Editing Plücker coordinates (section)

==Bijection between lines and Klein quadric==
=== Plane equations ===
If the point <math>\mathbf z = (z_0:z_1:z_2:z_3)</math> lies on {{mvar|L}}, then the columns of

: <math> \begin{bmatrix} x_0 & y_0 & z_0 \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \\ x_3 & y_3 & z_3 \end{bmatrix} </math>

are [[linearly dependent]], so that the rank of this larger matrix is still 2. This implies that all 3×3 submatrices have determinant zero, generating four (4 choose 3) plane equations, such as

: <math>
\begin{align}
0 & = \begin{vmatrix} x_0 & y_0 & z_0 \\ x_1 & y_1 & z_1 \\ x_2 & y_2 & z_2 \end{vmatrix} \\[5pt]
& = \begin{vmatrix} x_1 & y_1 \\ x_2 & y_2 \end{vmatrix} z_0 - \begin{vmatrix} x_0 & y_0 \\ x_2 & y_2 \end{vmatrix} z_1 + \begin{vmatrix} x_0 & y_0 \\ x_1 & y_1 \end{vmatrix} z_2 \\[5pt]
& = p_{12} z_0 - p_{02} z_1 + p_{01} z_2 . \\[5pt]
& = p^{03} z_0 + p^{13} z_1 + p^{23} z_2 .
\end{align}
</math>

The four possible planes obtained are as follows.

: <math> \begin{matrix}
0 & = & {}+ p_{12} z_0 & {}- p_{02} z_1 & {}+ p_{01} z_2 & \\
0 & = & {}- p_{31} z_0 & {}- p_{03} z_1 & & {}+ p_{01} z_3 \\
0 & = & {}+p_{23} z_0 & & {}- p_{03} z_2 & {}+ p_{02} z_3 \\
0 & = & & {}+p_{23} z_1 & {}+ p_{31} z_2 & {}+ p_{12} z_3
\end{matrix} </math>

Using dual coordinates, and letting {{math|(''a''{{sup|0}} : ''a''{{sup|1}} : ''a''{{sup|2}} : ''a''{{sup|3}})}} be the line coefficients, each of these is simply {{math|1=''a<sup>i</sup>'' = ''p<sup>ij</sup>''}}, or

: <math> 0 = \sum_{i=0}^3 p^{ij} z_i , \qquad j = 0,\ldots,3 . </math>

Each Plücker coordinate appears in two of the four equations, each time multiplying a different variable; and as at least one of the coordinates is nonzero, we are guaranteed non-vacuous equations for two distinct planes intersecting in {{mvar|L}}. Thus the Plücker coordinates of a line determine that line uniquely, and the map α is an [[Injective function|injection]].

=== Quadratic relation ===
The image of {{math|α}} is not the complete set of points in {{tmath|\mathbb P^5}}; the Plücker coordinates of a line {{mvar|L}} satisfy the quadratic Plücker relation

: <math>
\begin{align}
0 & = p_{01}p^{01}+p_{02}p^{02}+p_{03}p^{03} \\
& = p_{01}p_{23}+p_{02}p_{31}+p_{03}p_{12}.
\end{align}
</math>

For proof, write this homogeneous polynomial as determinants and use [[Laplace expansion]] (in reverse).

: <math>
\begin{align}
0 & = \begin{vmatrix}x_0&y_0\\x_1&y_1\end{vmatrix}\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}+
\begin{vmatrix}x_0&y_0\\x_2&y_2\end{vmatrix}\begin{vmatrix}x_3&y_3\\x_1&y_1\end{vmatrix}+
\begin{vmatrix}x_0&y_0\\x_3&y_3\end{vmatrix}\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix} \\[5pt]
& = (x_0 y_1-y_0 x_1)\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}-
(x_0 y_2-y_0 x_2)\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+
(x_0 y_3-y_0 x_3)\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix} \\[5pt]
& = x_0 \left(y_1\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}-
y_2\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+
y_3\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}\right)
-y_0 \left(x_1\begin{vmatrix}x_2&y_2\\x_3&y_3\end{vmatrix}-
x_2\begin{vmatrix}x_1&y_1\\x_3&y_3\end{vmatrix}+
x_3\begin{vmatrix}x_1&y_1\\x_2&y_2\end{vmatrix}\right) \\[5pt]
& = x_0 \begin{vmatrix}x_1&y_1&y_1\\x_2&y_2&y_2\\x_3&y_3&y_3\end{vmatrix}
-y_0 \begin{vmatrix}x_1&x_1&y_1\\x_2&x_2&y_2\\x_3&x_3&y_3\end{vmatrix}
\end{align}
</math>

Since both 3×3 determinants have duplicate columns, the right hand side is identically zero.

Another proof may be done like this:
Since vector

: <math> d = \left( p_{01}, p_{02}, p_{03} \right) </math>

is perpendicular to vector

: <math> m = \left( p_{23}, p_{31}, p_{12} \right) </math>

(see above), the scalar product of {{mvar|d}} and {{mvar|m}} must be zero. q.e.d.

=== Point equations ===
Letting {{math|(''x''{{sub|0}} : ''x''{{sub|1}} : ''x''{{sub|2}} : ''x''{{sub|3}})}} be the point coordinates, four possible points on a line each have coordinates {{math|1=''x<sub>i</sub>'' = ''p<sub>ij</sub>''}}, for {{math|1=''j'' = 0, 1, 2, 3}}. Some of these possible points may be inadmissible because all coordinates are zero, but since at least one Plücker coordinate is nonzero, at least two distinct points are guaranteed.

=== Bijectivity ===
If <math>(q_{01}:q_{02}:q_{03}:q_{23}:q_{31}:q_{12})</math> are the homogeneous coordinates of a point in {{tmath|\mathbb P^5}}, without loss of generality assume that {{math|''q''<sub>01</sub>}} is nonzero. Then the matrix

: <math> M = \begin{bmatrix} q_{01} & 0 \\ 0 & q_{01} \\ -q_{12} & q_{02} \\ q_{31} & q_{03} \end{bmatrix} </math>

has rank 2, and so its columns are distinct points defining a line {{mvar|L}}. When the {{tmath|\mathbb P^5}} coordinates, {{mvar|q<sub>ij</sub>}}, satisfy the quadratic Plücker relation, they are the Plücker coordinates of {{mvar|L}}. To see this, first normalize {{math|''q''<sub>01</sub>}} to 1. Then we immediately have that for the Plücker coordinates computed from {{mvar|M}}, {{math|1=''p<sub>ij</sub>'' = ''q<sub>ij</sub>''}}, except for

: <math> p_{23} = - q_{03} q_{12} - q_{02} q_{31} . </math>

But if the {{mvar|q<sub>ij</sub>}} satisfy the Plücker relation 
:<math>q_{23} + q_{02}q_{31} + q_{03}q_{12} = 0,</math> 
then {{math|1=''p''<sub>23</sub> = ''q''<sub>23</sub>}}, 
completing the set of identities.

Consequently, {{math|α}} is a [[surjection]] onto the [[algebraic variety]] consisting of the set of zeros of the quadratic polynomial

: <math> p_{01}p_{23}+p_{02}p_{31}+p_{03}p_{12} . </math>

And since {{math|α}} is also an injection, the lines in {{tmath|\mathbb P^3}} are thus in [[bijection|bijective]] correspondence with the points of this [[quadric]] in {{tmath|\mathbb P^5}}, called the Plücker quadric or [[Klein quadric]].