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I followed a course in projective geometry and I'm not sure about 2 things:

  1. If I have 6 lines in projective space (IP³) with commun secant, why are the 6 corresponding tensors linearly dependent?
  2. Why do 4 lines in complex projective space always have a line which intersects all 4 lines?

Thanks in advance

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    $\begingroup$ For 2, you have the result that 3 general lines in $\mathbb P^3$ determine a smooth quadric hypersurface, and by definition a quadric has degree $2$, i.e. it meets a general line (the fourth!) in two points. Hence there are two lines meeting four general lines in $\mathbb P^3$. $\endgroup$ – Jack Jun 27 '13 at 12:56
  • $\begingroup$ When you talk about the associated tensors, are you thinking of the decomposable $2$-forms, i.e., $G(2,4)\hookrightarrow \mathbb P(\Lambda^2 \mathbb C^4)$? $\endgroup$ – Ted Shifrin Jun 27 '13 at 13:34
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If I am correctly interpreting your first question, here's one way to look at it. Every line $\ell$ in $\mathbb P^3$ is given by its Plücker coordinates. That is, if the two points $P=\sum_{j=0}^3 s_je_j$ and $Q=\sum_{j=0}^3 t_je_j$ determine our line $\ell$, then it is given by the coordinates of $[P\wedge Q]\in\mathbb P(\Lambda^2\mathbb C^4)$. We can think of homogeneous coordinates $[x_{01},x_{02},x_{03},x_{12},x_{13},x_{23}]\in\mathbb P^5$.

Suppose $\ell$ intersects the line $x_0=x_1=0$. Thus, we can choose $P$ in this line and so $P= s_2e_2+s_3e_3$ for some $s_2,s_3$, not both $0$. This means that in $P\wedge Q$ the coefficient of $e_0\wedge e_1$ must equal $0$, and so the $01$-coordinate of $\ell$ in our $\mathbb P^5$ must be $0$. If all six of your lines have this property, that means that they all lie in the hyperplane $\{x_{01}=0\}\subset\mathbb P^5$.

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