This is a little question inspired from Hartshorne's Geometry, which I've been juggling around for a while.

Suppose that $\Pi$ is the Cartesian plane $F^2$ for some field $F$, with the set of ordered pairs of elements of $F$ being the points and lines those subsets defined by linear equations. Let $\Pi'$ be the associated projective plane. Here $\Pi'$ is just $\Pi$ together with the points at infinity, which are the pencils of sets of parallel lines. A line in $\Pi'$ will be the subset consisting of a line of $\Pi$ plus its unique point at infinity.

Also, let $V=F^3$ be a three-dimensional vector space over $F$. Let $\Sigma$ be the set of $1$-dimensional subspaces of $V$, and call them points. If $W\subseteq V$ is a $2$-dimensional subspace of $V$, then the set of all 'points' contained in $W$ will be called a 'line,' and $\Sigma$ forms a projective plane.

I would like to see how $\Pi'$ and $\Sigma$ are isomorphic. I figure I should construct a bijection from the points of $\Pi'$ to the points of $\Sigma$, however, I don't see a natural mapping. At first I figured if $(a,b)\in\Pi'$, then maybe $(a,b)\mapsto\text{span}(a,b,0)$, but this isn't injective or surjective. Not only that, but I figure if $A$ and $B$ are points in $\Pi'$ on a line, then the images of all points in that line in $\Pi'$ should be in the same $2$-dimensional subspace in $\Sigma$.

I don't quite see why they should be isomorphic. Suppose to lines $l$ and $m$ intersect in $\Pi'$, shouldn't their intersection point map to the intersection of the lines in $\Sigma$? But what if this intersection doesn't even pass through the origin, and hence isn't even a $1$-dimensional subspace? Thank you for any insight on how to solve this.

Edit: For my last question above, I just realized that all the $2$-dimensional subspaces pass through the origin (whoops!), and thus the intersection will contain the origin as well, so that's not an issue.


Maybe picking $(a,b) \mapsto \text{span}(a,b,1)$ would be a better choice.

This map seem awfully arbitrary and unmotivated, and I find it easier to explain how there can be an isomorphism if I go the other way : start from $\Sigma$ and go to $\Pi'$ :

In order to see the points of $\Sigma$ as points on a normal plane, pick a 2-dimensional affine subspace $P$ of $V$ that doesn't contain the origin and put a coordinate system on it so that we can call it $\Pi$ (for example the plane z=1 if you want to compare with the map I gave).

You can construct an isomorphism as follows :

Pick $W$ a line in $\Sigma$ (a 2-dimensional linear subspace of V). If $W$ intersects $P$ on a line $w$ of $P$, we want to take $f(W)$ = the line of $\Pi'$ containing $w$. So in $\Pi'$, $f(W) = w$ plus the pencil of parallel lines that contains $w$.

Pick $A$ a point if $\Sigma$ (a 1-dimensional linear subspace of V). If $A$ intersects $P$ on a point $a$ of $P$, we define $f(A) = a$. It is consistent with our choice of images of lines, because if a $W$ contains $A$ then $W \cap P$ contains $A \cap P$.

If $A$ is parallel to $P$, consider a line $W$ containing $A$ and intersecting $P$. $f(W) = $ the intersection of $P$ and $W$ = a line of $P$ "parallel" to $A$. Thus our only choice is to pick $f(A) =$ the point lying at the intersection of those $f(W)$ = the pencil of parallel lines that are the lines "parallel" to $A$.

Finally, if $W$ is the 2-dimensional subspace "parallel" to $P$, we have to chose $f(W) =$ the reunion for the points $A$ in $W$ of $f(A)$ = the line of $\Pi'$ whose points are the pencils of parallel lines, that is, the "line at infinity".

  • $\begingroup$ Aargh, you beat me by 10 seconds :-). $\endgroup$ – t.b. Feb 4 '11 at 10:30
  • $\begingroup$ Thanks chandok, this was a helpful walk through, much appreciated! $\endgroup$ – yunone Feb 4 '11 at 22:11

This is too long for a comment [I hope that this time I'm not using anything I'm not allowed to use ;-)]

I would start by identifying $\Pi$ with the affine plane $(x,y,1) \subset F^{3}$. Obviously, every $1$-dimensional linear subspace of $F^{3}$ that doesn't lie in the $(x,y)$-plane intersects this affine plane in a unique point and every $2$-dimensional linear subspace that isn't the $(x,y)$-plane intersects this affine plane in a $1$-dimensional affine subspace. Conversely, every point in $\Pi$ determines a $1$-dimensional linear subspace of $F^{3}$ and every line in $\Pi$ determines a $2$-dimensional linear subspace of $F^{3}$. Now it should be obvious what happens with the points at infinity: they determine (and are determined by) a $1$-dimensional linear subspace $\ell$ of the $(x,y)$-plane and they themselves lie on the line at infinity in $\Pi'$ (which you've forgotten in your description of lines of $\Pi'$).

This construction extends of course to higher dimensions. Hope that helps.

  • $\begingroup$ Many thanks Theo! So for example, if a point in $F^3$ intersects the affine plane at $(a,b,1)$, then we identify it with the point $(a,b)$ in $F^2$. And if a line in $F^3$ intersects the affine plane on $[x,x,1]$, where $x$ ranges over all values, we identify it with with the line $x=y$ in $F^2$? Then for the points in $F^3$ that lie in $z=0$, we identify with the pencil of lines that essentially lie in the same way in $F^2$? And this whole line $z=0$ in $F^3$ is identified with the line at infinity in $\Pi'$? Just clarifying, thanks. $\endgroup$ – yunone Feb 4 '11 at 22:15
  • $\begingroup$ @yunone: I'm not sure I understand your question. The affine plane parameterizes all $1$ and $2$-dimensional subspaces of $F^3$ that aren't parallel to it (via points and lines in $\Pi$). The stuff at infinity happens in the $(x,y)$-plane where the 'directions' are given by the $1$-dim subspaces and the plane itself is the line at infinity of $\Pi'$. $\endgroup$ – t.b. Feb 4 '11 at 23:42
  • $\begingroup$ ah ok, thanks, I see it now. $\endgroup$ – yunone Feb 5 '11 at 0:19

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