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, . Because they satisfy a quadratic constraint, they establish a one-to-one correspondence between the 4-dimensional space of lines in and points on a quadric in (projective 5-space). A predecessor and special case of Grassmann coordinates (which describe -dimensional linear subspaces, or flats, in an -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.
A line in 3-dimensional Euclidean space is determined by two distinct points that it contains, or by two distinct planes that contain it. Consider the first case, with points
x=(x1,x2,x3)
y=(y1,y2,y3).
Although neither nor alone is sufficient to determine, together the pair does so uniquely, up to a common (nonzero) scalar multiple which depends on the distance between and . That is, the coordinates
(d:m)=(d1:d2:d3 : m1:m2:m3)
may be considered homogeneous coordinates for, in the sense that all pairs, for, can be produced by points on and only, and any such pair determines a unique line so long as is not zero and . Furthermore, this approach extends to include points, lines, and a plane "at infinity", in the sense of projective geometry. In addition a point
x
x x d=m
Example. Let and . Then .
Alternatively, let the equations for points of two distinct planes containing be
\begin{align} 0&=a+a ⋅ x,\ 0&=b+b ⋅ x. \end{align}
Then their respective planes are perpendicular to vectors and, and the direction of must be perpendicular to both. Hence we may set, which is nonzero because are neither zero nor parallel (the planes being distinct and intersecting). If point satisfies both plane equations, then it also satisfies the linear combination
\begin{align} 0&=a(b+b ⋅ x)-b(a+a ⋅ x)\\ &=(ab-ba) ⋅ x \end{align}
That is,
m=ab-ba
is a vector perpendicular to displacements to points on from the origin; it is, in fact, a moment consistent with the previously defined from and .
Proof 1: Need to show that
m=ab-ba=r x d=r x (a x b).
Without loss of generality, let
a ⋅ a=b ⋅ b=1.
Point is the origin. Line passes through point and is orthogonal to the plane of the picture. The two planes pass through and and are both orthogonal to the plane of the picture. Points and are the closest points on those planes to the origin, therefore angles and are right angles and so the points lie on a circle (due to a corollary of Thales's theorem). is the diameter of that circle.
\begin{align} &a:=
BE | |
||BE|| |
, b:=
BC | |
||BC|| |
, r:=BD;\\[4pt] &-a=||BE||=||BF||, -b=||BC||=||BG||;\\[4pt] &m=ab-ba=FG\\[4pt] &||d||=||a x b||=\sin\angleFBG \end{align}
Angle is a right angle due to the following argument. Let . Since (by side-angle-side congruence), then . Since, let . By the inscribed angle theorem,, so . ;, ; therefore, . Then must be a right angle as well.
Angles are right angles, so the four points lie on a circle, and (by the intersecting secants theorem)
||BF||||BC||=||BH||||BD||
\begin{align} &ab\sin\angleFBG=||BH||||r||\sin\angleFBG,\\[4pt] &2Area\triangle=ab\sin\angleFBG=||BH||||FG||=||BH||||r||\sin\angleFBG,\\[4pt] &||m||=||FG||=||r||\sin\angleFBG=||r||||d||,\\[4pt] &m=r x d.\blacksquare \end{align}
Proof 2:
Let
a ⋅ a=b ⋅ b=1.
This implies that
a=-||BE||, b=-||BC||.
According to the vector triple product formula,
r x (a x b)=(r ⋅ b)a-(r ⋅ a)b.
Then
\begin{align} r x (a x b) &=a||r||||b||\cos\angleDBC-b||r||||a||\cos\angleDBE\\[4pt] &=a||r||\cos\angleDBC-b||r||\cos\angleDBE\\[4pt] &=a||BC||-b||BE||\\[4pt] &=-ba-(-a)b\\[4pt] &=ab-ba \blacksquare \end{align}
When
||r||=0,
||r||>0,
||r||.
Example. Let, and, . Then .
Although the usual algebraic definition tends to obscure the relationship, are the Plücker coordinates of .
In a 3-dimensional projective space, let be a line through distinct points and with homogenous coordinates and .
The Plücker coordinates are defined as follows:
pij=\begin{vmatrix}xi&yi\ xj&yj\end{vmatrix}=xiyj-xjyi.
(the skew symmetric matrix whose elements are is also called the Plücker matrix)
This implies and, reducing the possibilities to only six (4 choose 2) independent quantities. The sextuple
(p01:p02:p03:p23:p31:p12)
is uniquely determined by up to a common nonzero scale factor. Furthermore, not all six components can be zero.Thus the Plücker coordinates of 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 be the 4×2 matrix with the point coordinates as columns.
M=\begin{bmatrix}x0&y0\ x1&y1\ x2&y2\ x3&y3\end{bmatrix}
The Plücker coordinate is the determinant of rows and of .Because and are distinct points, the columns of are linearly independent; has rank 2. Let be a second matrix, with columns a different pair of distinct points on . Then the columns of are linear combinations of the columns of ; so for some 2×2 nonsingular matrix,
M'=MΛ.
In particular, rows and of and are related by
\begin{bmatrix}x'i&y'i\\x'j&y'j\end{bmatrix}=\begin{bmatrix}xi&yi\\xj&yj\end{bmatrix}\begin{bmatrix}λ00&λ01\ λ10&λ11\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, . Furthermore, all six 2×2 subdeterminants in cannot be zero because the rank of is 2.
Denote the set of all lines (linear images of) in by . We thus have a map:
\begin{align} \alpha\colonG1,3& → P5\\ L&\mapstoL\alpha, \end{align}
L\alpha=(p01:p02:p03:p23:p31:p12).
Alternatively, a line can be described as the intersection of two planes. Let be a line contained in distinct planes and with homogeneous coefficients and, respectively. (The first plane equation is for example.) The dual Plücker coordinate is
pij=\begin{vmatrix}ai&aj\ bi&bj\end{vmatrix}=aibj-ajbi.
Dual coordinates are convenient in some computations, and they are equivalent to primary coordinates:
(p01:p02:p03:p23:p31:p12)= (p23:p31:p12:p01:p02:p03)
pij=λpk\ell.
To relate back to the geometric intuition, take as the plane at infinity; thus the coordinates of points not at infinity can be normalized so that . Then becomes
M=\begin{bmatrix}1&1\ x1&y1\ x2&y2\ x3&y3\end{bmatrix},
and setting
x=(x1,x2,x3)
y=(y1,y2,y3)
d=(p01,p02,p03)
m=(p23,p31,p12)
Dually, we have
d=(p23,p31,p12)
m=(p01,p02,p03).
If the point
z=(z0:z1:z2:z3)
\begin{bmatrix}x0&y0&z0\ x1&y1&z1\ x2&y2&z2\ x3&y3&z3\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
\begin{align} 0&=\begin{vmatrix}x0&y0&z0\ x1&y1&z1\ x2&y2&z2\end{vmatrix}\\[5pt] &=\begin{vmatrix}x1&y1\ x2&y2\end{vmatrix}z0-\begin{vmatrix}x0&y0\ x2&y2\end{vmatrix}z1+\begin{vmatrix}x0&y0\ x1&y1\end{vmatrix}z2\\[5pt] &=p12z0-p02z1+p01z2.\\[5pt] &=p03z0+p13z1+p23z2. \end{align}
The four possible planes obtained are as follows.
\begin{matrix} 0&=&{}+p12z0&{}-p02z1&{}+p01z2&\\ 0&=&{}-p31z0&{}-p03z1&&{}+p01z3\\ 0&=&{}+p23z0&&{}-p03z2&{}+p02z3\\ 0&=&&{}+p23z1&{}+p31z2&{}+p12z3 \end{matrix}
Using dual coordinates, and letting be the line coefficients, each of these is simply, or
0=
3 | |
\sum | |
i=0 |
pijzi, 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 . Thus the Plücker coordinates of a line determine that line uniquely, and the map α is an injection.
The image of is not the complete set of points in ; the Plücker coordinates of a line satisfy the quadratic Plücker relation
\begin{align} 0&=p01p01+p02p02+p03p03\\ &=p01p23+p02p31+p03p12. \end{align}
For proof, write this homogeneous polynomial as determinants and use Laplace expansion (in reverse).
\begin{align} 0&=\begin{vmatrix}x0&y0\\x1&y1\end{vmatrix}\begin{vmatrix}x2&y2\\x3&y3\end{vmatrix}+ \begin{vmatrix}x0&y0\\x2&y2\end{vmatrix}\begin{vmatrix}x3&y3\\x1&y1\end{vmatrix}+ \begin{vmatrix}x0&y0\\x3&y3\end{vmatrix}\begin{vmatrix}x1&y1\\x2&y2\end{vmatrix}\\[5pt] &=(x0y1-y0x1)\begin{vmatrix}x2&y2\\x3&y3\end{vmatrix}- (x0y2-y0x2)\begin{vmatrix}x1&y1\\x3&y3\end{vmatrix}+ (x0y3-y0x3)\begin{vmatrix}x1&y1\\x2&y2\end{vmatrix}\\[5pt] &=x0\left(y1\begin{vmatrix}x2&y2\\x3&y3\end{vmatrix}- y2\begin{vmatrix}x1&y1\\x3&y3\end{vmatrix}+ y3\begin{vmatrix}x1&y1\\x2&y2\end{vmatrix}\right) -y0\left(x1\begin{vmatrix}x2&y2\\x3&y3\end{vmatrix}- x2\begin{vmatrix}x1&y1\\x3&y3\end{vmatrix}+ x3\begin{vmatrix}x1&y1\\x2&y2\end{vmatrix}\right)\\[5pt] &=x0\begin{vmatrix}x1&y1&y1\\x2&y2&y2\\x3&y3&y3\end{vmatrix} -y0\begin{vmatrix}x1&x1&y1\\x2&x2&y2\\x3&x3&y3\end{vmatrix} \end{align}
Since both 3×3 determinants have duplicate columns, the right hand side is identically zero.
Another proof may be done like this:Since vector
d=\left(p01,p02,p03\right)
is perpendicular to vector
m=\left(p23,p31,p12\right)
(see above), the scalar product of and must be zero. q.e.d.
Letting be the point coordinates, four possible points on a line each have coordinates, for . 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
(q01:q02:q03:q23:q31:q12)
M=\begin{bmatrix}q01&0\ 0&q01\ -q12&q02\ q31&q03\end{bmatrix}
has rank 2, and so its columns are distinct points defining a line . When the coordinates,, satisfy the quadratic Plücker relation, they are the Plücker coordinates of . To see this, first normalize to 1. Then we immediately have that for the Plücker coordinates computed from,, except for
p23=-q03q12-q02q31.
But if the satisfy the Plücker relation
q23+q02q31+q03q12=0,
Consequently, is a surjection onto the algebraic variety consisting of the set of zeros of the quadratic polynomial
p01p23+p02p31+p03p12.
And since is also an injection, the lines in are thus in bijective correspondence with the points of this quadric in, called the Plücker quadric or Klein quadric.
Plücker coordinates allow concise solutions to problems of line geometry in 3-dimensional space, especially those involving incidence.
Two lines in are either skew or coplanar, and in the latter case they are either coincident or intersect in a unique point. If and are the Plücker coordinates of two lines, then they are coplanar precisely when
d ⋅ m'+m ⋅ d'=0,
as shown by
\begin{align} 0&=p01p'23+p02p'31+p03p'12+p23p'01+p31p'02+p12p'03\\[5pt] &=\begin{vmatrix}x0&y0&x'0&y'0\\ x1&y1&x'1&y'1\\ x2&y2&x'2&y'2\\ x3&y3&x'3&y'3\end{vmatrix}. \end{align}
When the lines are skew, the sign of the result indicates the sense of crossing: positive if a right-handed screw takes into, 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
0=(m ⋅ d')x0+(d x d') ⋅ x,
where
x=(x1,x2,x3).
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
(x0:x)=(d ⋅ m':m x m').
To handle lines not meeting this restriction, see the references.
Given a plane with equation
0=
0x | |
a | |
0 |
+
1x | |
a | |
1 |
+
2x | |
a | |
2 |
+
3x | |
a | |
3 |
,
or more concisely,
0=
0x | |
a | |
0 |
+a ⋅ x;
(x0:x)=(a ⋅ d:a x m-a0d).
The point coordinates,, can also be expressed in terms of Plücker coordinates as
xi=\sumjajpij, i=0\ldots3.
Dually, given a point and a line not containing it, their common plane has equation
0=(y ⋅ m)x0+(y x d-y0m) ⋅ x.
The plane coordinates,, can also be expressed in terms of dual Plücker coordinates as
ai=\sumjyjpij, i=0\ldots3.
Because the Klein quadric is in, it contains linear subspaces of dimensions one and two (but no higher). These correspond to one- and two-parameter families of lines in .
For example, suppose are distinct lines in determined by points and, respectively. Linear combinations of their determining points give linear combinations of their Plücker coordinates, generating a one-parameter family of lines containing and . 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 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 .
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 inIntroduction to Plücker Coordinates written for the Ray Tracing forum by Thouis Jones.