(0,1,0) \} -\span\{(1,0,0)\})/\sim\right] \approx \mathbb{K} $$ $$\left[\left(\span\{(1,0,0) \}\right)/\sim\right] \approx \mathbb{K}^0 $$ The third is the "point at infinity" completing the projective "line at infinity" which is the union of the second and the third; then the whole projective space $\mathbb{KP}^2$ is $\mathbb{K}^2$ with the (projective) line at infinity attached to its "boundary", leading to a two-dimensional space.

Similar to (or exactly the same as?) the decomposition mentioned in https://math.stackexchange.com/a/172799/327486

My full overly long thought process: Of course, the above setup in my question ignores the underlying equivalence relation on the space. If one quotiented the plane $\operatorname{span}\{(1,0,0),(0,1,0) \}$ by setting as equivalent any two points with the same magnitude/distance from the origin (in other words $(a,b,0) \sim (c,d,0) \iff \sqrt{a^2 + b^2} = \sqrt{c^2 +d^2}$) I could see how the resulting space would be a one-dimensional manifold (in fact it would be homeomorphic to the unbounded interval $[0,\infty)$).

I suppose it would be simple to set-up a bijection between $\mathbb{RP}^1$/$\mathbb{CP^1}$ and the line at infinity; in fact, this might even be an inclusion, since we can compose the inclusion $\mathbb{K}^2 \hookrightarrow \mathbb{K}^3$ with the quotient map defining the projective space. And I do see how $\mathbb{RP}^1$ is one-dimensional; it's essentially an "infinite circle", which means that the $\mathbb{CP}^1$ should be an infinite sphere. My notes (Algebraic Geometry: A Problem Solving Approach) do confirm that $\mathbb{CP}^1$ is essentially the Riemann sphere, and Wikipedia also says that the complex line at infinity is a Riemann sphere: https://en.wikipedia.org/wiki/Line_at_infinity

I guess is the idea the following? Even if one does have the line at infinity is equal to $$\left[ \operatorname{span}\{(1,0,0),(0,1,0) \} \right] / \sim$$ where $\sim$ is the equivalence relation is the one defining projective space, this isn't two-dimensional, because $$\left[\operatorname{span}\{(1,0,0) \}\right] / \sim$$ is a point, i.e. 0-dimensional, so "adding" the span of $(0,1,0)$ to it only leads to a one-dimensional manifold, because all of the dimensionality comes from all of the different possible linear combinations of $(1,0,0)$ and $(0,1,0)$ which are not related by a scalar multiple, i.e. by parametrizing the set $\{(a,0,0)+(0,b,0) : a,b \in \mathbb{K}, a>0 \} / \sim$ (the equivalence classes of rays starting from the origin which are to the right of the vertical axis, i.e. the first and fourth quadrants) by the angle $(0,\pi)$ they make with the ray $\{0,b,0): b\ge 0\}$, and at least then in the case $\mathbb{K=\mathbb{R}}$ applying the cotangent function gives us back an unbounded object homeomorphic to $\mathbb{R}$, i.e. why $$\left(\left[ \operatorname{span}\{(1,0,0),(0,1,0) \} \right] / \sim\right) \setminus \left(\left[\operatorname{span}\{(1,0,0) \}\right] / \sim\right)$$ is homeomorphic to $\mathbb{R}$ and why then $\left[\operatorname{span}\{(1,0,0) \}\right] / \sim$ is the "point at infinity", and thus the line at infinity corresponds to $\mathbb{RP}^1$, which itself can be decomposed into a line and a point at infinity.

In other words, even though it may not necessarily be clear from its preimage under the quotient map, the "line at infinity" is a projective line "at infinity".

I can image easily analogous arguments for complex projective space, although it is not as easy for more to construct as explicit of a homeomorphism as I did above. (Update: I think the generalization to complex projective spaces could come from considering slopes rather than angles -- we can still define slope for complex lines even if not angles.)

Note: Here what the cotangent function is doing geometrically is essentially giving a one-to-one correspondence between the angle formed by a line and the $y$ axis and the slope of that line. Since characterizing slopes via their lines generalizes better to higher dimensions, it makes more sense to think of the projective coordinate on the line at infinity as corresponding to slope than it does to think of it as corresponding to angle.

Let's try to see how we can quotient out $\mathbb{R}^3$ to become $\mathbb{RP}^2$. "Cut out" the $yz$-plane and quotient it to become $\mathbb{RP}^1$ in the manner mentioned above -- i.e. consider the right half-plane and assign to each line its slope (which was the end result of my angle plus cotangent method above).

Now let's focus on the side of $\mathbb{R}^3$ to the right of the $yz$ plane (i.e. $\{x>0 \}$ ) and show that we can create a one-to-one correspondence between rays starting at the origin and $\mathbb{R}^2$, in a manner analogous to how quotienting out the right side of $\mathbb{R}^2$ gives something homeomorphic to $\mathbb{R}^1$.

For any such ray, we can consider its projections onto the $xy$ plane and the $xz$ plane. Then each of the two projections has a well-defined slope because $x>0$ (for the one in the $xy$ plane, the change in $y$ over the change in $x$ and not vice versa, and for the one in the $xz$ plane, the change in $z$ over the change in $x$ and not vice versa). Moreover, the pair of these two slope values characterizes the entire ray uniquely; thus we get a bijection between $\mathbb{R}^2$ and equivalence class of lines through the origin in $\mathbb{R}^3$ minus the $yz$ plane.

The key to this construction was removing the set $\{x=0 \}$, which is the $yz$ plane, everything else falls into line from this, allowing us to generalize this process to complex projective spaces. (Because the notion of slope also exists for complex lines.)

Thus, if we want to construct the decomposition $\mathbb{P}^n = \mathbb{A}^n \cup \mathbb{P}^{n-1}$, for a real or complex projective space, the recursive procedure is simply as follows: given a coordinate $i=1,\dots,n+1$, split $\mathbb{A}^{n+1}$ into the sets $\{x_i =0 \}$ and $\{x_i \not=0 \}$. Recursively, use this procedure to identify the quotient of $\{x_i =0 \}$ with $\mathbb{P}^{n-1}$. For each ray (i.e. equivalence class in $\{x_i \not=0 \}$), identify it with the $n-$ tuple of the slopes of its projections in the $x_1 x_i$ plane, in the $x_2 x_i$ plane, $\dots$, in the $x_{i-1} x_i$ plane, in the $x_{i+1}x_i$ plane, $\dots$, in the $x_{n-1}x_i$ plane, and the $x_n x_i$ plane, which characterizes the ray uniquely. (And since slopes also are defined for "lines" in $\mathbb{C}^2$ this also works for complex projective spaces.) Thus we can identify the quotient of $\{x_i \not=0 \}$ with $\mathbb{A}^n$. Therefore, projective spaces $\mathbb{P}^n$ can really be thought of as spaces of lines through the origin in $\mathbb{A}^{n+1}$, with their homogeneous coordinates identifying the slopes of these lines in various planes of $\mathbb{A}^{n+1}$.


There is another way to make it more obvious: Points on the real projective plane correspond to lines through the origin in ordinary space, and lines on the projective plane correspond to planes through the origin in ordinary space.

This can be understood as follows: Take first the affine plane, and embed it in space so that it does not contain the origin. Then obviously, for each point on the affine plane, there exists exactly one line connecting this point with the origin. Also for each line on the affine plane, there exists exactly one plane in space containing exactly that line and the origin.

Now looking at the reverse, you find that each line through the origin intersects the affine plane unless that line is parallel to that plane. However if you choose a line on the affine plane, and consider a point on that plane that goes to infinity, you'll find that the corresponding line through the origin approaches the line parallel to that line on the affine plane; therefore it makes sense to identify that parallel line with corresponding the point at infinity. Obviously all lines that are parallel in the affine plane are also parallel to the same line through the origin, so they all intersect in that point at infinity.

Now the lines through the origin that are parallel to the affine plane all lie in a single plane, the plane through the origin parallel to the affine plane. This is also the only plane that does not correspond to a line on the affine plane. Therefore obviously it is the line at infinity, since as plane through the origin it should correspond to a line, but it contains only points in infinity, and it contains all of them. Also, if you move a line on the affine plane to infinity, the corresponding plane through that origin approaches the parallel plane, and that is independent of the slope of the line.

Now zero-dimensional points on the plane correspond to one-dimensional lines through the origin and one-dimensional lines on the plane correspond to two-dimensional planes through the origin. The line at infinity corresponds to the two-dimensional plane parallel to the affine plane, and therefore is is one-dimensional.

Note that this construction is directly related to the homogeneous coordinates: The set of points on a line through the origin is exactly given by the points $(\alpha x, \alpha y, \alpha z):\alpha\in\mathbb R$ for some $(x,y,z)\ne (0,0,0)$. And in reverse, any such point other than the origin uniquely determines a line through the origin, and thus, a projective point. The origin is exactly the point on the line with $\alpha=0$, therefore you've got the condition $\alpha\ne 0$ in the equivalence relation of the harmonic coordinates.


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.
  • $\begingroup$ This is a really nice answer which constructs the bijection with $\mathbb{R}/\mathbb{C}$ in a very direct and clear fashion. Just a followup -- for the points of the form $(r:1:0)$, can we think of $r$ as the slope of the line? And then $(1:0:0)$ is the "point at infinity" because it corresponds to vertical lines whose slope is undefined? $\endgroup$ – Chill2Macht Sep 4 '16 at 21:31
  • 1
    $\begingroup$ @William - Yes, although it might make more sense to think of the points of the form $(1:r:0)$ as corresponding to line with slope $r$, and then $(0:1:0)$ corresponds to vertical lines. $\endgroup$ – mweiss Sep 5 '16 at 1:18

In projective geometries, two lines have one intersection point. If two lines already intersect in the finite part of the plane, they can't also have a common point at infinity. So outside of rather trivial geometries, infinity must have several points.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.