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Is it true that there is a bijection between planes through origin in $\mathbf R^3$ and the projective plane $P^2 (\mathbf R)$? It is a remark in a lecture note I am reading. I tried to prove it but it seems to be a mistake. My idea was to intersect such plane with disc at origin but then many planes map to same equivalence class in $P^2 (\mathbf R)$.

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    $\begingroup$ Sure, they're both the cardinality of $\mathbb{R}$. (I get annoyed when people say "there is a bijection" instead of "this particular important map I'm thinking of is a bijection, and also it has a bunch of other properties I don't want to discuss right now.") $\endgroup$ Feb 28, 2013 at 18:12

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This is certainly true, and in fact more is true.

$P^2(\mathbb R)$ is the set of lines through the origin in $\mathbb R^3$. Now, think about the planes through the origin in $\mathbb R^3$. All planes are completely characterized by a normal vector (which is nonzero). This normal vector spans a subspace of dimension 1. Can you see the bijection now?

For fun, think about this. $G(k,n)$ is the Grassmannian of $k$-planes through the origin in a vector space $V$ of dimension $n$. Can you find a general duality such as the one you came across, but between $G(k,n)$ and and $G(n-k,n)$?

Hint: Think about $V^*$ the dual vector space

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    $\begingroup$ Great answer. Let me just add that the correspondence suggested here between $G(k,n)$ and $G(n-k,n)$ is not only a bijection, but can be made into a diffeomorphism of smooth manifolds. $\endgroup$ Feb 28, 2013 at 15:35
  • $\begingroup$ Many thank you, your answer helped me greatly. To answer question about Grassmanian: if $G(k,n)$ is $k$-dimensional plane in $V^n$ then map the $p \in G(k,n)$ to the normal $n$ of $p$? $\endgroup$
    – goobie
    Mar 4, 2013 at 14:24
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A plane $P$ in $\mathbb{R}^3$ passing through $(0,0,0)$ is given by $ax+by+cz=0$. Now, let $\varphi : (a,b,c) \mapsto P$ from $P^2\mathbb{R}$. You can show that $\varphi$ is well defined and bijective.

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Planes in $\mathbb{R}^3$ through the origin are in one-to-one correspondence with lines in $\mathbb{R}^3$ through $0$. The correspondence is given by taking normal directions. The lines in $\mathbb{R}^3$ through $0$ are exactly $P^2(\mathbb{R})$.

(Just for fun, to see how this links with the answer by Seirios, note that if the equation of a plane is $ax + by + cz = 0$, then $(a,b,c)$ is a normal direction to the plane. Two equations $ax + by + cz = 0$ and $a'x + b'y + c'z = 0$ describe the same plane if and only if $[a:b:c] = [a':b':c']$ in $P^2(\mathbb{R})$.)

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