In geometry, Plücker coordinates, introduced by Julius Plücker in the 19th century, are a way to assign six homogeneous coordinates to each line in projective 3-space, P 3 {\displaystyle \mathbb {P} ^{3}} . Because they satisfy a quadratic constraint, they establish a one-to-one correspondence between the 4-dimensional space of lines in P 3 {\displaystyle \mathbb {P} ^{3}} and points on a quadric in P 5 {\displaystyle \mathbb {P} ^{5}} (projective 5-space). A predecessor and special case of Grassmann coordinates (which describe k-dimensional linear subspaces, or flats, in an n-dimensional Euclidean space), Plücker coordinates arise naturally in geometric algebra. They have proved useful for computer graphics, and also can be extended to coordinates for the screws and wrenches in the theory of kinematics used for robot control.
Geometric intuition
A line L in 3-dimensional Euclidean space is determined by two distinct points that it contains, or by two distinct planes that contain it (a plane-plane intersection). Consider the first case, with points x = ( x 1 , x 2 , x 3 ) {\displaystyle x=(x_{1},x_{2},x_{3})} and y = ( y 1 , y 2 , y 3 ) . {\displaystyle y=(y_{1},y_{2},y_{3}).} The vector displacement from x to y is nonzero because the points are distinct, and represents the direction of the line. That is, every displacement between points on the line L is a scalar multiple of d = y − x. If a physical particle of unit mass were to move from x to y, it would have a moment about the origin of the coordinate system. The geometric equivalent to this moment is a vector whose direction is perpendicular to the plane containing the line L and the origin, and whose length equals twice the area of the triangle formed by the displacement and the origin. Treating the points as displacements from the origin, the moment is m = x × y, where "×" denotes the vector cross product. For a fixed line, L, the area of the triangle is proportional to the length of the segment between x and y, considered as the base of the triangle; it is not changed by sliding the base along the line, parallel to itself. By definition the moment vector is perpendicular to every displacement along the line, so d ⋅ m = 0, where "⋅" denotes the vector dot product. Although neither direction d nor moment m alone is sufficient to determine the line L, together the pair does so uniquely, up to a common (nonzero) scalar multiple which depends on the distance between x and y. That is, the coordinates
( d : m ) = ( d 1 : d 2 : d 3 : m 1 : m 2 : m 3 ) {\displaystyle (\mathbf {d} :\mathbf {m} )=(d_{1}:d_{2}:d_{3}\ :\ m_{1}:m_{2}:m_{3})}
may be considered homogeneous coordinates for L, in the sense that all pairs (λd : λm), for λ ≠ 0, can be produced by points on L and only L, and any such pair determines a unique line so long as d is not zero and d ⋅ m = 0. Furthermore, this approach extends to include points, lines, and a plane "at infinity", in the sense of projective geometry. In addition a point x {\displaystyle x} lies on the line L if and only if x × d = m {\displaystyle x\times d=m} .
Example. Let x = (2, 3, 7) and y = (2, 1, 0). Then (d : m) = (0 : −2 : −7 : −7 : 14 : −4). Alternatively, let the equations for points x of two distinct planes containing L be
0 = a + a ⋅ x , 0 = b + b ⋅ x . {\displaystyle {\begin{aligned}0&=a+\mathbf {a} \cdot \mathbf {x} ,\\0&=b+\mathbf {b} \cdot \mathbf {x} .\end{aligned}}}
Then their respective planes are perpendicular to vectors a and b, and the direction of L must be perpendicular to both. Hence we may set d = a × b, which is nonzero because a, b are neither zero nor parallel (the planes being distinct and intersecting). If point x satisfies both plane equations, then it also satisfies the linear combination
0 = a ( b + b ⋅ x ) − b ( a + a ⋅ x ) = ( a b − b a ) ⋅ x {\displaystyle {\begin{aligned}0&=a(b+\mathbf {b} \cdot \mathbf {x} )-b(a+\mathbf {a} \cdot \mathbf {x} )\\&=(a\mathbf {b} -b\mathbf {a} )\cdot \mathbf {x} \end{aligned}}}
That is,
m = a b − b a {\displaystyle \mathbf {m} =a\mathbf {b} -b\mathbf {a} }
is a vector perpendicular to displacements to points on L from the origin; it is, in fact, a moment consistent with the d previously defined from a and b.
Algebraic definition
Primal coordinates In a 3-dimensional projective space P 3 {\displaystyle \mathbb {P} ^{3}} , let L be a line through distinct points x and y with homogeneous coordinates (x0 : x1 : x2 : x3) and (y0 : y1 : y2 : y3). The Plücker coordinates pij are defined as follows:
p i j = | x i y i x j y j | = x i y j − x j y i . {\displaystyle p_{ij}={\begin{vmatrix}x_{i}&y_{i}\\x_{j}&y_{j}\end{vmatrix}}=x_{i}y_{j}-x_{j}y_{i}.}
(the skew symmetric matrix whose elements are pij is also called the Plücker matrix ) This implies pii = 0 and pij = −pji, reducing the possibilities to only six (4 choose 2) independent quantities. The sextuple
( p 01 : p 02 : p 03 : p 23 : p 31 : p 12 ) {\displaystyle (p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12})}
is uniquely determined by L up to a common nonzero scale factor. Furthermore, not all six components can be zero. Thus the Plücker coordinates of L may be considered as homogeneous coordinates of a point in a 5-dimensional projective space, as suggested by the colon notation. To see these facts, let M be the 4×2 matrix with the point coordinates as columns.
M = [ x 0 y 0 x 1 y 1 x 2 y 2 x 3 y 3 ] {\displaystyle M={\begin{bmatrix}x_{0}&y_{0}\\x_{1}&y_{1}\\x_{2}&y_{2}\\x_{3}&y_{3}\end{bmatrix}}}
The Plücker coordinate pij is the determinant of rows i and j of M. Because x and y are distinct points, the columns of M are linearly independent; M has rank 2. Let M′ be a second matrix, with columns x′, y′ a different pair of distinct points on L. Then the columns of M′ are linear combinations of the columns of M; so for some 2×2 nonsingular matrix Λ,
M ′ = M Λ . {\displaystyle M'=M\Lambda .}
In particular, rows i and j of M′ and M are related by
[ x i ′ y i ′ x j ′ y j ′ ] = [ x i y i x j y j ] [ λ 00 λ 01 λ 10 λ 11 ] . {\displaystyle {\begin{bmatrix}x'_{i}&y'_{i}\\x'_{j}&y'_{j}\end{bmatrix}}={\begin{bmatrix}x_{i}&y_{i}\\x_{j}&y_{j}\end{bmatrix}}{\begin{bmatrix}\lambda _{00}&\lambda _{01}\\\lambda _{10}&\lambda _{11}\end{bmatrix}}.}
Therefore, the determinant of the left side 2×2 matrix equals the product of the determinants of the right side 2×2 matrices, the latter of which is a fixed scalar, det Λ. Furthermore, all six 2×2 subdeterminants in M cannot be zero because the rank of M is 2.
Plücker map Denote the set of all lines (linear images of P 1 {\displaystyle \mathbb {P} ^{1}} ) in P 3 {\displaystyle \mathbb {P} ^{3}} by G1,3. We thus have a map:
α : G 1 , 3 → P 5 L ↦ L α , {\displaystyle {\begin{aligned}\alpha \colon \mathrm {G} _{1,3}&\rightarrow \mathbb {P} ^{5}\\L&\mapsto L^{\alpha },\end{aligned}}}
where
L α = ( p 01 : p 02 : p 03 : p 23 : p 31 : p 12 ) . {\displaystyle L^{\alpha }=(p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12}).}
Dual coordinates Alternatively, a line can be described as the intersection of two planes. Let L be a line contained in distinct planes a and b with homogeneous coefficients (a0 : a1 : a2 : a3) and (b0 : b1 : b2 : b3), respectively. (The first plane equation is ∑ k a k x k = 0 , {\textstyle \sum _{k}a^{k}x_{k}=0,} for example.) The dual Plücker coordinate pij is
p i j = | a i a j b i b j | = a i b j − a j b i . {\displaystyle p^{ij}={\begin{vmatrix}a^{i}&a^{j}\\b^{i}&b^{j}\end{vmatrix}}=a^{i}b^{j}-a^{j}b^{i}.}
Dual coordinates are convenient in some computations, and they are equivalent to primary coordinates:
( p 01 : p 02 : p 03 : p 23 : p 31 : p 12 ) = ( p 23 : p 31 : p 12 : p 01 : p 02 : p 03 ) {\displaystyle (p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12})=(p^{23}:p^{31}:p^{12}:p^{01}:p^{02}:p^{03})}
Here, equality between the two vectors in homogeneous coordinates means that the numbers on the right side are equal to the numbers on the left side up to some common scaling factor λ. Specifically, let (i, j, k, ℓ) be an even permutation of (0, 1, 2, 3); then
p i j = λ p k ℓ . {\displaystyle p_{ij}=\lambda p^{k\ell }.}
Geometry To relate back to the geometric intuition, take x0 = 0 as the plane at infinity; thus the coordinates of points not at infinity can be normalized so that x0 = 1. Then M becomes
M = [ 1 1 x 1 y 1 x 2 y 2 x 3 y 3 ] , {\displaystyle M={\begin{bmatrix}1&1\\x_{1}&y_{1}\\x_{2}&y_{2}\\x_{3}&y_{3}\end{bmatrix}},}
and setting x = ( x 1 , x 2 , x 3 ) {\displaystyle x=(x_{1},x_{2},x_{3})} and y = ( y 1 , y 2 , y 3 ) {\displaystyle y=(y_{1},y_{2},y_{3})} , we have d = ( p 01 , p 02 , p 03 ) {\displaystyle d=(p_{01},p_{02},p_{03})} and m = ( p 23 , p 31 , p 12 ) {\displaystyle m=(p_{23},p_{31},p_{12})} . Dually, we have d = ( p 23 , p 31 , p 12 ) {\displaystyle d=(p^{23},p^{31},p^{12})} and m = ( p 01 , p 02 , p 03 ) . {\displaystyle m=(p^{01},p^{02},p^{03}).}
Bijection between lines and Klein quadric
Plane equations If the point z = ( z 0 : z 1 : z 2 : z 3 ) {\displaystyle \mathbf {z} =(z_{0}:z_{1}:z_{2}:z_{3})} lies on L, then the columns of
[ x 0 y 0 z 0 x 1 y 1 z 1 x 2 y 2 z 2 x 3 y 3 z 3 ] {\displaystyle {\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}}}
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
0 = | x 0 y 0 z 0 x 1 y 1 z 1 x 2 y 2 z 2 | = | x 1 y 1 x 2 y 2 | z 0 − | x 0 y 0 x 2 y 2 | z 1 + | x 0 y 0 x 1 y 1 | z 2 = p 12 z 0 − p 02 z 1 + p 01 z 2 . = p 03 z 0 + p 13 z 1 + p 23 z 2 . {\displaystyle {\begin{aligned}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{aligned}}}
The four possible planes obtained are as follows.
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