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I am currently studying projective geometry. I have trouble understanding the point-line duality concept.

Why is the cross product of two points a line and the cross product of two lines a point?

Kevin Wu
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  • Since points are lines (in the underlying vector space), the cross product of two points in the p.s. defines a normal on a plane (namely the plane spanned by the two lines (v.s) which correspond to the two points (p.s.)) and thus a plane in the v.s., which is a line in the p.s.. Does that help a bit? – B. Pasternak May 08 '16 at 07:49
  • Thanks. What about the line joining the two points and therefore on the plane of projection? Is that line not significant? – Kevin Wu May 08 '16 at 08:11
  • Well I would say that the line (p.s.) corresponding to the plane (v.s.) you get when taking the cross product of the two lines (v.s.) corresponding to the two points (p.s.) is precisely the line (p.s.) between the two points (p.s.). – B. Pasternak May 08 '16 at 08:23
  • Isn't that the line normal to the plane? The line connecting the points is on the plane though. – Kevin Wu May 08 '16 at 08:29

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I'm confused as to how you want the duality here.

Let $V$ be the underlying vector space, let $k$ ($=\mathbb{R}$ or $\mathbb{C}$) be the field over which the vector space is taken (so $V=k^3$). Let $(x_1:x_2:x_3)$ be homogeneous coordinates in the projective plane $\mathbb{P}^2$ (defined by the standard basis of $V$).

Let $p_1=(\alpha_1:\alpha_2:\alpha_2)$ and $p_2=(\beta_1:\beta_2:\beta_3)$ be two points in $\mathbb{P}^2$, and let $l_1$ and $l_2$ be the corresponding lines in the underlying vector space $V$. Then $\overline{p_1p_2}$ is a line in $\mathbb{P}^2$, which is a plane in $V$; the normal to this plane is given by $l_1\times l_2=:l_3$, and the equation of the line in $\mathbb{P}^2$ is given by $(l_3)_1x_1+(l_3)_1x_2+(l_3)_3x_3=0$, where $(l_3)_i$ is the $i$-th component of $l_3$, $1\le i\le 3$.

Let $l_1$ and $l_2$ be two lines in $\mathbb{P}^2$; there are numbers $\gamma_1,\gamma_2,\gamma_3,\delta_1,\delta_2,\delta_3\in k$ such that \begin{align*} l_1&=\{(x_1:x_2:x_3)\in\mathbb{P}^2:\gamma_1x_1+\gamma_2x_2+\gamma_3x_3=0\}, \\ l_2&=\{(x_1:x_2:x_3)\in\mathbb{P}^2:\delta_1x_1+\delta_2x_2+\delta_3x_3=0\}, \end{align*} so $\gamma:=(\gamma_1,\gamma_2,\gamma_3)$ and $\delta:=(\delta_1,\delta_2,\delta_3)$ are the corresponding normals to the planes $L_1$ and $L_2$ to which $l_1$ and $l_2$ correspond in $V$. The cross product of $l_1$ and $l_2$ is the simply the cross product of $\gamma$ and $\delta$, which is a vector normal to both $\gamma$ and $\delta$; this vector necessarily lies in both planes and thus equals the intersection $L_1\cap L_2$, which is projectively the intersection $l_1\cap l_2$, and is a point in $\mathbb{P}^2$.

B. Pasternak
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  • Why is The cross product of l1 and l2 equal to the cross product of y and epsilon? – Kevin Wu May 08 '16 at 10:23
  • By definition I would say; it's also the only definition that makes sense, since only the normal to a plane (through the origin) defines the plane completely. – B. Pasternak May 08 '16 at 11:19
  • @KevinWu If you consider the answer I gave helpful, consider accepting the answer so that this question doesn't stay unanswered. – B. Pasternak May 29 '16 at 08:19