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For example in three dimensional projective space. I thought lines in projective space corresponded to planes in euclidian space. So the space is just the dimension of the Grassmanian $G(2,3)$. Which is $3$. But in a book I'm reading it is said that the dimension must be $4$. What did I do wrong?

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You are only counting lines through some fixed point (because that Grassmannian is only counting planes through the origin). – user29743 Nov 13 '12 at 21:16
up vote 2 down vote accepted

You are right that lines in $\mathbb{P}^3$ correspond to planes through the origin in an euclidian space. This is because $\mathbb{P}^3$ is a quotient space of such an euclidian space.

However, remember that $\mathbb{P}^3$ is a quotient of an euclidian space of one dimension higher, i.e. a 4-dimensional euclidian space in this case.

Hence the grassmanian you are looking for is $G(2,4)$ instead of $G(2,3)$.

In general lines in $\mathbb{P}^n$ correspond to $G(2,n+1)$.

Even more general linear spaces of dimension $l$ in $\mathbb{P}^n$ correspond to $G(l+1,n+1)$.

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$G(2,4)$ has dimension $6$. I quote Miles Reid in Undergraduate Algebraic Geometry, page 102.: "lines of $P^3$ are parametrised by a 4-dimensional variety,". – mayonnaise Nov 13 '12 at 21:31
I believe that $G(2,4)$ is actually a subset of $\mathbb{P}^5$ given as the zeroes of $x_0x_1 - x_2x_3 + x_4x_5$. Hence the dimension is 4. – Joachim Nov 13 '12 at 21:32
You are right, I thought the whole time that the dimension of $G(2,4)$ is just 4 choose 2, which is wrong, it is 4 by the Plucker embedding, right? – mayonnaise Nov 13 '12 at 21:37
The polynomial i gave you is the result of the plucker embedding: realizing the grassmanian as an algebraic subspace of a projective space. So yes. – Joachim Nov 13 '12 at 21:39
For the record: $\dim G(k,n)=k(n-k)$. – Fran Burstall Oct 7 '13 at 20:43

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