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7 More on Bracket Algebra 7 More on Bracket Algebra Algebra is generous; she often gives more than is asked of her. D’Alembert The last chapter demonstrated, that determinants (and in particular multi- homogeneous bracket polynomials) are of fundamental importance to express projectively invariant properties. In this chapter we will alter our point of view. What if our “first class citizens” were not the points of a projective plane but the values of determinants generated by them? We will see that with the use of Grassmann-Plu¨cker-Relations we will be able to recover a projective configuration from its values of determinants. We will start our treatment by considering vectors in R3 rather than con- 2 sidering homogeneous coordinates of points in RP . This gives us the advan- tage to neglect the (only technical) difficulty that the determinant values vary when the coordinates of a point are multiplied by a scalar. While reading this chapter the reader should constantly bear in mind that all concepts presented in this chapter generalize to arbitrary dimensions. Still we will concentrate on the case of vectors in R2 and in R3 to keep things conceptually as simple as possible. 7.1 From Points to Determinants. Assume that we are given a configuration of n vectors in R3 arranged in a matrix: . | | | | P = p1 p2 p3 . pn . | | | | 114 7 More on Bracket Algebra Later on we will consider these vectors as homogeneous coordinates of a point configuration in the projective plane. The matrix P may be considered 3 n n as an Element in R · . There is an overall number of 3 possible 3 3 matrix minors that could be formed from this matrix, since there are t×hat many % & ways to select 3 points from the configuration. If we know the value of the corresponding determinants we can reconstruct the value of any bracket by using permutations of the points and applying the appropriate sign changes since we have [a, b, c] = [b, c, a] = [c, a, b] = [b, a, c] = [a, c, b] = [c, b, a]. − − − We consider the index E set for the points in P by E = 1, 2, . , n { } and a corresponding index set for the index triples 3 Λ(n, 3) := (i, j, k) E i < j < k . { ∈ | } Calculating the determinants for our vector configuration can now be consid- ered as a map 3 n (n) Γ : R · R d → P ([p1, p2, p3], [p1, p2, p4], . , [pn 2, pn 1, pn]) %→ − − We can consider the vector of determinants Γ (P ) itself as a map that assigns to each element (i, j, k) Λ(n, 3) the value of the corresponding determinant ∈ [pi, pj, pk]. By applying the alternating rule the values of all brackets can be recovered from the values of Γ (P ). In order to avoid that the whole infor- mation about a bracket is captured by the subscripts we make the following typographical convention. If P is a matrix consisting of columns p1, p2, . , pn, then we may also write [i, j, k]P instead of [pi, pj, pk]. The reader should not be scared of the high dimension of the spaces in- volved. This high dimensionality comes from the fact that we consider an n 3 entire collection of n vectors now as a single object in R · . Similarly, an (n) element in the space R d may be considered as one object that carries the values of all determinants, simultaneously. It will be our aim to show that both spaces carry in principle the same information if we are concerned with projective properties. One of the fundamental relations between P and Γ (P ) is given by the following lemma: Lemma 7.1. Let P (R3)n be a configuration of n vectors in R3. and let T be an invertible 3 3∈matrix. Then Γ (T P ) = λ Γ (P ) for a suitable λ = 0. × · · & Proof. Let (i, j, k) Λ(n, 3) then we have det(T pi, T pj, T pk) = det(T ) [p , p , p ]. Thus we∈have Γ (T P ) = det(T ) Γ (P·). · · · i j k · · '( 7.1 From Points to Determinants. 115 6 G PSfrag replacements PSfrag replacements 6 3 4 3 4 G 5 5 1 2 1 2 1.0 1.0 F L Fig. 7.1. A configuration an a projective image of it. The last lemma states that up to a scalar multiple the vector Γ (P ) is invariant under linear transformations. On the level of point configurations in the pro- jective plane this means that if two configurations of points are projectively equivalent, we can assign matrices of homogeneous coordinates P and Q to them such that Γ (P ) = Γ (Q). A little care has to be taken. Since the homo- geneous coordinates of each point are only determined up to a scalar factor, we have to adjust these factors in the right way, to get the above relation. We can get rid of the λ in Lemma 7.1 by an overall scaling applied to all homogeneous coordinates. Example 7.1. Consider the two pictures in Figure 7.1. They are related by a projective transformation. We get homogeneous coordinates of the points by simply extending the euclidean coordinates by a 1. The two coordinate matrices are: 8 0 4 0 4 2 2 2 4 0 4 3 0 P = 0 0 4 4 2 6 Q = 0 0 3 3 1 9 1 1 1 1 1 1 1 1 1 1 1 1 For P the corresponding determinants indexed by Λ(6, 3) are [1, 2, 3] = 16 [1, 3, 5] = 8 [2, 3, 4] = 16 [2, 5, 6] = 8 P P − P − P − [1, 2, 4]P = 16 [1, 3, 6]P = 8 [2, 3, 5]P = 0 [3, 4, 5]P = 8 [1, 2, 5] = 8 [1, 4, 5] = 0− [2, 3, 6] = 16 [3, 4, 6] = 8− P P P − P [1, 2, 6]P = 24 [1, 4, 6]P = 16 [2, 4, 5]P = 8 [3, 5, 6]P = 8 [1, 3, 4] = 16 [1, 5, 6] = 8 [2, 4, 5] = 8 [4, 5, 6]F = 8 P − P P P − The corresponding values for Q are L 116 7 More on Bracket Algebra [1, 2, 3] = 6 [1, 3, 5] = 4 [2, 3, 4] = 16 [2, 5, 6] = 8 Q Q − Q − Q − [1, 2, 4]Q = 6 [1, 3, 6]Q = 12 [2, 3, 5]Q = 0 [3, 4, 5]Q = 8 [1, 2, 5] = 2 [1, 4, 5] = 0− [2, 3, 6] = 24 [3, 4, 6] = 2−4 Q Q Q − Q [1, 2, 6]Q = 18 [1, 4, 6]Q = 24 [2, 4, 5]Q = 4 [3, 5, 6]Q = 16 [1, 3, 4] = 12 [1, 5, 6] = 8 [2, 4, 5] = 12 [4, 5, 6] = 16 Q − Q Q Q − At first sight these two collections of values seem not to be very related. However if we simply choose different homogeneous coordinates for the point in the second picture like: r = 4 q , r = 4 q , r = 2 q , r = 2 q , r = 3 q , r = 1 q , 1 · 1 2 · 2 3 · 3 4 · 4 5 · 5 6 · 6 we get: [1, 2, 3] = 192 [1, 3, 5] = 96 [2, 3, 4] = 192 [2, 5, 6] = 96 R R − R − R − [1, 2, 4]R = 192 [1, 3, 6]R = 96 [2, 3, 5]R = 0 [3, 4, 5]R = 96 [1, 2, 5] = 96 [1, 4, 5] = 0− [2, 3, 6] = 192 [3, 4, 6] = 9−6 R R R − R [1, 2, 6]R = 288 [1, 4, 6]R = 192 [2, 4, 5]R = 96 [3, 5, 6]R = 96 [1, 3, 4] = 192 [1, 5, 6] = 96 [2, 4, 5] = 96 [4, 5, 6] = 96 R − R R R − This is exactly 12 times the values of the determinants obtained from P . It is also instructive to check the Grassmann-Plu¨cker-Relations for a few special cases. For instance we should have [1, 2, 3] [4, 5, 6] [1, 2, 4] [3, 5, 6] +[1, 2, 5] [3, 4, 6] [1, 2, 6] [3, 4, 5] = 0. P P − P P P P − P P This can be easily verified: 16 ( 8) 16 8 + 8 8 24 ( 8) = 128 128 + 64 + 192 = 0. · − − · · − · − − − 7.2 . .and back We will now focus on the other direction. To what extend do the values of Γ (P ) already determine the entries of P . Let us first start with two observations. Observation 1: The elements in Γ (P ) are not independent of each other. This comes from the fact that the entries at least have to satisfy the Grassmann-Plu¨cker-Relations. Observation 2: The elements in Γ (P ) can determine P only up to a linear transformation. This is the statement of Lemma 7.1. These two observations extract the essence of the relation of Γ (P ) and P . (n) 3 n We will prove that for Every Γ R 3 0 there is a P R · such that Γ = Γ (P ). This P is uniquely de∈termine−d {up}to a linear tra∈nsformation. 7.2 . .and back 117 We again consider Γ as a map Λ(n, 3) R. For the rest of this chapter we will denote the value Γ ((i, j, k)) by [i, j→, k]. Thus the problem of Finding a suitable P can be restated in the following way: Find a P such that [i, j, k] = [i, j, k] for all (i, j, k) Λ(n, 3).
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