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I read in an article that the collection of lines going through a point in RP3 form a projective plane. I can't understand why . we know that a point in RP3 is a line going through the origin of the vector space R4 and a line going through this point is actually a plane containing the line and going through origin. so how can we think of the collection of lines going through a point in RP3 as a projective plane?

david
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  • Let us imagine the ordinary plane: a line going through a point could be parameterised by only one parametre (the angle) in the ordinary plane. And such a line could be parameterised by two parametres in 3-dimensions. Now, adding to each term "projective", it yields the result. Of course this is intuitive, and is not a real proof. – awllower Feb 03 '13 at 06:55

2 Answers2

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You can embed $\mathbb R^3$ into $\mathbb R\mathrm P^3$. In this case, every affine line becomes a projective line. All the lines passing through a given point become projective lines, which means they gain one additional point but otherwise stay the same set of lines.

You can also embed $\mathbb R\mathrm P^2$ into $\mathbb R^3$. All the lines passing through the origin of $\mathbb R^3$ represent points of $\mathbb R\mathrm P^2$. Your question indicated that you have a good understanding of this.

Now you can combine these embeddings:

  1. You start with a given point in $\mathbb R\mathrm P^3$. For the sake of simplicity, you apply a projective transformation (or a change of basis, if you prefer to not move “actual objects” around) which change the coordinates of that point to $(0,0,0,1)$, i.e. the “origin” of $\mathbb R\mathrm P^3$. The set of lines through that point is a set of projective lines which meet in that point.
  2. Next you strip away infinity. Every line looses one point, but otherwise stays the same line. All your finite points get dehomogenized, so your common point now has coordinates $(0,0,0)$. You end up describing all one-dimensionsl subspaces of $\mathbb R^3$.
  3. Then you interpret this $\mathbb R^3$ as $\mathbb R\mathrm P^2$. Every linear subspace represents a point. You can intersect these with a drawing plane not passing through the origin.

So each line through the point in the original 3-space gets mapped to a point in the projective plane. The mapping is bijective.

MvG
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  • I am sorry, but I don't think there is an embedding from $\mathbb{R}\mathbb{P}^2$ to $\mathbb{R}^3$, you can only immerse it (the image $i(\mathbb{R}\mathbb{P}^2) \subset \mathbb{R}^3$ of such an immersion will always have some self-intersections. – Willem Noorduin Feb 04 '13 at 16:23
  • @WillemNoorduin: You cannot immerse the points of $\mathbb{RP}^2$ as a surface of points in $\mathbb R^3$. But you can use lines through the origin of $\mathbb R^3$ as representants of points in $\mathbb{RP}^2$. That's the common construction of equivalence classes: $$\mathbb{RP}^2=\frac{\mathbb R^3\setminus\left{\begin{pmatrix}0\0\0\end{pmatrix}\right}}{\mathbb R\setminus{0}}$$ – MvG Feb 04 '13 at 16:37
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Fix a point $P\in\Bbb P^3$ and a plane $\pi\subset\Bbb P^3$ such that $P\notin\Bbb \pi$. Also, let $\cal L$ be the set of lines through $P$.

Then there is a bijection $$ \cal L\longleftrightarrow\pi $$ as follows. To each $L\in\cal L$ associate the unique point $P_L\in\pi$ given by $P_L=L\cap\pi$. In the other direction, for each $Q\in\pi$ consider the unique $L\in\cal L$ such that $\{P,Q\}\subset L$. The two constructions are inverse of each other, obviously.

But $\pi$ is a $\Bbb P^2$, so you're basically done.

daltonb
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Andrea Mori
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