Lines on the Plucker quadric

Consider the Grassmanian of projective lines in $\mathbb P^3$, namely $\mathbb G(1,3)$ with its Plucker embedding in $\mathbb P^5$. Now let $p$ be a point in $\mathbb P^3$ and $H$ be a hyperplane in $\mathbb P^3$ such that $p\in H.$ Consider the set $\Sigma_{p,H}$ of $l\in \mathbb G(1,3)$ such that $p\in l\subset H$. Show that $\Sigma_{p,H}$ is a line contained in $\mathbb G(1,3)$. Conversely, show that any line in $\mathbb P^5$ lying on $\mathbb G(1,3)$ appears in this way.

Now, using the action of $\mathrm{PGL}(4)$, I can prove the first part (one simply checks the case when $p=[1:0:0:0]$ and $H: x_3=0$), however I have no ideas for the converse. I have tried parametrizations of lines, but this leads me nowhere so far.

• Is it exercice 6.5 page 67 of ''Algebraic Geometry: A First Course'' by Joe Harris ? – Jean Marie Sep 23 '16 at 19:50
• You are right, it is from there I think. – user223794 Sep 23 '16 at 19:55
• Hint: think how the tautological bundles restrict to the line. – Sasha Sep 23 '16 at 22:37

Maybe one can do this using the more conceptual construction of wedge product and exterior algebra. Consider $\mathbb{C}^4$ and as usual $\mathbb{P}^3 = \big\{[u] \, : \, u \in \mathbb{C}^4 \big\}$ with $[u] = \big\{\lambda \, u \, \, : \,\, \lambda \in \mathbb{C}\setminus\{0\} \, \big\}$. Projective lines in the projective three space $\mathbb{P}^3$ correspond to the projectivisation of two-dimensional vector subspaces in $\mathbb{C}^4$. If I fix a two-dimensional subspace in $\mathbb{C}^4$ I can define it as the span of two linearly independent vectors $u,v \in \mathbb{C}^4$ lying on it and I can form the wedge product $q = u \wedge v \, \in \, \wedge^{2} \,\mathbb{C}^4 = \mathbb{C}^4 \wedge \mathbb{C}^4$. Moreover, if $u_1, v_1$ are two different vectors spanning the same subspace as $u, v$, then there is $\lambda \in \mathbb{C}$ such that $u_1\wedge v_1 = \lambda (u \wedge v)$. Thus prompts us to projectivize the wedge product: $$\mathbb{P}(\wedge^{2} \,\mathbb{C}^4 ) = \big\{ [\omega] \,\, : \,\, \omega \in \wedge^2 \, \mathbb{C}^4\big\}$$ $$[\omega] = \big\{\lambda \, \omega \in \wedge^{2} \,\mathbb{C}^4 \,\, : \,\, \lambda \in \mathbb{C}\setminus\{0\}\big\}$$ and the elements that describe the lines in $\mathbb{P}^3$, which is the same as the two dimensional subspaces of $\mathbb{C}^4$, have the form $[u \wedge v]$. However, not all elements of $\mathbb{P}(\wedge^{2} \,\mathbb{C}^4 )$ are of this form. How to distinguish the ones that define lines from the one that do not, i.e. when is an element of $\wedge^{2} \,\mathbb{C}^4$ of the form $u \wedge v$? As it turns out, $\omega \in \wedge^{2} \,\mathbb{C}^4$ has the form $\omega = u \wedge v$ if and only if $\omega \wedge \omega = 0$, where $\omega \wedge \omega \in \wedge^{4} \,\mathbb{C}^4$. Since there exists an isomorphism (non-canonical) $\phi : \wedge^{4} \,\mathbb{C}^4 \to \mathbb{C}$, we can define the complex bilinear dot product (nondegenerate bilinear form) $(\omega \cdot \sigma) = \phi(\omega \wedge \sigma)$ for $\omega, \sigma \in \wedge^2 \, \mathbb{C}^4$. Then, this dot product has two very important properties:

1. $(\omega \cdot \omega) = \phi(\omega \wedge \omega) = 0$ if and only if $\omega \wedge \omega = 0$ if and only if $\omega=u \wedge v$;

2. If two lines in $\mathbb{P}^3$ intersect at a common point, then their corresponding two two-dimensional subspaces of $\mathbb{C}^4$ intersect at a common one dimensional subspace, spanned by a vector $v_0$, so the two lines will be represented by two wedge products $q_1 = u_1 \wedge v_0$ and $q_2 = u_2 \wedge v_0$ and so $(q_1 \cdot q_2) = \phi(q_1 \wedge q_2) = \phi(0) = 0$. The converse is also true, if two wedge products $q_1 = u_1 \wedge v_1$ and $q_2 = u_2 \wedge v_2$ are such that $(q_1 \cdot q_2) = \phi(q_1 \wedge q_2) = \phi(0) = 0$, then there is at least one vector $v_0 \in \mathbb{C}^4$ such that $q_1 = \lambda_1 (w_1 \wedge v_0)$ and $q_2 = \lambda _2 (w_2 \wedge v_0)$.

With this information at hand, define the quadric $$\hat{Q} = \big\{\omega \in \wedge^2 \, \mathbb{C}^4 \,\, : \,\, (\omega \cdot \omega) = 0 \big\}.$$ Then it's projectivization $$Q =\big\{\, [\omega] \in \mathbb{P}(\wedge^2 \, \mathbb{C}^4) \,\, : \,\, \omega \in \hat{Q} \, \big\} = \big\{\, [\omega] \in \mathbb{P}(\wedge^2 \, \mathbb{C}^4) \,\, : \,\, (\omega \cdot \omega) = 0 \, \big\}.$$ So thus $\mathbb{G}(1,3) \cong Q$. If you have a plane $H$ and a point on it $p \in H$ one can pick two different lines $l_1$ and $l_2$ lying on $H$ and passing through $p$. Then $H$ is spanned by these two lines. Let $l_1 = [u_1 \wedge v_0]$ and $l_2 = [u_2 \wedge v_0]$, where $p = [v_0]$. Denote $\hat{l}_1 = u_1 \wedge v_0$ and $\hat{l}_2 = u_2 \wedge v_0$. Then any other line on $H$ passing through $p$ should look like $$l = [\lambda_1 \hat{l}_1 + \lambda_2 \hat{l}_2] = [\lambda_1 u_1 \wedge v_0 + \lambda_2 u_2 \wedge v_0] = [(\lambda_1 u_1 + \lambda_2 u_2) \wedge v_0]$$ for any $\lambda_1, \lambda_2 \in \mathbb{C}$. Thus the set $\Sigma_{p,H}$ of all lines lying on $H$ and passing through $p$ is the projectivisation of a two-dimensional subspace of $\wedge^2 \, \mathbb{C}^2$ so it is a one-dimensional line on $\mathbb{P}(\wedge^2 \, \mathbb{C}^2)$ composed of elements $l = [\lambda_1 \hat{l}_1 + \lambda_2 \hat{l}_2] \in \Sigma_{p,H}$ with the property that $$\big((\lambda_1 \hat{l}_1 + \lambda_2 \hat{l}_2)\cdot (\lambda_1 \hat{l}_1 + \lambda_2 \hat{l}_2) \big) = 0$$ which means that $l \in Q \cong \mathbb{G}(1,3)$ and hence $\Sigma_{p,H} \subset Q \cong \mathbb{G}(1,3)$. Observe that $\dim_{\mathbb{C}} \big(\wedge^2 \, \mathbb{C}^4\big) = 6$, so $\wedge^2 \, \mathbb{C}^4 \cong \mathbb{C}^6$ and so $\mathbb{P}(\wedge^2 \, \mathbb{C}^4) \cong \mathbb{P}^5$ and thus $Q = \mathbb{G}(1,3)$ is a four dimensional projective quadric in $\mathbb{P}^5$.

Edit. The converse. Let $\Sigma$ be a line lying on the conic $Q$. Then there are two different points $[l_1]$ and $[l_2]$ lying on $\Sigma$, where $l_i = u_i\wedge v_i$ for $i=1,2$. So these two points span $\Sigma$, i.e. $[l] \in \Sigma$ if and only if $[l] = [\lambda \, l_1 + \lambda_2 \, l_2] \, \in \, \Sigma \, \subset \, Q$ for any complex numbers $\lambda_1, \lambda_2 \in \mathbb{C}$ not simultaneously zero. Then since $[l] \in \Sigma$ and $\Sigma \subset Q$, $$0 = (l \cdot l) = \big((\lambda \, l_1 + \lambda_2 \, l_2)\cdot(\lambda \, l_1 + \lambda_2 \, l_2)\big) = \lambda_1^2 (l_1 \cdot l_1) + 2\lambda_1 \lambda_2 (l_1 \cdot l_2) + \lambda_2^2 (l_2 \cdot l_2) = 2\lambda_1 \lambda_2 (l_1 \cdot l_2)$$ because $(l_1\cdot l_1) = (l_2 \cdot l_2) = 0$ as $[l_1], [l_2]$ are points lying on the quadric $Q$. Hence $(l_1 \cdot l_2) = 0$. But this means that $\phi(l_1 \wedge l_2)=0$ i.e. $u_1 \wedge v_1 \wedge u_2 \wedge v_2 = 0 \in \wedge^4 \, \mathbb{C}^4$. The latter statement means that the four vectors $u_1, v_1, u_2, v_2 \in \mathbb{C}^4$ cannot span the whole space and are linearly dependent. Hence there is a vector $v_0 \in \mathbb{C}^4$ such that $$\text{span}\{u_1, v_1\} \cap \text{span}\{u_2, v_2\} = \text{span}\{v_0\}$$ Hence $l_1 = [u_1 \wedge v_0]$ and $l_2 = [u_2 \wedge v_0]$ and $[\lambda_1 \, l_1 + \lambda_2\, l_2] = [\,(\lambda_1 v_1 + \lambda_2 u_2) \wedge v_0]$ represents the two-space $\text{span}\{\lambda_1 v_1 + \lambda_2 u_2, v_0\}$ in $\mathbb{C}^4$ which means that all the lines from $\Sigma$ pass through the same point $p = [v_0] \in \mathbb{P}^3$ all of them are contained in the plane $$H = \mathbb{P}\big(\text{span}\{u_1, u_2, v_0\}\big) \, \subset \, \mathbb{P}^3$$

• I think this only shows the forward direction, and not the converse as it was asked. – user223794 Sep 30 '16 at 10:43
• @user223794 The converse can be derived from the construction. See my edit. – Futurologist Sep 30 '16 at 12:52
• Now it's all good. Great proof! – user223794 Sep 30 '16 at 13:24