The Plücker matrix is a special skew-symmetric 4 × 4 matrix, which characterizes a straight line in projective space. The matrix is defined by 6 Plücker coordinates with 4 degrees of freedom. It is named after the German mathematician Julius Plücker.
A straight line in space is defined by two distinct points
A=\left(A0,A1,A2,
\top | |
A | |
3\right) |
\inRl{P}3
B=\left(B0,B1,B2,
\top | |
B | |
3\right) |
\inRl{P}3
[L] x \proptoAB\top-BA\top= \left(\begin{array}{cccc} 0&-L01&-L02&-L03\\ L01&0&-L12&-L13\\ L02&L12&0&-L23\\ L03&L13&L23&0 \end{array}\right)
L\propto(L01,L02,L03,L12,L13,L23)\top
with
Lij=AiBj-BiAj.
Plücker coordinates fulfill the Grassmann–Plücker relations
L01L23-L02L13+L03L12=0
and are defined up to scale. A Plücker matrix has only rank 2 and four degrees of freedom (just like lines in
R3
A
B
The Plücker matrix allows us to express the following geometric operations as matrix-vector product:
0=[L] x E
X=[L] x E
L
E
0=[\tilde{L
E=[\tilde{L
E
X
L
[L] x \piinfty=[L] x (0,0,0,1)\top=\left(-L03,-L13,-L23,0\right)\top
X0\cong[L] x [L] x \piinfty.
Two arbitrary distinct points on the line can be written as a linear combination of
A
B
A\prime\proptoA\alpha+B\betaandB\prime\proptoA\gamma+B\delta.
Their Plücker matrix is thus:
\begin{align}
\prime{]} | |
{[}L | |
x |
&=A\primeB\prime-B\primeA\prime\\[6pt] &=(A\alpha+B\beta)(A\gamma+B\delta)\top-(A\gamma+B\delta)(A\alpha+B\beta)\top\\[6pt] &=\underbrace{(\alpha\delta-\beta\gamma)}λ[L] x , \end{align}
up to scale identical to
[L] x
Let
E=\left(E0,E1,E2,E3\right)\top\inRl{P}3
E0x+E1y+E2z+E3=0.
which does not contain the line
L
X=[L] x E =A\underset{\alpha}{\underbrace{B\topE
which lies on the line
L
A
B
X
E
E\topX =E\top[L] x E =\underset{\alpha}{\underbrace{E\topA
and must therefore be their point of intersection.
In addition, the product of the Plücker matrix with a plane is the zero-vector, exactly if the line
L
\alpha=\beta=0\iffE
L.
In projective three-space, both points and planes have the same representation as 4-vectors and the algebraic description of their geometric relationship (point lies on plane) is symmetric. By interchanging the terms plane and point in a theorem, one obtains a dual theorem which is also true.
In case of the Plücker matrix, there exists a dual representation of the line in space as the intersection of two planes:
E=\left(E0,E1,E2,
\top | |
E | |
3\right) |
\inRl{P}3
and
F=\left(F0,F1,F2,
\top | |
F | |
3\right) |
\inRl{P}3
in homogeneous coordinates of projective space. Their Plücker matrix is:
\left[\tilde{L
and
G=\left[\tilde{L
describes the plane
G
X
L
As the vector
X=[L] x E
E
\forallE\inRl{P}3: X=[L] x EliesonL \iff\left[\tilde{L
Thus:
\left([\tilde{L
The following product fulfills these properties:
\begin{align} &\left(\begin{array}{cccc} 0&L23&-L13&L12\\ -L23&0&L03&-L02\\ L13&-L03&0&L01\\ -L12&L02&-L01&0 \end{array}\right) \left(\begin{array}{cccc} 0&-L01&-L02&-L03\\ L01&0&-L12&-L13\\ L02&L12&0&-L23\\ L03&L13&L23&0 \end{array}\right)\\[10pt] ={}&\left(L01L23-L02L13+L03L12\right) ⋅ \left(\begin{array}{cccc} 1&0&0&0\\ 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1 \end{array}\right) =0, \end{align}
due to the Grassmann–Plücker relation. With the uniqueness of Plücker matrices up to scalar multiples, for the primal Plücker coordinates
L=\left(L01,L02,L03,L12,L13,L23\right)\top
we obtain the following dual Plücker coordinates:
\tilde{L
The 'join' of two points in the projective plane is the operation of connecting two points with a straight line. Its line equation can be computed using the cross product:
l\proptoa x b= \left(\begin{array}{c} a1b2-b1a2\\ b0a2-a0b2\\ a0b1-a1b0\end{array}\right)=\left(\begin{array}{c} l0\\ l1\\ l2\end{array}\right).
Dually, one can express the 'meet', or intersection of two straight lines by the cross-product:
x\proptol x m
The relationship to Plücker matrices becomes evident, if one writes the cross product as a matrix-vector product with a skew-symmetric matrix:
[l] x =ab\top-ba\top= \left(\begin{array}{ccc} 0&l2&-l1\\ -l2&0&l0\\ l1&-l0&0 \end{array}\right)
and analogously
[x] x =lm\top-ml\top
Let
d=\left(-L03,-L13,-L23\right)\top
m=\left(L12,-L02,L01\right)\top
[L] x =\left(\begin{array}{cc} [m] x &d\\ -d&0 \end{array}\right)
and
[\tilde{L
where
d
m