In my algebraic geometry course, we study projective spaces. We jumped right in with the definition and then gave some examples. I think I understand the definition itself, but I'm really struggling to get why it matters at all.

The definition we have is the standard one: The n-dimensional projective space over a field K, $\mathbb{P}^n_k$, is the set of all one-dimensional subspaces of $K^{n+1}$.

I get that that as a set, $\mathbb{P}^n_k$ is just all the lines through the origin in the vector space $K^{n+1}$. We considered this specific example: $\mathbb{P}^1_\mathbb{R}$. This is the set of all lines in $\mathbb{R}^2$ through the origin. However, we 'identified' this set with the circle; we showed that every line through the origin intersects the top half of a hemisphere in $\mathbb{R}^2$, except for the horizontal line, which intersects two antipodal points. So we say by identifying these points together, the space is essentially, the circle.

My confusion is the following: In the example of $\mathbb{P}^1_\mathbb{R}$, we say it's a circle. The only reason I see we did this, is because there was a bijection of lines through the origin, to points (antipodal points) on the circle.

But I don't get why we say this, and what this circle represents? For example, if by 'identify with' we mean merely 'is in bijection with', I can identify $\mathbb{P}^1_\mathbb{R}$ with any set of equal cardinality. But why is this useful? For example, in algebra, we're not typically interested in bijections, if they don't also preserve the algebraic structure. A bijection that is not an isomorphism is generally not as enlightening as a bijection that respects the ring structure for example.

So in the case of projective spaces, what is the 'geometric structure' that $\mathbb{P}^1_\mathbb{R}$ comes endowed with? As far as I can tell, the $\mathbb{P}^1_\mathbb{R}$ is just a set of equivalence classes that can be put in bijection with a circle; but why is this interesting? Am I completely misunderstanding?

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    $\begingroup$ Please read Wikipedia projective geometry for some answers. $\endgroup$
    – Somos
    Jul 22 '21 at 19:44
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    $\begingroup$ The real projective line is really just the usual affine line with a point „added at infinity“. $\endgroup$
    – Qi Zhu
    Jul 23 '21 at 5:17
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    $\begingroup$ I think of it as an ambient space that is ‘complete’, in the sense that it contains infinity. In Algebraic Geometry, its geometric structure comes from the Zariski topology, mirroring the structure of homogeneous prime ideals in the ring $k[x_0, \dots , x_n]$. $\endgroup$
    – Daniel
    Aug 9 '21 at 1:12

This is not an answer to your question, but a motivation of why this is interesting and important. You can also read the introduction of the book "Rational Points on Elliptic Curves" by Silverman and Tate, which I highly recommend. It is a nice mix between math book and novel, if I am allowed to judge.

I want to explore on the example of rational points on conics why these projective lines and bijections are interesting to us. As I said, I basically shorten the introduction of the named book.

So suppose we are interested in rational solutions to some conic. A conic is something that looks like this: $ax^2+bxy+cy^2+dx+ey+f=0$.

For example $x^2+y^2-2=0$ (a conic with rational coefficients, or geometrically a circle). One can easily see four solutions, namely combinations of $(\pm 1,\pm 1)$. A natural question is if there are more, and if there are how do we find them?

This is where a projective line comes in very handy.

When we take a rational line, what happens if we intersect the line with the conic? In general the intersection does not have to be rational.

But if we take a rational point on the conic $x^2+y^2-2=0$ for example, and some line, where we now project the rational point onto this line, the second intersection will also be a rational point!

enter image description here

This is true because when you solve a quadratic equation either both solutions are rational, or both solutions are irrational (and one solution is known to be rational).

So a though algebraic question "Are there more rational solutions for a conic equation", becomes an easy geometric question. So there you have a reason why to care about such a bijection.

The rational points on a line correspond in bijection to the rational points on such a conic as $x^2+y^2-2=0$, and that is nothing more then amazing in my opinion.

Keep in mind, that we have to know at least one rational point to perform this. But once you have one, you can find every single one, by just projecting the conic on a line.

I hope this is helpful to you.

[The picture is of course taken from the named book.]

  • $\begingroup$ "The rational points on the line correspond in bijection..." - this is clear for any curve with infinitely many rational points - $\mathbb{Q}$ is countable, and the set of rational points on a curve has cardinality less that of the field. The thing which is notable is the fact that the bijection is given by a birational morphism of varieties. $\endgroup$ Jul 23 '21 at 9:33

In case anyone ever comes across the same question as me; I hope this answer will be useful.

I came across an example from a different course on smooth manifolds discussing the real projective space.

We proved why it is a smooth manifold and that $\mathbb{R}P^n$ and $S^1$ are homeomorphic (and it generalises to higher dimensions). In other words, the structure I was looking for actually given by the topological structure, and the argument presented in the algebraic geometry course was essentially saying this (by providing the bijection and the equivalence class of identifying antipodal points together).


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