Lines on grassmannian Given a projective space $\mathbb{P}^n(\mathbb{C})$, I can consider the Grasmannian of lines $G(2,n+1)$, which has a structure  of projective variety inside $\mathbb{P}^N$, where $N=\binom{2+n+1}{2}-1$ thanks to the Pl$\ddot{u}$cker embedding.
It has been told me -literally, I have no references, it was a speech- that while a  point $p\in G(2,n+1)$ obviously represents a line in $\mathbb{P}^n$ by definition, a line inside $G(2,n+1)$ -viewd as a projective variety in $\mathbb{P}^N$-corresponds to a pencil of planes.
Unfortunately I'm still trying to properly understand why this work -because unfortunately it is not crystal for me -, but I was also wondering if it makes sense -that is, if there is a geometrical meaning-, also for quadrics contained in $G(2,n+1)$.
Therefore I'd like to have an idea of what happens for quadrics, and moreover even a little help for lines in $G(2,n+1)$.
Since this question comes primarily from my curiosity, I apologize for the vagueness: any comment, reference or answer would be much appreciate!
 A: You have that lines in $\mathbb{P}^{n}(\mathbb{C})$ correspond to points of the Grassmannian $G(2,n+1)$. Now when two lines $\ell_{1}$, $\ell_{2}$ of the projective space intersect each other at a point $x$, you can check the Plücker coordinates and verify that the corresponding points of the Grassmannian span a line that is completely contained in the Grassmannian. Furthermore, under the inverse map, all of the points on this line of the Grassmannian correspond to lines in $\mathbb{P}^{n}(\mathbb{C})$ lying in a common plane (you should verify this using the Plücker map).
So what you actually should be saying is that a line completely contained in $G(2,n+1)$ corresponds to a pencil of lines in $\mathbb{P}^{n}(\mathbb{C})$, that is, the collection of all lines through a fixed point $x$ lying in a common plane.
These intersection properties carry over to other kinds of quadrics, though I am not an expert at all of the correspondences. I do know that for example a conic in $G(2,4)$ will correspond to one of the rulings of a hyperbolic quadric in $\mathbb{P}^{3}(\mathbb{C})$.
A: I don't know if I'm understanding your question right, but here are some ideas. The Grassmanian $G(1,n)$ can be seen as the projective $n-1$ dimensional space, because lines are seen as points. Then a line in projective space comes from a plane in $n$-dimensional space, perhaps that's what they meant? That plane can be understood as a pencil of lines.
What is unclear to me is your claim that the embeding is in $\mathbb{P}^N$ with $N={n+3\choose 2}-1$. The embedding should go $i: G(1,n)\rightarrow \mathbb{P}(\Lambda^1 k^n)\simeq \mathbb{P}^{n-1}$, so the Plücker embedding would be just the isomorphism of the Grassmanian $G(1,n)$ with $\mathbb{P}^{n-1}$.
A: End up here with the same question, but I could not understand the answers very well. I'll let a proposal of an answer out here, and see what's up.
Plucker relations for $k=2$: An element $\omega \in \bigwedge^2 V$ is decomposable, meaning of type $v \wedge w$ (in the image of the Plucker embedding $G(2,V) \hookrightarrow \mathbb{P}(\bigwedge^2 V)$, $\langle v,w \rangle \mapsto [v \wedge w]$) iff $\omega \wedge \omega = 0$.
Now, a line in $\mathbb{P}(\bigwedge^2 V)$ is a 2-plane through the origin in $\bigwedge^2 V$. Take the image of two non-trivial Grassmannians $\alpha \doteq a_{1} \wedge a_{2}$ and $\beta \doteq b_{1} \wedge b_{2}$, not a scalar multiple one of each other. The projective line $\mathfrak{L}$, given by $[\lambda \alpha + (1-\lambda) \beta]\subset \mathbb{P}(\bigwedge^2 V)$ is inside the image of the Plucker embedding iff
$$ (\lambda \alpha + (1-\lambda) \beta) \wedge (\lambda \alpha + (1-\lambda) \beta)  = 0 $$
Notice that $\alpha \wedge \alpha = \beta \wedge \beta = 0$ and you need an even number of permutations to transform $\beta \wedge \alpha$ into $\alpha \wedge \beta$. We conclude $\alpha \wedge \beta = 0$. But then, we have one of the following:

*

*some $a_i \in \mathrm{span}(b_1,b_2)$ or

*some $b_i \in \mathrm{span}(a_1,a_2)$,

but then this vector represents a common vector $v \in \mathrm{span}(a_1,a_2) \cap \mathrm{span}(b_1,b_2)$.
Now the point $[v] \in \mathbb{P}V$ is common to both projective lines $\ell_1 = [\mathrm{span}(a_1,a_2)]$ and $\ell_2 = [\mathrm{span}(b_1,b_2)]$, and all of the lines in $\mathfrak{L}$ can be considered to be, w.l.o.g., inside the projective plane $\pi \doteq [\mathrm{span}(a_1,a_2,b_1)]$ (when $b_2$ is the L.D. one).
