Plücker coordinates

In a 3-dimensional projective space ${\displaystyle \mathbb {P} ^{3}}$, let L be a line through distinct points x and y with homogenous coordinates (x₀ : x₁ : x₂ : x₃) and (y₀ : y₁ : y₂ : y₃).

The Plücker coordinates p_ij are defined as follows:

${\displaystyle p_{ij}={\begin{vmatrix}x_{i}&y_{i}\\x_{j}&y_{j}\end{vmatrix}}=x_{i}y_{j}-x_{j}y_{i}.}$

(the skew symmetric matrix whose elements are p_ij is also called the Plücker matrix )
This implies p_ii = 0 and p_ij = −p_ji, reducing the possibilities to only six (4 choose 2) independent quantities. The sextuple

${\displaystyle (p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12})}$

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.

${\displaystyle M={\begin{bmatrix}x_{0}&y_{0}\\x_{1}&y_{1}\\x_{2}&y_{2}\\x_{3}&y_{3}\end{bmatrix}}}$

The Plücker coordinate p_ij 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,

${\displaystyle M'=M\Lambda .}$

In particular, rows i and j of M′ and M are related by

${\displaystyle {\begin{bmatrix}x'_{i}&y'_{i}\\x'_{j}&y'_{j}\end{bmatrix}}={\begin{bmatrix}x_{i}&y_{i}\\x_{j}&y_{j}\end{bmatrix}}{\begin{bmatrix}\lambda _{00}&\lambda _{01}\\\lambda _{10}&\lambda _{11}\end{bmatrix}}.}$

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.

Denote the set of all lines (linear images of ${\displaystyle \mathbb {P} ^{1}}$) in ${\displaystyle \mathbb {P} ^{3}}$ by G_1,3. We thus have a map:

${\displaystyle {\begin{aligned}\alpha \colon \mathrm {G} _{1,3}&\rightarrow \mathbb {P} ^{5}\\L&\mapsto L^{\alpha },\end{aligned}}}$

where

${\displaystyle L^{\alpha }=(p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12}).}$

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 ${\textstyle \sum _{k}a^{k}x_{k}=0,}$ for example.) The dual Plücker coordinate p^ij is

${\displaystyle p^{ij}={\begin{vmatrix}a^{i}&a^{j}\\b^{i}&b^{j}\end{vmatrix}}=a^{i}b^{j}-a^{j}b^{i}.}$

Dual coordinates are convenient in some computations, and they are equivalent to primary coordinates:

${\displaystyle (p_{01}:p_{02}:p_{03}:p_{23}:p_{31}:p_{12})=(p^{23}:p^{31}:p^{12}:p^{01}:p^{02}:p^{03})}$

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

${\displaystyle p_{ij}=\lambda p^{k\ell }.}$

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

${\displaystyle M={\begin{bmatrix}1&1\\x_{1}&y_{1}\\x_{2}&y_{2}\\x_{3}&y_{3}\end{bmatrix}},}$

and setting ${\displaystyle x=(x_{1},x_{2},x_{3})}$ and ${\displaystyle y=(y_{1},y_{2},y_{3})}$, we have ${\displaystyle d=(p_{01},p_{02},p_{03})}$and ${\displaystyle m=(p_{23},p_{31},p_{12})}$.

Dually, we have ${\displaystyle d=(p^{23},p^{31},p^{12})}$ and ${\displaystyle m=(p^{01},p^{02},p^{03}).}$

If the point ${\displaystyle \mathbf {z} =(z_{0}:z_{1}:z_{2}:z_{3})}$ lies on L, then the columns of

${\displaystyle {\begin{bmatrix}x_{0}&y_{0}&z_{0}\\x_{1}&y_{1}&z_{1}\\x_{2}&y_{2}&z_{2}\\x_{3}&y_{3}&z_{3}\end{bmatrix}}}$

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

${\displaystyle {\begin{aligned}0&={\begin{vmatrix}x_{0}&y_{0}&z_{0}\\x_{1}&y_{1}&z_{1}\\x_{2}&y_{2}&z_{2}\end{vmatrix}}\\[5pt]&={\begin{vmatrix}x_{1}&y_{1}\\x_{2}&y_{2}\end{vmatrix}}z_{0}-{\begin{vmatrix}x_{0}&y_{0}\\x_{2}&y_{2}\end{vmatrix}}z_{1}+{\begin{vmatrix}x_{0}&y_{0}\\x_{1}&y_{1}\end{vmatrix}}z_{2}\\[5pt]&=p_{12}z_{0}-p_{02}z_{1}+p_{01}z_{2}.\\[5pt]&=p^{03}z_{0}+p^{13}z_{1}+p^{23}z_{2}.\end{aligned}}}$

The four possible planes obtained are as follows.

${\displaystyle {\begin{matrix}0&=&{}+p_{12}z_{0}&{}-p_{02}z_{1}&{}+p_{01}z_{2}&\\0&=&{}-p_{31}z_{0}&{}-p_{03}z_{1}&&{}+p_{01}z_{3}\\0&=&{}+p_{23}z_{0}&&{}-p_{03}z_{2}&{}+p_{02}z_{3}\\0&=&&{}+p_{23}z_{1}&{}+p_{31}z_{2}&{}+p_{12}z_{3}\end{matrix}}}$

Using dual coordinates, and letting (a⁰ : a¹ : a² : a³) be the line coefficients, each of these is simply aⁱ = p^ij, or

${\displaystyle 0=\sum _{i=0}^{3}p^{ij}z_{i},\qquad j=0,\ldots ,3.}$

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 L. Thus the Plücker coordinates of a line determine that line uniquely, and the map α is an injection.

The image of α is not the complete set of points in ${\displaystyle \mathbb {P} ^{5}}$; the Plücker coordinates of a line L satisfy the quadratic Plücker relation

${\displaystyle {\begin{aligned}0&=p_{01}p^{01}+p_{02}p^{02}+p_{03}p^{03}\\&=p_{01}p_{23}+p_{02}p_{31}+p_{03}p_{12}.\end{aligned}}}$

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

${\displaystyle {\begin{aligned}0&={\begin{vmatrix}x_{0}&y_{0}\\x_{1}&y_{1}\end{vmatrix}}{\begin{vmatrix}x_{2}&y_{2}\\x_{3}&y_{3}\end{vmatrix}}+{\begin{vmatrix}x_{0}&y_{0}\\x_{2}&y_{2}\end{vmatrix}}{\begin{vmatrix}x_{3}&y_{3}\\x_{1}&y_{1}\end{vmatrix}}+{\begin{vmatrix}x_{0}&y_{0}\\x_{3}&y_{3}\end{vmatrix}}{\begin{vmatrix}x_{1}&y_{1}\\x_{2}&y_{2}\end{vmatrix}}\\[5pt]&=(x_{0}y_{1}-y_{0}x_{1}){\begin{vmatrix}x_{2}&y_{2}\\x_{3}&y_{3}\end{vmatrix}}-(x_{0}y_{2}-y_{0}x_{2}){\begin{vmatrix}x_{1}&y_{1}\\x_{3}&y_{3}\end{vmatrix}}+(x_{0}y_{3}-y_{0}x_{3}){\begin{vmatrix}x_{1}&y_{1}\\x_{2}&y_{2}\end{vmatrix}}\\[5pt]&=x_{0}\left(y_{1}{\begin{vmatrix}x_{2}&y_{2}\\x_{3}&y_{3}\end{vmatrix}}-y_{2}{\begin{vmatrix}x_{1}&y_{1}\\x_{3}&y_{3}\end{vmatrix}}+y_{3}{\begin{vmatrix}x_{1}&y_{1}\\x_{2}&y_{2}\end{vmatrix}}\right)-y_{0}\left(x_{1}{\begin{vmatrix}x_{2}&y_{2}\\x_{3}&y_{3}\end{vmatrix}}-x_{2}{\begin{vmatrix}x_{1}&y_{1}\\x_{3}&y_{3}\end{vmatrix}}+x_{3}{\begin{vmatrix}x_{1}&y_{1}\\x_{2}&y_{2}\end{vmatrix}}\right)\\[5pt]&=x_{0}{\begin{vmatrix}x_{1}&y_{1}&y_{1}\\x_{2}&y_{2}&y_{2}\\x_{3}&y_{3}&y_{3}\end{vmatrix}}-y_{0}{\begin{vmatrix}x_{1}&x_{1}&y_{1}\\x_{2}&x_{2}&y_{2}\\x_{3}&x_{3}&y_{3}\end{vmatrix}}\end{aligned}}}$

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

Another proof may be done like this: Since vector

${\displaystyle d=\left(p_{01},p_{02},p_{03}\right)}$

is perpendicular to vector

${\displaystyle m=\left(p_{23},p_{31},p_{12}\right)}$

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

Letting (x₀ : x₁ : x₂ : x₃) be the point coordinates, four possible points on a line each have coordinates x_i = p_ij, for 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.

If ${\displaystyle (q_{01}:q_{02}:q_{03}:q_{23}:q_{31}:q_{12})}$ are the homogeneous coordinates of a point in ${\displaystyle \mathbb {P} ^{5}}$, without loss of generality assume that q₀₁ is nonzero. Then the matrix

${\displaystyle M={\begin{bmatrix}q_{01}&0\\0&q_{01}\\-q_{12}&q_{02}\\q_{31}&q_{03}\end{bmatrix}}}$

has rank 2, and so its columns are distinct points defining a line L. When the ${\displaystyle \mathbb {P} ^{5}}$ coordinates, q_ij, satisfy the quadratic Plücker relation, they are the Plücker coordinates of L. To see this, first normalize q₀₁ to 1. Then we immediately have that for the Plücker coordinates computed from M, p_ij = q_ij, except for

${\displaystyle p_{23}=-q_{03}q_{12}-q_{02}q_{31}.}$

But if the q_ij satisfy the Plücker relation

${\displaystyle q_{23}+q_{02}q_{31}+q_{03}q_{12}=0,}$

then p₂₃ = q₂₃, completing the set of identities.

Consequently, α is a surjection onto the algebraic variety consisting of the set of zeros of the quadratic polynomial

${\displaystyle p_{01}p_{23}+p_{02}p_{31}+p_{03}p_{12}.}$

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

Two lines in ${\displaystyle \mathbb {P} ^{3}}$ are either skew or coplanar, and in the latter case they are either coincident or intersect in a unique point. If p_ij and p′_ij are the Plücker coordinates of two lines, then they are coplanar precisely when

${\displaystyle \mathbf {d} \cdot \mathbf {m} '+\mathbf {m} \cdot \mathbf {d} '=0,}$

as shown by

${\displaystyle {\begin{aligned}0&=p_{01}p'_{23}+p_{02}p'_{31}+p_{03}p'_{12}+p_{23}p'_{01}+p_{31}p'_{02}+p_{12}p'_{03}\\[5pt]&={\begin{vmatrix}x_{0}&y_{0}&x'_{0}&y'_{0}\\x_{1}&y_{1}&x'_{1}&y'_{1}\\x_{2}&y_{2}&x'_{2}&y'_{2}\\x_{3}&y_{3}&x'_{3}&y'_{3}\end{vmatrix}}.\end{aligned}}}$

When the lines are skew, the sign of the result indicates the sense of crossing: positive if a right-handed screw takes L into L′, else negative.

The quadratic Plücker relation essentially states that a line is coplanar with itself.

In the event that two lines are coplanar but not parallel, their common plane has equation

${\displaystyle 0=(\mathbf {m} \cdot \mathbf {d} ')x_{0}+(\mathbf {d} \times \mathbf {d} ')\cdot \mathbf {x} ,}$

where ${\displaystyle x=(x_{1},x_{2},x_{3}).}$

The slightest perturbation will destroy the existence of a common plane, and near-parallelism of the lines will cause numeric difficulties in finding such a plane even if it does exist.

Dually, two coplanar lines, neither of which contains the origin, have common point

${\displaystyle (x_{0}:\mathbf {x} )=(\mathbf {d} \cdot \mathbf {m} ':\mathbf {m} \times \mathbf {m} ').}$

To handle lines not meeting this restriction, see the references.

Given a plane with equation

${\displaystyle 0=a^{0}x_{0}+a^{1}x_{1}+a^{2}x_{2}+a^{3}x_{3},}$

or more concisely,

${\displaystyle 0=a^{0}x_{0}+\mathbf {a} \cdot \mathbf {x}$ ;}

and given a line not in it with Plücker coordinates (d : m), then their point of intersection is

${\displaystyle (x_{0}:\mathbf {x} )=(\mathbf {a} \cdot \mathbf {d}$ :\mathbf {a} \times \mathbf {m} -a_{0}\mathbf {d} ).}

The point coordinates, (x₀ : x₁ : x₂ : x₃), can also be expressed in terms of Plücker coordinates as

${\displaystyle x_{i}=\sum _{j\neq i}a^{j}p_{ij},\qquad i=0\ldots 3.}$

Dually, given a point (y₀ : y) and a line not containing it, their common plane has equation

${\displaystyle 0=(\mathbf {y} \cdot \mathbf {m} )x_{0}+(\mathbf {y} \times \mathbf {d} -y_{0}\mathbf {m} )\cdot \mathbf {x} .}$

The plane coordinates, (a⁰ : a¹ : a² : a³), can also be expressed in terms of dual Plücker coordinates as

${\displaystyle a^{i}=\sum _{j\neq i}y_{j}p^{ij},\qquad i=0\ldots 3.}$

Because the Klein quadric is in ${\displaystyle \mathbb {P} ^{5}}$, it contains linear subspaces of dimensions one and two (but no higher). These correspond to one- and two-parameter families of lines in ${\displaystyle \mathbb {P} ^{3}}$.

For example, suppose L, L′ are distinct lines in ${\displaystyle \mathbb {P} ^{3}}$ determined by points x, y and x′, y′, respectively. Linear combinations of their determining points give linear combinations of their Plücker coordinates, generating a one-parameter family of lines containing L and L′. This corresponds to a one-dimensional linear subspace belonging to the Klein quadric.

If three distinct and non-parallel lines are coplanar; their linear combinations generate a two-parameter family of lines, all the lines in the plane. This corresponds to a two-dimensional linear subspace belonging to the Klein quadric.

If three distinct and non-coplanar lines intersect in a point, their linear combinations generate a two-parameter family of lines, all the lines through the point. This also corresponds to a two-dimensional linear subspace belonging to the Klein quadric.

A ruled surface is a family of lines that is not necessarily linear. It corresponds to a curve on the Klein quadric. For example, a hyperboloid of one sheet is a quadric surface in ${\displaystyle \mathbb {P} ^{3}}$ ruled by two different families of lines, one line of each passing through each point of the surface; each family corresponds under the Plücker map to a conic section within the Klein quadric in ${\displaystyle \mathbb {P} ^{5}}$.

During the nineteenth century, line geometry was studied intensively. In terms of the bijection given above, this is a description of the intrinsic geometry of the Klein quadric.

Line geometry is extensively used in ray tracing application where the geometry and intersections of rays need to be calculated in 3D. An implementation is described in Introduction to Plücker Coordinates written for the Ray Tracing forum by Thouis Jones.